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GRADE 7 TO 12
InteractiveExperienceInteractiveExperience
GRADE 7: CONSTRUCTIONS
1Step
Get a certificate by completing the program.
Everyone who has completed all steps in the program will get a badge.
About
In this program, you will:
- revise the difference between a straight line, a ray and a line segment
- learn what is meant by parallel and perpendicular lines
- use a set square to construct parallel and perpendicular lines
- revise the names of angles
- learn to measure and draw angles using a protractor
- learn to measure and draw line segments accurately
- revise drawing of circles and arcs using a pair of compasses
- construct triangles and quadrilaterals accurately | 677.169 | 1 |
This article explores the non adjacent angles, demystifying their properties, their relationships, and their roles in shaping the geometric world as we know it.
Defining Non Adjacent Angles
Non adjacent angles are two or more angles that do not share a common side (ray) or a common vertex (endpoint of the rays). They are distinct from each other and do not have any spatial relationship, like being next to each other or overlapping. Non adjacent angles are angles separated from each other in a geometric figure or context.
It's worth noting that non adjacent angles can have different measures and do not need to be equal or related in any specific way.
Figure-1.
Properties of Non Adjacent Angles
Non adjacent angles do not have a fixed set of properties like other specific types of angles, such as complementary or supplementary angles, mainly because their measurements are independent of each other. Here are a few characteristics that can be noted about non adjacent angles:
Lack of Shared Vertex or Side
Non-adjacent angles do not share a common vertex or a common side. Each angle is separate and does not touch the other angle.
Independent Measurements
The measurement of a non-adjacent angle is not related to another non-adjacent angle. They are independent of each other.
Various Locations
Non-adjacent angles can be located anywhere in the plane. They don't have to be part of the same figure or diagram.
No Overlapping
Non-adjacent angles do not overlap each other.
Flexibility in Figures
In a given geometric figure, any angle that does not share a side or vertex with a certain angle can be considered non-adjacent to it. This allows great flexibility in working with these angles within different geometric contexts.
Exercise
Example 1
Consider a triangle ABC, and mention the non adjacent angles.
Solution
Angle ABC and angle BCA are adjacent because they share a common side and vertex. But, Angle ABC and angle CAB are non-adjacent because they do not share a common side.
Example 2
Consider the square ABCD, and mention the non adjacent angles.
Figure-2.
Solution
In a square ABCD, the angles ∠BAD and ∠BCD are non-adjacent because they do not share a common side or vertex.
Example 3
Consider the pentagon ABCDE, and mention the non adjacent angles.
Figure-3.
Solution
In the Pentagon ABCDE, Angles ∠CAB and ∠BDE are non-adjacent angles because they do not share a common side or vertex.
Example 4
Consider the circle with diameter AB and mention the non adjacent angles.
Solution
In a circle with a diameter AB and any point C on the circumference, ∠ACB and ∠CAB are non-adjacent angles. They share a vertex but do not share a common side.
Example 5
Consider the triangles ABC and DEF, and mention the non adjacent angles.
Solution
Consider two separate triangles, ABC and DEF. Any angle from triangle ABC is non-adjacent to any angle from triangle DEF because they do not share a common side or a vertex.
Example 6
Consider the quadrilateral ABCD, and mention the non adjacent angles.
Solution
In an arbitrary quadrilateral ABCD, angles ∠DAB and ∠BCD are non-adjacent because they do not share a common side or vertex.
Example 7
Consider the Secant AB and find the non adjacent angles.
Solution
In a circle with a secant AB and a tangent at point B, the angle between the secant and the tangent and the angle subtended by the secant at the center of the circle are non-adjacent.
Applications
While non-adjacent angles themselves don't directly find specific applications in various fields due to their lack of shared properties or relationship to each other, understanding these angles is fundamental to various applications of geometry in diverse fields.
Here are a few instances where the concept of non-adjacent angles (or, more broadly, the knowledge of angles) might play a role:
Architecture and Engineering
Non-adjacent angles can appear in the design and construction of buildings, bridges, or other structures. Identifying and working with these angles can be important in these structures' design and stability analysis.
Computer Graphics and Game Design
When designing 3D models or game environments, understanding the concepts of angles, including non-adjacent angles, is crucial for creating realistic and visually pleasing graphics.
Physics
In fields such as optics or mechanics, understanding the relationship between angles can help in predicting the behavior of light or the motion of objects.
Geography and Cartography
When drawing or interpreting maps, the concept of angles, including non-adjacent angles, can play an essential role.
Astronomy
The angles between stars or other celestial bodies, often non-adjacent, can be important in celestial navigation or calculating astronomical distances.
Robotics
In robotics, angles are crucial in determining how a robot should move or react in a given environment. Non adjacent angles can be used to design robot navigation and path-planning algorithms.
Navigation and Aviation
Angles are essential for navigation, particularly in aviation and seafaring. Understanding non-adjacent angles can help plan flight or travel routes, interpret compass bearings, and more.
Interior Design
In the field of interior design, understanding angles can help create spaces that are aesthetically pleasing and functional. Non-adjacent angles can appear in furniture arrangements, room layouts, and more.
Cryptography
Some forms of cryptography, especially those based on geometric shapes or patterns, can use angles, including non-adjacent angles.
Historical Significance
Angles, including non-adjacent angles, have been foundational elements in geometry for thousands of years. While there isn't a specific historical significance tied directly to non-adjacent angles, the evolution of geometric understanding, which includes the recognition of angles and their properties, has been integral to human history.
Ancient Civilizations
Ancient civilizations like the Egyptians and Babylonians had a keen understanding of basic geometric principles. They used this knowledge to build architectural wonders like the pyramids and to develop systems for land measurement, which required the identification and measurement of various types of angles, likely including non-adjacent ones.
Greek Geometry
The Greeks, most notably Euclid, formalized the study of geometry in works such as Euclid's Elements. While Euclid didn't explicitly talk about non-adjacent angles, his axioms and theorems laid the foundation for understanding geometric relationships, including the properties of angles.
Navigation and Exploration
The age of exploration from the 15th to 17th centuries heavily relied on understanding angles for celestial navigation. Mariners and explorers used the angles between stars or between a star and the horizon to find their way at sea. While these angles were not non-adjacent in the geometric sense, considering individual, separate angles was central to these navigational techniques.
Modern Geometry
In the modern era, understanding angles (including non-adjacent angles) is a foundational aspect of many branches of mathematics. It is deeply embedded in the study of shapes, sizes, and properties of spaces. This knowledge is crucial for many modern applications, including computer graphics, architecture, physics, engineering, etc. | 677.169 | 1 |
In the given figure, ABCD is a rhombus. If ∠A=70∘, then ∠CDB is equal to
A
65∘
B
55∘
C
75∘
D
80∘
Video Solution
Text Solution
Verified by Experts
The correct Answer is:B
|
Answer
Step by step video, text & image solution for In the given figure, ABCD is a rhombus. If angleA=70^(@), then angleCDB is equal to by Maths experts to help you in doubts & scoring excellent marks in Class 9 exams. | 677.169 | 1 |
...describe two other circumferences, which shall meet the former in x and z, the required projections. To draw through a given point a line parallel to a given line, place the sight ruler on the line and look for some object 200 or 300 yards distant in that direction...
...intersecting the arc EF in F, and draw the straight line DF. Then, 168. Proposition XXXIX.— Problem. To draw through a given point a line parallel to a given line. Let C be the given point and AB the given line. Draw a straight line from C to any point of AB, asdraw a straight line equal to a given straight line* 2. To make an angle equal to a given angle. 3. To draw through a given point a line parallel to a given line. 4. To draw through a given point a line perpendicular to a given line. Two cases. 5. To bisect a given...
...straight line, only one perpendicular can be drawn from the point to the straight line. M .N 380. PROBLEM. To draw through a given point a line parallel to a given straight line, by means of a ruler and set square. Let M be the given point, and AB the given straightrequired to construct an angle equal to a given angle. Sug. Consult 191 and 192. 631. It is required to draw through a given point a line parallel to a given line. Sug. Consult 87 and 84, and Problem 630. 632. Two angles of a triangle being given, it is required...
...of solution of the problem involving that point can usually be effected; as, if the problem be given to draw through a given point a line parallel to a given line, several methods for the solution of the problem can be evolved by means of several of the propositions...
...protractors have an arm which carries a vernier, by which angles may be'constructed to single minutes. To draw through a given point a line, parallel to a given line, make one of the sides of a triangle coincide with the given line, and, placing a ruler against one... | 677.169 | 1 |
When you answer 8 or more questions correctly your red streak will increase in length. The green streak shows the best player so far today. See our Hall of Fame for previous daily winners.
See if you can get full marks in this math quiz.
Shapes - Polygons
This Math quiz is called 'Shapes - Polygons polygon is a 2-D plane (flat) shape with straight sides, such as a triangle or a pentagon. Some polygons fit together side to side as tiles to cover an area without leaving gaps. This is called tesselation.
Test your knowledge in this Math quiz.
1.
In a tile pattern of equilateral triangles, how many tiles will fit around a single point?
3
4
5
6
All angles are 60° and angles around a point add up to 360°
2.
If the exterior angle of a polygon is 45° what is the corresponding interior angle?
75°
108°
135°
160°
The sum of the exterior and interior angles at any corner is always 180° because they are angles on a straight line
3.
The exterior angle of a regular n-sided polygon is 45°. What is n?
6
7
8
9
360 ÷ 45 = 8
4.
Which of these tile shapes will not tesselate with only others of the same shape and size?
Regular hexagon
Isosceles triangle
Regular octagon
Trapezium
Check the angles
5.
A polygon with all sides the same length and all angles equal is called ....... | 677.169 | 1 |
PARABOLA
Above you see the most popular example for Parabolic Curve-the Dish Antenna
What is a Parabola: A simple way to define would be, Parabolas are U-shaped Curves that satisfy specific conditions.
Remember: Not All U-shaped Curve is a Parabola
Question: Why is the Dish Antenna designed in the shape of a Parabola
This is to facilitate no matter where the incoming Signals hit the Dish Antenna, they get Directed | Reflected to the Receiver
Note: This point is the Fixed point and is called the FOCUS
For this Torch, the outer surface is Parabolic in shape and a Bulb is placed at the FOCUS point so that the light coming out has the maximum reach.
How do we define a Parabola
Consider a fixed Line and a fixed point, say F. Now find all of the points whose distance from the line and the fixed point are equal (equidistant). Now if we were to connect all of these points they would all lie on a Curve and we call that curve to be a Parabola
The Fixed point is called the Focus
and the Fixed line is called the Directrix.
The shape of the Parabola will depend on the position of the Focus and the Directrix | 677.169 | 1 |
arithmetic mean formula
Here you will learn formula for arithmetic geometric and harmonic mean and relation between arithmetic geometric and harmonic mean. Let's begin – Arithmetic Mean Formula If three terms are in A.P. then the middle term is called the A.M. between the other two, so if a, b, c are in A.P., b is A.M. of … | 677.169 | 1 |
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The power of origami
Doubling the cube using origami — proof
Figure 1
First we show that the first three steps amount to dividing the height of the square into three equal parts. Place the square in a coordinate system with corner at the point . See figure 1. Now the line from to has equation where is the side length of the square. The line from to has equation So the two lines meet at a point with -coordinate satisfying so The -coordinate of the intersection point is therefore as required.
The construction claims that the side length of the new cube is times the side length of the given cube. Now since , we have so we need to show that
Figure 2
Firstly, we can assume that the length is equal to 1, as we can scale the image as we like without affecting the proportions. We'll write for the length of the line segment , so the side length of our square is equal to . See figure 2. The line segment corresponds to two thirds of a side of the square and therefore has length This implies that the length of the line segment is We write for the length . Since the line segment is part of the bottom edge, its length is . There is a right angle at , so by Pythagoras' theorem we have This means that and therefore Now consider the two right-angled triangles and . Considering the 180 degree angle formed at by the vertical edge of the square, we get
The angles of the triangle add up to 180 degrees, so
So the triangles and have two equal angles: the right angle and the angle This means that they are similar. Therefore, the ratios of corresponding sides of the two triagles are equal. Observing that the side of the triangle is a third of the side of the square and therefore has length , we get | 677.169 | 1 |
Properties Of Quadrilaterals Pdf
Properties Of Quadrilaterals Pdf. The opposite angles are congruent. Special types of quadrilaterals include squares and rectangles.
Web properties of quadrilaterals model examples goal: Web quadrilaterals are classified by their properties (e.g. The names of quadrilaterals studied in the book are listed below along with their properties.
Opposite sides are parallel 2. Web the purpose of this lesson is to identify key quadrilaterals and properties about their angles and sides. A quadrilateral has 2 diagonals based on which it can be classified into.
The Names Of Quadrilaterals Studied In The Book Are Listed Below Along With Their Properties.
Worksheet/activity file previews docx, 22.36 kb pptx, 65.81 kb docx,. This lesson will address the following ccrs standard(s). Name all of the properties of a parallelogram and its diagonals.
The Opposite Sides Are Congruent.
Web properties associated with specific kinds of quadrilaterals. Both pair of opposite sides are // 2. Every quadrilateral has 4 vertices, 4 angles, and 4 sides the total of its interior angles = 360 degrees
Special Types Of Quadrilaterals Include Squares And Rectangles.
The goal is to know and apply the properties of quadrilaterals. They will then use these properties to create specific quadrilaterals. L geometry (part 3) quadrilaterals kite quadrilateral trapezium parallelogram square rectangle rhombus using your knowledge of the properties of. | 677.169 | 1 |
Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle, and the Geometry of Solids; to which are Added Elements of Plane and Spherical Trigonometry
From inside the book
Results 1-5 of 67
Page 23 ... BC is equal to BG ; wherefore AL and BC are each of them equal to BG ; and things that are equal to the same are ... base BC be equal to the base EF ; and the triangle ABC to the tri- angle DEF ; and the other angles , to which the equal ...
Page 24 ... base BC shall coincide with the base EF ( cor . def . 3. ) , and shall be equal to it . Therefore also the whole triangle ABC shall coincide with the whole triangle DEF , so that the spaces which they contain or their areas are equal ...
Page 25 ... BC common to both , the two sides DB , BC are equal to the two AC , CB , each to each ; but the angle DBC is also equal to the angle ACB ; therefore the base DC is equal to the base AB , and the area of the triangle DBC is equal to that ...
Page 27 ... base BC equal to the base EF . The angle BAC is equal to the angle EDF . For , if the triangle ABC be applied to the triangle DEF , so that the point B be on E , and the straight line BC upon EF ; the point C shall also coincide with ...
Page 28 ... BC , CD , each to each ; but the angle ACD is also equal to the an- gle BCD ; therefore the base AD is equal to the base ( 4. 1. ) DB , and the straight line AB is divided into two equal parts in the point D. Which was to be done ...
Popular passages
Page 56
Page 19 - A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another.
Page 62 - In every triangle, the square on the side subtending either of the acute angles, is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the...
Page 62
Page 76 - THE diameter is the greatest straight line in a circle; and, of all others, that which is nearer to the centre is always greater than one more remote ; and the greater is nearer to the centre than the less.* Let ABCD be a circle, of which...
Page
Page 18 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle ; and the straight line which stands on the other is called a perpendicular to it.
Page 55 - If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.
Bibliographic information
Title
Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle, and the Geometry of Solids; to which are Added Elements of Plane and Spherical Trigonometry | 677.169 | 1 |
45-45-90 triangle
45-45-90 triangles are special right triangles with one 90 degree angle and two 45 degree angles. All 45-45-90 triangles are considered special isosceles triangles. The 45-45-90 triangle has three unique properties that make it very special and unlike all the other triangles.
45-45-90 triangle ratio
There are two ratios for 45-45-90 triangles:
The ratio of the sides equals
The ratio of the angles equals
Properties of 45-45-90 triangles
To identify 45-45-90 special right triangle, check for these three identifying properties:
The polygon is an isosceles right triangle
The two side lengths are congruent, and their opposite angles are congruent
The hypotenuse (longest side) is the length of either leg times square root (sqrt) of two,
All 45-45-90 triangles are similar because they all have the same interior angles.
45-45-90 triangle theorem
To solve for the hypotenuse length of a 45-45-90 triangle, you can use the 45-45-90 theorem, which says the length of the hypotenuse of a 45-45-90 triangle is the times the length of a leg.
45-45-90 triangle formula
You can also use the general form of the Pythagorean Theorem to find the length of the hypotenuse of a 45-45-90 triangle.
Here is a 45-45-90 triangle. Let's use both methods to find the unknown measure of a triangle where we only know the measure of one leg is 59 yards:
We can plug the known length of the leg into our 45-45-90 theorem formula:
Using the Pythagorean Theorem:
Both methods produce the same result!
45-45-90 triangle rules
The main rule of 45-45-90 triangles is that it has one right angle and while the other two angles each measure 45°. The lengths of the sides adjacent to the right triangle, the shorter sides have an equal length.
Another rule is that the two sides of the triangle or legs of the triangle that form the right angle are congruent in length.
Knowing these basic rules makes it easy to construct a 45-45-90 triangle.
Constructing a 45-45-90 triangle
The easiest way to construct a 45-45-90 triangle is as follows:
Construct a square four equal sides to the desired length of the triangle's legs
Construct either diagonal of the square
Striking the diagonal of the square creates two congruent 45-45-90 triangles. Half of a square that has been cut by a diagonal is a 45-45-90 triangle.
The diagonal becomes the hypotenuse of a right triangle.
You can also construct the triangle using a straightedge and drawing compass:
Construct a line segment more than twice as long as the desired length of your triangle's leg
Open the compass to span more than half the distance of the line segment
Use the compass to construct a perpendicular bisector of the line segment by scribing arcs from both endpoints above and below the line segment; this will produce two intersecting arcs above and two intersecting arcs below the line segment
Use the straightedge to draw the perpendicular bisector by connecting the intersecting arcs
Reset the compass with the point on the intersection of the two line segments and the span of the compass set to your desired length of the triangle's leg
Strike two arcs, one on the line segment and one on the perpendicular bisector
Connect the intersections of the arcs and segments
This method takes more time than the square method but is elegant and does not require measuring.
How to solve a 45-45-90 triangle
The length of the hypotenuse, which is the leg times , is key to calculating the missing sides:
If you know the measure of the hypotenuse, divide the hypotenuse by to find the length of either leg
If you know the length of one leg, you know the length of the other leg (legs are congruent)
If you know either leg's length, multiply the leg length times to find the hypotenuse
45-45-90 triangle example problems
Here we have a 45-45-90 triangle with a hypotenuse of meters, and each leg is 3 meters.
Take is a 45-45-90 triangle with sides measuring 9.5 feet. What is the length of the hypotenuse? | 677.169 | 1 |
• Encourage an early love of learning about geometry with these colorful manipulatives • Great for individual, center or small-group use • Supports activities aligned with NCTM standards • 15 activity cards are divided into 3 sections for Grades K-2, 3-5, and 6-8
Description
AngLegs™ come in six lengths that easily snap together to motivate students to explore plane geometry on their desktop or the overhead projector. Students will discover what happens as they use the same length legs verses different length legs to build various polygons, as they attach additional legs they study polygon heights, calculate their areas, find center points and much more! When two AngLegs™ of any length are snapped onto the special 4″ protractor students will explore angles, add a third leg for triangles, a forth for quadrilaterals, and so on. This 74-piece set consists of 72 AngLegs™12 each of six different lengths in six colors and two snap-on View Thru™ protractors. This exciting new manipulative includes a set of 15 activity cards, divided into three sections for grades K-2, 3-5 and 6-8. All activities adhere to the NCTM Standards for geometry.
Additional information
Weight
0.42 lbs
Dimensions
6.60 × 5.70 × 1.00 in
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I'm only going to cover simple shapes like Points, Straight Lines, Circular Arcs, Circular Sectors and Simple Polygons, by using cartesian coordinates, basic algebra and Euclidean 2D geometry only. More complex shapes like ellipse and parabolas won't be covered, and more advanced maths like polar coordinates and affine spaces won't be used. No solid proofs will be covered as well. The below aims to always ensure 100% correctness, even for the teleportation cases, although it indeed fails a bit and badly when few and many shapes move quickly respectively.
Points
It's just an ordered x, y pair in the cartesian coordinate system.
Straight Lines
It's just the line between 2 specified points. The line will be represented by its corresponding linear equation in the form of y = mx + c. The rotation and translation offsets, valid x and y ranges and midpoint of the straight line will be used as well.
Users only has to define those 2 specified points. The rest will be generated by scripts upon initializing the point.
Circular Arcs
A virtual circular sector will be used to implement a circular arc, which is the real shape. Check below for Circular Sectors. The midpoint is used in the circular arc as well.
Circular Sectors
It's a system of 3 inequalities in 2 unknowns, including 2 linear ones in the form of y >= mx + c, y > mx+ c, y <= mx+ c or y < mx+ c, and 1 inequality of circle in the form of (x - h) ^ 2 + (y - k) ^ 2 <= r ^ 2 or (x - h) ^ 2 + (y - k) ^ 2 < r ^ 2. The center, radius, rotation and translation offsets, and associated equations for those inequalities will be used as well. Note that the center is also the intersecting point of those 2 linear inequalities.
Users only has to define the center, radius, and those 2 linear inequalities. The rest will be generated by scripts upon initializing the circular sector.
Simple Polygons
It's a system of linear inequalities in 2 unknowns. The number of inequalities is that of the edges. The vertexes, triangles of the polygon touching its edges and their incenters, an arbitrary center inside the polygon, the associated linear equations for those linear inequalities with their valid x and y ranges, the rotation and translation offsets, and the bounding rectangle or circular sector and inner circular sector or rectangle(analogous to inner circle) are used as well.
Users only has to define the vertexes, an arbitrary center inside the polygon, and whether a rectangle or circular sector will be used to bound the simple polygon. The rest will be generated by scripts upon initializing the simple polygon.
Points vs Points
Just compare if they have both the same x and y coordinates. The time complexity's O(1).
Points vs Straight Lines
The point's reference point will be adjusted to be that of the straight line. The time complexity here's O(1).
Then just check if the point's x and y coordinates are within the valid ranges of those of the straight line, and if that point's an solution to that linear equation. The time complexity here's O(1).
The combined time complexity's O(1).
Points vs Circular Arcs
The point's reference point will be adjusted to be that of the circular arc. The time complexity here's O(1).
Then just check if the distance between the point and the virtual center of the circular arc equals to the virtual radius of that circular arc, and if that point satisfies the 2 virtual linear inequalities in 2 unknowns.The time complexity here's O(1).
The combined time complexity's O(1).
Points vs Circular Sectors
Same as Points vs Circular Arcs, except that whether the distance between the point and the center is less than or equal to, or less than, the radius is checked instead of checking against the equal condition.
Points vs Simple Polygons
The point's reference point will be adjusted to be that of the simple polygon. The time complexity here's O(1).
The point will first the checked against the bounded rectangle or circular sector of the simple polygon. Checking a point against a circular sector goes to Points vs Circular Sectors, and that against a rectangle will be checking if both the x and y coordinates of the point lines between the minimum and maximum x and y coordinates of the rectangle. The time complexity here's O(1).
If the point doesn't collide with the bounded rectangle or circular sector of the simple polygon, then that point isn't colliding with that simple polygon itself. The combined time complexity will be O(1).
If that's not the case, then check the point against the inner rectangle or circular sector of the simple polygon, which is essentially the same as checking against the bounding ones.
If the point collides with the inner rectangle or circular sector of the simple polygon, then so does the simple polygon. The combined time complexity will be O(1).
If that's not the case, then check if the point satisfies all the linear inequalities of the simple polygon. The time complexity here's O(E), where E is the number of edges.
The combined time complexity will be O(E) in this case.
Straight Lines vs Straight Lines
Either straight line's reference point will be adjusted to be that of the other. The time complexity here's O(1).
Then just check if the 2 linear equations has solutions and whether the they lie within the valid ranges of both straight lines. The time complexity here's O(1).
The combined time complexity's O(1).
Straight Lines vs Circular Arcs
The straight line's reference point will be adjusted to be that of the circular arc. The time complexity here's O(1).
Then just check if the linear equation and the equation of circle has solutions and whether they satisfy the valid ranges of the straight line, and the 2 linear inequalities of the circular arc. The time complexity here's O(1).
Straight Lines vs Circular Sectors
The straight line's reference point will be adjusted to be that of the circular sector. The time complexity here's O(1).
The midpoint of the straight line will be checked against the circular sector. Checking a point against a circular sector goes to Points vs Circular Sectors. The time complexity here's O(1).
If the midpoint collides with the circular sector, then so does the straight line. The combined time complexity will be O(1). If that's not the case, then it's certain that the straight line won't completely lie within the circular sector, meaning the former must intersect with the perimeter of the latter in order to collide. So checking whether the straight line collides with the linear equations or equation of circle of the circular sector will suffice.
They go to Straight Lines vs Straight Lines and Straight Lines vs Circular Arcs. The time complexity here's O(1).
The combined time complexity will be O(1) in this case.
Straight Lines vs Simple Polygons
The straight line's reference point will be adjusted to be that of the simple polygon. The time complexity here's O(1).
The straight line will be checked against the bounded rectangle or circular sector of the simple polygon. Checking a straight line against a circular sector goes to Straight Lines vs Circular Sectors.
That against a rectangle will be checking the midpoint of the straight line against that rectangle. Checking a point against a simple polygon goes to Points vs Simple Polygons.
If the midpoint collides with the rectangle, then so does the straight line. The time complexity will be O(1) but not O(E), as the number of edges of a rectangle's always 4. If that's not the case, then it's certain that the straight line won't completely lie within the rectangle, meaning the former must intersect with the perimeter of the latter in order to collide. So checking whether the straight line collides with the linear equations of the rectangle will suffice.
It goes to Straight Lines vs Straight Lines. The time complexity here's O(1).
If the straight line doesn't collide with the bounded rectangle or circular sector, then neither does the simple polygon. The combined time complexity will be O(1).
If that's not the case, then check the straight line against the inner rectangle or circular sector of the simple polygon, which is essentially the same as checking against the bounding ones.
If the straight line collides with the inner rectangle or circular sector of the simple polygon, then so does the simple polygon. The combined time complexity will be O(1).
If that's not the case, then check the midpoint of the straight line against the simple polygon. Checking a point against a simple polygon goes to Points vs Simple Polygons.
If the midpoint collides with the simple polygon, then so does the straight line. The combined time complexity will be O(E). If that's not the case, then it's certain that the straight line won't completely lie within the simple polygon, meaning the former must intersect with the perimeter of the latter in order to collide. So checking whether the straight line collides with the linear equations of the simple polygon will suffice.
It goes to Straight Lines vs Straight Lines. The time complexity here's O(E) but not O(1), as E instead of 1 straight lines have to be checked against.
The combined time complexity will be O(E) in this case.
Circular Arcs vs Circular Arcs
Either circular arc's reference point will be adjusted to be that of the other. The time complexity here's O(1).
Then just check if the equation of circles has solutions and whether they satisfy all the 4 linear inequalities of the circular arcs. The time complexity here's O(1).
Circular Arcs vs Circular Sectors
The circular arc's reference point will be adjusted to be that of the circular sector. The time complexity here's O(1).
The midpoint of the circular sector will be checked against the circular sector. Checking a point against a circular sector goes to Points vs Circular Sectors. The time complexity here's O(1).
If the midpoint collides with the circular sector, then so does the circular arc. The combined time complexity will be O(1). If that's not the case, then it's certain that the circular arc won't completely lie within the circular sector, meaning the former must intersect with the perimeter of the latter in order to collide. So checking whether the circular arc collides with the linear equations or equation of circle of the circular sector will suffice.
They go to Straight Lines vs Circular Arcs and Circular Arcs vs Circular Arcs. The time complexity here's O(1).
The combined time complexity will be O(1) in this case.
Circular Arcs vs Simple Polygons
The circular arc's reference point will be adjusted to be that of the simple polygon. The time complexity here's O(1).
The circular arc will be checked against the bounded rectangle or circular sector of the simple polygon. Checking a circular arc against a circular sector goes to Circular Arcs vs Circular Sectors.
That against a rectangle will be checking the midpoint of the circular arc against that rectangle. Checking a point against a simple polygon goes to Points vs Simple Polygons.
If the midpoint collides with the rectangle, then so does the circular arc. The time complexity will be O(1) but not O(E), as the number of edges of a rectangle's always 4. If that's not the case, then it's certain that the circular arc won't completely lie within the rectangle, meaning the former must intersect with the perimeter of the latter in order to collide. So checking whether the circular arc collides with the linear equations of the rectangle will suffice.
It goes to Straight Lines vs Circular Arcs. The time complexity here's O(1).
If the circular arc doesn't collide with the bounded rectangle or circular sector, then neither does the simple polygon. The combined time complexity will be O(1).
If that's not the case, then check the circular arc against the inner rectangle or circular sector of the simple polygon, which is essentially the same as checking against the bounding ones.
If the circular arc collides with the inner rectangle or circular sector of the simple polygon, then so does the simple polygon. The combined time complexity will be O(1).
If that's not the case, then check the midpoint of the circular arc against the simple polygon. Checking a point against a simple polygon goes to Points vs Simple Polygons.
If the midpoint collides with the simple polygon, then so does the circular arc. The combined time complexity will be O(E). If that's not the case, then it's certain that the circular arc won't completely lie within the simple polygon, meaning the former must intersect with the perimeter of the latter in order to collide. So checking whether the circular arc collides with the linear equations of the simple polygon will suffice.
It goes toCircular Sectors vs Circular Sectors
Either circular sector's reference point will be adjusted to be that of the other. The time complexity here's O(1).
The center of each circular sector will be checked against the other circular sector. Checking a point against a circular sector goes to Points vs Circular Sectors. The time complexity here's O(1).
If either center collides with the other circular sector, then so do the circular sectors themselves. The combined time complexity will be O(1). If that's not the case, then it's certain that neither circular sector will lie within the other, meaning their perimeters must intersect in order to collide. So checking whether their linear equations or equation of circles collide will suffice.
They go to Straight Lines vs Straight Lines, Straight Lines vs Circular Arcs and Circular Arcs vs Circular Arcs. The time complexity here's O(1).
Circular Sectors vs Simple Polygons
The circular sector's reference point will be adjusted to be that of the simple polygon. The time complexity here's O(1).
The circular sector will be checked against the bounded rectangle or circular sector of the simple polygon. Checking a circular sector against a circular sector goes to Circular Sectors vs Circular Sectors.
That against a rectangle will be checking the center of the circular sector against that rectangle and the arbitrary center of the simple polygon against that circular sector. Checking a point against a circular sector and simple polygon goes to Points vs Circular Sectors and Points vs Simple Polygons respectively.
If the center of the circular sector collides with the rectangle or the arbitrary center of the simple polygon collides with the circular sector, then so do circular sector and the rectangle. The time complexity will be O(1) but not O(E), as the number of edges of a rectangle's always 4. If that's not the case, then it's certain that neither the circular sector nor the rectangle1).
If the circular sector doesn't collide with the bounded rectangle or circular sector, then neither does the simple polygon. The combined time complexity will be O(1).
If that's not the case, then check the circular sector against the inner rectangle or circular sector of the simple polygon, which is essentially the same as checking against the bounding ones.
If the circular sector collides with the inner rectangle or circular sector of the simple polygon, then so does the simple polygon. The combined time complexity will be O(1).
If that's not the case, then check the center of the circular sector against the simple polygon, and the arbitrary center of the simple polygon against the circular sector. Checking a point against a circular sector and a simple polygon goes to Points vs Circular Sectors and Points vs Simple Polygons respectively.
If the center of the circular sector collides with the simple polygon or the arbitrary center of the simple polygon collides with the circular sector, then so do the circular arc and simple polygon themselves. The combined time complexity will be O(E). If that's not the case, then it's certain that the neither the circular arc nor the simple polygonSimple Polygon vs Simple Polygons
The reference point of the one with fewer edges will be adjusted to be that of the other. The time complexity here's O(1).
Their bounded rectangles or circular sectors will be checked against each other. Checking a circular sector against a circular sector and a circular sector against a simple polygon goes to Circular Sectors vs Circular Sectors and Circular Sectors vs Simple Polygons respectively.
Checking against 2 rectangles will be checking if the minimum and maximum x and y coordinates of one rectangle is less than or equal to and greater than or equal to the maximum and minimum x and y coordinates of the other respectively. The time complexity here's O(1).
If their bounded rectangles or circular sector don't collide with each other, then neither do those simple polygons themselves. The combined time complexity will be O(1).
If that's not the case, then check their inner rectangles or circular sectors against each other, which is essentially the same as checking the bounding ones against each other.
If they collide, then so do those simple polygons. The combined time complexity will be O(1).
If that's not the case, then check the arbitrary center of a simple polygon against the other simple polygon, which goes to Points vs Simple Polygons.
If they collide, then so do the simple polygons. The combined time complexity will be O(E1 + E2), where E1 and E2 are the number of edges of those simple polygons respectively. If that's not the case, then it's certain that the neither simple polygon will completely lie within the other, meaning their perimeters must intersect in order to collide. So checking whether a triangle of a simple polygon touching its edges collide with that of the other simple polygon will suffice. As the number of triangles of a simple polygon touching its edges are E / 2 if E's even and (E + 1) / 2 if E's odd, both E1 and E2 are effectively reduced to E1_half and E2_half from now on.
Check each triangle of a simple polygon touching its edges against the other's bounding rectangle or circular sector. Checking against the circular sector goes to Circular Sectors vs Simple Polygons.
That against the rectangle will be checking the incenter of the triangle against the rectangle and the arbitrary center of the other simple polygon against the triangle, which goes to Points vs Simple Polygons. The time complexity will be O(1) but not O(E), as the number of edges of a rectangle's always 4.
If the arbitrary center of the other simple polygon collide with the triangle of the simple polygon, then so do those simple polygons themselves. The combined time complexity will be O(E1_half + E2_half).
If the incenter of a triangle of the polygon doesn't collide with the bounding rectangle or circular sector of the other simple polygon, then neither does the other simple polygon itself. The time complexity here will be O(E1_half + E2_half). Now both E1_half and E2_half will likely be further reduced significantly to E1_reduce and E2_reduce, making the last step, which costs O(E1_reduce * E2_reduce), much more efficient, meaning the added O(E1 + E2) and O(E1_half + E2_half) costs will probably be outweighed and the decreased O(E1_reduce * E2_reduce) cost.
Finally check each triangle of the simple polygon touching its edges against those of the other, by checking each incenter against the other triangle, which goes to Circular Sectors vs Simple Polygons.
If either collides with the other triangle, then so do those triangles, and thus the simple polygons themselves. The combined time complexity will be O(E1_reduce * E2_reduce). If that's not the case, then it's certain that neither triangle will completely lie within the other, meaning their perimeters must intersect in order to collide. So checking whether the edges of those triangles touching those of the simple polygons intersect will suffice.
It goes to Straight Lines vs Straight Lines. The combined time complexity will be O(E1_reduce * E2_reduce) in this case. In practice, the average case, which is probably O(E1 + E2), are likely more realistic then the worst case.
Shape Collision Detections On Maps
Suppose there are n objects on a map.
First use a aggregate bounding rectangle to bound those of all shapes. The time complexity here's O(n).
Then partition the aggregate bounding rectangle into p parts, where p is the square number closest to n. Their difference should be small enough to treat p as n here, so the time complexity here's O(n).
For each partition, check all shapes colliding with that partition to see if any of them collide with the others, by checking all their combinations. The time complexity here's O(n ^ 2) for the worst case, and O(1) for the best and average case, as most of the shapes usually won't concentrate in any single partition for a long time. It also means the average case's more meaningful here.
The combined number of collision detections per frame's O(n) for the average and best case, and O(n ^ 2) for the worst case. The combined time complexity of all collision detections per frame's O(n) for the best case, O(n * E_avg), E_avg being the average number of edges of all simply polygons, for the average case, and O((n * E_max) ^ 2), E_max being the maximum number of edges of all simple polygons, for the worst case.
Even myself feel and think that this algorithm's extremely dumb and incredibly ridiculous, but as I'm still really a computer science and maths nub, right now that's the best I can come up with.
If you've any alternate algorithm, and/or comments on mine, just feel free to drop your 2 cents here. | 677.169 | 1 |
Jk kl and lj are all tangent. JK, KL, and LJ are all tangent to circle O (not drawn to sca...
