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GM2-4: Identify and describe the plane shapes found in objects be able to identify plane (flat) shapes in objects and structures around them and consider why the given shape is suitable for its purpose, for example wheels are circular so they roll freely, floors are usually rectangles because they are easier to build and things fit efficiently, etc. They should consider how three dimensional objects are built from flat shapes through pulling packets apart and constructing solids of their own, for example nets for cubes.
In this unit students sort and explore two-dimensional and three-dimensional geometric shapes, identify and describe their distinguishing features and come to appreciate the efficiency of the tessellating hexagon in meeting the needs of honeybees.
Identify distinguishing features of 2D (plane) shapes using the language of sides and corners.
Identify distinguishing features of 3D shapes using the language of faces, edges, vertex/vertices.
The purpose of this activity is to support students using their understanding of properties to establish the differences and similarities between and among shapes. Similarity is the basis of classifying shapes by their defined properties and difference is important for recognising non-examples. | 677.169 | 1 |
isosceles and equilateral triangles worksheet answers practice 4-5 Triangle Worksheet Answers – Triangles are one of the most basic shapes found in geometry. Understanding triangles is important for learning more advanced geometric concepts. In this blog post we will look at the different kinds of triangles and triangle angles, as well as how to determine the extent and perimeter of any triangle, and also provide some examples to illustrate each. Types of Triangles There are three types of triangles: equal isosceles, and scalene. | 677.169 | 1 |
Rays In Math Ray - Math.net What Is A Ray In Math - The Smarter Learning Guide What is a Ray in Geometry? - Definition & Examples - Tutors.com Learn what rays are and how to identify them in geometry, including lines, angles, polygons and real world shapes. See how to use raymath to graph inequalities, find vectors and solve problems. Find worksheets, quizzes and FAQs to practice raymath skills. Lines | Geometry (all content) | Math | Khan Academy Understanding Rays in Mathematics | Definition, Properties, and Uses What is Ray in Math ⭐ Meaning, Definition, Examples, Facts - Brighterly Line Segment and Ray (Definition) - BYJUu0027S A rayinmath is a fundamental geometric concept that kids encounter as they learn about various shapes and figures. Simply put, a ray is a portion of a line that has a starting point and extends infinitely in one direction. Itu0027s like a straight path that goes on forever, but only from its starting point. Lines, Segments, and Rays - Varsity Tutors What Is A Ray In Geometry? Definition & Examples | Learnt A ray is a part of a line with a fixed starting point but that has no endpoint. It is infinite and can be named using its origin point. Learn how to teach and learn about rays, angles, and how they are related to each other. Find out the difference between a ray and a line, and the common topics and struggles for students. A ray is a geometrical concept described as a sequence of points with one endpoint or point of origin extending infinitely in one direction. Rays should not be confused with other geometric... Learn how to identify raysin geometry, which are shapes that start at one point and extend forever in one direction. Watch a video example and see how to name rays using two points, one on the ray and one beyond it. Ray in Geometry | Definition & Examples - Lesson | Study.com Lines, line segments, and rays review (article) | Khan Academy Lines, Rays and Line Segments - Math . info 1. Write if each figure is a line, ray, line segment, or an angle, and name it. 2. a. Find the angle formed by the rays DE and DF. How do we name it? b. Find the angle formed by the rays CA and CE. How do we name it? c. What is BD? (a line, a line segment, or a ray)? 3. a. Draw two points, D and E. Then draw line DE. b. Lines, segments, and rays: rays. A ray is part of a line with a clear endpoint in one direction that stretches indefinitely in the other. Since one side is infinitely long, we cannot measure the length of a ray. The diagram below illustrates what a ray looks like: Rays are named by their endpoints first and any other named point after that. A ray is a part of a line that has a fixed starting point but no endpoint. It can extend infinitely in one direction. Learn how to name, identify, and form angles using rays with examples and practice problems. Find out the difference between a ray and a line, the endpoint, and the thickness of a ray. Rayin Mathematics is any line that has a fixed endpoint on one side and on the other side, it can be extended infinitely. The fixed endpoint is depicted by a point whereas the infinitely extandible side has an arrow on it. Example of Rayin Real Life. Geometry. Line Geometry. Lines. Ray. There are several definitions of a ray. When viewed as a vector, a ray is a vector from a point to a point . In geometry, a ray is usually taken as a half-infinite line (also known as a half-line) with one of the two points and taken to be at infinity . See also. Interval, Line , Line Segment, Vector. Understanding Rays in Mathematics | Definition, Characteristics, and ... Lines, Rays, and Angles - Homeschool Math Learn the basics of lines, line segments, and raysin fourth grade math. Identify and draw them in practice problems, and practice your skills with exercises and quizzes. See examples of how to use them in plane figures and other contexts. Ray - Math Steps, Definition, Examples & Questions - Third Space Learning Identifying rays (video) | Lines | Khan Academy A ray is a line that has a fixed starting point and no end point, extending in only one direction infinitely. Learn how to draw a ray, symbolize and label it, and see examples of real-life objects that are raysin geometry. In mathematics, a ray is a straight line that extends infinitely in one direction from a specific starting point called the endpoint. A ray is named based on its endpoint and any other point lying on the line. The endpoint is denoted with a capital letter followed by an arrowhead pointing towards the infinite direction. Illustrated definition of Ray: A part of a line with a start point but no end point (it goes to infinity) Try moving... Ray -- from Wolfram MathWorld Ray Definition (Illustrated Mathematics Dictionary) - Math is Fun About this unit. Learn what lines, line segments, and rays are and how to use them. Lines, line segments, and rays. Learn. Euclid as the father of geometry. Terms & labels in geometry. Lines, line segments, & rays. Identifying rays. Lines, line segments, and rays review. Practice. Identify points, lines, line segments, rays, and angles. The first letter represents the endpoint while the second letter represents another point on the ray. A line segment is the portion of a line between two points (reference depiction below): Line segments are represented by a single overbar with no arrowheads over the letters representing the two endpoints. Consequently, the line segment above ... What is Ray in Math? Meaning, Definition, Examples, Facts - SplashLearn What are Lines, Line Segment, Rays, and Opposite Rays - Maths for Kids What is a Ray? Ray is another part of a line. It is a combination of a line and a line segment that has an infinitely extending end and one terminating end. As its one end is non-terminating, its length cannot be measured. A ray is represented by. (begin {array} {l}overrightarrow {AB}end {array} ) What is a Line Segment, Ray, Point & Line in Math? (Understanding Geometry with Definitions & Examples) - BYJUS. Home / United States / Math Classes / 4th Grade Math / Concept of Points, Lines and Rays. Concept of Points, Lines and Rays. Points, lines, and rays form the basic building blocks of geometry. Science and mathematics. Ray (geometry), half of a line proceeding from an initial point. Ray (graph theory), an infinite sequence of vertices such that each vertex appears at most once in the sequence and each two consecutive vertices in the sequence are the two endpoints of an edge in the graph. Ray (optics), an idealized narrow beam of light. Ray - Wikipedia A ray is a geometric object that has one endpoint and extends infinitely in one direction. Learn how to name, identify and measure angles formed by two rays, and see how rays are used in various contexts such as coordinate geometry, laser pointers and number lines. What is a Rayin Geometry? A ray has a start point and it continues indefinitely to one direction only. It cannot expand to both directions like Line does. Take an example of a Torch Beam. The start point is the Torch, but the endpoint is not certain and it lights everything it touches. Rays are used to describe lines that originate from a specific point and extend infinitely in one direction. In this article, we will explore the meaning of raysin geometry, how to identify and name them, their properties, real-life examples, and their applications in various fields. Background-checked tutors you can trust. Top 10 Geometry Tutors. ray. In mathematics, a ray is a geometric object that consists of a single point (called the endpoint) and extends infinitely in one direction. A ray is represented by a starting point and an arrowhead, indicating the direction in which it extends. The starting point is called the endpoint of the ray, and it is the only point that is specified. What is a Line Segment, Ray, Point & Line in Math? (Understanding ... What is a Ray in Maths? - Definition, Representation & Examples | 677.169 | 1 |
The quadrant is surely relevant, so it isn't being pedantic to bring it up.
But if we knew we were operating in Q1, I would ask, "Is there a triangle here? If so, what do you know about it?" I think it's easy to forget the context and get lost in algebraic manipulation mindlessly. And yes, asking how cosine (theta) could be > 1 is another good query.
The student seems to remember how the trig functions relate to one another but lacks conceptual understanding of what tan @ = 3/5 actual means with respect to right triangles. I would give the student a picture of the right triangle quadrant 1 with angle @ and then ask to find all 3 sides(and explain how they know) given that the tan @ = 3/5. How could you use this information to find sec @?
seems likely to be a lapse in understanding of ratios. I would be curious to see algebraic ratios of n * y versus m * x from this student, for enumerated constant values of n and m and values of y and x that would violate the apparent ratio between n and m.
As for actual written feedback to this kid, I'd write, "great hypothesis — I can follow your reasoning about how different trig functions are related" (because it is a great hypothesis). And then follow up with, "Can you draw and label the sides of a real triangle to show that your answer works i.e. where tan @ = 3/5 and sec @ = 1/5?"
That task acknowledges the kid's thinking so far and teaches a general strategy to recover from this kind of tantalizing algebraic manipulation.
In the process of building that triangle, the kid is likely to encounter both where their thinking about proportions went wrong, and why the cosine is actually going to be less than one, and have the experience Bob outlines (but coming from the perspective of continuing their thinking by checking it vs. starting over with a new strategy).
If this were non-written feedback, I'd tend to ask, "great, how can we check that?" and be prepared to offer a strategy, but since checking is not most students' strong suit, in written feedback I present the challenge right out, as a follow-up challenge. If the student is successful the next written prompt would be to reflect on why I asked them to draw and label a real triangle.
Good thoughts to chew on here. I wrote 3/5=6/10 on the paper of any kid who did this. I may have written 4=2+2 though. Kids all called me over and said "yeah, so?" Though it did get them to call me over so then I could have the conversation of fractions being about relationships, not values. I didn't think to mention some of the great suggestions here, but asked students for a way to relate tangent to secant, they looked to their list of identities and found the relevant pythagorean one. Same end result as drawing the triangle but less conceptual understanding perhaps… | 677.169 | 1 |
Friday 20 February 2015
MFM2P - Day 13 (Triangles out of Squares)
Today was our first Always-Sometimes-Never warm up:
They quickly thought of 2 + 2 = 2 * 2. I liked hearing students saying that since they found one case that worked, it couldn't be always nor never. Someone also said that it worked for 0 and 0. I asked if it could work with two numbers that were different from each other. I left them thinking about that one. Back to our 26 squares. I first asked them to make something with their squares. I saw a rocket, a caterpillar, a cat and many towers. They didn't seem to be moving toward making triangles so I asked if they could use their squares to make a mathematical shape. It took a bit, but someone finally made a triangle so I asked others to follow suit. They got the idea of how to make a triangle and that the corners needed to match up. I asked if they could make a triangle with any three squares. Most said yes. I then chose three of their squares that could not form a triangle and asked them to make one. This was followed up by me asking how you could know whether it would work. We eventually all came to the conclusion that the sum of the two shorter sides had to be longer than the long side. On to right triangles. We defined a right triangle and I asked them to use their squares to determine whether the triangles listed (see below) seemed to contain a 90 degree angle. Here are a couple of pictures - the first one worked, the second one didn't.
They had to come up to the SMARTboard and either cross out or place a check mark next to each triangle. Here is the list (updated to remove those that were not obvious).
and I asked what they noticed about the areas. Many blank stares... I asked a student what he noticed about the triangle with areas 100, 576 and 676. They are all even numbers (true, but not all areas in the table were even numbers - what else do you notice?). The 3rd number is bigger than the other too (yes, by how much?)... Eventually we got to noticing the the third area was the sum of the other two. | 677.169 | 1 |
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Sphere–cylinder intersection
In the theory of analytic geometry for real three-dimensional space, the intersection between a sphere and a cylinder can be a circle, a point, the empty set, or a special type of curve.
For the analysis of this situation, assume (without loss of generality) that the axis of the cylinder coincides with the z-axis; points on the cylinder (with radius r) satisfy
\( x^2 + y^2 = r^2. \)
We also assume that the sphere, with radius R is centered at a point on the positive x-axis, at point (a, 0, 0). Its points satisfy
\( (x-a)^2 + y^2 + z^2 = R^2.\)
The intersection is the collection of points satisfying both equations.
Trivial cases
Sphere lies entirely within cylinder
If a+R < r, the sphere lies entirely in the interior of the cylinder. The intersection is the empty set.
Sphere touches cylinder in one point
If the sphere is smaller than the cylinder (R < r) and a+R = r, the sphere lies in the interior of the cylinder except for one point. The intersection is the single point (r, 0, 0).
Sphere centered on cylinder axis
If the center of the sphere lies on the axis of the cylinder, a = 0. In that case, the intersection consists of two circles of radius r. These circles lie in the planes
\( z = \pm\sqrt{R^2 - r^2};\)
If r = R, the intersection is a single circle in the plane z = 0.
Non-trivial cases
Subtracting the two equations given above gives
\( z^2 + (r^2 - R^2 + a^2) = 2ax.\)
Since x is a quadratic function of z, the projection of the intersection onto the xz-plane is the section of an orthogonal parabola; it is only a section due to the fact that -r < x < r. The vertex of the parabola lies at point (-b, 0, 0), where
\(b = \frac{R^2 - r^2 - a^2}{2a}.\)
Intersection consists of two closed curves
If R > r + a, the condition x < r cuts the parabola into two segments. In this case, the intersection of sphere and cylinder consists of two closed curves, which are mirror images of each other. Their projection in the xy-plane are circles of radius r.
Each part of the intersection can be parametrized by an angle \( \phi \):
If R < r + a, the intersection of sphere and cylinder consists of a single closed curve. It can be described by the same parameter equation as in the previous section, but the angle \( \phi\) must be restricted to -\( \phi_0 < \phi < +\phi_0 \), where \( \cos\phi_0 = -b/r\).
Limiting case
Viviani's curve as intersection of a sphere and a cylinder
In the case R = r + a, the cylinder and sphere are tangential to each other at point (r, 0, 0). The intersection resembles a figure eight: it is a closed curve which intersects itself. The above parametrization becomes | 677.169 | 1 |
Lesson
Lesson 15
15.1: Part Way: Points (5 minutes)
Warm-up
Students review the definition of midpoint by finding the midpoint of a horizontal segment, the midpoint of a vertical segment, and the midpoint of the diagonal formed by these two segments. Identify students who found the midpoint visually as well as students who found the midpoint algebraically.
Student Facing
For the questions in this activity, use the coordinate grid if it is helpful to you.
What is the midpoint of the segment connecting \((1,2)\) and \((5,2)\)?
What is the midpoint of the segment connecting \((5,2)\) and \((5,10)\)?
What is the midpoint of the segment connecting \((1,2)\) and \((5,10)\)?
Student Response
Anticipated Misconceptions
If students don't remember the term midpoint, remind them that it is the point that partitions the segment exactly in half.
Activity Synthesis
The purpose of the discussion is to introduce notation for segment partitioning.
Display a graph of the three segments for all to see. Invite the previously selected students to share their methods. If anyone used the midpoint formula, ask them to share last. (This isn't the "best" method, but it's the most connected to the new notation introduced here.)
Ask students how the word average relates to finding midpoints. (The coordinates of the midpoint are the averages of the corresponding pairs of the coordinates of the endpoints of the segment.)
Now ask students to consider what the notation \(\frac 12 (A+C)\) or \(\frac 12 A + \frac 12 C\) could mean. (These both represent the midpoint. The midpoint is halfway between the 2 \(x\)-coordinates and halfway between the 2 \(y\)-coordinates. It is like the average of the two points.)
Invite students to describe the calculations indicated by the notation \(\frac 12 A + \frac 12 C\). (First, calculate half of each coordinate for both points. Point \(A\) has coordinates \((1,2)\), so \(\frac12 A = \left(\frac12, 1\right)\). Point \(C\) has coordinates \((5,10)\), so \(\frac12(C)=\left(\frac52, 5\right)\). Now add them together to get \(\left(\frac12,1\right)+\left(\frac52,5\right)=(3,6)\).)
15.2: Part Way: Segment (15 minutes)
Activity
Students find the point that partitions a segment in a given ratio. The activity starts with an introduction to the concept of partitioning a segment. Then students use informal methods to find a point that partitions a particular segment in a \(2:1\) ratio. Next, students compute a weighted average and connect that to the first prompt. Finally, they generalize the process for a \(3:1\) ratio.
Launch
Display this image for all to see.
Ask students what point would partition segment \(AB\) in a \(1:2\) ratio. That is, if we call the point \(C\), the ratio \(AC:CB\) should be \(1:2\).
If students struggle, remind students that this notation means that the whole segment is divided into 3 equal sections. To the left of point \(C\) will be 1 of those equal sections, and to the right of \(C\) will be 2 equal sections. The point \((4,2)\) meets this description.
Action and Expression: Develop Expression and Communication. Maintain a display of important terms and vocabulary. During the launch, take time to review the following terms that students will need to access for this activity: midpoint, partition. Supports accessibility for: Memory; Language
Student Facing
Point \(A\) has coordinates \((2,4)\). Point \(B\) has coordinates \((8,1)\).
Find the point that partitions segment \(AB\) in a \(2:1\) ratio.
Calculate \(C=\frac 13 A + \frac 23 B\).
What do you notice about your answers to the first 2 questions?
For 2 new points \(K\) and \(L\), write an expression for the point that partitions segment \(KL\) in a \(3:1\) ratio.
Student Response
Anticipated Misconceptions
Some students might write the ratios for the fourth question with the fractions reversed. Ask them to choose a pair of points—perhaps the points \((0,2)\) and \((12,2)\) from the activity launch—and use their expression to calculate the coordinates of the desired point to see if it's correct. It's okay if students continue to struggle; the activity synthesis will help all students gain intuition about the placement of the fractions.
Activity Synthesis
The goal of the discussion is to make sure students understand why the point that partitions segment \(AB\) in a \(2:1\) ratio can be expressed as \(\frac13A + \frac23B\). Invite students to consider just the \(x\)-coordinate, or the horizontal component. Display this image for all to see.
Tell students that the midpoint notation \(\frac12A + \frac12B\) represents an average. The expression \(\frac13A + \frac23 B\) is called a weighted average because the 2 points have different weights. Ask students these questions:
"What is the horizontal distance from \(A\) to \(B\)?" (6 units)
"How would you calculate this distance if you couldn't just count it?" (Subtract the coordinates. We can write \(B-A\).)
"What fraction of this distance do we need to add to \(A\) to get to \(B\)?" (We need to add \(\frac23\) of this distance to get to \(C\).)
"In light of the previous answers, what does \(A+\frac23(B-A)\) mean?" (Take \(A\), and add \(\frac23\) of the distance from \(A\) to \(B\).)
"How can we rewrite this expression to look more streamlined?" (Distribute the fraction and combine like terms to get \(\frac13A + \frac23 B\).)
Now invite students to describe all of this in common sense language. The key idea that should surface is that point \(B\) needs to be more heavily weighted in the calculation, because \(C\) is closer to \(B\) than it is to \(A\). On the segment connecting \(A\) and \(B\), point \(C\) is \(\frac23\) of the way towards \(B\).
Speaking: MLR8 Discussion Supports. As students share how they would calculate the horizontal distance between \(A\) and \(B\), press for details by asking students how they know that the distance is \(B-A\). Also ask how they know that the distance from \(A\) to \(C\) is \(\frac{2}{3}\) of the distance from \(A\) to \(B\). Show concepts multi-modally by drawing and labeling the horizontal lengths of \(A\), \(B\), and \(B-A\). This will help students justify why the value of \(C\) is equivalent to the expression \(A+\frac{2}{3}(B-A)\). Design Principle(s): Support sense-making; Optimize output (for justification)
15.3: Part Way: Quadrilateral (15 minutes)
Activity
Students apply partitioning to a quadrilateral to see that segment partitioning is another method for building similar figures on the coordinate plane. Monitor for students who use a weighted average.
Launch
Writing, Listening, Conversing: MLR1 Stronger and Clearer Each Time. Use this routine to help students improve their written responses for the last question. Give students time to meet with 2–3 partners to share and receive feedback on their responses. Display feedback prompts that will help students strengthen their ideas and clarify their language. For example, "How do you know that \(AB'C'D'\) is a dilation of \(ABCD\)?" and "What is the center and scale factor of the dilation?" Invite students to go back and revise or refine their written responses based on the feedback from peers. This will help students justify why partitioning the segments of \(ABCD\) results in a dilation of the figure. Design Principle(s): Optimize output (for justification); Cultivate conversation
Student Facing
Here is quadrilateral \(ABCD\).
Find the point that partitions segment \(AB\) in a \(1:4\) ratio. Label it \(B'\).
Find the point that partitions segment \(AD\) in a \(1:4\) ratio. Label it \(D'\).
Find the point that partitions segment \(AC\) in a \(1:4\) ratio. Label it \(C'\).
Is \(AB'C'D'\) a dilation of \(ABCD\)? Justify your answer.
Student Response
Anticipated Misconceptions
If students struggle to begin, ask them how many parts total we have if we are looking at a ratio of \(1:4\) (5 parts total). Ask students how dividing by this number might help them. Tell students that it's okay if their answers don't come out to integers. Decimal values are valid answers.
Activity Synthesis
Invite students to share their approaches, including the previously selected student(s) who used weighted averages. Ask the class which method seems easiest. (Any answer with support is valid. Students might find weighted averages most efficient in this case since they are repeating the same ratio three times.)
Tell students this process was a second coordinate version of dilation. Ask students how the process they completed here matches the definition of a dilation on their reference chart. (We did exactly what the definition says: We found the points along each ray \(AB,AC,\) and \(AD\) whose distance from \(A\) was \(\frac15\) the original distance from \(A\) to points \(B,C,\) or \(D\).)
Lesson Synthesis
Lesson Synthesis
Ask students to use segment partitioning to dilate triangle \(END\) using center \(E\) and scale factor \(\frac 34\).
After 2 minutes of quiet work time, ask students to share their plans. Students should recognize that dilating by a scale factor of \(\frac34\) is equivalent to partitioning the figure's segments in a \(3:1\) ratio. Give students a few more minutes of work time to complete the task. (\(N'=(2.25, 9), D'=(2.25, 3)\))
15.4: Cool-down - Part Way: Triangle (5 minutes)
Cool-Down
Student Lesson Summary
Student Facing
To find the midpoint of a line segment, we can average the coordinates of the endpoints. For example, to find the midpoint of the segment from \(A=(0,4)\) to \(B=(6,7)\), average the coordinates of \(A\) and \(B\): \(\left(\frac{0 + 6}{2}, \frac{4+7}{2}\right) = (3,5.5)\). Another way to write what we just did is \(\frac12 (A+B)\) or \(\frac12 A + \frac12 B\).
Now, let's find the point that is \(\frac23\) of the way from \(A\) to \(B\). In other words, we'll find point \(C\) so that segments \(AC\) and \(CB\) are in a \(2:1\) ratio.
In the horizontal direction, segment \(AB\) stretches from \(x=0\) to \(x=6\). The distance from 0 to 6 is 6 units, so we calculate \(\frac23\) of 6 to get 4. Point \(C\) will be 4 horizontal units away from \(A\), which means an \(x\)-coordinate of 4.
In the vertical direction, segment \(AB\) stretches from \(y=4\) to \(y=7\). The distance from 4 to 7 is 3 units, so we can calculate \(\frac23\) of 3 to get 2. Point \(C\) must be 2 vertical units away from \(A\), which means a \(y\)-coordinate of 6.
It is possible to do this all at once by saying \(C = \frac13 A + \frac23 B\). This is called a weighted average. Instead of finding the point in the middle, we want to find a point closer to \(B\) than to \(A\). So we give point \(B\) more weight—it has a coefficient of \(\frac23\) rather than \(\frac12\) as in the midpoint calculation. To calculate \(C = \frac13 A + \frac23 B\), substitute and evaluate | 677.169 | 1 |
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are externally in the ratio 1 : 2. Also, show that P is the mid point of the line segment RQ.
A.
B.
C.
D.
Detailed Solution for Test: Vector Algebra- 2 - Question 13
Let position vector of point R be . As point R divides externally the line segment PQ in the ratio 1:2 .therefore ,
Also , mid point of the line segment RQ is :
Which is the position vector of point P.
Therefore , P is the mid point of line segment RQ.
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Vector Analysis
Geometry of Curves and Surfaces - Basics of Vectors
Q.01
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Q.02
'(1) Let A(-3), B(7), C(2). Find the distances between points A and B, B and C, C and A, respectively. (2) Find the coordinates of the points R which divides the line segment PQ connecting two points P(-4) and Q(8) at a ratio of 1:3, the points S which divides the line segment at a ratio of 3:1 externally, and the midpoint M of segment RS.'
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Q.22
'In an equilateral triangle ABC with side length 2, let L, M, and N be the midpoints of the sides AB, BC, and CA respectively. Find all of the following vectors represented by the 6 points A, B, C, L, M, N:'
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Q.24
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Q.25
'When two linearly independent vectors \ \\overrightarrow{\\mathrm{OA}}, \\overrightarrow{\\mathrm{OB}} \ are defined on a plane, any point \ \\mathrm{P} \ can be uniquely represented as \ \\overrightarrow{\\mathrm{OP}}=s \\overrightarrow{\\mathrm{OA}}+t \\overrightarrow{\\mathrm{OB}} \\quad(s, t \ are real numbers \\( ) \\ldots \\ldots \\cdot(A) \\). In this case, the pair of real numbers \\( (s, t) \\) is called the oblique coordinates, and the point defined by (A) is denoted as \\( \\mathrm{P}(s, t) \\). In particular, when \ \\overrightarrow{\\mathrm{OA}} \\perp \\overrightarrow{\\mathrm{OB}},|\\overrightarrow{\\mathrm{OA}}|=|\\overrightarrow{\\mathrm{OB}}|=1 \, the oblique coordinates will become the \ x y \ coordinates with the extension of \ \\overrightarrow{\\mathrm{OA}} \ as the \ x \ axis and the extension of \ \\overrightarrow{\\mathrm{OB}} \ as the \ y \ axis. \\n【Basic Example 38(1)】\ \\overrightarrow{\\mathrm{OP}}=s \\overrightarrow{\\mathrm{OA}}+t \\overrightarrow{\\mathrm{OB}}, s+2 t=3 \\ldots \\ldots \ That is, the points \ \\mathrm{P} \ that satisfy \\( \\mathrm{P}(s, t), s+2 t=3 \\) lie on the line in the Cartesian coordinate plane given by \ x+2 y=3 \. The intersections of this line with the coordinate axes are \\( \\mathrm{C}(3,0) \\) and \\( \\mathrm{D}\\left(0, \\frac{3}{2}\\right) \\). Considering the points C, D with the same coordinates in the oblique coordinate plane, we have \ \\overrightarrow{\\mathrm{OA}}=\\frac{1}{3} \\overrightarrow{\\mathrm{OC}}, \\overrightarrow{\\mathrm{OB}}=\\frac{2}{3} \\overrightarrow{\\mathrm{OD}} \Thus, the condition equation (*) for point \ \\mathrm{P} \ becomes \ \\overrightarrow{\\mathrm{OP}}=\\frac{s}{3} \\overrightarrow{\\mathrm{OC}}+\\frac{2}{3} t \\overrightarrow{\\mathrm{OD}},\\frac{s}{3}+\\frac{2}{3} t=1 \, and the range for the existence of point \ \\mathrm{P} \ is the line CD.'
Q.27
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Q.28
'When a point P moves on the plane and its coordinates (x, y) are functions of time t, answer the following questions:\n1. Derive the vector equation representing velocity.\n2. Derive the vector equation representing acceleration.'
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Q.30
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Q.31
'Collinearity condition\nWhen two points A, B are different\nWhen point P is on the line segment AB\n\ \\Leftrightarrow \\overrightarrow{\\mathrm{AP}}=k \\overrightarrow{\\mathrm{AB}} \ for some real number k'
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Q.34
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Q.35
'(2) Let D, E, and F be points on the line segments OA, OB, and OC respectively, such that OD = 1/2 · OA, OE = 2/3 · OB, and OF = 1/3 · OC. If the plane containing the three points D, E, F intersects the line OQ at point R, then express vector OR in terms of vectors a, b, and c.'
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Q.36
'Let the midpoints of the sides AB, BC, CD, DA of quadrilateral ABCD be K, L, M, N respectively, and the midpoints of the diagonals AC, BD be S, T respectively. (1) If the position vectors of vertices A, B, C, D are a, b, c, d respectively, express the position vector of the midpoint of segment KM using a, b, c, d. (2) By expressing the position vectors of the midpoints of segments LN, ST using a, b, c, d, prove that the three segments KM, LN, ST intersect at a single point.'
Q.37
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Q.38
'Conditions for being collinear, coincident\n(1) Collinear conditions\nWhen two different points A, B, if a point P lies on the line AB, then there exists a real number k such that vector AP = k vector AB.'
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Q.47
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Q.48
'Basic Concepts\n3. Position Vector of the Centroid of a Triangle\nLet points A(𝑎⃗), B(𝑏⃗), C(𝑐⃗) be the vertices of triangle ABC, and let G be the position vector of the centroid. Then\n𝑔⃗=1/3(𝑎⃗+𝑏⃗+𝑐⃗)'
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Q.52
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Q.53
"In triangle ABC with vertices A(a), B(b), C(c), let D be the point dividing side BC in the ratio 2:3, and E be the point dividing side BC externally in the ratio 1:2. Let G be the centroid of triangle ABC and G' be the centroid of triangle AED. Express the following vectors in terms of a, b, and c.\n(1) Position vectors of points D, E, G'\n(2) GG'"
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Q.64
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Q.65
'In rectangle ABCD, AB = 3 and AD = 4. Let AB vector be b and AC vector be c. (1) If E is the midpoint of side AD, express the vector DE using b and c. (2) Express a unit vector d in the same direction as c using c.'Find the equation of a line that satisfies the following conditions using vectors:\n1. Passes through point A(-3,5) and is parallel to vector d=(1,-sqrt(3))\n2. Passes through points A(-7,-4) and B(5,5)'
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Q.82
'In tetrahedron OABC, let OA=a, OB=b, and OC=c. Let M be the midpoint of AB, N be the point that divides BC in the ratio 3:1, and G be the centroid of triangle OAB. Express vectors MN and GN in terms of a, b, and c.'
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Q.83
"Let's consider a straight line passing through a point and having a given slope (direction). Let g be the line passing through point A(\\\vec{a}\) and parallel to a non-zero vector \\\vec{d}\. For any point P(\\\vec{p}\) on the line g (excluding point A), the following holds true."
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Q.86
'For the points \\( \\mathrm{A}(1,2,3), \\mathrm{B}(-3,2,-1), \\mathrm{C}(-4,2,1) \\), find the following:\n(1) Distance between the points \ \\mathrm{B}, \\mathrm{C} \\n(2) Coordinates of the point \ \\mathrm{P} \ that divides the segment \ \\mathrm{BC} \ in the ratio 1:3\n(3) Coordinates of the point \ \\mathrm{Q} \ that divides the segment \ \\mathrm{AB} \ externally in the ratio 2:3\n(4) Coordinates of the midpoint R of the segment CA\n(5) Coordinates of the centroid G of the triangle \ \\triangle \\mathrm{PQR} \'
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Q.87
'For points A(0,3,7), B(3,-3,1), C(-6,2,-1), find the following:\n(1) Distance between points A and B\n(2) Coordinates of a point that divides the line segment AB in the ratio 2:1\n(3) Coordinates of a point that divides the line segment AB externally in the ratio 3:2\n(4) Coordinates of the midpoint of the line segment BC\n(5) Coordinates of the centroid of triangle ABC'
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Q.88
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Q.89
"The line perpendicular to vector \ \\vec{n} \\nFinally, let's consider expressing the line using dot product.\nPassing through point \\( \\mathrm{A}(\\vec{a}) \\), and a non-zero vector \ \\overrightarrow{0} \ perpendicular to vector \ \\vec{n} \ is denoted as line \ g \, and any point on line \ g \ is denoted as \\( \\mathrm{P}(\\vec{p}) \\) such that \ \\vec{n} \\perp \\overrightarrow{\\mathrm{AP}} \ or \ \\overrightarrow{\\mathrm{AP}}=\\overrightarrow{0} \\n\\[\n\egin{array}{l}\n\\Longleftrightarrow \\vec{n} \\cdot \\overrightarrow{\\mathrm{AP}}=0 \\\\\n\\Longleftrightarrow \\vec{n} \\cdot(\\vec{p}-\\vec{a})=0\n\\end{array}\n\\]\n(D) represents the vector equation of a line passing through point \ \\mathrm{A} \ and perpendicular to vector \ \\vec{n} \. Furthermore, \ \\vec{n} \ is referred to as the normal vector of line \ g \.\n\ -\\overrightarrow{\\mathrm{AP}}=\\overrightarrow{0} \ holds true only when point P coincides with point A.\nThe normal vector of line \ g \ is perpendicular to it.\nNow, let's solve a vector equation problem."
