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Centroid of a Trapezoid Formulas - Find the Centroid of a Trapezium
Consider the following trapezium in which AB || CD. The trapezium OABC is placed such that the origin coincides with one of its vertices. G is its centroid. The lengths of its parallel sides are AB = a and OC = b and its height is h.
The coordinates of the centroid of the trapezium are given by the following formula.
\[G\left ( \frac{h}{2},\frac{b+2a}{3(a+b)}h \right )\]
Let's look at an example to see how to use this formula.
Some Solved Questions by Vedantu
Question: Find the centroid of a trapezium of height 5 cm whose parallel sides are 6 cm and 8 cm.
What Exactly is a Centroid? Definition of a Centroid
A centroid, also known as a geometric center, is the center of mass of a uniformly dense object. To make it easier to understand, think of it as the point where you should place the tip of a pin in order to balance your geometric shape on it.
Formula for the Centroid of a Trapezoid
The centroid of a trapezoid formula can be used to determine the position of a trapezoid's centroid. A quadrilateral with two parallel sides is known as a trapezoid. A trapezoid's centroid is located halfway between the two bases. Let's look at a few examples of the centroid of a trapezoid formula.
A trapezoid is a four-sided quadrilateral with two parallel sides. A trapezoid's centroid is located halfway between the two bases. Use the formula below to find any trapezoid with parallel sides a and b.
(Image will be added soon)
Find the formula for the centroid of a trapezoid at a distance of x in the table below.
\[x=\frac{b+2a}{3(a+b)}h\]
Where,
h = Trapezoid height
a, b = Parallel side lengths
Calculator for Centroid
Simply enter the vertices of your shape as Cartesian coordinates to utilize this centroid calculator. Let's look at how to find the trapezoid's centroid:
Select the sort of form for which you want the centroid to be calculated. In this scenario, we'll go with an N-sided polygon.
Fill in the N parameter (if required). For our example, we'll need to know how many sides a polygon has. We type 4 into the N box since the trapezoid is a quadrilateral.
After that, the fields for entering coordinates will display. Enter the vertices of your shape's coordinates. Assume that the vertices of our trapezoid are:
A = (1,1)
B = (2,4)
C = (5,4)
D = (11,1)
(Image will be added soon)
The result will be displayed on our centroid calculator! The choice of a trapezoid's centroid is (4.974,2.231).
Example: Find the centroid of the trapezoid with the following dimensions: a = 12′, b = 5′, and h = 5′.
FAQs on Centroid of a Trapezoid Formula
1. What is a trapezoid?
A trapezium or a trapezoid is a quadrilateral with at least one pair of parallel sides. The sides that are parallel to each other are called bases. The other sides are called legs. The legs may or may not be parallel to each other. Trapezoids are called Trapezium in the UK. You can download the free pdf of Centroid of a Trapezoid Formula | Find the Centroid of a Trapezium from Vedantu and learn the concepts and meaning of the Trapezium in a better way. Regular learning can make you understand the concepts properly.
2. How to calculate the area and perimeter of a trapezoid?
The area and perimeter of a trapezoid can be calculated by using the formulas given below:
Perimeter: the sum of lengths of sides of a trapezoid. Students are required to solve problem equations sums based on the area and perimeter of a trapezoid. You can learn with the help of pdf of Centroid of a Trapezoid Formula | Find the Centroid of a Trapezium.
3. Which website explains the centroid of a trapezoid formula clearly with examples?
The centroid of a trapezoid formula can be found on Vedantu's website. Vedantu has provided a simple explanation to calculate the centroid of a trapezium. All a student needs to do is follow the instructions line by line and practice example problems. Students can be assured that the formula is accurate as it is provided by top experts at Vedantu. The formulations of the topics are easy to read and understandable. You can download the PDF and start learning with Vedantu.
4. What is the formula to find the centroid of a trapezium?
The centroid of a trapezium lies between the two bases. The centroid of a trapezoid can be found out by using the formula given below:\[x=\frac{b+2a}{3(a+b)}\times h\]. Here, h= height of the trapezoid, a, b= length of the parallel sides. These are also the important Questions for CBSE Class 10 Math Chapter 9 - Some Applications of Trigonometry. Students should be thorough with the formula to find the centroid of a trapezium. You can learn better with the help of experts at Vedantu.