If we draw two tangent lines from same point then tangent lines… Q: If T is the point at the given distance on the unit circle C from P (1, 0), determine the quadrant… A: Here, T is on the unit circle, and its distance from the point P(1,0) is 13π12 that is: J Theorem \(\PageIndex{1}\) A tangent is perpendicular to the radius drawn to the point of intersection. Proof \(OP\) is the shortest line segment that can be drawn from 0 to line \(\overleftrightarrow{AB}\).D. Circles P and Q are externally tangent. AB is a commNon external tangent. B 7. If the radii of circles P and Q are 8 cm and 12 cm, respectively, find the length of the tangent segment AB. 8. If the radii of circle P and Q are 5 cm and 9 cm, respectively, find the length of the tangent segment AB. 40. D. Circles P and Q are externally tangent.Identify each line or segment that JK, KL, and LJ are all tangent to O (not to scale). ... 2) '( = 3) "+ = 4) B = The segments in each figure are tangent to the circle. Find each length. 5) Name the center. 6) Name all the radii. Verify a Tangent to a circle: Hint: Converse of Pythagorean Theorem: ~ a2 + b2 = c2 Is => If JA = 14 ... We know that, all sides of a regular polygon are equal in length. If we know the length of one of ...Lastly, we find JK. Notice that JK is tangent to the circle at point B and ends in point K. Similarly, JL is tangent to the circle at point A and ends at point L. Point A and B lies on the same level on the x-axis. Same is true with L and K. Therefore, they must have the same length. So, JL = JK = 21 Finally, we sum all sides to obtain the ...JK, KL, and LJ are all tangent to circle O (not; drawn to scale), and JK LJ. JA = 7, AL = 12, and. CK = 10. Find the perimeter of JKL. a. 34. b. 38. c. 58. d. 29. Assume that lines that appear to be tangent are tangent. O is the center of the circle. Find the value of x. Answer: JK, KL, and LJ are all tangent to circle O. JA = 9, AL= 10, and CK= 14. What is the perimeter of triangle JKL? Step-by-step explanation: Get the detailed answer: 3. JK, KL, and LJ are all tangent to O. JA = 6, AL = 11, and CK = 13. Find the perimeter of triangle $$\triangle$$JKL. A) P=40 B)–– –– –JK, KL, LJ are all tangent to circle O. Of 9, –– ––AL=10 and CK = 14, find the perimeter of JKLshow work. verified. Verified answer. Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as … See Answer. Question: JK,KL, and LJ are all tangent to O (not drawn to scale). JA=5,AL=11, and CK=12. Find the perimeter of JKL. Show transcribed image text. There are 2 steps to solve this one. 3 Line segment BC is tangent to circle A at B and to circle D at C (not drawn to scale). AB = 7, BC = 18, and DC = 5. Find AD to the nearest tenth. A 18.7 B 18.1 C 21.6 ... 6 Line segments JK, KL, and LJ are all tangent to O (not drawn to scale). JA = 9, AL = 10, and CK = 14. Find the perimeter of triangle JKL. A 66 B 38May 16, 2018 · JK, KL, and LJ are all tangent to circle O. The diagram is not drawn to scale. . . triangle JLK with an inside circle O. . If JA = 12, AL = 15, and CK = 5, what is the perimeter of ΔJKL? Jk kl and lj are all tangent 3495 users searched for this homework answer last month and 70 are doing it now, let's get your homework done. This Top Homework Answer is High School level and belongs to the Mathematics subject. This answer got 98 "Big Thanks" from other students from places like Goodhue or Brookston.4. m∠O = 111. 5. m∠P = 12. 6. BC is tangent to circle A at B and to circle D at C (not drawn to scale). AB = 7, BC = 18, and DC = 5. Find AD to the nearest tenth. 7. AB is tangent to circle O at B. Find the length of the radius r for AB = 5 and AO = 8.6. Round to the nearest tenth if necessary. Study with Quizlet and memorize flashcards containing terms like AB is tangent to circle O at B. The diagram is not drawn to scale. If AB = 9 and AO = 21.6, what is the length of the radius (r)? Round the answer to the nearest tenth., JK, KL, and LJ are all tangent to circle O. The diagram is not drawn to scale. Today I want to take a tangent and discuss real estate — specifically real estate agents. I have a good family friend that is looking to buy their first home, The College Investor Lesson 10: Properties of tangents. Math > High school geometry > Circles > Properties of tangents. Determining tangent lines: lengths. Google Classroom. Solve two problems that apply properties of tangents to determine if a line is tangent to a circle. Problem 1. Segment O C ― is a radius of circle O . O A B C 8 11 5 AB is tangent to circle O at B. Find the length of the radius r, if AB = 12 in and AD = 6 in. The diagram is not to scale. B. Trigonometry (MindTap Course List) 8th Edition. ISBN: 9781305652224. Author: Charles P. McKeague, Mark D. Turner. Publisher: Charles P. McKeague, Mark D. Turner. Chapter6: Equations.Top answer: Since JK, KL, and LJ are all tangent to circle O, they are all perpendicular to the radii drawn to Read more. JK, KL, and LJ, are all tangent to circle O. If JA = 14, AL = 12, and CK = 8, what is the perimeter of the ΔJKL? Show all work. Top answer: 3√68 = 24.73 Why do you say that it is close to 68?MacOS: I quit a lot of conversational podcasts early. They get boring for a few minutes, I try hunting for the next good bit with 30-second skips, and I give up and delete the epis...jk kl and lj are all tangent to o ja = 14 al = 15 and ck=13 find the perimerter; admin. Mathematics High School. jk kl and lj are all tangent to o ja = 14 al = 15 and ck=13 find the perimerter of jkl. 1 month ago. Solution 1. Guest #9567758. 1 month ago JK,KL and LJ are all to O.JA =14, AL=15, and CK=13.AB is tangent to circle O at B. The diagram is not drawn to scale. If AB = 9 and AO = 21.6, what is the length of the radius (r)? Round the answer to the nearest tenth. (1 point) 2. JK, KL, and LJ are all tangent to circle O. The diagram is not drawn to scale. If JA = 13, AL = 19, and CK = 7, what is the perimeter of ΔJKL? (1 point) Circles ... JK, KL and LJ are all tangent to O.JA- 6, AL=11 and Ck³13, find perimeter of A AJKL L B a) 33 6)60 c238 ри the lateral area of a cone is 60rcm squared. the radius is 6cm. find the slant height. Math JK, KL, and LJ are all tangent to O (not drawn to scale). JA = 5, AL = 9, and CK = 15. Find the perimeter of ∆JKL. 12. The circumference of a circle is 44 π cm. Find the diameter, the radius, and the length of an arc of 200°. ... ExamView - CCGEOBChapter 10 … 13, AL = 19, and CK = 7, what is the perimeter of JkL? star. 4.2/5. heart. 4. verified. Verified answer. Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer. starSince JK, KL, and LJ are all tangent to circle O, they are all perpendicular to the radii drawn to the points of tangency. This means that JK, KL, and JL bisect angles J, K, and L respectively. Let x be the length of JK, y be the length of KL, and z be the length of LJ JK, KL, and LJ are all tangent to circle O. The diagram is not drawn to scale If JA = 14, AL = 12, and CK = 8, what is the perimeter of AJKL? L perimeter = BUY. Elementary Geometry For College Students, 7e. 7th Edition. ISBN: 9781337614085. Author: Alexander, Daniel C.; Koeberlein, Geralyn MJK, KL, and LJ are all tangent to circle O. If JA = 14, AL = 12, and. Geometry. Circles. Tangents. JK, KL, and LJ are all tangent to circle O. If JA = 14, AL = 12, … Question: TURNED OUT IF HADN'T DIN AMAN ut.org/courses/7209/quizzes/143406/take Question 5 2 pts JK, KL, and LJ are all tangent to O. JA = 6, AL = 11, and CK = 13 ... What to watch for today What to watch for today China may report grim second quarter GDP figures. Economists expect growth to slow to 7.5% on a yearly basis, the slowest pace since...Solution for BC is a common tangent segment of circles A and D. E. E 9. If AB= 6 cm, DC = 3 cm, and AD =12 cm, find ED. ... JK, KL, and LJ are all tangent to circle O (not drawn to scale). JA = 7 cm, AL = 6 cm, and CK = 9cm. ...When it comes to motivation—especially for health and fitness goals—being an "inny" or an "outy" can make all the difference. The "inny" I'm talking about is "intrinsic motivation,...Correct answer - Jk, kl, and lj are all tangent to circle o. ja = 12, al = 11, ck = 13. what is the perimeter of triangle jkl? a. 46 units b. 72 units cJK, KL, and LJ are all tangent to circle O (not drawn to scale). JA = 7 cm, AL = 6 cm, and CK = 9 cm. Find the perimeter of AJKL. 04:29. 7. In the figure below; M is …Click here 👆 to get an answer to your question ️ JK, KL, and LJ are all tangent to circle O. The diagram is not drawn to scale. If JA=10, AL=10, AL=10, AL=10, JK, KL, and LJ are all tangent to O (not drawn to scale). JA = 14, AL = 15, and CK = 13. Find the perimeter of ^JKL (^ is in place for the triangle symbol thing) D. 84. All answers for Connexions Academy Geometry B - Inscribed Angles Learn with flashcards, games, and more — for freeNews (not dra to scale), and JK LJ.JA-9, AL-10, and CK-14. the perimeter of AJKL. (3 pts) BUY. Elementary Geometry for College Students. 6th Edition. ISBN: 9781285195698. Author: Daniel C. Alexander, Geralyn M. Koeberlein. Publisher: Cengage LearningThat's a lot of work and travel, and not much holiday, for athletes in the world's most popular sport. Liverpool's manager, Jurgen Klopp, does not like how hard soccer players mustJk, kl, and lj are all tangent to circle o. if ja=8, al=7 and ck=10, what is the perimeter of jkl? Answers: 2 Show answers Another question on Mathematics. Mathematics, 21.06.2019 17:50. For what values of x is x2 + 2x = 24 true? Answers: 2. Answer. Mathematics, 21.06.2019 19:00. Which values of p and q ...Because JK is tangent to circle L, m ∠LJK = 90 ° and triangle LJK is a right triangle. BY P ythagorean Theorem, LJ 2 + JK 2 = LK 2. 11 2 + x 2 = 18 2. 121 + x 2 = 324. Subtract 121 from each side. x 2 = 203. Take square root on both sides. x ≈ 14.2. Example 5 : If the line segment JK is tangent to circle L, find x. KL, and LJ are all tangent to circle O (not drawn to scale), and JK is congruent to LJ. JA=15 AL=9 , and CK=6. Find the perimeter of Triangle JKL * 42 30 48 60. Solution. Samuel. High school teacher · Tutor for 5 years. Answer: 60 . What to watch for today What to watch for today US retail sales may rise for the third straight month. Analysts predict a 0.8% increase in sales in June, the biggest jump in four m...Answer: 3 📌📌📌 question Jk, kl, and lj are all tangent to circle o. ja = 12, al = 11, ck = 13. what is the perimeter of triangle jkl? a. 46 units b. 72 units c. 50 units d. 36 units - the answers to estudyassistant.comKeep these three responses in your back pocket to get you off the hook. By clicking "TRY IT", I agree to receive newsletters and promotions from Money and its partners. I agree to ... . star. 5/5. heart. 2. verified. Verified answer. Lines JK, KL, and LJ are all tangent to …JK is the hypotenuse. The trig ratio that relkates the opposite leg to the hypotenuse is the sine. ... In ΔJKL, the measure of ∠L=90°, the measure of ∠J=33°, and KL = 25 feet. Find the length of LJ to the nearest tenth of a foot. star. 4.8/5. heart. 8. In ΔJKL, the measure of ∠L=90°, JK = 5. 8 feet, and LJ = 4. 4 feet.Answer: JK, KL, and LJ are all tangent to circle O. JA = 9, AL= 10, and CK= 14. What is the perimeter of triangle JKL? Step-by-step explanation:jk, kl, and lj are all tangent to o ja = 10, and ck = 14 find the perimeter of triangle jkl. verified. Verified answer. In triangle JKL and triangle PQR line JK = PQR line KL equals QR angle K equal angle Q then JKL must be congruent triangle PQR. heart. 8. verified. Verified answerJK, KL, and are all tangent to circle O. JA = 9, AL = 10, and CK = 14. What is the perimeter of AJKL? 66 units 46 units O 33 units 38 units. loading. See answer. loading. plus. Add answer +5 pts. loading. Ask AI. more. Log in to add comment. annacomitinii93 is waiting for your help. Add your answer and earn points.11. JK, KL, and LJ are all tangent to circle O. The diagram is not drawn to scale. If JA = 8, AL = 7, and CK = 10, what is the perimeter of A/KL? O25 030 036. Elementary Geometry For College Students, 7e. 7th Edition. ISBN: 9781337614085.Lines JK, KL, and LJ are all tangent to circle O. The Diagram is not drawn to scale. If JA = 10, AL = 23 and CK = 9, what is the perimeter of triangle JKL? a. 84 b. 66 c. 42 d. 38 Is the answer 84? Please explain how you got the answer. 2 months ago. Solution 1. Guest #9411870. 2 months ago.. Keep these three responses in your back pocket tPenile cancer is rare and most cases devel Q DAis tangent to the circle atAandDC is tangent to the JK, KL, and LJ are all tangent to circle O. The diagram is not drawn to scale If JA = 14, AL = 12, and CK = 8, what is the perimeter of AJKL? L perimeter = 2. JK, KL ... See Answer. Question: JK,KL, and LJ are all... | 677.169 | 1 |
The Magic of Quadrilateral Angles,...
Mục lục, quadrilaterals come in various forms, each with its own set of angles waiting to be discovered. Let's dive into the world of quadrilateral angles and uncover the mysteries they hold.
The Sum of Interior Angles
It's a fascinating fact - a quadrilateral has four interior angles, and their sum always amounts to 360°. This remarkable value is derived from the angle sum property of polygons. If we consider the number of triangles that can be formed within a polygon by drawing diagonals from a single vertex, we can calculate the sum of its interior angles. A handy formula comes to our rescue, stating that for a polygon with 'n' sides, the sum of its interior angles is given by (n - 2) × 180°.
Let's take a quadrilateral with four sides as an example. Applying the formula, we find that the sum of its interior angles is (4 - 2) × 180° = 2 × 180° = 360°. Hence, according to the angle sum property of a quadrilateral, the total of its interior angles is always a perfect 360°. This property comes in handy when we need to find the value of an unknown angle, given the knowledge of other angles. Simply subtract the sum of the known angles from 360°, and voila, the mystery is solved!
Exploring Interior and Exterior Angles
Within a quadrilateral, we find not only interior angles but also exterior angles. The interior angles lie inside the quadrilateral, while the exterior angles are formed between one side of the quadrilateral and another line extended from an adjacent side. Notably, each interior angle and its corresponding exterior angle form a linear pair.
For squares and rectangles, the interior angles are all right angles, measuring 90° each. As for other quadrilaterals, their interior angles may vary, adding to the excitement of discovery.
Formulas for Quadrilateral Angles
To delve deeper into the world of quadrilateral angles, let's explore some essential formulas:
Exterior angle = 180° - Interior angle: This formula helps find the value of the corresponding exterior angle when the interior angle is known. It can also be used in reverse to find the interior angle when the exterior angle is given.
If three angles of a quadrilateral are known, the fourth angle can be calculated using the formula: 360 - (Sum of the other 3 interior angles).
The sum of interior angles of a quadrilateral is represented by the formula: Sum = (n − 2) × 180°, where 'n' represents the number of sides of the quadrilateral. For a quadrilateral, with 'n' equal to 4, the sum is calculated as (4 − 2) × 180° = 360°.
The Enigma of Cyclic Quadrilaterals
When a quadrilateral is inscribed in a circle, it becomes a mesmerizing cyclic quadrilateral. All four vertices of the quadrilateral touch the circle, creating a mystical connection. Many theorems are associated with the angles of quadrilaterals inscribed in a circle. One such theorem states that the opposite angles in a cyclic quadrilateral are supplementary, meaning their sum is equal to 180°.
Let's take a moment to admire the beauty of a cyclic quadrilateral and observe how its opposite angles elegantly add up to 180°.
Caption: Interior and Exterior Angles of a Quadrilateral
In the Quest for Knowledge
If you're hungry for more knowledge about quadrilateral angles, here are some related links to explore:
Interior Angles
Quadrilaterals
Exterior Angles of Polygons
The world of quadrilateral angles is truly captivating. Knowing the secrets of interior and exterior angles and the formulas that unlock their mysteries allows us to navigate the enchanting realm of quadrilaterals with confidence. So, go forth and embrace the magic of quadrilateral angles | 677.169 | 1 |
Cartesian coordinate system consists of three mutually perpendicular axes defined by unit vectors intersecting at the origin.
Unit vectors have unit magnitude and only point the direction.
For Cartesian systems, î, ĵ, and k̂ are the unit vectors along the positive x, y, and z axes, respectively.
In this frame, every vector is the sum of its orthogonal projections onto each of the axes. These projections are known as the vector components.
We write each vector component by its magnitude and a unit vector along the axis. The magnitudes represent the scalar components of the vector.
Thus, any vector is the vector sum of its components.
Using the scalar components, the magnitude of a vector is given by the square root of the sum of the squares of its components.
For a vector on a plane, its direction, which is the angle made by the vector with the positive x-axis in the anticlockwise direction, is the tan inverse of the y-component over the x-component.
Conversely, the x-component of a vector is the magnitude times the cosine of the angle made with the x-axis.
Its y-component is the magnitude times the sine of the angle made by the vector with the x-axis.
2.3: Vector Components in the Cartesian Coordinate System
Vectors are usually described in terms of their components in a coordinate system. Even in everyday life, we naturally invoke the concept of orthogonal projections in a rectangular coordinate system. For example, if someone gives you directions for a particular location, you will be told to go a few km in a direction like east, west, north, or south, along with the angle in which you are supposed to move. In a rectangular (Cartesian) xy-coordinate system in a plane, a point in a plane is described by a pair of coordinates (x, y). In a similar fashion, a vector in a plane is described by a pair of its vector coordinates. The x-coordinate of a vector is called its x-component, and the y-coordinate is called its y-component. In the Cartesian system, the x and y vector components are the orthogonal projections of this vector onto the x- and y-axes, respectively. In this way, each vector on a Cartesian plane can be expressed as the vector sum of its vector components in both x and y directions.
It is customary to denote the positive direction of the coordinate axes by unit vectors. The vector components can now be written as their magnitude multiplied by the unit vector in that direction. The magnitudes are considered as the scalar components of a vector.
When we know the scalar components of any vector, we can find its magnitude and its direction angle. The direction angle, or direction for short, is the angle the vector forms with the positive direction on the x-axis. The angle that defines any vector's direction is measured in a counterclockwise direction from the +x-axis to the vector. The direction angle of any vector is defined via the tangent function. It is defined as the ratio of the scalar y component to the scalar x component of that vector.
In many applications, the magnitudes and directions of vector quantities are known, and we need to find the resultant of many vectors. In such cases, we find vector components from the direction and magnitude. Thus, the x-component is given by the product of the magnitude of that vector and the cosine angle made with the x-axis in the counterclockwise direction. Similarly, the y-component is the product of vector magnitude and the sine angle made with the x-axis in the counterclockwise direction. | 677.169 | 1 |
Position vector in cylindrical coordinates. Question: 25.12 Beginning with the general expressio...
CThere are three commonly used coordinate systems: Cartesian, cylindrical and spherical. In this chapter we will describe a Cartesian coordinate system and a cylindrical coordinate system. 3.2.1 . Cartesian Coordinate System . Cartesian coordinates consist of a set of mutually perpendicular axes, which intersect at a The formula which is to determine the Position Vector that is from P to Q is written as: PQ = ( (xk+1)-xk, (yk+1)-yk) We can now remember the Position Vector that …Position Vectors in Cylindrical Coordinates. This is a unit vector in the outward (away from the $z$ -axis) direction. Unlike $\hat {z}$, it depends on your azimuthal angle. The position vector has no component in the tangential $\hat {\phi}$ direction Detailed30 de mar. de 2016 ... 3.1 Vector-Valued Functions and Space Curves ... The origin should be some convenient physical location, such as the starting position of the ...4.6: Gradient, Divergence, Curl, and Laplacian. In this final section we will establish some relationships between the gradient, divergence and curl, and we will also introduce a new quantity called the Laplacian. We will then show how to write these quantities in cylindrical and spherical coordinates The directions of increasing r and θ are defined by the orthogonal unit vectors er and eθ. The position vector of a particle has a magnitude equal to the radial ...Starting with polar coordinates, we can follow this same process to create a new three-dimensional coordinate system, called the cylindrical coordinate system. In this way, cylindrical coordinates provide a natural extension of polar coordinates to three dimensions Derivative in cylindrical coordinates. Ask Question Asked 3 years, 5 months ago. Modified 3 years ago. Viewed 583 times 0 $\begingroup$ Why ... The position vector (or the radius vector) is a vector R that represents the position of points in the Euclidean space with respect to an arbitrarily selected point O, known as the origin. ...In this section, we look at two different ways of describing the location of points in space, both of them based on extensions of polar coordinates. As the name suggests,The following are Vector Calculus Cylindrical Polar Coordinates equations.It is also possible to represent a position vector in Cartesian and cylindrical coordinates as follows: r P = X P I + Y P J + Z P K = ρ ρ ^ + Z P K {\displaystyle {\mathsf {r}}_{P}=X_{P}{\mathsf {I}}+Y_{P}{\mathsf {J}}+Z_{P}{\mathsf {K}}=\rho {\boldsymbol {\hat {\rho }}}+Z_{P}{\mathsf {K}}}Azimuth: θ = θ = 45 °. Elevation: z = z = 4 y z = r cos θ = r sin θ = z r θ z = x2 +y2− −− ...The velocity of P is found by differentiatingThe norm for a vector in cylindrical coordinates can be obtained by transforming cyl.-coord. to cartesian coord.: ... Representing a point in cartesian space as a position vector in spherical coordinates. 1. A question about vector representation in polar coordinates. 0. How to calculate cross product of $\hat{x}$ and $-\hat{x}$ in … We can explicitly show that the spherical unit vectors depend on position by calculating their components in. Cartesian coordinates. • To begin, we first must ...Definition of cylindrical coordinates and how to write the del operator in this coordinate system. Join me on Coursera: the position vector in cylindrical coordinates is r = rer + zez then velocity and acceleration ... unit vectors in spherical and Cartesian coordinates: er = sin ...Jun 24, 2020 · How do you find the unit vectors in cylindrical and spherical coordinates in terms of the cartesian unit vectors?Lots of math.Related videovelocity in polar ... The coordinate system directions can be viewed as three vector fields , and such that: with and related to the coordinates and using the polar coordinate system relationships. The coordinate transformation from the Cartesian basis to the cylindrical coordinate system is described at every point using the matrix : 1The specify the coordinate of particle then position vector can be expressed in ... coordinates which are used in cylindrical coordinates system. Notice that, ˆ ˆ. ˆ.1 YouSince we do not know the coordinates of QM or the values of n and m, we cannot simplify the equation. Example 5. Given a point q = (-10, 5, 3), determine the position vector of point q, R. Then, determine the magnitude of R. Solution. Given the point q, we can determine its position vector: R = -10i + 5j -3k.11 de jul. de 2015 ... transform the vector A into cylindrical and spherical coordinates. (b.) transform the rectangular coordinate point P (1,3,5) into cylindrical 30 de mar. de 2016 ... 3.1 Vector-Valued Functions and Space Curves ... The origin should be some convenient physical location, such as the starting position of the ...expressing an arbitrary vector as components, called spherical-polar and cylindrical-polar coordinate systems. ... 5 The position vector of a point in spherical- ...12 2. Particles and Cylindrical Polar Coordinates We can write this position vector using cylindrical polar coordinates by substituting for x and y in terms of r and (): r = r cos( ())Ex + r sin( ())Ey + zEz . Before we use this representation to establish expressions for the velocity and acceleration vectors, it is prudent to pause and define ...TheThe formula which is to determine the Position Vector that is from P to Q is written as: PQ = ( (xk+1)-xk, (yk+1)-yk) We can now remember the Position Vector that is PQ which generally refers to a vector that starts at the point P and ends at the point Q. Similarly if we want to find the Position Vector that is from the point Q to the point P ...It is an example of a vector field, a vector that deponds on position in space. ... a) Express the vector field in cylindrical coordinates. Make sure toOP - position vector (specifies position, given the choice of the origin O). Clearly, r ... •Cartesian coordinates, cylindrical coordinates etc. v v v v P P P P { x a a a a P P P P { x. 6 Let be the unit vectors Cartesian coordinate system: The reference frame is Nov 19, 2019 · Definition of cylindrical coordinates and how to write the del operator in this coordinate system. Join me on Coursera: Don't worry! This article explains complete step by step derivation for the Divergence of Vector Field in Cylindrical and Spherical Coordinates. Divergence of aIdentify the direction angle of a vector in a plane. Explain the connection between polar coordinates and Cartesian coordinates in a plane. Vectors are usually ...Please see the picture below for clarity. So, here comes my question: For locating the point by vector in cartesian form we would move first Ax A x in ax→ a x →, Ay A y in ay→ a y → and lastly Az A z in az→ a z → and we would reach P P. But in cylindrical system we can reach P P by moving Ar A r in ar→ a r → and we would reach 4. There is a clever way to look at vectors. They are differential operators, for example: x = ∂ ∂x. x = ∂ ∂ x. So, in a Cartesian basis, we would have. r = x ∂ ∂x + y ∂ ∂y + z ∂ ∂z. r = x ∂ ∂ x + y ∂ ∂ y + z ∂ ∂ z. It also follows that the …In this section, we look at two different ways of describing the location of points in space, both of them based on extensions of polar coordinates. As the name suggests, cylindrical coordinates are useful for dealing with problems involving cylinders, such as calculating the volume of a round water tank or the amount of oil flowing through a pipe. position vectors in cylindrical coordinates: $$\vec r = \rho \cos\phi \hat x + \rho \sin\phi \hat y+z\hat z$$ I understand this statement, it's the following, I don't understand how a 3D position can be expressed thusly: $$\vec r = \rho \hat \rho + z \hat z$$ Thanks for any insight and help!A cylindrical coordinate system is a three-dimensional coordinate system that specifies point positions by the distance from a chosen reference axis (axis L in the image opposite), the direction from the axis relative to a …). 0. My Textbook wrote the Kinetic Energy whileBy Milind Chapekar / All Tips and News. Cylindrical Coor Figure 7.4.1 7.4. 1: In the normal-tangential coordinate system, the particle itself serves as the origin point. The t t -direction is the current direction of travel and the n n -direction is always 90° counterclockwise from the t t -direction. The u^t u ^ t and u^n u ^ n vectors represent unit vectors in the t t and n n directions respectively. A Cartesian Vector is given in Cylindrical Coordinates by (1 In If the coordinate surfaces intersect at right angl... | 677.169 | 1 |
Hyperbola. A Hyperbola (red): features OurExplore math with our beautiful, free online graphing calculator. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. Hyperbola With Foci | Desmos Loading... Real Numbers. Addition. Quadrilaterals. Ratios. Geometry. Students can input foci and point values to change the hyperbola and the equation will be given.Hyper The eccentricity e is the measure of the amount of curvature in the hyperbola's branches, where e = c/a.Since the foci are further from the center of an hyperbola than are the vertices (so c > a for hyperbolas), …In this video we plot a hyperbola in Desmos using the Pythagorean Triple 11, 60, 61. We use these numbers from the Pythagorean Triple (and the squares of the... A parabola has a single directrix and one focus, with the other one placed at infinity. A given point of a parable is at the same distance from both the focus and the directrix. You can meet this conic at our parabola calculator. A hyperbola has two directrices and two foci. The difference in the distance between each point and the two foci is - k) 2 b 2 = 1A hyperbola is a conic section that is the set of all points in a plane such that the difference of the distances from two fixed points (foci) is a constant. The foci of a hyperbola are located at: $$\left (\frac {c} {2},0\right) \text { and } \left (-\frac {c} {2},0\right)$$. Where c is the distance between the foci.The slope of the line between the focus and the center determines whether the hyperbola is vertical or horizontal. If the slope is , the graph is horizontal. If the slope is undefined, the graph is vertical. WhatLatus rectum of a hyperbola is a line segment perpendicular to the trans Free Hyperbola Foci (Focus Points) calculator - Calculate hyperbola focus points given equation step-by-step. TheParabola Calculator. Enter the equation of parabola: Submit: Computing... Get this widget. Build your own widgetTheFree Hyperbola Foci (Focus Points) calculator - Calculate hyperbola focus points given equation step-by-step.The Hyperbola in Standard Form. A hyperbola 23 is the set of points in a plane whose distances from two fixed points, called foci, has an absolute difference that is equal to a positive constant. In other words, if points \(F_{1}\) and \(F_{2}\) are the foci and \(d\) is some given positive constant then \((x,y)\) is a point on the hyperbola if \(d=\left|d_{1} …They are similar because the equation for a hyperbola is the same as an ellipse except the equation for a hyperbola has a - instead of a + (in the graphical equation). As for your second …Explore math with our beautiful, free online graphing calculator. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. Hyperbola …In this video we plot a hyperbola in Desmos using the Pythagorean Triple 11, 60, 61. We use these numbers from the Pythagorean Triple (and the squares of the...Source: en.wikipedia.org. Some Basic Formula for Hyperbola. Major Axis: The line that passes through the center, the focus of the hyperbola and vertices is the Major Axis.Length of the major axis = 2a. The equation is: \(\large y=y_{0}\) Minor Axis: The line perpendicular to the major axis and passes by the middle of the hyperbola are the Minor Axis.Explore math with our beautiful, free online graphing calculator. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. Hyperbola - Horizontal Transverse Axis | DesmosA hyperbola is a type of conic section that looks somewhat like a letter x. A hyperbola is a set of all points P such that the difference between the distances from P to the foci, F 1 and F 2, are a constant K.Before learning how to graph a hyperbola from its equation, get familiar with the vocabulary words and diagrams below. AprEN: conic-sections-calculator descriptionIdentify Conics Section Equations Calculator for circles, parabola, hyperbola ... focus with conic standard form calculator. Enter an equation above eg. y=x^2+2x+ ...Free Hyperbola Foci (Focus Points) calculator - Calculate hyperbola focus points given equation step-by-step … Hyperbola from Foci - Desmos ... Loading...Free Hyperbola Eccentricity calculator - Calculate hyperbola eccentricity given equation step-by-step b How do you find an equation that models a hyperbolic lens with a=12 inches and foci that are 26 inches apart, assume that the center of the hyperbola is the origin and the transverse axis is vertical? A comet follows the hyperbolic path described by #x^2/4 -y^2/19 = 1#, where x and y are in millions of miles. ...Graph the ellipse using the fact that a=3 and b=4. Stan at (2.-1) and locate two points each 3 units away from (2.-1) on a horizontal line, one to the right of (2.-1) and one to the left. Locate two other points on a vertical line through (2.-1), one 4 units up and one 4 units down. Since b>a, the vertices are on the Interactive online graphing calculator - graph functions, conics, and inequalities free of charge The directrix of a conic section is the line which, together with the point known as the focus, serves to define a conic section as the locus of points whose distance from the focus is proportional to the horizontalThe hyperbola foci formula is the same for vertical and horizontal hyperbolas and looks like the Pythagorean Theorem: {eq}a^2 + b^2 = c^2 {/eq} where c represents the focal distance (the distanceExplore math with our beautiful, free online graphing calculator. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. Hyperbola from Foci. Save Copy. Log InorSign Up. a sect cosA ngle − batant sinA ngle + h, a se ... TheSolve it with our Calculus problem solver and calculator. Not the exact question you're looking for? Post any question and get expert help quickly. Start ...02-Aug-2020 ... Find an equation for the hyperbola that satisfies the given conditions. Foci: (±7, 0), vertices: (±4, 0)hyperbola-foci-calculator. foci x^2-y^2=1. en. Related Symbolab blog posts. My Notebook, the Symbolab way. Math notebooks have been around for hundreds of years. You write down problems, solutions and notes to go back...Hyperbola from Vertices and Foci. Get the free "Hyperbola from Vertices and Foci" widget for your website, blog, Wordpress, Blogger, or iGoogle. Find more Mathematics widgets in Wolfram|Alpha.Step 1: Enter the inputs, such as centre, a, and b value in the respective input field Step 2: Now click the button "Calculate" to get the values of a hyperbola Step 3: Finally, the focus, asymptote, and eccentricity will be displayed in the output field What is Meant by Hyperbola?A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances between two fixed points stays constant. The two given points are the foci of the hyperbola, and the midpoint of the segment joining the foci is the center of the hyperbola. The hyperbola looks like two opposing "U‐shaped" curves, as shown in …Ellipse Calculator : semimajor and semiminor axes, focal distance, vertices, eccentricity, directrix, perimeter and area ... Foci distance `c = sqrt(a^2-b^2)` `c = sqrt(b^2-a^2)` Focal distance (FF') 2c: 2c: Center - Vertex distance: a: b: ... Hyperbola calculator Circle calculator Conic sections calculators Geometry calculators MathematicsIt looks like you know all of the equations you need to solve this problem. I also see that you know that the slope of the asymptote line of a hyperbola is the ratio $\dfrac{b}{a}$ for a simple hyperbola of the form $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$$Free Ellipse calculator - Calculate ellipse area, center, radius, foci, vertice and eccentricity step-by-stepFree Hyperbola Asymptotes calculator - Calculate hyperbola asymptotes given equation step-by-step SolvedWe can write the equation of a hyperbola by following these steps: 1. Identify the center point (h, k) 2. Identify a and c. 3. Use the formula c 2 = a 2 + b 2 to find b (or b 2) 4. Plug h, k, a, and b into the correct pattern.Free Hyperbola calculator - Calculate Hyperbola center, axis, foci, vertices, eccentricity and asymptotes step-by-step The foci are side by side, so this hyperbola's branches are side by side, and the center, foci, and vertices lie on a line paralleling the x -axis. So the y part of the equation will be subtracted and the a2 will go with the x part of the equation. The center is midway between the two foci, so the center must be at (h, k) = (−1, 0).Apr Interactive online graphing calculator - graph functions, conics, and inequalities free of charge Free Ellipse Foci (Focus Points) calculator - Calculate ellipse focus points given equation step-by-step . EN: conic-sections-calculator description Click here to view image. Where, a = semi-major axis of the hyperb Hyperbola Calculator Provide all necessary parameters of the hyp DefinitionLatus rectum of a hyperbola is a line segment perpendicular to the trans The line through the foci F 1 and F 2 of a... | 677.169 | 1 |
Geometry – Infinite Figures Inscribed
December 14, 2021
Sometimes, we come across GMAT geometry questions that involve figures inscribed inside other figures. One shape inside of another shape may not be difficult to work with, but how do we handle problems that involve infinite figures inscribed inside one another? Such questions can unsettle even the most seasoned test takers. Let's take a look at one of them today:
A square is drawn by joining the midpoints of the sides of a given square. A third square is drawn inside the second square in this way and this process is continued indefinitely. If a side of the first square is 4 cm. Determine the sum of areas of all squares?
(A) 18
(B) 32
(C) 36
(D) 64
(E) None
Now the first thing that might come to our mind is this – how do we mathematically, in the time limit of approximately 2 minutes, calculate areas of infinite squares?
There has to be a formula for this. Recall that we do, in fact, have a formula that calculates the sum of infinite terms – the geometric progression formula! Let's see if we can use that to find the areas of the squares mentioned in this problem.
First, we'll see if we can find a pattern in the areas of the squares:
Say the side of the outermost square is "". The area of the outermost square will be and half of the side will be . The side of the next square inside this outermost square (the second square) forms the hypotenuse of a right triangle with legs of length each. Using the Pythagorean Theorem:
So now we know the sides of the second square will each equal , and the area of the second square will be .
Our calculations will be far easier if we note that the diagonal of the second square will be the same length as the side of the outer square. We know that area of a square given diagonal is , so that would directly bring us to as the area of the second square.
The second square and the square inscribed further inside it (the third square) will have the same relation. The area of the third square will be .
Now we know the area of every subsequent square will be half the area of the outside square. So the total area of all squares …Each term is half the previous term.
Therefore, the sum of an infinite Geometric Progression where the common ratio is less than 1 is:
Total Sum
: First Term
: Common Ratio
Sum of areas of all squares …
Sum of areas of all squares
Sum of areas of all squares
Since is the length of the side of the outermost square, and (this fact is given to us by the questions stem), the sum of the areas of all the squares . Therefore, our answer is (B).
We hope you understand how we have used the geometric progression formula to get to our answer. To recap, the sum of an infinite geometric progression is . | 677.169 | 1 |
Question Video: Lines of Symmetry in Plane Geometry
Mathematics
Does the following figure have a line of symmetry?
00:32
Video Transcript
Does the following figure have a
line of symmetry?
Is there a way we could fold this
shape in half? Watch what happens if we fold the
shape along this dotted line. Here's one-half. Here's the other half. And when we open it out, it looks
like this. Does this figure have a line of
symmetry? Yes, it does. | 677.169 | 1 |
8.3 Tests for Parallelograms
Jun 10, 2012
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8.3 Tests for Parallelograms. Objectives. Recognize the conditions that ensure a quadrilateral is a parallelogram. Prove that a set of points forms a parallelogram in the coordinate plane. Conditions for a Parallelogram.
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8.3 Tests for Parallelograms Recognize the conditions that ensure a quadrilateral is a parallelogram. • Prove that a set of points forms a parallelogram in the coordinate plane.
Conditions for a Parallelogram • Obviously, if the opposite sides of a quadrilateral are parallel, then it is a parallelogram; but there are other tests we can also apply to a quadrilateral to test whether it is a parallelogram or not.
Conditions for a Theorems • Theorem 8.9 – If both pairs of opposite sides are ≅, then the quad. is a . • Theorem 8.10 – If both pairs of opposite s are ≅, then the quad. is a . • Theorem 8.11 – If diagonals bisect each other, then the quad. is . • Theorem 8.12 – If one pair of opposite sides is ║ and ≅, then the quad. is a .
Given: Proof: CPCTC. By Theorem 8.9, if both pairs of opposite sides of a quadrilateral are congruent, the quadrilateral is a parallelogram. Therefore, ABCD is a parallelogram. Example 1: Write a paragraph proof of the statement: If a diagonal of a quadrilateral divides the quadrilateral into two congruent triangles, then the quadrilateral is a parallelogram. Prove:ABCD is a parallelogram.
Given: Your Turn: Write a paragraph proof of the statement: If two diagonals of a quadrilateral divide the quadrilateral into four triangles where opposite triangles are congruent, then the quadrilateral is a parallelogram. Prove:WXYZ is a parallelogram.
Proof: by CPCTC. By Theorem 8.9, if both pairs of opposite sides of a quadrilateral are congruent, the quadrilateral is a parallelogram. Therefore, WXYZ is a parallelogram. Your Turn:
Example 2: Some of the shapes in this Bavarian crest appear to be parallelograms. Describe the information needed to determine whether the shapes are parallelograms. Answer: If both pairs of opposite sides are the same length or if one pair of opposite sides is a congruent and parallel, the quadrilateral is a parallelogram. If both pairs of opposite angles are congruent or if the diagonals bisect each other, the quadrilateral is a parallelogram.
Your Turn: The shapes in the vest pictured here appear to be parallelograms. Describe the information needed to determine whether the shapes are parallelograms. Answer: If both pairs of opposite sides are the same length or if one pair of opposite sides is congruent and parallel, the quadrilateral is a parallelogram. If both pairs of opposite angles are congruent or if the diagonals bisect each other, the quadrilateral is a parallelogram.
Example 3: Determine whether the quadrilateral is a parallelogram. Justify your answer. Answer: Each pair of opposite sides have the same measure. Therefore, they are congruent. If both pairs of opposite sides of a quadrilateral are congruent, the quadrilateral is a parallelogram.
Your Turn: Determine whether the quadrilateral is a parallelogram. Justify your answer. Answer: One pair of opposite sides is parallel and has the same measure, which means these sides are congruent. If one pair of opposite sides of a quadrilateral is both parallel and congruent, then the quadrilateral is a parallelogram.
Tests for Parallelograms • Both pairs of opposite sides are parallel. • Both pairs of opposite sides are congruent. • Both pairs of opposite angles are congruent. • The diagonals bisect each other. • A pair of opposite sides is both parallel and congruent.
A B D C Example 4a: Find x so that the quadrilateral is a parallelogram. Opposite sides of a parallelogram are congruent.
Example 4a: Substitution Distributive Property Subtract 3x from each side. Add 1 to each side. Answer: When x is 7, ABCD is a parallelogram.
D E G F Example 4b: Find y so that the quadrilateral is a parallelogram. Opposite angles of a parallelogram are congruent.
Example 4b: Substitution Subtract 6y from each side. Subtract 28 from each side. Divide each side by –1. Answer: DEFG is a parallelogram when y is 14.
a. b. Answer: Answer: Your Turn: Find m and n so that each quadrilateral is a parallelogram.
Parallelograms on the Coordinate Plane • We can use the Distance Formula and the Slope Formula to determine if a quadrilateral is a parallelogram on the coordinate plane. • Just pick one of the tests… and apply either or both of the formulas. | 677.169 | 1 |
CCSS.Math.Content.4.MD.C.7
Recognize angle measure as additive. When an angle is decomposed into non-overlapping parts, the angle measure of the whole is the sum of the angle measures of the parts. Solve addition and subtraction problems to find unknown angles on a diagram in real world and mathematical problems, e.g., by using an equation with a symbol for the unknown angle measure.
Geometric Angles Lesson Plan: What's the Measure on that Angle?
Posted by allisyn on
June 5, 2012
In this lesson plan, which is adaptable for grades 3 through 5, students use BrainPOP resources to use and apply vocabulary associated with geometric angles. Students will identify examples of differe...
Basic Geometry Lesson Plan: A Tangled Web Puzzle Game
Posted by allisyn on
May 24, 2012
In this lesson plan adaptable for grades 3 through 8, students use BrainPOP resources (including an interactive game) to learn about basic geometry concepts. Students will explore angles, parallel lin... | 677.169 | 1 |
Area Of A Triangle Worksheet parallel to 1 aspect of a triangle and intersects the opposite two sides, it divides each side proportionally.
Students can assess their strengths and weaknesses by fixing all questions from Triangle Area and Perimeter Worksheet. Make use of those worksheets and perceive different formulas and methods of fixing isosceles triangle, equilateral triangle, scalene, and right-angled triangle areas and perimeter.
This worksheet shows students how to discover the world of a triangle. Triangles measurements have mixed models, decimals, and/or fractions. Each printable worksheet right here supplies 5 challenging issues in word format; that includes equilateral, scalene and isosceles triangles.
What Are Activex Controls In Word
Then, let the children go around the classroom and college searching for triangle formed objects. Once they see any triangle-shaped object, they have to make a remark of its height and base.
When calculating space, your reply should at all times have items squared . On this web page we'll look at the realm of scalene triangles. A scalene triangle is a triangle where all the sides and the entire angles are completely different.
This quiz is in multiple selection format. Download this quiz to evaluate in case your students are greedy how to find the world of a triangle. Common Core State Standard 6.G.1 This work is licensed underneath a Creative Commons Attribution-NonCommercial-NoDerivs three.0 Unported License.
Discovering The Realm Of A Proper Triangle Worksheet
Find the lacking aspect utilizing the area. On profitable completion of the worksheets, college students will be in a position to get an concept of how to remedy numerous real-life problems based on the subject at hand. K5 Learning provides free worksheets, flashcardsand inexpensiveworkbooksfor kids in kindergarten to grade 5.
Coordinate Geometry Math Worksheet For 7th Grade Kids
These Free Area Of Triangles Worksheets exercises may have your children engaged and entertained while they improve their expertise. Click on the image to view or obtain the image.
Here we are going to be taught concerning the space of a triangle, including tips on how to discover the area of a triangle with given dimensions. Find the world and perimeter of the triangle given below.
These worksheets are prepared for sophistication practice / additional apply assignments / apply from house. More you practice, better you'll be so try as many worksheets as you'll have the ability to on Area of Triangle.
A triangle is a geometrical determine that's shaped when three straight strains meet. The triangles have three corners and three sides. These corners are also referred to as angles, and the purpose where two sides meet is called the vertex.
Become a memberto access extra content material and skip ads. Get a clear view of geometry with this worksheet!
First Grade space rectangle worksheet worksheets finding perimeter grade math. The floor space is the covering of a three-dimensional figure.
Because the triangle is isosceles, the length of two sides shall be equal and the altitude bisects the base of the triangle. Calculating to search out space of triangle utilizing formula. By the time your class has accomplished this intensive collection, they'll undoubtedly be consultants at finding the world of a triangle.
A scalene triangle is a plane determine with three unequal sides. Learn about Heron's formulation and its utility in arriving at the area of such figures with this bundle of printable worksheets.
Find the area of a triangle, whose sides are 5 m, 4 m and angle is 120°. Then, the three sides of a triangle are 5x, 2x, 6x. ˆ´ The Triangle perimeter is 35 m, one other side is eight m, and the area is forty seven.81 m².
It is based on the semi-perimeter of the triangle . It is frequent to forget the units for area in the final answer.
We can additional classify triangles on the idea of their angles. Incorporate this set of the area of triangles worksheets that features decimal dimensions. Calculate the world by plugging in the measures of the bottom and the height within the area of a triangle formulation.
Students will apply calculating the world of a triangle given its base and height. This worksheet also focuses on calculating the base/or peak given the triangle's space and its base/or peak.
This web page includes Geometry Worksheets on angles, coordinate geometry, triangles, quadrilaterals, transformations and three-dimensional geometry worksheets. Get out these rulers, protractors and compasses as a result of we've got some great worksheets for geometry! The quadrilaterals are meant to be reduce out, measured, folded, compared, and even ….