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Q.98
'Important Example 63 | Length of Common Perpendicular\nIn the coordinate space, let point A(1,3,0) lie on a line l parallel to vector a=(-1,1,-1), and point B(-1,3,2) lie on a line m parallel to vector b=(-1,2,0). Let P be a point on line l and Q be a point on line m. Find the minimum value of the magnitude |PQ| of vector PQ, and the coordinates of points P and Q at that instance.'
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Q.15
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Q.16
'A directed line segment AB is represented as vector →AB. Additionally, vectors can also be represented using a single letter with an arrow, such as →a, →b. How do we represent the magnitude of vectors →AB, →a?'
Q.28
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Q.29
'Find the components of each vector, where 21 \\\overrightarrow{AC}=\\vec{a}+\\vec{b}\, \\\overrightarrow{AG}=\\vec{a}+\\vec{b}+\\vec{c}\, \\\overrightarrow{BH}=-\\vec{a}+\\vec{b}+\\vec{c}\, and \\\overrightarrow{CP}=-\\frac{1}{2}\\vec{a}-\\frac{1}{2}\\vec{b}+\\frac{1}{2}\\vec{c}\.'
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Q.36
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Q.37
'In triangle ABC with vertices A(a), B(b), and C(c), let point P divide side AB in the ratio 2:1, point Q divide side BC externally in the ratio 3:2, and point R divide side CA externally in the ratio 1:3. Let G be the centroid of triangle PQR. Express the following vectors in terms of a, b, and c: (1) Position vectors of points P, Q, R (2) Vector PQ (3) Position vector of point G'
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Q.45
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Q.46
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Q.47
'Consider an equilateral triangle ABC with side length 1 on the plane. For a point P, let vector v(P) be defined as v(P)=→PA−3→PB+2→PC. Prove: (1) v(P) is a constant vector independent of P. (2) For which figure does point P lie when |→PA+→PB+→PC|=|v(P)|.'
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Q.51
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Q.52
'When four points O, A, B, C are not in the same plane, if overrightarrowOA=veca\\overrightarrow{OA}=\\vec{a}overrightarrowOA=veca, overrightarrowOB=vecb\\overrightarrow{OB}=\\vec{b}overrightarrowOB=vecb, overrightarrowOC=vecc\\overrightarrow{OC}=\\vec{c}overrightarrowOC=vecc, then any vector vecp\\vec{p}vecp can be uniquely represented as vecp=sveca+tvecb+uvecc\\vec{p}=s\\vec{a}+t\\vec{b}+u\\vec{c}vecp=sveca+tvecb+uvecc, where sss, ttt, uuu are real numbers.'
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Q.55
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Q.56
'(1) Find the coordinates of point B after rotating point A(2,1) around the origin O by π/4 radians.\n(2) Point P was the center of rotation when point A(2,1) was rotated by π/4 radians to the coordinates (1-√2, -2+2√2). Find the coordinates of point P.'
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Q.63
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Q.64
'Show that the following equations hold for the tetrahedron ABCD: (1) overrightarrowAB+overrightarrowBD+overrightarrowDC+overrightarrowCA=overrightarrow0\\overrightarrow{AB}+\\overrightarrow{BD}+\\overrightarrow{DC}+\\overrightarrow{CA}=\\overrightarrow{0}overrightarrowAB+overrightarrowBD+overrightarrowDC+overrightarrowCA=overrightarrow0 (2) overrightarrowBC−overrightarrowDA=overrightarrowAC−overrightarrowDB\\overrightarrow{BC}-\\overrightarrow{DA}=\\overrightarrow{AC}-\\overrightarrow{DB}overrightarrowBC−overrightarrowDA=overrightarrowAC−overrightarrowDB'
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Q.82
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Q.83
'In triangle ABC with vertices A(a), B(b), and C(c), where point P divides side AB internally in the ratio 2:1, point Q divides side BC externally in the ratio 3:2, and point R divides side CA externally in the ratio 1:3, and G is the centroid of triangle PQR. Express the following vectors in terms of a, b, and c: (1) Position vectors of points P, Q, and R (2) Vector PQ (3) Position vector of point G'
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Q.84
'Consider a circle with center O. There are 3 points A, B, and C on the circumference of this circle such that OA vector + OB vector + OC vector = 0. Prove that triangle ABC is an equilateral triangle.'
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Q.85
'Vector decomposition In parallelogram ABCD, point E divides side BC internally in the ratio 2:1, point F is the intersection of the diagonals AC and BD, and point G is the intersection of segments AE and BD. Let vector AB=b and vector AD=d. (1) Express vectors AE, AF, GC in terms of b and d. (2) If vector AE=e and vector AF=f, then express vector BD in terms of e and f.'
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Q.92
'Example 38 Problem on angles formed by vectors\n(1) In space, there are fixed points A(0,4,2) and B(2√3, 2,2), and a moving point P(0,0,p). Find the maximum value of ∠APB denoted as θ(0° ≤ θ ≤ 180°) and the corresponding value of p.\n(2) For vectors a=(3,-4,12), b=(-3,0,4), and c=a+tb, find the real value of t when the angle between c and a is equal to the angle between c and b.'
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Q.93
'Important Example 61: Equations of Lines\nFind the equations of the following lines:\n(1) Passing through point A(1,3,-2) and parallel to vector d=(3,2,-4)\n(2) Passing through points A(0,1,1) and B(-1,3,1)\n(3) Passing through point A(-3,5,2) and parallel to vector d=(0,0,1)'
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Q.98
'The vector equation of a plane defined by three non-collinear points A(⃗a), B(⃗b), C(⃗c) and an arbitrary point P(⃗p), with s, t, u as real numbers, is given by ⃗p=s⃗a+t⃗b+u⃗c, s+t+u=1 or ⃗p=s⃗a+t⃗b+(1-s-t)⃗c'
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Q.01
'Find the complex numbers representing the following points. (1) The midpoint of the line segment AB connecting the points A(-3+6i) and B(5-8i) (2) The point P which divides the line segment AB connecting the points A(2-3i) and B(-7+3i) in the ratio 2:1, and the point Q which divides externally.'
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Q.03
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Q.04
'Let point P divide side OA of acute-angled triangle OAB in the ratio k:(1-k), and point Q divide side OB in the ratio l:(1-l). Also, let R be the intersection of AQ and BP. Let OA be represented by vector a, and OB be represented by vector b. 1) Express OP vector and OQ vector in terms of vectors a and b. 2) Express OR vector in terms of vectors a and b.'
Q.05
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Q.06
'Position vector and internal division points・external division points\nLet the position vector be \ \\vec{p} \ and denote the point as \\( \\mathrm{P}(\\vec{p}) \\).\nIn space, just like in a plane, the following holds:\n\nProblem 1: For points \\( \\mathrm{A}(\\vec{a}), \\mathrm{B}(\\vec{b}), \\mathrm{C}(\\vec{c}) \\), derive the following equations.\n1. \ \\overrightarrow{\\mathrm{AB}} = \\vec{b} - \\vec{a} \\n2. Find the position vector of a point dividing the line segment \ \\mathrm{AB} \ in the ratio of \ m: n \.\n3. Find the position vector of the midpoint of the line segment \ \\mathrm{AB} \.\n4. Find the position vector of the centroid G of \ \\triangle \\mathrm{ABC} \.'
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Q.14
"A vector is a quantity with both magnitude and direction. This chapter discusses vectors based on directed line segments on a plane, representing vectors with pairs of numbers (components), operations like 'dot product,' and applying vectors to geometry. Understanding the concept of 'linear independence' in vectors is crucial for preparing for future studies in mathematics, physics, economics, and other fields."
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Q.17
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Q.18
'In triangle OAB, let C be the midpoint of OA and D be the point that divides OB in a 1:3 ratio externally. If OA vector is a and 35 times the OB vector is b, find the vector equation of the following line.'
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Q.30
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Q.31
'Let N be the point that divides AB in the ratio of 2:3. Let P be the intersection point of the line segment LM and ON. If a is the vector OA and b is the vector OB, express ON and OP in terms of vectors a and b.'
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Q.33
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Q.34
'Consider a circle C with radius r and center position vector →OA on the coordinate plane with EXO as the origin. Let the position vector of a point P on the circumference be →OP. Also, consider a point B outside circle C with position vector →OB. Furthermore, let Q be the midpoint of points B and P, with position vector →OQ. Define D as the shape traced by point Q as point P moves along the circumference.\n(1) Find the vector equation representing circle C.'
Q.52
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Q.53
'If ∣veca∣=2,∣vecb∣=3,∣2veca−vecb∣=sqrt13|\\vec{a}|=2,|\\vec{b}|=3,|2\\vec{a}-\\vec{b}|=\\sqrt{13}∣veca∣=2,∣vecb∣=3,∣2veca−vecb∣=sqrt13, find the range of real numbers for kkk such that ∣kveca+tvecb∣>sqrt3|k\\vec{a}+t\\vec{b}|>\\sqrt{3}∣kveca+tvecb∣>sqrt3 holds for all real numbers ttt.'
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Q.55
'Consider a plane α determined by three non-collinear points A (vector a), B (vector b), and C (vector c) that are not on the same line. When point P (vector p) lies on the plane α, the following vector equation holds true. Prove this.'
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Q.56
'Point C divides the side \\\mathrm{OA}\ of triangle \\\triangle OAB\ in the ratio 3:1, and point D divides the side \\\mathrm{OB}\ in the ratio 4:1. Let P be the intersection of the segments \\\mathrm{AD}\ and \\\mathrm{BC}\, and Q be the intersection of the segments \\\mathrm{OP}\ and \\\mathrm{AB}\. Given that \\\overrightarrow{\\mathrm{OA}}=\\vec{a}\ and \\\overrightarrow{\\mathrm{OB}}=\\vec{b}\, express \\\overrightarrow{\\mathrm{OP}}\ in terms of \\\vec{a}\ and \\\vec{b}\ and find the ratio of BP to CP. Also, express \\\overrightarrow{\\mathrm{OQ}}\ in terms of \\\vec{a}\ and \\\vec{b}\ and determine the ratio of \\\mathrm{OP}\ to \\\mathrm{PQ}\.'
Q.57
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Q.58
'In the line segment AB, when the direction is specified from point A to point B, it is called directed line segment AB. In the directed line segment AB, A is called its initial point and B is called its terminal point. The length of the line segment AB is called the magnitude or length of the directed line segment AB. Ignoring the difference in position, focusing only on the direction and magnitude is called a vector. Write down the vector represented by the directed line segment AB.'
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Q.60
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Q.61
The condition for point P(\vec{p}) to be on the plane defined by three points A(\vec{a}), B(\vec{b}), and C(\vec{c}) is
CP→=sCA→+tCB→Longleftrightarrowp⃗=sa⃗+tb⃗+uc⃗,s+t+u=1
\overrightarrow{\mathrm{CP}}=s \overrightarrow{\mathrm{CA}}+t \overrightarrow{\mathrm{CB}}\ \ Longleftrightarrow \vec{p}=s \vec{a}+t \vec{b}+u \vec{c}, \ s+t+u=1
CP=sCA+tCBLongleftrightarrowp=sa+tb+uc,s+t+u=1
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Q.62
■ The position vector of the centroid of triangle ABC \mathrm{ABC} ABC, the position vector of the centroid G of triangle ABC, is established as follows. The position vector of the centroid of Triangle ABC, with vertices \( \mathrm{A}(\vec{a}), \mathrm{B}(\vec{b}), \mathrm{C}(\vec{c}) \), is
a⃗+b⃗+c⃗3\frac{\vec{a}+\vec{b}+\vec{c}}{3}3a+b+c
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Q.64
Let lpha=x-2 i and eta=3-6 i . When the two points \( \mathrm{A}(lpha) \) and \( \mathrm{B}(eta) \) lie on the same straight line with the origin O \mathrm{O} O, find the value of the real number x x x.
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Q.67
Parallel Vectors
Two non-zero vectors ec{a}, \ec{b} are said to be parallel if they have the same or opposite direction, and it is written as a⃗/b⃗\vec{a}/\vec{b}a/b. From the definition of real multiples of vectors, the following holds true.
Next, demonstrate that sample vectors ecа= лучший ), \вейвек same and equal to \( 2 \vec {b}。
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Q.73
For the vectors shown in the right diagram, list all pairs of vector numbers that meet the following criteria74
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Q.75
There are fixed points O, A, and a moving point P. Given \overrightarrow{\mathrm{OA}}=ec{a} and \overrightarrow{\mathrm{OP}}=ec{p} , when |6 ec{p}-3 ec{a}|=2 , point P lies on the circumference of a certain circle. Find the center and radius of that circle. Assume ec{a}
eq \overrightarrow{0} .
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Q.76
In the quadrilateral ABCD A B C D ABCD shown on the right, which is a rhombus, point O O O is the intersection of the diagonals AC A C AC and BD \mathrm{BD} BD. Given that OA→=a⃗,AB→=b⃗,CD→=c⃗ \overrightarrow{\mathrm{OA}}=\vec{a}, \overrightarrow{\mathrm{AB}}=\vec{b}, \overrightarrow{\mathrm{CD}}=\vec{c} OA=a,AB=b,CD=c,
(1) Draw the vectors a⃗−b⃗ \vec{a}-\vec{b} a−b and a⃗−c⃗ \vec{a}-\vec{c} a−c. (2) What kind of vector is b⃗+c⃗ \vec{b}+\vec{c} b+c?
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Q.82
Example Question 35 Minimum Magnitude of Vector (Space) Let \(ec{a}=(2,-4,-3)\) and \(ec{b}=(1,-1,1)\). Find the minimum magnitude of ec{a}+t ec{b} (where ttt is a real number) and the value of ttt at that minimum. [Chiba Institute of Technology]
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Q.86
The coordinates of point A are (2,-4), the coordinates of point B are (-2,2), and the coordinates of point C are (0,-4). For the vectors ec{a}, ec{b}, ec{c}, please answer the following questions:
(1) Represent the vectors ec{a}, ec{b}, ec{c} in component form.
(2) Calculate the magnitudes of |ec{a}|,|ec{b}|,|ec{c}|.
Q.90
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Q.91
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Q.92
In each of the following situations, find the angle θ \theta θ between a⃗ \vec{a} a and b⃗ \vec{b} b.
(1) When ∣a⃗∣=2,∣b⃗∣=3,∣2a⃗+b⃗∣=13 |\vec{a}|=2,|\vec{b}|=3,|2 \vec{a}+\vec{b}|=\sqrt{13} ∣a∣=2,∣b∣=3,∣2a+b∣=13
(2) When ∣a⃗∣=2,∣b⃗∣=3 |\vec{a}|=2,|\vec{b}|=\sqrt{3} ∣a∣=2,∣b∣=3 and a⃗−b⃗ \vec{a}-\vec{b} a−b is perpendicular to 6a⃗+7b⃗ 6 \vec{a}+7 \vec{b} 6a+7b
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Q.94
Find the centroid coordinates of points A, B, and C. If the coordinates of point A are (a1, a2, a3), the coordinates of point B are (b1, b2, b3), and the coordinates of point C are (c1, c2, c3), what are the centroid coordinates?
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Q.03
Regarding the vectors shown in the diagram on the right, list all pairs of vector numbers as follows06
Given vectors \vec{a} and \vec{b}. If |\vec{a}| = 2\sqrt{10}, |\vec{b}| = \sqrt{5}, and \vec{a} \cdot \vec{b} = -10, answer the following questions. (1) For any real number t, find the minimum value of |\vec{a} + t\vec{b}| and the value of t at that time. (2) For the value of t obtained in (1), prove that \vec{a} + t\vec{b} is perpendicular to \vec{b}.
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Q.07
In space, a directed line segment from point A to point B, represented by AB→ \overrightarrow{\mathrm{AB}} AB, has a magnitude represented by ∣AB→∣ |\overrightarrow{\mathrm{AB}}| ∣AB∣. Space vectors are also often represented by lowercase letters like ec{a} and ec{b} . Space vectors are defined in exactly the same way as plane vectors. Please answer the following questions about the basic properties of vectors.
1. If ec{a} and ec{b} have the same direction and magnitude, how can they be represented?
2. How is the inverse vector of ec{a} represented?
3. What are the vectors with a magnitude of 0 and a magnitude of 1 called, respectively?
4. Give examples of vector addition, subtraction, and scalar multiplication.
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Q.09
(1) Find the value of x x x such that \( \vec{a}=(x+2,1) \) and \( \vec{b}=(1,-6) \) are perpendicular.
(2) Find the vector d⃗ \vec{d} d, which is perpendicular to \( \vec{c}=(2,1) \) and has a magnitude of 25 2 \sqrt{5} 25.
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Q.10
Find the equation of a line that satisfies the following conditions using vectors:
(1) Passes through point \( \mathrm{A}(-2,3) \) and is parallel to vector \( ec{d}=(2,1) \)
(2) Passes through two points \( \mathrm{A}(-1,2) \) and \( \mathrm{B}(3,1) \)
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Q.13
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Q.14
Find the coordinates of the point P that internally divides the line segment AB in the ratio m:n. Given that the coordinates of point A are (a1, a2, a3) and the coordinates of point B are (b1, b2, b3), what are the coordinates of point P?
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Q.15
The coordinates of the midpoint of line segment AB
left(\raca1+b12,\raca2+b22,\raca3+b32\ \ left(\rac{a_{1}+b_{1}}{2}, \rac{a_{2}+b_{2}}{2}, \rac{a_{3}+b_{3}}{2}left(\raca1+b12,\raca2+b22,\raca3+b32
Q.16
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Q.17
When points O, P, and C are collinear in this order, P is the point closer to O among the intersection points of line OC and sphere S. Given that OC = √(0^2+1^2+2^2) = √5, what is the y-coordinate of point P when points O, P, and C are collinear in this order? | 677.169 | 1 |
Hint: Since the problem is based on applications of trigonometry hence, it is necessary to draw the figure for better visualization of the problem. Here, we will use proper trigonometric identities and ratios and then equate them to get the required dimension of a pillar and find out the solution to a given problem.
Note: In the problems based on the application of trigonometry, apply the required trigonometric ratios to get the accurate solution to the problem. Sometimes, the problem seems to be complex while analyzing the problem hence, it is advised to draw a figure and calculate ratios one by one carefully. | 677.169 | 1 |
Tucker Hexagon
A Tucker hexagon is a hexagon inscribed in a reference triangle that has sides which are alternately parallel and antiparallel to the
corresponding sides of the triangle. Tucker hexagons are always cyclic,
and the corresponding circumscribing circle is called a Tucker
circle.
Thomsen's figure is similar to a Tucker hexagon; Thomsen's hexagon closes after six parallels, while a Tucker hexagon closes after
alternately three parallels and three antiparallels. | 677.169 | 1 |
In this tutorial, we will learn all about similar triangles – what similarity of triangles means, how to identify similar triangles, and also how the concept is useful.
Similar Triangles - Meaning
Two triangles are said to be similar if they have the exact same shape. They may or may not have the same size.
One way to think of similarity is – if one triangle can be turned into another by scaling it up or down (zooming in or out) and adjusting its orientation.
For example, if you flip the green triangle sideways and scale it up by 50% (1.5x), it will overlap perfectly with the yellow one.
How to Identify Similar Triangles
To determine whether two triangles are similar, we have three sets of conditions to check against. If we can prove that the two triangles fulfill any one set of conditions, we can be sure they are similar.
Here are the three criteria.
If you need help with hash marks and bands and terms like corresponding parts or the included angle, I have explained them in the footnotes.
Angle-Angle (AA)
By the AA condition, two triangles are similar if two angles in one triangle are the same as two angles in the other triangle.
Side-Side-Side (SSS)
By the SSS condition, two triangles are similar if the three sides in one triangle are proportional to the three sides in the other triangle.
Side-Angle-Side (SAS)
By the SAS condition, two triangles are similar if two sides in one triangle are proportional to two sides in the other triangle and the included angles (between the two sides in each triangle) are equal.
Similar Triangles - Notation
The symbol for similarity is ~. However, there's more to it than just the symbol.
When we write the names of the two triangles, corresponding vertices must be in the same order/position.
For example, in the figure below vertices A, B, and C correspond to Q, P, and R respectively (corresponding angles are equal).
So the area of △CDE\hspace{0.2em} \triangle CDE \, \hspace{0.2em}△CDE is 277.12in2\hspace{0.2em} \,277.12 \text{ in}^2 \hspace{0.2em}277.12in2.
Similarity and Congruency
Similarity and congruency are two closely related concepts. So let's see what's common and what's different between them.
The idea is simple. Similarity requires two triangles (or any geometric figures) to have exactly the same shape. They may or may not have the same size. Congruency, on the other hand, requires them to have exactly the same shape and size.
So if two triangles are congruent, they must be similar too. But the converse is not true.
And that brings us to the end of this tutorial on similar triangles. Until next time.
Corresponding Parts
Corresponding parts (sides or angles) refer to parts in two triangles that have the same relative position - sort of matching parts.
Here's an example.
In the figure below, the second triangle is a scaled-up (1.5x) and flipped version of the first triangle. Also, if you imagine the transformation, vertex C in the first triangle turned into P in the second, and side BC turned into PR.
So which side in the second triangle corresponds to AB in the first? And which angle corresponds to B?
Well, AB is opposite C. And C and P are corresponding angles. So QR – the side opposite P – must correspond to AB.
Now, side BC corresponds to side PR. And C (one angle on BC) corresponds to P (one angle on PR). So B (the other angle on BC) and R (the other angle on PQ) must be corresponding parts.
In the figure below, the corresponding sides/angles are in the same color.
Included Side/Angle
For any two sides in a triangle (or a polygon), the included angle is the one formed between those sides – at their intersection. For example, here B is the included angle between AB and BC.
Similarly, the included side is the side that's common to two angles. In the figure below, AC is the included side for angles A and C.
Hash Marks and Bands
When comparing two triangles or even two sides or angles within a triangle, hash marks and/or bands are used to show which sides or angles are equal. For example – | 677.169 | 1 |
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Question 16: The area of the triangle formed by the points $A(2,0) B(6,0)$ and $C(4,6)$ is
(a) 24 sq. units
(b) $12 \mathrm{sq}$. units
(c) 10 sq. units
(d) none of these
Solution: Given that points $\mathrm{A}(2,0), \mathrm{B}(6,0)$ and $\mathrm{C}(4,6)$ form a triangle which is shown in the figure. We are asked to find the area of the triangle $\triangle \mathrm{ABC} .$
Given that
$\mathrm{OA}=2$
and $\mathrm{OB}=6$
Hence:
$A B=O B-O A$
$=6-2$
$=4$
$C D=6$
By using formula,
$\Delta A B C=\frac{1}{2} \times A B \times C D$
$=\frac{1}{2} \times 4 \times 6$
$=12$ squnits
Thus the correct answer is (b).
Question 17: The area of the triangle formed by the points $P(0,1), Q(0,5)$ and $R(3,4)$ is
(a) $16 \mathrm{sq}$. units
(b) 8 sq. units
(c) 4 sq. units
(d) 6 sq. units
Solution: Given that the points $\mathrm{P}(0,1), \mathrm{Q}(0,5)$ and $\mathrm{R}(3,4)$ form a triangle.
We are asked to find the area of the triangle $\triangle \mathrm{PQR}$ which is shown in the figure. | 677.169 | 1 |
Step three: Locating the Sin and Cos regarding a certain Position Introduction: Trigonometry. That it Instructable is in the first place intended for brand new ninth people within DIS, but people is actually thank you for visiting realize about Trigonometry. Contained... | 677.169 | 1 |
Demonstrate an understanding of circles by: • describing the relationships among radius, diameter and circumference of circles; • relating circumference to π; • determining the sum of the central angles; • constructing circles with a given radius or diameter; • solving problems involving the radii, diameters and circumferences of circles.
Students who have achieved this outcome should be able to: A. Illustrate and explain that the diameter is twice the radius in a given circle. B. Illustrate and explain that the circumference is approximately three times the diameter in a given circle. C. Explain that, for all circles, π is the ratio of the circumference to the diameter, (C/d), and its value is approximately 3.14. D. Explain, using an illustration, that the sum of the central angles of a circle is 360°. E. Draw a circle with a given radius or diameter with and without a compass. F. Solve a given contextual problem involving circles.
Develop and apply a formula for determining the area of: • triangles; • parallelograms; • circles.
Students who have achieved this outcome should be able to: A. Illustrate and explain how the area of a rectangle can be used to determine the area of a triangle. B. Generalize a rule to create a formula for determining the area of triangles. C. Illustrate and explain how the area of a rectangle can be used to determine the area of a parallelogram. D. Generalize a rule to create a formula for determining the area of parallelograms. E. Illustrate and explain how to estimate the area of a circle without the use of a formula. F. Apply a formula for determining the area of a given circle. G. Solve a given problem involving the area of triangles, parallelograms and/or circles.
Students who have achieved this outcome should be able to: A. Describe examples of parallel line segments, perpendicular line segments, perpendicular bisectors and angle bisectors in the environment. B. Identify line segments on a given diagram that are parallel or perpendicular. C. Draw a line segment perpendicular to another line segment and explain why they are perpendicular. D. Draw a line segment parallel to another line segment and explain why they are parallel. E. Draw the bisector of a given angle using more than one method and verify that the resulting angles are equal. F. Draw the perpendicular bisector of a line segment using more than one method and verify the construction.
Identify and plot points in the four quadrants of a Cartesian plane using integral ordered pairs. Students who have achieved this outcome should be able to: A. Label the axes of a four quadrant Cartesian plane and identify the origin. B. Identify the location of a given point in any quadrant of a Cartesian plane using an integral ordered pair. C. Plot the point corresponding to a given integral ordered pair on a Cartesian plane with units of 1, 2, 5 or 10 on its axes. D. Draw shapes and designs, using given integral ordered pairs, in a Cartesian plane. E. Create shapes and designs, and identify the points used to produce the shapes and designs in any quadrant of a Cartesian plane.
Perform and describe transformations (translations, rotations or reflections) of a 2-D shape in all four quadrants of a Cartesian plane (limited to integral number vertices). Students who have achieved this outcome should be able to: A. Identify the coordinates of the vertices of a given 2-D shape on a Cartesian plane. B. Describe the horizontal and vertical movement required to move from a given point to another point on a Cartesian plane. C. Describe the positional change of the vertices of a given 2-D shape to the corresponding vertices of its image as a result of a transformation or successive transformations on a Cartesian plane. D. Determine the distance between points along horizontal and vertical lines in a Cartesian plane. E. Perform a transformation or consecutive transformations on a given 2-D shape and identify coordinates of the vertices of the image. F. Describe the positional change of the vertices of a 2-D shape to the corresponding vertices of its image as a result of a transformation or a combination of successive transformations. G. Describe the image resulting from the transformations of a given 2-D shape on a Cartesian plane by identifying the coordinates of the vertices of the image.
Solve problems involving percents from 1% to 100%. Students who have achieved this outcome should be able to: A. Express a given percent as a decimal or a fraction. B. Solve a given problem that involves finding a percent. C. Determine the answer to a given percent problem where the answer requires rounding and explain why an approximate answer is needed; e.g., total cost including taxes.
Demonstrate an understanding of the relationship between positive repeating decimals and positive fractions, and positive terminating decimals and positive fractions. Students who have achieved this outcome should be able to: A. Predict the decimal representation of a given fraction using patterns, e.g., 1/11 = 0.09, 2/11 = 0.18, 3/11 = ?, ... B. Match a given set of fractions to their decimal representations. C. Sort a given set of fractions as repeating or terminating decimals. D. Express a given fraction as a terminating or repeating decimal. E. Express a given repeating decimal as a fraction. F. Express a given terminating decimal as a fraction. G. Provide an example where the decimal representation of a fraction is an approximation of its exact value.
Demonstrate an understanding of adding and subtracting positive fractions and mixed numbers, with like and unlike denominators, concretely, pictorially and symbolically (limited to positive sums and differences). Students who have achieved this outcome should be able to: A. Model addition and subtraction of a given positive fraction or a given mixed number using concrete representations, and record symbolically. B. Determine the sum of two given positive fractions or mixed numbers with like denominators. C. Determine the difference of two given positive fractions or mixed numbers with like denominators. D. Determine a common denominator for a given set of positive fractions or mixed numbers. E. Determine the sum of two given positive fractions or mixed numbers with unlike denominators. F. Determine the difference of two given positive fractions or mixed numbers with unlike fractions. G. Simplify a given positive fraction or mixed number by identifying the common factor between the numerator and denominator. H. Simplify the solution to a given problem involving the sum or difference of two positive fractions or mixed numbers. I. Solve a given problem involving the addition or subtraction of positive fractions or mixed numbers and determine if the solution is reasonable.
Students who have achieved this outcome should be able to: A. Order the numbers of a given set that includes positive fractions, positive decimals and/or whole numbers in ascending or descending order, and verify the result using a number of strategies. B. Identify a number that would be between two given numbers in an ordered sequence or on a number line. C. Identify incorrectly placed numbers in an ordered sequence or on a number line. D. Position fractions with like and unlike denominators from a given set on a number line and explain strategies used to determine order. E. Order the numbers of a given set by placing them on a number line that contains benchmarks, such as 0 and 1 or 0 and 5. F. Position a given set of positive fractions, including mixed numbers and improper fractions, on a number line and explain strategies used to determine position.
Demonstrate an understanding of oral and written patterns and their equivalent linear relations. Students who have achieved this outcome should be able to: A. Formulate a linear relation to represent the relationship in a given oral or written pattern. B. Provide a context for the given linear relation that represents a pattern. C. Represent a pattern in the environment using a linear relation
Explain the difference between an expression and an equation. Students who have achieved this outcome should be able to: A. Identify and provide an example of a constant term, a numerical coefficient and a variable in an expression and an equation. B. Explain what a variable is and how it is used in a given expression. C. Provide an example of an expression and an equation, and explain how they are similar and different.
Evaluate an expression given the value of the variable(s). Students who have achieved this outcome should be able to: A. Substitute a value for an unknown in a given expression and evaluate the expression.
Model and solve problems that can be represented by one-step linear equations of the form x + a = b , concretely, pictorially and symbolically, where a and b are integers. Students who have achieved this outcome should be able to: A. Represent a given problem with a linear equation and solve the equation using concrete models, e.g., counters, integer tiles. B. Draw a visual representation of the steps required to solve a given linear equation. C. Solve a given problem using a linear equation. D. Verify the solution to a given linear equation using concrete materials and diagrams. E. Substitute a possible solution for the variable in a given linear equation into the original linear equation to verify the equality.
Students who have achieved this outcome should be able to: A. Model a given problem with a linear equation and solve the equation using concrete models; e.g., counters, integer tiles. B. Draw a visual representation of the steps used to solve a given linear equation. C. Solve a given problem using a linear equation and record the process. D. Verify the solution to a given linear equation using concrete materials and diagrams. E. Substitute a possible solution for the variable in a given linear equation into the original linear equation to verify the equality.
Demonstrate an understanding of central tendency and range by: • determining the measures of central tendency (mean, median, mode) and range; • determining the most appropriate measures of central tendency to report findings.
Students who have achieved this outcome should be able to: A. Determine mean, median and mode for a given set of data, and explain why these values may be the same or different. B. Determine the range of given sets of data. C. Provide a context in which the mean, median or mode is the most appropriate measure of central tendency to use when reporting findings. D. Solve a given problem involving the measures of central tendency.
Determine the effect on the mean, median and mode when an outlier is included in a data set. Students who have achieved this outcome should be able to: A. Analyse a given set of data to identify any outliers. B. Explain the effect of outliers on the measures of central tendency for a given data set. C. Identify outliers in a given set of data and justify whether or not they are to be included in the reporting of the measures of central tendency. D. Provide examples of situations in which outliers would and would not be used in reporting the measures of central tendency.
Construct, label and interpret circle graphs to solve problems. Students who have achieved this outcome should be able to: A. Identify common attributes of circle graphs, such as: • title, label or legend; • the sum of the central angles is 360°; • the data is reported as a percent of the total and the sum of the percents is equal to 100%. B. Create and label a circle graph, with and without technology, to display a given set of data. C. Find and compare circle graphs in a variety of print and electronic media, such as newspapers, magazines and the Internet. D. Translate percentages displayed in a circle graph into quantities to solve a given problem. E. Interpret a given circle graph to answer questions.
Express probabilities as ratios, fractions and percents. Students who have achieved this outcome should be able to: A. Determine the probability of a given outcome occurring for a given probability experiment, and express it as a ratio, fraction and percent. B. Provide an example of an event with a probability of 0 or 0% (impossible) and an event with a probability of 1 or 100% (certain).
Identify the sample space (where the combined sample space has 36 or fewer elements) for a probability experiment involving two independent events. [C, ME, PS] Students who have achieved this outcome should be able to: A. Provide an example of two independent events, such as: • spinning a four-section spinner and an eight-sided die; • tossing a coin and rolling a twelve-sided die; • tossing two coins; • rolling two dice, and explain why they are independent. B. Identify the sample space (all possible outcomes) for each of two independent events using a tree diagram, table or another graphic organizer.