5. Is the Centroid of a Trapezoid Formula | Find the Centroid of a Trapezium helpful?
Yes, the Centroid of a Trapezoid Formula | Finding the Centroid of a Trapezium is helpful. It can help students to know the concepts of the centroid of a trapezium, its formula and how to find the centroid of the trapezium. We have drafted a set of complex questions on the trapezium formula. Students are required to learn and practice these problem equations daily and it can help them to understand the concepts in a better way. Regular practice can make you an expert in the chapter on trapezium centroid. | 677.169 | 1 |
2 Answers
2
A line is a simple line between two points and is described by four required attributes.
x1: The x position of the first end of the line as measured from the
left of the screen.
y1: The y position of the first end of the line
as measured from the top of the screen.
x2: The x position of the
second end of the line as measured from the left of the screen.
y2:
The y position of the second end of the line as measured from the top
of the screen.
The following is an example of the code section required to draw a line;
holder.append("line") // attach a line
.style("stroke", "black") // colour the line
.attr("x1", 100) // x position of the first end of the line
.attr("y1", 50) // y position of the first end of the line
.attr("x2", 300) // x position of the second end of the line
.attr("y2", 150); // y position of the second end of the line
I like to think that an attribute of an element is something that is a characteristic of the object without defining it, and/or it may affect the object's position or orientation on the page. There could be a strong argument to say that styles could be seen to cross-over into attributes and I agree. I like to think that an element can be selected and arranged on a web page with select and attr, but once it's there, changes to how it looks are a matter for style.
To draw a line we need TWO points, in a graph if we want to refer any point we use co-ordinates, (x1,y1) is the start point of a line (x2,y2) is the end point of a line, these two points are connected. | 677.169 | 1 |
How to Find How Many Diagonals Are in a Polygon
8 2 minutes read
To find the number of diagonals in a polygon, use the formula: n(n-3)/2 where n is the number of sides.
To find how many diagonals are in a polygon, first count the vertices. Next, use the formula n(n-3)/2. For example, a pentagon has 5 vertices, so the formula would be 5(5-3)/2 = 5. Therefore, a pentagon has 5 diagonals. This method works for any regular polygon. Remember to subtract 3 from the number of vertices before multiplying by the total number of vertices. Practice with different shapes to determine the number of diagonals in each polygon. This formula will help you calculatediagonals with ease. | 677.169 | 1 |
A non-convex polygon is any polygon with at least one interior
angle measuring at least 180 degrees. . A polygon is defined as an
area enclosed by 3 or more line segments connected at their
endpoints. Basically, the figure needs to be completely closed in,
have no extra lines or part of lines coming out of it, and no
curves. For a polygon to be convex all of its interior angles need
to measure less than 180 degrees.
How do you use the word convex in a sentence?
Regular polygons are always convex by definition.
What letters in the alphabet are convex polygons?
G
What is non-convex and convex?
These terms describe polygons. To identify a polygon as convex,
we draw a segment from any vertex to any other vertex. This segment
cannot go outside of the polygon.
Non-convex is concave. If we draw a segment from a vertex to any
other vertex, at least one of the segments will go outside of the
polygon. | 677.169 | 1 |
What Is a Milliradian?
A milliradian is equal to 1/1,000 of a radian.
One milliradian is equal to 0.0573 degrees and there are 6,283 milliradians in a full revolution/circle.
The milliradian is also a commonly used unit in mathematics.
The milliradian is a multiple of the radian, which is the SI derived unit for angle. In the metric system, "milli" is the prefix for thousandths, or 10-3. A milliradian is sometimes also referred to as a mil. Milliradians can be abbreviated as mrad; for example, 1 milliradian can be written as 1 mrad. | 677.169 | 1 |
Exemplar Solution Class 10 Co-ordinate Geometry Exercise 7.3
Co-ordinate Geometry is a branch of mathematics that focuses on the study of the position and movement of points, lines, and shapes in two-dimensional (2D) and three-dimensional (3D) space. It is a fundamental tool for understanding the fundamentals of physics, engineering, and many other scientific and engineering disciplines. In CBSE Class 10, co-ordinate geometry is an important topic that students must master in order to be successful in their future studies.