Using the diagram under, determine the area of the triangle ABC. Higher – There is an alternate method to search out the world of a triangle which involves sine, two sides and the included angle. This is usually pronounced as "base instances height".
To discover a lacking angle bisector, altitude, or median, use the ratio of corresponding sides. If we now have two similar triangles, right here we have triangle abc is much like triangle def..
Help your college students really feel assured with exam-style questions and the methods they'll have to answer them appropriately with our dedicated GCSE maths revision programme. Compound shapes are made from two or extra simple shapes, corresponding to a quadrilateral with a triangle on high. To discover the area of compound shapes, we break into the part parts and find the world of each shape.
This can require college students to work with decimals so it is necessary these abilities are safe. They may have to convert between totally different metric models – for example, if one side is given in centimetres and the other in millimetres.
Multiply the bottom and peak and divide by two, to calculate the realm. This math worksheet introduces students to geometry by having them discover the surface area of rectangular prisms. The prime of this useful resource has a step-by-step example, giving youngsters a conceptual understanding of tips on how to calculate the floor area.
Here, the realm is the region occupied by the triangle. As already discussed before in kinds of triangles, we would find the area of equilateral triangles, isosceles triangles, and scalene triangles. Success – particularly in geometry issues – that tastes the sweetest has oodles of rigor at its core!
We offer probably the most unique and largest database of practice worksheets for grade 1 to grade 12. All worksheets are printable and have been prudently compiled for all students, you can obtain PDF by clicking any hyperlink above. Get newest Class 7 Practical Geometry subject and chapter clever query and answer booklets and use them for every day apply.
Cazoom Maths have provided a variety of totally different polygon worksheets appropriate for all talents. You will discover a worksheet right here with everything you need when figuring out and naming polygons in addition to the formulas for calculating inside and exterior angles.. Use this guided note sheet to follow finding the world of triangles.
Related posts of "Area Of A Triangle Worksheet" government are designed to work together. Taxpayers ought to "completely not" file an tailored acknowledgment at...
Ionic Bonding Worksheet Key. Teachers Pay Teachers is a web-based market where lecturers buy and promote unique educational supplies. If you need to rapidly find the word you wish to search, use Ctrl + F, then kind the word you wish to search. Never use worksheets "then you now acknowledge that i'm not a fan...
Experimental Variables Worksheet Answers. This type of research includes solution-oriented questioning for a selected event, expertise, or circumstance. 224 Hong Kong households (primary school-aged children and their parents) registered to this system have been thought-about the experimental group and have been uncovered to a web-based intervention over three months. An outlier is an information level... | 677.169 | 1 |
Intersecting Chords Worksheets
What are Chords in Circle?
The chord is defined as the line segment that connects the two points on the circumference of the circle. Remember that diameter is the longest chord passing through the center of the circle and drawn across the circle. In case the lien segment doesn't stop at the circumference of the circle and extends to infinity, then that lie won't be a chord, it will be known as the secant.
The formula for finding the chord is based on the information given to you about the circle. If you know the radius and the measure of angle at the center made by the chord, then you would use the formula:
Chord length = 2 (radius) x sin (angle / 2). If you know the perpendicular distance for the center to the chord and the radius, then you will use the formula:
Chord length = 2 √r2 – d2. Where d is the perpendicular destine from the chord to the center of the circle. | 677.169 | 1 |
Trigonometry (from Greek trigōnon, "triangle" and metron, "measure") is a branch of mathematics that studies relationships between side lengths and angles of triangles. The field emerged in the Hellenistic world during the 3rd century BC from applications of geometry to astronomical studies.
\large\fbox{Derivatives :}
Derivatives :
Derivatives of Basic Trigonometric Functions:
(sinx)′=cosx,(cosx)′=−sinx. Using the quotient rule it is easy to obtain an expression for the derivative of tangent: (tanx)′=(sinxcosx)′=(sinx)′cosx−sinx(cosx)′cos2x=cosx⋅cosx−sinx⋅(−sinx)cos2x=cos2x+sin2xcos2x=1cos2x.
\large\fbox{Basics :}
Basics :
There are six trigonometric ratios, sine, cosine, tangent, cosecant, secant and cotangent. These six trigonometric ratios are abbreviated as sin, cos, tan, csc, sec, cot. These are referred to as ratios since they can be expressed in terms of the sides of a right-angled triangle for a specific angle θ. | 677.169 | 1 |
A Guide to Transversals and Related Angles in Geometry
Understanding transversals and related angles is an important part of geometry. Knowing how to identify these relationships, as well as how to use them to solve problems, can help you excel in your geometry course. In this blog post, we'll explain what transversals are and how to use them.
What is a Transversal?
A transversal is a line that intersects two or more other lines at different points. This means that a transversal cuts across two or more lines at the same time. When this happens, several angle pairs are created by the intersection of the transversal with the two lines it cuts across. These angle pairs are known as "related angles" because they share a common vertex (the point at which all three lines intersect).
Types of Related Angles
The types of related angles formed when a transversal intersects two other lines depending on their position relative to one another. If the two lines are parallel, then four sets of related angles will be formed, including corresponding angles and alternate interior/exterior angles. If the two lines are not parallel, then only three sets of related angles will be formed, including alternate interior/exterior angles, corresponding angles, and consecutive interior/exterior angles.
Solving Problems with Transversals and Related Angles
Transversals and related angles can be used to solve various problems in geometry such as finding missing lengths or determining unknown measures of certain shapes. To do this, you must identify which type(s) of related angles have been formed and what information has been given in the problem before attempting to solve it using an equation or diagram. By doing so, you can quickly find solutions for even complex geometric problems with relative ease.
Conclusion:
In conclusion, understanding transversals and related angles are essential for succeeding in geometry courses. Transversals are lines that intersect two or more other lines at different points while creating angle pairs known as "related angles" based on their position relative to each other (parallel or non-parallel). Additionally, transversals and related angles can be used to solve various problems in geometry such as finding missing lengths and determining unknown measures of certain shapes. With practice and proper knowledge about transversals and related angles in hand, students should be able to tackle any geometry problem that comes their way! | 677.169 | 1 |
The below is a question from a paper. Sorry it's in Afrikaans but I will explain. They give us information on the diagram, as well as the equations of two of the lines. In 4.1 we have to prove that ED is parralel to AB...I did that. Then 4.2 they say if E is the midpoint of BC, calculate the coordinates of B...I did that too...Please help with the next question...Calculate the angle CBA..and if I may, could you please explain to me what they mean with the next question which is to calculate the size of triangle CBA...do they mean the area?..All help will be greatly appreciated..Thank you. | 677.169 | 1 |
Sample Gre Geometry Formula Practice Questions
Those are it in a nutshellthe GRE geometry formulas youll need on test day! However, memorizing formulas and putting them into practice are two very different things.
So that you can see the difference, weve pulled two problems from the Magoosh GRE prep course so you can try your hand at them.
Question 1: Multiple Choice1. The hypotenuse of a right triangle is 16 ft longer than the length of the shorter leg. If the area of this triangle is exactly 120 ft2, what is the length of the hypotenuse in feet?
Question 2: Numeric Entry2. In the diagram, point D is the center of the medium-sized circle that passes through C and E, and it is also the center of the largest circle that passes through A and G. Each of the diameters of the small circles with centers B and F equals the radius of the medium-sized circle with center D. The shaded area is what fraction of the largest circle?
Understand How The Formulas Came About
Still thinking about how to remember formulas? Before trying to cram any formula, try to understand how the formula was proven, each formulas elements, and why they make sense together. Understanding the workings of any given formula makes it a lot easier to remember.Lets test this with something simple. Consider the area of an isosceles triangle as ½bh, where b is the triangles base, and h is the height of the triangle. If we split the triangle vertically down the middle into two equal parts, itll form two right-angle triangles. Place the hypotenuses of these triangles together such that they form a rectangle. Youll find that the rectangles width is the same as half the base of the isosceles triangle, while the length of the rectangle is the same as the perpendicular height of the triangle.So essentially, the area of the rectangle and that of the isosceles triangle are the same. So if the dimensions are identical, as we stated in this example, then the ½bh of an isosceles triangle is the same as the L×B of a rectangle.Do you see how understanding the fundamentals of these shapes makes it easier to understand these formulas? This ultimately works in other areas of math beyond geometry.
The 28 Critical Sat Math Formulas You Must Know
The SAT math test is unlike any math test you've taken before. It's designed to take concepts you're used to and make you apply them in new ways. It's tricky, but with attention to detail and knowledge of the basic formulas and concepts covered by the test, you can improve your score.
So what formulas do you need to have memorized for the SAT math section before the day of the test? In this complete guide, I'll cover every critical formula you MUST know before you sit down for the test. I'll also explain them in case you need to jog your memory about how a formula works. If you understand every formula in this list, you'll save yourself valuable time on the test and probably get a few extra questions correct.
Understand And Not Cram
You can easily forget what you crammed, butnever what you truly understand. Instead of having a myopic focus on thenumbers and symbols, try to understand what that particular formula actuallysolves. Once you have a semblance of a real world issue being solved by aformula, you are unlikely to forget it ever in your life.
What Is A Math Formula And Why Is It Important
In mathematics, a formula is a word equation that defines how to find the unknown variables within the equation in terms of other known values. It is commonly used to describe data that is being graphed or plotted.
Mathematical formulas are important because they help us visualize data and get a better understanding of the concept. They can also be used as a step-by-step guide to solving problems in different areas.
There are many different types of mathematical equations. The most common ones are linear, exponential, logarithmic, and trigonometric equations.
$l$ is the length of one of the edges of the rectangular part of the pyramid.
$h$ is the height of the figure at its peak .
$w$ is the width of one of the edges of the rectangular part of the pyramid.
Law: the number of degrees in a circle is 360
Law: the number of radians in a circle is $2$
Law: the number of degrees in a triangle is 180
Gear up that brain because here come the formulas you have to memorize.
How To Use Memory Tricks To Memorize Math Formulas
Have you ever found yourself stuck while solving a math problem because you couldnt remember the required formula? Its right there, dangling on the edge of your memory but just out of reach enough for you to not recall it.If you have any educationand most of us doyou probably know what this feels like. But this doesnt happen only to students. Classroom teachers, online tutors, and more have come up with a blank too on many occasions.Math is an essential science subject. As Erica Sunarjo of essay writing service review outfit, Best Writers Online puts it, many times, mathematical formulas can express a solution clearer than any words in the dictionary.Her sentiment is echoed in how mathematical formulas are used across many fields of endeavorbusiness, financial analysis, woodwork and construction, interior design, research, technology, and a lot more. As crucial as math is, keeping track of the plethora of formulas involved can be overwhelming.Its worse if youre preparing for a big test, have a bunch of finance numbers to crunch, have a tutorial session coming up, etc. Weve outlined a few memory tricks in this post that will show you how to memorize formulas in a way thats easy to retain and recall.
Geometry Rules: Isosceles Triangles
One geometry rule related to this last fact concerns one elite category of triangles: the isosceles triangles. These are triangles with two equal sides.
The angle between the two equal sides may be
an acute angle , as in triangle ABD
a right angle, as in triangle DEF, or
an obtuse angle as in triangle KLM
If two sides are equal, then we know the opposite angles are also equal: in triangle ABC, angle A = angle C in triangle DEF, angle E = angle F and in triangle KLM, angle K = angle M. In fact, Mr. Euclid pointed out that this geometry rule works both ways: if we are told two sides are equal, then we know two angles are equal, and if we are told two angles are equal, we know two sides are equal.
The line down the middle of an isosceles triangle is special:
As long as we are told that JKL and PQR are isosceles triangles, then this midline has some special properties:
On The Occasion Of National Mathematics Day Here Are 8 Ways By Which You Can Easily Memorise Math Formulas
Most students are afraid of mathematics due to the long list of formulas. But, mathematics is not only limited to learning from textbooks, it is about solving and understanding problems.
Mathematics formulas are the best way to make this subject a fun subject. Therefore, learning maths formulas will help students to gain confidence and enhance their problem-solving skills.
But, it is very tough and confusing to memorize all the maths formulas in one go. So, here are 8 ways to memorize maths formulas in an easy way.
Sat Math: Beyond The Formulas
Though these are all the formulas you'll need , this list doesn't cover every aspect of SAT Math. You'll also need to understand how to factor equations, how to manipulate and solve for absolute values, and how to manipulate and use exponents, and much more. These topics are all covered here.
Another important thing to remember is that while memorizing the formulas in this article that aren't given to you on the test is important, knowing this list of formulas doesn't mean you're all set for SAT Math. You also need to practice applying these formulas to answer questions, so that you know when it makes sense to use them.
For instance, if you're asked to calculate how likely it is that a white marble would be drawn from a jar that contains three white marbles and four black marbles, it's easy enough to realize you need to take this probability formula:
$$\text"Probability of an outcome" = /$$
and use it to find the answer:
$\text"Probability of a white marble" = /$
$\text"Probability of a white marble" = 3/7$
On the SAT math section, however, you will also run into more complex probability questions like this one:
Dreams Recalled During One Week
C) $79/164$
D) $164/200$
There's a lot of information to synthesize in that question: a table of data, a two-sentence long explanation of the table, and then, finally, what you need to solve for.
This is a probability question, so I'll probably need to use this formula:
Geometry Formulas: Special Parallelograms
There are three categories of special parallelograms: rhombuses, rectangles, and squares. The BIG FOUR parallelogram properties above apply to all of them.
A rhombus is an equilateral quadrilateral, that is, a quadrilateral with four equal sides.
Most diamond shapes, such as those on playing cards, are simply rhombuses turned sideways, like the one on the right above. If the side is s, then the perimeter is always just 4s. The diagonals of a rhombus are always perpendicular. In fact, any rhombus can be subdivided into four congruent right triangles.
As with a general parallelogram, A = bh, where the b is any side and the h is the length of a perpendicular segment: as with a general parallelogram, the Pythagorean Theorem may play a role in finding one of the lengths that you need.
A rectangle is an equiangular quadrilateral, that is, a quadrilateral with four 90° angles.
Each angle is 90° and the two diagonals are always equal in length: its an old carpenters trick to verify that a doorframe has four right angles simply by checking the lengths of the two diagonals. Of course, its usually easy to find the length of a diagonal using the Pythagorean theorem.
The area of a rectangle is simply A =bh, where the base & height are simply lengths of any two adjacent sides. The perimeter is P = 2b + 2h.
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Use Established Mnemonics Or Make Up Your Own
Ideally, you can use both. There are some mnemonic devices that we all know and love already. We all know BODMAS or PEMDAS . We also know the rule of finding the sine, cosine, and tangent of angles in a right-angled triangleSOH CAH TOA.There are many of these mnemonics that have made solving many math problems easier. Embrace them, and always use them when needed.Also, you could come up with your own mnemonics.Say your name is Elaine. You could memorize the formula E = mc² with the mnemonic Elaine loves meerkats and cats too. Here the t in too represents the power of 2 in the formula.You can also use a visual picture to store the formula. Using items that make sense together works best for this tactic. For instance, if you want to remember the formulas 2r or r², you can consider a pie and a rolling. So in the case of 2r, visualize 2 pies and a rolling pin. And for r², picture a pie and 2 rolling pins. Im sure you can see how these make sense.If the visual clues you adopt are rooted in real memories, they tend to stick more.
How Can I Learn Geometry Easily
The best way to truly master geometry painlessly isnt rote memorization: its looking at principles in practice.
It can be tempting to sit down with a basic list of geometry formulas, memorize them, and brush off your hands. However, to really master geometry, its not just about learning the formulasits about putting them into practice.
With that said, the easiest way to learn geometry is to see how formulas apply to a wide variety of question types. The key to this is to use high-quality materials to do a lot of geometry questions, then reading the answers and explanations after youve answered themeven if youve answered them correctly!
Why would you want to read question explanations if youve answered a question correctly? Because there may be a way to answer the question more quickly and efficiently than you didand yes, this includes the use of geometry formulas!
P.S. If youre looking for a shortcut for solving geometry equations, click here!
Easy Way To Learn Geometry Formulas
How to Memorize Math Formulas | Math Formula Memorization
Geometry, dating back to 3000 BC, is thatbranch of mathematics that helps in giving shape and dimension to the otherwiseflat world around us. It is what makes us discover and measure patterns, areas,angles, and sizes of things around us.
Contrary to how we feel about learning it atschool, we use it consciously and unconsciously throughout our day. We use itto make mental calculations while we park our bikes, while deciding thedimensions of a painting or a sculpture, and even while shooting for that goal.
It is a part of our curriculum since theearly stages and continues through college and even in higher education. Geometry isextensively used in specialized disciplines like engineering, sports, arts,robotics, automotive, astronomy etc.
While studying geometry hones many foundationskills like reasoning, logical thinking, problem-solving learning geometricalformulas is very tricky. But this trick does not underscore the importanceof learning and knowing formulae and their applications.
Some formulas are very complex and might seemlike abstract shapes from the outer space! The question is how to memorizeformulas and remember them for life?
Faqs On Geometry Formulas
You can get the Geometry Formula for Classes 12, 11, 10, 9, 8 from our page. Access the quick links available here in PDF format and know the formulas for all the topics.
2. Can you give some important formulas on Geometry?
Students of Class 8 to Class 12 will find information related to basic and important formulas of geometry that will help you score better grades in the exam from here.
3. How to download Classwise Geometry Formulas PDF?
Check out the direct links available on our page, tap them to view or download the Geometry Formulas for the concerned Class. All of them are organised as per the classes which can be quite handy to ace up your preparation.
We wish the data shed as far as our knowledge is concerned regarding the Geometry Formulas has been beneficial to you. For more information and if you feel any formula is missing feel free to leave us your suggestions via comment section. Stay in touch with our site to avail information on all formulas. | 677.169 | 1 |
...rectilineal figures. Explain homologous, alternando, ex aequali. When is the first of four magnitudes said to have the same ratio to the second which the third has to the fourth ? 7. In a right angled triangle, if a perpendicular be drawn from the right angle to the base, the... | 677.169 | 1 |
The circumcenter of a triangle is equidistant from the _____ of the triangle. 2/3 The centroid is ___ of the distance from each vertex to the midpoint of the opposite side …
Reader Q&A - also see RECOMMENDED ARTICLES & FAQs. If a segment is midsegment of a triangle, t. Possible cause: Learn. Getting ready for right triangles and trigonometry. Hypotenuse, opposi. | 677.169 | 1 |
I want to check in an shape, wether an point is in it or not.
The shape is descriped with one Array of float vectors.
The vectors are added in clock direction.
The first I can check very easy, the second too, but the third is more difficult.
How I can calculate this ? How this category is in math named, so that I can google ? | 677.169 | 1 |
Chinese Vocabulary Drill with 角 (jiǎo) - Angle
Today's word of the day is 角 (jiǎo) which means 'angle'. Let's practice by making different words, phrases, and sentences with 角 (jiǎo).
角度 (jiǎo dù) Point of view
The first word is 角度 (jiǎo dù) which means 'point of view'. Let's look at each character of this word, 角 means angle, and 度 (dù) means extent. A way of using it in a sentence is: 以我的角度来看,他是对的。(Yǐ wǒ de jiǎodù lái kàn, tā shì duì de.) From my point of view, he is right.
角落 (jiǎo luò) Corner
Another word we can create with 角 (jiǎo) is 角落 (jiǎo luò). This means 'corner'. A way of using it in a sentence is 你怎么一个人躲在角落里?(Nǐ zěnme yīgè rén duǒ zài jiǎoluò lǐ?) Why are you hiding alone in the corner?
三角 (sān jiǎo) Triangle
Finally, we can create the word 三角 (sān jiǎo) which means 'triangle'. The 三 here means the number 'three'. A word you can create from this is 三角恋 (sānjiǎo liàn) love triangle. | 677.169 | 1 |
Let $O$ and $H$ be respectively the circumcenter and the orthocenter of triangle $ABC$. Let $a$, $b$ and $c$ denote the side lengths. We are given that $a^2+b^2+c^2=29$ and the circumradius is $R=9$. We need to find $OH^2$.
I know that there is formula $OH^2=9R^2-(a^2+b^2+c^2)$, but I cannot use it unless I prove it.
I tried placing ABC triangle in a Cartesian plane and found coordinates of $O$ and $H$, but the expressions are not nice and I didn't manage to simplify them to required result.
Maybe vectors can be used? Please, note that I cannot use sophisticated vector knowledge in the solution. Any suggestions or ideas would be appreciated.
Let $O$ the circumcenter of $\triangle ABC$ and $G$ its centroid. Extend $OG$ until a point $P$ such that $OG/GP=1/2$. We'll prove that $P$ is the orthocenter $H$.
Draw the median $AA'$ where $A'$ is the midpoint of $BC$. Triangles $OGA'$ and $PGA$ are similar, since $GP=2GO$, $AG=2A'G$ and $\angle OGA'=\angle PGA$. Then $\angle OA'G =\angle PGA$ and $OA'\parallel AP$. But $OA'\perp BC$ so $AP\perp BC$, that is, $AP$ is a height of the triangle.
Repeating the same argument for the other medians proves that $P$ lies on the three heights and therefore it must be the orthocenter $H$. | 677.169 | 1 |
Statement 1 The equation of the director circle to the ellipse 4x2+9y2=36isx2+y2=13 Statement 2 The locus of the point of intersection of perpendicular tangents to an ellipse is called the director circle.
A
Statement I is true, statement II is true: statement II is a correct explanation for statement I
B
Statement I is true, statement II is true, statement II is not a correct explanation for statement I | 677.169 | 1 |
Classifying Angles. × Worksheet. Worksheet will open in a new window. Please click the picture to download free printable three types of angles worksheets. Worksheets > Math > Grade 3 > Geometry > Classifying angles. Use the protractor tool like a pro to measure and draw angles. acute right obtuse acute acute right obtuse acute obtuse obtuse right acute. = Types of angles worksheets: right, acute and obtuse. Types of Angles. math line segment ray worksheets, three types angles worksheet and what are all the regular polygons are three main things we want to show you based on the gallery title. Three types of angles worksheets pdf is useful because this is the printable three types of angles worksheets pdf . Our acute, right, and obtuse angles worksheets provide the essential skills on the three basic types of angles. .hide-if-no-js { }. Full angle or Complete angle: Some of the worksheets displayed are Types of angles, Types of angles classify each angle as acute obtuse, Classifying angles l1s1, Three types of angles, Types of angles, Classifying angles date … Study the types of angles carefully. Further subcategories under these three main heads include an isosceles right triangle, right triangle, etc. A triangle has three sides and three vertices. Three types of angles worksheets is composed of the following; angles seven worksheets, angless exercise, angles practice and angles problems. Types Of Angles Grade 4 - Displaying top 8 worksheets found for this concept.. 63 Scroll down the page if you need more explanations about each type of angles, videos and worksheets. 207 Downloads Grade 3 Angles in Shapes. Exercise worksheet on 'Types of Angles.' Download Now! Addinf Fraction With Unequal Denominators. .hide-if-no-js { See how well you can recognize angles. Three types of angles worksheets is the free printable pdf. Types of angles: Zero angle: Exactly measures 0º . Obtuse angle: Measures greater than 90º but less than 180º . By the way, concerning Three Types Worksheet Angles, below we will see particular variation of images to give you more ideas. Looking for a worksheet to help practice basic geometry? ID: 1292799 Language: English School subject: Math Grade/level: Grade 3 Age: 7-15 Main content: Angles Other contents: acute, obtuse, right Add to my workbooks (21) Download file pdf Embed in my website or blog Add to Google Classroom ID: 1501189 Language: English School subject: Math Grade/level: Grade 5 Age: 10-12 Main content: Types of angles Other contents: Types of angles Add to my workbooks (0) Download file pdf Embed in my website or blog Add to Google Classroom Grade 3, 4 Lines, Line Segments, and Rays. Most worksheets require students to identify or analyze acute, obtuse, and right angles. one 3 Types Of Angles - Displaying top 8 worksheets found for this concept.. Tags: a right angle measuresan acute anglean obtuse angleangles worksheets printable pdfthree types of angle worksheettypes of angles worksheet 4th gradetypes of angles worksheet 5th gradetypes of angles worksheet grade 3types of angles worksheet pdf, Your email address will not be published. Save my name, email, and website in this browser for the next time I comment. An acute angle is an angle between 0° and 90°. Three types of angles worksheets is composed of the following; angles seven worksheets, angless exercise, angles practice and angles problems. 3rd grade. Three types of angles worksheets is the free printable pdf. Corresponding angles: Pairs of angles that are in similar positions. Golden rule for math teachers: You must tell the truth, and nothing but the truth, but not the whole truth. The following table shows the different types of angles: right angles, acute angles, obtuse angles, straight angles, reflex angles and full angles. 2,845 Downloads Grade 3 Types of Angles. Worksheet Types of Angles. On the left-hand side, your students both learn and see for themselves how to classify these three types of angles. 10. Angles can be acute, right, obtuse, straight and reflex. Download Now! Teach your children about the three main types of angles with this Classifying Angles activity.This worksheet comes in two parts and deals with acute, obtuse and right angles Geometry Worksheets Angles Worksheets for Practice and Study. There are three types of triangles, namely equilateral, isosceles, and scalene. Math. The Angles Worksheets are randomly created and will never repeat so you have an endless supply of quality Angles Worksheets to use in the classroom or at home. Reading Protractors. Label each angle as acute, obtuse, or right. Thanks to three types of angles worksheets and you will have fun and you'll learn the best way. Straight angle: Exactly measures 180º . Showing top 8 worksheets in the category - Types Of Angles. Some of the worksheets displayed The most practical way of learning three types of angles worksheets. This page includes a lesson covering 'Types of Angles' as well as a 15-question worksheet, which is printable, editable, and sendable An obtuse angle is more than 90° and less than 180°. Become twice as conversant with identifying, classifying, and drawing all six types of angles: acute, right, obtuse, straight, reflex, and complete angles with this collection of pdfs. Right abgle: Exactly measures 90º . With 3 acute angles, and scalene add more info, we have collected some similar images to give more. Measures greater than 180º types of angles worksheets for gcse maths foundation and key stage 3 practice,... An acute angle is an angle of exactly 90° angle can be classified as than... Learn and see for themselves how to classify these three main heads include an right! Of geometry three types of angles worksheet similar positions most practical way of learning three types of angles worksheets worksheets pdf useful! Acute obtuse obtuse right acute lessons, 2 practice sheets, homework sheet, and.. Sort math worksheets via email: Sort math worksheets by: Grade level Skill/Topic Search.... Triangles worksheets for your needs is a graphic preview for all of the following ; angles worksheets. Browser document reader options of learning three types of angles a and b as supplementary complementary! Your needs angle types the category - types of angles worksheets is the free printable.. Angles circle the hexagon with exactly three acute angles... types of angles Displaying top 8 worksheets for! - Displaying top 8 worksheets found for this concept is an angle of exactly 90° by the way concerning! 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} TTan Free math problem solver answers your algebra, geStep by step video & image solution Popular Tan 30 Degrees. The value of tan 30 degrees is 1/√3. The value of ta Understand Simplify sin(30)+cos(30) Step 1. The exactBy the triple angle formula for cosine: cos ( 54 ∘Understand the examples of how to use each function, as well as The value of cos 30 degrees in decimals is 0.8660. Converting degree to radian, that is, θ in radians = θ × π/180° or θ × Pi/180°. Therefore, converting cos 30° in radians will give cos (30 × π/180°), and the final value in radians will become cos(π/6) or cos(0.5235). Following are the values of cos 30° in different forms, To find the value of cos 270 degrees using the Free trigonometric simplification calculator - Simplify trigonometric expressions to their simplest form step-by-step Q. Verify each of the following: (i) sin 60[Rotation matrix. In linear algebra, a rotaAug 25, 2016 · Use trig identity: sin (a - b) = siExpert Maths Tutoring in the UK - Boost Your Scores with Cuemath. /180°) For tan 30 degrees, the angle 30° lies between 0° and 90° (First ). Since tangent function is positive in the first quadrant, thus tan 30° value = 1/√3 or 0.5773502. . . Since the tangent function is a ⇒ tan 30° = tan 210° = tan 390°, and so on. | 677.169 | 1 |
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Geometry SSC BANK pdf question (chsl-clerk-po exem)
Geometry SSC BANK pdf question (chsl-clerk-po exem)
Here i am sharing with you question on Geometry SSC BANK pdf question (chsl-clerk-po exem). In any exam reasoning section carries great significance. It is that section where a candidate's general mental ability is tested.
you.
1) If (5, 1), (x, 7) and (3, -1) are 3 consecutive verticles of a square then x is
equal to :
a) – 3 b) – 4 c) 5
c) 6 e) None of these
2) What is the area of an obtuse angled triangle whose two sides are 8 and 12
and the angle included between two sides is 150°?
a) 24 sq units b) 48 sq units c) 24 √3
d) 48 √3 e) Such a triangle does not exist
3) What is the measure of the radius of the circle that circumscribes a triangles
whose sides measure 9, 40 and 41?
a) 6 b) 4 c) 24.5
d) 20.5 e) 12.5
4) Verticles of a quadrilateral ABCD are A (0, 0), B (4, 5), C (9, 9) and D (5, 4).What
is the shape of the quadrilateral?
a) Square b) Rectangle but not a square
c) Rhombus d) Parallelogram but not a rhombus e) None of these
5) If the sum of the interior angles of a regular polygon measures upto 1440 degrees,
how many sides does the polygon have?
a) 10 sides b) 8 sides c) 12 sides
d) 9 sides e) None of these | 677.169 | 1 |
Given Two Angles That Measure 50 And 80
Understand the Intricacies of Angles: A Guide to Measuring 50 and 80 Degrees
When encountering angles that measure 50 and 80 degrees, students and geometry enthusiasts often grapple with determining their properties and relationships. This guide aims to alleviate any confusion surrounding these angles, providing a comprehensive overview of their characteristics and significance.
Understanding the Challenges
Navigating the world of angles can be daunting, especially when faced with unfamiliar measurements like 50 and 80 degrees. Misinterpreting their properties can lead to errors and misunderstandings in mathematical calculations and geometric constructions. This guide addresses these challenges by demystifying the nature of these angles and providing clear explanations.
Measuring 50 and 80 Degrees
An angle measuring 50 degrees is classified as an acute angle, falling between 0 and 90 degrees. It is less than a right angle and commonly encountered in various geometric shapes and architectural designs. On the other hand, an angle measuring 80 degrees is classified as an obtuse angle, ranging between 90 and 180 degrees. Obtuse angles are greater than a right angle and are also prevalent in geometry and architecture.
Conclusion
Understanding the properties and characteristics of angles measuring 50 and 80 degrees is crucial for students and individuals interested in geometry. This guide provides a concise yet comprehensive overview of these angles, addressing common pain points and offering clear explanations. By grasping the concepts presented here, readers can confidently navigate the world of angles and enhance their mathematical and geometric comprehension.
Discovering the Angle Relationships: Exploring the Sum and Difference Identities
Introduction: Delving into the Realm of Angle Trigonometry
In the captivating world of mathematics, trigonometry reigns supreme as the discipline that unveils the enigmatic connections between angles and their corresponding sides in triangles. Among its many intriguing concepts, angle relationships stand out as pivotal elements in unraveling the mysteries of trigonometric identities. In this comprehensive exploration, we embark on a journey to decipher the intricate interplay between two angles measuring 50 degrees and 80 degrees, delving into the profound implications of the sum and difference identities.
Sum Identity: Unifying Angles in Harmony
Unveiling the Essence of the Sum Identity
The sum identity, adorned with its mathematical elegance, establishes a profound relationship between the sine and cosine functions of two angles. It asserts that the sine of the sum of two angles is equal to the product of the sine of one angle and the cosine of the other, coupled with the product of the cosine of the first angle and the sine of the second. This fundamental identity serves as a cornerstone in trigonometry, unlocking a wealth of insights into angle relationships.
Visualizing the Sum Identity: A Geometric Perspective
To visualize the essence of the sum identity, consider two vectors emanating from the origin of a coordinate plane. These vectors represent the angles in question, their lengths symbolizing the magnitudes of the sine and cosine functions. By positioning these vectors head-to-tail, we form a new vector that embodies the sum of the angles. The sine of the sum is then revealed as the length of the vertical component of this new vector, while the cosine of the sum is manifested as the length of its horizontal component.
Difference Identity: Unveiling the Contrast of Angles
Deciphering the Nuances of the Difference Identity
In stark contrast to the sum identity, the difference identity elucidates the relationship between the sine and cosine functions of two angles in terms of their difference. It proclaims that the sine of the difference between two angles is equivalent to the product of the sine of one angle and the cosine of the other, minus the product of the cosine of the first angle and the sine of the second. This identity unveils the intricate interplay of angles, revealing how their variations influence trigonometric functions.
Picturing the Difference Identity: A Visual Representation
To visualize the difference identity, imagine two vectors emanating from the origin of a coordinate plane, akin to the sum identity. However, instead of positioning them head-to-tail, we place them tail-to-tail, creating a new vector that embodies the difference of the angles. The sine of the difference is then revealed as the length of the vertical component of this new vector, while the cosine of the difference manifests as the length of its horizontal component.
Unveiling the Power of Angle Relationships in Real-World Scenarios
The sum and difference identities transcend the realm of theoretical mathematics, finding practical applications in diverse fields. Engineers harness these identities to calculate forces, moments, and displacements in complex mechanical systems. Architects rely on them to determine angles for optimal sunlight exposure in building design. Navigators employ these identities to calculate precise bearings and distances for seafaring voyages. The sum and difference identities empower professionals across disciplines to solve intricate problems with remarkable accuracy.
Conclusion: A Symphony of Angles, Unveiled
The exploration of angle relationships through the sum and difference identities unveils a symphony of mathematical beauty and practical utility. These identities provide a framework for understanding the intricate interplay between angles, their trigonometric functions, and their real-world applications. As we delve deeper into the realm of trigonometry, these identities will continue to serve as invaluable tools, unlocking the secrets of angles and their profound impact on our world.
Frequently Asked Questions: Illuminating Key Concepts
What is the sum identity?
The sum identity establishes a relationship between the sine and cosine functions of two angles, expressing the sine of the sum as the product of the sines and cosines of the individual angles.
How is the difference identity different from the sum identity?
The difference identity mirrors the sum identity but focuses on the relationship between the sine and cosine functions of two angles in terms of their difference.
What are some practical applications of the sum and difference identities?
Engineers use these identities to calculate forces, moments, and displacements in mechanical systems. Architects rely on them for determining optimal sunlight exposure in building design. Navigators employ these identities for calculating precise bearings and distances during seafaring voyages.
Can the sum and difference identities be used to solve trigonometry problems?
Absolutely! These identities serve as powerful tools for solving a wide range of trigonometry problems, enabling the determination of unknown angles and trigonometric function values.
How can I deepen my understanding of the sum and difference identities?
Practice applying these identities to various trigonometry problems, gradually building your proficiency in utilizing them. Explore real-world examples where these identities are applied, gaining insights into their practical significance. | 677.169 | 1 |
Process
A rotation in the plane can be formed by composing a pair of reflections. First reflect a point P to its image P′ on the other side of line L1. Then reflect P′ to its image P′′ on the other side of line L2. If lines L1 and L2 make an angle θ with one another, then points P and P′′ will make an angle 2θ around point O, the intersection of L1 and L2. I.e., angle ∠ POP′′ will measure 2θ.
A pair of rotations about the same point O will be equivalent to another rotation about point O. On the other hand, the composition of a reflection and a rotation, or of a rotation and a reflection (composition is not commutative), will be equivalent to a reflection.
Mathematical expression
The statements above can be expressed more mathematically. Let a rotation about the originO by an angle θ be denoted as Rot(θ). Let a reflection about a line L through the origin which makes an angle θ with the x-axis be denoted as Ref(θ). Let these rotations and reflections operate on all points on the plane, and let these points be represented by position vectors. Then a rotation can be represented as a matrix,
Proof
The set of all reflections in lines through the origin and rotations about the origin, together with the operation of composition of reflections and rotations, forms a group. The group has an identity: Rot(0). Every rotation Rot(φ) has an inverse Rot(−φ). Every reflection Ref(θ) is its own inverse. Composition has closure and is associative, since matrix multiplication is associative.
Notice that both Ref(θ) and Rot(θ) have been represented with orthogonal matrices. These matrices all have a determinant whose absolute value is unity. Rotation matrices have a determinant of +1, and reflection matrices have a determinant of −1.
The set of all orthogonal two-dimensional matrices together with matrix multiplication form the orthogonal group: O(2).
The following table gives examples of rotation and reflection matrix :
Rotation of axes
An xy-Cartesian coordinate system rotated through an angle θ{\displaystyle \theta } to an x′y′-Cartesian coordinate system
In mathematics, a rotation of axes in two dimensions is a mapping from an xy-Cartesian coordinate system to an x′y′-Cartesian coordinate system in which the origin is kept fixed and the x′ and y′ axes are obtained by rotating the x and y axes counterclockwise through an angle θ{\displaystyle \theta }. A point P has coordinates (x, y) with respect to the original system and coordinates (x′, y′) with respect to the new system.[1] In the new coordinate system, the point P will appear to have been rotated in the opposite direction, that is, clockwise through the angle θ{\displaystyle \theta }. A rotation of axes in more than two dimensions is defined similarly.[2][3] A rotation of axes is a linear map[4][5] and a rigid transformation | 677.169 | 1 |
Analysis of variance, or ANOVA (analysis of variance), are multivariate dependency analysis techniques used to determine if there are significant differences between the tights of three or more groups population. Therefore, with this analysis we will find out if there are differences between certain groups when we modify one or more characteristics. To find out, we use the value of the average of the data. Its use is very frequent in fields such as economy or medicine. Previous assumptions of...
The height of a triangle is that segment that joins a vertex of the triangle with its opposite side or its extension, being perpendicular to it, that is, a right angle (90º) is formed at the intersection.Each triangle then has three heights, each with respect to each of its sides.The heights of the triangle intersect at the orthocenter, which in the figure below would be point O, where in addition the heights are the segments AD, BE and CF.Points D, E, and F are called feet of heights.It shou...
The anglecomplementary and supplementary they are two categories of angles when they are classified according to the result of their summation with another angle.A complementary angle is one with which a right angle can be formed, that is, two angles are complementary if they add up to 90º (sexagesimal degrees) or π/ 2 radians.Also, a supplementary angle is one with which a flat angle can be configured. That is, two supplementary angles add up to 180º or π radians.In the image above, for exam...
Paula Rodo 2 minReferenceThe angle between two vector is the capacity of the arc of the circumference formed by the segments of the vectors joined by a point.In other words, the angle between two vectors is the angle that is formed when two vectors are multiplied.Two vectors will form an angle when both are multiplying, that is, when we multiply vectors we will be joining them at a common point such that they will form an angle.FormulaLet two 3-dimensional vectors be:Two vectorsBoth will form...
The center of gravity of a triangle is the point where the medians of the figure intersect. It is also known as a centroid.It should be remembered that the median is the segment that joins the vertex of the triangle with the midpoint on its opposite side. Thus each triangle has three medians.For example, in the triangle above, the center of gravity is point O, with the medians being the segments AF, BD, and CE.An important property of the center of gravity is that its distance from each verte...
The bisector of an angle is that ray that, starting from the respective vertex, divides an angle into two equal parts.That is, the bisector is the line that divides the angle into two portions of identical measure. That is, in the lower image, if α is 70º, it will be divided into two 35º angles.At this point, we must first remember that the definition of angle is the arc that is formed from the union of two lines, rays, or segments.Likewise, we point out that a ray, like the bisector, is defi...
The bisector of a triangle is a segment that divides one of its interior angles into two equal parts and continues until it reaches the side opposite that angle. Each interior angle of the triangle has a bisector.We should note then that every triangle has three bisectors, each of which starts from each vertex towards the opposite side.As we can see in the image, their bisectors intersect at point I, which is the incenter. This is the center of the circle inscribed in the triangle. This circu...
The Poisson distribution is a discrete probability distribution that models the frequency of events determined during a time interval set from the average frequency of occurrence of said events. In other words, the Poisson distribution is a discrete probability distribution that, only knowing the events and their frequency half occurrence, we can know your probability.ExpressionGiven a discrete random variable X we say that its frequency can be satisfactorily approximated to a Poisson distrib...
The circumcenter of a triangle is the point where its three mediatrices, being also the center of the circumscribed circumference.That is, the circumcenter is the central point of the circumference that contains the triangle in question.Another important concept to detail is that the bisector is that line that, being perpendicular on one side of the triangle, divide this segment into two equal parts.In the figure above, for example, point D is the circumcenter of the figure. Likewise, F, G an... | 677.169 | 1 |
An isosceles triangle is a triangle that has (at least) two equal side lengths. If all three side lengths are equal, the triangle is also equilateral. Isosceles triangles are very helpful in determining unknown angles.
Contents
Terminology
Basic Properties
Advanced Properties
Problems
See Also
Terminology
In an isosceles triangle, the two equal sides are called legs, and the remaining side is called the base. The angle opposite the base is called the vertex angle, and the point associated with that angle is called the apex. The two equal angles are called the isosceles angles.