Conduct a probability experiment to compare the theoretical probability (determined using a tree diagram, table or another graphic organizer) and experimental probability of two independent events. Students who have achieved this outcome should be able to: A. Determine the theoretical probability of a given outcome involving two independent events. B. Conduct a probability experiment for an outcome involving two independent events, with and without technology, to compare the experimental probability to the theoretical probability. C. Solve a given probability problem involving two independent events. | 677.169 | 1 |
Private: Learning Math: Geometry
Dissections and Proof Part A: Tangrams (15 minutes)
Session 5, Part A
In this part
Exploring Tangrams
Moving Tangrams
Exploring Tangrams
A tangram is a seven-piece puzzle made from a square. A typical tangram set contains two large isosceles right triangles, one medium isosceles right triangle, two small isosceles right triangles, a square, and a parallelogram.
Use a set of tangrams, or print and cut out the set provided here, to work on the following problems.
Think about the angles you need to create in order for a shape to be a square.
Problem A1
Given that the tangram puzzle is made from a square, can you recreate the square using all seven pieces?
Problem A2
Use all seven tangram pieces to make a rectangle that is not a square.
Part A (continued): Moving Tangrams
One set of tangram pieces can make more than one shape. Geometric language helps you give clear descriptions of how to move pieces to change one shape into another. Roll your cursor over each of the three images below to see an animation of that move.
Slide or Translation
Here, a parallelogram made of two small triangles becomes a square as you slide one of the small triangles parallel to the base of the parallelogram. This type of movement is known as a translation. Try this with your tangram pieces.
Flip or Reflection
Here, the parallelogram becomes a triangle when you flip one of its halves (small triangles) at its horizontal midline. This type of movement is called a reflection. Try it.
Rotation
Here, a square becomes a triangle when you rotate one of its two halves (small triangles) 270° around vertex A. This type of movement is called a rotation. Try it.
Problem A3
For each of the pairs of figures below, do the following:
Build the shape on the left with your tangram set.
Turn it into the shape on the right by reflecting, rotating, or translating one of two of the pieces. (This may take more than one step.)
Write a description, telling which piece or pieces you moved and how you moved them. | 677.169 | 1 |
In the figure below, the circles O and Q intersect at A and B.
A line through A intersects circle O at C and circle Q at D. If M is the
midpoint of OQ prove that circle of center M and radius MA bisects CD at
E. | 677.169 | 1 |
Pythagoras Theorem
Pythagoras Theorem (also called Pythagorean Theorem) is an important topic in Mathematics, which explains the relation between the sides of a right-angled triangle. The sides of the right triangle are also called Pythagorean triples. The formula and proof of this theorem are explained here with examples.
Pythagoras theorem is basically used to find the length of an unknown side and the angle of a triangle. By this theorem, we can derive the base, perpendicular and hypotenuse formulas. Let us learn the mathematics of the Pythagorean theorem in detail here.
Pythagoras Theorem Statement
Pythagoras theorem states that "In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides". The sides of this triangle have been named Perpendicular, Base and Hypotenuse. Here, the hypotenuse is the longest side, as it is opposite to the angle 90°. The sides of a right triangle (say a, b and c) which have positive integer values, when squared, are put into an equation, also called a Pythagorean triple.
History
The theorem is named after a Greek Mathematician called Pythagoras.
Pythagoras Theorem Formula
Consider the triangle given above:
Where "a" is the perpendicular,
"b" is the base,
"c" is the hypotenuse.
According to the definition, the Pythagoras Theorem formula is given as:
Hypotenuse2 = Perpendicular2 + Base2
c2 = a2 + b2
The side opposite to the right angle (90°) is the longest side (known as Hypotenuse) because the side opposite to the greatest angle is the longest.
Consider three squares of sides a, b, c mounted on the three sides of a triangle having the same sides as shown.
By Pythagoras Theorem –
Area of square "a" + Area of square "b" = Area of square "c"
Example
The examples of theorem and based on the statement given for right triangles is given below:
Consider a right triangle, given below:
Find the value of x.
X is the side opposite to the right angle, hence it is a hypotenuse.
Now, by the theorem we know;
Hypotenuse2 = Base2 + Perpendicular2
x2 = 82 + 62
x2 = 64+36 = 100
x = √100 = 10
Therefore, the value of x is 10.
Pythagoras Theorem Proof
Given: A right-angled triangle ABC, right-angled at B.
To Prove- AC2 = AB2 + BC2
Construction: Draw a perpendicular BD meeting AC at D.
Proof:
We know, △ADB ~ △ABC
Therefore,
\(\begin{array}{l}\frac{AD}{AB}=\frac{AB}{AC}\end{array} \)
(corresponding sides of similar triangles)
Or, AB2 = AD × AC ……………………………..……..(1)
Also, △BDC ~△ABC
Therefore,
\(\begin{array}{l}\frac{CD}{BC}=\frac{BC}{AC}\end{array} \)
(corresponding sides of similar triangles)
Or, BC2= CD × AC ……………………………………..(2)
Adding the equations (1) and (2) we get,
AB2 + BC2 = AD × AC + CD × AC
AB2 + BC2 = AC (AD + CD)
Since, AD + CD = AC
Therefore, AC2 = AB2 + BC2
Hence, the Pythagorean theorem is proved.
Note:Pythagorean theorem is only applicable to Right-Angled triangle.
Video Lesson on Pythagoras Theorem
Applications of Pythagoras Theorem
To know if the triangle is a right-angled triangle or not.
In a right-angled triangle, we can calculate the length of any side if the other two sides are given.
To find the diagonal of a square.
Useful For
Pythagoras theorem is useful to find the sides of a right-angled triangle. If we know the two sides of a right triangle, then we can find the third side.
How to use Pythagoras Theorem?
To use Pythagoras theorem, remember the formula given below:
c2 = a2 + b2
Where a, b and c are the sides of the right triangle.
For example, if the sides of a triangles are a, b and c, such that a = 3 cm, b = 4 cm and c is the hypotenuse. Find the value of c.
We know,
c2 = a2 + b2
c2 = 32+42
c2 = 9+16
c2 = 25
c = √25
c = 5 cm
Hence, the length of hypotenuse is 5 cm.
How to find whether a triangle is a right-angled triangle?
If we are provided with the length of three sides of a triangle, then to find whether the triangle is a right-angled triangle or not, we need to use the Pythagorean theorem.
Let us understand this statement with the help of an example.
Suppose a triangle with sides 10cm, 24cm, and 26cm are given.
Clearly, 26 is the longest side.
It also satisfies the condition, 10 + 24 > 26
We know,
c2 = a2 + b2 ………(1)
So, let a = 10, b = 24 and c = 26
First we will solve R.H.S. of equation 1.
a2+ b2 = 102 + 242 = 100 + 576 = 676
Now, taking L.H.S, we get;
c2 = 262 = 676
We can see,
LHS = RHS
Therefore, the given triangle is a right triangle, as it satisfies the Pythagoras theorem.
Related Articles
Pythagorean Theorem Solved Examples
Problem 1: The sides of a triangle are 5, 12 & 13 units. Check if it has a right angle or not.
Solution: From Pythagoras Theorem, we have;
Perpendicular2 + Base2 = Hypotenuse2
P2 + B2 = H2
Let,
Perpendicular (P) = 12 units
Base (B)= 5 units
Hypotenuse (H) = 13 units {since it is the longest side measure}
LHS = P2 + B2
⇒ 122 + 52
⇒ 144 + 25
⇒ 169
RHS = H2
⇒ 132
⇒ 169
⇒ 169 = 169
L.H.S. = R.H.S.
Therefore, the angle opposite to the 13 units side will be a right angle.
Problem 2: The two sides of a right-angled triangle are given as shown in the figure. Find the third side.
Solution: Given;
Perpendicular = 15 cm
Base = b cm
Hypotenuse = 17 cm
As per the Pythagorean Theorem, we have;
Perpendicular2 + Base2 = Hypotenuse2
⇒152 + b2 = 172
⇒225 + b2 = 289
⇒b2 = 289 – 225
⇒b2 = 64
⇒b = √64
Therefore, b = 8 cm
Problem 3: Given the side of a square to be 4 cm. Find the length of the diagonal.
Solution- Given;
Sides of a square = 4 cm
To Find- The length of diagonal ac.
Consider triangle abc (or can also be acd)
(ab)2 +(bc)2 = (ac)2
(4)2 +(4)2= (ac)2
16 + 16 = (ac)2
32 = (ac)2
(ac)2 = 32
ac = 4√2.
Thus, the length of the diagonal is 4√2 cm.
Practice Problems on Pythagoras Theorem
In a right triangle ABC, right-angled at B, the lengths of AB and BC are 7 units and 24 units, respectively. Find AC.
If the length of the diagonal of a square is 10 cm, then find the length of the side of the square.
A triangle is given whose sides are of length 11 cm, 60 cm, and 61 cm. Check whether these are the sides of a right-angled triangle.
Stay tuned with BYJU'S – The Learning App to learn all the important mathematical concepts and also watch interactive videos to learn with ease.
Frequently Asked Questions on Pythagoras Theorem
Q1
What is the formula for Pythagorean Theorem?
The formula for Pythagoras, for a right-angled triangle, is given by; P2 + B2 = H2
Q2
What does Pythagoras theorem state?
Pythagoras theorem states that, in a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.
Q3
What is the formula for hypotenuse?
The hypotenuse is the longest side of the right-angled triangle, opposite to right angle, which is adjacent to base and perpendicular. Let base, perpendicular and hypotenuse be a, b and c respectively. Then the hypotenuse formula, from the Pythagoras statement will be; c = √(a2 + b2)
Q4
Can we apply the Pythagoras Theorem for any triangle?
No, this theorem is applicable only for the right-angled triangle.
Q5
What is the use of Pythagoras theorem?
The theorem can be used to find the steepness of the hills or mountains. To find the distance between the observer and a point on the ground from the tower or a building above which the observer is viewing the point. It is mostly used in the field of construction.
Q6
Can the diagonals of a square be found using Pythagoras theorem?
Yes, the diagonals of a square can be found using the Pythagoras theorem, as the diagonal divides the square into right triangles.
Q7
Explain the steps involved in finding the sides of a right triangle using Pythagoras theorem.
Step 1: To find the unknown sides of a right triangle, plug the known values in the Pythagoras theorem formula.
Step 2: Simplify the equation to find the unknown side.
Step 3: Solve the equation for the unknown side.
Q8
What are the different ways to prove Pythagoras theorem?
There are various approaches to prove the Pythagoras theorem. A few of them are listed below:
Proof using similar triangles
Proof using differentials
Euclid's proof
Algebraic proof, and so on.
Test your Knowledge on Pythagoras TheoremAccording to this theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Check the proof of Basic Proportionality Theorem converse theorem here. | 677.169 | 1 |
This game challenges students to calculate the value of the legs or the hypotenuse of a right triangle. Students must also be able to determine whether a given set of numbers could be the lengths of the sides in a right triangle. SEE MORE
In this interactive, students use logic and mathematical skill to place animals at the correct points on a Cartesian graph representing the cardinal directions. Then, they use the Pythagorean theorem to determine the distances between points. The riddles in the interactive, including one requiring an understanding of rate, have randomized values so that students can place points at different locations and calculate different distances. SEE MORE | 677.169 | 1 |
PROBLEM # 1 Find the equation of the line that passes through the points (-1 , 0) and (-4 , 12).
Solution: y = -4x - 4. FOR GROWNUPS
PROBLEM # 2 A triangle XYZ is equilateral. - all three sides have the same length. Locate a point P so that the sum of the three perpendicular distances from P to the sides of triangle XYZ is as small as possible.
Solution: Any point inside the triangle.
Congratulations to Sitaram Chandawarkar, Somnath Chattopadhyay, Venki Nagesha, Mukund Seshadri and Niharika Yerneni, who were winners of the last set of puzzles. | 677.169 | 1 |
What is the measurement of each of the exterior angles of a regular pentegon?
What is a irregular pentegon?
An irregular pentagon is a closed 2-dimensional shape formed by
five straight sides such that at least one of the sides is of a
different length form the others or at least one of the angles is
different from the others. | 677.169 | 1 |
...angle AFE is equal to the angle BFE. [I. 8. But 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 ; [I. Definition 10. therefore each of the angles AFE, BFE is a right...
...; name also two vertically opposite angles. H 7. 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 each of the lines is perpendicular to the other, or at right angles...
...named from three or one letters, as ABC or B. X. 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...
...the following definition of a right angle : — ' 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...
...instead of it. See Note 2 on Prop. 9. DEF. 10. — 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 said to be perpendicular,...same straight line as in Fig. 1. Fig. 1. Fig. 2. 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...
...same straight line as in Fig. 34. Fi(. 34. Fig 35. 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...
...are not in the same straight line as in Fig. 34. 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... | 677.169 | 1 |
The Elements of Euclid with Many Additional Propositions and Explanatory Notes
DEMONSTRATION. Describe the circle ACB about the triangle (a), and draw its diameter AE, and join EC: because the right angle BDA is equal to the angle ECA in a semicircle (6), and the angle ABD equal to the angle AEC in the same segment (c); the triangles ABD, AEC are equiangular: therefore, as BA is to AD, so is EA to AC (d); and consequently the rectangle BA, AO is equal to the rectangle EA, AD (e).
PROPOSITION D.
THEOREM.-The rectangle under the diagonals of a quadrilateral figure inscribed in a circle, is equal to both the rectangles contained by its opposite sides.
DEMONSTRATION. Let ABCD be any quadrilateral figure inscribed in a circle, and join AC, BD: the rectangle contained by AC, BD shall be equal to the two rectangles contained by AB, CD, and by AD, BC.
Make the angle ABE equal to the angle DBC (a); add to each of these the common angle EBD, then the angle ABD is equal to the angle EBC: and the angle BDA is equal to the angle BCE, because they are in the same segment (6); therefore the triangle ABD is equiangular to the triangle BCE: wherefore, as BC is to CE, so is BD to DÁ (c); and consequently the rectangle BC, AD is equal to the rectangle BD, CE (d): again, because the angle ABE is equal to the angle DBC, and the angle BAE to the angle BDC (6), the triangle ABE is equiangular to the triangle BCD; therefore as BA is to AE, so is BD
B
I. 23. (b) III. 21. (c) VI. 4. (d) VI. 16.
to DC (c); wherefore the rectangle BA, DC is equal to the rectangle BD, AE (đ): but the rectangle BC, AD has been shown equal to the rectangle BD, CE; therefore the rectangles BC, AD, and BA, DC are together equal to the rectangles BD, CE, and BD, AE; that is, to the whole rectangle BD, AC (e); therefore the whole rectangle AC, BD is equal to the rectangle AB, DC, together with the rectangle AD, BC.
SCHOLIUM. This proposition is a Lemma of Cl. Ptolomæus, in page 9 of the Meydan Zuvragis, or " Great Construction."
THE
ELEMENTS OF EUCLID.
BOOK XI.
DEFINITIONS.
1. A SOLID is a magnitude, having length, breadth, and thickness. COROLLARY. All solids are bounded by superficies, or surfaces.
2. A straight line AB is said to be perpendicular to a plane, when it makes right angles with all straight lines which meet it in that place.
A
3. A plane is said to be perpendicular to a plane, when any straight line AB, drawn in one of the planes perpendicular to the common section of the two planes, is perpendicular to the other plane.
SCHOLIUM. The common section of two planes is the line in which they mutually cut or intersect each other.
4. The inclination of a straight line AC to a plane is the acute angle Č formed by that straight line, and another CB drawn from the point C, in which the first line meets the plane, to the point B in which a perpendicular AB to the plane drawn from any point A of the first line above the plane, meets the same plane.
A
5. The inclination of one plane to another is the acute angle ABC, formed by two straight lines drawn from any the same point B of their common section at right angles to it, one AB upon one plane, and the other BC upon the other plane.
C
6. PARALLEL PLANES are such as do not meet one another, though produced ever so far in every direction.
7. A SOLID ANGLE is that which is made by the meeting in one point of more than two plane angles, which are not in the same plane.
9. SIMILAR SOLID FIGURES are such as have all their solid angles equal, each to each, and are contained by the same number of planes similarly situated.
10. A PYRAMID is a solid figure contained by planes that are constituted between one plane figure and a point above it.
SCHOLIUM. The last-named plane figure is called the base, and the point above it the vertex of the pyramid; and all the planes meeting together in the vertex are triangles. The altitude of a pyramid is the perpendicular drawn from its vertex to its base.
11. A PRISM is a solid figure contained by plane figures, of which two that are opposite are equal, similar, and parallel to one another; and the others are parallelograms.
SCHOLIA. 1. The opposite ends are termed the bases of the prism, and the parallelograms its sides; but the term base is sometimes applied to any side upon which it is supposed to
stand. The altitude of a prism is a perpendicular from one of its ends or bases to the other.
2. A prism, the ends or bases of which are perpendicular to its sides, is said to be a right prism; any other is an oblique prism.
3. Pyramids and prisms are said to be triangular, quadrangular, pentagonal, or polygonal, according as their bases are triangles, quadrangles, pentagons, or polygons.
12. A SPHERE is a solid figure described by the revolution of a semicircle (ABC) about its diameter (AC), which remains unmoved.
13. The AXIS OF A SPHERE is the fixed straight line (AC) about which the semicircle revolves.
14. The CENTER OF A SPHERE is the same with that of the generating semicircle.
15. The DIAMETER OF A SPHERE is any straight line which passes through its center, and is terminated both ways by the superficies of the sphere."
16. A CONE is a solid figure described by the revolution of a right-angled triangle about one of the sides containing the right angle, which side remains fixed. If the fixed side (AB) be equal to the other side containing the right angle (CB), the cone is said to be right-angled; if it (DF) be less than the other side (EF), obtuse-angled; and if greater (as GH and HI) acute-angled.
17. The AXIS OF A CONE is the fixed straight line about which the triangle revolves.
18. The BASE OF A CONE is the circle described by that side containing the right angle, which revolves.
19. A CYLINDER is a solid figure described by the revolution of a right-angled parallelogram (ABC) about one of its sides (AB), which remains fixed.
20. The AXIS OF A CYLINDER is the fixed straight line (AB) about which the parallelogram revolves.
21. The BASES OF A CYLINDER are the circles described by the two revolving opposite sides of the parallelogram. | 677.169 | 1 |
1. Begin by coloring the triangles in the different colors you have chosen. 2. Look at the angles and lengths of each triangle to determine if they are | 677.169 | 1 |
Angle Practice solving problems using angle relationships with this seventh grade geometry worksheet Using their knowledge of complementary supplementary vertical and adjacent angles students will need to set up and solve 12 unique equations in order to find the value of the given variable in each problem
Grade 7 Angle Relationships 7 G B 5 Teaching Identifying the relationship between angles Vertical Angles The angles that are opposite to each other when two lines cross each other Transversal When two straight lines are crossed by another line the angles formed in matching corners are known as transversal Under the 7th grade angle relationships worksheet students can learn about different angles and about how to draw relationships between these angles Here s why you should use 7th grade angle relationships worksheets for your students It helps students to understand the concept of angles and identify relationships between lines and angles
More picture related to Angle Relationships Worksheet 7th Grade Pdf
Eighth Grade I NEED HELP WITH MATH
Angle Relationships Date Period Name the relationship complementary supplementary vertical or adjacent 1 a b vertical 2 a b supplementary 3 a b vertical 4 a b complementary 5 a b complementary 6 a b adjacent Name the relationship alternate interior corresponding or alternate exterior 7 a b corresponding 8 a b Angle Relationships NOTES 9 What are vertical angles a Angles that add up to 180 c Angles that add up to 360 b Angles that add up to 90 d Angles that are opposite of each other when lines intersect and are equal 3 7 19 Two angles are complementary if the sum of their angles is 180o a True 20 Vertical angles are always acute False
Use this one page geometry handout to help students determine angle relationships with confidence and ease This helpful resource describes and illustrates four different angle relationships complementary angles supplementary angles vertical angles and adjacent angles Students will learn the properties of the four angle relationships and Explore math program Angles Worksheets for grade 7 Worksheets prove to be the best resources to refine concepts through various types of questions Find exciting 7th Grade Angles worksheets here
Angle Relationships Worksheet 7th Grade Pdf - Angle | 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.
The angles of a triangle total 180o.
Position (Year 6)
In KS2 Maths, Year Six is when you get super good at playing with positions and directions! You'll know all about degrees and be a pro at spotting acute, obtuse, and right angles. Did you know a triangle has 180 degrees? Cool, right?
Now, let's talk about coordinates on a grid – it's like playing treasure hunt on a map! You'll learn how to describe where things are on the grid and even figure out what happens when you turn or flip shapes. It's like magic!
Knowing positions means reading coordinates, which are like secret codes for maps. Imagine being a map explorer! But, oh, figuring out new coordinates after turning a shape can be a bit tricky – like a puzzle you solve with a pencil and paper!
Test your brain with a fun quiz all about positions – let's see how much you remember from your super cool maths adventures! | 677.169 | 1 |
ANGLE PAIR RELATIONSHIP S Angles An angle is
Angles An angle is a figure formed by two noncollinear rays that have a common endpoint. Symbols: D Definition of Angle E DEF FED 2 E F 2
Angles 1) Name the angle in four ways. ABC C A CBA 1 B 2) Identify the vertex and sides of this angle. vertex: Point B sides: BA and BC
Angles 1) Name all angles having W as their vertex. X 1 2 W 1 2 XWZ Y 2) What are other names for XWY or 1 ? YWX 3) Is there an angle that can be named No! Z W?
Adjacent Angles When you "split" an angle, you create two angles. The two angles are called _______ adjacent angles adjacent = next to, joining. A B D 2 1 1 and 2 are examples of adjacent angles. They share a common ray. Name the ray that 1 and 2 have in common. C ____
Adjacent Angles Determine whether 1 and 2 are adjacent angles. No. They have a common vertex B, but no common side _______ 2 1 B 1 Yes. They have the same vertex G and a common side with no interior points in common. 2 G N L J 2 1 No. They do not have a common vertex or a______ common side The side of 1 is ____ The side of 2 is ____
Linear Pairs of Angles Two angles form a linear pair if and only if (iff): A) they are adjacent and B) their noncommon sides are opposite rays A D B 1 Definition of Linear Pairs C 2 1 and 2 are a linear pair.
Linear Pairs of Angles In the figure, and are opposite rays. 1) Name the angle that forms a linear pair with 1. ACE H T A 2 1 ACE and 1 have a common side the same vertex C, and opposite rays 3 4 C M and 2) Do 3 and TCM form a linear pair? Justify your answer. No. Their noncommon sides are not opposite rays. E
Complementary and Supplementary Angles If two angles are complementary, each angle is a complement of the other. ABC is the complement of DEF and DEF is the complement of ABC. E A B D 30° C 60° F Complementary angles DO NOT need to have a common side or even the same vertex.
Complementary and Supplementary Angles If the sum of the measure of two angles is 180, they form a special pair of angles called supplementary angles. Two angles are supplementary if and only if (iff) the sum of their degree measure is 180. D C Definition of Supplementary Angles 50° A 130° B E F m ABC + m DEF = 50 + 130 = 180
Congruent Angles To show that 1 is congruent to 2, we usearcs ____. 1 2 To show that there is a second set of congruent angles, X and Z, we use double arcs. This "arc" notation states that: X Z X Z m X = m Z | 677.169 | 1 |
Vectors addition online
Vectors addition
can be accomplished in several different ways, depending on their representation form. If vectors represented by coordinates, their sum is also the vector with coordinates which are the sum of corresponding coordinates of initial vectors. If vectors are given in geometrical form, to find their sum one can use parallelogram rule, which shown on the picture below.
From the picture it follows, that one need to match the initial points of vectors being summed, and on their final points - complete corresponding sides of parallelogram. The sum of the vectors is the vector which is the parallelogram's diagonal.
Our calculator can find
step by step solution
for the sum of vectors in coordinate form. As vector's coordinates, one can use coordinates of initial and final points of vector. To find vectors sum one should choose dimension, representation form and input vectors coordinates. As coordinates of vector one can use not only numbers (
ect.) and fractions (
ect.), but also the parameters (
ect.) | 677.169 | 1 |
Tell Harmal Mathematical Tablets
Left- Algebraic-geometrical tablet involving triangles described by a perpendicular drawn from the right angle to the hypotenuse (similar to Euclid's theory), from Tell Harmal, datable to the early 2nd millennium BCE.
Right- Clay tablet deciphered as an algebraic-geometrical problem involving a rectangle whose diagonal and area are given and it is required to find its length and width. It is solved in the same way reputed as the Pythagorean theorem, found at Tell Harmal, datable to the early 2nd millennium BCE. | 677.169 | 1 |
Midpoint And Distance Formula Worksheet
Exploring the Basics of the Midpoint and Distance Formula Worksheet
The midpoint and distance formula are two of the most basic and important formulas in geometry. Understanding these formulas can be the difference between getting a good grade in a math class and struggling to keep up. For that reason, it is essential to explore the basics of the midpoint and distance formula worksheet in order to gain a better understanding of these formulas.
The midpoint formula is used to find the exact center point between two given points. To use this formula, all one needs to do is take the x and y coordinates of the two points and add them together and then divide by two. Doing this will give the exact midpoint between the two points.
The distance formula is used to find the exact distance between two points. To use this formula, all one needs to do is take the x and y coordinates of the two points and subtract them from each other and then square each of the differences. After this, one must then take the square root of the sum of the squared differences. Doing this will give the exact distance between the two points.
Both of these formulas are incredibly useful when it comes to solving many types of problems in geometry. By understanding the basics of the midpoint and distance formula worksheet, one will be able to use them to solve a variety of problems quickly and accurately.
In conclusion, it is essential to understand the basics of the midpoint and distance formula worksheet in order to gain a better understanding of these formulas. Doing so will enable one to solve a variety of problems quickly and accurately.
How to Use the Midpoint and Distance Formula Worksheet to Solve Complex Math Problems
The midpoint and distance formula worksheet is a powerful tool for solving complex math problems. It provides students with the ability to work out the midpoint and distance between two points, as well as the length of a line segment. This worksheet can be used to solve a variety of problems from geometry, algebra, and calculus.
When solving a problem with the midpoint and distance formula worksheet, the first step is to determine the coordinates of the two points. The coordinates of the first point should be written in the designated area for the first point on the worksheet. The coordinates of the second point should be noted in the designated area for the second point. Once these coordinates are noted, the midpoint and distance between the two points can be calculated.
To calculate the midpoint, the coordinates of both points should be added together and then divided by two. The result of this calculation is the midpoint. To calculate the distance between the two points, the coordinates of the two points should be subtracted from each other and then the result should be squared. The result of this calculation is the distance.
The midpoint and distance formula worksheet can be used to help students understand how to solve more complex math problems. For example, it can be used to calculate the length of a line segment or to calculate the area of a triangle. By understanding how to use the midpoint and distance formula worksheet, students will be able to quickly and accurately solve complex math problems.
The midpoint and distance formula worksheet can also be used to help students understand how to graph equations. By understanding how to use the midpoint and distance formula worksheet, students will be able to quickly and accurately graph equations. This will allow them to better understand the relationship between the points on the graph and the equation.
The midpoint and distance formula worksheet can also be used to help students understand how to solve equations. By understanding how to use the midpoint and distance formula worksheet, students will be able to quickly and accurately solve equations. This will help them better understand the relationship between the equations and the points on the graph.
Overall, the midpoint and distance formula worksheet is a powerful tool for solving complex math problems. By understanding how to use this worksheet, students will be able to quickly and accurately solve a variety of problems from geometry, algebra, and calculus. This worksheet can also be used to help students understand how to graph equations, solve equations, and understand the relationship between the equations and the points on the graph.
Tips and Tricks for Making the Most of the Midpoint and Distance Formula Worksheet
1. Before beginning the worksheet, it is important to ensure that students have a thorough understanding of the midpoint and distance formula. To ensure this understanding, students should review any relevant notes or practice problems prior to beginning the worksheet.
2. When solving the problems on the worksheet, students should take their time and be sure to double-check their work. It is important to be thorough and accurate when solving these types of problems.
3. When solving problems, students should draw diagrams to help them visualize the problems and better understand how to apply the midpoint and distance formula.
4. As students work through the worksheet, they should take the time to explain their work and reasoning out loud. This will help them to better understand the concepts and strengthen their problem-solving skills.
5. When students encounter questions they don't understand, they should take the time to ask their teacher or classmates for help. This will ensure that they have a better understanding of the material and can complete the worksheet successfully.
6. After completing the worksheet, students should review their answers and double-check their work. This will help them to identify any mistakes they have made and understand the material better.
By following these tips and tricks, students can make the most of the midpoint and distance formula worksheet and ensure that they have a thorough understanding of the material.
Step-by-Step Guide to Working with the Midpoint and Distance Formula Worksheet
Step 1: Read the introduction to the Midpoint and Distance Formula worksheet. This will introduce you to the concepts of midpoints and distances, as well as provide an example of how to use the worksheet.
Step 2: Carefully review the questions given on the worksheet. This will help you understand which parts of the worksheet need your attention.
Step 3: Solve the midpoint and distance questions given on the worksheet. The midpoint questions involve finding the coordinates of the midpoint of two given points, while the distance questions involve finding the distance between two given points.
Step 4: Read the explanation for each question given on the worksheet. This will help you understand the answer to each question.
Step 5: Check your answers. Make sure that your answers are correct and that you have followed all the steps given in the worksheet.
Step 6: Review the worksheet. This will help you understand how the midpoint and distance formulas work and how to use them in real-world scenarios.
Step 7: Take the time to practice the midpoint and distance formulas by working through the examples given in the worksheet. This will help you become familiar with the formulas and how to use them correctly.
Step 8: If you have any questions, reach out to your instructor or a math tutor for help. This will ensure that you understand the concepts and can confidently use the midpoint and distance formulas in the future.
Conclusion
The Midpoint and Distance Formula Worksheet is a great tool for students to practice applying the midpoint and distance formulas to find the midpoint and distance between two points. By practicing with this worksheet, students can become more comfortable with the formulas and will be better prepared to use them in real-world applications. | 677.169 | 1 |
Trigonometry Problem Solver
This tool combines the power of mathematical computation engine that excels at solving mathematical formulas with the power of artificial intelligence large language models to parse and generate natural language answers. This creates a math problem solver that's more accurate than ChatGPT, more flexible than a math calculator, and provides answers faster than a human tutor.
Problem Solver Subjects
Our math problem solver that lets you input a wide variety of trigonometry math problems and it will provide a step by step answer. This math solver excels at math word problems as well as a wide range of math subjects.
Here are example math problems within each subject that can be input into the calculator and solved. This list is constanstly growing as functionality is added to the calculator.
Basic Math Solutions
Below are examples of basic math problems that can be solved.
Long Arithmetic
Rational Numbers
Operations with Fractions
Ratios, Proportions, Percents
Measurement, Area, and Volume
Factors, Fractions, and Exponents
Unit Conversions
Data Measurement and Statistics
Points and Line Segments
Math Word Problem Solutions
Math word problems require interpreting what is being asked and simplifying that into a basic math equation. Once you have the equation you can then enter that into the problem solver as a basic math or algebra question to be correctly solved. Below are math word problem examples and their simplified forms.
Word Problem: Rachel has 17 apples. She gives some to Sarah. Sarah now has 8 apples. How many apples did Rachel give her?
Simplified Equation: 17 - x = 8
Word Problem: Rhonda has 12 marbles more than Douglas. Douglas has 6 marbles more than Bertha. Rhonda has twice as many marbles as Bertha has. How many marbles does Douglas have?
Variables: Rhonda's marbles is represented by (r), Douglas' marbles is represented by (d) and Bertha's marbles is represented by (b)
Simplified Equation: {r = d + 12, d = b + 6, r = 2 �� b}
Word Problem: if there are 40 cookies all together and Angela takes 10 and Brett takes 5 how many are left?
Simplified: 40 - 10 - 5
Pre-Algebra Solutions
Below are examples of Pre-Algebra math problems that can be solved.
Variables, Expressions, and Integers
Simplifying and Evaluating Expressions
Solving Equations
Multi-Step Equations and Inequalities
Factors, Fractions, and Exponents
Operations with Fractions
Rational Numbers
Ratios, Proportions, and Percents
Points and Line Segments
Linear Equations and Inequalities
Unit Conversions
Measurement, Area, and Volume
Data Measurement and Statistics
Algebra Solutions
Below are examples of Algebra math problems that can be solved.
Algebra Concepts and Expressions
Points, Lines, and Line Segments
Simplifying Polynomials
Factoring Polynomials
Linear Equations
Absolute Value Expressions and Equations
Radical Expressions and Equations
Systems of Equations
Quadratic Equations
Inequalities
Relations
Functions
Complex Numbers and Vector Analysis
Logarithmic Expressions and Equations
Exponential Expressions and Equations
Matrices
Conic Sections
Vectors
Vector Spaces
3d Coordinate System
Eigenvalues and Eigenvectors
Linear Transformations
Number Sets
Tables
Analytic Geometry
Trigonometry Solutions
Below are examples of Trigonometry math problems that can be solved.