In this blog post, we will discuss the Exemplar Solution Class 10 Co-ordinate Geometry and some of the key concepts that students should understand in order to be successful in their CBSE Class 10 exams. We will also provide some tips and strategies for mastering this topic and preparing for the exam.
To begin, let's define some important terms related to coordinate geometry. The coordinate system is used to represent points in space using numerical values. A point is identified by its coordinates, which are the numbers that represent its position in the coordinate system. The coordinates of a point are written in the form (x,y). The x-coordinate tells us the point's distance from the y-axis, and the y-coordinate tells us the point's distance from the x-axis.
Once we have the coordinates of a point, we can draw lines and shapes in the coordinate system. Lines are represented by their equations, which are written in the form y=mx+c, where m is the slope of the line and c is the constant term. Shapes are represented by their equations as well, but these equations are usually more complex than those used to represent lines.
Now that we have the basics of coordinate geometry out of the way, let's discuss some of the key concepts that students should understand in order to be successful in their CBSE Class 10 exams. First, students should be familiar with the basic equations used to represent lines and shapes in the coordinate system. They should also understand how to calculate the slope of a line and the area of a shape.
In addition, students should understand how to calculate the distance between two points and the midpoint of a line. They should also know how to calculate the equation of a line given two points and the equation of a circle given its center and radius.
Finally, students should be familiar with the concept of transformations, which are used to move shapes around in the coordinate system. This includes understanding how to translate, rotate, and reflect shapes.
These are just some of the key concepts that students should understand in order to be successful in their CBSE Class 10 exams. With the right approach and a thorough understanding of the material, students can master coordinate geometry and be well-prepared for their exams. Good luck!
Q6. Find the coordinates of the point Q on the x–axis which lies on the perpendicular bisector of the line segment joining the points A (–5, –2) and B(4, –2). Name the type of triangle formed by the points Q, A and B. | 677.169 | 1 |
What do the numbers mean on a survey map?
The units of measure for property survey bearings are degrees, minutes, and seconds. degrees. The "legs" of the boundary lines are listed directly below the name of the surveyor.
States, the most common unit of measurement is the foot, which is defined by the International System of Units (SI) as the distance between the top of a person's foot and the ground. In other countries, such as Canada and Australia, feet are measured in millimeters (mm) or centimeters (cm).
The metric system is based on the SI system, but it uses a different system of units.
What do survey measurements mean?
Measure distances and angles from a reference. (3-D) coordinates for each point on the surface of the earth are established through the use of trigonometric calculations. The distance between two points on a surface is measured by measuring the difference between the two distances.
For example, if you are standing at the intersection of a street and a highway, and you want to know how far you have to walk to get from one point to the other, you can use the following formula to determine your distance: Distance = Distance x Distance/2. This formula can be used for any distance, but it is most useful for measuring distances between points that are close together, such as intersections.
If you need to measure a distance that is greater than a certain value, then you must use a formula that takes the value into account. In this case, your formula must take the square root of this value. You can find the formula for this formula by using the Pythagorean Theorem, which states that the sum of two squares is the same as the product of three squares.
What do the coordinates mean on a survey?
Survey coordinates are the x-coordinates and y-coordinates which define the horizontal position of a point on a ground system, and the z-coordinate, which is the elevation of the point relative to the ground surface. For example, in the figure below, the vertical axis is measured from the top of an object to its bottom.
The horizontal axis measures the distance between the object's center of mass and its top. This equation is often used to determine the location of objects in 3D space, but it can also be used for other purposes, such as determining the position and orientation of points in a 2D coordinate system.
What is a good survey score?
A good response rate is between 5% and 30%. An excellent response rate is more than 50%. The best way to do this is to use a tool like SurveyMonkey, which allows you to see how many people have responded to your survey.