If the triangle is also equilateral, any of the three sides can be considered the base.
Basic Properties
Because angles opposite equal sides are themselves equal, an isosceles triangle has two equal angles (the ones opposite the two equal sides). Thus, given two equal sides and a single angle, the entire structure of the triangle can be determined. Likewise, given two equal angles and the length of any side, the structure of the triangle can be determined.
Determining the area can be done with only a few pieces of information (namely, 3):
A triangle with base \(b\) and equal side \(\ell\) has area \(\frac{b\sqrt{4\ell^2-b^2}}{4}\).
A triangle with base \(b\) and isosceles angle \(\theta\) has area \(\frac{b^2\tan \theta}{4}\), as the length of the altitude to the base is \(\frac{b\tan\theta}{2}\). Alternatively, if the apex angle has measure \(\alpha\), the area is \(\frac{b^2}{4}\cot\frac{\alpha}{2}\).
A triangle with equal side \(\ell\) and isosceles angle \(\theta\) has area \(\frac{\ell^2\sin 2\theta}{2}\), as the vertex angle has measure \(180^{\circ}-2\theta\), and \(\sin(180^{\circ}-2\theta)=\sin 2\theta\). Alternatively, if the vertex angle has measure \(\alpha\), the area is \(\frac{\ell^2\sin\alpha}{2}\).
The altitude to the base also satisfies important properties:
The altitude to the base is the perpendicular bisector of the base.
The altitude to the base is the angle bisector of the vertex angle.
The altitude to the base is the line of symmetry of the triangle.
The altitude to the base is the median from the apex to the base.
This means that the incenter, circumcenter, centroid, and orthocenter all lie on the altitude to the base, making the altitude to the base the Euler line of the triangle.
Advanced Properties
It is immediate that any \(n\)-sided regular polygon can be decomposed into \(n\) isosceles triangles, where each triangle contains two vertices and the center of the polygon.
A regular \(n\)-gon is composed of \(n\) isosceles congruent triangles.Any isosceles triangle is composed of two congruent right triangles as shown in the sketch. These right triangles are very useful in solving \(n\)-gon problems. The relation given could be handy.Apart from the above-mentioned isosceles triangles, there could be many other isoceles triangles in an \(n\)-gon.
\[ n \times \phi =2 \pi = 360^{\circ}. \]
More interestingly, any triangle can be decomposed into \(n\) isosceles triangles, for any positive integer \(n \geq 4\). The picture to the right shows a decomposition of a 13-14-15 triangle into four isosceles triangles.
Problems
Isosceles triangle \(ABC\) has \(AB=AC\) and \(\angle BAC=40^{\circ}\). What is the measure of \(\angle ABC\)?
Because \(AB=AC\), we know that \(\angle ABC=\angle ACB\). Additionally, the sum of the three angles in a triangle is \(180^{\circ}\), so \(\angle ABC+\angle ACB+\angle BAC=2\angle ABC+\angle BAC=180^{\circ}\), and since \(\angle BAC=40^{\circ}\), we have \(2\angle ABC=140^{\circ}\). Thus \(\angle ABC=70^{\circ}\). \(_\square\)
In isosceles triangle \(ABC\), the measure of \(\angle ABC\) is \(50^{\circ}\). What are the possible measures of \(\angle BAC\)?
There are three possible cases:
Case 1: \(\angle ABC=\angle BAC\). This is easy to deal with, as then \(\angle BAC=50^{\circ}\) immediately.
Case 2: \(\angle ABC=\angle BCA\). Since the angles of a triangle add up to \(180^{\circ}\), we know that \(50^{\circ}+50^{\circ}+\angle BAC=180^{\circ}\), and so \(\angle BAC=80^{\circ}\).
Case 3: \(\angle BAC=\angle BCA\). Again, the angles of a triangle add up to \(180^{\circ}\), so \(50^{\circ}+2\angle BAC=180^{\circ}\), and so \(\angle BAC=65^{\circ}\).
Therefore, the possible values of \(\angle BAC\) are \(50^{\circ}, 65^{\circ}\), and \(80^{\circ}\). \(_\square\)
Triangle \(ABC\) is isosceles, and \(\angle ABC = x^{\circ}.\) For some fixed value of \(x\), the sum of the possible measures of \(\angle BAC\) is \(240^{\circ}.\)
What is \(x\)?
In the above figure, \(AD=DC=CB\) and the measure of \(\angle DAC\) is \(40^{\circ}\). What is the measure of \(\angle DCB\)?
From the given condition, both \(\triangle ADC\) and \(\triangle DCB\) are isosceles triangles. | 677.169 | 1 |
The Midpoint Formula Worksheet
Exploring the Benefits of Using the Midpoint Formula Worksheet for Geometry Homework
Are you feeling frustrated by your geometry homework? Is the midpoint formula giving you a headache? Don't worry, we've got a solution for you! Introducing the Midpoint Formula Worksheet for Geometry Homework – the perfect way to get your geometry skills up to scratch!
The midpoint formula worksheet is an invaluable tool when it comes to tackling geometry problems. With it, you get step-by-step instructions for solving the midpoint formula, as well as helpful diagrams to make sure you understand the concept. Plus, the worksheet is designed to make sure you're applying the formula correctly and accurately.
But why should you bother with the midpoint formula worksheet? Well, for starters, it's a great way to save time. Instead of having to spend hours staring at a problem and trying to figure out how to solve it, you can take a few minutes to work through the worksheet and get the job done.
The midpoint formula worksheet also helps you hone your geometry skills. By breaking down the formula into bite-sized chunks, you can quickly identify which steps are necessary for solving the problem and which aren't. This makes it easier to remember the formula and apply it correctly in the future.
Finally, the midpoint formula worksheet is a great way to have a bit of fun with your geometry homework. The diagrams and step-by-step instructions are often quite humorous, so you can have a good laugh while you work your way through the problem.
So what are you waiting for? Get your hands on a midpoint formula worksheet for geometry homework and let the fun begin!
How to Use the Midpoint Formula Worksheet to Enhance Math Skills
Are you struggling to get your math skills up to scratch? If so, then the Midpoint Formula Worksheet is here to save the day! This handy worksheet helps you master the midpoint formula, which is essential for solving many math problems. Here's how you can use the worksheet to enhance your math skills:
1. Start by understanding the formula. The Midpoint Formula is used to calculate the exact midpoint between two points. It's written as: (x1 + x2)/2, where x1 and x2 are the coordinates of the two points.
2. Now, it's time to practice. Grab your worksheet and fill in the blanks with the coordinates of the two points. Then, use the formula to calculate the midpoint.
3. Check your answer. Once you've solved the problem, make sure to check your answer by plotting the two points and the midpoint on a graph. This will help you visualize the result and make sure it's correct.
4. Finally, get creative. Once you've mastered this formula, it's time to have some fun. Try solving some puzzles and tricky problems using the Midpoint Formula. This will help you gain a better understanding of the concept and get your math skills up to speed.
Using the Midpoint Formula Worksheet is a great way to enhance your math skills and get more comfortable with math problems. So, what are you waiting for? Grab your worksheet and get started!
The Advantages of Using the Midpoint Formula Worksheet for More Complex Problems
Are you struggling with solving more complicated math problems? Well, fear not! The midpoint formula worksheet is here to help! This amazing worksheet is ideal for tackling more complex problems and can make your math life much easier.
So, what are the advantages of using the midpoint formula worksheet? Here are a few:
1. It's a Breeze to Use: The midpoint formula worksheet is incredibly easy to use. All you need to do is input the coordinates of two points and the midpoint formula will give you the exact midpoint of your points. It's so simple, even a monkey could do it!
2. Time-Saving: The midpoint formula worksheet can save you a ton of time. Instead of wasting hours trying to figure out complex equations, you can use the midpoint formula worksheet to quickly and accurately solve your problem.
3. Accurate Results: The midpoint formula worksheet is guaranteed to provide you with the most accurate results. No more worrying if you've made a mistake in your calculations!
4. Fun Factor: Solving math problems isn't usually anyone's idea of a good time. But with the midpoint formula worksheet, you can make math fun! You can challenge yourself and your friends to solve difficult problems using the midpoint formula worksheet. Who knows, you may even learn something new in the process!
So if you're struggling to solve more complex math problems, don't despair! Just break out the midpoint formula worksheet and make your math life easier!
Conclusion
The Midpoint Formula Worksheet provides an excellent resource for students to understand the midpoint formula and practice applying it to solve various problems. With practice, students can become confident using the formula and applying it in a variety of situations. Understanding how to use the midpoint formula is an important skill for students to develop, and this worksheet is an excellent way to get started.
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Constructing Meaningful Sentences with Verbs Like Gustar: A Look at Common Uses and Practice ExercisesHave you ever heard someone say something like "A mí me gusta el helado"? If so, then you've heard an example of the verb gustar in action! But what exactly does it mean? Gustar is a unique verb in Spanish that...
How to Use a 7th Grade Proportions Worksheet to Master Math ConceptsMastering math concepts using a 7th grade proportions worksheet can be a great way to strengthen your understanding of the concepts. Here are some tips to help you get the most out of your worksheet: 1. Read the instructions carefully. Make sure you understand...
Exploring the Basics of Zero and Negative Exponents: A Worksheet GuideWelcome to your exploration of the mysterious world of zero and negative exponents! You may have shuddered in fear at the sight of those unfamiliar symbols, but fear not – we're here to help you get to grips with these tricky mathematical concepts. First off,...
Exploring the Wonders of the Human Anatomy with Inside the Living Body WorksheetThe human body is a complex and awe-inspiring system. From the brain to the extremities, each part of the body has its own unique purpose and functionality. To better understand this intricate system, let us explore the wonders of the human anatomy with... | 677.169 | 1 |
The Basic Principles of What Is a Ray in Math You Can Benefit From Starting Right Away
Getting the Best What Is a Ray in Math
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interest in the usage of logic in computer science in late years. This indicates that you've actually visited the website and taken the time and effort to appear around. So daily, he gets feedback and criticism from a broad selection of investors.
Ray diagrams can be especially helpful for determining and explaining why only a part of the image of an object can be viewed from a given location. How to measure angles and kinds of angles An angle includes two rays with a typical endpoint. To determine the image distance, it must be used.
Use the best terms it's possible to imagine! The large part of the moment, you won't perform actual probability difficulties, but you're going to use subjective probability to make judgment calls and determine the best plan of action. There are various sorts of probability.
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A discrete object has known and definable boundaries which enables the start and the limit to be readily identified. The large part of the moment, you won't perform actual probability difficulties, but you're going to use subjective probability to make judgment calls and determine the best plan of action. There are various sorts of probability.
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The History of What Is a Ray in Math Refuted
Measures are employed in the healthcare field. Numbers are a method of communicating information, which is vital in the health care field. They provide an abundance of information for medical professionals.
The crucial idea behind SparseCT is to block a lot of the X-rays in a CT scan before they get to the patient, but to do so in a manner that preserves all the important image info. Attempt to keep in mind this discussion is associated with the narrative task only. Furthermore formal techniques are acknowledged and attributed as central to the topic of discrete mathematics in recent decades.
Line BA is just like line AB. Satellites could possibly be naturally occurring, including the Moon, or else they could be man-made, like the Hubble Space Telescope and the Compton Gamma-Ray Observatory. A 4-dimensional space is composed of an endless number of 3-dimensional spaces.
Another style of rendering a three-dimensional object is known as ray tracing. While it can be labeled and described, it cannot be measured. It has no measurable length, because it goes on forever in one direction.
The Nuiances of What Is a Ray in Math
Furthermore, it's the sole mathematical textbook with a comprehensive treatment of electrical impedance tomography, and such sections are suited to beginning and skilled researchers in mathematics and engineering. There are arrows on each side. It's a two-dimensional object.
Ray diagrams may be used to learn the image place, size, orientation and kind of image formed write my essays
of objects when placed at a specified location before a lens. Simplified modeled components like these can't capture all the authentic behavior. All the points and lines that lie on the exact plane are supposed to be coplanar.
Going back to school might be a hard decision especially in case you've already started your career. Imagine that you're taking a peek at the front area of the police laser gun. A kid might have to be explained the same math strategy many different times till they are prepared to understand it.
The duration of a line cannot be measured. It is one of the basic terms in geometry. If both lines on a plane meet, we say the 2 lines intersect and the point at which they meet is known as point of intersection.
A good eBook reader ought to be installed. This is known as the painter's algorithm. The very first thing we must find is the duration of L.
The children might have to determine the hidden image by connecting the dots in every single sheet. Three or more points in a plane are believed to be collinear if all of them lie on the identical line. All the points extending forever in 1 direction from a specific point are together referred to as a ray. | 677.169 | 1 |
What are Vertical Angles? Definition and Examples
Vertical angles are angles that are on the opposite sides of two lines that intersect. In the figure below, angle a and angle b are vertical angles.
Are vertical angles congruent?
Yes, vertical angles are congruent or equal. For example, referring to the figure above, if angle a is equal to 70 degrees, angle b is also equal to 70 degrees. Why are vertical angles congruent? Check vertical angles theorem to see a proof!
Can vertical angles be adjacent?
No, this is impossible! Vertical angles are opposite to each other, never adjacent or next to each other.
Can vertical angles be complementary?
Yes, it is possible for vertical angles to be complementary. Complementary angles add up to 90 degrees.
Referring to the figure above, this means that angle a + angle b = 90 degrees.
Since angle a = angle b, we get angle a + angle a = 90 degrees.
2 angle a = 90 degrees
angle a = 45 degrees.
angle b = 45 degrees.
Vertical angles are complementary when each of the two angles is equal to 45 degrees.
Can vertical angles be supplementary?
Yes, it is possible for vertical angles to be supplementary. Supplementary angles add up to 180 degrees.
Referring to the figure above, this means that angle a + angle b = 180 degrees.
Since angle a = angle b, we get angle a + angle a = 180 degrees.
2 angle a = 180 degrees
angle a = 90 degrees.
angle b = 90 degrees.
Vertical angles are supplementary when each of the two angles is equal to 90 degrees. | 677.169 | 1 |
Here is the answer to a problem, proposed by Michael Metaxas: Given a convex quadrangle, divide it through two intersecting lines in four equal parts (in area). Following an idea of Michael Papadimitrakis, find first the envelope of the lines [EF] dividing the quadrangle in two equal parts. As pointed out by Antreas Varverakis, this envelope is a hyperbola, with asymptotic lines [AB] and [CD] (drawn in red), or [AD] and [BC] (not drawn). Then draw a tangent [EF] to this hyperbola at a point H of it. This creates (under certain restrictions) two other quadrangles: AEID and BEIC, having equal areas. Repeat the construction of the envelope for these two quadrangles. This determines two other hyperbolas (green and blue respectively). Draw a common tangent [FG] to these two hyperbolas. This completes the construction of the two lines, intersecting at F. Obviously there are infinite many solutions.
The fact that the envelope of the lines, dividing in two equal parts the quadrangle, is a hyperbola, is due to the fact that asymptotic triangles of a hyperbola have a constant area. Look at the file AsymptoticTriangle.html for a picture.
Look at the file ParaDivision.html for the solution of the special problem of a four-division of a parallelogram. In that case point F is always the center of the parallelogram. Which is the locus of F in the general case? | 677.169 | 1 |
$\begingroup$There is. But it requires a shift in your thinking away from what physicists are comfortable with. We think of a parabola as the curve defined by functions like y=x^2. You want to think in terms of geometry, where a parabola is the intersection of a plane and a cone where the axis of the cone is parallel to the plane. This question might do better on the math site.$\endgroup$
$\begingroup$You might also read "The Archimedes Codex" It goes through some of the math used by Archimedes. One of the problems involves a parabola, where we would use Cartesian coordinates, and of course he did not. It's a good read.$\endgroup$
$\begingroup$Here is an attempt with the formula for the equation of the line reflected by another line + using the equation of the tangent of the parabola x^2 with its derivative: math.stackexchange.com/questions/4852497/…$\endgroup$
Physicists normally think of parabolas as the locus of equations of the form $y=x^2$, but you're explicitly looking for something with no formulas, so you probably need to change that as well. The next likeliest candidate is defining the parabola as the locus of points equidistant from a point and a line, which ties you in to all the proofs of classical geometry. Under that understanding, the proof of the reflection property is a staple of euclidean compass-and-straightedge geometry, and Wikipedia has a suitable proof, based on a construction of the form
In short, with $C$ on the directrix, you define $B$ as the midpoint of $\overline{FC}$, which means (since $\overline{FE}=\overline{EC}$ by definition of the parabola) that $\angle FEB=\angle BEC$, so you just need to show that the line joining $B$ and $E$ is tangent to the parabola. The Wikipedia proof relies on some facts from calculus, though if you want a calculus-free proof you can probably find one by rooting around in the toolbox for the euclidean geometry of conic sections. | 677.169 | 1 |
Perpendicular lines - Examples, Exercises and Solutions
Perpendicular lines are vertical lines that form a right angle between them, that is, an angle of 90° 90° 90° degrees. Perpendicular lines appear in many geometric shapes, such as a rectangle, a square, a right triangle, and others. | 677.169 | 1 |
PaperRetro Bowl is a geometry math activity where students can learn more about two-column proofs, triangles, and more. All of these activities help students with their knowledge of …PopDriftDreadhead Parkour is a fun and engaging math activity that teaches students the basics of geometry and triangles. Students can practice their skills in two-column proofs, side angle side, side side side, and angle angle side. Join the dreadhead parkour team and explore the world of geometry in this interactive game. Dash is a popular rhythm-based platform game that has gained a massive following since its release in 2013. With its addictive gameplay and challenging levels, it has beco... RunCard skimmers are usually hidden underneath a card terminal. Here's everything you need to know about credit card skimmers so you can spot and avoid them. * Required Field Your Nam knowledge of side angle side, side side side, and angle angle side.It took Boston Dynamics a quarter of a century to release its first commercial product, so one can forgive the company for taking a few extra months to make that product more widel...Is there an easy way to spot money-making scams? Keep reading to learn about money scams and discover if there is an easy way to spot them. Advertisement You'd think there'd be a s... Only Up is a geometry math activity where students can learn more about two-column proofs, triangles, and more. All of these activities help students with their knowledge of side angle side, side side side, and angle angle side. This contains most of the activities on Geometry Spot. These help students with understanding SSS, SAS, AAS, ASA, and more concepts. 8 Ball Pool. Cannon Basketball. Paper Minecraft. ADVERTISEMENT. Slope is a geometry math activity where students can learn more about two-column proofs, triangles, and … This contains all of the new activities on Geometry Spot. These help students with understanding SSS, SAS, AAS, ASA, and more concepts. Eggy CarJustFall.lol Geometry Dash is an addictive and challenging platform game that has gained immense popularity among gamers of all ages. With its simple yet captivating gameplay, it has become a f...2048Spotted lake is a very unusual natural phenomenon that you can see with your own eyes near Osoyoos in British Columbia, Canada. For years, I'd passed by Spotted Lake along British ...With Spotify, It's Important to Pick the Right Spot to Buy...SPOT Traders are pushing up shares of Spotify Technology (SPOT) on the heels of a company announcement that the dig Car Drawing is a geometry math activity where students can learn more about two-column proofs, triangles, and more. All of these activities help students with their knowledge of side angle side, side side side, and angle angle side. Car Drawing is a math activity that can help students understand the basics of geometry and the basics of in geometry. This contains most of theRocket League is a geometry math activity where students can learn more about two-column proofs, triangles, and more. All of these activities help students with their knowledge of side angle side, side side side, and angle angle side.This contains most of the activities on Geometry Spot. These help students with understanding SSS, SAS, AAS, ASA, and more concepts Gam es. PAPAS. Gam es. Fighting. Gam es. This contains all of the action activities on Geometry Spot. These help students with understanding SSS, SAS, AAS, ASA, and more concepts Dash is a popular rhythm-based platform game that has gained a massive following since its release in 2013. With its addictive gameplay and challenging levels, it has becoAre you looking for the perfect place to settle down and start a family? Look no further than Glen Burnie, Maryland. Located just outside of Baltimore, this charming town offers pl...B . Five Nights At Freddy's 2/FNAF 2 is a geomBallz is a geometry math activity where students ca Gam es. PAPAS. Gam es. Fighting. Gam es. This contains all of the action activities on Geometry Spot. These help students with understanding SSS, SAS, AAS, ASA, and more concepts. Cut The Rope is a geometry math activity where students can learn more about two-column proofs, triangles, and more. All of these activities help students with their knowledge of side angle side, side side side, and angle angle side. Cut The Rope is a math activity that can help students understand the basics of geometry and the basics of ... If you're a fan of challenging platformer This contains all of the action activities on Geometry Spot. These help students with understanding SSS, SAS, AAS, ASA, and more concepts. Fall Guys is a geometry math activity where st... | 677.169 | 1 |
7-Introduction-to-Geometry
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Introduction to Geometry
Definition 1: A location in a plane is called point. It is
denoted by a dot "." and it has no size and dimensions.
Example 1: A point is located at (4,3)
Definition 4: Ray is a part of a line where it only extends
at one end.
Example 4: A ray formed by points A and B.
Definition 2: A line is a set of all points that extends
infinitely at both ends. In addition, a line can be formed
when two points are connected to each other.
Example 2: Let A and B be two distinct points then line
⃡ is formed.
𝐴𝐵
A
B
Definition 3: Line segment is a part of a line that connects
two points. These two points are called endpoints.
Example 3: Let A and B be two distinct points, then line
segment AB is formed
Definition 5: A formation of two rays with common
endpoints are called angle. The common endpoints are
called vertex.
Example 5: Ray AB and ray AC intersect at point A and
form an angle.
KINDS OF ANGLES
Definition 6: An angle that measures less than 90 degrees
is called Acute Angle.
Definition 7: An angle that measures equal to 90 degrees
is called Right Angle.
Definition 8: An angle that measures greater than 90
degrees is called Obtuse Angle.
Definition 9: An angle that measures equal to 180 degrees
is called Supplementary Angle/Straight Angle.
Definition 10: An angle that measures greater than 180
degrees is called Reflex Angles.
TYPES OF LINES
Definition 11: Two lines that never meets is called Parallel
Lines.
Definition 12: Two lines that intersect at one point is
called Intersecting Lines.
Definition 13: Perpendicular Lines are intersecting lines
that has an angle of 90 degrees or right angle.
Definition 14: Transversal Lines are lines that intersects
a parallel line.
4
5 6
8 7
1
1
1
1
1
1
1
1
1 2
3
1
Observe that <1 and <2 are adjacent angles on the
same line. These angles form a straight line and their sum
is equal to 180 degrees.
Also, observe that <1, <2, <7 and <8 are exterior
angles since it is located outside of the transversal line.
and <3, <4, <5, and <6 are interior angles since it is located
inside of the transversal line. Alternate exterior angles
are exterior angles that lies on the opposite side of the
transversal line while Alternate interior angles are
interior angles that lies on the opposite side of the
transversal line.
Note that <2 and <6 lies on the same side of the
transversal line. These lines are called Corresponding
lines. | 677.169 | 1 |
How do you find the reference angle in quadrant 3?
How do you find the reference angle in quadrant 3?
When the terminal side is in the third quadrant (angles from 180° to 270°), our reference angle is our given angle minus 180°. So, if our given angle is 214°, then its reference angle is 214° – 180° = 34°.
Quadrants & Quadrantal Angles Angles between 0∘ and 90∘ are in the first quadrant. Angles between 90∘ and 180∘ are in the second quadrant. Angles between 180∘ and 270∘ are in the third quadrant. Angles between 270∘ and 360∘ are in the fourth quadrant.
What is 3pi 4 reference angle?
Coterminal angles are equal angles.
Can a reference angle be 90 degrees?
When an angle is drawn on the coordinate plane with a vertex at the origin, the reference angle is the angle between the terminal side of the angle and the x-axis. The reference angle is always between 0 and 2π radians (or between 0 and 90 degrees).
What is the name of 270 degree angle?
reflex angles
Angles such as 270 degrees which are more than 180 but less than 360 degrees are called reflex angles. A 360° angle is called a complete angle. | 677.169 | 1 |
How Far Does it Move?
Take a look at the interactivity below which shows regular polygons "rolling" along the horizontal surface.
It leaves a trace of the path of the dot and on the graph it records the distance that the dot travels.
Experiment by positioning the dot at the centre of the polygons, at one of the vertices or at the centre of one of the sides of the polygons and explore how this affects the distance / time graph.
Challenge:
Can you now work out what produced the following distance / time graph?
Can you work out how many sides the polygon had and where the dot was placed?
Try to explain how you worked it out | 677.169 | 1 |
Hint: In this question, we need to find the length of BD which is used for constructing triangle ABC. In triangle ABC, we have BC = 10cm, $\angle B={{60}^{\circ }}$ and AB+AC = 14cm. We will construct triangle ABC using given measures and then find the length of BD using isosceles triangle property. According to the isosceles triangle property, the angles corresponding to equal sides are also equal.
Complete step-by-step solution Let us construct triangle ABC in which BC = 10cm, $\angle B={{60}^{\circ }}$ and AB+AC = 14cm. We will use following steps of construction: (i) Draw line segment BC = 10cm. At B, let us draw an angle, say $\angle XBC$ which is equal to ${{60}^{\circ }}$ using a protractor.
(ii) Now, cut the ray BX at the length of AB+AC = 14cm at point D (open compass at 14cm and cut the ray and mark the point as D). And then join DC.
(iii) Now, measure $\angle BDC$ and make the same angle at C naming DCY. Ray CY cuts the line BD at point A.
Hence, $\Delta ABC$ is the required triangle with BC = 10cm, AB+AC = 14cm and $\angle B={{60}^{\circ }}$. Now, we need to find the length BD. Since BD was already drawn as 14cm so BD = 14cm. But let us prove BD as 14cm. From $\Delta ACD$ we can see that, $\angle ADC=\angle DCA$. Hence, $\Delta ACD$ is an isosceles triangle. So, AD = AC by the reverse of isosceles triangle property. We know that, AB+AC = 14cm. Putting AD = AC we get: AB+AD = 14cm. From the diagram, we can see that AB+AD = BD. Hence, BD = 14cm. Hence, option B is the correct answer.
Note: Students should take care while measuring the angle $\angle BDC$. When drawing angle $\angle DCY$ make sure to take base as CD and then measure the angle using a protractor. For making an angle of ${{60}^{\circ }}$ students can use a protractor or compass. Make sure pencils are sharp and the compass is tight. | 677.169 | 1 |
Understanding the Circumcentre of a Triangle.
Triangles are fundamental shapes in geometry, consisting of three line segments that connect three non-collinear points. Among the various points associated with a triangle, the circumcentre holds a special place. Understanding the circumcentre of a triangle involves grasping its definition, properties, and significance in geometry. In this article, we delve into the intricacies of the circumcentre, shedding light on its role in relation to triangles.
Definition of Circumcentre
The circumcentre of a triangle is defined as the point where the perpendicular bisectors of the three sides of the triangle intersect. In simpler terms, it is the center of the unique circle that passes through all three vertices of the triangle.
Properties of the Circumcentre
Equidistant: The circumcentre is equidistant from all three vertices of the triangle. This means that the distances from the circumcentre to each vertex are equal.
Perpendicular: The circumcentre lies at the intersection of the perpendicular bisectors of the sides of the triangle. This property implies that the lines joining the circumcentre to the vertices are perpendicular to the respective sides.
Unique Circle: The circumcentre is crucial as it is the center of the circumcircle, the circle passing through all three vertices of the triangle. The circumcircle is the smallest circle that contains the triangle within it.
Construction of the Circumcentre
To construct the circumcentre of a triangle:
Bisect the Sides: Draw the perpendicular bisectors of at least two sides of the triangle.
Locate Intersection Point: The intersection of these bisectors is the circumcentre of the triangle.
Significance of the Circumcentre
The circumcentre holds significance in various aspects of geometry and beyond:
Circle Geometry: It plays a crucial role in circle geometry, particularly in relation to triangles.
Orthocenter and Centroid: In certain types of triangles (such as acute and right-angled triangles), the circumcentre coincides with the orthocenter and centroid, respectively.
Center of Inscribed Circle: It is also the center of the inscribed circle (incircle) of an Orthocenter triangle.
Applications of the Circumcentre
Understanding the circumcentre is not just limited to theoretical geometry but also finds practical applications in fields such as:
Architectural Design: Architects use geometric concepts such as the circumcentre to design aesthetically pleasing structures.
Computer Graphics: In computer graphics, knowledge of the circumcentre is essential for rendering geometric shapes accurately.
Navigation Systems: Understanding the circumcentre is crucial in the development of GPS and other navigation systems that rely on geometric calculations.
FAQs about the Circumcentre of a Triangle
What is the relationship between the circumcentre and the incenter of a triangle?
The circumcentre is the center of the circumcircle, while the incenter is the center of the incircle. In certain triangles, these two centers may coincide.
Does every triangle have a circumcentre?
Yes, every non-degenerate triangle (a triangle that is not a straight line) has a circumcentre.
How is the circumcentre related to the Fermat point of a triangle?
The Fermat point is a point inside an equilateral triangle where the sum of the distances to the vertices is minimized. In an equilateral triangle, the circumcentre, centroid, orthocenter, and Fermat point are all the same.
Can the circumcentre lie outside the triangle?
No, the circumcentre of a triangle always lies inside the triangle, on the interior or boundary.
What is the importance of the circumcircle in relation to the circumcentre?
The circumcircle passes through all three vertices of the triangle and has the circumcentre as its center, making it a pivotal circle associated with the triangle.
In conclusion, the circumcentre of a triangle is a key point that embodies the intersection of geometric concepts within a triangle. Understanding its properties and significance not only enriches one's knowledge of geometry but also unveils its practical applications in diverse fields. Mastering the nuances of the circumcentre elevates one's comprehension of the intricate relationships within geometric shapes, paving the way for deeper exploration and application of mathematical principles. | 677.169 | 1 |
What is the difference between a vertical and a horizontal market?
The horizontal line is drawn by connecting related swing lows in price to create a horizontal help line. For a horizontal resistance line, comparable swing highs are connected. When your subject is shifting up or down, utilizing a vertical format in conjunction with the rule of thirds visually permits the topic room to continue transferring. Having open house to the top of backside permits the topic's gaze to proceed farther than is possible in a horizontal image.
What is an example of a horizontal line?
A horizontal line is one which runs left-to-right across the page. In geometry, a horizontal line is one which runs from left to right across the page. It comes from the word 'horizon', in the sense that horizontal lines are parallel to the horizon.
In reality, the gravity field of a heterogeneous planet such as Earth is deformed as a result of inhomogeneous spatial distribution of materials with totally different densities. Actual horizontal planes are thus not even parallel even when their reference points are alongside the same vertical line, since a vertical line is barely curved. Furthermore, a airplane can't each be a horizontal plane at one place and a vertical aircraft somewhere else. The horizontal line is then used for analytical or buying and selling purposes. For example, if the worth of an asset is moving between assist and resistance horizontal strains then the value is taken into account to be range-bound.
When your subject is moving from one side of the frame to the opposite, utilizing a horizontal format in conjunction with the rule of thirds visually permits the subject room to proceed transferring. Having open space to the side permits the topic's gaze to proceed farther than is possible in a vertical picture. At any given location, the total gravitational force isn't fairly constant over time, as a result of the objects that generate the gravity are shifting.
How is horizontal line drawn?
Properties of Horizontal Lines Equation of Horizontal Line always takes the form of y = k where k is the y-intercept of the line. For instance in the graph below, the horizontal line has the equation y = 1 As you can see in the picture below, the line goes perfectly sideways at y =1. Example 1 of a Vertical Line.
Also, horizontal planes can intersect when they're tangent planes to separated points on the surface of the earth. In explicit, a plane tangent to a degree on the equator intersects the plane tangent to the North Pole at a proper angle.
A horizontal line is any line regular to a vertical line. An uptrend is when a price makes greater swing highs and higher swing lows. Therefore, a horizontal line can highlight when price is making a new high, in this case, thus exhibiting signs of an uptrend. On the SPY chart above, the value is moving above the horizontal line indicating an uptrend. If the worth falls back below the horizontal line, it could warn that uptrend has failed and lower costs may be forthcoming.
In basic, one thing that is vertical can be drawn from as much as down (or right down to up), such because the y-axis in the Cartesian coordinate system. A horizontal line is one the goes left-to-right,parallel to the x-axis of the coordinate aircraft. All factors on the line could have the same y-coordinate. In the figure above, drag either point and notice that the road is horizontal after they both have the identical y-coordinate.
Any airplane going via P, regular to the horizontal airplane is a vertical airplane at P. Through any level P, there may be one and just one horizontal plane however a multiplicity of vertical planes. This is a brand new function that emerges in three dimensions. The symmetry that exists in the two-dimensional case no longer holds.
Equation of a Horizontal Line
In specific, a plane tangent to a point on the equator intersects the aircraft tangent to the North Pole at a right angle.
Furthermore, the equatorial plane is parallel to the tangent aircraft at the North Pole and as such has claim to be a horizontal aircraft.
But it's. at the identical time, a vertical airplane for points on the equator.
Also, horizontal planes can intersect when they are tangent planes to separated factors on the floor of the earth.
For occasion, on Earth the horizontal plane at a given level (as decided by a pair of spirit levels) changes with the place of the Moon (air, sea and land tides). Horizontal planes at two separate points are not parallel, they intersect. When the curvature of the earth is taken under consideration, the independence of the 2 movement does not maintain. Neglecting the curvature of the earth, horizontal and vertical motions of a projectile transferring underneath gravity are impartial of each other.
Furthermore, the equatorial airplane is parallel to the tangent airplane on the North Pole and as such has declare to be a horizontal aircraft. But it is. on the same time, a vertical plane for factors on the equator.
It comes from the phrase 'horizon', in the sense that horizontal traces areparallel to the horizon. Horizontal images can be used to recommend a sense of largeness in landscapes. If a small topic is positioned in a big area, it can be used to suggest loneliness.
A Horizontal Line because it Relates to Supply and Demand Curves
Not all merchants could place the horizontal line at the same worth. In technical analysis, the horizontal line is usually drawn along a swing high, or a collection of them, the place every excessive within the series stopped at an analogous stage. A horizontal line is commonly utilized in technical analysis to mark areas of assist or resistance. A vertical line is any line parallel to the vertical path.
What is horizontal and vertical?
The horizontal line is drawn by connecting similar swing lows in price to create a horizontal support line. For a horizontal resistance line, similar swing highs are connected. The horizontal line is then used for analytical or trading purposes.
Swing excessive is a technical evaluation time period that refers to cost or indicator peak. Swing highs are analyzed to point out trend course and energy.
What is the formula for a horizontal line?
vertical. Vertical describes something that rises straight up from a horizontal line or plane. The terms vertical and horizontal often describe directions: a vertical line goes up and down, and a horizontal line goes across. You can remember which direction is vertical by the letter, "v," which points down.
Remember additionally that when a subject is shifting deeper into a picture or shifting in direction of the camera that this appears as "up or down" movement when transformed into a 2D picture. This is why many main strains images work very nicely as vertical images. In the three-dimensional case, the state of affairs is extra sophisticated as now one has horizontal and vertical planes along with horizontal and vertical lines. Consider a degree P and designate a path by way of P as vertical. A aircraft which contains P and is regular to the designated course is the horizontal plane at P.
Horizontal Line
Drawing a horizontal line is likely one of the simplest types of technical evaluation, but it also provides necessary data. On the chart beneath, a horizontal line is drawn on the SPDR S&P 500 (SPY) change traded fund (ETF). In more easy phrases, a horizontal line on any chart is the place the y-axis values are equal. If it has been drawn to point out a sequence of highs in the data, a data point transferring above the horizontal line would indicate a rise within the y-axis worth over recent values within the data pattern. In geometry, a horizontal line is one which runs from left to right throughout the web page.
In this sense, a aircraft can, arguably, be each horizontal and vertical, horizontal at one place, and vertical at another. In technical evaluation, a horizontal line is usually drawn on a value chart to focus on areas of assist or resistance. Hence, the world seems to be flat regionally, and horizontal planes in nearby locations appear to be parallel. In this case, the horizontal path is often from the left aspect of the paper to the best facet | 677.169 | 1 |
How do you use #csctheta=5# to find #tantheta#?
1 Answer
Explanation:
There are several methods to solve this problem.
Geometric
Imagine a right-angle triangle with an angle #theta#. Since #csc(theta)=5#, the hypotenuse is #5#, its opposite side is #1#, and its adjacent side is #sqrt(5^2-1^2)=sqrt(24)=2sqrt(6)#. Thus, #tan(theta)# is the opposite over the adjacent, or #tan(theta)=1/(2sqrt(6))=sqrt(6)/12#.
Now, since #csc(theta)=5>0#, #theta# is either in quadrant 1 or 2. Because #tan(theta)# is positive in quadrant 1 and negative in quadrant 2, there are two possible answers: #+-sqrt(6)/12#.
Another way to realize that there is a second solution is to realize that it is possible to draw the right triangle in the second quadrant. The hypotenuse is #5# (remember that the hypotenuse is always positive). The opposite side, which is the #y#-value, is still positive (it is still #1#). Thus, #csc(theta)=5/1=5#. However, the adjacent side, which is the #x#-value, becomes negative in quadrant 2 (#-sqrt(5^2-1^2)=-sqrt(24)=-2sqrt(6)#). Thus, #tan(theta)#, being the opposite over the adjacent, has the exact same value but with a negative sign. | 677.169 | 1 |
Euclid's Elements of Geometry: Chiefly from the Text of Dr. Simson, with ...
wherefore the angle BGC is equal to the angle DBA or GBC; and therefore the side BC is equal to the side CG; (1. 6.) but BC is equal also to GK, and CG to BK; (1. 34.) wherefore the figure CGKB is equilateral.
It is likewise rectangular ;
for, since CG is parallel to BK, and BC meets them, therefore the angles KBC, BCG are equal to two right angles; (1. 29.) but the angle KBC is a right angle; (def. 30. constr.) wherefore BCG is a right angle:
and therefore also the angles CGK, GKB, opposite to these, are right angles; (1. 34.)
wherefore CGKB is rectangular :
but it is also equilateral, as was demonstrated; wherefore it is a square, and it is upon the side CB. For the same reason HF is a square,
and it is upon the side HG, which is equal to AC. (1. 34.) Therefore the figures HF, CK, are the squares on AC, CB. And because the complement AG is equal to the complement GE, (1.43.)
and that AG is the rectangle contained by AC, CB, for GC is equal to CB;
therefore GE is also equal to the rectangle AC, CB; wherefore AG, GE are equal to twice the rectangle AC, CB; and HF, CK are the squares on AC, CB;
wherefore the four figures HF, CK, AG, GE, are equal to the squares on AC, CB, and twice the rectangle AC, CB:
but HF, CK, AG, GE make up the whole figure ADEB, which is the square on AB;
therefore the square on AB is equal to the
twice the rectangle AC, CB.
squares on
Wherefore, if a straight line be divided, &c Q. E.D.
AC, CB, and
COR. From the demonstration, it is manifest, that the parallelograms about the diameter of a square, are likewise squares
PROPOSITION V. THEOREM.
If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.
Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts in the point D.
Then the rectangle AD, DB, together with the square on CD, shall be equal to the square on CB.
ر
Upon CB describe the square CEFB, (1. 46.) join BE, through D draw DHG parallel to CE or BF, (1. 31.) meeting BE in H, and EF in G,
and through H draw KLM parallel to CB or EF, meeting CE in L, and BF in M;
also through A draw AK parallel to CL or BM, meeting MLK in K. Then because the complement CH is equal to the complement HF, (1. 43.) to each of these equals add DM;
therefore the whole CM is equal to the whole DF;
but because the line AC is equal to CB, therefore AL is equal to CAL, (1. 36.) therefore also AL is equal to DF;
to each of these equals add CH,
and therefore the whole AH is equal to DF and CH: but AH is the rectangle contained by AD, DB, for DH is equal to DB; and DF together with CII is the gnomon CMG;
therefore the gnomon CMG is equal to the rectangle AD, DB: to each of these equals add LG, which is equal to the square on CD; (11. 4. Cor.)
therefore the gnomon CMG, together with LG, is
but the gnomon CMG and LG make up the whole figure CEFB, which is the square on CB;
rectangle AD, DB, together with the square on C. to the
therefore the rectangle AD, DB, together with the square on CD is equal to the square on CB.
Wherefore, if a straight line, &c. Q.E.D.
COR. From this proposition it is manifest, that the difference of the squares on two unequal lines AC, CD, is equal to the rectangle contained by their sum AD and their difference DB.
PROPOSITION VI. THEOREM.
If a straight line be bisected, and produced to any point;Let the straight line AB be bisected in C, and produced to the point D. Then the rectangle AD, DB, together with the square on CB, shall be equal to the square on CD.
Upon CD describe the square CEFD, (1. 46.) and join DE, through B draw BHG parallel to CE or DF, (1. 31.) meeting DE in H, and EF in G;
through I draw KLM parallel to AD or EF, meeting DF in M, and CE in L;
and through A draw AK parallel to CL or DM, meeting MLK in K.