Algebra Concepts and Expressions Review
Right Triangle Trigonometry
Radian Measure and Circular Functions
Graphing Trigonometric Functions
Simplifying Trigonometric Expressions
Verifying Trigonometric Identities
Solving Trigonometric Equations
Complex Numbers
Analytic Geometry in Polar Coordinates
Exponential and Logarithmic Functions
Vector Arithmetic
Vectors
Precalculus Solutions
Below are examples of Precalculus math problems that can be solved.
Algebra Concepts and Expressions
Relations
Functions
Operations on Functions
Points, Lines, and Line Segments
Absolute Value Expressions and Equations
Rational Expressions and Equations
Polynomial and Rational Functions
Factoring Polynomials
Conic Sections
Exponential and Logarithmic Functions
Trigonometry
Analytic Trigonometry
Inequalities
Linear Equations
Systems of Equations
Quadratic Equations
Matrices
Sequences and Series
Analytic Geometry in Rectangular Coordinates
Analytic Geometry in Polar Coordinates
Limits and an Introduction to Calculus
Vectors
Number Sets
Calculus Solutions
Below are examples of Calculus math problems that can be solved.
Algebra Concepts and Expressions
Functions
Operations on Functions
Polynomial and Rational Functions
Exponential and Logarithmic Functions
Sequences and Series
Evaluating Limits
Derivatives
Applications of Differentiation
Integrals
Applications of Integration
Techniques of Integration
Parametric Equations and Polar Coordinates
Differential Equations
Statistics Solutions
Below are examples of Statistics problems that can be solved.
Algebra Review
Average Descriptive Statistics
Dispersion Statistics
Probability
Probability Distributions
Frequency Distribution
Normal Distributions
t-Distributions
Hypothesis Testing
Estimation and Sample Size
Correlation and Regression
Finite Math Solutions
Below are examples of Finite Math problems that can be solved.
Polynomials and Expressions
Ratios, Proportions, and Percents
Equations and Inequalities
Linear Functions and Points
Functions
Relations
Matrices
Systems of Linear Equations
Mathematics of Finance
Average Descriptive Statistics
Dispersion Statistics
Statistical Distributions
Frequency Distribution
Normal Distributions
t-Distributions
Hypothesis Testing
Estimation and Sample Size
Correlation and Regression
Vectors
Linear Algebra Solutions
Below are examples of Linear Algebra math problems that can be solved.
Introduction to Matrices
Complex Numbers
Matrices
Systems of Linear Equations
Vectors
Linear Independence and Combinations
Vector Spaces/li>
Eigenvalues and Eigenvectors
Linear Transformations
Number Sets
Chemistry Solutions
Below are examples of Chemistry problems that can be solved.
Unit Conversion
Atomic Structure
Molecules and Compounds
Chemical Equations and Reactions
Behavior of Gases
Solutions and Concentrations
Physics Solutions
Below are examples of Physics math problems that can be solved.
Static Equilibrium
Dynamic Equilibrium
Kinematics Equations
Electricity
Energy
Waves
Thermodymanics
Geometry Graphing Solutions
Below are examples of Geometry and graphing math problems that can be solved. | 677.169 | 1 |
The flattened angle is divided into three parts so that the first is 3 times the second and 2
The flattened angle is divided into three parts so that the first is 3 times the second and 2 times smaller than the third. Find the degree measure of each angle.
Let the second angle be x, then the first is 3x, and the third is 2 * 3x or 6x. You can make an equation
x + 3x + 6x = 180 °.
Here are similar
10x = 180 °;
x = 180 ° / 10.
We calculate the unknown
x = 18 ° – the first angle.
3 * 18 ° = 54 ° is the second angle.
6 * 18 ° = 108 ° is the third angle.
Answer: 18 °, 54 °, 108 | 677.169 | 1 |
Geometry
Using the vector cross product formula and combining it with the trigonometric dot product, we find that we can express the sine of the non-reflex angle between two vectors in terms of the cross product. Another way of thinking about it is that the magnitude of the cross product is related to the sine […]
Students in their senior high school year often learn the formula for the cross product of two three-dimensional vectors by heart. In this video, we derive the cross product formula from first principles and show an amazing way of remembering it using the determinant of a formal matrix.
The dot product is a natural function that can be defined on a pair of Euclidean vectors. Interestingly, it can be combined with the cosine law to provide insight into the angle (or that angle's explementary angle) between two vectors. Thus, we can derive what can be called the trigonometric form of the dot
After the regular tetrahedron, the regular octahedron is the next simplest of the Platonic solids. In fact, the octahedron can be constructed by gluing together two square-based pyramids on their square faces. In this video, we present and prove simple formulas for calculating the surface area and volume of a regular octahedron in terms
The regular tetrahedon is the simplest of the Platonic solids, otherwise known as regular polyhedra. In this video, show how to find the surface area and volume of a regular tetrahedron in terms of its edge length. The formulas $$V=frac{s^3}{6sqrt{2}} text{ and } S = sqrt{3} s^2$$ are surprisingly simple!
The Pythagorean theorem tells us how to compute the length of the hypotenuse of a right triangle in terms of its legs. The 3-dimensional analogue of this problem is to determine the distance between two opposite corners of a rectangular prism. Such a line segment is called a space diagonal and we show that
Almost everyone knows that the area of a triangle is base times height divided by 2. But there are numerous other ways of finding the area of a triangle. For example, if we are given only the three sides, we could use Heron's formula. In this video, we show Heron, along with methods involving
The law of sines is well-known, but its proof is not. It turns out that the expressions in the law of sines can be proven to all be equal to twice the circumradius, which not only proves the law of sines but extends it. We prove this result here, which says that $$frac{a}{sin A}=frac{b}{sin
Interestingly, it is possible to explicitly compute the lengths of the diagonals of any cyclic quadrilateral in terms of the side lengths of the cyclic quadrilateral. We derive these formulas here. This is not possible for arbitrary quadrilaterals.
Stewart's theorem gives us an explicit formula for the length of a triangle cevian in terms of the ratio in which it splits the opposite side and the side lengths of the triangle. It can be applied to find formulas for the lengths of well-known cevians such as medians and angle bisectors. We prove | 677.169 | 1 |
Which quadrilateral can have exactly 2 right angles?
Which quadrilateral can have exactly 2 right angles?
trapezoid
The quadrilateral that can have only two right angles is a trapezoid. Not all trapezoids have right angles, but we can construct one that does.
Can you draw a quadrilateral with four right angles?
Rectangles. A rectangle is one kind of quadrilateral with four right angles. The definition of a rectangle is a shape with four sides and four right angles. This means each angle in a rectangle measures 90 degrees.
Is there a shape with only two angles?
A regular digon has both angles equal and both sides equal and is represented by Schläfli symbol {2}. It may be constructed on a sphere as a pair of 180 degree arcs connecting antipodal points, when it forms a lune. The digon is the simplest abstract polytope of rank 2. A truncated digon, t{2} is a square, {4}.
Can a kite have 2 right angles?
Thus the right kite is a convex quadrilateral and has two opposite right angles. If there are exactly two right angles, each must be between sides of different lengths.
Can a parallelogram have 2 right angles?
A parallelogram is a quadrilateral with 2 pair of opposite sides parallel. However, a trapezoid could have one of the sides connecting the two parallel sides perpendicular to the parallel sides which would yield two right angles.
Which shape has only right angles?
Squares, rectangles, and right triangles all have right angles.
Does a diamond have 4 right angles?
They'll probably say you have a diamond on your wall. But a diamond also has four equal sides and right angles at the corners.
Can a parallelogram have exactly 2 right angles?
A parallelogram is a quadrilateral with 2 pair of opposite sides parallel. A square is a special rectangle that has all four sides congruent. A kite has two consecutive sides congruent. The angle between these two sides could be a right angle, but there would only be one right angle in the kite.
Are opposite angles in a kite equal?
Kite. A kite has two pairs of equal sides. It has one pair of equal angles.
What shape has all right angles?
A rectangle is a four-sided shape where every angle is a right angle (90°). Also opposite sides are parallel and of equal length. A rhombus is a four-sided shape where all sides have equal length.
What shapes have one right angle?
The only regular shape that has right angles is a square. A regular shape is one where all of its sides are the same length and all of its angles are the same angle.
What shapes are quadrilaterals?
Yes, it is. Quadrilaterals are four (quad) -lined (lateral) shapes. Quadrilaterals include squares, rectangles, rhombuses, diamonds, trapeziums and trapezoids. All shapes wi…th four sides are quadrilaterals and they all have four angles inside the | 677.169 | 1 |
সমদ্বিবাহু ত্রিভুজ সমাচার
In $\triangle ABC$ with $AB=AC$, point $D$ lies strictly between $A$ and $C$ on side $\overline{AC}$, and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC$. The degree measure of $\angle ABC$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 677.169 | 1 |
How to work with Shapes: Part 1
We all know our basic shapes, like squares, circles and triangles. However, scientists are still discovering new shapes, and there are other unusual shapes you may not know have names!
In 2018, a group of scientists modelled a new shape called the scutoid. Epithelial cells form protective layers in our bodies, for example around organs and in our skin. To do this they must pack together closely so there are no gaps between them. Where there is a need for curved layers, 3D shapes such as cuboids leave gaps. As we, and our organs are not completely flat – there must be a different shape to allow this to happen. Scientists named this shape the scutoid, after a part of the beetle's thorax, which looks similar to it.
You will definitely have seen a squircle. It is a cross between a circle and a square. Basically, they are squares with round edges.
Or what about a Reuleaux triangle? They look just like triangles, apart from having curved sides. These shapes are used to make wheels and pencils – as they provide a better grip!
Our "How to work with…" guide this week is all about shapes. If you need to know what a parallelogram or trapezium is, or how to tell the difference between an isosceles and scalene triangle – this is the guide for you. It describes the properties of regular 2D shapes, including lines of symmetry and rotational symmetry. There are some questions to try, and answers are included so you can | 677.169 | 1 |
In Class 6 ICSE Mathematics, the topic of construction involves creating various geometric figures using a compass, ruler, and other tools. It includes constructions like drawing perpendicular bisectors, angle bisectors, constructing triangles, and more. These constructions are fundamental in understanding geometric concepts and developing problem-solving skills. | 677.169 | 1 |
Exploring Isosceles Triangle Angles | Evaluating Statement Options
If isosceles triangle ABC has a 130° angle at vertex B, which statement must be true?
In an isosceles triangle, two sides have equal lengths, and consequently, two angles are also equal
In an isosceles triangle, two sides have equal lengths, and consequently, two angles are also equal. Let's use this information to evaluate the statement options.
Option 1: Angle A is also 130°.
This statement cannot be true because in an isosceles triangle, the base angles (the angles opposite the equal sides) are equal, but they are not necessarily equal to the vertex angle.
Option 2: Angle A is 25°.
This statement cannot be true because in an isosceles triangle, the base angles are equal, meaning that if one base angle is 25°, the other base angle should also be 25°. However, the vertex angle being 130° means that the base angles are larger.
Option 3: Angle C is 25°.
This statement must be true. Since angles A and C are the base angles of the isosceles triangle, they are congruent. Therefore, if angle A is not 130°, it must be smaller. This means that angle C, being one of the base angles, should also be smaller. The only option provided that satisfies this condition is angle C being 25° | 677.169 | 1 |
Question ID - 55871 | SaraNextGen Top Answer
1 Answer
127 votes
Answer Key / Explanation : (c) -
(c)
represents interior region of circle, where on taking any two points the midpoint of that segment will also lie inside that circle
and
Which represents the interior region of a square with its sides and in which for any two points, their midpoint also lies inside the region
represents the exterior region of hyperbola in which we take two points (4, 3) and . Then their midpoint (4, 0) does not lie in the same region (as shown in the figure)
represents interior region of parabola in which for any two points, their midpoint also lies inside the region | 677.169 | 1 |
Measuring And Naming Angles WorksA protractor can be used by students to practice measuring Even the slightest difference can change the answer. It is important to practice as often as you can. These skills can be learned with the help of many online worksheets.
Avoid angles that aren't right
Angles that are not quite right can be problematic. They can cause a discontinuity of traced signal integrity. An acute angle's corner contains more copper and the impedance changes suddenly. This change can lead to resonant circuits. Angles that are not right are not recommended. | 677.169 | 1 |
Solving Right Triangles
Solving right triangles involves determining the lengths of sides and the measures of angles using trigonometric ratios. The Pythagorean theorem is essential: \\(a^2 + b^2 = c^2\\), where \\(c\\) is the hypotenuse. Key trigonometric functions—sine, cosine, and tangent—relate the angles to the sides, aiding in solving these triangles effectively.
Understanding Right Triangles
Understanding right triangles is fundamental in geometry and trigonometry. Right triangles are unique because they have one angle that is exactly 90 degrees.
Definition of a Right Triangle
Right Triangle: A right triangle is a triangle in which one of the angles is a right angle (90 degrees). The side opposite the right angle is called the hypotenuse, and the other two sides are called the legs.
Properties of Right Triangles
Right triangles have some specific properties that you should be aware of for solving problems:
The sum of the two non-right angles is 90 degrees.
The hypotenuse is always the longest side.
Pythagoras' theorem applies to right triangles.
Using these properties, you can solve right triangles by finding unknown sides and angles.
Pythagoras' Theorem
Pythagoras' Theorem: In a right triangle, the square of the length of the hypotenuse (\textit{c}) is equal to the sum of the squares of the lengths of the other two sides (\textit{a} and \textit{b}). This can be written as: \( c^2 = a^2 + b^2 \).
Example: If the lengths of the legs of a right triangle are 3 and 4, you can find the length of the hypotenuse by using Pythagoras' theorem. \( c^2 = 3^2 + 4^2 \) \( c^2 = 9 + 16 \) \( c^2 = 25 \) \( c = 5 \) Hence, the hypotenuse is 5 units long.
The history of Pythagoras' theorem dates back to ancient civilizations. The Egyptians used a rope loop divided into 12 equal spaces, forming a 3-4-5 triangle, which they used to create right angles for their constructions.
Hint: Always make sure to identify the right angle first; it helps in recognising the hypotenuse and applying the theorem correctly.
Trigonometric Ratios in Right Triangles
Trigonometric ratios are essential tools for solving right triangles. These ratios relate the angles of a right triangle to the lengths of its sides. The primary trigonometric ratios are sine (\textit{sin}), cosine (\textit{cos}), and tangent (\textit{tan}). They are defined as follows: \( \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \) \( \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} \) \( \tan\theta = \frac{\text{opposite}}{\text{adjacent}} \)
Steps to Solve Right Triangles
Solving right triangles involves finding the unknown sides and angles. By following a systematic approach, you can solve these triangles effectively.
Identifying Triangle Parts
The first step in solving a right triangle is identifying its parts. A right triangle consists of three main parts: the hypotenuse, the opposite side, and the adjacent side.
Hypotenuse: The longest side opposite the right angle.
Opposite side: The side opposite the angle you are examining.
Adjacent side: The side next to the angle you are examining and not the hypotenuse.
Recognising these parts is crucial for applying various formulas.
Hint: Label each side clearly when analysing the triangle to avoid confusion.
Using Pythagoras Theorem
Pythagoras' Theorem: In a right triangle, the square of the length of the hypotenuse \( c \) is equal to the sum of the squares of the lengths of the other two sides \( a \) and \( b \). This can be written as: \( c^2 = a^2 + b^2 \).
Example: If the lengths of the legs of a right triangle are 5 and 12, you can find the length of the hypotenuse by using Pythagoras' theorem. \( c^2 = 5^2 + 12^2 \) \( c^2 = 25 + 144 \) \( c^2 = 169 \) \( c = 13 \) Hence, the hypotenuse is 13 units long.
Pythagoras' theorem also has a rich historical background. It was not only known to Greek mathematicians but was also used in Babylon and ancient India, showing the universality of mathematical principles across different cultures and eras.
Example: Suppose you know one angle (other than the right angle) of a right triangle is 30 degrees and the length of the hypotenuse is 10 units. To find the length of the opposite side, use the sine ratio. \( \sin 30^\circ = \frac{\text{opposite}}{10} \) \( \frac{1}{2} = \frac{\text{opposite}}{10} \) \( \text{opposite} = 10 \times \frac{1}{2} \) \( \text{opposite} = 5 \) Hence, the length of the opposite side is 5 units.
Finding Missing Angles
To find the missing angles in a right triangle, you can use the trigonometric ratios or the sum of angles in a triangle property. The sum of angles in any triangle is 180 degrees. Since one angle is already 90 degrees, the sum of the other two angles must be 90 degrees. If you know one of the non-right angles, subtract it from 90 degrees to find the other angle.
Example: If one of the non-right angles in a right triangle is 45 degrees, you can find the other angle as follows: \(90^\circ - 45^\circ = 45^\circ \) Where both angles are 45 degrees, you have an isosceles right triangle.
Hint: Knowing just one angle other than the right angle can help you determine the remaining angle and side lengths quickly.
Right Triangle Trigonometry Techniques
In this section, you'll learn about various techniques to solve right triangles using trigonometry. These methods will help in finding unknown sides and angles efficiently.
Sine, Cosine, and Tangent
Sine, cosine, and tangent are fundamental trigonometric ratios utilised to solve right triangles. Each ratio links an angle of a right triangle to the lengths of its sides.
Special Right Triangles
Special right triangles have particular angle measures that make calculations easier. The two main types are the 45-45-90 triangle and the 30-60-90 triangle. Each has specific properties and side ratios.
45-45-90 Triangle:
Each leg is congruent.
The hypotenuse is \( \sqrt{2} \) times the length of each leg.
30-60-90 Triangle:
The length of the hypotenuse is twice the length of the shorter leg.
The length of the longer leg is \( \sqrt{3} \) times the length of the shorter leg.
Example: For a 45-45-90 triangle with each leg measuring 5 units, find the hypotenuse: Using the properties of the 45-45-90 triangle: \( \text{hypotenuse} = 5 \times \sqrt{2} \) Hence, the hypotenuse is approximately 7.07 units.
Inverse Trigonometric Functions
Inverse trigonometric functions are used to find the angles when the side lengths are known. The main inverse functions are arcsine (\( \sin^{-1} \)), arccosine (\( \cos^{-1} \)), and arctangent (\( \tan^{-1} \)). These functions are the inverses of the basic trigonometric ratios.
Hint: Inverse trigonometric functions are helpful when you need to determine an angle from side lengths.
Radians and degrees are two units to measure angles. In trigonometry, radians often replace degrees because they make calculations involving \( \pi \) simpler. One complete revolution is \(2\pi \) radians or 360 degrees.
How to Solve a Right Triangle: Problem Examples
Solving right triangles involves using various methods and mathematical theorems. In this section, you'll explore specific problem examples to enhance your understanding.
Using Given Angles
When working with given angles in a right triangle, you can use trigonometric ratios to find the unknown side lengths. The primary trigonometric ratios are sine, cosine, and tangent.
Hint: Remember that a 45-degree right triangle is isosceles. Both legs are equal.
Different approaches can be used to verify your results, such as checking that the sum of the squares of the two legs equals the square of the hypotenuse. This approach helps to affirm the accuracy of your calculations.
Working with Given Side Lengths
When you know the lengths of certain sides in a right triangle, you can use various methods, such as Pythagoras' Theorem and trigonometric ratios, to find the remaining sides and angles.
Hint: Ensure your calculator is in the correct mode (degree/radian) when working with trigonometric functions.
Understanding the historical context of Pythagoras' Theorem enriches your comprehension. Ancient mathematicians, including Pythagoras, developed this fundamental theorem that has wide applications in geometry and trigonometry.
Mixed Information Problems
Sometimes, you may encounter problems where both angles and side lengths are provided. Use a combination of Pythagoras' Theorem and trigonometric ratios to solve these problems.
In some complex problems, you might need to solve for more than one unknown variable. Using systems of equations or simultaneous trigonometric functions can help achieve this. These problems require careful analysis and a step-by-step approach to ensure accuracy.
Learn with 12 Solving Right Triangles flashcards in the free StudySmarter app
Frequently Asked Questions about Solving Right Triangles
What are the primary methods for solving right triangles?
The primary methods for solving right triangles are using the Pythagorean theorem, trigonometric ratios (sine, cosine, and tangent), and the properties of special right triangles (30-60-90 and 45-45-90). Additionally, inverse trigonometric functions can find angles when side lengths are known.
What are the trigonometric ratios used in solving right triangles?
The trigonometric ratios used in solving right triangles are sine (sin), cosine (cos), and tangent (tan). These ratios are defined as follows: sin(θ) = opposite/hypotenuse, cos(θ) = adjacent/hypotenuse, and tan(θ) = opposite/adjacent.
What is the Pythagorean Theorem and how is it used to solve right triangles?
The Pythagorean Theorem states that in a right-angled triangle, the square of the hypotenuse's length is equal to the sum of the squares of the other two sides' lengths: \\(c^2 = a^2 + b^2\\). To solve right triangles, use this formula to find an unknown side length when the other two side lengths are known.
What are the common mistakes to avoid when solving right triangles?
Common mistakes include: misidentifying the hypotenuse, incorrectly applying trigonometric ratios, neglecting to use the Pythagorean theorem, and errors in angle measurement or conversion between degrees and radians. Always double-check assignments of sides and angles, and verify calculations.
How do you determine which trigonometric ratio to use when solving a right triangle?
To determine which trigonometric ratio to use, identify the given information and what you need to find. Use sine (sin) for the ratio of opposite side to hypotenuse, cosine (cos) for adjacent side to hypotenuse, and tangent (tan) for opposite side to adjacent side. Choose the ratio involving the known and unknown sides or angles.
Test your knowledge with multiple choice flashcards
How do you find the opposite side's length if you know an angle of 30° and the hypotenuse is 10 units?
Learn with 12 Solving Right Triangles | 677.169 | 1 |
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...that space will be a maximum when the segments are equal. PROP. V. 8. Theorem. Of all triangles having two sides of the one equal to two sides of the other, each to each, that which has the two sides perpendicular to each other is the greatest. Let the triangle ACB have...
...FAG is made equal to the given rectilineal angle DCE. Which was to be done. PROP. XXIV. THEOR. If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of the one greater than the angle contained by the two sides...
...the greater of the two given figures above the less. PKOP. LXXIII. 96. THEOREM. If two right-angled triangles have two sides of the one equal to two sides of the other, each to each, the triangles shall be equal, and similar to each other. If the two sides about the right-angle of...
...Moon. TUESDAY EVENING. — MR. PEACOCK. 1. If two spherical triangles have two sides of one triangle •equal to two sides of the other, each to each, and the included angles equal, the triangles are equal in every respect. 2. The modulus of tabular logarithms or M = 4342944819;...
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...rectilineal angle equal to a giren rectilineal angle. Prop. XXIV. Theor. If two triangles have two ¿in of the one equal to two sides of the other, each to each, but the angle contained by the two ;ides of one of them greater than the angle contained by the two...
...the given rectilineal angle DCE. Which was to be done. PROP. XXIV. THEOR. IF two triangles havftwo sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two...
...= 0. Tuesday Evening. — Mr. PEACOCK. 1. If two spherical triangles have two sides of one triangle equal to two sides of the other, each to each, and the included angles equal, the triangles are equal in every respect. 2. The modulus of tabular logarithms or M=. 43429448... | 677.169 | 1 |
Each question is a choice-summary multiple choice question that will present you with a linear equation and then make 4 statements about that equation. You must determine which of the 4 statements are true (if any) in regards to the equation.
Add various inputs until you have created anequilateraltriangle, a right triangle and a triangle that is neither right nor equilateral. Verify that the proper output is displayed for each.
i) Closing the application.
10820 Green's TheoremEvaluatedIntegral Apply Green's Theorem to evaluate the integralover C of 2(x^2+y^2)dx + (x+y)^2 dy, where C is the boundary of the triangle with vertices (1,1), (2,2) and (1,3) oriented in the counterclockwise direction.
Given anequilateraltriangle, what is the area of the inscribed circle
2. What is the side length of three little equilateral triangles in the three corners (see attachment)
1. Let the side length of anequilateraltriangle be L.
480070 Finding the Double Integral 1. Use a double integral to find the area of the region bounded by the graphs of
y= x^2 and y= 8-x^2. Provide a sketch and use fubini'stheorem to determine the order of integration.
2.
Verify Stokes' Theoremfor the vector field F = (x + y) + (2x − z) + (y + z) by performing the surface integral and line integral separately. Clearly show the surface and the positive direction of the line integral on a sketch. | 677.169 | 1 |
This category includes the 13-atom icosahedron, which can be decomposed into twenty tetrahedra sharing a common vertex.
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Several of these arise naturally as crystals, and the truncated icosahedron occurs in real life as a football.
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The energy is measured relative to the energy of the global minimum icosahedron.
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Structure 69C has a vertex atom missing from the underlying Mackay icosahedron like 38A.
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Further growth then leads to the next Mackay icosahedron.
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In order to see some of these more clearly, 64 of the 280 water molecules have been removed from the water icosahedron.
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The "regular icosahedron" is one of the Platonic solids; the "great icosahedron" is a Kepler-Poinsot solid; and the "truncated icosahedron" is an Archimedean solid (see Polyhedron).
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In crystallography the icosahedron is a possible form, but it has not been observed; it is closely simulated by a combination of the octahedron and pentagonal dodecahedron, which has twenty triangular faces, but only eight are equilateral, the remaining twelve being isosceles (see Crystallography).
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The distance between adjacent vertices of the icosahedron is 5% longer than the distance between a vertex and the center.
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Nevertheless, holding that every dimension has a principle of its own, he rejected the derivation of the elemental solids - pyramid, octahedron, icosahedron and cube - from triangular surfaces, and in so far approximated to atomism.
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The first three were certainly known to the Egyptians; and it is probable that the icosahedron and dodecahedron were added by the Greeks.
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The equilateral triangle is the basis of the tetrahedron, octahedron and icosahedron.'
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The great dodecahedron is determined by the intersections of the twelve planes which intersect the Platonic icosahedron in five of its edges; or each face has the same boundaries as the basal sides of five covertical faces of the icosahedron.
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The great icosahedron is the reciprocal of the great stellated dodecahedron.
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Each of the twenty triangular faces subtend at the centre the same angle as is subtended by four whole and six half faces of the Platonic icosahedron; in other words, the solid is determined by the twenty planes which can be drawn through the vertices of the three faces contiguous to any face of a Platonic icosahedron.
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It is enclosed by 20 triangular faces belonging to the original icosahedron, and 12 pentagonal faces belonging to the coaxial dodecahedron.
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The truncated icosahedron is formed similarly to the icosidodecahedron, but the truncation is only carried far enough to leave the original faces hexagons.
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It is therefore enclosed by 20 hexagonal faces belonging to the icosahedron, and 12 pentagonal faces belonging to the coaxial dodecahedron.
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The truncated dodecahedron is formed by truncating the vertices of a dodecahedron parallel to the faces of the coaxial icosahedron so as to leave the former decagons.
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It is enclosed by 20 triangular faces belonging to the icosahedron and 12 decagons belonging to the dodecahedron.
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In the " small rhombicosidodecahedron " there are 12 pentagonal faces belonging to the dodecahedron, 20 triangular faces belonging to the icosahedron and 30 square faces belonging to the triacontahedron.
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The pentagons belong to a dodecahedron, and 20 triangles to an icosahedron; the remaining 60 triangles belong to no regular solid.
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Thus the faces of the cuboctahedron, the truncated cube, and truncated octahedron, correspond; likewise with the truncated dodecahedron, truncated icosahedron, and icosidodecahedron; and with the small and great rhombicosidodecahedra. | 677.169 | 1 |
How do you find two missing sides of a right triangle with a right angle and a 50 degree angle, the hypotenuse is 2√2?
1 Answer
Explanation:
Let hypotenuse of right triangle is #c=2sqrt2#. Base of right triangle is #a# and perpendicular of right triangle is #b# and angle between sides #a and c# is #/_B=50^0 ; sinB=b/c or sin50=b/(2sqrt2) or b=2sqrt2sin50=2.17(2dp) # | 677.169 | 1 |
Trigonometric point determination
With trigonometric point determination the survey points are determined from the ground and calculated in accordance with the rules of trigonometry. These triangulation points – familiar metal pyramids at lookout points that serve as witnesses to the earliest national survey – are the result of this method. | 677.169 | 1 |
GRE Math: Geometry If you measure all exterior angles in a polygon - they all equal 2(pi)r
D=Square root of (x2-x1)²+(y2-y1)² 360°
(n-2)x180/n
2. For any given perimeter - the rectangle with the largest AREA is a 180
A=½(base1+base2)(height)
N-2
Square (versus a rectangle) | 677.169 | 1 |
Consider the triangle of reference ABC and a point P. The pedal triangle of P with respect to ABC is the triangle PaPbPc of the projections of P on the sides of ABC (see Pedal.html ).
[1] The area of the pedal triangle is area(PaPbPc) = (R2-|PO|2)*sin(A)*sin(B)*sin(C)/2, where R is the circumradius of ABC and O is the circumcenter (R2-|PO|2 is the power of P w.r. to the circumcircle).
[2] For P moving on circles concentric to the circumcircle, the pedal triangles have constant area and vice versa. In particular, for P on the circumcircle the corresponding area is zero, the three projection-points being on the Simson line of P.
[3] Denoting by (x,y,z) the trilinear coordinates of P, and considering signed (oriented) areas: area(PaPbPc) = (1/2)*(sin(A)*y*z+sin(B)*z*x+sin(C)*x*y)
is a quadratic form in the trilinear coordinates.
[1] The formula follows by extending PC to cut the circumcircle of ABC at D and observing that angle(PcPaPb) = angle(DBP). Also angle(PDB) equals angle A of the triangle and by the sinus theorem applied to triangle DBP we have: sin(<(DBP))/|PD| = sin(A)/|PB|.(*)
In addition, since APbPcP is cyclic and PA is a diameter we have |PA| = |PbPc|*sin(A) and analogous equations for the segments PB and PC.(**)
Now using (*) and (**) the area of the pedal triangle is:
area(PaPbPc) = |PaPb|*|PaPc|*sin(<(PcPaPb))/2 = |PC|*sin(C)*|PB|*sin(B)*sin(DBP)/2 = |PC|*|PD|*sin(A)*sin(B)*sin(C)/2 = (R2-|PO|2)*sin(A)*sin(B)*sin(C)/2.
[2] Follows immediately from [1].
[3] Follows by dividing the area of the triangle in the sum: area(PaPbPc) =area(PPbPc)+area(PaPPc)+area(PaPbP).
Remarks
[1] Formula (1) is general valid, even when point P is outside the triangle, provided we use oriented areas. Next figure illustrates the corresponding proof.
[2] By equating the expressions in (1) and (3) we get:
sin(A)*y*z+sin(B)*z*x+sin(C)*x*y = (R2-|PO|2)*sin(A)*sin(B)*sin(C). <==>
a*y*z+b*z*x+c*x*y = (R2-|PO|2)*a*b*c/(4*R2).
By taking P on the circle we get its equation in trilinears (see also CircumcircleInTrilinears.html ).
[3] The previous equation shows that the quadratic form f(x,y,z) = a*y*z+b*z*x+c*x*y, where a, b, c denote the lengths of the sides of the triangle, is positive inside the circumcircle, zero on the circumcircle and negative outside. Notice that (x,y,z) are not independent, but satisfy the equation a*x+b*y+c*z = 2*area(ABC). Points at infinity satisfy a*x+b*y+c*z = 0 and fall "far out" of the circumcircle of ABC, there where f is negative. This is in accordance with the formula for the distance of two points in trilinears (see last formula in BarycentricsFormulas.html ). | 677.169 | 1 |
You are a polygon you are a quadrangle all your corners are square corners what are you?
Since you are a quadrangle with square corners, you are a
4-sided polygon with 90° angles, which means you are a rectangle.
If all the sides are equal length, you are a square, which is a
specific type of rectangle. | 677.169 | 1 |
Get Answers to all your Questions
In a huge park, people are concentrated at three points (see Fig.): A : where there are different slides and swings for children, B : near which a man-made lake is situated, C : which is near to a large parking and exit.
3. In a huge park, people are concentrated at three points (see Fig.):
A : where there are different slides and swings for children,
B : near which a man-made lake is situated,
C : which is near to a large parking and exit.
Where should an icecream parlour be set up so that maximum number of persons can approach it?
(Hint : The parlour should be equidistant from A, B and C)
Answers (1)
The three main points form a triangle ABC. Now we have to find a point which is equidistant from all the three points.
Thus we need to find the circumcenter of the .
We know that circumcenter is defined as the point as the intersection point of the perpendicular bisectors of the sides of the triangle.
Hence the required point can be found out by drawing perpendicular bisectors of . | 677.169 | 1 |
Geometry Transformations Worksheet Pdf
Geometry Transformations Worksheet Pdf. If you appetite added responses, your finest wager is to dig up a make clear architecture equipment online to do the algebraic for you. The following steps are to be adopted whereas we do transformations on a graph. The use of mirrors utterly revolutionized how telescopes and microscopes are used. Triangles, 4-sided polygons and field shaped objects is also chosen.
That polygon is then given a scale factor and the dilation may be created. If the scale factor is 2, for instance, then every of the coordinate pairs naming an endpoint can be doubled or multiplied by two. If the size issue was ½, then each coordinate paid could be multiplied by ½.
The analog band-aid is to do away with the nuisance alerts anon by clarification them away, and sampling aloof sooner than alert the abundance of the adapted sign. Middle school kids ought to decide out the proper transformations undergone.
Equation Freak: Transformations Summative Evaluation Project
When one shape can turn into one other utilizing only Turns, Flips and/or Slides, then the 2 shapes are Congruent. To see the Review answers, open this PDF file and search for part 12.6.
Scaling the cap bottomward by ten and the resistor up by ten is aloof about proper. Because we're absorbed within the DC values, artlessly bead the ascribe capacitor and bent resistors and the achievement blocking cap. Let's chase on with the archetype of a ablaze sensor in the attendance of ablaze autogenous lighting.
This Transformations Worksheet will produce issues for practicing reflections of objects. Practice utilizing and apply tips for the composition of transformations on the coordinate airplane. A transformation is an operation that moves, flips, or in any other case modifications a determine to create a brand new figure.
Geometry Transformations Worksheet Pdf
Common kinds of worksheets utilized in enterprise embody financial statements, similar to revenue and loss reviews. Analysts, merchants, and accountants monitor an organization's monetary statements, steadiness sheets, and other info on worksheets. Most of the MS Excel display display screen is devoted to the present of the worksheet, which consists of rows and columns.
A complete good, if a bit superior, capability on clarify architecture is Chapter Eight of Analog Devices' Linear Ambit Architecture Handbook. If you are not exercise the mathematics, you can skip the realm on s transforms after accident out a lot.
Let the orange triangle be the pre-image and the blue triangle be the reworked picture. Members have unique services to obtain an individual worksheet, or an entire stage. Rotate, reflect and translate each point following the given rules.
Line segments are named by two endpoints that might again fall throughout the coordinate grid. Students may then be given directions to maneuver the line phase under the interpretation (7, -3). Students would have to take every endpoint of the original line section and move it seven spots to the right and three spots down.
Translation, reflection, rotation, and dilation are the 4 types of transformations. This Transformations Worksheet will produce points for training translations, rotations, and reflections of objects. Triangles, 4-sided polygons and field shaped objects is also chosen.
Geometry Transformations Worksheet Pdf
An achievement capacitor decouples again, so the arresting is accessible to bung into whatever amplifier you've received. The ethics of those capacitors don't bulk a lot, however they accept to be so much bigger than C.
The worksheets are visually appealing to have interaction college students within the work. For a small month-to-month fee, the worksheets also come with reply keys in order that students and fogeys can verify the work as quickly as it's completed.
The length of each aspect can be multiplied by the dimensions issue to create the dilation. A dilation is a transformation that modifications the size of the determine, but the shape stays the same. One of one of the best real-world connections to this talent is how a movie show is in a position to project such a large picture from such a small system.
The line of reflection is usually the x or y-axis, however that doesn't all the time need to be the case. To make a new shape, all of the pairs of factors that name the endpoints of the shape need to be moved to the precise opposite aspect of the road of reflection. A level can be named with a given coordinate pair, and college students could be given instructions on the way to translate it.
The translation is in a course parallel to the line of reflection. Learn recommendations on tips on how to compose transformations of a determine on a coordinate plane, and understand the order throughout which to apply them. Let the highschool college students translate each quadrilateral and graph the picture on the grid.
An Excel spreadsheet contains 16,384 rows which may be labeled numerically. As in opposition to to the PWM software, the TL074 fails us actuality whereas the LM324 shines.
These math worksheets must be practiced frequently and are free to download in PDF codecs. In these worksheets determine the image which best describes the transformation of the given figure. You can then transfer to the worksheet you want by clicking it inside the list.
Let the excessive school students translate each quadrilateral and graph the image on the grid. Each grid has the determine and the image obtained after transformation. From this, the time period was prolonged to designate a single, two-dimensional array of knowledge inside a computerized spreadsheet program.
Another useful skill that could be practiced and mastered using the cazoommaths.com transformation worksheets is creating dilations. Reflection A reflection is a metamorphosis that flips a decide on the coordinate airplane throughout a given line with out changing the form or measurement of the determine. Glide Reflection A reflection followed by a translation where the road of reflection is parallel to the path of translation known as a glide reflection or a walk.
Back may urge for food to clarify out the highs. In the PWM case, the high-frequency agreeable comes from the basal aboveboard beachcomber whose project aeon is actuality articulate — all you appetite is the boilerplate bulk over a few cycles.
In audio purposes, you normally alone use the centermost allocation of the voltage ambit anyway, in order that's OK. The LM324/LM358 is slower, but uses beneath capability and alcove bottomward to the abrogating abuse on input, which makes it acceptable for low-frequency, battery-powered, DC arresting conditioning. In explicit, the LM324 and LM358 accept unhealthy crossover distortion, so back the PWM ascribe switches rapidly from on to off, the op-amp fails to clue it.
We'll use it beneath for apathetic ADC pattern filtering, but its poor crossover achievement makes it inappropriate for PWM filtering. A all-encompassing make clear architecture begins with at atomic the bristles choices mentioned above.
When a certain curve is rotated, translated, or reflected alongside totally different axes, it is stated to have undergone a change. Some of the other popular kinds of transformations used in engineering are Fourier remodel, Laplace remodel, and so forth.
Here, we're aiming at an audio PWM, so alluringly we would set the blow of the make clear about 20 kHz, at the high of animal listening to. In an ideal world, our PWM abundance could be so abundant school than 20 kHz that we would not accept to compromise. In the complete world, area we capacity be energetic PWM frequencies as little as kHz, a blow of about 10 kHz is possibly a bigger objective.
You can change which cell is the energetic cell by clicking the left mouse button as quickly as or utilizing the arrow keys on the keyboard. The current energetic cell could be recognized as being the one which has a darker black border round it. Also, the energetic cell reference is listed within the Name Box instantly above the spreadsheet's column headings.
This batch of worksheets is extremely recommended for the scholars of grade 2 through grade eight. Translations, rotations and reflections are all important parts of mastering geometry and we'll discuss reflections and rotations in additional element now. The energetic cell is the cell in the spreadsheet that's presently chosen for data entry.
When we look in the mirror and see the reflection, we see a model of the original image that has been flipped.
Once students have mastered factors and line segments, then polygons ought to be very comprehensible.
If needed, limit the area on the perform so that the inverse exists.
Here, we're aiming at an audio PWM, so alluringly we would set the blow of the make clear about 20 kHz, at the top of animal hearing.
Clarify Architecture in Thirty Seconds is a affair information, and complete plentiful within the spirit of this collection. It's a ample refresher already you have been through mixture already or twice, nevertheless it artlessly leaves out lots of the appropriate stuff.
The type of transformation to be performed is described above every query. Full textual content search our database of 156,200 titles for Spreadsheet Program to look out associated analysis papers. Under this mannequin, journals will turn out to be primarily obtainable beneath electronic format and articles might be immediately on the market upon acceptance.
Once students have mastered points and line segments, then polygons should be very understandable. Then translated horizontally 6 unit to the correct and vertically 2 fashions up. A glide reflection is a composition of a mirrored image and a translation.
In accounting, a worksheet is, or was, a sheet of ruled paper with rows and columns on which an accountant might document data or perform calculations. These are generally known as columnar pads, and usually green-tinted.
Reflect \(\Delta DEF\) from Question 2 over the \(x\)-axis, followed by the \(y\)-axis. Find the coordinates of \(\Delta D′′E′′F′′\) and the one transformation this double reflection is the same as. Be an undisputed grasp of sliding a form on a coordinate aircraft with this set of printable translation worksheets.
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Question 4.
Draw the front view, side view and top view of the given objects.
Solution:
Exercise 10.2
Question 1.
Look at the given map of a city.
Answer the following.
(a) Colour the map as follows: Blue-water, red- fire station, orange-library, yellow-schools, Green-park, Pink-College, Purple-Hospital, Brown-Cemetery.
(b) Mark a green 'X' at the intersection of Road 'C' and Nehru Road, Green 'Y' at the intersection of Gandhi Road and Road A.
(c) In red, draw a short street route from Library to the bus depot.
(d) Which is further east, the city park or the market?
(e) Which is further south, the primary school or the Sr. Secondary School?
Exercise 10.3
Question 1.
Can a polyhedron have for its faces
(i) 3 triangles?
(ii) 4 triangles?
(iii) a square and four triangles?
Solution:
(i) No, because polyhedron must have edges meeting at vertices which are points.
(ii) Yes, because all the edges are meeting at the vertices.
(iii) Yes, because all the eight edges meet at the vertices.
Question 2.
Is it possible to have a polyhedron with any given number of faces?
(Hint: Think of a pyramid)
Solution:
Pyramid
Yes, it is possible if the number of faces is greater than or equal to 4.
Example: Pyramid which has 4 faces.
Question 3.
Which are prisms among the following?
Solution:
Only (ii) unsharpened pencil and (iv) a box are the prism.
Question 4.
(i) How are prisms and cylinders alike?
(ii) How are pyramids and cones alike?
Solution:
(i) If the number of sides in a prism is increased to certain extent, then the prism will take the shape of cylinder.
(ii) If the number of sides of the pyramid is increased to same extent, then the pyramid becomes a cone.
Question 5.
Is a square prism same as a cube? Explain.
Solution:
Every square prism cannot be cube. It may be cuboid also. | 677.169 | 1 |
Accurately make three copies of the given tangram. Number the pieces as in the given tangram above. Cut out the pieces and paste them in your answer book to make each of the following figures:congruen...
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NCERT Class 9 Maths Solutions -Complete PDF
"Areas of Parallelograms and Triangles", "Constructions", and "Probability" has been rationalized from the Class 9 NCERT books for 2025 session.
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With Educart, students can download their copy of NCERT Class 9 Math solutions for free, although OTP validation might be required sometimes for security purposes. Student's one-stop destination to find detailed explanations and self-assessment questions curated by renowned experts and a panel of Educart school HODs that assists in step-by-step explanation that provides students with a clear understanding.
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Tan 1290 Degrees
The value of tan 1290 degrees is 0.5773502. . .. Tan 1290 degrees in radians is written as tan (1290° × π/180°), i.e., tan (43π/6) or tan (22.514747. . .). In this article, we will discuss the methods to find the value of tan 1290 degrees with examples.
Tan 1290°: 1/√3
Tan 1290° in decimal: 0.5773502. . .
Tan (-1290 degrees): -0.5773502. . . or -1/√3
Tan 1290° in radians: tan (43π/6) or tan (22.5147473 . . .)
What is the Value of Tan 1290 Degrees?
The value of tan 1290 degrees in decimal is 0.577350269. . .. Tan 1290 degrees can also be expressed using the equivalent of the given angle (1290 degrees) in radians (22.51474 . . .)
Methods to Find Value of Tan 1290 Degrees
The tangent function is positive in the 3rd quadrant. The value of tan 1290° is given as 0.57735. . .. We can find the value of tan 1290 degrees by:
Using Unit Circle
Using Trigonometric Functions
Tan 1290 Degrees Using Unit Circle
To find the value of tan 1290 degrees using the unit circle, represent 1290° in the form (3 × 360°) + 210° [∵ 1290°>360°] ∵ The angle 1290° is coterminal to 210° angle and also tangent is a periodic function, tan 1290° = tan 210°.
Rotate 'r' anticlockwise to form 210° or 1290° angle with the positive x-axis.
The tan of 1290 degrees equals the y-coordinate(-0.5) divided by x-coordinate(-0.866) of the point of intersection (-0.866, -0.5) of unit circle and r.
Hence the value of tan 1290° = y/x = 0.5774 (approx).
Tan 1290° in Terms of Trigonometric Functions
Using trigonometry formulas, we can represent the tan 1290 degrees as:
sin(1290°)/cos(1290°)
± sin 1290°/√(1 - sin²(1290°))
± √(1 - cos²(1290°))/cos 1290°
± 1/√(cosec²(1290°) - 1)
± √(sec²(1290°) - 1)
1/cot 1290°
Note: Since 1290° lies in the 3rd Quadrant, the final value of tan 1290° will be positive.
FAQs on Tan 1290 Degrees
What is Tan 1290 Degrees?
Tan 1290 degrees is the value of tangent trigonometric function for an angle equal to 1290 degrees. The value of tan 1290° is 1/√3 or 0.5774 (approx).
How to Find the Value of Tan 1290 Degrees?
The value of tan 1290 degrees can be calculated by constructing an angle of 1290° with the x-axis, and then finding the coordinates of the corresponding point (-0.866, -0.5) on the unit circle. The value of tan 1290° is equal to the y-coordinate(-0.5) divided by the x-coordinate (-0.866). ∴ tan 1290° = 1/√3 or 0.5774
What is the Value of Tan 1290 Degrees in Terms of Cos 1290°?
We know, using trig identities, we can write tan 1290° as -√(1 - cos²(1290°))/cos 1290°. Here, the value of cos 1290° is equal to -0.866025.
What is the Value of Tan 1290° in Terms of Cosec 1290°?
Since the tangent function can be represented using the cosecant function, we can write tan 1290° as 1/√(cosec²(1290°) - 1). The value of cosec 1290° is equal to -2.
How to Find Tan 1290° in Terms of Other Trigonometric Functions?
Using trigonometry formula, the value of tan 1290° can be given in terms of other trigonometric functions as: | 677.169 | 1 |
If the position vectors of three points A, B, C are respectively i+j+k,2i+3j−4kand7i+4j+9k, then the unit vector perpendicular to the plane of triangle ABCis
A
(31i−38j−9k)
B
31i−38j−9k√2486
C
31i+38j+9k√2486
D
none of these
Video Solution
Text Solution
Verified by Experts
The correct Answer is:B
|
Answer
Step by step video, text & image solution for If the position vectors of three points A, B, C are respectively i+j+k, 2i + 3j -4k and 7i+4j+9k, then the unit vector perpendicular to the plane of triangle ABCis by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams. | 677.169 | 1 |
Exploring Quotient Trig Identities: Simplify, Solve, and Understand the Ratios in Right Triangles
Quotient Trig Identities
Quotient trig identities are a set of formulas that express the trigonometric functions in terms of the ratios of their respective sides in a right triangle
Quotient trig identities are a set of formulas that express the trigonometric functions in terms of the ratios of their respective sides in a right triangle. These identities can be very useful when simplifying trigonometric expressions or solving trigonometric equations.
The three main quotient trig identities are:
1. Cotangent (cot):
The cotangent of an angle is the ratio of the adjacent side to the opposite side in a right triangle. The quotient identity for cotangent is:
cot(x) = cos(x) / sin(x)
This means that the cotangent of an angle is equal to the cosine of the angle divided by the sine of the angle.
2. Secant (sec):
The secant of an angle is the reciprocal of the cosine of the angle. The quotient identity for secant is:
sec(x) = 1 / cos(x)
This means that the secant of an angle is equal to 1 divided by the cosine of the angle.
3. Cosecant (csc):
The cosecant of an angle is the reciprocal of the sine of the angle. The quotient identity for cosecant is:
csc(x) = 1 / sin(x)
This means that the cosecant of an angle is equal to 1 divided by the sine of the angle.
These quotient trig identities can be derived from the basic trigonometric definitions and the Pythagorean identity (sin^2(x) + cos^2(x) = 1). They are helpful in simplifying trigonometric expressions, rewriting complex trigonometric functions in terms of simpler ones, or solving trigonometric equations.
It is important to understand these identities and their applications in various math and science fields, such as calculus, physics, and engineering. Practicing with trigonometric problems and exercises will help to reinforce your understanding of the quotient trig identities and their usage | 677.169 | 1 |
IB DP Maths AA SL Study Notes
3.1.1 Introduction to Radians
What is a Radian?
A radian is a standard unit of angular measurement. Adopted under the Système International d'Unités (SI), it's extensively utilised in mathematics, physics, and engineering disciplines. The definition of a radian is intrinsically tied to the properties of a circle. Specifically, one radian is the angle created at the centre of a circle by an arc whose length matches the radius of the circle. For more on how radians relate to arc lengths, see Arc Length.
To visualise, consider a circle with a radius 'r'. If you were to take a segment of the circle's circumference equivalent to the radius and measure the angle it creates at the circle's centre, that angle would be one radian. In terms of degrees, 1 radian is roughly equivalent to 57.295 degrees.
The Circle and Radians
The bond between radians and circles is profound. A circle, with its continuous curvature, has a circumference described by 2πr, where 'r' is its radius. When we discuss the angles within a circle in terms of radians:
A full circle encompasses an angle of 2π radians.
Half a circle, representing a straight angle, measures π radians.
A quarter of a circle, or a right angle, measures π/2 radians.
This intrinsic relationship between the circle's circumference and radians leads to several insightful observations:
The angle for a complete rotation or a full circle is 2π radians, equivalent to 360°.
Halfway around the circle, or 180°, is equivalent to π radians.
To understand how radians are used in calculating sector areas, check out Sector Area.
Conversion between Degrees and Radians
Switching between degrees and radians is a crucial skill, especially when tackling trigonometric problems. Here's a detailed breakdown of the conversion:
From Degrees to Radians
To convert an angle from degrees to radians, utilise the relationship that 180° is equivalent to π radians. The conversion formula is: Angle in radians = Angle in degrees x (π/180)
In essence, understanding radians is pivotal for anyone delving into maths, especially topics related to geometry and trigonometry. It offers a natural way to comprehend and measure angles, especially concerning the properties of circles. Whether you're diving into advanced maths or just embarking on your mathematical journey, a robust understanding of radians and their relationship with degrees will be invaluable. For visual learners, the Graphs of Sine can provide a clear depiction of these concepts.
FAQ
The value of π is fundamental in the relationship between degrees and radians because it represents the ratio of a circle's circumference to its diameter. When we define angles in terms of radians, we are essentially relating the angle to portions of the circle's circumference. Since a full circle's circumference is 2π times its radius and a full circle also encompasses 360°, it naturally follows that 360° is equivalent to 2π radians. This relationship makes π the bridge between the two units of angular measure, and it's the reason why π frequently appears in formulas and conversions between degrees and radians.
Yes, besides degrees and radians, there are other units of angular measure, though they are less commonly used. One such unit is the gradian (or gon). In the gradian system, a right angle is 100 gradians, making a full circle 400 gradians. This system is sometimes used in surveying. Another unit is the turn, where a full circle is considered one turn. This unit is intuitive as it directly relates to the idea of completing rotations. However, in most mathematical, scientific, and engineering applications, degrees and radians remain the predominant units of angular measure due to their historical significance and practicality.
The unit circle in trigonometry is a circle with a radius of 1 unit, centred at the origin of a coordinate plane. Radians play a crucial role in the unit circle's definition and understanding. When we measure angles in radians on the unit circle, the angle's measure directly corresponds to the arc length subtended by that angle. For example, an angle of π/2 radians on the unit circle corresponds to an arc length of π/2 units. This direct relationship between angle measure and arc length in the unit circle simplifies the understanding and computation of trigonometric values, making it a foundational concept in trigonometry.
The radian is considered a "natural" unit of angular measure because it directly relates the angle measurement to the radius of a circle. When an angle is measured in radians, it represents the length of the arc subtended by that angle, divided by the radius of the circle. This relationship simplifies many mathematical expressions, especially in calculus and higher-level mathematics. For instance, when working with trigonometric functions in calculus, derivatives and integrals become more straightforward when angles are measured in radians. Moreover, many mathematical relationships and formulas, such as those in Fourier series or Taylor series expansions of trigonometric functions, are naturally expressed in terms of radians.
Radians play a pivotal role in understanding and describing circular motion in physics. When an object moves in a circular path, its angular displacement, angular velocity, and angular acceleration are often measured in radians. Using radians simplifies equations and relationships in circular motion. For instance, the linear velocity (v) of an object moving in a circle of radius (r) with an angular velocity (ω) is given by v = rω. Here, if ω is in radians per unit time, the equation directly gives the linear velocity without the need for conversion factors. Similarly, many formulas in rotational dynamics, such as torque and moment of inertia, are more straightforward and intuitive when angles are measured in radians.
Practice Questions
A circular pond has a diameter of 10 metres. A duck swims from one point on the edge of the pond to another point, covering an arc length of 8 metres. Calculate the angle, in radians, subtended by the arc at the centre of the pond.
To determine the angle in radians subtended by the arc at the centre, we can use the formula: Angle in radians = Arc length / Radius Given the diameter is 10 metres, the radius is half of that, which is 5 metres. Using the formula: Angle in radians = 8 metres / 5 metres = 1.6 radians. Thus, the angle subtended by the arc at the centre of the pond is 1.6 radians.
A wheel with a radius of 2 metres is rotating. If the wheel turns through an angle of π/4 radians, how long is the arc travelled by a point on the edge of the wheel?
To determine the arc length travelled by a point on the edge of the wheel, we can use the formula: Arc length = Radius x Angle in radians Given the radius is 2 metres and the angle is π/4 radians, using the formula: Arc length = 2 metres x π/4 = π/2 metres. Thus, the arc length travelled by a point on the edge of the wheel is π/2 metres, which is approximately 1.57 metres | 677.169 | 1 |
D1
Explanation: To find the distance (d) between two points on a coordinate graph, we can draw a line between the two points, and then draw a triangle with this line as the hypotenuse. Using algebra, we can determine that the two legs of the triangle have lengths and
Now we can use the Pythagorean Theorem to determine the length of the hypotenuse.
Then we can rearrange the equation to solve for d
This is the formula for finding the distance between two points.
To find the midpoint (M) of two points, we need to find the x value that is halfway between the x values of the two points, and the y value that is halfway between the y values of the two points. This will give us the coordinates of the midpoint. To find these halfway values, we can find the average of the two x values and the average of the two y values.
For example, if the two points were and , then:
the average of the x values would be,
and the average of the y values would be,
so to find the coordinates of M we can use this formula:
Problem:
Bob and Tim are on a jungle gym. Bob is at point (1,3) and Tim is at point (3,1)
a) how far far would bob have to go to get to tim if he traveled a straight line?
b) if they both traveled an equal distant, at what point would they meet up? | 677.169 | 1 |
There are dotes on the plane $(x,y)$ connected with directed edges. The distance $\rho(A,B)$ is standard euclidean: $|\overrightarrow{AB}|$. Except the distance cost we pay for rotation: $k\alpha$, $\alpha$ is the angle. We need to find the shortest ("cheapest") path between to vertexes $s,\,t$.
It's sad, but triangle inequality doesn't hold true anymore. I think about smth like launching BFS from $s$ and keeping for each vertex a list of pairs (predecessor, distance). It's just an idea and I'm not sure if it'll work. Maybe it's a well known problem?
1 Answer
1
You could replace each vertex $v$ by vertices $v_e$, one for each edge $e$ incident at $v$, with $e$ now incident at $v_e$ instead, and connect all pairs of vertices $v_e$, $v_f$ by edges reflecting the rotation cost. Or you could reduce the number of edges by only connecting $v_e$ and $v_f$ if $e$ and $f$ are at adjacent angles. For $s$ and $t$, you could either leave them unchanged, or, if that's simpler, treat them like all other vertices and then add new vertices $s'$, $t'$ with zero-cost edges to all the vertices that $s$ and $t$ were replaced by. In any case, the problem becomes a standard problem of finding the shortest path in the resulting graph.
$\begingroup$@Igor: Sure, the number of vertices after replacement is twice the number of edges before replacement. If you do the variant where you only add between new vertices corresponding to edges at adjacent angles, there's one new edge for every new vertex, so the total number of edges after replacement is three times the number of edges before replacement.$\endgroup$ | 677.169 | 1 |
Q. Define the collections {E1, E2, E3, ...} of ellipses and {R1, R2, R3, ...} of rectangles as follows: E1:x29+y24=1; R1: rectangle of largest area, with sides parallel to the axes, inscribed in E1; En: ellipse x2a2n+y2b2n=1 of largest area inscribed in Rn−1, n>1; Rn: rectangle of largest area, with sides parallel to the axes, inscribed in En, n>1;
Then which of the following option is/are correct?
If (x, y, z) be an arbitrary point lying on a plane P which passes through the points (42, 0, 0), (0, 42, 0)and(0, 0, 42), then the value of the expression 3+x-11(y-19)2(z-12)2+y-19(x-11)2(z-12)2+z-12(x-11)2(y-19)2–x+y+z14(x-11)(y-19)(z-12) is equal to : | 677.169 | 1 |
Pages
Monday, March 14, 2011
Archimedes angle trisection or tripling?
Pi-day today, so a nice occasion to write something about circles. π is the ratio between the circles' perimeter (περίμετρος in Greek) and diameter.
In the previous post, I explored some angular tricks using circles. The circle is a key geometrical object when dealing with angles. For example, Archimedes' construction to trisect an angle uses a circle and a straight line with a marked unit. But for clarity's sake, I prefer to draw a set of circles which show how a straight line can mark odd multiples of a unit angle α on intersecting circles, see Figure 1. In that way, it is easier to see how Archimedes' trisection circle relates to the multiplication and division of angles by odd integers, see alternative Figure 2.
This makes me think that Archimedes' angle trisection is in fact the inverse of "Archimedes' angle tripling".
4 comments:
Arjen, Awesome stuff.. I especially like the idea from the last post that the two arcs cut out on any circle will always have a constant difference. These two posts could make a really good exploration in a HS geometry class.. Thanks Pat
It would be nice indeed to ask HS students, or even children, to explore such patterns. I am sure some would discover new ones faster than we would. My 9-year old daughter sometimes draws geometric figures next to me. I try to teach her some recursive drawings with circles or triangles. It's amazing how fast she catches it up. | 677.169 | 1 |
Diagonaux Demystified Bridging Math, Design, and Technology
In the worlds of mathematics, design, and technology, certain concepts serve as the fundamental building blocks for innovation and creativity. One such concept is the diagonaux. Though it may sound complex, understanding diagonaux can open up a myriad of possibilities across various fields. This blog post aims to demystify diagonaux, providing insights into its historical roots, mathematical properties, and practical applications.
In this article, we will explore what diagonaux are, why they matter, and how they are used in different disciplines. By the end, you will have a comprehensive understanding of this fascinating concept and its potential to influence various aspects of your work and daily life.
Historical Background
Origins
The origins of diagonal lines can be traced back to ancient geometry. These lines often appeared in ancient Greek and Egyptian designs, where they were used to create complex patterns and structures. The Greeks, in particular, were fascinated by the properties of diagonal lines and incorporated them into their geometric studies.
Evolution
Over time, the concept of diagonaux has evolved significantly. During the Renaissance period, artists and architects began to use diagonal lines more frequently to create perspective and depth in their works. This marked the beginning of a new era where diagonaux became essential in both art and architecture.
With the advent of modern technology, the application of diagonal lines has expanded even further. Today, diagonaux are used in various software programs, enabling designers and engineers to create intricate patterns and structures with ease.
Applications in Geometry
Basic Geometry
In basic geometry, diagonaux play a crucial role. They are often used to connect opposite corners of polygons, creating a line that bisects the shape. This simple yet powerful concept helps in dividing shapes into smaller, more manageable parts, making it easier to understand their properties.
Advanced Geometry
In more advanced geometric constructs, diagonaux serve as the foundation for various theorems and formulas. For instance, in polyhedra, diagonaux are used to connect non-adjacent vertices, forming complex three-dimensional shapes. Understanding these connections is essential for solving high-level geometric problems and creating intricate designs.
Diagonaux in Design and Aesthetics
Architecture
In the realm of architecture, diagonal lines are often used to create dynamic and visually appealing structures. Famous examples include the Eiffel Tower in Paris and the Gherkin in London, both of which incorporate diagonal elements in their designs. These lines not only add aesthetic value but also contribute to the structural integrity of the buildings.
Art and Graphics
In art and graphic design, diagonaux are used to create movement and depth. Artists often use diagonal lines to guide the viewer's eye through the composition, making the artwork more engaging. Graphic designers use diagonaux to create dynamic layouts and visually striking designs, enhancing the overall appeal of their work.
Practical Applications
Engineering
In engineering, diagonal lines are used in structural designs to distribute weight and provide support. For example, in bridge construction, diagonal beams are used to create trusses, which help distribute the load evenly across the structure. Understanding the principles of diagonaux is essential for designing safe and efficient structures.
Technology
In modern technology, diagonaux are used in various design software programs. These lines help in creating complex patterns and structures, enabling designers and engineers to bring their ideas to life. Software programs like AutoCAD and SketchUp use diagonaux to create accurate and detailed designs, making them indispensable tools in the industry.
Mathematical Properties
Formulas and Theorems
Diagonaux have several key mathematical properties and theorems associated with them. One of the most well-known is the Pythagorean theorem, which states that the square of the length of the diagonal in a right-angled triangle is equal to the sum of the squares of the lengths of the other two sides. This theorem is fundamental in both geometry and trigonometry.
Problem-Solving
Understanding the properties of diagonaux can help solve various mathematical problems. For instance, in coordinate geometry, the equation of a diagonal line can be used to determine the points of intersection with other lines. This is particularly useful in fields like physics and engineering, where precise calculations are essential.
Diagonaux in Everyday Life
Common Uses
Diagonaux are present in many everyday objects and scenarios. From the layout of city streets to the design of household items, these lines are used to create efficient and aesthetically pleasing structures. Understanding diagonaux can help in recognizing their applications in daily life and appreciating their significance.
Impact
Knowing how to identify and use diagonaux can benefit daily activities and problem-solving. For example, when arranging furniture in a room, using diagonal lines can create a more dynamic and spacious layout. In addition, understanding diagonaux can improve one's ability to analyze and interpret various forms of visual information.
Future Trends
Innovations
Emerging trends and innovations involving diagonaux are shaping the future of design and technology. Researchers are exploring new ways to use diagonal lines in fields like robotics and artificial intelligence, where they can help create more efficient and adaptive systems.
Research
Current research is focused on finding new applications for diagonaux in various fields. From developing advanced materials to creating innovative architectural designs, the potential for diagonaux is vast and continues to grow. Staying informed about these developments can provide valuable insights and opportunities for professionals in related fields.
Conclusion
In conclusion, diagonaux are a fundamental concept that bridges the gap between math, design, and technology. Their applications are vast, ranging from basic geometry to advanced engineering and art. By understanding diagonaux, professionals in various fields can enhance their work, solve complex problems, and create innovative designs.
Whether you are a math enthusiast, a designer, or a tech innovator, exploring the world of diagonaux can open up new possibilities and inspire creativity. We encourage you to continue learning about this fascinating concept and discover how it can benefit your work and daily life.
FAQs
What are diagonaux?
Diagonaux are diagonal lines that connect opposite corners of a polygon or non-adjacent vertices of a polyhedron.
Why are diagonaux significant in various fields?
Diagonaux are essential in fields like geometry, design, engineering, and technology because they help create efficient and aesthetically pleasing structures.
How can understanding diagonaux benefit daily life?
Understanding diagonaux can help in recognizing their applications in daily life, improving problem-solving skills, and enhancing one's ability to analyze and interpret visual information.
What are some common uses of diagonaux in everyday objects?
Diagonaux are used in the design of household items, city layouts, and various structures to create efficient and visually appealing designs.
What are some emerging trends and innovations involving diagonaux?
Researchers are exploring new applications for diagonaux in fields like robotics and artificial intelligence, where they can help create more efficient and adaptive systems. | 677.169 | 1 |
As long as you have a minimum of two pairs of congruent angles you can prove the triangles similar by using .
Cheri and Roberta noticed their similarity statements for part (b) were not the same. Cheri had stated , while Roberta maintained that . Who is correct? Or are they both correct? Explain your reasoning.
Hint (c):
Cheri's similarity statement says the second triangle is a reflection and dilation of the first, while Roberta's statement says it is a rotation and dilation of the original triangle. Who is correct? | 677.169 | 1 |
Similar triangles definition
Mint chocolate chip ice cream and chocolate chip ice cream are similar, but not the same. This is an everyday use of the word "similar," but it not the way we use it in mathematics.
In geometry, two shapes aresimilarif they are the same shape but different sizes. You could have a square with sides 21 cm and a square with sides 14 cm; they would be similar. An equilateral triangle with sides 21 cm and a square with sides 14 cm would not be similar because they are different shapes.
Similar trianglesare easy to identify because you can apply three theorems specific to triangles. These three theorems, known asAngle-Angle (AA),Side-Angle-Side (SAS), andSide-Side-Side (SSS), are foolproof methods for determining similarity in triangles.
Angle-Angle (AA)
Side-Angle-Side (SAS)
Side-Side-Side (SSS)
Corresponding angles
In geometry,correspondencemeans that a particular part on one polygon relates exactly to a similarly positioned part on another. Even if two triangles are oriented differently from each other, if you can rotate them to orient in the same way and see that their angles are alike, you can say those angles correspond.
The three theorems for similarity in triangles depend upon corresponding parts. You look at one angle of one triangle and compare it to the same-position angle of the other triangle.
Proportion
Similarity is related to proportion. Triangles are easy to evaluate for proportional changes that keep them similar. Their comparative sides are proportional to one another; their corresponding angles are identical.
You can establish ratios to compare the lengths of the two triangles' sides. If the ratios are congruent, the corresponding sides are similar to each other.
Included angle
Theincluded anglerefers to the angle between two pairs of corresponding sides. You cannot compare two sides of two triangles and then leap over to an angle that is not between those two sides.