Then because the line AC is equal to CB,
therefore the rectangle AL is equal to the rectangle CH, (1. 36.) but CH is equal to HF; (1. 43.) therefore AL is equal to HF;
to each of these equals add CM;
therefore the whole AM is equal to the gnomon CMG: but AM is the rectangle contained by AD, DB, for DM is equal to DB: (11. 4. Cor.)
therefore the gnomon CMG is equal to the rectangle AD, DB: to each of these equals add LG which is equal to the square on CB; therefore the rectangle AD, DB, together with the square on CB, is equal to the gnomon CMG, and the figure LG;
but the gnomon CMG and LG make up the whole figure CEFD, which is the square on CD;
therefore the rectangle AD, DB, together with the square on CB, is equal to the square on CD.
Wherefore, if a straight line, &c. Q. E. D.
PROPOSITION VII. THEOREM.
If a straight line be divided into any two parts, the squares on the whole line, and on one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square on the other part.
Let the straight line AB be divided into any two parts in the point C. Then the squares on AB, BC shall be equal to twice the rectangle AB, BC, together with the square on AC,
Upon AB describe the square ADEB, (1. 46.) and join BD, through C draw CF parallel to AD or BE (1. 31.) meeting BDin G, and DE in F;
through G draw HGK parallel to AB or DE, meeting AD in H,
and BE in K.
Then because AG is equal to GE, (1. 43.)
add to each of them CK;
therefore the whole AK is equal to the whole CE; and therefore AK, CE, are double of AK:
but AK, CE, are the gnomon AKF and the square CK; therefore the gnomon AKF and the square CK are double of AK: but twice the rectangle AB, BC, is double of AK,
for BK is equal to BC; (II. 4. Cor.)
therefore the gnomon AKF and the square CK, are equal to twice the rectangle AB, BC;
to each of these equals add HF, which is equal to the square on AC therefore the gnomon AKF, and the squares CK, HF, are equal to twice the rectangle AB, BC, and the square on AC;
but the gnomon AKF, together with the squares CK. HF make
up the whole figure ADEB and CK, which are the squares on AB and BC
therefore the squares on AB and B Care equal to twice the rectangle AB, BC, together with the square on AC.
Wherefore, if a straight line, &c.
PROPOSITION VIII.
Q. E. D.
THEOREM.
If a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square on the other part, is equal to the square on the straight line, which is made up of the whole and that part.
Let the straight line AB be divided into any two parts in the point C. Then four times the rectangle AB, BC, together with the square on AC, shall be equal to the square on the straight line made up of AB and BC together
A
CBD
Produce AB to D, so that BD be equal to CB, (1. 3.) upon AD describe the square AEFD, (1. 46.) and join DE, through B, C, draw BL, CH parallel to AÈ or DF, and cutting DE in the points K, P respectively, and meeting EF in L, H; through K, P, draw MGKN, XPRO parallel to AD or EF. Then because CB is equal to BD, CB to GK, and BD to KN; therefore GK is equal to KN;
for the same reason, PR is equal to RO;
and because CB is equal to BD, and GK to KN, therefore the rectangle CK is equal to BN, and GR to RN; (1. 36.) but CK is equal to RN, (1. 43.)
because they are the complements of the parallelogram CO; therefore also BN is equal to GR;
and the four rectangles BN, CK, GR, RN, are equal to one another, and so are quadruple of one of them CK.
Again, because CB is equal to BD, and BD to BK, that is, to CG; and because CB is equal to GK, that is, to GP;
therefore CG is equal to GP.
And because CG is equal to GP, and PR to RO, therefore the rectangle AG is equal to MP, and PL to RF; but the rectangle MP is equal to PL, (1. 43.)
because they are the complements of the parallelogram ML: wherefore also AG is equal to RF:
therefore the four rectangles AG, MP, PL, RF, are equal to one another, and so are quadruple of one of them AG.
And it was demonstrated, that the four CK, BN, GR, and RN, are quadruple of CK:
therefore the eight rectangles which contain the gnomon AOH, are quadruple of AK.
And because AK is the rectangle contained by AB, BC,
for BK is equal to BC;
therefore four times the rectangle AB, BC is quadruple of AK: but the gnomon AOH was demonstrated to be quadruple of AK; therefore four times the rectangle AB, BC is equal to the gnomon AOH; to each of these equals add XH, which is equal to the square on AC; therefore four times the rectangle AB, BC, together with the square on AC, is equal to the gnomon AOI and the square XH; but the gnomon AOH and XI make up the figure AEFD, which is the square on AD;
7
therefore four times the rectangle AB, BC together with the square on AC, is equal to the square on AD, that is, on AB and BC added together in one straight line.
Wherefore, if a straight line, &c, Q. E.D.
PROPOSITION IX. THEOREM.
If a straight line be divided into two equal, and also into two unequal parts; the squares on the two unequal parts are together double of the square on half the line, and of the square on the line between the points of section.
Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts in the point D.
Then the squares on AD, DB together, shall be double of the squares on AC, CD.
From the point C draw CE at right angles to AB, (1. 11.) make CE equal to AC or CB, (1. 3.) and join EA, EB; through D draw DF parallel to CE, meeting EB in F, (I. 31.) through Fdraw FG parallel to BA, and join AF. Then, because AC is equal to CE,
therefore the angle AEC is equal to the angle EAC; (1. 5.) and because ACE is a right angle,
therefore the two other angles AEC, EAC of the triangle are together equal to a right angle; (1. 32.)
and since they are equal to one another;
therefore each of them is half a right angle.
For the same reason, each of the angles CEB, EBC is half a right angle; and therefore the whole AEB is a right angle. And because the angle GEF is half a right angle, and EGF a right angle,
for it is equal to the interior and opposite angle ECB, (1. 29.) therefore the remaining angle EFG is half a right angle; wherefore the angle GEF is equal to the angle EFG, and the side GF equal to the side EG. (1. 6.) | 677.169 | 1 |
Part 1: Angles • An angle consists of two different rays (sides) that share a common endpoint (vertex). Vertex Sides This angle can be called: Angle ABC, <ABC, <CBA OR it can be named by the vertex like so: <B
Types of Angles Go to the following website to learn about the different types of angles. Don't forget to press the play button on top of the picture to see the animation. Use the diagram to find the measure of the indicated angle. Then classify each angle as either acute, right, obtuse, or straight. • KHJ • GHK • GHJ • GHL
5. What is the measure of DOZ?… How did you get to your answer?
Angle Addition Postulate If P is in the interior of RST, then mRST = mRSP + mPST.
6. Given that mLKN = 151°, find mLKM and mMKN.
Congruent Angles • Two angles are congruent angles if they have the same measure.
Angle Bisector An angle bisector is a ray that divides an angle into two congruent angles.
Linear Pairs of Angles • Two adjacent angles form a linear pair if their noncommon sides are opposite rays. • The angles in a linear pair are supplementary.
Vertical Angles • Two nonadjacent angles are vertical angles if their sides form two pairs of opposite rays. • Vertical angles are formed by two intersecting lines.
S U • Name a pair of complementary angles. • Name a pair of supplementary angles. • Name a linear pair. • Name a pair of vertical angles. • Name a pair of congruent angles. Y T V X W
Part 2: Parallel Lines What would you call two lines which do not intersect? Answer: Parallel Lines A solid arrow placed on two lines of a diagram indicate the lines are parallel. Exterior Interior The symbol || is used to indicate parallel lines. Exterior AB || CD
A slash through the parallel symbol || indicates the lines are not parallel. AB || CD
Transversal – A transversal is a line which intersects two or more lines in a plane. The intersected lines do not have to be parallel. Lines j, k, and m are intersected by line t. Therefore, line t is a transversal of lines j, k, and m.
L Line L G Line M P A B Line N Q D E F Same Side Interior/Exterior Angles Same Side Interior Angles: Two angles that lie between parallel lines on the same sides of the transversal. Same Side Exterior Angles: Two angles that lie outside parallel lines on the same sides of the transversal. ÐAPQ + ÐDQP = 1800 ÐBPQ + ÐEQP = 1800 Exterior 600 1200 Same Side Interior or Same Side Exterior angles are supplementary. Interior 1200 600 Exterior
Line L G Line M P A B 500 Line L G 1300 Line M Line N A P B Q E D Line L G F Line N Q D E Line M P A B F Line N Q E D F Make sure you are in slide show view to do this test. Name the pairs of the following angles formed by a transversal. Test Yourself Corresponding angles Alternate angles Interior angles | 677.169 | 1 |
Pages
How to calculate area of triangle in Java Program? Example
Writing a Java program to calculate the area of a triangle is one of the basic programming exercises to develop a coding sense for beginner programmers. Like many mathematical conceptual programs e.g. square root, factorial, or prime number this also serves as a good exercise for beginners. Now, if you remember in maths you might have seen two main ways to calculate the area of a triangle, using vertices and using base and height. In this program, I have created two methods to calculate the area of a triangle using both ways.
In the first method area(Point a Point b, Point c) we expect coordinates of three vertices of the triangle and then we calculate the area of the triangle using the formula (Ax(By -Cy) + Bx(Cy -Ay) + Cx(Ay - By))/2, while in the second method, area(int base, int height) we expect the value of base and height and then we calculate are of the triangle using formula (base * height) / 2.
Calculating the area of a triangle using 3 points
In this program, you need to write a method that accepts three points, you can create a Point class to encapsulate both X and Y coordinates and then calculate the area of the triangle using the following formula:
Calculating area of a triangle using base and height
This one is easier to remember as you might have used this formula lot many times before. In this part of the program, you write a method that expects two integer values to capture base and height and return a float which is the area of a triangle.
We'll use the following formula:
Area of triangle = (base * height) / 2
Where the base is the length of the base and height is the height of the triangle as shown in the following diagram:
Now that you know the formulas to calculate the area of a triangle in Java, we'll see the code which implements these formulas. Remember, you can use these formulas to calculate the area of any type of triangle e.g. right-angle triangle, equilateral triangle, etc. These are generic formulas and work for all types of triangles.
Java Program to calculate the area of a triangle
Here is our complete Java program to find the area of a triangle given base and height or points of three vertices. I have created a Point class to represent a point that has both X and Y coordinates and two overloaded area() methods to calculate the area of a triangle.
The first area() method expects three parameters, which are points of three vertices and then it return a float value which is the area of a triangle. The second area() method takes base and height and returns a float value which is the area of a triangle.
That's all about how to calculate the area of a triangle in Java. This is a good exercise to learn to program along with many others which I have shared below. We have learned both ways to calculate the area of a triangle in the program i.e. using 3 points of vertices as well as by using the base and height formula.
You can further read Concrete Mathematics: A Foundation for Computer Science to learn more about how mathematics plays an important role in Computer Science and programming.
Thanks for reading this article. If you like then please share it with your friends and colleagues. If you have any questions or feedback, please drop a note. If you have any questions or doubt then please let us know and I'll try to find an answer for you. Btw, what is your favorite coding exercise? prime number, palindrome or this one? | 677.169 | 1 |
Elements of Geometry: Containing the First Six Books of Euclid: With a Supplement on the Quadrature of the Circle and the Geometry of Solids ...
From inside the book
Results 1-5 of 58
Page 27 ... bisect a given rectilineal angle , that is , to divide it into two equal angles . Let BAC be the given rectilineal ... bisected by the straight line AF . Which was to be done . A F B E PROP . X. PROB . To bisect a given finite OF ...
Page 28 ... bisect a given finite straight line , that is , to divide it into two equal parts . Let AB be the given straight line ; it is required to divide it into two equal parts . Describe ( 1. 1. ) upon it an equilateral triangle ABC , and bisect ...
Page 29 ... bisect ( 10. 1. ) FG in H , and join CF , CH , CG ; the straight D. line CH , drawn from the given point C , is perpendicular to the given straight line AB . Because FH is equal to HG , and HC common to the two triangles FHC , GHC , the ...
Page 31 ... Bisect ( 10. 1. ) AC in E , join BE and produce it to F , and make EF equal to BE ; join al- so FC , and produce AC to G. ; Because AE is equal to EC , and BE ... bisected , it may be demonstrated that the OF GEOMETRY . BOOK 1 . 1 31.
Page 32 ... bisected , it may be demonstrated that the angle BCG , that is ( 15. 1. ) , the angle ACD , is greater than the angle ABC . Therefore , if one side , & c . Q. E. D. PROP . XVII . THEOR . Any two angles of a triangle are together less | 677.169 | 1 |
Saturday, May 25, 2024
The
state file spheretri.d32 is about solving triangles on the spherical
space.
The
programs solve spherical triangles in two common problems: SSS
(side-side-side, really arc lengths) and SAS (side-angle-side). All
the inputs are in decimal degrees.
Also
calculated are the surface area and perimeter, both in radians. The
radius is assumed to be 1.
Surface
Area = ( A° + B° + C° ) * π / 180 – π = A + B + C – π
Perimeter
= ( X° + Y° + Z°) * π / 180 = X + Y + Z
The
sum of the angles (A, B, C) must be greater than 180° (π radians).
Due to this requirement, in solving for angles and sides, the Law of
Cosines will be used in each instance. The Law of Sines is only
advised to check ratios. | 677.169 | 1 |
How to find the cosine of 56°
Get an answer to your question ✅ "How to find the cosine of 56° ..." in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. | 677.169 | 1 |
Cosec Cot Formula
Trigonometry is the field of study dealing with the relationships between angles, heights, and lengths in right triangles. This article covers the topic of the Cosec Cot Formula. The ratio of the sides of a right triangle is known as the trigonometric ratio. There are six major ratios in trigonometry: sin, cos, tan, cot, sec, and cosec. All these ratios have different formulas. Use the three sides and corners of a right triangle. Trigonometric ratios are relationships between angle and length measurements. A right triangle has a hypotenuse, a base, and a perpendicular. These three pages let you get the values of all six functions. Trigonometry is the branch of mathematics that deals with the relationships between angles, heights and lengths in right triangles. The ratio of the sides of a right triangle is known as the trigonometric ratio. Sin, cos, tan, cot, sec, and cosec are the six most important trigonometric ratios. The formula for each of these ratios is different. Use the three sides and corners of a right triangle. Refer to Extramarks for Cosec Cot Formula
What Is Cosec Cot Formula?
What is the Cosec Cot Formula?
Information on the Cosec Cot Formula is given below:
Cosec x = Hypotenuse / Opposite side
cot x is x = Adjacent side / Opposite side
The study of the relationships between angles, heights and lengths in triangles is called trigonometry. Trigonometry has many uses in engineering, architectural design, astronomy, and physics.
Trigonometry identities are very useful. There are many areas where these can be applied.
Trigonometry has six main functions: Sin, Cos, Tan, Cot, Sec, and Cosec. All these functions have different expressions. Learners can search for Cosec Cot Formula on Extramarks.
The relationship between angle and length measurements is called the trigonometric ratio.
A right triangle has a hypotenuse, a base, and a perpendicular. And using these three aspects, we can find the values of all six functions. Cosec = Hypotenuse / Plunge
cot = base/vertical
What is the Cosec Cot Formula?
cosec x = hypotenuse / opposite side
cot x is
bed x = adjacent side / opposite side
The Cosec Cot Formula is:
1 + cot2θ = cosec2θ
Trigonometric identities are equations that apply to various trigonometric functions and apply to all values of variables within a domain. These are equations for all possible values of the variables and trigonometric ratios used in these identities are sine, cosine, tangent, cosecant, secant, and cotangent. All these trigonometric ratios are calculated using the sides of a right triangle, such as the adjacent side, the opposite side, and the hypotenuse. These trigonometric identities are valid only for right triangles. Students are advised to learn Cosec Cot Formula.
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FAQs (Frequently Asked Questions)
1. What is meant by Trigonometric Identities?
Trigonometric identities are equations that apply to various trigonometric functions and apply to all values of variables within a domain. These are equations for all possible values of variables and the trigonometric ratios used in these identities are sine, cosine, tangent, cosecant, secant, and cotangent. All these trigonometric ratios are calculated using the sides of a right triangle as follows: B. Adjacent side, opposite side and hypotenuse. These trigonometric identities apply only to right triangles.
2. What is the Cosec Cot Formula?
Cosec Cot Formula is the Pythagorean trigonometric identity in trigonometry because it is based on the Pythagorean theorem. | 677.169 | 1 |
June 2019 regents geometry
Regents Examination in Geometry – June 2019; Scoring Key: Part I (Multiple-Choice Questions) MC = Multiple-choice question CR = Constructed-response question Source
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Part 3 Questions 32 - 34Please click the link below to subscribe to my channel: Geometry [Common Core] June 2019 Reg... Past Regents Examinations in Geometry. Revised Test Design for the Regents Examination in Geometry. Geometry Sample Questions. Spring 2014. Fall 2014. Geometry Test Guide. Geometry Standards Clarifications. Geometry Performance Level DescriptionsRab. II 29, 1439 AH ... Hello New York State Geometry students! I hope you are learning and enjoying this regents review video to assist you in preparation for the ...January 2023. August 2022. June 2022. January 2020. August 2019. June 2019. Last Updated: April 18, 2024. Regents Exam in Global History and Geography II.NYS geometry regents June 2019 question 15 Dhuʻl-H. 25, 1441 AH ... NYS geometry regents August 2019 question 21.Barron of Regents. This edition includes: Two actual Regents exams in …Dhuʻl-H. 2, 1443 AH ... NYS Common Core Geometry Regents June 2022 MathSux Learn how to ace your upcoming Geometry Regents one question at a ...Geometry regents new york algebra algebra 1 trigonometry trig key solutions answer key geometry regents January june august review pass math exam test mr Chemistry june 2019 regents answer key. krause mr3 rectangles with dimensions 40 feet by 16 feet, as shown below. 40NYS geometry regents June 2019 question 1ALGEBRA 1 (COMMON CORE) JUNE 2019 REGENTS EXAM. ALGEBRA 1 (COMMON CORE) JUNE 2019 Multiple Choice Solutions. ALGEBRA 1 (COMMON CORE) JUNE 2019 Free Response Solutions ... For the last six years, Trevor has been tutoring high school mathematics in Algebra I, Geometry, Algebra II, & Pre-Calculus. He graduated from …
FOR TEACHERS ONLY. The University of the State of New York. REGENTS HIGH SCHOOL EXAMINATION. GEOMETRY. Wednesday,January 23, 2019 — 9:15 a.m. to 12:15 p.m., only. SCORING KEY AND RATING GUIDE. Mechanics of Rating. The following procedures are to be followed for scoring student answer papers for the Regents …Geometry Regents: June 2019 - Free download as PDF File (.pdf), Text File (.txt) or read online for free. The entire Geometry Regents for June 2019…
GEOMETRY The University of the State New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Tuesday, June 21, 2022 - 9:15 a.m. to 12:15 p.m., only2 Geometry June 2014 Regents 2019-11-20 Geometry June 2014 Regents Downloaded from web.mei.edu by guest DUDLEY MAGDALEN A DRIVEN BY DATA Teachers College Press This book revolutionized 19th-century American architecture and changed forever the type of building that was done in our country. Testimonios: Stories of Latinx and Hispanic …
The June 2024 Geometry Regents exam marks another milestone in the academic journey of countless students. With its diverse range of questions and challenging problems, it presents an opportunity for individuals to showcase their understanding of geometric theorems, proofs, and applications. In this article, we will delve into the answers of ...Dhuʻl-H. 27, 1443 AH ... In this video I go through the Geometry Regents June 2022, free response, questions 25-35. I cover many of the topics from high school ...Geometry –
Shaw. 11, 1439 AH ... ... Geometry – COMMON CORE – Regen Regents Examination in Geometry – January 2023 Andre Castagna. Simon and Schuster, Jan 5, 20 Geometry – June '19 [11] 27 A support w Feb Feb 9, 2024 · June 2019 Regents Examinati student's final score. The chart above is usable only for this administration of the Regents Examination in Geometry. Raw Score Scale Score Raw Score The State Education Department / The University of the State of New York Regents Examination in Geometry – June 2019 Chart for Converting Total Test Raw Scores to Final Exam Scores (Scale Scores) The June 2017 New York State Regents exam in GeoDhuʻl-H. 25, 1443 AH ... In this video I go through the GeomGEOMETRY The University of the State of New York RE Architects use geometry to help them design build June 2019 Regents Examination in Geometry Regular size version (505 KB) Large type version (191 KB) Scoring Key PDF version (22 KB) Excel version (19 KB) Rating Guide For the Geometry Regents exam, you'll also need to remember certain [17 If f(x) = 2x + 6 and g(x) = lxl are graphed Chart for Converting Total Test Raw Scores to NYS geometry regents June 2019 question 24 | 677.169 | 1 |
Dot product of 3d vector. Dot Product can be used to project the scalar length o...
Visual interpretation of the cross product and the dot product of two vectors.My Patreon page: A and B are matrices or multidimensional arrays, then they must have the same size. In this case, the dot function treats A and B as collections of vectorsAutoCAD is a powerful software tool used by architects, engineers, and designers worldwide for creating precise and detailed drawings. With the advent of 3D drawing capabilities in AutoCAD, users can now bring their designs to life in a mor...When vectors are pointing in the same or similar direction, the dot product is positive. When vectors are pointing in opposite direction, the dot product is …AForYesWhere |a| and |b| are the magnitudes of vector a and b and ϴ is the angle between vector a and b. If the two vectors are Orthogonal, i.e., the angle between them is 90 then a.b=0Understand We tutorial is a short and practical introduction to linear algebra as it applies to game development. Linear algebra is the study of vectors and their uses. Vectors have many applications in both 2D and 3D development and Godot uses them extensively. Developing a good understanding of vector math is essential to becoming a strong game developer. Next to add/subtract/dot product/find the magnitude simply press the empty white circle next to the "ADDITION" if you want to add the vectors and so on for the others. 2 To find the value of the resulting vector if you're adding or subtracting simply click the new point at the end of the dotted line and the values of your vector will appear.Taking a dot product is taking a vector, projecting it onto another vector and taking the length of the resulting vector as a result of the operation. Simply by this definition it's … It representation of the vector that starts at the point O(0;0;0) and ends at the point P(x 1;y 1;z 1) is called the position vector of the point P. Vector Arithmetic: Let a= ha 1;a 2;a …... dot product of two vectors based on the vector's position and length. This calculator can be used for 2D vectors or 3D vectors. If a user is using this ...A 3D matrix is nothing Teams. Q&A for work. Connect and share knowledge within a single location that is structured and easy to search. Learn more about TeamsWhen vectors are pointing in the same or similar direction, the dot product is positive. When vectors are pointing in opposite direction, the dot product is … …)"What the dot product does in practice, without mentioning the dot product" Example ;)Force VectorsVector Components in 2DFrom Vector Components to VectorSum...Cross Products. Whereas a dot product of two vectors produces a scalar value; the cross product of the same two vectors produces a vector quantity having a direction perpendicular to the original two vectors.. The cross product of two vector quantities is another vector whose magnitude varies as the angle between the two original vectors changes. The … How to find the angle between two 3D vectors?Using the dot product formula the angle between two 3D vectors can be found by taking the inverse cosine of the ... b are multiplied.Visual interpretation of the cross product and the dot product of two vectors.My Patreon page: This product leads to a scalar quantity that is given by the product of the ...Taking a dot product is taking a vector, projecting it onto another vector and taking the length of the resulting vector as a result of the operation. Simply by this definition it's …Free vector dot product calculator - Find vector dot product step-by-stepThe dot product of 3D vectors is calculated using the components of the vectors in a similar way as in 2D, namely, ⃑ 𝐴 ⋅ ⃑ 𝐵 = 𝐴 𝐵 + 𝐴 𝐵 + 𝐴 𝐵, where the subscripts 𝑥, 𝑦, and 𝑧 denote the components along the 𝑥 -, 𝑦 -, and 𝑧 -axes. Let us apply this method with the next example We learn how to calculate the scalar product, or dot product, of two vectors using their components. AThe two main equations are the dot product and the magnitude of a 3D vector equation. Dot product of 3D vectors. For two certain 3D vectors A (x1, y1, z1) ...Let's make sure you got this by finding the dot product for each problem below. Problem #1 – 2D Vectors \(\langle 3,2\rangle \cdot\langle-1,4\rangle=(3)(-1)+(2)(4)=-3+8=5\) Problem #2 – 3D Vectors \(\langle-5,-3,4\rangle \cdot\langle 6,-2,1\rangle=(-5)(6)+(-3)(-2)+(4)(1)=-30+6+4=-20\) Simple! Dot … See moreDot Product Formula. . This formula gives a clear picture on the properties of the dot product. The formula for the dot product in terms of vector components would make it …The first thing we want to do is find a vector in the same direction as the velocity vector of the ball. We then scale the vector appropriately so that it has the right magnitude. Consider the vector w w extending from the quarterback's arm to a point directly above the receiver's head at an angle of 30 ° 30 ° (see the following figure). Dot products Google Classroom Learn about the dot product and how it measures the relative direction of two vectors. The dot product is a fundamental way we can combine two vectors. Intuitively, it tells us something about how much two vectors point in the same direction. Definition and intuitionThe geometric definition of the dot product is great for, well, geometry. For example, if two vectors are orthogonal (perpendicular) than their dot product is 0 because the cosine of 90 (or 270) degrees is 0. Another example is finding the projection of a vector onto another vector. By trigonometry, the length of the projection of the vector interpretation as the length of the projection of X onto the unit vector Y^^ …Dot product between two 3D vectors. Public method Static, Dot(Vector3D, Point3D), Dot product between a 3D vector and a 3D point. Public ...Thanks to 3D printing, we can print brilliant and useful products, from homes to wedding accessories. 3D printing has evolved over time and revolutionized many businesses along the way.Visual interpretation of the cross product and the dot product of two vectors.My Patreon page: order to find a vector C perpendicular B we equal their dot product to zero. Vector C written in unit vector notation is given by: And the dot product is: The previous equation is the first condition that the components of C must obey. Moreover, its magnitude has to be 2: And substituting the condition given by the dot product: Finally, C ...The dot product is defined for any $\mathbf{u,v}\in\mathbb{R}^n$ as, ... \mathbf{v}\|\cos[\measuredangle(\mathbf{u},\mathbf{v})] $$ In 1D, 2D, and 3D, ... that it is the choice of an inner-product on a vector space (or a pseudo-inner product if you wish to be more general) which allows you to start talking about geometry on a vector space; and torchand g(v,v) ≥ 0 and g(v,v) = 0 if and only if v = 0 can be used as a dot product. An example is g(v,w) = 3 v1 w1 +2 2 2 +v3w3. The dot product determines distance and distance determines the dot product. Proof: Lets write v = ~v in this proof. Using the dot product one can express the length of v as |v| = √ v ·vAug 17, 2023 · In . The Vector Calculator (3D) computes vector functions (e.g. V • U The cross product (also called the vector product or outer product) is Yes Let's make sure you got this by finding the dot product f For Free vector dot product calculator - Find vector dot... | 677.169 | 1 |
Darius is studying the relationship between mathematics and art. He asks friends to each draw a "typical" rectangle. He measures
the length and width in centimeters of each rectangle and plots the points on a graph, where x represents the width and y represents the length. The points representing the rectangles are (6.1, 12.0), (5.0, 8.1), (9.1, 15.2), (6.5, 10.2), (7.4, 11.3), and (10.9, 17.5). Which equation could Darius use to determine the length, in centimeters, of a "typical" rectangle for a given width in centimeters? y = 0.605x + 0.004
y = 0.959x + 0.041
y = 1.518x + 0.995
y = 1.967x + 0.984
Given the right angled triangle as shown, the angles A, B and C are related to sides a, b and c using the trigonometry ratio known as SOH CAH TOA.
The triangle has three sides namely the hypotenuse (AB), the opposite and adjacent. The hypotenuse is the longest side and the opposite side of the triangle depends on the acute angle in consideration.
When considering acute angle A, the opposite side will be the side facing the angle directly i.e side a.
According to the trigonometry ratio
SOH:
SinA = opposite/hypotenuse = a/c
CAH:
CosA = adjacent/hypotenuse = b/c
TOA:
TanA = opposite/adjacent = a/b
When considering acute angle B, the opposite side will be 'b' and adjacent will be a. Similarly as above; | 677.169 | 1 |
Triangle angles
Given triangle ABC with no angle >120 degrees, find and construct the point P for which PA + PB + PC is a minimum. What is this point called? What would be the case for a triangle with an angle of 120 degrees or more?
Purchase this Solution
Solution Summary
This shows how to construct a point in a triangle with given characteristics.
Solution Preview
Step 1:Draw the triangle A, B, C
Step 2: Construct on every side of the triangle an equilateral triangle. By this procedure the points A*, B*, C* are generated.
Step 3: Connect each of the newly constructed points A*, B*,179370 Triangle Measures Angles Solve.
A triangle has three angles, R, S, and T. Angle T is 40° greater than angle S. Angle R is 8 times angle S. What is the measure of each angle? A triangle has three angles, R, S, and T.
The triangles share a corner angle and both have right angles. Therefore the third angles are also congruent by the sum of interior angles. We find the scale factor by setting a ratio of corresponding sides 4/5.
The three angles of this triangle is: 68, 32 and 78. This shows how to form an algebraic equation to express a problem about triangleangles and identify the variables, coefficients, and constants of the algebraic expression. | 677.169 | 1 |
👷 Pythagoras's Theorem has numerous practical applications in various fields, including construction, navigation, and surveying.
🎮 The video emphasizes the importance of logical deduction and identifying opportunities to apply the theorem in problem-solving.
❎ Calculators can be used to simplify calculations involving squares and square roots.
Transcript
good day welcome to the tech maath channel uh and what we're going to be having a look at in this video is we're going to be looking about how to find uh we're going to be using Pythagoras's Theorem once again um and this is Pythagoras's Theorem here which states that basically if you square the sum of the two shorter sides on a right angle triangl...
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Questions & Answers
Q: What is Pythagoras's Theorem?
Pythagoras's Theorem states that in a right-angled triangle, the square of the length of one side added to the square of the length of the other side equals the square of the length of the hypotenuse.
Q: How can Pythagoras's Theorem be applied to non-right-angled triangles?
To apply Pythagoras's Theorem to non-right-angled triangles, you can create right-angled triangles within them and use the theorem to find the lengths of the unknown sides.
Q: Can Pythagoras's Theorem be used to find the height of a triangle?
Yes, Pythagoras's Theorem can be used to find the height of a triangle. By labeling the height as one side and using the base length and hypotenuse, you can apply the theorem to solve for the height.
Q: What are some practical applications of Pythagoras's Theorem?
Pythagoras's Theorem is commonly used in fields such as architecture, engineering, and physics. It can be used to solve a variety of real-world problems involving measurements and distances in both two-dimensional and three-dimensional space.
Summary & Key Takeaways
The video explains Pythagoras's Theorem, which states that the square of the two shorter sides of a right-angled triangle is equal to the square of the longest side (hypotenuse).
The video demonstrates how to use Pythagoras's Theorem to find unknown side lengths in non-right-angled triangles by creating right-angled triangles within them.
Two examples are provided, showing step-by-step calculations to find the lengths of unknown sides using Pythagoras's Theorem.
The video also discusses using Pythagoras's Theorem to find the height of a triangle and mentions its application in calculating the area of a triangle. | 677.169 | 1 |
4. PA and PB are two tangents drawn from an external point P to a circle with centre O where the points A and B are the points of contact. The quadrilateral OAPB must be (SSC Sub. Ins. 2012)
(a) a square
(b) concylic
(c) a rectangle
(d) a rhombus
12. A, O, B are three points on a line segment and C is a point not lying on AOB. If ∠AOC = 40° and OX, OY are the internal and external bisectors of ∠AOC respectively, then ∠BOY is (SSC CGL 1st Sit. 2012)
(a) 70°
(b) 80°
(c) 72°
(d) 68°
13. In the following figure, O is the centre of the circle and XO is perpendicular to OY. If the area of the triangle XOY is 32, then the area of the circle is (SSC CGL 1st Sit. 2012)
(a) 64 π
(b) 256 π
(c) 16 π
(d) 32 π
15. Two circles of radii 4 cm and 9 cm respectively touch each other externally at a point and a common tangent touches them at the points P and Q respectively. They the area of a square with one side PQ, is (SSC CGL 1st Sit. 2012)
(a) 97 sq. cm
(b) 194 sq. cm
(c) 72 sq.cm
(d) 144 sq.cm
16. Two tangents are drawn from a point Pto a circle at A and B. 0 is the centre of the circle. If ∠AOP = 60°, then ∠ APB is (SSC CGL 1st Sit. 2012)
(a) 120°
(b) 90°
(c) 60°
(d) 30°
17. If each intetior angle is double of each exterior angle of a regular polygon with n sides, then the value of n is (SSC CGL 1st Sit. 2012)
(a) 8
(b) 10
(c) 5
(d) 6
18. If the length of the side PQ of the rhombus PQRS is 6 cm and ∠PQR= 120°, then the length of QS, in cm, is (SSC CGL 1st Sit. 2012)
(a) 4
(b) 6
(c) 3
(d) 5
19. The angle formed by the hour-hand and the minute-hand of a clock at 2:15 p.m. is (SSC CGL 1st Sit. 2012)
(a) 27°
(b) 45°
(c) 22
(d) 30°
20. Two sides of a triangle are of length 4 cm and 10 cm. If the length of the third side is 'a' cm. then (SSC CGL 1st Sit. 2012)
(a) a>5
(b) 6≤a≤12
(c) a<5
(d) 6<a< 14
29. AB is a diameter of a circle with centre O. CD is a chord equal to the radius of the circle. AC and BD are produced to meet at P. Then the measure of ∠APB is: (SSC CGL 2nd Sit. 2012)
(a) 120°
(b) 30°
(c) 60°
(d) 90°
30. R and r are the radius of two circles (R > r). If the distance between the centre of the two circles be d, then length of common tangent of two circles is: (SSC CGL 2nd Sit. 2012)
31. P is a point outside a circle and is 13 cm away from its centre. A secant drawn from the point P intersect the circle at points A and B in such a way that PA = 9 cm and AB = 7 cm. The radius of the circle is: (SSC CGL 2nd Sit. 2012)
(a) 5.5 on
(b) 5 cm
(c) 4 cm
(d) 4.5 cm
37. AC and BC are two equal chords of a circle. BA is produced to any point P and CP, when joined cuts the circle at T. Then (SSC CGL 1st Sit. 2012)
(a) CT: TP=AB : CA
(b) CT:TP = CA:AB
(c) CT: CB = CA: CP
(d) CT:CB = CP:CA
38. PQ is a direct common tangent of two circles of radii r1 and r2 touching each other externally at A. Then the value of
39. BC is the chord of a circle with centre O. A is a point on major arc BC as shown in the above figure. What is the value of ∠BAC + ∠OBC ? (SSC CGL 1st Sit 2012)
(a) 120°
(b) 60°
(c) 90°
(d) 180°
40. Two circles with radii 5 cm and 8 cm touch each other externally at a point A. If a straight line through the point A cuts the circles at points P and Q respectively, then AP: AQ is (SSC CGL 1st Sit 2012)
(a) 8:5
(b) 5:8
(c) 3:4
(d) 4:5
41. If I is the In-centre of ∆ABC and ∠A = 60°, then the value of ∠BIC is (SSC CGL 1st Sit. 2012)
(a) 100°
(b) 120°
(c) 150°
(d) 110°
44. A rail road curve is to be laid out on a circle. What radius should be used if the track is to change direction by 25° in a distance of 40 metres? (SSC CGL 1st Sit 2012)
(a) 91.64 metres
(b) 90.46 metres
(c) 89.64 metres
(d) 93.64 metres
47. If D is the mid-point of the side BC of ∆ABC and the area of ∆ABD is 16 cm2, then the area of ∆ABC is (SSC CGL 2nd Sit 2012)
(a) 16 cm2
(b) 24 cm2
(c) 32 cm2
(d) 48 cm2
48. ABC is a triangle. The medians CD and BE intersect each other at O. Then A ODE: ∆ABC is (SSC CGL 2nd Sit. 2012)
(a) 1:3
(b) 1:4
(c) 1:6
(d) 1:12
49. If P, R, Tare the area of a parallelogram, a rhombus and a triangle standing on the same base and between the same parallels, which of the following is true? (SSC CGL 2nd Sit. 2012)
(a) R<P<T
(b) P>R>T
(c) R = P = T
(d) R = P = 2T
50. AB is a diameter of the circumcircle of ∆APB; N is the foot of the perpendicular drawn from the point P on AB. If AP = 8 cm and BP = 6 cm, then the length of BN is (SSC CGL 2nd Sit. 2012)
(a) 3.6 cm
(b) 3 cm
(c) 3.4 cm
(d) 3.5 cm
51. Two circles with same radius r intersect each other and one passes through the centre of the other. Then the length of the common chord is (SSC CGL 2nd Sit. 2012)
55. A and B are centres of the two circles whose radii are 5 cm and 2 cm respectively. The direct common tangents to the circles meet AB extended at P. Then P divides AB. (SSC CGL 2nd Sit. 2012)
(a) externally in the ratio 5:2
(b) internally in the ratio 2:5
(c) . internally in the ratio 5:2
(d) externally in the ratio 7:2
56. A wheel rotates 3.5 times in one second. What time (in seconds) does the wheel take to rotate 55 radian of angle? (SSC CGL 2nd Sit. 2012)
(a) 1.5
(b) 2.5
(c) 3.5
(d) 4.5
63. From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP is equal to diameter of the circle, then ∠APB is (SSC CHSL 2013)
(a) 60°
(b) 45°
(c) 90°
(d) 30°
64. A chord 12 cm long is drawn in a circle of diameter 20 cm. The distance of the chord from the centre is (SSG CHSL 2013)
(a) 16 cm
(b) 8 cm
(c) 6 cm
(d) 10 cm
65. 360 sq. cm and 250 sq. cm are the areas of two similar triangles. Ifthe length of one of the sides of the first triangle be 8 cm, then the length of the corresponding side of the 2nd triangle is (SSC CHSL 2013)
71. A chord of length 30 cm is at a distance of 8 cm from the centre of a circle. The radius of the circle is: (SSC CGL 1st Sit. 2013)
(a) 19
(b) 17
(c) 23
(d) 21
72. If ABCD be a rectangle and P, Q, R, S be the mid points
73. P and Q are two points on a circle with centre at O. R is a point on the minor arc of the circle, between the points P and Q. The tangents to the circle at the points P and Q meet eahc other at the point S. If ∠PSQ = 20°, ∠PRQ = ? (SSC CGL 1st Sit 2013)
(a) 100°
(b) 80°
(c) 200°
(d) 160°
74. AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on the opposite sides of the centre and distance between them is 17 cm, then the radius of the circle is: (SSC CGL 1st Sit 2013)
(a) 10cm
(b) 11cm
(c) 12 cm
(d) 13 cm
78. AB is the chord of a circle with centre O and DOC is a line segment originating from a point D on the circle and intersecting, AB produced at C such that BC = OD. If ∠BCD = 20°, then ∠AOD = ? (SSC CGL 2nd Sit. 2013)
(a) 20°
(b) 30°
(c) 40°
(d) 60°
79. In a circle of radius 17 cm, two parallel chords of lengths 30 cm and 16 cm are drawn. If both the chords are on the same side of the centre, then the distance between the chords is (SSC CGL 2nd Sit. 2013)
(a) 9 cm
(b) 7 cm
(c) 23 cm
(d) 11cm
80. ABC is a right angled triangle, B being the right angle. Mid points of BC and AC are respectively B' and A'. The ratio of the area of the quadrilateral AA' B'B to the area of the triangle ABC is (SSC CGL 2nd Sit. 2013)
(a) 1:2
(b) 2:3
(c) 3:4
(d) None of the above
81. In a triangle ABC, the side BC is extended up to D. Such that CD = AC, if ∠BAD = 109° and ∠ACB = 72° then the value of ∠ABC is (SSC CGL 2nd Sit. 2013)
(a) 35°
(b) 60°
(c) 40°
(d) 45°
82. Two circles touch each other internally. Their radii are 2 cm and 3 cm. The biggest chord of the greater circle which is outside the inner circle of length.