Proving triangles similar
Here are two congruent triangles. To make your life easy, we made them both equilateral triangles.
The two equilateral triangles are the same except for their letters. They are the same size, so they areidentical triangles. If they both were equilateral triangles but sideENwas twice as long as sideHE, they would besimilar triangles.
Triangle similarity theorems
Angle-Angle (AA) theorem
Angle-Angle (AA)says that two triangles are similar if they have two pairs of corresponding angles that are congruent. The two triangles could go on to bemorethan similar; they could be identical. For AA, all you have to do is compare two pairs of corresponding angles.
Trying angle-angle
Here are two scalene triangles△JAMand△OUT. We have already marked two of each triangle's interior angles with the geometer's shorthand for congruence: the little slash marks.
A single slash for interior∠Aand the same single slash for interior∠Umean they are congruent. Notice∠Mis congruent to∠Tbecause they each have two little slash marks.
Since∠Ais congruent to∠U, and∠Mis congruent to∠T, we now have two pairs of congruent angles, so the AA Theorem says the two triangles are similar.
Tricks of the trade
Watch for trickery from textbooks, online challenges, and mathematics teachers. Sometimes the triangles are not oriented in the same way when you look at them. You may have to rotate one triangle to see if you can find two pairs of corresponding angles.
Another challenge: two angles are measured and identified on one triangle, but two different angles are measured and identified on the other one.
Because each triangle has only three interior angles, one each of the identified angles has to be congruent. By subtracting each triangle's measured, identified angles from 180°, you can learn the measure of the missing angle. Then you can compare any two corresponding angles for congruence.
Side-Angle-Side (SAS) theorem
The second theorem requires an exact order: a side, then the included angle, then the next side. TheSide-Angle-Side (SAS) theoremstates if two sides of one triangle are proportional to two corresponding sides of another triangle, and their corresponding included angles are congruent, the two triangles are similar.
Trying side-angle-side
Here are two triangles, side by side and oriented in the same way.△RAPand△EMOboth have identified sides measuring 37 inches on△RAPand 111 inches on△EMO, and also sides 17 on△RAPand 51 inches on△EMO. Notice that the angle between the identified, measured sides is the same on both triangles:47°.
Is the ratio37111\frac{37}{111}11137the same as the ratio1751\frac{17}{51}5117? Yes; the two ratios are proportional, since they each simplify to13\frac{1}{3}31. With their included angle the same, these two triangles are similar.
Side-Side-Side (SSS) theorem
The last theorem isSide-Side-Side, or SSS. This theorem states that if two triangles have proportional sides, they are similar. This might seem like a big leap that ignores their angles, but think about it: the only way to construct a triangle with sides proportional to another triangle's sides is to copy the angles.
Trying side-side-side
Here are two triangles,△FLOand△HIT. Notice we have not identified the interior angles. The sides of△FLOmeasure 15, 20, and 25 cm in length. The sides of△HITmeasure 30, 40, and 50 cm in length.
You need to set up ratios of corresponding sides and evaluate them:
1530=12\frac{15}{30}=\frac{1}{2}3015=21
2040=12\frac{20}{40}=\frac{1}{2}4020=21
2550=12\frac{25}{50}=\frac{1}{2}5025=21
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They all are the same ratio when simplified. They all are12\frac{1}{2}21. So even without knowing the interior angles, we know these two triangles are similar, because their sides are proportional to each other.
Lesson summary
Now that you have studied this lesson, you are able to define and identify similar figures, and you can describe the requirements for triangles to be similar (they must either have two congruent pairs of corresponding angles, two proportional corresponding sides with the included corresponding angle congruent, or all corresponding sides proportional).
You also can apply the three triangle similarity theorems, known as Angle-Angle (AA), Side-Angle-Side (SAS) or Side-Side-Side (SSS), to determine if two triangles are similar. | 677.169 | 1 |
...and at the fame time denied, of the fame : alfo, that which is employed by geometry, viz. that thofe things which are equal to the fame thing are equal to each other; that which the natural philofopher ufes, that nothing can be generated from that which is not; and...
...just proved that AHG=HG D. Consequently FHB=HGD ; each being equal to AHG ; for it is an axiom that things which are equal to the fame thing are equal to each other. 35 . THEOREM. When two parallels are crossed by a straight line, the sum of the two interior angles...
...connexion, their use, their restriction, or their extension. Axioms. I. Things which are equal to the same thing, are equal to each other. 2 If equals be added to equals, the wholes will he equal. 3. If equals be taken from equals, the remainders will be equal. 4. If equals be added to...
...may be described from any centre, and with any radius. AXIOMS. 1. Things that are equal to the same thing are equal to each other. 2. If equals be added to equals, the sums are equals. 3. If equals be taken from equals, the remainders are equals. 4. If equals be added...
...proposition, or in the course of a demonstration. AXIOMS. — 1. Things which are equal to the same thing are equal to each other. 2. If equals be added to equals, the whole will be equal. 3. If equals be taken from equals, the remainders will be equal. 4. If equals...
...54, or Part I. AXIOMS An axiom is a self-evident proposition. 1. Things which are equal to the same thing are equal to each other. 2. If equals be added to equals, the wholes will be equal. 8. If equals be taken from equals, the remainders will be equal. 4. If equals be added to unequals,...
...tests of equality between two quantities : 1. Things which are equal to the tame or to equal things, are equal to each other. 2. If equals be added to equals the sums will be equal. 3. If equals be subtracted from equals the remainders will be equal. 4. Like parts...
...— neither admitting nor requiring any demonstration; such as, truth trath | 677.169 | 1 |
math4finance
Meg constructed triangle POQ and then used a compass and straightedge to accurately construct line s...
7 months ago
Q:
Meg constructed triangle POQ and then used a compass and straightedge to accurately construct line segment OS, as shown in the figure below:Which could be the measures of angle POS and angle POQ? Measure of angle POS is 25 degrees, measure of angle POQ is 40 degrees Measure of angle POS is 25 degrees, measure of angle POQ is 30 degrees Measure of angle POS is 20 degrees, measure of angle POQ is 40 degrees Measure of angle POS is 20 degrees, measure of angle POQ is 30 degrees
Accepted Solution
A:
In order to understand this question, one needs to understand the construction first. After creating POQ, Meg picks a radius r and makes a circle around O. There are 2 points where the circle meets the triangle. Then, she meets another radius and she makes a circle of radius s around each of the points on the triangle. The two circles cross at one point, as can be seen in the figure. Let's name the first two points A and B and let's name the last point S. We have that for the triangles OAS, OBS we have OA=OB=r, BS=AS=s and that OS is common to both. Thus by the S-S-S triangle equality theorem, the triangles are equal. Hence for the angles AOS=BOS, namely POS=QOS. Since POQ=POS+QOS, we have that POS has to be half of POQ. The only consistent choice is the third one, POS=20 degrees and POQ=40 degrees. | 677.169 | 1 |
Ray-Plane and Ray-Disk Intersection
Reading time: 4 mins.
Ray-Plane Intersection
Figure 1: Ray-plane intersection.
In this chapter, we explore how to calculate the intersection of a ray with a plane and a disk. From our Geometry lesson, we know that the dot (or scalar) product of two vectors perpendicular to each other always equals 0:
$$A \cdot B = 0$$
This holds true specifically when vectors \(A\) and \(B\) are perpendicular. A plane is characterized by a point \(p_0\), indicating its distance from the world's origin, and a normal \(n\), which defines the plane's orientation. We can derive a vector on the plane from any point \(p\) on it by subtracting \(p_0\) from \(p\). Since this resultant vector lies within the plane, it is perpendicular to the plane's normal. Leveraging the property that the dot product of two perpendicular vectors equals 0, we have (equation 1):
$$(p - p_0) \cdot n = 0$$
Likewise, a ray is described using the parametric form (equation 2):
$$l_0 + l * t = p$$
Here, \(l_0\) represents the ray's origin, and \(l\) denotes the ray's direction. This implies we can pinpoint any position along the ray using the ray's origin, its direction, and the scalar \(t\), a positive real number signifying the parametric distance from the ray's origin to the point of interest. If the ray intersects the plane, they share a point \(p\) at the intersection. Substituting equation 2 into equation 1 gives us:
$$(l_0 + l * t - p_0) \cdot n = 0 $$
Our goal is to find a value for \(t\) that allows us to compute the intersection point's position using the ray's parametric equation. Solving for \(t\), we obtain:
It's worth noting that if the plane and ray are parallel (i.e., when \(l \cdot n\) approaches 0), they either perfectly coincide, offering an infinite number of solutions, or they do not intersect at all. In practical C++ implementations, we typically return false (indicating no intersection) when the denominator is less than a very small threshold.
Ray-Disk Intersection
Figure 2: Ray-disk intersection.
The procedure for determining a ray-disk intersection is straightforward. Typically, a disk is defined by its position (the center of the disk), a normal, and a radius. The initial step involves checking if the ray intersects the plane where the disk resides. For this ray-plane intersection phase, we can employ the code developed for the ray-plane intersection test. If there's an intersection with the plane, the next step is to calculate the intersection point and measure the distance from this point to the disk's center. The ray intersects the disk if this distance is less than or equal to the disk's radius. As an optimization, instead of directly calculating the distance, you can compare the square of the distance against the square of the disk's radius. This squared distance can be derived from the dot product of vector \(v\) (in the code) with itself. Computing the actual distance would necessitate taking the square root of this dot product. However, for efficiency, we can compare the dot product result directly against the square of the radius (often precalculated), thereby avoiding the costly square root operation. | 677.169 | 1 |
geometry angles in circles partner worksheet answers
Angles Geometry Worksheet Answers – Angle worksheets can be helpful when teaching geometry, especially for children. These worksheets contain 10 types of questions on angles. These questions include naming the vertex, arms, and location of an angle. Angle worksheets are a key part of a student's math curriculum. They help students understand the different parts … Read more | 677.169 | 1 |
They [the students] also cameuponnew and unusualmathematicalfigures: the digon, a two-sidedpolygon on a sphericalspace, and the apeirogon, an openpolygon with infinitelymanysides [ …] . All these discoveriesbrought upevenmorequestions. Is a circle a polygon? What makes an octagon an octagon – its eightvertices, its eightsides, or both? Can a polygoncrossitself? Does a polygonneed to be closed?
A directed graph (A,R) is a set of vertices A together with an incidencerelation R: if aRb then there is an edgegoing from A to B. | 677.169 | 1 |
hangobw6h
Answered question
2023-03-24
Answer & Explanation
l0st4w0rdsx3hyfw
Beginner2023-03-25Added 5 answers
Each pair of alternative angles and each pair of matching angles are equal if a transversal cuts two parallel lines.These two propositions, however, do not stand alone. By supposing the opposite and employing propositions, this may be simply demonstrated. So, answers is each pair of alternate angles are equal and each pair of corresponding angles is equal. | 677.169 | 1 |
19, Asset beta, 0.55, Calculated as an average during five years for the Equity premium - government bonds 1900-2016, Geometric mean
n. The n th root, usually the positive n th root, of a product of n factors. American Heritage® Dictionary of the English Language, Fifth Edition. Kemna and Vorst (1990) proposed an analytic expression for Asian options with geometric average using the partial differential equation (PDE) approach, on this basis, geometric average as control variable employed in the Monte Carlo simulation method (Boyle, 1977) was used to obtain satisfactory result for pricing Asian options with arithmetic
2019-03-01
Geometric average - math problems Number of problems found: 29. Geometric mean Calculate the geometric mean of numbers a=15.2 and b=25.6. Determine the mean by construction where a and b are the length of the lines.
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gmean3 is a 3-by-5 array. Find the geometric mean of each page of X by specifying the first and second dimensions using the vecdim input argument. mpage = geomean (X, [1 2]) mpage = mpage (:,:,1) = 6.4234 mpage (:,:,2) = 22.5845. For example, mpage (1,1,2) is the geometric mean of the elements in X (:,:,2). The geometric mean, sometimes referred to as geometric average of a set of numerical values, like the arithmetic mean is a type of average, a measure of central tendency. Due to the formula used to calculate it, all values in the dataset must have the same sign, that is, they must be all positive or all negative.
When the means of a proportion are the same number, that number is called the geometric mean of the extremes. So if. px
ge′omet′ric mean′, [Math.] Mathematicsthe mean of n positive numbers obtained by taking the n th root of the product of the numbers:The geometric mean of
The arithmetic mean, or simple average, is the unbiased measure of the expected value of repeated observa- tions of a random variable, not the geometric mean.
If you calculate this geometric mean you get approximately 1.283, so the average rate of return is about 28% (not 30% which is what the arithmetic mean of 10%, 60%, and 20% would give you). Any time you have a number of factors contributing to a product, and you want to find the "average" factor, the answer is the geometric mean.
df["returns"] = 1 + .01*df
15 Nov 2018 Geometric mean involves roots and multiplication, not addition and division. You get geometric mean by multiplying numbers together and then
Length Of Stay: What is the difference between "Average" and "Geometric Mean" ? One focus of every hospital case management department or utilization
This handy geometric average return (GAR) calculator can be used with investments that undergo compounding over a number of timespans to calculate the
26 Apr 2020 Quick Summary of Geometric Average.
av M Miki · 2005 · Citerat av 155 — For rocket-triggered lightning the geometric mean ( GM) values of the three overall charge transfer, and average current, are similar to their counterparts for the
Geometric mean has the specific definitions below, and has utility in science, finance, and statistics. Mathematical definition: The nth root of the product of n numbers. Practical definition: The average of the logarithmic values of a data set, converted back to a base 10 number. The geometric mean can be used to calculate average rates of return in finances or show how much something has grown over a specific period of time. In order to find the geometric mean, multiply all of the values together before taking the nth root, where n equals the total number of values in the set. You can also use the logarithmic functions on your calculator to solve the geometric mean if you want. The geometric mean, sometimes referred to as geometric average of a set of numerical values, like the arithmetic mean is a type of average , a measure of central tendency. Michael bratton bgsu
Yet, the average return is 25%! The geometric average is the annual return that would have made $10 grow into $10 over 2 years with compounding. The only return that can do that is 0%, so 0% is the geometric average return for Explosive. Mutual fund companies are required by law to report the compound average, not the arithmetic average.
It's the average return rate for a set of values that is calculated using the products of the terms. In other words, the geometric average takes several values (the return rates), multiplies them all together, and sets them to the 1/nth power. 2020-04-02
The Geometric Mean is a special type of average where we multiply the numbers together and then take a square root (for two numbers), cube root (for three numbers) etc. Example: What is the Geometric Mean of 2 and 18 ? | 677.169 | 1 |
Set of 4 artificial extra angles with a diameter of 12 mm and a length of 75 mm. Packed in 4 pieces (MIX gradations) 1, 2, 3, 4. They are mainly used for larger surfaces. They are an equivalent substitute for natural angle. They are suitable for artistic and hobby creation. It is advisable to stabilize the resulting drawing with a fixative. | 677.169 | 1 |
The development of any software program, including, but not limited to, training a machine learning or artificial intelligence (AI) system, is prohibited using the contents and materials on this website.
Too easy. I'll call 'r' the distance between the lines; x,y,z as the distances between the lines.
Here's what you do:
project the lines into the x,y plane, and find the distances (treating it as a 2D line, I take it you know how to do that, yes?). This should get you x and y. Then project it into the x,z or y,z plane (your preferacne) to get z. | 677.169 | 1 |
Find the position of the point P on seg AB of length 8 cm, so that AP2+BP2 is minimum.
Video Solution
Text Solution
Verified by Experts
The correct Answer is:P is the midpoint of seg AB
|
Answer
Step by step video, text & image solution for Find the position of the point P on seg AB of length 8 cm, so that "AP"^(2)+"BP"^(2) is minimum. by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams. | 677.169 | 1 |
However, I think I'm confusing it with the Forward Vector. I know I can get a rotation matrix and use also cross products to obtain them, but they specify that I should use yaw and pitch and I'm kinda lost here since cannot find a clear answer with the setup I'm mentioning.
Could you help me to find the Up, Right and Direction vectors? (I suppose the direction is the Forward one.)
1 Answer
1
You don't specify your reference frame, but based on your
$$V_{Right} = \lbrace \cos(90−yaw),0,\sin(90−yaw) \rbrace $$
I'm guessing it's {forward, up, right} with yaw about the y-axis and pitch about the z-axis. To get the yaw rotation matrix, consider rotation in the x-z plane: | 677.169 | 1 |
Re: In the triangle ABC shown above, what is the value of ∠ACB
[#permalink]
23 Apr 2018, 03:59Re: In the triangle ABC shown above, what is the value of ∠ACB
[#permalink]
23 Apr 2018, 04:10Re: In the triangle ABC shown above, what is the value of ∠ACB
[#permalink]
23 Apr 2018, 05:07
rohitf23 wrote:IMHO, answer is B -- statement 2 alone is sufficient.
You dont know that angle b is 90. only from option a it can be decipehred that angle b is right angle
gmatclubot
Re: In the triangle ABC shown above, what is the value of ∠ACB [#permalink] | 677.169 | 1 |
This cylindrical coordinates calculator will allow you to convert Cartesian to cylindrical coordinates, as well as the other way around. It is a more complex version of the polar coordinates calculator that allows you to analyze an arbitrary point in a 3D space.
Cylindrical and Cartesian coordinates
Source: Wikimedia
Coordinates are used to describe the position of any point in space. When given the set of three coordinates, you can uniquely define a point located in a three-dimensional space.
The Cartesian coordinate system is created by choosing three lines, all perpendicular to each other, and crossing a common point known as the origin. These lines are called the x-axis, y-axis, and z-axis. Each pair of axes defines a plane. The coordinates of any arbitrary point are defined as the distance between that point and each plane.
The cylindrical coordinate system is an extension of the polar system in three dimensions. It means that the system, just like the polar one, has a pole – a central point (most often overlapping with the origin of the Cartesian system) and a polar axis – a ray that starts in the pole as in located on the CY plane (that is, at a level where z=0z = 0z=0). Additionally, the z-axis is used to describe the position of the point.
In the cylindrical system, every point has three coordinates (ρ\rhoρ, θ\thetaθ, zzz). First, you have to project the analyzed point to the XY plane. Then, draw a line between this projected point and the pole. ρ\rhoρ is the distance between the pole and the projected point, measured on the XY plane only. θ\thetaθ is the angle created by the line you drew and the polar axis. The third coordinate, zzz, is the vertical distance between the original and projected points.
You can, of course, use the cylindrical coordinates calculator to find the polar coordinates in a 2D space. Then, the z-coordinate will always be equal to 0.
Converting Cartesian to cylindrical
Let's say you want to convert the Cartesian coordinates of a point to cylindrical ones. (Our calculator assumes that the origin overlaps with the pole, and the z-axes are the same for both systems.) You need to apply the following equations:
r=x2+y2r = \sqrt{x^2 + y^2}r=x2+y2
θ=arctan(y/x)\theta = \arctan(y/x)θ=arctan(y/x)
z2=z1z_2 = z_1z2=z1
where:
(x,y,z1)(x,\ y,\ z_1)(x,y,z1) – Cartesian coordinates; and
(ρ,θ,z2)(\rho,\ \theta,\ z_2)(ρ,θ,z2) – Cylindrical coordinates.
The cylindrical coordinates are subject to the following constraints:
ρ\rhoρ must be equal or greater than 000; and
θ\thetaθ has to lie within the range (−π,π](−\pi,\pi](−π,π].
Converting cylindrical to Cartesian
You can also use our cylindrical coordinates calculator to convert cylindrical coordinates to Cartesian. In order to do that, simply use the formulas below: | 677.169 | 1 |
In order to solve this problem, you need to know the definition of the tangent. The tangent of a right triangle is the ratio of the opposite leg to the adjacent leg.
By the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the legs. Find the AC leg.
AC ^ 2 = AB ^ 2 – BC ^ 2 = 13 – 9 = 4;
AC = 2;
tgA = BC / AC = 3/2.
Answer: tgA = 3/ | 677.169 | 1 |
Man in Hat Facing Right
Tangrams, invented by the Chinese, are used to develop geometric thinking and spatial sense. Seven figures consisting of triangles, squares, and parallelograms are used to construct the given shape. This tangram depicts a man in hat facing right. | 677.169 | 1 |
Hint: Here we will use the permutation and combination method to find the total number of the triangle. The total number of vertices in regular polygons is 20 and a triangle has 3 vertices. So, using permutation and combination, the total number of the triangle can be made by 20 side polygon is ${}^{20}{{C}_{3}}$. Now, we will have to remove the unwanted triangle which is made of sides. That is to remove triangles made by using one side of polygon and triangles which are made by using two sides of the polygon.
Complete step-by-step answer:
We know that the regular polygon has 20 sides and a triangle has 3 sides. So, using permutation and combination method, the total number of triangles can be drawn is ${}^{20}{{C}_{3}}$ which we can solve by the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Applying the formula, $\begin{align} & {}^{20}{{C}_{3}}=\dfrac{20!}{3!17!} \\ & =\dfrac{20\times 19\times 18\times 17!}{3\times 2\times 17!} \\ & =\dfrac{20\times 19\times 18}{6} \\ & =20\times 19\times 3 \\ & =1140 \\ \end{align}$ Now, we have to remove the unwanted triangle that is triangles can be drawn using one side of polygon and triangles can be drawn using two side of polygon. Triangles can be drawn using one side of the polygon. If we take one side, then there are a total of 16 triangles that can be drawn and we have 20 sides. So, total $20\times 16=320$ triangles. Now, there will be a total 20 triangles that can be drawn using two sides of the polygon. So, total triangle can be drawn by vertices not using side, $\begin{align} & =1140-320-20 \\ & =1140-340 \\ & =800\ \text{triangles} \\ \end{align}$
Note: Students may make a mistake that is they do not remove the unwanted triangle which can lead to the wrong answer i.e. 1140 triangles. Also, sometimes students will not consider both cases, i.e. triangle formed using 1 side and 2 sides. So, they might get a different answers. | 677.169 | 1 |
Press ESC to closeSimilarly, Which quadrilateral has only one pair of parallel sides?
trapezoids
Also, Is rectangle a rhombus?What shape has 4 sides with different lengths?
rhombus
What are the 7 types of quadrilaterals?
Here are the seven quadrilaterals: Parallelogram: A quadrilateral that has two pairs of parallel sides. Rhombus: A quadrilateral with four congruent sides; a rhombus is both a kite and a parallelogram. Rectangle: A quadrilateral with four right angles; a rectangle is a type of parallelogram.
What is a 4 sided shape with 2 parallel sides called?
Parallelogram: A quadrilateral that has two pairs of parallel sides. Rhombus: A quadrilateral with four congruent sides; a rhombus is both a kite and a parallelogram.How many pairs of parallel sides does a trapezoid have?
A trapezoid has one pair of parallel sides and a parallelogram has two pairs of parallel sides.
What shape has 2 pairs of parallel sides?
Parallelogram
How many parallel sides does a rectangle have?
A parallelogram has two parallel pairs of opposite sides. A
Is rhombus a parallelogram?
Diagonals are equal only in the special case of trapezium called trapezoid ( or isosceles trapezium). It is a characteristic property of trapezoid, that the diagonals are equal. It is not necessary that diagonals of all types of trapezium are equal.
What does it mean to be congruent?
The adjective congruent fits when two shapes are the same in shape and size. If you lay two congruent triangles on each other, they would match up exactly. Congruent comes from the Latin verb congruere "to come together, correspond with." Figuratively, the word describes something that is similar in character or type.
What is a perpendicular side?
Perpendicular means "at right angles". A line meeting another at a right angle, or 90° is said to be perpendicular to it. In the figure above, the line AB is perpendicular to the line DF. If they met at some other angle we would say that AB meets DF 'obliquely'.
What is a quadrilateral with 2 parallel sides?
Parallelogram: a quadrilateral with two pairs of parallel sides. Equivalent conditions are that opposite sides are of equal length; that opposite angles are equal; or that the diagonals bisect each other shape is a trapezoid?Are rectangles Rhombuses?
ExplanIs a rectangle always a parallelogram?
It is true that every rectangle is a parallelogram, but it is not true that every parallelogram is not a rectangle. For instance, take a square. It's a parallelogram — it is a quadrilateral with two pairs of parallel faces. But it is also a rectangle — it is a quadrilateral with four right angles.
Does a rectangle have one pair of parallel sides?
A No, because a rhombus does not have to have 4 right angles.
Is square a rhombus?
A rhombus is a quadrilateral with all sides equal in length. A square is a quadrilateral with all sides equal in length and all interior angles right angles. Thus a rhombus is not a square unless the angles are all right angles. A square however is a rhombus since all four of its sides are of the same length.
What shape is a trapezoid?
A trapezoid is a four-sided shape with at least one set of parallel sides. It can have two and be a parallelogram. But, if two sides aren't parallel, then it's just the lowly trapezoid. So, in a trapezoid, the parallel sides are called the bases.
Is rhombus a parallelogram?
Definition of Parallel
Two sides or lines are parallel if they are lines that are always the same distance from each other and will never intersect or touch. That is what it means to be parallel.
What is a rectangle but not a parallelogram
Are squares trapezoids?
Explanation: A trapezoid is a quadrilateral with at least one pair of parallel sides. In a square, there are always two pairs of parallel sides, so every square is also a trapezoid. Conversely, only some trapezoids are squares.
What is a four sided shape with unequal sides called?
A quadrilateral is a square if and only if it is both a rhombus and a rectangle (four equal sides and four equal angles). Oblong: a term sometimes used to denote a rectangle that has unequal adjacent sides (i.e. a rectangle that is not a square). Kite: two pairs of adjacent sides are of equal length.
What is a four sided shape with unequal sides called?
DEFINITION: A rhombus is a parallelogram with four congruent sides. THEOREM: If a parallelogram is a rhombus, each diagonal bisects a pair of opposite angles. THEOREM Converse: If a parallelogram has diagonals that bisect a pair of opposite angles, it is a rhombus.
What is a rhombus look like?
A rhombus looks like a diamond
Opposite sides are parallel, and opposite angles are equal (it is a Parallelogram). And the diagonals "p" and "q" of a rhombus bisect each other at right angles.
Is a kite tangential quadrilateral. That is, it has an inscribed circle that is tangent to all four sides.
How many pairs of parallel sides does a trapezoid have?
A trapezoid has one pair of parallel sides and a parallelogram has two pairs of parallel sides. So a parallelogram is also a trapezoid is the difference between a square and tangential quadrilateral. That is, it has an inscribed circle that is tangent to all four sides.
What are the different Quadrilaterals?
A quadrilateral is a four-sided polygon with four angles. There are many kinds of quadrilaterals. The five most common types are the parallelogram, the rectangle, the square, the trapezoid, and the rhombus.
What type of quadrilateral is ABCD | 677.169 | 1 |
Hint: Sketch the diagram first to visualize the question better, then you could arrive at the answer merely by finding the perpendicular distance between the two lines.
Let's sketch the situation given to us in the question. So we have a circle that has the line $2x-y+1=0$ acting as a tangent to it at the point $(2,5)$, and we have the centre of the circle lying on the line $x-2y=4$. Drawing the figure while keeping all these features in mind, the diagram we'll get is:
The lines do look parallel in the diagram, but from the equations of the lines, we can clearly see that they are not parallel. Let's assume the $y$ coordinate of the centre of the circle to be $=t$. Since we're talking about the centre, the point will definitely satisfy the equation $x-2y=4$. Thus, we can find the $x$ coordinate of the centre simply by substituting for $y=t$ in the equation of the line. Doing so, we get : $\begin{align} & x-2y=4 \\ & \Rightarrow x=4+2y \\ & \Rightarrow x=4+2\left( t \right)=4+2t \\ \end{align}$ So, the coordinates of the centre are : $\therefore c\equiv \left( 4+2t,t \right)$ Now, let's find out the distance between the centre and the tangent line. That will be found out by applying the formula using which we can find the perpendicular distance of any point $(X,Y)$ from the line $Ax+By+C=0$. So, if we assume the perpendicular distance of point $(X,Y)$ from the line $Ax+By+C=0$ is $d$, then : $d=|\dfrac{AX+BY+C}{\sqrt{{{A}^{2}}+{{B}^{2}}}}|$ Thus, here our point $(X,Y)=(4+2t,t)$ and our line $Ax+By+C=0$ is equal to the line $2x-y+1=0$. Therefore, the perpendicular distance $d$ of point $(4+2t,t)$ from line $2x-y+1=0$ is : \[\begin{align} & d=\left| \dfrac{AX+BY+C}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right| \\ & \Rightarrow d=\left| \dfrac{2\left( 4+2t \right)-\left( t \right)+1}{\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}}} \right| \\ & \Rightarrow d=\left| \dfrac{8+4t-t+1}{\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}}} \right| \\ & \Rightarrow d=\left| \dfrac{\left( 3t+9 \right)}{\sqrt{5}} \right|...............(1) \\ \end{align}\] And since, from the figure, we can see that this distance will also be equal to the radius of the circle, $d=r=|\dfrac{3t+9}{\sqrt{5}}|$ Now, since the tangent touches the circle at the point $(2,5)$, the distance of the centre from this point will also be equal to the radius. So, by distance formula $\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}$, we get : \[\begin{align} & r=\sqrt{{{\left( (4+2t)-2 \right)}^{2}}+{{\left( t-5 \right)}^{2}}} \\ & r=\sqrt{{{\left( 2+2t \right)}^{2}}+{{\left( t-5 \right)}^{2}}}............(\text{2)} \\ \end{align}\] By equating $(1)$ and $(2)$ with each other, we get; $|\dfrac{\left( 3t+9 \right)}{\sqrt{5}}|=\sqrt{{{\left( 2+2t \right)}^{2}}+{{\left( t-5 \right)}^{2}}}$ Squaring both sides, we get : $\begin{align} & \Rightarrow \dfrac{9{{t}^{2}}+81+54t}{5}={{(2+2t)}^{2}}+{{(t-5)}^{2}} \\ & \Rightarrow 9{{t}^{2}}+81+54t=5[4+4{{t}^{2}}+8t+{{t}^{2}}+25-10t] \\ \end{align}$ $\begin{align} & \Rightarrow 9{{t}^{2}}-54t+81=5\left( 5{{t}^{2}}-2t+29 \right) \\ & \Rightarrow 16{{t}^{2}}-64t+64=0 \\ \end{align}$ Dividing by $16$ on both sides, we get : $\Rightarrow {{t}^{2}}-4t+4=0$ | 677.169 | 1 |
NCERT Solutions for Class 6 Math Chapter 14 - Practical Geometry
NCERT Solutions for Class 6 Math Chapter 14 Practical Geometry are provided here with simple step-by-step explanations. These solutions for Practical Geometry are extremely popular among class 6 students for Math Practical Geometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 6 Math Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation's NCERT Solutions. All NCERT Solutions for class 6 Math are prepared by experts and are 100% accurate.
Page No 276:
Question 1:
Draw a circle of radius 3.2 cm.
Answer:
The required circle can be drawn as follows.
Step 1
First, open the compasses for the required radius 3.2Page No 276:
Question 2:
With the same centre O, draw two circles of radii 4 cm and 2.5 cm.
Answer:
The required circle can be drawn as follows.
Step 1
First, open the compasses for the required radius 4Step 5
Now, open the compasses for 2.5 cm.
Step 6
Again put the pointer of the compasses on point 'O'
and turn the compasses slowly to draw the circle.
Page No 276:
Question 3:
Draw a circle and any two of its diameters. If you join the ends of these diameters, what is the figure obtained? What figure is obtained if the diameters are perpendicular to each other? How do you check your answer?
Answer:
A circle can be drawn of any convenient radius, also having its centre as O. Let AB and CD be two diameters of this circle. When we join the ends of these diameters, a quadrilateral ACBD is formed.
As we know that the diameters of a circle are equal in length, therefore, the quadrilateral so formed will have its diagonals of equal length.
Also, OA = OB = OC = OD = radius r and if a quadrilateral has its diagonals of same length which are bisecting each other, then it will be a rectangle.
Let DE and FG be two diameters of this circle such that these are perpendicular to each other. A quadrilateral is formed by joining the ends of these diameters.
Here, OD = OE = OF = OG = radius r
In this quadrilateral DFEG, the diagonals are equal and perpendicular to each other. Also, since these are bisecting each other, it will be a square.
The length of the sides of the quadrilateral so formed can be measured to check our answers.
Video Solution for Practical Geometry (Page: 276 , Q.No.: 3)
NCERT Solution for Class 6 math - Practical Geometry 276 , Question 3
Page No 276:
Question 4:
Draw any circle and
mark points A, B and C such that
(a) A is on the
circle. (b) B is in the interior of the circle.
(c) C is in the
exterior of the circle.
Answer:
A circle and three
required points A, B, C can be drawn as follows.
Page No 276:
Question 5:
Let
A, B be the centres of two circles of equal radii; draw them so that
each one of them passes through the centre of the other. Let them
intersect at C and D.
Examine
whether
and
are at right angles.
Answer:
Let
us draw two circles of same radius which are passing through the
centres of the other circle.
Here,
point A and B are the centres of these circles and these circles are
intersecting each other at point C and D.
In
quadrilateral ADBC,
AD
= AC (Radius of circle centered at A)
BC
= BD (Radius of circle centered at B)
As
radius of both circles are equal, therefore, AD = AC = BC = BD
Hence,
is
a rhombus and in a rhombus, the diagonals bisect each other at 90°.
Hence,
and
are at right angles.
Page No 278:
Question 1:
Draw a line segment of
length 7.3 cm using a ruler.
Answer:
A line segment of
length 7.3 cm can be drawn using a ruler as follows.
(1) Mark a point A on
the sheet.
(2) Put 0 mark of ruler
at point A.
(3) Mark a point B on
the sheet at 7.3 cm on ruler.
(4) Join A and B.
is
the required line segment.
Page No 278:
Question 2:
Construct a line
segment of length 5.6 cm using ruler and compasses.
Answer:
A line segment of length 5.6 cm can be drawn using a ruler and
compasses as follows.
(1) Draw a line l
and mark a point A on this line.
(2) Place the compasses on the zero mark of the ruler. Open it to
place the pencil up to the 5.6 cm mark.
(3) Place the pointer of compasses on point A and draw an arc to cut
l at B. AB is the line segment of 5.6 cm length.
Question 4:
Answer:
A line segment
can be drawn such that the length of
is twice that of
as follows.
(1) Draw a line l and mark a point P on it and let AB be the
given line segment of 3.9 cm.
(2) By adjusting the
compasses up to the length of AB, draw an arc to cut
the
line at X, while taking the pointer of compasses at point P.
(3) Again put the
pointer on point X and draw an arc to cut line l again at
Q.
is the required line segment. By ruler, the length of
can be measured which comes to 7.8 cm.
Page No 278:
Question 5:
Given of length 7.3 cm and of length 3.4 cm, construct a line segment such that the length of is equal to the difference between the lengths of and. Verify by measurement.
Answer:
(1) Given that, =7.3 cm and = 3.4 cm
(2) Adjust the compasses up to the length of CD and put the pointer of the compasses at A. Draw an arc to cut AB at P.
(3) Adjust the compasses up to the length of PB. Now draw a line l and mark a point X on it.
(4) Now, putting the pointer of compasses at point X, draw an arc to cut the line at Y.
is the required line segment.
Video Solution for Practical Geometry (Page: 278 , Q.No.: 5)
NCERT Solution for Class 6 math - Practical Geometry 278 , Question 5
Page No 279:
Question 1:
Draw any line segment.
Without measuring,
construct a copy of.
Answer:
The following steps
will be followed to draw the given line segment
and to construct a copy of.
(1) Let
be the given line segment.
(2) Adjust the
compasses up to the length of
.
(3) Draw any line l
and mark a point A on it.
(4) Put the pointer on point A, and without changing the setting of
compasses, draw an arc to cut the line segment at point B.
is the required line segment.
Page No 279:
Question 2:
Given some line segment, whose length you do not know, construct such that the length of is twice that of .
Answer:
The following steps will be followed to construct a line segment such that the length of is twice that of .
(1) Let be the given line segment.
(2) Adjust the compasses up to the length of .
(3) Draw any line l and mark a point P on it.
(4) Put the pointer on P and without changing the setting of compasses, draw an arc to cut the line segment at point X.
(5) Now, put the pointer on point X and again draw an arc with the same radius as before, to cut the line l at point Q.
is the required line segment.
Video Solution for Practical Geometry (Page: 279 , Q.No.: 2)
NCERT Solution for Class 6 math - Practical Geometry 279 , Question 2
Page No 284:
Question 1:
Draw any line segment. Mark any point M on it. Through M, draw a perpendicular to. (Use ruler and compasses)
Answer:
(1) Draw the given line segment and mark any point M on it.
(2) With M as centre and a convenient radius, construct an arc intersecting the line segmentat two points C and D.
(3) With C and D as centres and a radius greater than CM, construct two arcs. Let these be intersecting each other at E.
(4) Join EM. is perpendicular to .
Video Solution for Practical Geometry (Page: 284 , Q.No.: 1)
NCERT Solution for Class 6 math - Practical Geometry 284 , Question 1
Page No 284:
Question 2:
Draw any line segment.
Take any point R not on it. Through R, draw a perpendicular to.
(Use ruler and set-square)
Answer:
(1) Take the given line
segment
and mark any point R outside.
(2) Place a set square on
such that one arm of its right angle aligns along.
(3) Place the ruler
along the edge opposite to the right angle of the set square.
(4) Hold the ruler fixed. Slide the set square along the ruler till
the point R touches the other arm of the set square.
(5) Draw a line along this edge of the set square which will be
passing through R. It is the required line, which is perpendicular
to.
Page No 284:
Question 3:
Draw a line l and point X on it. Through X, draw a line segment perpendicular to l.
Now draw a perpendicular to at Y. (use ruler and compasses)
Answer:
(1) Draw a line l and mark a point X on it.
(2) Taking X as centre and with a convenient radius, draw an arc intersecting line l at two points A and B.
(3) With A and B as centres and a radius more than AX, construct two arcs intersecting each other at Y.
(4) Join XY. is perpendicular to l.
Similarly, a perpendicular to at the point Y can be drawn. The line is perpendicular to at Y.
Video Solution for Practical Geometry (Page: 284 , Q.No.: 3)
NCERT Solution for Class 6 math - Practical Geometry 284 , Question 3
Page No 286:
Question 1:
Draw
of
length 7.3 cm and find its axis of symmetry.
Answer:
The below given steps
will be followed to construct
of
length 7.3 cm and to find its axis of symmetry.
(1) Draw a line
segment
of 7.3 cm.
(2) Taking A as centre, draw a circle by using compasses. The radius
of circle should be more than half the length of.
(3) With the same radius as before, draw another circle using
compasses while taking point B as centre. Let it cut the previous
circle at C and D.
(4) Join.
is the axis of symmetry.
Page No 286:
Question 2:
Draw a line segment of
length 9.5 cm and construct its perpendicular bisector.
Answer:
The below given steps
will be followed to construct a line segment of length 9.5 cm and its
perpendicular bisector.
(1) Draw a line
segment
of 9.5 cm.
(2) Taking P as centre, draw a circle by using compasses. The radius
of circle should be more than half the length of
.
(3) With the same radius as before, draw another circle using
compasses while taking point Q as centre. Let it cut the previous
circle at R and S.
(4) Join RS.
is the axis of symmetry i.e., the perpendicular bisector of line .
Page No 286:
Question 3:
Draw the perpendicular
bisector ofwhose
length is 10.3 cm.
(a) Take any point P on
the bisector drawn. Examine whether PX = PY.
(b) If M is the mid
point of,
what can you say about the lengths MX and XY?
Answer:
(1) Draw a line
segment
of 10.3 cm.
(2) Taking point X as centre, draw a circle by using compasses. The
radius of circle should be more than half the length of
.
(3) With the same radius as before, draw another circle using
compasses while taking point Y as centre. Let it cut the previous
circle at A and B.
(4) Join.
is the axis of symmetry.
(a) Take any point P on
.
We will find that the measures of the lengths of PX and PY are same.
It
is because
is
the axis of symmetry. Hence, any point lying on
will
be at the same distance from both the ends of.
(b) M is the mid-point of.
Perpendicular bisector
will
be passing through point M. Hence, length ofis
just double of.
Or, 2MX = XY
Page No 286:
Question 4:
Draw a line segment of
length 12.8 cm. Using compasses; divide it into four equal parts.
Verify by actual measurement.
Answer:
(1) Draw a line
segmentof
12.8 cm.
(2) Draw a circle,
while taking point X as centre and radius more than half of XY.
(3) With same radius and taking centre as Y, again draw arcs to cut
the circle at A and B. Join AB which intersectsat
M.
(4) Taking X and Y as centres, draw two circles with radius more than
half of.
(5) With same radius and taking M as centre, draw arcs to intersect
these circles at P, Q and R, S.
(6) Join PQ and RS.
These are intersectingat
T and U.
(7) Now,
.
These are 4 equal parts of.
By
measuring these line segments with the help of ruler, we will find
that each is of 3.2 cm.
Page No 286:
Question 5:
With
of
length 6.1 cm as diameter draw a circle.
Answer:
(1) Draw a line
segmentof
6.1 cm.
(2) Taking point P as
centre and radius more than half of,
draw a circle.
(3) With same radius and taking Q as centre, draw arcs to intersect
this circle at points R and S.
(4) Join RS which
intersects
at T.
(5) Taking T as centre and with radius TP, draw a circle which will
also pass through Q. It is the required circle.
Page No 286:
Question 6:
Draw a circle with
centre C and radius 3.4 cm. Draw any chord.
Construct the perpendicular bisector of
and
examine if it passes through C Now, mark any
chordin
the circle.
(4) Taking A and B as centres, draw arcs on both sides of.
Let these intersect each other at D and E.
(5) Join DE, which is
the perpendicular bisector of AB.
Whenis
extended, it will pass through point C.
Page No 286:
Question 7:
Repeat question 6, if
happens
to be a diameter Mark any
diameterin
the circle.
(4) Now, taking A and B as centres, draw arcs on both sides of
taking radius more than.
Let these intersect each other at D and E.
(5) Join DE, which is
the perpendicular bisector of AB.
It
can be observed that
is
passing through the centre C of the circle.
Page No 286:
Question 8:
Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?
Answer:
(1) Mark any point C on the sheet. Now, by adjusting the compasses up to
4 cm and by putting the pointer of compasses at point C, turn the compasses slowly to draw the circle. It is the required circle of 4 cm radius.
(2) Take any two chordsandin the circle.
(3) Taking A and B as centres and with radius more than half of, draw arcs on both sides of AB, intersecting each other at E, F. Join EF which is the perpendicular bisector of AB.
(4) Taking C and D as centres and with radius more than half of, draw arcs on both sides of CD, intersecting each other at G, H. Join GH which is the perpendicular bisector of CD.
Now, we will find that when EF and GH are extended, they meet at the centre of the circle i.e., point O.
Video Solution for Practical Geometry (Page: 286 , Q.No.: 8)
NCERT Solution for Class 6 math - Practical Geometry 286 , Question 8
Page No 286:
Question 9:
Draw any angle with
vertex O. Take a point A on one of its arms and B on another such
that OA = OB. Draw the perpendicular bisectors of
and.
Let them meet at P. Is
PA = PB?
Answer:
(1)Draw any
angle whose vertex is O.
(2) With a convenient radius, draw arcs on both rays of this angle
while taking O as centre. Let these points be A and B.
(3) Taking O and A as centres and with radius more than half of OA,
draw arcs on both sides of OA. Let these be intersecting at C and D.
Join CD.
(4) Similarly, we can find the perpendicular bisectorof.
These perpendicular bisectorsandwill
intersect each other at P.
Now,
PA and PB can be measured. These are equal in length.
Page No 291:
Question 1:
Draw ÐPOQ
of measure 75° and find its
line of symmetry.
Answer:
The below given steps
will be followed to construct an angle of 75°
and its line of symmetry.
(1) Draw a line l and mark two points O and Q on it, as shown
in the figure. Draw an arc of convenient radius, while taking point O
as centre. Let it intersect line l at an arc of same radius to intersect
each other at U.
(5) Join OU. Let it intersect the arc at V. Now, taking S and V as
centres, draw arcs with radius more than
SV.
Let those intersect each other at P. Join OP, which is the ray making
75° with the line l.
(6) Let this ray be intersecting our major arc at point W. Now,
taking R and W as centres, draw arcs with radius more than
RW
in the interior of angle of 75º. Let these be intersecting each
other at X. Join OX.
OX
is the line of symmetry for ÐPOQ
= 75°.
Page No 291:
Question 2:
Draw an angle of
measure 147° and construct its
bisector.
Answer:
The below given steps
will be followed to construct an angle of 147 147º. Join OA. OA is the required ray
making 147º with line l.
(3) Draw an arc of convenient radius, while taking point O as centre.
Let it intersect both rays of angle 147º at point A and B.
(4) Taking A and B as centres, draw arcs of radius more than
AB
in the interior of angle of 147º. Let those intersect each other
at C. Join OC.
OC
is the required bisector of 147º angle.
Page No 291:
Question 3:
Draw a right angle and construct its bisector.
Answer:
The below given steps will be followed to construct a right angle and its bisector.
(1) Draw a line l and mark a point P on it. Draw an arc of convenient radius, while taking point P as centre. Let it intersect line l at R.
(2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(4) Taking S and T as centres, draw arcs of same radius to intersect each other at U.
(5) Join PU. PU is the required ray making 90º with line l. Let it intersect the major arc at point V.
(6) Now, taking R and V as centres, draw arcs with radius more than RV to intersect each other at W. Join PW.
PW is the required bisector of this right angle.
Video Solution for Practical Geometry (Page: 291 , Q.No.: 3)
NCERT Solution for Class 6 math - Practical Geometry 291 , Question 3
Page No 291:
Question 4:
Draw an angle of
measure153° and divide it into
four equal parts.
Answer:
The below given steps
will be followed to construct an angle of 153 153º. Join OA. OA is the required ray
making 153º with line l.
(3) Draw an arc of convenient radius, while taking point O as centre.
Let it intersect both rays of angle 153º at point A and B.
(4) Taking A and B as centres, draw arcs of radius more than
AB
in the interior of angle of 153º. Let those intersect each other
at C. Join OC.
(5) Let OC intersect the major arc at point D. Now, with radius more
than
AD,
draw arcs while taking A and D as centres, and D and B as centres.
Let these be intersecting each other at point E and F respectively.
Join OE, OF.
OF,
OC, OE are the rays dividing 153º angle in 4 equal parts.
Page No 291:
Question 5:
Construct with ruler
and compasses, angles of following measures:
(a) 60° (b) 30° (c) 90°
(d) 120° (e) 45° (f) 135°
Answer:
(a) 60°
The
below given steps will be followed to construct an angle of 60°.
(1) Draw a line l and mark a point P on it. Now, Join PR which is the required ray making 60°
with line l.
(b) 30°
The
below given steps will be followed to construct an angle of 30°.
(1) Draw a line l and mark a point P on it. Now taking P as
centre and with Now, taking Q and R as centre and with radius more than
RQ,
draw arcs to intersect each other at S. Join PS which is the required
ray making 30° with line l.
(c) 90º
The
below given steps will be followed to construct an angle of 90 centre, draw an arc of same radius to intersect
each other at T.
(5) Join
PT, which is the required ray making 90°
with line l.
(d) 120º
The
below given steps will be followed to construct an angle of 120 Join
PS, which is the required ray making 120°
with line l.
(e)45º
The
below given steps will be followed to construct an angle of 45 centres, draw arcs of same radius to intersect
each other at T.
(5) Join PT. Let it intersect the major arc at point U.
(6) Taking Q and U as centres, draw arcs with radius more than
QU
to intersect each other at V. Join PV.
PV is
the required ray making 45°
with the given line l.
(f) 135°
The
below given steps will be followed to construct an angle of 135°.
(1) Draw a line l and mark a point P on it. Now taking P as
centre and with a convenient radius, draw a semi-circle which
intersects line l at Q and arcs of same radius to intersect
each other at U.
(5) Join PU. Let it intersect the arc at V. Now taking Q and V as
centres and with radius more than
QV,
draw arcs to intersect each other at W.
(6) Join
PW which is the required ray making 135°
with line l.
Page No 291:
Question 6:
Draw an angle of measure 45° and bisect it.
Answer:
The below given steps will be followed to construct an angle of
45º and its
bisector.
(1) ∠POQ of 45º
measure can be formed on a line l by using the protractor.
(2) Draw an arc of a convenient radius, while taking point O as
centre. Let it intersect both rays of angle 45º at point A and
B.
(3) Taking A and B as centres, draw arcs of radius more than
AB
in the interior of angle of 45º. Let those intersect each other
at C. Join OC.
OC is the required bisector of 45º angle.
Page No 291:
Question 7:
Draw an angle of
measure 135° and bisect it.
Answer:
The below given steps
will be followed to construct an angle of 135°
and its bisector.
(1) ÐPOQ
of 135º measure can be formed on a line l by using the
protractor.
(2) Draw an arc of a convenient radius, while taking point O as
centre. Let it intersect both rays of angle 135º at point A and
B.
(3) Taking A and B as centres, draw arcs of radius more than
AB
in the interior of angle of 135º. Let those intersect each other
at C. Join OC.
OC
is the required bisector of 135º angle.
Page No 291:
Question 8:
Draw an angle of 70°.
Make a copy of it using only a straight edge and compasses.
Answer:
The below given steps
will be followed to construct an angle of 70º measure and its
copy.
(1) Draw a line l and mark a point O on it. Place the centre
of the protractor at point O and the zero edge along line l.
(2) Mark a point A at 70º. Join OA. OA is the ray making 70º
with line l. Draw an arc of convenient radius in the interior
of 70º angle, while taking point O as centre. Let it intersect
both rays of angle 70º at point B and C.
(3) Draw a line m and mark a point P on it. With the same
radius as used before, again draw an arc while taking point P as
centre. Let it cut the line m at point D.
(4) Now, adjust the compasses up to the length of BC. With this
radius, draw an arc while taking D as centre, which will intersect
the previously drawn arc at point E.
(5) Join PE. PE is the required ray which makes the same angle (i.e.
70º) with line m.
Page No 291:
Question 9:
Draw an angle of 40°.
Copy its supplementary angle.
Answer:
The below given steps
will be followed to construct an angle of 40º measure and the
copy of its supplementary angle.
(1) Draw a line segment
and mark a point O on it. Place the centre of the protractor at point
O and the zero edge along line segment.
(2) Mark a point A at 40º. Join OA. OA is the required ray
making 40º with.
Ð POA is the supplementary
angle of 40º.
(3) Draw an arc of convenient radius in the interior of Ð
POA, while taking point O as centre. Let it intersect both rays of Ð
POA at point B and C.
(4) Draw a line m and mark a point S on it. With the same
radius as used before, again draw an arc while taking point S as
centre. Let it cut the line m at point T.
(5) Now, adjust the compasses up to the length of BC. With this
radius, draw an arc while taking T as centre, which will intersect
the previously drawn arc at point R.
(6) Join RS. RS is the required ray which makes the same angle with
line m, as the supplementary of 40º is 140º. | 677.169 | 1 |
Plane and Solid Geometry
From inside the book
Results 1-5 of 56
Page 17 ... vertices of the three angles are the vertices of the triangle . The lines AB , BC , Thus ABC is a triangle . and CA are the sides of the triangle . The angles A , B , and C are the angles of the triangle . The points A , B , and C are ...
Page 35 ... vertices at the point A , making ≤z = 2Zx and Zy = 3Zx , when it is true that 2x + 2y + Zz = 180 ° . y z x A FIG . 5 61. From the definition of a right angle , § 37 , it follows that all right angles are equal . 62. From the definition ...
Page 43 ... vertices A and F C K D ДДД A BA FE F H falling on opposite sides of HK . Then ZKHF and ZKHA are right angles . AHF is a straight line . Given § 51 ( If two adjacent angles are supplementary , their exterior sides Then are in the same ...
Page 44 ... - celes triangle from the vertices of the opposite angles are equal . 9. In the square ACDE , BC = FE , and Zx = Lz . the middle point of FB . Prove that K is AXIOMS OF EQUALITY 103. Axiom . An axiom is a 44 PLANE GEOMETRY. | 677.169 | 1 |
What is the angle between the lines x−21=y+1−2=z+21andx−11=2y+33=z+52
A
π2
B
π3
C
π6
D
None of these
Video Solution
Text Solution
Verified by Experts
The correct Answer is:A
|
Answer
Step by step video, text & image solution for What is the angle between the lines (x-2)/1=(y+1)/(-2)=(z+2)/1 and (x-1)/1=(2y+3)/3=(z+5)/2 by Maths experts to help you in doubts & scoring excellent marks in Class 14 exams. | 677.169 | 1 |
Measuring an Angle
We will discuss here about measuring an angle with the help of
a protractor.
The measurement of an angle or its size depends upon the amount of opening between its sides.
In the above Fig., an ∠ABC has a measurement denoted by b or an arc of a circle drawn in the inside region.
Measuring Angles Using Protractor:
A protractor is a geometrical instrument that looks like the letter D.
You can find it in your Geometry Box. It is used to measure angle and draw angle of required magnitude.
It has the semi-circular shape. The curved edge is divided into 180 equal parts.
i.e., Its
semi-circular edge is divided into 180 divisions. Each division represents a
degree denoted by °.
Each part is equal to a 'degree'. The markings start from 0° on the right side and end with 180° on the left side and vice- versa.
It is used to measure angle and draw angle of required magnitude.
There are two scales marked on the protractor – outer and
inner. The outer scale is marked from 0° to 180° in clockwise direction.
Let us follow the Working Rules given below to measure an angle ABC, using the protractor.
Working Rules for Measurement of an Angle:
Measuring ∠ABC using the Protractor
Step I: Place the protractor such that the mid- point (M in the figure) of its straight-edge lies on the vertex B of the angle ABC.
Step II: Adjust the protractor such that BC is along the straight-edge of the protractor.
Step III: There are two 'scales' on the protractor: read that scale which has the 0° make coinciding with the straight-edge (i.e., with ray BC)
Step IV: The mark shown by BA on the curved edge gives the degree measure of the angle.
We write m∠ABC = 60° or ∠ABC = 60°.
Å To measure the given ∠MON, we place the protractor in such way that its centre is on the vertex O of ∠MON and its straight horizontal edge lies on the arm ON of the ∠MON as shown in figure.
Now we see which degree mark on the scale of the protractor,
the arm OM coincides with. In the given figure, OM coincides with 90°.
So, the ∠MON measures 90°.
Comparison of Angles:
The magnitude of an angle depends upon the opening or inclination between the two ray that form the angle. If two angles have different inclinations, then we say they have different magnitudes.
The magnitudes of two angles can be compared in the following manner.
Comparison by Inspection:
More is the opening between two arms, greater will be the magnitude of the angle.
Observe the opening of the two angles in the following figures
We can say that the angle measure of PQR is less than that of MNO.
Comparison by Using Trace Paper:
Trace an angle, say ∠ABC on a tracing paper. Place it on another angle, say ∠DEF such that the vertex E falls on the vertex B and arm EF along arm BC. Then, we may have any one of the following three situations for the other arm BA.
(i) Arm BA may fall beyond DE. In this case, ∠ABC > ∠DEF.
(ii) Arm BA may fall exactly along ED. In this case, ∠DEF = ∠ABC.
(iii) Arm BA may fall between DB and BC. In this case, ∠ABC < ∠DEF.
Example:
1. In the Figure below, find the largest angle. Also, arrange the angles in the descending order.
A simple closed curve or a polygon formed by three line-segments (sides) is called a triangle. The above shown shapes are triangles. The symbol of a triangle is ∆. A triangle is a polygon with three sides. In the given figure ABC is a triangle. AB, BC and CA are its sides.
In circle math the terms related to the circle are discussed here. A circle is such a closed curve whose every point is equidistant from a fixed point called its centre. The symbol of circle is O. We have learned to draw a circle, by tracing the outlines of objects like a
Symmetrical shapes are discussed here in this topic. Any object or shape which can be cut in two equal halves in such a way that both the parts are exactly the same is called symmetrical. The line which divides the shape is called the symmetry. So, if we place a mirror
We will discuss here how to find the perimeter of a triangle. We know perimeter of a triangle is the total length (distance) of the boundary of a triangle. Perimeter of a triangle is the sum of lengths of its three sides. The perimeter of a triangle ABC Perimeter
We will discuss here how to find the perimeter of a rectangle. We know perimeter of a rectangle is the total length (distance) of the boundary of a rectangle. ABCD is a rectangle. We know that the opposite sides of a rectangle are equal. AB = CD = 5 cm and BC = AD = 3 cm
We will discuss here how to find the perimeter of a square. Perimeter of a square is the total length (distance) of the boundary of a square. We know that all the sides of a square are equal. Perimeter of a Square Perimeter of the square ABCD = AB+BC+CD+AD=2 cm+2cm+2cm+2cm | 677.169 | 1 |
circumference of the circle above?
[#permalink]
30 Jan 2011, 12:08
3
Kudos
1
Bookmarks
Q: What is the radius of circle. \(Circumference=2*\pi*radius\) \(\pi\) is a constant.
1. \(\stackrel{\frown}{XYZ}=18\) \(\stackrel{\frown}{XYZ}=(\theta/360)*2*\pi*radius\) \(\stackrel{\frown}{XYZ}\)=length of the arc XYZ \(18=(\theta/360)*2*\pi*radius\) \(\theta\) angle subtended by major arc \(\stackrel{\frown}{XYZ}\) at the center of the circle \(\theta\) unknown NOT SUFFICIENT.
2. \(\angle r = \angle s\) The triangle is an equilateral triangle and \(\angle r = 60^{\circ}\) Minor arc \(\stackrel{\frown}{XZ}\) makes an angle of \(60^{\circ}\) with the point Y, which lies on the circumference. Theorem,The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. So if the minor arc XZ is making an angle of r with a point on the circumference of the circle, minor arc XZ will make an angle of 2r with the centre of the circle. Now we know angle of minor arc XZ is \(2r=120^{\circ}\) And angle subtended by major arc XYZ at the center is \(\theta=360-120=240\) However, with this information also we won't be able to find the radius of the circle because the length of the arc is not known. NOT SUFFICIENT.
Combining both of the above; We know, length of major arc \(\stackrel{\frown}{XYZ}\) and angle made by the arc from the center i.e. \(\theta\)
Re: What is the circumference of the circle above?
[#permalink]
11 Jun 2023, 11 | 677.169 | 1 |
The Elements of Euclid, books i. to vi., with deductions, appendices and historical notes, by J.S. Mackay. [With] Key
From inside the book
Results 6-10 of 78
Page 24 ... less . F E B Let AB and C be the two given straight lines , of which AB is the greater : it is required to cut off from AB a part = C. = C ; I. 2 From A draw the straight line AD with centre A and radius AD , describe the O DEF , Post ...
Page 25 ... less straight line . PROPOSITION 4 . THEOREM . If two sides and the contained angle of one triangle be equal to two sides and the contained angle of another triangle , the two triangles shall be equal in every respect - that is , ( 1 ) ...
Page 26 ... less than D , where would AC fall ? 3. If AB = DE , LA = LD , but AC greater than DF , where would fall ? 4. If AB DE , LA = LD , but AC less = would C fall ? than DF , where 5. Prove the proposition beginning the superposition with the ...
Page 45 ... less than two right angles . A Д B Let ABC be a triangle : -D it is required to prove the sum of any two of its angles less than 2 rt . LS . Produce BC to D. Then ABC is less than △ ACD . I. 16 L ABC + ACB is less than △ ACD + △ ACB ...
Popular passages
Page 147 - 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 276
Page 331 - If the vertical angle of a triangle be bisected by a straight line which also cuts the base, the segments of the base shall have the same ratio which the other sides of the triangle have to one another...
Page 112 - 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.
Page 87 254 - If there be four magnitudes, and if any equimultiples whatsoever be taken of the first and third, and any equimultiples whatsoever of the second and fourth, and if, according as the multiple of the first is greater than the multiple of the second, equal to it or less, the multiple of the third is also greater than the multiple of the fourth, equal to it or less ; then, the first of the magnitudes is said to have to the second the same ratio that the third has to the fourth.
Page 138 - RULE. from half the sum of the three sides, subtract each side separately; multiply the half sum and the three remainders together, and the square root of the product will be the area required. | 677.169 | 1 |
Unjustified Assumptions Based on Diagrams in Geometry
Abstract
abstract = "",
N2AB | 677.169 | 1 |
3 2 Angles and Parallel Lines Worksheet Answers
If you have some question in your mind regarding how to draw 3 2 Angles and Parallel Lines, then this is the right article that you should read. This is the perfect guide for you so that you can get more information about how to draw these lines.
This is also a guide for you if you are asking yourself if it is possible to draw 3 parallel lines. This is because the answer is yes, you can do this. You will learn the right technique to draw these lines and how to make them look like they have angles on them.
When you are trying to learn how to draw this, the first thing that you have to remember is that lines are not flat. They have different curvatures when you are drawing these lines and the angle of these lines is not always going to be straight. This is why this is not easy to draw but you can still do it.
Another important thing that you need to know about these lines is that they have angles on them that makes it possible to tell if a line has an angle on it. The way that you can tell if it has an angle is by looking at the angle of the curve on the curve.
If you want to learn how to draw these lines work, then you should read on. In this article, you are going to learn some important things about these lines. This is how you can do it properly when you are drawing these lines.
The first thing that you have to remember is that you have two angles on these lines and these are called the top and bottom angles. The top angle is the angle that is perpendicular to the top of the line. You have to remember that if the top of the line is pointing in the opposite direction, then you have to say that the bottom angle is also going to be perpendicular to the line. This is what we call a right-angled angle.
The next thing that you have to know is that the second angle of the lines are parallel to the other two angles that are on the lines. These are the left and right-angled angles.
You can learn how to draw these lines if you take note that the angles of these lines are perpendicular to each other. And also parallel to each other. The top and bottom angles of the lines are always perpendicular to the line that you are drawing.
The left side of the line is always going to be the lower part of the line because it is always going to have a downward slope. This means that the right side of the line is going to have a rising slope as well.
Then you can see that the top and bottom of the line are also going to be parallel to each other. This means that the right and left angles of the lines are going to be equal to each other. Therefore, when the line is drawn, you will have the same right and left angles on the line.
You also need to learn that there is another way that you can tell if the line has a different angle on it. You can learn how to do this by looking at the vertical and horizontal lines on the lines.
When you look at the horizontal line, you are going to see that it has a different angle than the vertical line. This is because the horizontal line has an up and down angle.
You should also learn that the vertical line has a different angle than the other lines. This is because the vertical line has an angle that is going to the left and going to the right of the other lines. | 677.169 | 1 |
( (b). The top view in this case comes above the front view.
.
4.2 Projections of the points located in same quadrant and different quadrants
[fig. (i)] shows a point A situated above the H.P. And in front of the V.P., i.e. in the first quadrant. a' is its front view and the top view. After rotation of the plane, these projections will be shown in fig. (ii). the distance of the top view from xy = the distance of A from the V.P. Viz. d.
If a point is situated in second quadrant,
A point B (fig.) (fig.) (fig. 17)
4.3 Projections of line with its inclination to one reference plane and with two reference planes
A straight line is the shortest distance between two points. Hence, the projections of a straight line may be drawn by joining the respective projections of its ends which are points.
The position of a straight line may also be described with respect to the two reference planes. It may be:
1. Parallel to one or both the planes.
2. Contained by one or both the planes.
3. Perpendicular to one of the planes.
4. Inclined to one plane and parallel to the other.
5. Inclined to both the planes.
6. Projections of lines inclined to both the planes.
7. Line contained by a plane perpendicular to both the reference planes.
8. True length of a straight line and its inclinations with the reference planes.
9. Traces of a line.
10. Methods of determining traces of a line.
11. Traces of a line, the projections of which are perpendicular to xy.
12. Positions of traces of a line.
Line parallel to one or both the planes:
(a) Line AB is parallel to the H.P.
a and b are the top views of the ends A and B respectively. It can be clearly seen that the figure ABba is a rectangle. Hence, the top view ab is equal to AB.
a'b' is the front view of AB and is parallel to xy.
(b) Line CD is parallel to the V.P. The line c'd' is the front view and is equal to CD; the top view cd is parallel to xy.
(c) Line ff is parallel to the H.P. And the V.P. Ef is the top view and e'f' is the front view; both are equal to ff and parallel to xy.
Hence, when a line is parallel to a plane, its projection on that plane is equal to its true length; while its projection on the other plane is parallel to the reference line.
Line contained by one or both the planes:
Line AB is in the H.P. Its top view ab is equal to AB; its front view a' b' is in xy.
Line CD is in the V.P. Its front view c'd' is equal to CO; its top view cd is in xy.
Line ff is in both the planes. Its front view e' f' and the top view ef coincide with each other in xy.
Hence, when a line is contained by a plane, its projection on that plane is equal to its true length; while its projection on the other plane is in the reference line.
Line perpendicular to one of the planes:
When a line is perpendicular to one reference plane, it will be parallel to the other.
(a) Line AB is perpendicular to the H.P. The top views of its ends coincide in the point a. Hence, the top view of the line AB is the point a. Its front view a' b' is equal to AB and perpendicular to xy.
(b) Line CD is perpendicular to the V.P. The point d' is its front view and the line cd is the top view. Cd is equal to CD and perpendicular to xy.
Hence, when a line is perpendicular to a plane its projection on that plane is a point; while its projection on the other plane is a line equal to its true length and perpendicular to the reference line.
In first-angle projection method, when top views of two or more points coincide, the point which is comparatively farther away from xy in the front view will be visible; and when their front views coincide, that which is farther away from xy in the top view will be visible.
In third-angle projection method, it is just the reverse. When top views of two or more points coincide the point, which is comparatively nearer xy in the front view will be visible; and when their front views coincide, the point which is nearer xy in the top view will be visible.
Line inclined to one plane and parallel to the other:
The inclination of a line to a plane is the angle which the line makes with its projection on that plane.