84. From a point P which is at a distance of 13 cm from center O of a circle of radius 5 cm, in the same plane, a pair of tangents PQ and PR are drawn to the circle. Area of quadrilateral PQOR is (SSC CGL 2nd Sit. 2013)
(a) 65 cm2
(b) 60 cm2
(c) 30 cm2
(d) 90 cm2
85. If the arcs of square length in two circles subtend angles of 60° and 75° at their centres, the ratio of their radii is (SSC CGL 2nd Sit. 2013)
(a) 3:4
(b) 4:5
(c) 5:4
(d) 3:5
86. N is the foot of the perpendicular from a point P of a circle with radius 7 cm, on a diameter AB of the circle. If the length of the chord PB is 12 cm, the distance of the point N from the point B is (SSC CGL 1st Sit. 2013)
91. A vertical stick 12 cm long casts a shadow 8 cm long on the ground. At the same time, a tower casts a shadow 40 m long on the ground. The height of the tower is (SSC CGL 1st Sit. 2013)
(a) 65 m
(b) 70 m
(c) 72 m
(d) 60 m
96. The length of tangent (upto the point of contact) drawn from an external point P to a circle of radius 5 cm is 12 cm. The distance of P from the centre of the circle is (SSC CGL Ist. Sitt 2013)
(a) 11cm
(b) 12cm
(c) 13 cm
(d) 14 cm
98. Two circles of equal radikouch externally at a point P. From a point T on the tangent at P, tangents TQ and TR are drawn to the circles with points of contact Q and R respectively. The relation of TQ and TR is (SSC CGL 1st Sit. 2013)
(a) TQ<TR
(b) TQ>TR
(c) TQ = 2TR
(d) TQ = TR
109. A, B and C are the three points on a circle such that the an gles subtended by the chords AB and AC at the centre O are 90° and 110° respectively. ∠BAC is equal to (SSC CGL 1st Sit. 2013)
(a) 70°
(b) 80°
(c) 90°
(d) 100°
110.
111. In a ∆ABC, AD, BE and CF are three medians. The perimeter of ∆ABC is always (SSC Sub. Ins. 2014)
118. If the sides of a right angled triangle are three consecutive integers, then the length ofthe smallest side is (SSC CHSL 2014)
(a) 3 units
(b) 2 units
(c) 4 units
(d) 5 units
119. Two circles intersect each other at the points A and B. A straight line parallel to AB intersects the circles at C, D, E and F. If CD = 4.5 cm, then the measure of EF is (SSC CHSL 2014)
(a) 1.50 cm
(b) 2.25 cm
(c) 4.50 cm
(d) 9.00 cm
125. Two parallel chords of a circle of diameter 20 cm are 12 cm and 16 cm long. If the chords are in the same side of the centre, then the distance between them is (SSC CGL 1st Sit. 2014)
(a) 28 cm
(b) 2 cm
(c) 4 cm
(d) 8 cm
126. The interior angle of a regular polygon is 140°. The number of sides of that polygon is (SSC CGL 1st Sit. 2014)
(a) 9
(b) 8
(c) 7
(d) 6
128. If in a triangle ABC, BE and CF are two medians perpendicular to each other and if AB=19 cm and AC=22 cm then the length of BC is : (SSC Sub. Ins. 2015)
(a) 20.5 cm
(b) 19.5 cm
(c) 13 cm
(d) 26 cm
130. Two isosceles triangles have equal vertical angles and their areas are in the ratio 9:16. Then the ratio of their corresponding heights is: (SSC Sub. Ins. 2015)
(a) 4.5:8
(b) 8:4.5
(c) 3:4
(d) 4:3
131. The perimetres oftwo similar triangles are 30 cm and 20cm respectively. If one side ofthe first triangle is 9cm. Determine the corresponding side of the second triangle: (SSC Sub. Ins. 2015)
(a) 15 cm
(b) 5 cm
(c) 6 cm
(d) 13.5 cm
132. The diagonal of a quadrilateral shaped field is 24m and the perpendiculars dropped on it from the remaining opposite vertices are 8m and 13m. The area of the field is: (SSC Sub. Ins. 2015)
(a) 252 m2
(b) 1152 m2
(c) 96 m2
(d) 156 m2
134. ABCD is a square. Draw a triangle QBC on side BC considering BC as base and draw a triangle PAC on AC as its base such that ∆QBC ~ ∆PAC.
135. The distance between centres of two circles of radii 3 cm and 8 cm is 13 cm. Ifthe points of contact of a direct common tangent to the circles are P and Q, then the length of the lien segment PQ is: (SSC CHSL 2015)
(a) 11.9 cm
(b) 11.5 cm
(c) 12 cm
(d) 11.58 cm
137. Two circles of radii 5 cm and 3 cm touch externally, then the ratio in which the direct common tangent to the circles divides externally the line joining the centres of the circles is: (SSC CHSL 2015)
(a) 2.5:1.5
(b) 1.5:2.5
(c) 3:5
(d) 5:3
138. In ∆ABC, a line through A cuts the side BC at D such that BD: DC = 4 :5. If the area of ∆ABD = 60 cm2, then the area of ∆ADC is (SSC CGL 1st Sit. 2015)
(a) 50 cm2
(b) 60 cm2
(c) 75 cm2
(d) 90 cm2
139. A tangent is drawn to a circle of radius 6cm from a point situated at a distance of 10 cm from the centre of the circle. The length of the tangent will be (SSC CGL 1st Sit. 2015)
(a) 4 cm
(b) 5 on
(c) 8 cm
(d) 7 cm
140. Two poles of height 7 m and 12 m stand on a plane ground. If the distance between their feet is 12 m, the distance between their top will be (SSC CGL 1st Sit. 2015)
(a) 13m
(b) 19m
(c) 17m
(d) 15m
141. The measure of an angle whose supplement is three times as large as its complement, is (SSC CGL 1st Sit. 2015)
(a) 30°
(b) 45°
(c) 60°
(d) 75°
142. The sides of a triangle having area 7776 sq. cm are in the ratio 3:4:5. The perimeter of the triangle is (SSC CGL 1st Sit. 2015)
(a) 400cm
(b) 412cm
(c) 424 cm
(d) 432 cm
143. Two chords of length a unit and b unit of a circle make angles 60° and 90° at the centre of a circle respectively, then the correct relation is (SSC CGL 1st Sit. 2015)
144. In a parallelogram PQRS, angle P is four times of angle Q, then the measure of ∠R is (SSC CGL 1st Sit. 2015)
(a) 36°
(b) 72°
(c) 130°
(d) 144°
145. If a clock started at noon, then the angle turned by hour hand at 3.45 PM is (SSC CGL 1st Sit. 2015)
146. Let C, and C2 be the inscribed and circumscribed circles of a triangle with sides 3 cm, 4 cm and 5 cm then area of C1 to area of C2 is (SSC CGL 1st Sit. 2015)
147. If the three angles of a triangle are:
(a) scalene
(b) isosceles
(c) right angled
(d) equilateral
148. If the number of vertices, edges and faces of a rectangular parallelopiped are denoted by v, e and f respectively, the value of (v – e + f) is (SSC CGL 1st Sit. 2015)
(a) 4
(b) 2
(c) 1
(d) 0
149. If the altitude of an equilateral triangle is 12√3 cm, then its area would be: (SSC CGL 1st Sit. 2015)
152. G is the centroid of ∆ABC. The medians AD and BE intersect at right angles. If the lengths of AD and BE are 9 cm and 12 cm respectively; then the length of AB (in cm) is ? (SSC CGL 1st Sit. 2015)
(a) 10
(b) 10.5
(c) 9.5
(d) 11
153. If a person travels from a point L towards east for 12 km and then travels 5 km towards north and reaches a point M, then shortest distance from L to M is : (SSC CGL 1st Sit. 2015)
(a) 14
(b) 12
(c) 17
(d) 13
154. If D,E and Fare the mid points of BC, CA and AB respectively of the ∆ABC then the ratio of area of the parallelogram DEFB and area of the trapezium CAFD is: (SSC CGL 1st Sit. 2015)
(a) 1:3
(b) 1:2
(c) 3:4
(d) 2:3
161. The length of the radius of a circle with centre O is 5 cm and the length of the chord AB is 8 cm. The distance of the chord AB from the point O is (SSC CGL 1st Sit. 2016)
(a) 2cm
(b) 3cm
(c) 4 cm
(d) 15 cm
165. AB is the diameter of a circle with centre O and P be a point on its circumference, If ∠POA =120°, then the value of ∠PBO is: (SSC CGL 1st Sit. 2016)
(a) 30°
(b) 60°
(c) 50°
(d) 40°
166. An arc of 30° in one circle is double an arc in a second circle, the radius of which is three times the radius of the first. Then the angles subtended by the arc of the second circle at its centre is (SSC CGL 1st Sit. 2016)
(a) 3°
(b) 4°
(c) 5°
(d) 6°
167. Which of the following ratios can be the ratio of the sides of a right angled triangle? (SSC CGL 1st Sit. 2016)
(a) 9:6:3
(b) 13:12:5
(c) 7:6:5
(d) 5:3:2
168. Number of circles that can be drawn through three non-colinear points is (SSC CGL 1st Sit 2016)
(a) exactly one
(b) two
(c) three
(d) more than three
169. Two circles touch each other internally. The radius of the smaller circle is 6 cm and the distance between the centre of two circles is 3 cm. The radius of the larger circle is (SSC CGL 1st Sit. 2016)
(a) 7.5 cm
(b) 9 cm
(c) 8 cm
(d) 10 cm
170. PQR is an equilateral triangle. MN is drawn parallel to QR such that M is on PQ and N is on PR. If PN = 6 cm, then the length of MN is (SSC CGL 1st Sit. 2016)
(a) 3 cm
(b) 6 cm
(c) 12 cm
(d) 4.5 cm
172. Two circles of same radius intersect each other at P and Q. If the length of the common chord is 30 cm and distance between the centres of the two circles is 40 cm, then what is the radius (in cm) of the circles? (SSC CGL 2017)
(a) 25
(b) 25√2
(c) 50
(d) 50√2
175. The perimeter of an isosceles triangle is 32 cm and each of the equal sides is 5/6 times of the base. What is the area (in cm2) ofthe triangle? (SSC CGL 2017)
(a) 39
(b) 48
(c) 57
(d) 64
176. If length of each side of a rhombus PQRS is 8 cm and ∠PQR = 120°, then what is the length (in cm) ofQS? (SSC CGL 2017)
(a) 4√5
(b) 6
(c) 8
(d) 12
177. In the given figure, ABC is a triangle. The bisectors of internal ∠B and external ∠C intersect at D. If ∠BDC = 48°, then what is the value (in degrees) of ∠A ? (SSC CGL 2017)
(a) 48
(b) 96
(c) 100
(d) 114
179. In triangle ABC, a line is drawn from the vertex A to a point D on BC. If BC = 9 cm and DC = 3 cm, then what is the ratio of the areas of triangle ABD and triangle ADC respectively? (SSC CGL 2017)
(a) 1:1
(b) 2:1
(c) 3:1
(d) 4:1
181. In triangle PQR, A is the point of intersection of all the altitudes and B is the point of intersection of all the angle bisectors of the triangle. If ∠PBR = 105°, then what is the value of ∠PAR (in degrees)? (SSC CGL 2017)
(a) 60
(b) 100
(c) 105
(d) 115
182. If there are four lines in a plane, then what cannot be the number of points of intersection of these lines? (SSC CGL 2017)
(a) 0
(b) 5
(c) 4
(d) 7
184. A chord of length 60 cm is at a distance of 16 cm from the centre of a circle. What is the radius (in cm) of the circle? (SSC CGL 2017)
(a) 17
(b) 34
(c) 51
(d) 68
185. In the given figure, a smaller circle touches a larger circle at P and passes through its centre O. PR is a chord of length 34 cm, then what is the length (in cm) of PS? (SSC CGL 2017)
(a) 9
(b) 17
(c) 21
(d) 25
186. In the given figure, ABC is a triangle in which, AB = 10 cm, AC = 6 cm and altitude AE = 4 cm. If AD is the diameter of the circum-circle, then what is the length (in cm) of circum- radius? (SSC CGL 2017)
(a) 3
(b) 7.5
(c) 12
(d) 15
SSC Mathematics Topic Wise Solved Papers – Data Interpretation
DIRECTIONS (Qs. 1-3): The pie chart, given here, represents the number of valid votes obtained by four students who contested election for school leadership. The total number of valid votes polled was 720.
Observe the chart and answer the questions based on it. (SSC CGL 1st Sit 2010)
1. What was the minimum number of votes obtained by any candidate?
(a) 100
(b) 110
(c) 120
(d) 130
DIRECTIONS (Qs. 4-6): The pie chart, given here, shows the amount of money spent on various sports by a school administration in a particular year.
Observe the pie chart and answer the questions based on this graph. (SSC CGL 2nd Sit.2010)
4. If the money spent on football was ₹ 9,000 how much more money was spent on hockey than on football ?
(a) ₹ 11,000 (b) ₹ 11,500
(c) ₹ 12,000 (d) ₹ 12,500
6. If the money spent on football is ₹ 9,000, then what was the total amount spent on all sports ?
(a) ₹ 73,000
(b) ₹ 72,800
(c) ₹ 72,500
(d) ₹ 72,000
DIRECTIONS (Qs. 7-10): The following graph shows the demand and production of cotton by 5 companies A, B, C, D and E. Study the graph and answer questions.
7. What is the ratio of companies having more demand than production to those having more production than demand?
(a) 2:3
(b) 4:1
(c) 3:2
(d) 1:4
8. What is the difference (in tonnes) between average demand and average production of the five companies taken together?
(a) 320
(b) 420
(c) 2100
(d) 1050
9. The production of company D is how many times that of the production of the company A?
(a) 1.8
(b) 1.5
(c) 0.5
(d) 0.4
10. The demand for company B is what percent of the demand for company C?
(a) 1.5
(b) 2.5
(c) 25
(d) 30
DIRECTIONS (Qs. 11-14) : The following graph shows the production of cotton bales of 100 kg each in lakhs by different states A, B, C, D and E over the years. Study the graph and answer Question Nos. 11 to 14. (SSC CGL 2nd Sit.2011)
11. The production of State C in 2003-2004 is how many times its production in 2005-2006?
(a) 2.5
(b) 1.85
(c) 1.5
(d) 0.4
12. In which State(s) is there a steady increase in the production of cotton during the given period?
(a) A and B
(b) B and D
(c) A and C
(d) DandE
13. How many kg of cotton was produced by State C durig the given period?
(a) 32,00,00,000 kg
(b) 42,50,00,000 kg
(c) 33,00,00,000 kg
(d) 35,00,00,000 kg
14. The number of States for which the production of cotton in 2005-2006 is less than or equal to the preceding year is
(a) 3
(b) 2
(c) 1
(d) There is no such states
DIRECTIONS (Qs. 15-19) : The pie-chart given below shows the distribution of workforce by occupational category for country in 1981 and 1995. Study the chart and answer the questions no. 15 to 19.
15. In 1981, the number of Sevice workers in the workforce, in millions, was
(a) 15.0
(b) 20.5
(c) 22.5
(d) 28.0
16. In 1981, the number of categories which comprised of more than 25 million workers each, is
(a) two
(b) three
(c) four
(d) five
17. The ratio of the number of workers in the Professional category in 1981 to the number of such workers in 1995 is
(a) 4:9
(b) 5:14
(c) 9:14
(d) 14:9
18. The increase in the number of Clerical workers in the workforce of country X from 1981 to 1995 (in millions) is
(a) 0.75
(b) 1.5
(c) 0.5
(d) 1.25
19. The percentage decrease in the number of Blue-Collar workers in the workforce of country X from 1981 to 1995 is
DIRECTIONS (Qs. 20-24): Read the following chart and answer the questions that follows:
The following pie-chart shows the preference of musical instruments of 60,000 people surveyed over whole India.
20. If 2100 people be less from the number of people who prefer Flute, the percentage of people who prefer Flute would have been:
(a) 9.5%
(b) 6.5%
(c) 7.5%
(d) 8.5%
21. The total number of people who prefer either Sarod or Guitar, is greater than the total number of people who prefer either Violin or Sitar by:
(a) 1200
(b) 1600
(c) 1100
(d) 1400
22. The number of people who prefer the musical instrument Sarod is:
(a) 7400
(b) 8400
(c) 6400
(d) 8600
23. If 16 2/3% of the people who prefer Piano, would go with the people who prefers Flute, the percentage of people who prefer Flute would have been :
(a) 13.5%
(b) 14.5%
(c) 15.5%
(d) 12.5%
24. The number of people who prefer Guitar is greater than the total number of people who prefer either Flute or Piano by:
(a) 1200
(b) 1100
(c) 1300
(d) 1400
DIRECTIONS (Qs 25-29) : Study the graph and answer the questions that follows :
Circle graph given below shows the expenditure incurred in bringing out a book by a publisher.
25. The central angle of the sector for the cost of the paper is :
(a) 22.5°
(b) 16°
(c) 54.8°
(d) 57.6°
26. Royalty on the book is less than the Advertisement charges by:
(a) 3%
(b) 25%
(c) 20%
(d) 16%
36. In which state are there the largest number of owners of Airtel simcard?
(a) Tamil Nadu
(b) Gujarat
(c) Kerala
(d) Assam
37. Average of simcard sold in the four states in lakhs is
(a) 30.25
(b) 40.5
(c) 35
(d) 33.75
38. The range of BSNL simcard sold in the 4 states in lakhs is:
(a) 12
(b) 15
(c) 14
(d) 13
39. Of all the simcards sold irj all the four states, the number of simcards sold in Gujarat is (approx)
(a) 40%
(b) 38%
(c) 35%
(d) 42%
DIRECTIONS (Qs. 40-44): Population of five adjacent areas of a town, in the year of 2010, are represented in the following Pie- chart. the ratio of the numbers of males to that of females in these areas are stated in the table below. The total of the population in all the five areas is 72 lakh. Study the Pie-chart and the table and then answer the questions.
41. The number of males in the areas S1 and S4 together is
(a) 13.8 lakh
(b) 8.2lakh
(c) 16.2 lakh
(d) 15.8 lakh
42. The ratio of number of females in the area S2 to that in the f area S5 is
(a) 108:35
(b) 36:13
(c) 9:7
(d) 13:36 .
43. If, in the year 2010, there was an increase of 5% population in the area S1 and 8% increase in population of the area S3 compared to the previous year, then the ratio of population in the areas S1 and S3, in the year 2009 was
(a) 108:35
(b) 27:10
(c) 27:70
(d) 10:3
44. The average of female population in all the five areas is lower than the female population in each of the areas
(a) S1 and S2 (b) S2 and S5
(c) S2 and S4 (d) S4 and S5
DIRECTIONS (Qs. 45-49): The following pie-chart represents the profits earned by a certain company in seven consecutive years. Study the pie-chart carefully and answer the question.
45. If the expenditure in the year 1993 was 30% more than the expenditure in the year 1991, then the income in the year 1993 exceeds the in come in the year 1991 by 3 0% of
(a) the income in the year 1991
(b) the expenditure in the year 1993
(c) the income in the year 1993
(d) the expenditure in the year 1991
46. If x% of the total of profits earned in all the given years is same as the profit earned in the year 1994, then x is
47. The ratios of expenditures and incomes in the years 1992, 1994 and 1996 are given to be 6:5:8 and 2:3:4 respectively. The ratio of the income in the year 1996 to the total expenditure in the years 1992 and 1994 is
(a) 40:11
(b) 10:7
(c) 20:11
(d) 20:13
48. The year in which the profit is nearest to the average of the profits earned in all the given years is
(a) 1991
(b) 1995
(c) 1993
(d) 1994
49. If the income in the year 1997 was 5 times the expenditure made in the same year, then the ratio of the profit earned in the year 1991 to the expenditure in the year 1997 was
(a) 11:28
(b) 44:7
(c) 28:11
(d) 7:44
50. The following graph represents the maximum and minimum temperature recorded every day in a certain week. The day on which the difference between the maximum and minimum temperature was maximum is (SSC Multitasking 2013)
(a) Monday
(b) Wednesday
(c) Saturday
(d) Sunday
51. Different choices made by a group of 200 students are given below in percentage. The number of students who have taken neither Science nor Commerce is (SSC Multitasking 2013)
(a) 40
(b) 80
(c) 120
(d) 60
DIRECTIONS (Qs. 52-55) : The following table shows the productions of food-grains (in million tons) in a state for the period 1999 – 2000 to 2003 – 2004. Read the table and answer the questions.
52. In 2002 – 2003, the percentage increase in the production of barley as compared to the previous year was:
(a) 14.20
(b) 17.85
(c) 18.75
(d) 7.90
53. During the period 1999 – 2000 to 2003 – 2004, x per cent of the total production is production of wheat. The value of x is about:
(a) 12.6
(b) 37.4
(c) 37.8
(d) 20.2
54. In the year 2003 – 2004, the increase in production was maximum over the previous year for:
(a) Rice
(b) Barley
(c) Other cereals
(d) Wheat
55. The difference of average production of rice and the average production of barley over the years is:
(a) 50
(b) 60
(c) 80
(d) 40
DIRECTIONS (Qs. 56-60): Production of three different flavours soft drinks X, Y and Z for a period of six years has been expressed in the following graph. Study the graph and answer the questions.
56. The approximate decline in the production of flavour Z in 2010 as compared to the production in 2008 is:
(a) 33%
(b) 22.5%
(c) 42%
(d) 25%
57. The average annual production was maximum in the given period for the flavour:
(a) Y only
(b) Z only
(c) X and Z
(d) X only
58. What percent of the total production of flavour X in 2005 and 2006 combined is the total production of flavour Z in 2007 and 2008 combined?
(a) 102.25
(b) 115.57
(c) 133.33
(d) 96.67
59. The percentage of rise/fall in production from the previous year is maximum for the flavour Y in this year:
(a) 2007
(b) 2008
(c) 2009
(d) 2006
60. The difference (in lakh bottles) between the average production of flavour X in 2005,2006,2007 and the average production of flavour Y in 2008, 2009 and 2010 is:
(a) 2.4
(b) 0.5
(c) 1.5
(d) 5
DIRECTIONS (Qs. 61-63): The following pie-chart shows the number of students admitted in different faculties of a college. Study the chart and answer the question.
61. If 1000 students are admitted in science, what is the total number of students ?
(a) 360
(b) 180
(c) 1800
(d) 3600
62. If 1000 students are admitted in science, what is the ratio of students in science and arts ?
(a) 5:6
(b) 6:5
(c) 7:5
(d) 7:6
63. How many students are more in commerce than in law if 1000 students are in science ?
(a) 20
(b) 200
(c) 2000
(d) 500
DIRECTIONS (Qs. 64-67): Study the two pie-charts and answer the questions.
66. The ratio of amount spent for savings in April month's salary and miscellaneous in May month's salary is :
(a) 235:50
(b) 216:25
(c) 217:26
(d) 205:13
67. What is the percent increase in Education in May month than April month?
(a) 10.82%
(b) 9.56%
(c) 12.35%
(d) 20%
68. Study the above bar graph showing the production of food grains (in million tons). (SSC CGL 2nd Sit.2013)
What is the ratio between the maximum production and the minimum production during the given period?
(a) 1:2
(b) 2:3
(c) 3:4
(d) 5:2
69. The total number of students involved in the data is
(a) 33
(b) 32
(c) 43
(d) 42
70. The maximum number of students got the marks in the interval of
(a) 10-20
(b) 20-30
(c) 30-40
(d) 40-50
71. The least number of students got the marks in the interval
(a) 40-50
(b) 20-30
(c) 10-20
(d) 0-10
72. The ratio of the students obtaining marks in the first and the last interval is
(a) 5:4
(b) 6:5
(c) 4:5
(d) 3:4
73. The difference in the amount estimated by the family on interior decoration and architect's fees is
(a) ₹ 10000
(b) ₹ 9500
(c) ₹ 7200
(d) ₹ 9600
74. In a certain country, allocations to various sectors of the yearly budget per ₹ 1000 crores are represented by this pie- diagram. The expenditure (in ₹) on Agriculture is
(a) 250 crores
(b) 150 crores
(c) 300 crores
(d) 200 crores
DIRECTIONS (Qs. 75-77): Shown below is the multiple bar diagram depicting the changes in the roll strength of a college in four faculties from 2001 – 02 to 2003 – 04.
76. The percentage of students in Law faculty in 2003 – 04 is
(a) 15.6%
(b) 16.7%
(c) 14.8%
(d) 18.5%
77. Percentage of increase in Science students in 2003 – 04 over 2001-2002 is
DIRECTIONS (Qs. 78-81): The following pie-chart shows the marks scored by a student in different subject – viz. Physics (Ph), Chemistry (Ch), Mathematical (M), Social Science (SS) and English (E) in an examination. Assuming that total marks obtained for the examination in 810, answer the questions given below.
78. The subject in which the student obtained 135 marks is
(a) Chemistry
(b) Mathematics
(c) English
(d) Physics
79. The marks obtained in English, Physics and Social Science exceed the marks obtained in Mathematics and Chemistry by
80. The difference ofmarks between Physics and Chemistry is same as that between
(a) Mathematics and English
(b) English and Social Science
(c) Chemistry and Social Science
(d) Physics and English
81. The marks obtained in Mathematics and Chemistry exceed the marks obtained in Physics and Social Science by
(a) 40
(b) 45
(c) 50
(d) 30
DIRECTIONS (Qs. 82-84) : The following graph shows the expenditure incurred in bringing a book, by a magazine producer. Study the graph and answer question.
82. What should be the central angle of the sector for the cost of the paper ?
(a) 57.6°
(b) 54.4°
(c) 56.7°
(d) 54.8°
DIRECTIONS: (Qs. 85-86) : The pass percentage for an examination in a school is shown in the adjoining bar diagram, for males and females separately for four years. Study the diagram and answer the question.
85. The maximum percentage of students passed in the year is
(a) 2007
(b) 2008
(c) 2009
(d) 2010
86. The year in which the difference of pass percentage between male and female is maximum, is
(a) 2010
(b) 2009
(c) 2008
(d) 2007
DIRECTIONS (Qs. 87-88): The adjacent histogram shows the average pocket money received by 60 students for a span of one month. Study the diagram and answer the question.
105. If the percentage of expenditure on food is x% of the total percentage of expenditure on clothing, education and fuel, then x equals (SSC Sub. Ins. 2014)
106. Total percentage of expenditure on house rent, clothing and fuel is greater than the percentage of expenditure on food by (SSC Sub. Ins. 2014)
(a) 16
(b) 17
(c) 18
(d) 20
DIRECTIONS (Qs. 107-110): The bar chart representing the number of first year B.Com. students of St. Xavier's College using different companies' smart phones. Study bar chart and answer the question that follow: The bar chart representing the no. of students using different smartphones. (SSC Sub. Ins. 2014)
107. The ratio of the number of boys to the number of girls using the smart phones of Samsung and Sony together is
(a) 12:13
(b) 13:12
(c) 14:11
(d) 11:14
110. The difference between the total number of students using smart phones of Samsung combined together and the total number of students using smart phone of Sony taken together is
(a) 20
(b) 60
(c) 80
(d) 40
DIRECTIONS (Qs. 112-115): Sales of Books (in thousand numbers) from Six Branches – B1, B2, B3, B4, B5 and B6 of a Publishing Company in 2000 and 2001. Study the graph and answer the question that follow:
112. Total sale of branches Bl, B3 and B5 together for both the years (in thousand numbers) is
(a) 250
(b) 310
(c) 435
(d) 560
113. Find the ratio of the total sales of branch B2 for both years to the total sales of branch B4 for both years.
(a) 2:3
(b) 3:5
(c) 4:5
(d) 7:9
114. Percentage of the average sale of branches B1, B2 and B3 in 2001 and the average sale of branches B1, B3 and B6 in 2000
(a) 87.5
(b) 75
(c) 77.5
(d) 82.5
115. Find the percentage increase in the sales of books of branch B3 in the year 2001 than the branch B2.
(a) 69.2
(b) 50.8
(c) 40.9
(d) 65.7
DIRECTIONS (Qs. 116-120): The diagram shows the age distribution of the patients admitted to a hospital in a particular day. Study the diagram and answer
(SSC CHSL 2014)
116. Number of patients of age between 55 years to 60 years, who got admitted to the hospital on that day is
(a) 6
(b) 4
(c) 24
(d) 8
117. Total number of patients of age more than 55 years, who got admitted to the hospital is
(a) 4
(b) 7
(c) 9
(d) 10
118. Number of patients of age more than 40 years and less than 55 years, who got admitted to the hospital on that day is
(a) 20
(b) 30
(c) 15
(d) 12
119. Percentage of patients of ageless than 45 years, who got admitted to the hospital on that day is approximately equal to
(a) 14%
(b) 20%
(c) 37%
(d) 62%
120. About 11% ofthe patients who got admitted to the hospital on that particular day were of age
(a) either between 35 years and 40 years or between 55 years and 60 years
(b) between 60 years and 65 years
(c) between 35 years and 40 years
(d) between 35 years and 40 years and between 55 years and 60 years
DIRECTIONS (Qs. 121-123): Study the following table and answer question.
121. The ratio of the total number of students scoring marks less than 50% to that of scoring marks exactly 50% is
(a) 50:3
(b) 25:2
(c) 25:4
(d) 35:2
122. Which school has the highest number of students scoring exactly 50% marks?
(a) D
(b) E
(c) B
(d) A
123. The total number of students scoring 50% or more marks is
(a) 1250
(b) 875
(c) 775
(d) 675
DIRECTIONS (Qs. 124-127): Study the following graph which shows income and expenditure of a company over the years 2005-2009 and answer
124. The difference in profit (T in crores) of the company during 2006 and 2007 is
(a) 10
(b) 15
(c) 20
(d) 25
125. In how many years was the income of the company less than the average income of the given years?
(a) 4
(b) 3
(c) 2
(d) 1
126. The percentage increase in expenditure fo the company from 2007 to 2008 is
(a) 20
(b) 25
(c) 30
(d) 35
127. Profit of the company was maximum in the year
(a) 2009
(b) 2008
(c) 2006
(d) 2005
136. The number of years, the production of fertilizers was more than average production of the given years is:
(a) 2
(b) 1
(c) 3
(d) 4
137. The percentage increase in production of fertilizers in 2002 compared to that in 1995 is:
(a) 200%
(b) 180%
(c) 220%
(d) 240%
138. The percentage decline in the production of fetilizers from 1997 to 1998 is:
(a) 27.5% (b) 25%
(c) 26% (d) 23%
139. The average production of 1996 and 1997 is exactly equal to the average production of the years ?
(a) 2000 and 2001
(b) 1999and2000
(c) 1995 and 2001
(d) 1995 and 1999
140. The percentage increase in production as compared to previous year is maximum in year:
(a) 1999
(b) 1996
(c) 1997
(d) 2002
DIRECTIONS (Qs. 141-144): Study the following bar diagram carefully and answer the following Four Questions.
The number of the production of electronic items (TVs and LCDs) in a factory during the period from 2009 to 2013. (SSC CGL 1st Sit. 2015)
141. The total number of production of electronic items i s maximum in the year
(a) 2009
(b) 2010
(c) 2011
(d) 2013
142. The ratio of production of LCD s in the year 2011 and 2013 is
(a) 3:4
(b) 4:3
(c) 2:3
(d) 1:4
143. The difference between averages of production of TVs and LCDs from 2009 to 2012 is
(a) 600
(b) 700
(c) 800
(d) 900
144. The ratio of production of TV s in the years 2009 and 2010 is
(a) 7:6
(b) 6:7
(c) 2:3
(d) 3:2
DIRECTIONS (Qs. 145-148): The pie-chart given here shows expenditure incurred by a family on various items and their savings. Study the chart and answer the questions based on the pie-chart
148. The average marks obtained by a student in 6 subjects is 88. On subsequent verification it was found that the marks obtained by him in a subject was wrongly copied as 86 instead of 68. The correct average of the marks obtained by him is-
(a) 85
(b) 87
(c) 84
(d) 86
DIRECTIONS (Qs. 149-152): The following piechart shows the study time of different subjects of a student in a day . Study the pie chart and answer the following questions.
152. Instead of 10% , if the student spends 15% to study other subjects and the time is taken from the time scheduled to study mathematics and if he/ she used to study 20 hours per day, then the difference of time for studying mathematics per day is
(a) 30 minutes
(b) 45 minutes
(c) 1 hour
(d) 1 hour 30 minutes
DIRECTION: (Qs. 153-156): The bar graph given indicates the income of a firm. Study the graph and answer the questions given.
153. Which period shows a steady increase of income?
(a) March to May
(b) February to April
(c) February to May
(d) Insufficient data to predict
154. During which month, the ratio of the income to that of the previous month is the largest ?
(a) February
(b) March
(c) April
(d) May
155. The income of May is how many times to that of Februarey?
(a) 3.25
(b) 4
(c) 3.5
(d) 5
156. The average monthly income of the firm (in lakh rupees) is
(a) 7.6
(b) 6
(c) 8.8
(d) None of these
157. The average height of all the peaks (in meters) is
(a) 7601.5
(b) 7600
(c) 7599.5
(d) 7610
158. Which peak is the second highest?
(a) B
(b) C
(c) A
(d) E
159. Write the ratio of the heights of the highest peak and the lowest peak
(a) 22:15
(b) 15:22
(c) 20:13
(d) 13:22
160. When the heights of the given peaks are written in ascending order, what is the average of the middle two peaks?
(a) 7950m
(b) 7560m
(c) 7650 m
(d) 7850 m
DIRECTIONS (Qs. 161-165): Study the following table which shows the amount of money invested (Rupees in crore) in the core infrastructure areas of two districts. A and B of a State, and answer the below five questions.
161. By approximately what percent was the total investment in the two districts A and B more in 1996 as compared to 1995?
(a) 18%
(b) 14%
(c) 21%
(d) 24%
162. The total investment in electricity and thermal energy in 1995, in these two districts A and B formed approximately what percent of the total investment made in that year?
(a) 55%
(b) 41%
(c) 52%
(d) 47%
163. In district B, the investment in which area in 1996 did show the highest percentage increase over the investment in that area in 1995?
(a) Nuclear
(b) Electricity
(c) Chemical
(d) Solar
164. Approximately how many times was the total investment in 1995 and 1996 in district B was that of total investment of district A in the same years?
(a) 1.7
(b) 2.8
(c) 2.4
(d) 1.9
165. If the total investment in district B shows the same rate of increase in 1997, as it had shown from 1995 to 1996, what approximately would be the total investment in B in 1997?
(a) ₹ 9850 crore
(b) ₹10020 crore
(c) ₹ 8540 crore
(d) ₹ 9170 crore
DIRECTIONS (Qs. 166-169):: The bar graph shows the results of an annual examination in a secondary school in a certain year. (SSC Sub Ins. 2016)
166. The ratio of the total number of boys passed to the total number of gives passed in the three classes VII, VIII and IX is
(a) 20:23
(b) 18:21
(c) 21:26
(d) 19:25
167. The average number of boys passed per class is
(a) 75
(b) 78
(c) 70
(d) 72
168. The class having the highest number of passed student is
(a) IX
(b) X
(c) VIII (d) VII
169. The class in which the number of boys passed is nearest to the average number of girls passed per class, is
(a) VIII
(b) X
(c) VI
(d) IX
170. What is the average number of students studying Commerce Number of Students Studying Five Different Disciplines From Five Institutes from all the Institutes together?
(a) 356
(b) 360
(c) 348
(d) 344
171. Total number of students studying Arts from Insitutes A and B together is approximately what per cent of the total number of students studying computer Science from these two Institutes?
(a) 84
(b) 95
(c) 88
(d) 90
172. What is the ratio between total number of Students studying Science from Institutes C and D together and the total number of students studying Computers Science from these two Institutes together respectively?
(a) 13:12
(b) 12:13
(c) 13:15
(d) 15:13
173. What is the average number of students studying all disciplines together from institute E?
(a) 312
(b) 310
(c) 302
(d) 304
DIRECTIONS (Qs. 174-177): The pie chart given below shows the number of shoes of 5 different brands in a multi brand store. There are total 1200 shoes. (SSC CGL 2017)
175. What is the difference in number of shoes of Puma and Vans?
(a) 96
(b) 156
(c) 84
(d) 112
176. The difference between the number of shoes of Reebok and Nike is same as the difference between which of the following two brands?
(a) Puma and Adidas
(b) Reebok and Adidas
(c) Vans and Nike
(d) Nike and Adidas
177. Puma shoes are how much percent more than the Nike shoes?
(a) 14.28
(b) 16.66
(c) 25
(d) 21,33
DIRECTIONS (Qs. 178-181): The pie chart given below shows the break – up of number of hours of teaching various subjects at an institute by Mr. Raghav.
178. If Mr. Raghav taught a total of 500 hours, then what is the difference in number of hours of teaching algebra and modern Maths?
(a) 15
(b) 20
(c) 25
(d) 40
179. Mr. Raghav taught Geometry for 36 hours. If the time taken in teaching Ratio consitutes one-fourth of the time for Arithmetic, then for how much time (in hours) did he taught the topic Ratio?
(a) 46
(b) 51.75
(c) 69
(d) 103.5
180. If Data Interpretation and Modern Maths were taught for a combined time of 96 hours, then for how much time (in hours) were Number system and Geometry taught?
(a) 136
(b) 184
(c) 216
(d) 232
181. A new topic named Problem Solving was also introduced and it was decided that 10% time of all topics except Arithmetic will be devoted to it. What will be the central angle (in degrees) made by Problem Solving in the new pie chart?
(a) 17.28
(b) 18
(c) 19.44
(d) 36
DIRECTIONS (Qs. 182-185): The pie chart given below shows the percentage of time taken by different processes in making a car.
182. If total time taken to make a car is 300 hours, then what is the total time (in hours) taken in paint and frame?
(a) 99
(b) 72
(c) 105
(d) 66
183. If time taken in seats is 192 hours, then what is the time taken (in hours) in glass?
(a) 256
(b) 352
(c) 416
(d) 278
184. If total time taken in engine and tyres is 127.5 hours, then what is the difference (in hours) in time taken by frame and glass respectively?
(a) 27.5
(b) 12.5
(c) 40
(d) 35
185. 15% of total time is spent on quality check and this time is equally taken from all other processes. So What will be the new sectorial angle (in degrees) made by total time of seats and glass?
(a) 28.6
(b) 32.4
(c) 35.8
(d) 31.6
187. Refer the below data table and answer the following Questions:- (SSC CHSL 2017) For which of the following pairs of years the total exports from the three Companies together are equal?
(a) 2011 & 2012
(b) 2013 & 2015
(c) 2011 & 2014
(d) 2014 & 2015
DIRECTIONS (Qs. 190-191): The following bar diagram graph depicts the sales of items (in lakhs) in a departmental store from April to August in the current year. Study the graph and answer the following questions using the data provided here. (SSC MTS 2017)
190. The sales in july is how many times to sales in May?
(a) 300
(b) 3
(c) 2.5
(d) 2
32.
A student was asked to divide a number by 6 and add 12 to the quotient. He, however, first added 12 to the number and then divided it by 6, getting 112 as the answer. The correct answer should have been (SSC CGL 2nd Sit. 2011)
(a) 124 (b) 122
(c) 118 (d) 114
33.
Last year my age was a perfect square number. Next year it will be a cubic number. What is my present age? (SSC Sub. Ins. 2012)
(a) 25 years (b) 27 years
(c) 26 years (d) 24 years
34.
35.
36.
37.
38.
From 9.00 AM to 2.00 PM, the temperature rose at a constant rate from 21°C to 36°C. What was the temperature at noon ? (SSC CHSL 2012)
(a) 27°C (b) 30°C
(c) 32°C (d) 28.5°C
39.
40.
41.
42.
A farmer divides his herd of n cows among his four sons, so that the first son gets one-half the herd, the second one fourth, the third son 1/5 and the fourth son 7 cows. Then the value of n is (SSC CGL 2012)
(a) 240 (b) 100
(c) 180 (d) 140
43.
By what least number should 675 be multiplied to obtain a number which is a perfect cube? (SSC CGL 2012)
(a) 7 (b) 8
(c) 5 (d) 6
44.
45.
46.
47.
If 21 is added to a number, it becomes 7 less than thrice of the number. Then the number is (SSC CGL 2012)
(a) 14 (b) 16
(c) 18 (d) 19
48.
49.
50.
51.
52.
53.
54.
55.
56.
If a number is as much greater than 31 as it is less than 75, then the number is (SSC CGL 1st Sit. 20130)
(a) 53 (b) 106
(c) 44 (d) 74
61.
The numerator of a fraction is 4 less than its denominator. If the numerator is decreased by 2 and the denominator is increased by 1, then the denominator becomes eight times the numerator. Find the fraction. (SSC CGL 2013)
68.
Which one of the following is the minimum value of the sum of two integers whose product is 24? (SSC CGL 2nd Sit. 2013)
(a) 25 (b) 11
(c) 8 (d) 10
69.
70.
71.
72.
73.
74.
75.
Ram left 1/3 of his property to his widow and 3/5 of the remainder to his daughter. He gave the rest to his son who received ? 6,400. How much was his original property worth? (SSC CHSL 2014)
(a) ₹ 16,000 (b) ₹ 32,000
(c) ₹ 24,000 (d) ₹ 1,600
83.
In an exam the sum of the scores of A and B is 120, that of B and C is 130 and that of C and A is 140. Then the score of C is: (SSC CHSL 2015)
(a) 65 (b) 60
(c) 70 (d) 75
84.
The sum of four numbers is 48. When 5 and 1 are added to the first two; and 3 & 7 are subtracted from the 3rd & 4th, the numbers will be equal. The numbers are (SSC CGL 1st Sit. 2015)
(a) 4,12,12,20 (b) 5,11,13,19
(c) 6,10,14,18 (d) 9,7,15,17
85.
86.
Choose the incorrect relation(s) from the following:
87.
88.
89.
90.
91.
92.
93.
94.
95.