Line PQ1 [fig. 22 (i)] is inclined at an angle 8 to the H.P. And is parallelto the V.P. The inclination is shown by the angle 8 which PQ1 makes withits own projection on the H.P., viz. The top view pq1.
The projections [fig. 22 (ii)] may be drawn by first assuming the line to be parallel to both the H.P. And the V.P. Its front view p'q' and the top view pq will both be parallel to xy and equal to the true length. When the line is turned about the end P to the position PQ1 so that it makes the angle 8 with the H.P. While remaining parallel to the V.P., in the front view the point q' will move along an arc drawn with p' as center and p'q' as radius to a point q'1 so that p'q'1 makes the angle 8 with xy. In the top view, q will move towards p along pq to a point q1 on the projector through q'1. p'q'1 and pq1 are the front view and the top view respectively of the line PQ1.
Line inclined to both the planes:
A line AB (fig. 23) is inclined at θ to the H.P. And is parallel to the V.P.The end A is in the H.P. AB is shown as the hypotenuse of a right-angledtriangle, making the angle θ with the base.
The top view ab is shorter than AB and parallel to xy. The front view a'b' is equal to AB and makes the angle θ with xy.
Keeping the end, A fixed and the angle θ with the H.P. Constant, if the end B is moved to any position, say B1, the line becomes inclined to the V.P. Also.
In the top view, b will move along an arc, drawn with a as center and ab as radius, to a position b1. The new top view ab1 is equal to ab but shorter than AB.
In the front view, b' will move to a point b'1 keeping its distance from xy constant and equal to b'o; i.e. it will move along the line pq, drawn through b' and parallel to xy. This line pq is the locus or path of the end B in the front view. b'1 will lie on the projector through b1. The new front view a'b'1 is shorter than a'b' (i.e. AB) and makes an angle a with xy. a is greater than θ.
Thus, if the inclination θ of AB with the H.P.is constant, even when it is inclined to the V.P.
(i) its length in the top view, viz. Ab remains constant; and
(ii) the distance between the paths of its ends in the front view,viz. b'o remains constant.
b) The same line AB (fig. 24) is inclined at ϕ to the V.P. And is parallel to the H.P. Its end A is in the V.P. AB is shown as the hypotenuse of a right-angled triangle making the angle ϕ with the base.
The front view a'b'2 is shorter than AB and parallel to xy. The top view ab2 is equal to AB and makes an angle ϕ with xy.
Keeping the end, A fixed and the angle ϕ with the V.P. Constant, if B is moved to any position, say B3, the line will become inclined to the H.P. Also.
In the front view, b'2, will move along the arc, drawn with a' as center and a'b'2 as radius, to a position b'3. The new front view a'b'3 is equal to a'b'2 but is shorter than AB.
In the top view, b2 will move to a point b3 along the line rs, drawn through b2 and parallel to xy, thus keeping its distance from the path of a, viz. b2o constant. Rs is the locus or path of the end B in the top view. The point b3 lies on the projector through b'3. The new top view ab3 is shorter than ab2 (i.e. AB) and makes an angle β with xy. β is greater than ϕ.
Here also we find that, if the inclination of AB with the V.P. Does not change, even when it becomes inclined to the H.P.
(i) its length in the front view, viz. a'b'2 remains constant; and
(ii) the distance between the paths of its ends in the top view, viz. b2o remains constant.
Hence, when a line is inclined to both the planes, its projections are shorter than the twe length and inclined to xy at angles greater than the true inclinations. These angles viz. α and β are called apparent angles of inclination.
Traces of a line:
When a line is inclined to a plane, it will meet that plane, produced if necessary. The point in which the line or line-produced meets the plane is called its trace.
The point of intersection of the line with the H.P. Is called the horizontal trace, usually denoted as H.T. And that with the V.P. Is called the vertical trace or V.T.
Refer to fig. 25.
(i) A line AB is parallel to the H.P. And the V.P. It has no trace.
(ii) A line CD is inclined to the H.P. And parallel to the V.P. It has only the H.T. But no V.T.
(iii) A line ff is inclined to the V.P. And parallel to the H.P. It has only the V.T. But no H.T.
Thus, when a line is parallel to a plane it has no trace upon that plane.
(i) A line PQ is perpendicular to the H.P. Its H.T. Coincides with its top view which is a point. It has no V.T.
(ii) A line RS is perpendicular to the V.P. Its V.T. Coincides with its front view which is a point. It has no H.T.
Hence, when a line is perpendicular to a plane, its trace on that plane coincides with its projection on that plane. It has no trace on the other plane.
Figure 27
(i) A line AB has its end A in the H.P. And the end B in the V.P. Its H.T. Coincides with the top view of A and the V.T. Coincides with b' the front view of B.
(ii) A line CD has its end C in both the H.P. And the V.P. Its H.T. And V.T. Coincide with c and c' (the projections of C) in xy. Hence, when a line has an end in a plane, its trace upon that plane coincides with the projection of that end on that plane.
Methods of determining traces of a line:
Method I:
Fig. 28 (i) shows a line AB inclined to both the reference planes. Its end A is in the H.P. And 8 is in the V.P. a'b' and ab are the front view and the top view respectively [fig. 28 (ii)].
The H.T. Of the line is on the projector through a' and coincides with a. The V.T. Is on the projector through b and coincides with b'.
Let us now assume that AB is shortened from both its ends, its inclination with the planes remaining constant. The H.T. And V.T. Of the new line CD are still the same as can be seen clearly in fig. 29 (i). c'd' and cd are the projections of CD [fig. 29 (ii)]. Its traces may be determined as described below.
(i) Produce the front view c'd' to meet xy at a point h.
(ii) Through h, draw a projector to meet the top view cd-produced, at the H.T. Of the line.
(iii) Similarly, produce the top view cd to meet xy at a point v.
(iv) Through v, draw a projector to meet the front view c'd'-produced, at the V.T. Of the line.
Method II:
c'd' and cd are the projections of the line CD [fig. 30 (ii)]. Determine the true length C1D1 from the front view c'd' by trapezoid method. The point of intersection between c'd'-produced and C1D1-produced is the V.T. Of the line.
Similarly, determine the true length C2D2 from the top view ed. Produce them to intersect at the H.T. Of the line.
Examples:
1. A line AB 50 mm long, has its end A in both the H.P. And the V.P. It is inclined at 30° to the H.P. And at 45° to the V.P. Draw its projections.
As the end A is in both the planes, its top view and the front view will coincide in xy.
(i) Assuming AB to be parallel to the V.P. And inclined at θ (equal to 30°) to the H.P., draw its front view ab' (equal to AB) and project the top view ab.
(ii) Again, assuming AB to be parallel to the H.P. And inclined at ϕ (equal to 45°) to the V.P., draw its top view ab1 (equal to AB). Project the front view ab'1.
Ab and ab'1 are the lengths of AB in the top view and the front view respectively, and pq and rs are the loci of the end B in the front view and the top view respectively.
(iii) With a as center and radius equal to ab'1, draw an arc cutting pq in b'2. With the same center and radius equal to ab, draw an arc cutting rs in b2.
Draw lines joining a with b'2 and b2. Ab'2 and ab2 are the required projections.
Fig. Shows in pictorial and orthographic views, the solution obtained with all the above steps combined in one figure only.
4.4 True length and inclination with the reference planes
Plane figures or surfaces have only two dimensions, viz. Length and breadth. They do not have thickness. A plane figure may be assumed to be contained by a plane, and its projections can be drawn, if the position of that plane with respect to the principal planes of projection is known.
In this chapter, we shall discuss the following topics:
1. Types of planes and their projections.
2. Traces of planes.
Type of planes:
Planes may be divided into two main types:
(1) Perpendicular planes.
(2) Oblique planes.
Perpendicular planes:
These planes can be divided into the following sub-types:
(i) Perpendicular to both the reference planes.
(ii) Perpendicular to one plane and parallel to the other.
(iii) Perpendicular to one plane and inclined to the other.
Perpendicular to both the reference planes (fig 1):
A square ABCD is perpendicular to both the planes. Its H.T. And V.T. Are in a straight-line perpendicular to xy.
The front view b'c' and the top view ab of the square are both lines coinciding with the V.T. And the H.T. Respectively.
Perpendicular to one plane and parallel to the other plane:
a) Plane, perpendicular to the H.P. And parallel to the V.P. [fig. 2(i)]. A triangle PQR is perpendicular to the H.P. And is parallel to the V.P. Its H.T. Is parallel to xy. It has no V.T.
The front view p'q'r' shows the exact shape and size of the triangle. The top view pqr is a line parallel to xy. It coincides with the H.T.
(b) Plane, perpendicular to the V.P. And parallel to the H.P. [fig. 2(ii)]. A square ABCD is perpendicular to the V.P. And parallel to the H.P. Its V.T. Is parallel to xy. It has no H.T.
The top view abed shows the true shape and true size of the square. The front view a'b' is a line, parallel to xy. It coincides with the V.T.
Perpendicular to one plane and inclined to the other plane:
A square ABCD is perpendicular to the H.P. And inclined at an angle φ to the V.P. Its V.T. Is perpendicular to xy. Its H.T. Is inclined at φ to xy.
Its top view ab is a line inclined at φ to xy. The front view a'b'c'd' is smaller than ABCD.
(b) Plane, perpendicular to the V.P. And inclined to the H.P. (fig. 4).
A square ABCD is perpendicular to the V.P. And inclined at an angle θ to the H.P. Its H.T. Is perpendicular to xy. Its V.T. Makes the angle e with xy. Its front view a'b' is a line inclined at θ to xy. The top view abed is a rectangle which is smaller than the square ABCD.
Fig. 5 shows the projections and the traces of all these perpendicular planes by third-angle projection method.
Oblique planes:
Planes which are inclined to both the reference planes are called oblique planes.
Traces of planes:
A plane, extended if necessary, will meet the reference planes in lines, unless it is parallel to any one of them.
These lines are called the traces of the plane. The line in which the plane meets the H.P. Is called the horizontal trace or the H.T. Of the plane. The line in which it meets the V.P. Is called its vertical trace or the V.T. A plane is usually represented by its traces.
General conclusions:
(1) Traces:
(a) When a plane is perpendicular to both the reference planes, its traces lie on a straight-line perpendicular to xy.
(b) When a plane is perpendicular to one of the reference planes, its trace upon the other plane is perpendicular to xy (except when it is parallel to the other plane).
(c) When a plane is parallel to a reference plane, it has no trace on that plane. Its trace on the other reference plane, to which it is perpendicular, is parallel to xy.
(d) When a plane is inclined to the H.P. And perpendicular to the V.P., its inclination is shown by the angle which its V.T. Makes with xy. When it is inclined to the V.P. And perpendicular to the H.P., its inclination is shown by the angle which its H.T. Makes with xy.
(e) When a plane has two traces, they, produced if necessary, intersect in xy (except when both are parallel to xy as in case of some oblique planes).
(2) Projections:
(a) When a plane is perpendicular to a reference plane, its projection on that plane is a straight line.
(b) When a plane is parallel to a reference plane, its projection on that plane shows its true shape and size.
(c) When a plane is perpendicular to one of the reference planes and inclined to the other, its inclination is shown by the angle which its projection on the plane to which it is perpendicular, makes with xy. Its projection on the plane to which it is inclined, is smaller than the plane itself.
Problem 1. Show by means of traces, each of the following planes:
(a) Perpendicular to the H.P. And the V.P.
(b) Perpendicular to the H.P. And inclined at 30° to the V.P.
(c) Parallel to and 40 mm away from the V.P.
(d) Inclined at 45° to the H.P. And perpendicular to the V.P.
(e) Parallel to the H.P. And 25 mm away from it.
(a) The H.T. And the V.T. Are in a line perpendicular to xy.
(b) The H.T. Is inclined at 30° to xy; the V.T. Is normal to xy; both the traces intersect in xy.
(c) The H.T. Is parallel to and 40 mm away from xy. It has no V.T.
(d) The H.T. Is perpendicular to xy; the V.T. Makes 45° angle with xy; both intersect in xy.
(e) The V.T. Is parallel to and 25 mm away from xy. It has no H.T.
Projection of planes parallel to one of the reference planes:
The projection of a plane on the reference plane parallel to it will show its true shape. Hence, beginning should be made by drawing that view. The other view which will be a line, should then be projected from it.
Note: When the plane is parallel to the H.P.: The top view should be drawn first, and the front view projected from it.
Problem:
An equilateral triangle of 50 mm side has its V.T. Parallel to and 25 rnm above xy. It has no H.T. Draw its projections when one of its sides is inclined at 45° to the V.P.
As the V.T. Is parallel to xy and as there is no H.T. The triangle is parallel to the H.P. Therefore, begin with the top view.
Beginning should be made with the front view and the top view projected from it.
Problem:
A square ABCD of 40 mm side has a corner on the H.P. And 20 mm in front of the V.P. All the sides of the square are equally inclined to the H.P. And parallel to the V.P. Draw its projections and show its traces.
As all the sides are parallel to the V.P., the surface of the square also is parallel to it. The front view will show the true shape and position of the square.
(i) Draw a square a'b'c'd' in the front view with one corner in xy and all its sides inclined at 45° to xy.
(ii) Project the top view keeping the line ac parallel to xy and 30 mm below it. The top view is its H.T. It has no V.T.
Projections of planes inclined to one reference plane and perpendicular to other:
When a plane is inclined to a reference plane, its projections may be obtained in two stages. In the initial stage, the plane is assumed to be parallel to that reference plane to which it must be made inclined. It is then tilted to the required inclination in the second stage.
1) Plane, Inclined to the H.P. And perpendicular to the V.P.:
When the plane is inclined to the H.P. And perpendicular to the V.P., in the initial stage, it is assumed to be parallel to the H.P. Its top view will show the true shape. The front view will be a line parallel to xy. The plane is then tilted so that it is inclined to the H.P. The new front view will be inclined to xy at the true inclination. In the top view the corners will move along their respective paths (parallel to xy).
Problem:
A regular pentagon of 25 mm side has one side on the ground. Its plane is inclined at 45° to the H.P and perpendicular to the V.P. Draw its projections and show its traces.
Assuming it to be parallel to the H.P.
(i) Draw the pentagon in the top view with one side perpendicular to xy [fig. 9 (i)]. Project the front view. It will be the line a'c' contained by xy.
(ii) Tilt the front view about the point a', so that it makes 45° angle with xy.
(iii) Project the new top view ab1c1d1e upwards from this front view and horizontally from the first top view. It will be more convenient if the front view is reproduced in the new position separately and the top view projected from it, as shown in fig. 9 (ii). The V.T. Coincides with the front view and the H.T. Is perpendicular to xy, through the point of intersection between xy and the front view-produced.
2) Plane, inclined to the V.P. And perpendicular to the H.P.:
In the initial stage, the plane may be assumed to be parallel to the V.P. And then tilted to the required position in the next stage.
Problem:
Draw the projections of a circle of 50 mm diameter, having its plane vertical and inclined at 30° to the V.P. Its center is 30 mm above the H.P. And 20 mm in front of the V.P. Show also its traces.
A circle has no corners to project one view from another. However, many points, say twelve, equal distances apart, may be marked on its circumference.
(i) Assuming the circle to be parallel to the V.P., draw its projections. The front view will be a circle [fig. 10 (i)], having its center 30 mm above xy. The top view will be a line, parallel to and 20 mm below xy.
(ii) Divide the circumference into twelve parts (with a 30°-60° set- square) and mark the points as shown. Project these points in the top view. The centre O will coincide with the point 4.
(iii) When the circle is tilted, to make 30° angle with the V.P., its top view will become inclined at 30° to xy. In the front view all the points will move along their respective paths (parallel to xy). Reproduce the top view keeping the centre o at the same distance, viz. 20 mm from xy and inclined at 30° to xy [fig. 10 (ii)].
(iv) For the final front view, project all the points upwards from this top view and horizontally from the first front view. Draw a freehand curve through the twelve points 1'1, 2'1 etc. This curve will be an ellipse.
Projection of oblique planes:
When a plane has its surface inclined to one plane and an edge or a diameter or a diagonal parallel to that plane and inclined to the other plane, its projections are drawn in three stages.
(1) If the surface of the plane is inclined to the H.P. And an edge (or a diameter or a diagonal) is parallel to the H.P. And inclined to the V.P.,
(i) in the initial position the plane is assumed to be parallel to the H.P. And an edge perpendicular to the V.P.
(ii) It is then tilted to make the required angle with the H.P. As already explained, its front view in this position will be a line, while its top view will be smaller in size.
(iii) In the final position, when the plane is turned to the required inclination with the V.P., only the position of the top view will change. Its shape and size will not be affected. In the final front view, the corresponding distances of all the corners from xy will remain the same as in the second front view.
If an edge is in the H.P. Or on the ground, in the initial position, the plane is assumed to be lying in the H.P. Or on the ground, with the edge perpendicular to the V.P. If a corner is in the H.P. Or on the ground, the line joining that corner with the center of the plane is kept parallel to the V.P.
(2) Similarly, if the surface of the plane is inclined to the V.P. And an edge (or a diameter or a diagonal) is parallel to the V.P. And inclined to the H.P.,
(i) in the initial position, the plane is assumed to be parallel to the V.P. And an edge perpendicular to the H.P.
(ii) It is then tilted to make the required angle with the V.P. Its top view in this position will be a line, while its front view will be smaller in size.
(iii) When the plane is turned to the required inclination with the H.P., only the position of the front view will change. Its shape and size will not be affected. In the final top view, the corresponding distances of all the corners from xy will remain the same as in the second top view.
If an edge is in the V.P., in the initial position the plane is assumed to be lying in the V.P. With an edge perpendicular to the H.P. If a corner is in the V.P., the line joining that corner with centre of the plane is kept parallel to the H.P.
Problem:
1. A square ABCD of 50 mm side has its corner A in the H.P., its diagonal AC inclined at 30° to the H.P. And the diagonal BO inclined at 45° to the V.P. And parallel to the H.P. Draw its projections.
In the initial stage, assume the square to be lying in the H.P. With AC parallel to the V.P.
(i) Draw the top view and the front view. When the square is tilted about the corner A so that AC makes 30° angle with the H.P., BO remains perpendicular to the V.P. And parallel to the H.P.
(ii) Draw the second front view with a'c' inclined at 30° to xy, keeping a' or c' in xy. Project the second top view. The square may now be turned so that BO makes 45° angle with the V.P. And remains parallel to the H.P. Only the position of the top view will change. Its shape and size will remain the same.
(iii) Reproduce the top view so that b1d1 is inclined at 45° to xy. Project the final front view upwards from this top view and horizontally from the second front view.
2. Draw the projections of a circle of 50 mm diameter resting in the H.P. On a point A on the circumference, its plane inclined at 45° to the H.P. And
(a) the top view of the diameter AB making 30° angle with the V.P.;
(b) the diameter AB making 30° angle with the V. P.
Draw the projections of the circle with A in the H.P. And its plane inclined at 45° to the H.P. And perpendicular to the V.P. [fig.(i) and fig.(ii)].
(a) In the second top view, the line a1b1 is the top view of the diameter AB. Reproduce this top view so that a1b1 makes 30° angle with xy [fig. 12(iii)]. Project the required front view.
(b) If the diameter AB, which makes 45° angle with the H.P., is inclined at 30° to the V.P. Also, its top view a1b1 will make an angle greater than 30° with xy. This apparent angle of inclination is determined as described below.
Draw any line a1b2 equal to AB and inclined at 30° to xy [fig. (iv)]. With a1 as center and radius equal to the top view of AB, viz. a1b1, draw an arc cutting rs (the path of B in the top view) at b3. Draw the line joining a1 with b3, and around it, reproduce the second top view. Project the final front view. It is evident that a1b3 is inclined to xy at an angle ϕ which is greater than 30°.
3. A thin 30°-60° set-square has its longest edge in the V.P. And inclined at 30° to the H.P. Its surface makes an angle of 45° with the V.P. Draw its projections.
In the initial stage, assume the set-square to be in the V.P. With its hypotenuse perpendicular to the H.P.
(i) Draw the front view a'b'c' and project the top view ac in xy.
(ii) Tilt ac around the end a so that it makes 45° angle with xy and project the front view a'1b'1c'1.
(iii) Reproduce the second front view a'1b'1c'1 so that a'1b'1 makes an angle of 30° with xy. Project the final top view a1b1c1. | 677.169 | 1 |
Scale Factors of Similar Figures.VERSION 2
Description: **It has come to my attention that not all textbooks define scale factors in the same way. The scale factors in this deck are the reciprocals of the scale factors in my original scale factors set of Boom Cards.
For example, if triangle A is 3 times larger than triangle B, the scale factor of triangle A to triangle B is 3.
This 20-card deck is a great way to help your geometry students practice finding scale factors and identify features of similar shapes. | 677.169 | 1 |
Trig shortcuts for quick nav solutions
Ocean NavigatorJanuary 1, 2003
In navigation, trigonometry is a highly useful discipline; estimating set and drift of tides, courses to steer to make good a target, true winds from apparent winds, even tide heights, are all exercises in plane trigonometry.
The problem, of course, is that few people have either the ability or the desire to memorize a trig table. Fortunately, there is no reason to, at least for those cases where what one desires is a quick, "dirty" answer. With some practice, one can easily determine answers to acceptable precision (say, ±10%) using mental arithmetic and the three trig tricks presented below. These tricks may seem overly quantitative to some, or imprecise to others. However, the advantages while sailing or racing are many.
All three of the shortcuts are based on the mathematical properties of the trig functions, principally sine and cosine. Rule one allows easy calculation of the sine; rule two easy calculation of the cosine; and rule three eliminates ever having to calculate either sine or cosine for more than six angles. In all three cases, my goal is to be precise to better than 10%, which is certainly acceptable for mental arithmetic. If more precision is needed, the quick answer provided here will probably satisfy any pressing requirements while you spin the maneuvering board or crunch the calculator. Those readers familiar with trig may cringe at some of the shortcuts, but all answers will be within the magic 10% error, our goal.Rule one:
The sine of an angle equals the angle measured in radians, up to angles of 50°. Radians are an alternative, and more natural, way of measuring angles. In any circle, an arc as long as the circle's radius maps out one radian, which equals approximately 57°. To find the sine of an angle, divide the angle by the number of degrees per radian, or 60. (Use 60 rather than 57 because the division is easiermost navigators seem pretty proficient at dividing by 60 anyway.) For example, by this method the sine of 15° is 15/60 = 0.25. The actual value is 0.26, for an error of 4%. This method is in error by more than 10% for angles of 50° to 90°, but this isn't a problem either: remember that sin 60° = 0.9, and the sine of anything higher than 70° is 1. While this list may seem arbitrary, rule three will show that one only needs to know the sine of only six angles to figure out the rest.Rule two:
The cosine of an angle is the same as the sine of the angle's complement. (Brings you back to that high school geometry class, doesn't it?) An angle and its complement sum to 90°. Thus Cos 20° = Sin 70°. This rule is good for all angles, and is exact. Another (though less convenient) rule for the coSine is below. It is significantly less simple than the corresponding rule for the Sine, and isn't really needed once you've mastered rules one and three.Rule three:
Interpolate whenever possible, preferably by using angles easily divisible by 60. For example, to determine Sin 22°, I would go through the following process: Sin 15/60 = 0.25, Sin 30/60 = 0.5, so Sin 22° would be in the middle, or 0.375. The actual value is 0.3746. Interpolation is a learned skill, but one in which most navigators are well practiced.
We'll close with two examples of the rules in action. Case one: On course 000°, speed 6 knots, determine speed over the ground with tide setting 135°, drifting 2 knots. Speed along the course will be reduced by the drift multiplied by the coSine of the angle between the course and drift. First, determine the coSine of the angle: Cos 45° = Sin 90°- 45° = Sin 45° = 45°/60 = 0.75. So, our speed made good is (6 – 2 x 0.75) knots = 4.5 knots. The exact result is 4.6 knots, for an error of 6%.
Case two: How much northing and easting will a vessel on course 033°, speed 5 knots make in one hour? The northing is equal to the total distance traveled multiplied by the Sine of 33°. Approximate Sin 33° as one fifth of the way between Sin 30° and Sin 45°: Sin 30° = 30°/60 = 0.50 Sin 45° = 45°/60 = 0.75 thus, Sin 33° = 0.55.
Northing is then 5 miles x 0.55 = 2.75 miles. Easting is total distance traveled multiplied by the coSine of 33°. Cos 33° = Sin 57°, which is close enough to Sin 60° = 0.9 (rule three) for me. Thus, easting is 5 miles x 0.9 = 4.5 miles. Actual values are 2.7 and 4.2 miles, so we're within 1% and 7%.
These rules allow you to perform first-level navigation calculations while you're in the cockpit, the head, or eating dinner. Equally important, they provide a rapid way of getting an approximate answer to a calculation as a check of more complicated procedures, or answers provided by those black boxes on so many navigation tables. Do not rely only on these mental calculations, though. As always, the prudent navigator uses more than one tool to guide his or her boat through the water. With practice, these calculations become automatic, as will the savings in time and effort their use will provide.
Larry McKenna is an assistant professor of geology at the University of Kansas and a navigation enthusiast. | 677.169 | 1 |
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Vector Algebra (Part-4) | S N Dey | Class 12
0AdminJuly 20, 2022
In the previous article, we have discussed few Long Answer Type Questions (1-10) and solutions of S N De Vector Algebra Chapter. In this article , we will solve the problems of Long Answer Type questions (11-20) as given in the Chhaya Publication Book of Vector Chapter. So, let's start.
$~11.~$ By vector method show that the figure formed by joining the mid points of a quadrilateral is parallelogram.
Solution.
Let $~ABCD~$ be a parallelogram and the position vectors of $~A,~B,~C,~D~$ are $~\vec{a},~\vec{b},~\vec{c},~\vec{d}~$ respectively.
Further suppose that $~P,~Q,~R,~S~$ are the mid-points of $~AB,~BC,~CD,~DA~$ respectively.
Hence, $~R_1$ and $~R_2~$ denote the same point and so $~\overline{AC}~$ and $~\overline{DP}~$ meet in a common point of trisection.
$~14.~~ABCD~$ is a parallelogram ; $~P~$ and $~Q~$ are the mid-points of the sides $~\overline{AB}~$ and $~\overline{DC}~$ respectively. Show that $~\overline{DP}~$ and $~\overline{BQ}~$ trisect and are trisected by $~\overline{AC}.$
Solution.
Let $~\vec{a},~\vec{b},~\vec{c},~\vec{d}~$ be the position vectors of the points $~A,~B,~C,~D~$ respectively.
By $~(3),~$ we can say that $~R_2~$ divides $~\overline{AC}~$ in the ratio $~ 2:1.$
Hence, $~\overline{DP}~$ and $~\overline{BQ}~$ trisect and are trisected by $~\overline{AC}.$
$~15.~~ABCD~$ is a parallelogram and $~P~$ is the mid-point of $~\overline{DC}~$ . If $~Q~$ is a point on $~\overline{AP},~$ such that $~~\overline{AQ}=\frac 23\overline{AP},~$ show that $~Q~$ lies on the diagonal $~\overline{BD}~$ and $~\overline{BQ}=\frac23 \overline{BD}.$
Hence, by $~(2)~$ we can conclude that $~Q~$ divides $~\overline{BD}~$ in the ratio $~2:1~$ and so finally we can say that $~Q~$ lies on the diagonal $~\overline{BD}~$ and $~\overline{BQ}=\frac23 \overline{BD}.$
$~16.~~D,~E,~F~$ are the midpoints of the sides $~\overline{BC},~\overline{CA}~$ and $~\overline{AB}~$ respectively of the triangle $~ABC.~$ If $~P~$ is any point in the plane of the triangle, show that $~\vec{PA}+\vec{PB}+\vec{PC}=\vec{PD}+\vec{PE}+\vec{PF}.$
Solution.
Let the position vectors of the points $~A,~B,~C~$ with respect to the point $~P~$ be $~~\vec{a},~\vec{b},~\vec{c}~~$ respectively.
So, the position vectors of $~D,~E,~F~$ are $~\frac{\vec{b}+\vec{c}}{2},~\frac{\vec{a}+\vec{c}}{2},~\frac{\vec{a}+\vec{b}}{2}~$ respectively.
Hence, from $~(1)~$ we get the position vector of $~G~$ and so $~G~$ is the centroid of the triangle $~ABC.$
$~19.~$ The diagonals of the parallelogram $~ABCD~$ intersect at $~E.~$ If $~\vec{a},\vec{b},\vec{c}~$ and $~\vec{d}~$ be the position vectors of its vertices with respect to an arbitrary origin $~O,~$ then show that $~\vec{a}+\vec{b}+\vec{c}+\vec{d}=4\vec{OE}.$
Solution.
Since diagonals of the parallelogram $~\vec{AC},~\vec{BD}~$ bisect each other at $~E~$, so the position vector of the midpoint $~E~$ of the diagonal $~\vec{AC}~$ is given by :
$~20.~$ By vector method prove that the straight line joining the midpoints of two non-parallel sides of a trapezium is parallel to the parallel sides and half of their sum.
Solution.
Let $~ABCD~$ be a trapezium and position vectors of $~A,~B,~C,~D~$ with respect to origin ($~O~$) are $~\vec{a},~\vec{b},~\vec{c},~\vec{d}~$ respectively and $~\vec{AD}\parallel \vec{BC}~~\text{(say)}.$ | 677.169 | 1 |
Use an inverse tangent to find an angle measure
Use a calculator to approximate the measure of ∠ A to the nearest tenth of a degree.
Solution:
Tan A = 16/20
=4/5
= 0.8
tan–1 0.8 = m ∠ A (Using a calculator)
tan–1 0.8 ≈ 38.65980825
So, the measure of ∠A is approximately 38.7°.
∠ A = 38.7°
Example 2:
Using inverse sines and cosines:
Use a calculator to approximate the measure of ∠A to the nearest tenth of a degree.
Solution:
Cos A = 6/10
m ∠ A = 53.1°
Example 3:
Solve the right triangle. Approximate angles to the nearest tenth of a degree and sides to two decimal places.
Solution:
Step 1: Find m ∠ p by using the Triangle Sum Theorem.
ÐP
180 – 90 – 37
mÐP = 53°
Step 2: Approximate PQ by using a sin ratio.
PQ
sin 37° = PQ/22
22 × sin 37° = PQ
13.24 = PQ
Step 3: Approximate QR using a cosine ratio.
QR
cos 37° = QR/22
22 × cos 37° = QR
17.57 = QR
Example 4:
At the circus, a person in the audience watches the high-wire routine. A 5-foot-6-inch-tall acrobat is standing on a platform that is 25 feet off the ground. How far is the audience member from the base of the platform, if the angle of elevation from the audience member's line of sight to the top of the acrobat is 27°?
Exercise
Use a calculator to approximate the measure of ∠ A to the nearest tenth of a degree.
Using inverse sines and cosines:
Use a calculator to approximate the measure of ∠ A to the nearest tenth of a degree.
Solving right triangles:
Solve the right triangle. Round decimal answers to the nearest tenth.
Let ∠ A be an acute angle in a right triangle. Approximate the measure of ∠ A to the nearest tenth of a degree.
sin A =0.5
A soccer ball is placed 10 feet away from the goal, which is 8 feet high. You kick the ball and it hits the crossbar along the top of the goal. What is the angle of elevation of your kick?
You are standing on a footbridge in a city park that is 12 feet high above a pond. You look down and see a duck in the water 7 feet away from the footbridge. What is the angle of depression? Explain your reasoning.
Solve the right triangle. Round decimal answers to the nearest tenth.
Find m∠ D to the nearest tenth of a degree if sin D = 0.54.
Solve a right triangle that has a 40° angle and a 20-inch hypotenuse.
Solve a right triangle
Concept Map
What have we learned
Understand inverse trigonometric ratios.
Understand inverse tangent , inverse sine, inverse cosine.
Use an inverse tangent to find an angle measure
Use an inverse sine and an inverse cosine
Understand how to solve a right triangle.
Solve a real-world problem.
Frequently asked questions
1. How to find the angle using inverse tangent if you have the opposite and adjacent sides?
Ans: Suppose the angle is A. So tan A=Opposite side/Adjacent side.
tan-1(opp.side/adj.side) = A
2. If sin x = ½ and cos x = 3–√3/2, find the value of x?
Ans: Given sin x = ½ and cos x =√3/2
Therefore tan x = sin x/cos x = 1/√3,
X = tan-1(1/√3) = 30o
3. If the Inverse tangnet function is tan-1(opp.side/adj.side) = A, what is Inverse cotangent function?
Ans: cot-1(adj.side/opp.side) = A | 677.169 | 1 |
oneADVERTISEMENTS:ADVERTISEMENTS:
A. Plain Scales.
B. Diagonal Scales.
C. Comparative or Corresponding Scales.
D. Vernier Scales.
ADVERTISEMENTS:ADVERTISEMENTS: Angle | 677.169 | 1 |
Let A B C D be a convex quadrilateral with A B=C D=10, B C=14, and A D=2 \sqrt{65}. Assume that the diagonals of A B C D intersect at point P, and that the sum of the areas of \triangle A P B and \triangle C P D equals the sum of the areas of \triangle B P C and \triangle A P D. Find the area of quadrilateral A B C D. | 677.169 | 1 |
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