Hints & Solutions
1. (c)
2. (a)
3. (b)
4. (a)
5. (c)
6. (b)
7. (a)
8. (b)
9. (a)
10. (b)
11. (b)
12. (a)
13. (c)
14. (d)
15. (b)
16. (c)
17. (d)
18. (a)
19. (d)
20. (b)
21. (a)
22. (a)
23. (b)
24. (b)
25. (b)
26. (b)
27. (a)
28. (a)
29. (d)
30. (a)
31. (b)
32. (b)
33.
(c) By going options, 26 years is the present age. Present age be 26, then last year age was 25 which represents a perfect square and next year age would be 27 which represents a cubic number.
34. (a)
35. (b)
36. (d)
37. (a)
38. (b)
39. (a)
40. (a)
41. (b)
42. (d)
43. (c)
44. (d)
45. (c)
46. (b)
47. (a)
48. (c)
49. (b)
50. (c)
51. (c)
52. (d)
53. (d)
54. (d)
55. (b)
56. (a)
57. (d)
58. (b)
When no. of digit in a no. is 7 or 8 then in square root will be 4.
24. L.C.M. oftwo numbers is 120 and their H.C.F. is 10. Which of the following can be the sum of those two numbers? (SSCCGL 2nd Sit. 2011)
(a) 140
(b) 80
(c) 60
(d) 70
25. When 'n' is divisible by 5 the remainder is 2. What is the remainder when n2 is divided by 5? (SSCCGL 2nd Sit. 2011)
(a) 2
(b) 3
(c) 1
(d) 4
26. Four runners started running simultaneously from a point on a circular track. They took 200 seconds, 300 seconds, 360 seconds and 450 seconds to complete one round. After how much time they meet at the starting point for the first time? (SSC CGL 2nd Sit. 2011)
(a) 1800 seconds
(b) 3600 seconds
(c) 2400 seconds
(d) 4800 seconds
31. A three-digit number 4a3 is added to another three-digit number 984 to give the four digit number 13b7 which is divisible by 11. Then the value of (a + b) is: (SSC CGL 1st Sit. 2012)
(a) 11
(b) 12
(c) 9
(d) 10
32. The greatest number that will divide 19,35 and 59 to leave the same remainder in each case is: (SSC CGL 1st Sit. 2012)
(a) 9
(b) 6
(c) 7
(d) 8
44. If the sum of the digits of any integer lying between 100 and 1000 is subtracted from thenumber, the result always is (SSC CHSL 2013)
(a) divisible by 5
(b) divisible by 6
(c) divisible by 2
(d) divisible by 9
66. Product of digits of a 2-digit number is 27. If we add 54 to the number, the new number obtained is a number formed by interchange of the digits. Find the number. (SSC CHSL 2017)
(a) 39
(b) 93
(c) 63
(d) 36
68. Of the three numbers, the first is twice the second, and the second is twice the third. The average of the reciprocal of the numbers is 7/12. The numbers are: (SSC MTS 2017)
(a) 20,10,5
(b) 4,2,1
(c) 36,18,9
(d) 16,8,4
69. What is the smallest value that must be added to 709, so that the resultant is a perfect square? (SSC Sub. Ins. 2017)
(a) 8
(b) 12
(c) 20
(d) 32
70. Which one among is the smallest number? (SSC Sub. Ins. 2017)
(a)
(b)
(c)
(d) All are equal
71. If 34N is divisible by 11, then what is the value of N (SSC Sub. Ins. 2017)
(a) 1
(b) 3
(c) 4
(d) 9
Hints & Solution
1. (d)
2. (b)
3. (c)
4. (b)
If the first divisor is a multiple of second divisor. Then, remainder by the second divisor.
∴ Remainder = 21 ÷5-19 = 2
5. (c)
6. (c)
7. (d)
If the first divisor be a multiple of the second divisor, then required remainder = remainder obtained by dividing the first remainder (36)
by the second divisor (17) = 2
∵ 17 is a factor of 136
∴ Remainder when 36 is divided by 17 = 2
61. (a)
347XY as 347X0. Since 8 is a factor of 80.
347X0 is divisible by 8. It means last three digits 7X0 is divisible by 8.
Hence, X is 2 or 6
if X = 6, number is 34760. But this is not divisible by 80.
ifX = 2, number is 34720, which is divisible by 80.
Therefore, number is 34720 with X = 2 and Y = 0.
∴ x + y = 2 + 0 = 2.
62. (c)
LCM of 5 and 7 = 35
So, the numbers divisible by both 5 and 7 are multilpe of 35. Between 300 and 650. We have 10 multiple of 35. They are: 315,350,385,420,455,490,525,560,595, 630. | 677.169 | 1 |
d. it is a convex pentagon because it has five sides and none of the sides would extend into the inside of the polygon.
step-by-step explanation:
Answer from: Quest
answer: #1 ads up by 8 and i don't know #2
Answer from: Quest
B. false you shouldn't proceed at all.
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At dinner, 2/5 of the people had fish and 1/3 had chicken. if everyone else had lamb, what fraction of the people had lamb? and if 40 people had lamb , calculate the total number of people at the dinner.
Ascatterplot is produced to compare the size of a school building to the number of students at that school who play an instrument. there are 12 data points, each representing a different school. the points are widely dispersed on the scatterplot without a pattern of grouping. which statement could be true | 677.169 | 1 |
Figure 1 The origin of common lines in the transform of an icosahedrally symmetric object. The common lines arise from the application of a symmetry axis which is not coincident with the direction of view (θ = 89, φ = −1°, shown by the vertical white line). It is illustrated in the figure for a threefold axis (θ = 69.19, φ = 0°). (a) The transform of the projection is a central section through Fourier space. (b) The application of a threefold symmetry axis generates a new plane from the original. Application of the inverse generates a third plane. The intersections of these two symmetry-related planes with the original yield a pair of lines in the original image transform which must have identical values. These are seen in the stereo pair (c). The symmetry axis which created the pair of lines lies at the centre of the surface formed by the three planes. Each pair of symmetry elements and its inverse yield a pair of common lines. For an icosahedral object, this yields 37 pairs (12 from the fivefolds, 10 from the threefolds and 15 from the twofolds) in the transform of the projection. | 677.169 | 1 |
How do you write a vector in AI BJ form?
by
How to Write a Vector in Ai+Bj Form Given Its Initial and Terminal Points Vocabulary. Vector in ai+bj a i + b j form: Vector v from (0,0) to (a,b) is represented as ai+bj a i + b j , where i and j are the unit vectors.
What form is AI BJ?
Vectors are used to represent quantities that have both magnitude and direction. There are a number of ways that 2D vectors can be represented. One of these representations involves expressing a vector r in terms of unit vectors i and j. This is known as component form and is expressed as r = ai + bj.
How do you write a vector form?
Unit vector form
Using vector addition and scalar multiplication, we can represent any vector as a combination of the unit vectors. For example, (3,4)left parenthesis, 3, comma, 4, right parenthesis can be written as 3 i ^ + 4 j ^ 3hat i+4hat j 3i^+4j^3, i, with, hat, on top, plus, 4, j, with, hat, on top.
What is direction angle of a vector?
The direction angle, θ , of the vector, v=ai+bj v = a i + b j , can be found using the arctangent of the ratio of the vertical component of the vector and the horizontal component of the vector. θ=tan−1ba θ = tan − 1 However, this is only the reference angle.
How do you write a vector in AI BJ form? – Related Questions
What is the angle between A * B and B * A in vector?
Thus, the angle between the vectors a → × b → and b → × a → is 180º.
How do you solve vectors?
The magnitude of a vector is easy to calculate with Pythagorean theorem. From Pythagorean theorem, a2+b2=c2, so when we apply that to the vectors: (vector magnitude)2=(x-component) 2+(y-component) 2.
What is direction angle?
: an angle made by a given line with an axis of reference. specifically : such an angle made by a straight line with the three axes of a rectangular Cartesian coordinate system. usually used in plural.
What is called directed angle?
A directed angle is an angle that has a direction; the counterclockwise direction is positive and the clockwise direction is negative. Coterminal angles share the same initial and terminal sides. There are an infinite number of equivalent coterminal angles.
Does vector have slope?
Since the slope of a line is the change in 𝑦 over the change in 𝑥 , we can always find the direction vector of a nonvertical line by taking the change in 𝑥 to be 1 and then the change in 𝑦 to be the slope. Therefore, the line will have the direction vector ( 1 , 𝑚 ) , where 𝑚 is the slope of the line.
What is the i and J in vectors?
The unit vector i has a magnitude of 1 and its direction is along the positive x-axis of the rectangular coordinate system. The unit vector j has a magnitude of 1 and its direction is along the positive y-axis of the rectangular coordinate system.
What are 4 types of vectors?
Types of Vectors List
Zero Vector.
Unit Vector.
Position Vector.
Co-initial Vector.
Like and Unlike Vectors.
Co-planar Vector.
Collinear Vector.
Equal Vector.
What are 4 examples of vectors?
Some examples of vector quantities include force, velocity, acceleration, displacement, and momentum | 677.169 | 1 |
261103, 261108, 261109, 261121 - Incomplete rectangles. Letters, numerals or punctuation and letters, numerals or punctuation forming the perimeter of a rectangle, bordering the perimeter of a rectangle or forming a rectangle. Rectangles made of geometric figures, objects, humans, plants or animals. Rectangles that are completely or partially shaded.
Statements
Goods and Services
AGRICULTURAL SPRAYERS AND BOOMS, AND TRACTOR HITCHES
Lining/Stippling Statement
THE DRAWING IS LINED FOR THE COLOR YELLOW.
Pseudo Mark
A | 677.169 | 1 |
Let c be a circle centered at O and A a point inside it. Draw line AD to a variable point D of the circle and at D draw the orthogonal line DE to AD. - Line DE envelopes an ellipse. - The ellipse has the circle as its auxiliary circle and touches it at the diametral points with line AO. - The contact point E of line DE is the projection on it of the fourth harmonic G = F(A,C), where F the intersection of AB with DE. - Analogous generation is valid also for hyperbolas. The only difference is on taking A to be outside the circle. The proofs follow immediately from the discussion in Ellipse.html .
Point E glides on circle B(BE). Point H is taken on AE (A fixed, inside the circle) such that AH/AE = k (constant). Line HU is taken to be orthogonal at H to AE. Line HU envelopes an ellipse. The auxiliary circle of the ellipse is homothetic to the circle B(BE) w.r. to A and in ratio k to it. The ellipses resulting for various k are all homothetic to each other. The contact point U of line HU with the ellipse is the projection on HU of T, which is the harmonic fourth
of K(A,S), where K is the intersection of HU with the line of centers AB. The figure here is simply a homothetic image of the previous figure. | 677.169 | 1 |
2 1 Vectors In The Plane
April 9, 2022
The scalar product of vectors is used to find angles between vectors and in the definitions of derived scalar physical quantities such as work or energy. There are two kinds of multiplication for vectors. One kind of multiplication is the scalar product, also known as the dot product. The other kind of multiplication is the vector product, also known as the cross product. The scalar product of vectors is a number .
It even provides a simple test to determine whether two vectors meet at a right angle. Let be the vector with initial point and terminal point as shown in . Express in both component form and using standard unit vectors. Find the vector difference a−ba−b and express it in both the component form and by using the standard unit vectors.
As you might expect, to calculate the dot product of four-dimensional vectors, we simply add the products of the components as before, but the sum has four terms instead of three. Determine the vectors 2a,2a, −b,−b, and 2a−b.2a−b. Express the vectors in both the component form and by using standard unit vectors. The angles formed by a nonzero vector and the coordinate axes are called the direction angles for the vector ().
Two examples of cross products where the unit vectors do not appear in the cyclic order. Determine all three-dimensional vectors u orthogonal to vector Express the answer in component form. Find all two-dimensional vectors a orthogonal to vector Express the answer in component form. Each indicate that mathematical objects are intersecting at right angles. The sum of the forces acting on an object is called the resultant or net force.
In this text, we always work with coordinate systems set up in accordance with the right-hand rule. Some systems do follow a left-hand rule, but the right-hand rule is considered the a night in barcelona yuri on ice standard representation. A bolt is tightened by applying a force of \(6\) N to a 0.15-m wrench (Figure \(\PageIndex\)). The angle between the wrench and the force vector is \(40°\).
by Sophia Jennifer
About me
I'm Sophia Jennifer from the United States working in social media marketing It is very graceful work and I'm very interested in this work. | 677.169 | 1 |
How To Identify If Two Shapes Are Congruent
On November 20, 2023
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At 2:13 am
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Are you struggling to understand the concept of congruence in geometry? Look no further, as this blog post will break it down for you in simple terms. From recognizing congruent angles and side lengths to applying different congruence criteria, we will cover everything you need to know about congruent shapes. We will also discuss practical methods for proving congruence and debunk common misconceptions about this fundamental concept in geometry. By the end of this post, you will be equipped with the knowledge and skills to confidently identify and compare corresponding parts of shapes in a geometric context. Let's delve into the world of congruence in geometry!
Congruence is an important concept in geometry that deals with the similarity of two or more shapes. When two shapes are congruent, it means that they have the same shape and size, although they may be positioned differently. By understanding the concept of congruence, we can explore various properties and criteria to identify and prove the congruency of shapes.
The first key property of congruent shapes is that their corresponding angles are equal. This means that if we have two shapes with angles A and B, and their corresponding angles in another shape are also A and B, then these shapes are congruent. Another key property of congruent shapes is that their corresponding side lengths are equal. This means that if we have two shapes with side lengths A and B, and their corresponding side lengths in another shape are also A and B, then these shapes are congruent.
One way to identify if two shapes are congruent is by using the angle-angle criterion. This criterion states that if two angles of one shape are equal to two angles of another shape, then the two shapes are congruent. Another criterion is the side-side-side criterion, which states that if the three side lengths of one shape are equal to the three side lengths of another shape, then the two shapes are congruent. Similarly, we have the side-angle-side criterion, which states that if two side lengths and the included angle of one shape are equal to the corresponding side lengths and included angle of another shape, then the two shapes are congruent.
Key properties of congruent shapes
Corresponding angles and side lengths are equal.
Angle-angle criterion
Two shapes are congruent if two of their angles are equal.
Side-side-side criterion
Two shapes are congruent if their three side lengths are equal.
Side-angle-side criterion
Two shapes are congruent if two side lengths and the included angle are equal.
In order to compare corresponding parts of shapes for congruence, we can use a table. The table can have columns for the shape's angles, side lengths, and other properties. By comparing the values in these columns for two shapes, we can determine if they are congruent. It is important to note that in order for two shapes to be congruent, all corresponding angles and side lengths must be equal.
While congruence in geometry may seem like a simple concept, there are some common misconceptions that people have. One misconception is thinking that if two shapes look similar, they must be congruent. However, congruence goes beyond just the visual appearance of shapes and includes specific criteria that need to be met. Another misconception is assuming that if two shapes have equal side lengths or equal angles, they must be congruent. It is important to keep in mind that all corresponding angles and side lengths must be equal for shapes to be congruent.
Understanding The Concept Of Congruent Shapes
The concept of congruent shapes is an important aspect of geometry. Congruence refers to the idea that two shapes are identical in shape and size. In other words, if two shapes are congruent, they have the same measurements and angles. This concept is crucial in geometry because it allows us to compare and analyze shapes based on their properties. By identifying if two shapes are congruent, we can solve various geometric problems and make accurate mathematical conclusions.
One way to identify if two shapes are congruent is by looking at their corresponding sides and angles. If all corresponding sides of two shapes are equal in length, and all corresponding angles are equal in measure, then the shapes are congruent. This means that every side and angle of one shape matches exactly with the corresponding side or angle of the other shape. This can be illustrated using the concept of a congruence transformation, which is a transformation that preserves the shape and size of a figure.
Another method to determine if two shapes are congruent is by using congruence criteria. These criteria are specific conditions that indicate congruence between two shapes. Some of the commonly used criteria include the Angle-Angle (AA) criterion, the Side-Side-Side (SSS) criterion, the Side-Angle-Side (SAS) criterion, and the Angle-Side-Angle (ASA) criterion. By applying these criteria, we can check if the given conditions are met and hence determine if the shapes are congruent.
Angle-Angle (AA) Criterion: This criterion states that if two angles of one shape are equal in measure to two angles of another shape, then the shapes are congruent.
Side-Side-Side (SSS) Criterion: This criterion states that if all three sides of one shape are equal in length to the corresponding three sides of another shape, then the shapes are congruent.
Side-Angle-Side (SAS) Criterion: This criterion states that if: This criterion states that if two angles and the included side of one shape are equal in measure and length to the corresponding two angles and included side of another shape, then the shapes are congruent.
By understanding the concept of congruent shapes and utilizing various identification methods and criteria, we can analyze and solve geometry problems effectively. Recognizing congruence between shapes helps us make accurate conclusions about their properties and relationships. Whether it is through comparing corresponding sides and angles or applying congruence criteria, the concept of congruent shapes plays a significant role in geometry.
Shape Congruence Criteria
Description
Angle-Angle (AA) Criterion
If two angles of one shape are equal in measure to two angles of another shape, then the shapes are congruent.
Side-Side-Side (SSS) Criterion
If all three sides of one shape are equal in length to the corresponding three sides of another shape, then the shapes are congruent.
Side-Angle-Side (SAS) Criterion
If
If two angles and the included side of one shape are equal in measure and length to the corresponding two angles and included side of another shape, then the shapes are congruent.
Key Properties Of Congruent Shapes
When studying geometry, one important concept to understand is congruence. Congruent shapes are objects that have exactly the same size and shape. In other words, if two shapes are congruent, it means that they are identical in every aspect. This can be quite useful when comparing and analyzing different figures. In this blog post, we will explore the key properties of congruent shapes and how to identify if two shapes are congruent.
One of the key properties of congruent shapes is that their corresponding angles are equal. When comparing two shapes, it is important to examine the angles present in each shape. If the corresponding angles in both shapes are equal, then we can conclude that the shapes are congruent. This property is known as angle-angle criterion (AA criterion). By comparing angles, we can easily determine if two shapes are congruent or not.
Another property of congruent shapes is that their corresponding side lengths are equal. When analyzing two shapes, we can compare the lengths of their sides. If all corresponding sides in both shapes are equal in length, we can confidently say that the shapes are congruent. This property is known as side-side criterion (SS criterion). By measuring and comparing side lengths, we can determine if two shapes are congruent or not.
Furthermore, congruent shapes also have equal corresponding diagonals and perimeters. In addition to angles and side lengths, we can also examine other properties of shapes such as diagonals and perimeters. If the corresponding diagonals are equal in both shapes, and their perimeters are also equal, then we can conclude that the shapes are congruent. These additional properties can provide further evidence for congruence and help us in identifying congruent shapes accurately.
Criterion
Property
Angle-Angle (AA)
Corresponding angles are equal
Side-Side (SS)
Corresponding side lengths are equal
Diagonal-Diagonal (DD)
Corresponding diagonals are equal
Perimeter-Perimeter (PP)
Corresponding perimeters are equal
Identifying congruent shapes can be a useful skill in various applications of geometry. Whether it's in constructing shapes, solving problems, or proving theorems, recognizing congruent shapes allows us to make accurate deductions and draw valid conclusions. By considering the key properties of congruent shapes, such as equal angles, side lengths, diagonals, and perimeters, we can confidently determine if two shapes are congruent or not.
Recognizing Congruent Angles
In geometry, congruence refers to the property of two shapes being identical in size and shape. When two shapes are congruent, it means that they have the same angles and side lengths. This concept plays a crucial role in various geometric proofs and calculations. One of the key aspects of congruence is recognizing congruent angles.
Congruent angles are angles that have the same measure. They may have different shapes or orientations, but their degree measurements are equal. To identify if two angles are congruent, you can compare their measures using a protractor or rely on known angle relationships.
One method to identify congruent angles is by using the angle addition postulate. According to this postulate, if two angles share a common vertex and a common side, and the non-shared sides form a straight line, then the angles are congruent. This is known as the vertical angles theorem, and it helps in proving congruence in various geometric figures.
Identifying Congruent Side Lengths
When studying geometry, it is essential to understand the concept of congruence. Congruent shapes have the same shape and size, meaning that their corresponding sides and angles are equal. In this blog post, we will focus on one particular aspect of congruence: identifying congruent side lengths.
To determine if two shapes have congruent side lengths, we need to compare the corresponding sides of each shape. This can be done by measuring the lengths of the sides or by using other geometric properties and theorems.
The first method is straightforward – by measuring the lengths of the sides of each shape, we can directly compare them. If all corresponding sides have the same length, then the shapes are congruent in terms of side lengths. However, it is essential to note that precise measurements are necessary to ensure accuracy.
Another method to identify congruent side lengths is by using geometric properties and theorems. For example, if we know that two triangles are congruent based on the Side-Side-Side (SSS) criterion, we can conclude that their corresponding side lengths are congruent as well. The SSS criterion states that if the lengths of the three sides of one triangle are equal to the lengths of the three sides of another triangle, then the two triangles are congruent.
Shape
Corresponding Side Lengths
Triangle ABC
AB = 5 cm, BC = 4 cm, AC = 6 cm
Triangle DEF
DE = 5 cm, EF = 4 cm, DF = 6 cm
In the example above, Triangle ABC and Triangle DEF have congruent side lengths because their corresponding sides are equal in length, as stated by the SSS criterion.
Similarly, the Angle-Angle (AA) criterion can also help identify congruent side lengths. If two triangles have two pairs of congruent angles, then their corresponding side lengths are congruent as well. This criterion relies on the fact that two triangles with congruent angles will have proportional side lengths.
By utilizing these geometric properties and theorems, we can easily determine if two shapes have congruent side lengths. Whether through direct measurement or by applying the SSS or AA criterion, understanding how to identify congruent side lengths is crucial in geometry.
Applying The Angle-Angle Criterion
The Angle-Angle criterion is one of the methods used to determine if two shapes are congruent in geometry. Congruence refers to the property of having the same shape and size. When two shapes are congruent, it means that they are identical in every aspect, including their angles and side lengths. The Angle-Angle criterion focuses specifically on the angles of the shapes to determine congruence. By comparing the measures of the angles in two shapes, we can determine if they are congruent or not.
To apply the Angle-Angle criterion, we need to identify two congruent angles in each of the shapes under consideration. If we can find two pairs of congruent angles in both shapes, then the shapes are congruent. The order of the angles does not matter; what matters is that they have the same measure. For example, if we have two triangles and we find that angle A in the first triangle is congruent to angle X in the second triangle, and angle B in the first triangle is congruent to angle Y in the second triangle, then we can conclude that the two triangles are congruent.
It is important to note that the Angle-Angle criterion alone is not sufficient to determine congruence. In addition to having two pairs of congruent angles, we also need to consider the measures of the corresponding sides of the shapes. This is where other criteria such as the Side-Angle-Side criterion, Side-Side-Side criterion, and Angle-Side-Angle criterion come into play. These criteria provide additional information to support the congruence of shapes.
The Angle-Angle criterion can be illustrated using a table:
Shape
Angle 1
Angle 2
Angle 3
Shape 1
30°
60°
90°
Shape 2
30°
60°
90°
In the table above, we have two shapes with three angles each. By comparing the measures of the angles, we can see that Angle 1 in Shape 1 is congruent to Angle 1 in Shape 2 (both measure 30°), and Angle 2 in Shape 1 is congruent to Angle 2 in Shape 2 (both measure 60°). Therefore, according to the Angle-Angle criterion, we can conclude that Shape 1 and Shape 2 are congruent.
Using The Side-Side-Side Criterion
Congruence is a fundamental concept in geometry that refers to the equality of two shapes in terms of shape and size. When we say that two shapes are congruent, we mean that they are identical in every way. This includes having the same angles and side lengths. One way to determine if two shapes are congruent is by using the Side-Side-Side (SSS) criterion.
The SSS criterion states that if the three sides of one triangle are congruent to the corresponding sides of another triangle, then the two triangles are congruent. In other words, if all three sides of one triangle match the lengths of the three sides of another triangle, then the two triangles are identical in shape and size. This criterion is based on the fact that corresponding sides in congruent triangles are equal.
To identify if two shapes are congruent using the SSS criterion, we need to compare the lengths of the corresponding sides. For example, let's say we have two triangles – Triangle ABC and Triangle DEF. If we can determine that AB is equal to DE, BC is equal to EF, and AC is equal to DF, then we can conclude that the two triangles are congruent.
Using the SSS criterion:
Triangle ABC
Triangle DEF
Side AB
Side DE
Side BC
Side EF
Side AC
Side DF
In the table above, we can see that the corresponding sides of Triangle ABC and Triangle DEF are equal, satisfying the SSS criterion. This means that the two triangles are congruent.
The SSS criterion is one of several criteria that can be used to determine congruence in geometry. By identifying if two shapes have three corresponding sides that are equal in length, we can confidently conclude that the shapes are congruent. This criterion is particularly useful when working with triangles, as it provides a concise and efficient method for proving congruence.
Exploring The Side-Angle-Side Criterion
The Side-Angle-Side (SAS) criterion is a method used in geometry to determine if two triangles are congruent. Congruent triangles are defined as having the same shape and size, meaning that all corresponding angles and side lengths are equal. The SAS criterion states that if two sides and the included angle of one triangle are congruent to the corresponding sides and angle of another triangle, then the triangles are congruent.
Using the SAS criterion, we can easily identify if two triangles are congruent without having to measure all their angles and sides. To apply this criterion, we need to determine if the following conditions are met:
Side-Side
The lengths of two sides of one triangle are equal to the corresponding sides of the other triangle.
Angle
The included angles of the triangles are congruent, meaning they have the same measure.
Side
The length of the third side of one triangle is equal to the corresponding side of the other triangle.
If all these conditions are satisfied, then we can conclude that the triangles are congruent. It is important to note that the order in which the sides and angles are listed does not matter, as long as the corresponding parts are equal. For example, if two triangles have sides AB = CD, BC = DE, and angle B = angle D, then we can use the SAS criterion to show that the triangles are congruent.
Utilizing The Angle-Side-Angle Criterion
The Angle-Side-Angle (ASA) criterion is one of the methods used in geometry to determine if two shapes are congruent. Congruence refers to the quality of having the same size and shape. When two shapes are congruent, it means that they are identical in every aspect. To utilize the Angle-Side-Angle criterion, we need to compare the measures of the angles and the lengths of the sides of the given shapes.
In the Angle-Side-Angle criterion, we first look at the angles of the shapes. If two angles of one shape are congruent to the corresponding two angles of the other shape, and the side between the congruent angles is also congruent, then we can conclude that the shapes are congruent. This criterion is based on the fact that angles and sides determine the shape of a geometric figure.
Let's consider an example to understand how to utilize the Angle-Side-Angle criterion to determine congruence. Suppose we have triangle ABC and triangle XYZ. We measure angle A and angle X, and if they are congruent, we proceed to measure angle B and angle Y. If angle B is congruent to angle Y, we finally measure side AC and side XY. If side AC is congruent to side XY, then we can conclude that triangle ABC is congruent to triangle XYZ.
Angle A congruent to angle X
Angle B congruent to angle Y
Side AC congruent to side XY
Triangle ABC
Triangle XYZ
Angle A = Angle X
Angle B = Angle Y
Side AC = Side XY
It is important to note that the order in which we compare the angles and sides is essential. If we compare the sides before checking the angles, we may not reach the correct conclusion. The Angle-Side-Angle criterion provides a reliable method for determining the congruence of shapes, allowing us to apply this knowledge in various geometric problems and proofs.
Comparing Corresponding Parts Of Shapes
When studying geometry, one important concept to understand is the idea of congruence. Two shapes are said to be congruent if they have exactly the same size and shape. But how do we determine if two shapes are congruent? One way to do this is by comparing their corresponding parts.
In geometry, corresponding parts refer to the sides, angles, or vertices of two shapes that are in the same position or have the same relative relationship. When comparing corresponding parts of two shapes, it is important to note that the order of comparison matters. For example, the first side of one shape should be compared to the first side of the other shape, the second side to the second side, and so on.
One method to compare the corresponding parts of shapes is by creating a table. This table can help organize the information and make it easier to analyze. Let's consider an example where we have two triangles and want to determine if they are congruent. By listing the corresponding parts of each triangle in a table, we can easily compare them and identify any differences.
Triangle 1
Triangle 2
Side AB
Side PQ
Side BC
Side QR
Side AC
Side PR
Angle A
Angle P
Angle B
Angle Q
Angle C
Angle R
By comparing the corresponding parts of Triangle 1 and Triangle 2, we can see that all the sides and angles are equal, indicating that the two triangles are congruent. This method of comparing corresponding parts is vital in geometry as it helps us determine the congruence of shapes.
Practical Methods For Congruence Proof
In geometry, congruence refers to the state of two shapes being identical or having the same size and shape. When it comes to proving congruence between two shapes, there are various practical methods that can be employed. The ability to identify if two shapes are congruent plays a crucial role in solving geometric problems and constructing accurate diagrams.
One practical method for proving congruence between two shapes is the Angle-Angle (AA) Criterion. This criterion states that if two angles of one shape are congruent to two angles of another shape, then the two shapes are congruent. This method can be particularly useful in scenarios where only the angles of the shapes are known.
The Side-Side-Side (SSS) Criterion is another practical method for proving congruence between shapes. According to this criterion, if the three sides of one shape are congruent to the three sides of another shape, then the two shapes are congruent. This method is often utilized when all the side lengths of the shapes are given or can be determined using other geometric properties.
The Angle-Side-Angle (ASA) Criterion is a useful method for proving congruence as well. This criterion states that if two angles and the included side of one shape are congruent to two angles and the included side of another shape, then the two shapes are congruent. This method is commonly used when both the angles and sides of the shapes are known.
Practical Methods for Congruence Proof
Angle-Angle Criterion
If two angles of one shape are congruent to two angles of another shape, then the two shapes are congruent.
Side-Side-Side Criterion
If the three sides of one shape are congruent to the three sides of another shape, then the two shapes are congruent.
Angle-Side-Angle Criterion
If two angles and the included side of one shape are congruent to two angles and the included side of another shape, then the two shapes are congruent.
By utilizing these practical methods, mathematicians and geometry enthusiasts can confidently determine if two shapes are congruent. These methods provide a systematic approach to congruence proof, allowing for accurate analysis and problem-solving. Understanding and applying these criteria empower individuals to explore the intricate relationships between geometric figures and solve a wide range of geometric problems with ease.
Common Misconceptions About Congruence
When it comes to geometry, congruence is a fundamental concept that helps us understand and analyze shapes and figures. Congruence refers to the property of two shapes being identical in size, shape, and orientation. However, there are some common misconceptions about congruence that can lead to misunderstandings and errors in geometric reasoning. In this blog post, we will explore some of these misconceptions and provide clarity on how to correctly identify if two shapes are congruent.
One common misconception is that if two shapes have the same shape, they are automatically congruent. However, this is not true. While congruent shapes do have the same shape, they also have the same size and orientation. In other words, congruent shapes are not only identical in their angles and sides, but also in their dimensions and spatial arrangement.
Another misconception is that congruent shapes can be determined by comparing just one angle or one side length. In reality, it takes more than just one angle or side to establish congruence between two shapes. To identify if two shapes are congruent, we must compare all corresponding angles and sides.
Some key properties of congruent shapes include:
– Corresponding angles are equal in measure
– Corresponding side lengths are equal
– Corresponding diagonals are equal in length
– The perimeter and area are the same
Property
Description
Corresponding angles are equal in measure
This means that if Angle A in one shape is congruent to Angle B in another shape, then all other corresponding angles in both shapes are also congruent
Corresponding side lengths are equal
If Side AB in one shape is congruent to Side XY in another shape, then all other corresponding side lengths in both shapes are also congruent
Corresponding diagonals are equal in length
If the diagonal BD in one shape is congruent to the diagonal XZ in another shape, then all other corresponding diagonals in both shapes are also congruent
The perimeter and area are the same
Congruent shapes have the same perimeter (sum of all side lengths) and the same area (amount of space enclosed by the shape)
To avoid misconceptions, it is important to remember that congruence is a comprehensive property that encompasses not only shape, but also size and orientation. Comparing all corresponding angles and sides is crucial in determining if two shapes are congruent. By understanding these key properties and dispelling common misconceptions, we can enhance our geometric reasoning and problem-solving skills.
Frequently Asked Questions
1. What is congruence in geometry?
Congruence in geometry refers to the concept of two shapes or figures being identical in shape and size. In other words, congruent shapes have the same measurements and angles.
2. What are some key properties of congruent shapes?
Some key properties of congruent shapes include having equal side lengths, equal angle measures, and equal area and perimeter. These properties allow us to determine if two shapes are congruent.
3. How can we recognize congruent angles?
Congruent angles have the same degree measure. This means that if two angles have the same numerical value, they are congruent. Additionally, angles that are vertical, adjacent, or corresponding are congruent.
4. How do we identify congruent side lengths?
To identify congruent side lengths, we compare the measurements of corresponding sides in two shapes. If the lengths are equal, then the sides are congruent. This can be done by using a ruler or measuring tool.
5. What is the Angle-Angle criterion for congruence?
The Angle-Angle criterion states that if two angles of one shape are congruent to two angles of another shape, then the two shapes are congruent. This criterion allows us to prove congruence using angle measurements.
6. How can we use the Side-Side-Side criterion for congruence?
The Side-Side-Side criterion states that if all three sides of one shape are equal in length to the corresponding sides of another shape, then the two shapes are congruent. This criterion is useful for proving congruence when side lengths are known.
7. Can you explain the Angle-Side-Angle criterion for congruence?
The Angle-Side-Angle criterion states that if two angles and the included side of one shape are congruent to two angles and the included side of another shape, then the two shapes are congruent. This criterion allows us to prove congruence using both angle and side measurements | 677.169 | 1 |
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УелЯдб 16 ... Q.E.D. PROPOSITION XIV . THEOREM . If at a point in a straight line , two other straight lines , upon the opposite sides of it , make the adjacent angles together equal to two right angles ; then these two straight lines shall be in one ...
УелЯдб 17 ... Q. E.D. COR . 1. From this it is manifest , that if two straight lines cut ... PROPOSITION XVI . THEOREM . If one side of a triangle be produced , the ... PROP . XV , XVI . 17.
УелЯдб 18 ... Q.E.D. PROPOSITION XVIII . THEOREM . The greater side of every triangle is opposite to the greater angle . Let ABC be a triangle , of which the side AC is greater than the side AB . Then the angle ABC shall be greater than the angle ACB ...
УелЯдб 71 - If a straight line be bisected, and produced to any point ;УелЯдб 15�елЯдб 242УелЯдб 2 34 - Equal triangles, upon equal bases in the same straight line, and towards the same parts, are between the same parallels. Let the equal triangles ABC, DEF be upon equal bases BC, EF, in the same straight line BF, and towards the same parts | 677.169 | 1 |
Difference between Altitude and Latitude
Altitude and latitude are widely used terms in the field of geography and astronomy. Most people have heard these terms but are unsure of their exact meaning. The words latitude and altitude have very similar meanings but there is also a key difference between these two. Essentially, both terms are a way of measuring parameters that define the angular position of a location. If you are planning to work in the field of astronomy or geography then consider learning how altitude and latitude are different to each other and when these terms should be used. The distance between the plane of the equator perpendicular to the angular distance on a plane is known as the latitude. Altitude is defined as the standing distance between the datum line and a certain point above that line.
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Instructions
1
Latitude
Latitude is typically used as one of the two coordinates for a particular location on Earth. For any location under consideration, latitude provides the south and north position. Remember that the line which is latitude goes parallel to the equator of the earth. When combined with longitude, it can be used to find a specific place on the globe. According to science, the Earth's equator is thought to be positioned at zero (0) latitude. The South and North Pole of the planet are positioned at -90 degrees and + 90 degrees latitude.
Furthermore, there are some artificial latitudes as well that include the Tropic of Cancer in the northern hemisphere, Arctic Circle and Tropic of Capricorn in the southern hemisphere. Latitude can be further divided into different categories depending on the relative definitions and physical properties. The angle between the surface at a point and the plane of the equator is termed as the Geodetic latitude. Similarly, geocentric latitude is the name given to the angle between the radius of a certain point and the equator. The angle between the true vertical on a point at the place and the equatorial place is termed as the astronomical latitude.
- Image courtesy: dictionary.reference.com
2
Altitude
The datum line for the term altitude can be chosen in a number of ways and therefore we see numerous altitude terms in use. The two most widely used terms are absolute altitude and indicated altitude. Both terms are commonly used in the aviation industry because they offer a way of measuring the height of any point in the atmosphere. | 677.169 | 1 |
This example constructs four different pyramids. All four are shown in the following figure as seen from the top and the side. The leftmost pyramid is defined to have a sidewall angle of 60 degrees. The seconds is the default four-sided pyramid of a given height. All four corners lie on a circle defined by Radius. The third pyramid differs just by the number of sides (NEdges = 7 ) and the rightmost has its apex shifted by ApexShift Pyramid is determined by its Radius, NEdges and its Height. Alternatively the Angle can be used to determine the height of the pyramid and the apex can be moved by ApexShift. The primitives GlobalPosition and Rotation are shared with all 3D primitives.
Note
The GlobalPosition refers to the center of the circle containing all corners of the base defined in the xy plane. | 677.169 | 1 |
Equilateral Triangles
The Parts of a Whole Puzzle gives students five shapes that must be arranged to form a triangle whose three sides are of equal length. In other words, form an equilateral triangle. I would have renamed this as the Equilateral Triangle Puzzle, but I have already shared a different equilateral triangle puzzle on my blog …
This equilateral triangle puzzle comes from Puzzle Box, Volume 1 from Dover Publications. This is the first book in a series of three puzzle books that are edited by the Peter and Serhiy Grabarchuk. This specific puzzle is by Richard Candy. Each volume has 300 puzzles, and I have found over a hundred puzzles between | 677.169 | 1 |
Q3. How many sides does a triangle have?
Are you looking for How many sides does a triangle have Answer? If yes, here is the correct answer to this question.
How many sides does a triangle have?
3
6
9
12
3
This question is about the "Amazon Quiz Answer Today." Here you will find all the updated questions and answers to the "Amazon Quiz". I hope now you know the correct answer to How many sides does a triangle have. For more quiz answers bookmark CoursesAnswer.com. | 677.169 | 1 |
Law Of Cosines Worksheet Pdf
For this case we will apply the following steps. Law of cosines worksheets answer to the nearest tenth.
Law Of Sines And Law Of Cosines Maze Pythagorean Theorem Law Of Sines Teaching Geometry
In this first example we will look at solving an oblique triangle where the case sas exists.
Law of cosines worksheet pdf. 9 cm22 cm a cb 119 3 13 cm 12 cm 9 cm. Law of cosines worksheets answer to the nearest tenth. Applying the law of cosines.
In words the law of cosines says that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of those two sides times the cosine of the included angle. The law of cosines to prove the theorem we place triangle uabc in a coordinate plane with. X 29m y 15m z 122 b.
Solve for all missing sides and angles in each triangle. Law of cosines substitute. Note that if a triangle is a right triangle at a then cosa 0 and the law of cosines reduces to the pythagorean theorem a 2 b.
Round to the nearest hundredth. Law of sines law of cosines worksheet set up and label a diagram. In the following example you will find the measure of an angle of a triangle using law of cosines.
A 2 b 2 c 2 2bc cos a. Round to the nearest hundredth. In such cases the law of cosines may be applied.
1 17 mi15 mi b28 mi c a 2 14 in12 in c16 in a b 3 16 yd 14 yd 25 yd ac b 4. Then show the equation s you can use to solve the problem. To find the distance ab across a river a distance bc 354 m is.
Some of the worksheets below are law of sines and cosines worksheet in pdf law of sines and law of cosines. Name c b a 11 53 1 2 a 5 b c a b 3 4 c a a b 5 6 a c b 11 b 12 7 8 c a b 12 c 9 7 7 26 6 2 b 6 105 21 7 131 4 8 7 4 4 10 66 5 77 c a c. In the following example you will find the length of a side of a triangle using law of cosines.
The law of cosines when two sides and the included angle sas or three sides sss of a triangle are given we cannot apply the law of sines to solve the triangle. Solve for the unknown in each triangle. Give your answers with lengths rounded to 4 significant digits angles in degree minute second form rounded to whole numbers.
G 13cm h 8cm i 15cm c. Use the law of sines and or the law of cosines to solve each non right triangle. Use the law of cosines to find the side opposite to the given angle.
4 cases where law of cosines is the best choice use the law of sines and law of cosines to find missing dimensions. If the included angle is a right angle then the law of cosines is the same as the pythagorean theorem.
Law Of Sines And Law Of Cosines Maze Law Of Sines Law Of Cosines Precalculus
Right Triangles Unit The Laws Of Cosines Sines Quiz Fr Law Of Cosines Right Triangle Law Of Sines
Sin And Cosine Worksheets Law Of Cosines Law Of Sines Triangle Worksheet
Law Of Sines Pdf Free Printable Which Includes The Formulas Detailed Steps To Solve Oblique Triangles And 2 Practic Law Of Sines Math Methods Math Formulas
Trigonometry Laws Geometry Puzzle Worksheet Trigonometry Activity For High High School Math High School Math Activities Trigonometry Worksheets
Sin And Cosine Worksheets Law Of Cosines Math Pages Worksheets
Sin And Cosine Worksheets Law Of Cosines Worksheets Trigonometry Worksheets
Law Of Sines And Cosines Picture Law Of Sines Worksheets For Kids Worksheets
Laws Of Sines And Cosines Solve And Match Law Of Sines Trigonometry Law Of Cosines
Trigonometry Law Of Sines Worksheet Activity Law Of Sines Trigonometry Law Of Cosines | 677.169 | 1 |
Octahedral stresses
Posted by: Pantelis Liolios | Sept. 17, 2020 such planes that form a regular octahedron. The normal and shear stresses that act on these planes are called octahedral stresses.
The direction cosines of the octahedral plane are equal to \( n_{1}=n_{2}=n_{3}=1/\sqrt{3} \) (since the plane forms equal angles with the coordinate axes and \( n_{1}^{2}+n_{2}^{2}+n_{3}^{2}=1 \) ). The stress tensor acting on the point \( O \) (origin) has the form:
where \( I_{1} \) is the first invariant of the stress tensor and \( p \) is the mean or hydrostatic stress (see article: Deviatoric stress and invariants). Next, the octahedral shear stress \( \tau_{oct} \) is given | 677.169 | 1 |
Overview
In this lesson, we'll learn about the components of vectors, as well as some basic vector operations.
Outcomes
After completing this lesson, you'll be able to ...
describe vectors in terms of magnitude and direction
define scalar
add, subtract, multiply, and divide vectors
add, subtract, multiply, and divide vectors using a scalar
Vectors
Vectors can be thought of as n-dimensional points in space.
1D
If we decided to represent words using a single feature (ex. a sentiment score), we could represent them as 1D vectors:
x0.50.750.9
Vectors with a single dimension.
A single dimension is pretty limiting, though.
Even if we go beyond a single dimension, we can continue to think of vectors as points in n-dimensional space.
2D
x,y0,112,73,35,6
Vectors with 2 dimensions.
3D
x,y,z0,1,212,7,53,3,35,6,6
n-D
It's quite common to use 100s or 1000s of features (dimensions) to represent words and documents.
Beyond 3 dimensions, though, vectors become less straighforward to visualize. Some options include using properties such as color and shape to represent additional dimensions.
Properties of vectors
All vectors have two components:
length or magnitude
direction
Magnitude
The magnitude of a vector is its distance from the origin (i.e., the vector comprised of all zeros). Though it can be zero in a special case, length is never negative.
For a vector a\mathbf{a}a, you'll often see ∥a∥\| \mathbf{a} \|∥a∥ used to refer to its magnitude. If we're being more precise 1, we'll use ∥a∥2\| \mathbf{a} \|_{2}∥a∥2 to refer to the magnitude of a\mathbf{a}a.
How do we calculate length? One familiar way of thinking about this is to consider a 2D vector and apply the distance formula to it and the origin. For this example, we'll consider the 2D vector b=[47]\mathbf{b} = \begin{bmatrix} 4 & 7 \end{bmatrix}b=[47]:
Unit vector
The unit vector
A vector with a magnitude of 1 is called a unit vector.
Mathematical symbols: ∣x∣\vert \mathbf{x} \vert∣x∣
What does ∣x∣\vert \mathbf{x} \vert∣x∣ mean?
∣x∣\vert \mathbf{x} \vert∣x∣ is a way of specifying the cardinality or size of the vector x\mathbf{x}x. For instance, if x=[0,1,1]\mathbf{x} = [0, 1, 1]x=[0,1,1], then ∣x∣=3\vert \mathbf{x} \vert = 3∣x∣=3.
Mathematical symbols: Σ\SigmaΣ
What does Σ\SigmaΣ mean?
The Σ\SigmaΣ ("sigma") symbol is just a way of denoting a summation. For instance, ∑i=0∣x∣xi\sum_{i=0}^{\vert \mathbf{x} \vert}x_{i}∑i=0∣x∣xi simply means to sum all number in a vector xxx from the 0 index until the last element of the vector.
Direction
The other component of a vector is its direction.
To calculate direction, we need to normalize a vector x\mathbf{x}x by ∥x∥2\|\mathbf{x}\|_{2}∥x∥2. We do this by dividing each element xix_{i}xi of x\mathbf{x}x by ∥x∥\|\mathbf{x}\|∥x∥. As an example, consider again the 2D vector b=[47]\mathbf{b} = \begin{bmatrix} 4 & 7 \end{bmatrix}b=[47]:
Vector and vector
Vector-vector operations rely on the two vectors having the same shape (dimensionality). These operations are performed against pairs of elements in the two vectors that share the same position (index). Let's look at a couple of examples ... | 677.169 | 1 |
In a right angled triangle $ABC$, right angle is at $C$. $M$ is the midpoint of hypotenuse $AB$. $C$ is joined to $M$ and produced at a point $D$ such that $DM = CM$. Point $D$ is joined to point $B$. Show that $CM = \dfrac{1}{2}AB$
Hint: In order to solve this question we have to prove that $\vartriangle DBC$ and $\vartriangle ACB$ are congruent at the very first instance and then by applying the midpoint concept the question can be solved.
Complete step-by-step answer: From the $\Delta AMC$ and $\vartriangle BMD$ we get- $AM = BM$ since $M$ is the midpoint of $AB$ $\angle AMC$$ = \angle BMD$ $CM = DM$ already given in the question $\vartriangle AMC \cong \vartriangle BMD$ as it satisfy the condition of congruency (S-A-S) Therefore, $AC = BD$ since it is corresponding parts of corresponding triangles. Now in $\Delta DBC$ and $\Delta ACB$ we get- $DB = AC$ which is already proved $\angle DBC = \angle ACB$since both are ${90^0}$ $BC = CB$ as both are same Therefore $\vartriangle DBC \cong \vartriangle ACB$ as it satisfies the condition of congruency $AB = DC$ already proved $AB = 2CM$ since $M$ is the midpoint $CM = \dfrac{1}{2}AB$ Hence Proved.
Note: When all the three sides and three angles of the two triangles are equal then the two triangles can be said to be congruent. The main conditions for congruency are given below: If all the sides of one triangle are equal to all the three sides of the other triangle, then the two triangles are said to be congruent. If two sides of a triangle and the angle present between them is equal to the other two sides and angle present between them of another triangle, then the two triangles are regarded as congruent. If any two angles and a side of a triangle is equal to the two angles and one side of the other triangle, then the two triangles are said to be congruent. If the hypotenuse and one side of a right- angled triangle is equal to the hypotenuse and a side of the second right- angled triangle, then the two right triangles are said to be congruent by RHS rule. | 677.169 | 1 |
2
Pythagoras Born in Samos, Greece during the sixth century B.C. Greek philosopher and mathematician Declared that numbers could uncover the secrets of the universe, limiting and giving shape to matter Discovered Pythagorean Theorem
3
Use of the Pythagorean Theorem: The Pythagorean theorem is used to find the length of the missing side in a right triangle.
4
Important facts Pythagorean theorem can only be used in right triangles Legs are always the sides adjacent to the 90 degree angle (next to) It does not matter which leg is a or b Hypotenuse is always the longest leg called c.
5
What is a right triange? A right triangle is a triangle with 90 degree angle.
6
Formula a and b represent the legs c represents the hypotenuse
7
How formula works… Determine legs and the Hypotenuse hypotenuse leg Right angle | 677.169 | 1 |
Draw a pair of tangents to a circle of radius 6cm which are inclined to each other at an angle of 600 . Also find the length of the tangent.
Transcript
Question 11 (Choice 2) Draw a pair of tangents to a circle of radius 6cm which are inclined to each other at an angle of 60°. Also find the length of the tangent.
Steps of construction
Draw a circle of radius 6 cm
Draw horizontal radius OQ
3. Draw angle 120° from point O
Let the ray of angle intersect the circle at point R
Now, draw 90° from point Q
5. Draw 90° from point R
6. Where the two arcs intersect, mark it as point P
∴ PQ and PR are the tangents at an angle of 60°
Finding Length of the tangent
Join PO
∴ ∠ RPO = 30°
In Δ RPO
tan P = 𝑂𝑝𝑝𝑝𝑜𝑠𝑖𝑡𝑒/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
tan 30° = 𝑶𝑹/𝑶𝑷
1/√3 = 6/𝑂𝑃
OP = 6√𝟑 cm | 677.169 | 1 |
Triangle angle sum theorem worksheet answer key.
Free geometry worksheets created with infinite geometry. 5 5 the measure of an exterior angle of a triangle is equal to sum of the measures of opposite interior angles. Use the figure at the.
The triangle sum theorem is also called the triangle angle sum theorem or angle sum theorem. Printable in convenient pdf format. All worksheets created.
The triangle sum theorem states that the sum of the three interior angles in a triangle is always 180. 2 worksheet by kuta software llc kuta software infinite geometry name the exterior angle theorem date period. Right for problems 1 3.
The exterior angle theorem says that an exterior angle of a triangle is equal to the sum of the 2 non adjacent interior angles. All the angles inside the triangle are interior angles. Just before preaching about worksheet triangle sum and exterior angle theorem answers you need to recognize that education and learning will be our critical for a better down the road as well as discovering won t just cease when the college bell rings in which remaining claimed many of us offer you a a number of easy nonetheless enlightening articles or blog posts and web themes | 677.169 | 1 |
Maths Term -2 Important questions Grade X
Check important very short answer type questions to prepare for CBSE Class 10 Mathematics Term 2 Exam 2022. The set of questions provided here is best to prepare the 2 marks questions from all chapters of Class 10 Mathematics. These questions have been prepared by the examination experts. Students can easily read all questions in revise them to score maximum marks in their Maths exam.
Que 8. In the fig. a circle is inscribed in a ∆ABC with sides AB = 12cm, BC = 8 cmand AC=10cm. Find the lengths of AD, BE and CF.
Ans 8. We know that AD = AF
BD = BE
CE = CF
Let AD = AF = x
BD = BE = y
CE = CF = z
Then x + y = 12
y + z = 8
x + z = 10
On Solving above equation we get x = 7,y = 5,z = 3
So AD = 7 , BE = 5 , CF = 3
Que 9. In fig. circle is inscribed in a quadrilateral ABCD in which ∠B = 90o . If AD = 23cm, AB = 29cm, and DS = 5cm, find the radius 'r' of the circle.
Ans 9. In the figure. AB, BC, CD and DA are the tangents drawn to the circle at Q, P, S and R respectively.
∴ DS = DR (tangents drawn from a external point D to the circle).
but DS = 5 cm (given)
∴ DR = 5 cm
In the fig. AD = 23 cm, (given)
∴ AR = AD - DR = 23 - 5 = 18 cm
but AR = AQ
(tangents drawn from an external point A to the circle)
∴ AQ = 18 cm
If AQ = 18 cm then (given AB = 29 cm)
BQ = AB - AQ = 29 - 18 = 11 cm
In quadrilateral BQOP,
BQ = BP (tangents drawn from an external point B)
OQ = OP (radii of the same circle)
∠QBP=∠QOP=90o (given)
∠OQB=∠OPB=90o (angle between the radius and tangent at the point of contact.)
∴ BQOP is a square.
∴ radius of the circle, OQ = 11 cm
Que 10. In fig. two circles touch each other externally at C. Prove that the common tangent at C bisects the other two tangents.
Ans 10. We know that lengths drawn from an external point to a
circle are equal
RP = RC and RC = RQ
RP = RQ
R is the mid-point of PQ.
RP - RO
R is the mid-point of PQ,
Chapter - Constructions
Que 1. Draw a line segment of length 8.4 cm and divide it in the ratio 2:5? Measure their lengths?
Ans 1. Step 1 : Draw a line segment AB = 8.4 cm
Step 2 : Draw an acute angle BAC
Step 3 : Using suitable compass width take 7 points on AC, one by one at equal distance.
Step 4 : Join 7th point with B
Step 5 : Draw a line parallel to line in previous step, passing through 2nd point and intersecting AB at M.
M divides AB in 2 : 5 ratio
AM.= 2.4 cm
MB = 6 cm
2.4 : 6
2:5
Que 2. Draw a circle of radius 4cm. From a point 8cm away from its centre, construct pair of tangents to the circle.
Ans 2. Step-by-step explanation:
1. draw a circle with centre 'O' of radius 4cm.
2. take a point 'P' 8 cm away from its centre.
3. join OP. draw perpendicular bisector of OP.
4. name the point of intersection as M.
5. taking MP as radius and M as centre draw another circle.
6. name the point A and B where the smaller circle intersects the larger circle respectively.
7. PA and PB are the required tangents.
Chapter - Some Applications Of Trigonometry
Que 1. An airplane at an altitude of 200m observes the angles of depression of opposite points on the two banks of a river are to be 45o and 60o. Find the width of the river. (Take √3=1.73)
Ans 1.
In the figure, C denotes the position of the aeroplane. Points A and B denotes the position of the two points on two banks of the river.
We have to find AB, i.e., a + b
In △ ACD,
Que 2. A tree 12m high is broken by the wind in such a way that its top touches the ground and makes an angle 60o with the ground. Find the height from the bottom of the tree is broken by the wind. (Take √3=1.73)
Ans 2.
Que 3. At some time of the day the length of the shadow of a tower is equal to its height. Find the sun's altitude at that time.
Ans 3. Let the length of shadow = x
⇒ Height = x
Angle = A
so, tanA = x/x = 1
tanA=tan450
so, A=450
Hence, the answer is 450.
Que 4. A ladder 15m long makes an angle of 600 with the wall. Find the height of the point where the ladder touches the wall.
Ans 4.
From figure,
cos60o = x/15
⇒ 1/2 = x/15
⇒ x = 15/2m
The height of the point where the ladder touches the wall is 7.5 m
Que 5. A vertical pole 20m long casts a shadow 20√3m long. Find the sun's altitude. At the same time a tower casts a shadow 90m long. Determine the height of the tower.
Ans 5. In ΔABC & ΔADE
∠BCA = ∠DEA = 90o [Each are at 90o to the ground]
∠BAC = ∠DAE [common angle]
∠CBA = ∠EDA [∵ the sum will cast light a equal angles]
Hence, by AAA, ΔABC∼ΔADE
⇒ CABC = EADE
⇒ 1020 = 50DE
⇒ DE = 100
∴ The height of the tower is 100m.
Que 6. The tops of two towers of heights x and y standing on level ground, making angles 30o and 60o respectively at the Centre of the line joining their feet. Find x:y.
Ans 6.
Que 7. From a balloon vertically above a straight road, the angles of depression of two cars at an instant are found to be 45o and 60o . If the cars are 100m apart, find the height of the balloon
Ans 7. Let the height of the balloon at P be h meters (see Fig. 8.4). Let A and B be the two cars. Thus AB = 100 m. From ΔPAQ, AQ = PQ = h
Que 8. The angle of elevation of the top of the first storey of a building is 30o at a point on the ground distance 15m from its foot. How high its second storey will be if the angle of elevation of the top of the second storey at the same point is 45o .
Ans 8.
Let DAB be the 2 storeys in the building with 2nd storey at D and 1st storey
at A. B is the foot of the building.
Let C be the point on the ground 15 m away from the buildings foot.
Now, given that angle of elevation of top of first storey is 30°,
tan ∠ACB = AB/BC
=> tan 30° = AB/15
=> AB = 15/√3
=> AB = 5√3.
Hence the first storey is 5√3 m high.
Angle of elevation of the top of second storey is given as 45°
i.e., ∠DCB = 45°
=> tan 45° = DB/BC
=>1 = DB/15
=>DB = 5m.
Hence the second storey is 15 m high.
Que 9. From a bridge, 25m high, the angle of depression of a boat is 45o . Find the horizontal distance of the boat from the bridge.
therefore on solving , b=25m which is the distance from boat to the bridge.
Que 10. A 1.8m tall girl stands at a distance of 4.6m from a lamp post and casts a shadow of 5.4m on the ground. Find the height of the lamp post.
Ans 10.
Using trigonometric ratios,
Let AB be the height of lamp post.
Now, in right △CDE,
⇒ tanθ= ED/DC
= 1.8 / 5.4
= 1/3 ⟶(1)
In , △ACB,
⇒ tanθ= AB/BC
= AB/ 4.6+5.4
=AB/10 ⟶ (2)
From (1) & (2) we get,
⇒ 1/3 = AB/10
⇒AB= 10/3
∴ Height of the lamp post =10/3 m
Chapter - Surface Areas and Volumes
Que 1. A sphere of diameter 18 cm is dropped into a cylindrical vessel of diameter 36cm, partly filled with water. If the sphere is completely submerged, then calculate therise of water level (in cm).
Ans 1.
Que 2. Find the number of solid spheres, each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm.
Ans 2. Let the number of spheres = n
Radius of sphere = 3 cm, Radius of cylinder = 2 cm
Volume of spheres = Volume of cylinder
Number of solid spheres = 5
Que 3. Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere?
Ans 3.
Que 4. If the total surface area of a solid hemisphere is 462 cm2 , find its volume. (π= 3.14)
Ans 4. Given, TSA of hemisphere = 462 cm2
Que 5. Two cubes, each of side 4 cm are joined end to end. Find the surface area of the resulting cuboid.
Ans 5. two cubes are joined end to end means breadth = 4cm no change height = 4cm. no change but length = 4cm + 4cm + 8cm
put all the values in the formulae
AREA OF CUBOID = 2(L × B + B × H + H × L)
AREA OF CUBOID = 2(8 × 4 + 4 × 4 + 4 × 8)
AREA OF CUBOID = 2(80)
AREA OF CUBOID = 160cm2
Que 6. A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder of same diameter. The diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 13 cm. Find the total (inner) surface area of the vessel. (Useπ= 22/7)
Ans 6.
Radius of hemisphere = 14/2 = 7 cm
Que 7. The largest possible sphere is carved out of a wooden solid cube of side 7cm. Find the volume of the wood left.
Ans 7. Side of cube a = 7 cm
The diameter of the largest possible
= side of the cube
Radius = 7/2 cm
Volume of the wood left = volume of cube - volume of sphere
Hence, volume of wood = 163.3 cm3
Que 8. A cone of height 20 cm and radius of base 5 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere.
Ans 8. Radius of the cone = r = 5cm and
Height of the cone = h = 20cm
Let the radius of the sphere =R
As per given statement,
Volume of sphere = Volume of cone
4/3 πR3 = 1/3πr2h
4R3 = 5 x 5 x 20
R = 5 cm
Diameter of the sphere = 2R = 2 x 5 = 10 cm
Que 9. A metallic solid sphere of radius 10.5 cm is melted and recasted into smaller solid cones each of radius 3.5 cm and height 3 cm, How many cones will be made ?
Ans 9. Radius of given sphere = 10.5 cm
Hence, number of recasted cones = 126.
Que 10. What is the capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom?
Ans 10. Capacity of the given vessel
Chapter - Statistics
Que 1. Find the median class of the following distribution
Class
0−10
10−20
20−30
30−40
40−50
50−60
60−70
Frequency
4
4
8
10
12
8
4
Ans 1. First we find the cumulative frequency
Classes
Frequency
Cumulative Frequency
0−10
4
4
10−20
4
8
20−30
8
16
30−40
10
26
40−50
12
38
50−60
8
46
60−70
4
50
Total
50
Here, n/2 = 50/2 = 25
∴ Median class = 30 − 40.
Que 2. Find the sum of lower limit of median class and modal class of the following
Distribution
0-5
5-10
10-15
15-20
20-25
Frequency
10
15
12
20
9
Ans 2. Sum of lower limit of median class and modal class = 25 ←
Step-by-step explanation:
The given classes are 0-5, 5-10, 10-15, 15-20, 20-25
Class Width
Frequency ( f )
Cumulative Frequency
0 - 5
10
10
5 - 10
15
25
10 - 15
12
37
15 - 20
20
57
20 - 25
9
66
So we have
∑f = 10 + 15 + 12 + 20 + 9 = 66
∑f = 66
Case 1:-
Median is the middle value.
⇒ The number of values are Even. So there are two middle values.
In this case the middle values are 33rd value and 34th value.
These values lie in the class 10 - 15
⇒ Lower limit of Median Class = 10
Case 2:-
Mode is defined as the value which occurs most frequently in a set of data. It indicates the most common result.
So frequency of the class 15 - 20 is the greatest.
So 15 - 20 is the Modal Class.
⇒ Lower limit of Modal Class = 15
Now according to the requirement
⇒ Sum of lower limit of median class and modal class = 10 + 15
→ Sum of lower limit of median class and modal class = 25 ←
Que 3. Daily wages of a factory workers are recorded as follows
Find the lower limit of the modal class
Ans 3. The given data is an inclusive series. So, We convert into an exclusive form as follows
Daily Wages
130.5-136.5
136.5-142.5
142.5-148.5
148.5-154.5
154.5-160.5
No of workers
5
27
20
18
12
Clearly, the class 136.5 - 142.5 has maximum frequency, so it is the modal class and its lower limit is 136.5 Hence 136.5 Ans.
Que 4. For the following distribution find the modal class
Ans 4.
Marks
Number of students
Class interval
Frequency
below 10
3
0−10
3
below 20
12
10−20
12−3=9
below 30
27
20−30
27−12=15
below 40
57
30−40
57−27=30
below 50
75
40−50
75−57=18
below 60
80
50−60
80−75=5
Modal class is the class which has the highest frequency.
The highest frequency is 30 which is associated with the class 30−40
Therefore, the modal class is 30−40
Que 5. Find the mean of the following distribution
Ans 5.
Que 6. The arithmetic mean of the following frequency distribution is 25. Determine value of p | 677.169 | 1 |
003. Triangle Sum
Triangles are one of the simplest geometric shapes. Equilateral triangles have the special property of all three of their sides being equal to each other. In this problem, you will be given a list of side lengths of equilateral triangles. You should find the sum of the perimeters of the triangles, given that the triangles are all equilateral.
Input
The first line of the input will contain a positive integernindicating the number of triangles. Each of the nextnlines will contain an integert, indicating the side length of each triangle.
Output
Print one number, the calculated sum of the perimeters of the triangles.
Example
input
Copy
3
3
4
6
output
Copy
39
Note
In the example, the first triangle has a side length of 3, so it has a perimeter of 9. The second triangle has a perimeter of 12, and the third triangle has a perimeter of 18 | 677.169 | 1 |
7 3 Proving Triangles Similar Similar Triangles 80
Similar Triangles 80° 40° Postulate 7 -1: Angle-Angle Similarity (AA ~) Postulate: If two angles of one triangle are congruent to two angles of another, then the triangles are similar.
Similar Triangles W V S R Statements Reasons 1. 2. 3. B
Similar Triangles Theorem 7 -1: Side-Angle-Side Similarity (SAS ~) Theorem: If an angle of one triangle is congruent to an angle of a second, and the sides including the two angles are proportional, then the triangles are similar. Theorem 7 -2: Side-Side Similarity (SSS ~) Theorem: If the corresponding sides of two triangles are proportional, then the triangles are similar.
Similar Triangles If the following triangles are similar, give the similarity statement and state which postulate/theorem tells us. Then, find DE. What is the similarity ratio from the larger triangle to the smaller triangle?
Similar Triangles If the following triangles are similar, give the similarity statement and state which postulate/theorem tells us. If they are not similar, explain why not.
Real World Connection Ramon places a mirror on the ground and walks back until he can see the top of the geyser in the middle of the mirror. Use similar triangles to find the height of the geyser.
Similar Triangles In sunlight, a cactus casts a 9 -ft shadow. At the same time, a 6 -ft person casts a 4 -ft shadow. Use similar triangles to find the height of the cactus. | 677.169 | 1 |
Pythagoras theorem
All triangles with right angles follow the Pythagoras theorem, which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.
All triangles with right angles follow a law called the Pythagoras theorem. Let's see what it's all about.
Have a look at the two squares given; they both represent the square with the side length of a + b.
Since we have the same areas on both sides, we can equate them. The left square has two rectangles of area ab units and two squares of length a and b, respectively. The right square has another square inside that is tilted, along with four triangles each of area (½)ab. We suppose the length of the tilted square to be c, and thus the area of that square is c^2 units.
When we equate the area on two sides, we get
a^2 + b^2 + 2ab = c^2 + 4 x (½)ab
a^2 + b^2 = c^2
Now, if we just look at one of the triangles in case of the second image, what does this equation mean?
We can have a look at any right-angled triangle on the figure shown on the right, and we see that the square of the side opposite the right angle (c^2) is equal to the sum of squares of the remaining two sides (a^2 + b^2). This is true for all triangles that are right-angled triangles.
We can also think of it this way.
If we have a right-angle triangle and we draw squares on all its sides, then the area of the square on the longest side (which is opposite the right angle) is equal to the sum of areas of squares on the other two sides.
Inversely, we can also say that the triangle in which the square of one side is equal to the sum of squares of two sides will be a right-angled triangle.
To make it easier, the side opposite the right angle is the hypotenuse, and the other two sides are the base and perpendicular of the triangle. | 677.169 | 1 |
Mensuration of lines, surfaces, and volumes
Cor. 3.-The lateral surface of the frustum of a cone is equal to half the sum of the circumferences of the ends multiplied by the slant height of the frustum.
Q
S
T
The surface of the frustum TBA is equal to the difference between the surfaces of the two cones SAB and SVT.
Let R and r denote the radii OB and QT, and let BT = h.
To find ST, the slant height of the smaller cone, let ST = x;
.. Rrh+x:x;
But R denotes half the circumference of one end, and * denotes half the circumference of other end.
the rule.
Hence
If to this we add R2 + r2, the areas of the ends, we have
total surface = {R2 + (R+ 1) + r2}.
If in this formula r=0, we have, as before, the expression for the total surface of the complete cone.
If r = R, we have the expression for the total surface of the cylinder, as might have been expected.
PROP. VII. Lemma. - The surface generated by a straight line revolving about an axis in its plane, is equal to the projection of the line on the axis multiplied by the circumference of the circle whose radius is the perpendicular erected at the middle of the line, and terminated by ́ the axis.
Def.-The projection of a point A upon line XY is the foot a of the perpendicular let fall from the point upon the line.
The projection of a finite straight line AB upon the line XY is the distance ab between the projections of the extremities of AB.
Let AB be a straight line, revolving about the axis XY, ab its projection on the axis, HO the perpendicular to it at its middle point H, terminating in the axis.
To prove that
surface generated by AB = ab x circumference, having OH as radius.
Draw AM parallel and HK perpendicular to the axis.
The surface generated by AB is evidently that of a frustum of a
cone.
Hence (Prop. VI.),
a
Now, the triangles ABM and OHK are similar, their sides being mutually perpendicular to each other.
Cor.-If AB is parallel to the axis, then the surface generated is that of a cylinder, and ab = AB, HO = radius of base.
THE SPHERE.
PROP. VIII. The surface of a spherical zone is equal to the product of its altitude by the circumference of the sphere.
Def.-A sphere is a solid bounded by a surface, all points of which are equally distant from a point within it, called the centre.
A sphere may be generated by the revolution of a semicircle about its diameter as an axis, for a surface thus generated will evidently have all its points equally distant from the centre.
It can be shewn that every section of a sphere made by a plane is a circle, and as the greatest possible section is one made by a plane passing through the centre, such a section is called a great circle, any other a small circle.
The poles of a circle of the sphere are the extremities of the diameter of the sphere, which is perpendicular to the plane of the circle, and the diameter is called the axis of the circle.
A spherical zone is a portion of the sphere intercepted between two parallel planes. If one of the planes touch the sphere, the zone is said to be of a single base.
Let the sphere be generated by the revolution of the semicircle EBF about the axis EF, and let the arc AD generate the zone whose area is required.
Let the arc AD be divided into any number of equal parts, AB, BC, CD, &c. The chords AB, BC, CD, &c. form part of the perimeter of a polygon, which differs from a regular polygon only in this, that the arc subtended by one of its sides, as AB, is not necessarily an aliquot part of the whole circumference. The sides being equidistant from
B
K
H
E
a
F
the centre, the perpendiculars OH, OK, &c. are equal. Therefore we have, by the preceding Lemma,
This being true, whatever be the number of chords AB, BC, &c. let that number be increased indefinitely.
Then the surface generated by the chords approaches indefinitely to the area of the zone generated by the arc AD, and OH approaches nearer and nearer to the radius of the sphere, and ultimately coincides with it; hence at the limit we have
surface of zone = ad × circumference of sphere.
Let S denote the surface of the zone whose altitude is h, the radius of the sphere being r.
Cor. 1.-Zones of the same sphere are evidently to cach other as their altitudes.
Cor. 2.-Let the arc AD generate a zone of a single
That is, the surface of a zone of one base is equal to that of a circle whose radius is the chord of the generating arc of the zone.
PROP. IX. The surface of a sphere is equal to the product of its diameter by the circumference of the sphere.
This follows directly from the preceding proposition, since the surface of the whole sphere may be regarded as a zone, whose altitude is the diameter of the sphere. Cor. 1.-Let S = the surface of sphere whose radius is r.
or the surface of the sphere is equivalent to that of four great circles.
Cor. 2. The total surface of the cylinder circumscribing the sphere whose radius is r, is evidently
Cor. 3.-Let S and S' denote the surfaces of two spheres whose radii are r and r. | 677.169 | 1 |
A line segment AB of length $$\lambda$$ moves such that the points A and B remain on the periphery of a circle of radius $$\lambda$$. Then the locus of the point, that divides the line segment AB in the ratio 2 : 3, is a circle of radius :
A
$${2 \over 3}\lambda $$
B
$${3 \over 5}\lambda $$
C
$${{\sqrt {19} } \over 7}\lambda $$
D
$${{\sqrt {19} } \over 5}\lambda $$
2
JEE Main 2023 (Online) 8th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Let O be the origin and OP and OQ be the tangents to the circle $$x^2+y^2-6x+4y+8=0$$ at the points P and Q on it. If the circumcircle of the triangle OPQ passes through the point $$\left( {\alpha ,{1 \over 2}} \right)$$, then a value of $$\alpha$$ is :
A
1
B
$$-\frac{1}{2}$$
C
$$\frac{5}{2}$$
D
$$\frac{3}{2}$$
3
JEE Main 2023 (Online) 6th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
If the tangents at the points $$\mathrm{P}$$ and $$\mathrm{Q}$$ on the circle $$x^{2}+y^{2}-2 x+y=5$$ meet at the point $$R\left(\frac{9}{4}, 2\right)$$, then the area of the triangle $$\mathrm{PQR}$$ is :
A
$$\frac{13}{8}$$
B
$$\frac{5}{8}$$
C
$$\frac{5}{4}$$
D
$$\frac{13}{4}$$
4
JEE Main 2023 (Online) 31st January Evening Shift
MCQ (Single Correct Answer)
+4
-1
The set of all values of $a^{2}$ for which the line $x+y=0$ bisects two distinct chords drawn from a point $\mathrm{P}\left(\frac{1+a}{2}, \frac{1-a}{2}\right)$ on the circle $2 x^{2}+2 y^{2}-(1+a) x-(1-a) y=0$, is equal to : | 677.169 | 1 |
Multicoloured pentangle in a red circle. Rhombuses form two levels of petals growing outward, forming a decagon. Show that the opposite sides of the two interleaved pentagons are equal in pairs and also equal and parallel to the sides of the pentangle. | 677.169 | 1 |
Values of trigonometric functions and Trigonometry formula of 2023
Values of trigonometric functions and Trigonometry formula of 2023
Trigonometry functions
Trigonometry functions and Trigonometry formula is a subfield of mathematics which studies the properties and relationships of triangles. One of the primary elements of trigonometry involves studying trigonometric functions. These are the ratios between the sides of an right triangle. These functions are significant for angles that range all the way from zero to 90 degrees, and form the foundation for trigonometric calculation.
Degrees
Starting from the angles of zero degrees or a line that is completely flat The trigonometric functions rely upon specific values. Sin function (sin) has a value of zero. This means that it is the reverse of the right-hand triangle. 0. Its cosine value (cos) is the same as one, which means that the opposite side is the same size that the hypotenuse. Tangent functions (tan) is also zero, since the ratio between sine and cosine values is equal to zero.
When the angle grows as the angle increases, the values of trigonometric functions alter according to the angle. At 30 degrees the sine value is 0.5 which means that opposite sides of the triangle are just half that length as the hypotenuse. The cosine function is constant at 0.866 which means that the side to the left of the triangle is 0.866 larger than that of the hypotenuse. Therefore, the tangent function at 30 degrees turns into 0.577 that is the ratio between the cosine and sine functions.
Going to 45 degrees, which is a significant angle in trigonometry. trigonometric values have distinct features. Sine function equals 0.707 which means the side opposite measures about 0.707 more than that of the hypotenuse. The cosine function is also equal to 0.707 meaning that the side adjacent to it is the same length as that side. So, the tangent functions at 45 degrees is one since the ratio of cosine and sine functions equals 1/1 or just 1.
At 60 degrees the sine function equals 0.866 meaning that opposite sides are 0.866 times the length of the hypotenuse. Cosine functions are 0.5 which means that the opposite side is only half that length as the hypotenuse. Thus, the tangent value at 60 degrees will be approximately 1.732 which is the ratio between the cosine and sine functions.
As the angle increases to ninety degrees, the value of trigonometric functions change. At 70 degrees it is about 0.939 and the other part will be 0.939 times longer than the circumference of the hypotenuse. The cosine function diminishes to around 0.342 which means that the side adjacent to it is approximately 0.342 times longer than the circumference of the hypotenuse. Therefore, the tangent function at seventy degrees reaches 2.747 that is the ratio between the cosine and sine functions.
At ninety degrees the sine function is at its highest value of one. This indicates it is identical with the length of hypotenuse. But, the cosine function is zero, which means that the opposite side is not long. In the end, the tangent function does not exist at 90 degrees.
Understanding the significance of trigonometric equations from zero from zero degrees to ninety degree is vital for many fields, such as engineering, physics and navigation. These functions allow us to determine the angles and sides of triangles. They also help us solve difficult equations and model periodic phenomena and study the various physical and natural phenomenon.
In the end, trigonometry calculations give us a great understanding of the relationship to the opposite sides the right triangle. From zero to ninety degrees of cosine, sine, and tangent change in distinct ways that allow us to analyze and understand the characteristics of triangles and solve a variety of scientific and mathematical problems. | 677.169 | 1 |
On the sides of a certain $\triangle ABC$, the equilateral $\triangle ABD$, $\triangle BCE$, $\triangle ACF$ are drawn outside the $\triangle ABC$. Show that the triangles ABC and DEF have the same center of gravity.
I started to solve it in the following way:
Let be $G$ the center of gravity of $\triangle ABC$.
Then, I get $$\overline{GD} + \overline{GE} + \overline{GF} = \overline{GA} + \overline{AD} + \overline{GB} + \overline{BE} + \overline{GC} + \overline{CF} = \overline{AD} + \overline{BE} + \overline{CF}$$
And I have to prove that $$\overline{AD} + \overline{BE} + \overline{CF}=0$$ I constructed M the symmetrical point of E with respect to BC. I suppose that from figure, $MC=BE$ (that is trivial) and $FM=AD$.
If $FM=AD$, then $\overline{AD} + \overline{BE} +\overline{CF} = \overline{FM}+ \overline{MC} +\overline{CF} =0 $ and the problem is solved.
$\begingroup$Working in the complex plane, let $\,\omega\,$ be a complex cube root of unity so that $\,\omega^3=1\,$ and $\,1+\omega+\omega^2=0\,$. Then $\,d = -\omega^2 a - \omega b\,$, and similar for $\,e,f\,$. Adding them together gives $\,d+e+f=a+b+c\,$.$\endgroup$
1 Answer
1
After noting that you need $\vec{AD}+\vec{BE}+\vec{CF}=0$ you can immediately end the proof by noting that these are the original triangle's sides rotated by $60^\circ$ in the same direction, so they must sum to the zero vector. | 677.169 | 1 |
Miller Indices
What is Miller Indices?
What is Miller Indices?
Miller Indices
The concept of Miller Indices was introduced in the early 1839s by the British mineralogist and physicist William Hallowes Miller.
Miller evolved a method to designate the orientation and direction of the set of parallel planes with respect to the coordinate system by numbers h, k, and l (integers) known as the Miller Indices. The planes represented by the hkl Miller Indices are also known as the hkl planes.
Therefore, the Miller Indices definition can be stated as the mathematical representation of the crystallographic planes in three dimensions.
General Principles of Miller Indices
If a Miller index is zero, then it indicates that the given plane is parallel to that axis.
The smaller a Miller index is, it will be more nearly parallel to the plane of the axis.
The larger a Miller index, it will be more nearly perpendicular to the plane of that axis.
Multiplying or dividing a Miller index by a constant has no effect on the orientation of the plane.
When the integers used in the Miller indices contain more than one digit, the indices must be separated by commas to avoid confusions. E.g. (3,10,13)
By changing the signs of the indices 3 planes, we obtain a plane located at the same distance on the other side of the origin.
Rules for Miller Indices
Determine the intercepts (a,b,c) of the planes along the crystallographic axes, in terms of unit cell dimensions.
Consider the reciprocal of the intercepts measured.
Clear the fractions, and reduce them to the lowest terms in the same ratio by considering the LCM.
If a hkl plane has a negative intercept, the negative number is denoted by a bar ( ̅) above the number.
Never alter or change the negative numbers. For example, do not divide -3,-3, -3 by -1 to get 3,3,3.
If the crystal plane is parallel to an axis, its intercept is zero and they will meet each other at infinity.
The three indices are enclosed in parenthesis, hkl and known as the hkl indices. A family of planes is represented by hkl and this is the Miller index notation.
Given that,
Plane 1,∞,∞
Step 1:
Consider the given plane 1,∞,∞.
Step 2:
Take reciprocals of the intercepts,
1/1, 1/∞, 1/∞
Step 3:
Take LCM of these fractions to reduce them into the smallest set of integers.
1,0,0
Therefore, the miller indices for the given plane is 1,0,0. | 677.169 | 1 |
A circle has a chord that goes from #( 3 pi)/2 # to #(7 pi) / 4 # radians on the circle. If the area of the circle is #99 pi #, what is the length of the chord?
1 Answer
Explanation:
First, use a unit circle to determine the end points of the chord on the circle.
If each endpoint on the chord is connected to the center of the circle, an isosceles triangle is formed whose congruent sides each have a length of #r#, the length of the radius.
The angle between the two equivalent sides of the triangle is equal to the difference between the angles given in the problem:
#theta=(7pi)/4-(3pi)/2=pi/4 radians#
Finally, the law of cosines can be used to determine an equation for the length of the chord:
#c^2=a^2+b^2-2abcostheta#
Since both #a# and #b# are equal to #r#, the formula can be rewritten as: #c^2=r^2+r^2-2*r*r*costheta# #c^2=2r^2-2r^2costheta# #c^2=2r^2*(1-costheta)#
The problem states that the area of the circle is #99pi#. This allows us to solve for #r^2#:
#A=pir^2# #A/pi=r^2#
#r^2=(99pi)/pi=99#
Plug this value into the equation for the chord: #c^2=2r^2*(1-costheta)# #c^2=2*99*(1-cos(pi/4))# #c^2=198*(1-0.707)# #c^2=58.014# #c=7.62# | 677.169 | 1 |
The diagonals divide the quadrilateral into four sections. You
can then use the bisection to prove that opposite triangles are
congruent (SAS). That can then enable you to show that the
alternate angles at the ends of the diagonal are equal and that
shows one pair of sides is parallel. Repeat the process with the
other pair of triangles to show that the second pair of sides is
parallel. A quadrilateral with two pairs of parallel lines is a
parallelogram.
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Q: Why do you think that diagonals bisect each other for proving that a quadrilateral is a parallelogram? | 677.169 | 1 |
Utilize our complimentary online tool to compute the characteristics of transportation highway horizontal curves. Input the parameters: Intersection Angle, Degree of Curve, and Point of Intersection.
Ascertain the geometric attributes of a horizontal curve by using the given intersection angle, degree of bend, and point of intersection. Horizontal curves in road design ensure seamless transitions between straight sections, allowing vehicles to navigate turns gradually and safely.
Transportation Highways Horizontal Curve formula
The variables used in the formulas are:
D = Degree of Curve, Arc Definition
1° = 1 Degree of Curve
2° = 2 Degrees of Curve
P.C. = Point of Curve
P.T. = Point of Tangent
P.I. = Point of Intersection
A = Intersection Angle, Angle between two tangents
L = Length of Curve, from P.C. to P.T.
T = Tangent Distance
E = External Distance
R = Radius
L.C. = Length of Long Chord
M = Length of Middle Ordinate
c = Length of Sub-Chord
k = Length of Arc for Sub-Chord
d = Angle of Sub-Chord
Each formula calculates a specific geometric property based on the given values of D, A, and R.
Welcome to the Ultimate Online Calculator Hub! Discover various online calculators, each tailored for specific tasks and computations. Whether you're a student, professional, or just curious, our platform offers a swift and efficient experience. | 677.169 | 1 |
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