If you see a high number of responses, it means that people are interested in what you have to and are willing to take the time to fill out the survey, even if they don't have an immediate need for your product or service. You can also look at how long it takes for people to complete a survey and compare it to the average response time for similar products and services.
This will give you an idea of how much time you need to spend on your surveys to get the results you want.
How do you read a property map?
The lot number will be included in the center of each plot of land if you read a plat map in a property survey. Map of a Lot The plat map is the most common type of map used in property surveys. It is used to show the size and location of lots, roads, buildings, and other features on a property, as well as to identify the boundaries of property lines.
Plat maps can be found in many different formats, such as a sheet of paper, a computer printout, or a digital image file. A lot map can also be created by drawing a circle around the area of interest and then using a pen or pencil to draw a line through the circle.
What do the numbers on the bottom of a map mean?
The map's scale tells you how detailed your map is. You can also zoom in and out with the arrow keys on your keyboard, or use the mouse wheel to move the camera around. You can zoom into a particular area by clicking on it, and you can pan around by using the left and right mouse buttons. | 677.169 | 1 |
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Glosario
Euclidean GeometryIntroduction
Tiempo de leer: ~10 min
Mathematics has been studied for thousands of years – to predict the seasons, calculate taxes, or estimate the size of farming land.
Mathematicians in ancient Greece, around 500 BC, were amazed by mathematical patterns, and wanted to explore and explain them. For the first time, they began to study mathematics just "for fun", without a specific application in mind.
One of these mathematicians was Thales of Miletus, who made a surprising discovery when playing around with geometric shapes:
Start by picking two points anywhere in the box on the left. Let's draw a semicircle around these points.
Now pick a third point that lies somewhere on the circumference of the semicircle.
We can draw a triangle formed by the two corners of the semicircle, as well as the point you picked on the circumference.
Try moving the position of the three points and observe what happens to the angle at the top of the triangle. It seems like it is always °! This means that the triangle is .
For Thales, this was a pretty spectacular result. Why should semicircles and right-angled triangles, two completely different shapes, be linked in this fundamental way? He was so awed by his discovery that, according to legend, he sacrificed an entire ox to thank the gods.
However, simply observing a relationship like this was not enough for Thales. He wanted to understand why it is true, and verify that it is always true – not just in the few examples he tried.
An argument that logically explains, beyond any doubt, why something must be true, is called a proof.In the following courses you will learn a number of geometric techniques, that will eventually allow us to prove Thales' theorem.
But geometry is not just useful for proving theorems – it is everywhere around us, in nature, architecture, technology and design. We need geometry for everything from measuring distances to constructing skyscrapers or sending satellites into space. Here are a few more examples:
Sailors use sextants to determine their location while at sea, using angles formed by the sun or stars.
Geometry is needed to create realistic video game or movie graphics.
Geometry can help design and test new airplane models, making them safer and more efficient.
Geometry was key when designing this skyscraper in Beijing – and to make sure it won't fall over.
Geometry allows us to predict the position of stars, planets and satellites orbiting Earth.
In this and the following courses, you will learn about many different tools and techniques in geometry, that were discovered by mathematicians over the course of many centuries. We will also see how these techniques can be used to solve important problems in the real world. | 677.169 | 1 |
for b, you could try finding the point(s) where the gradient of the plane is equal to the gradient of the surface?
1. What is the definition of a surface normal?
A surface normal is an imaginary line that is perpendicular to a surface at a specific point. It is used to determine the orientation of a surface and is often represented by a vector.
2. How can I find the surface normal at a specific point on a plane?
To find the surface normal at a specific point on a plane, you will need to know the equation of the plane and the coordinates of the point. Using the equation of the plane, you can find the normal vector and then normalize it to get the surface normal.
3. What does it mean for a surface normal to be perpendicular to a plane?
When a surface normal is perpendicular to a plane, it means that the normal vector is at a 90-degree angle to the plane. This indicates that the surface is perfectly flat at that point, with no slope or curvature.
4. Why is it important to find all points where the surface normal is perpendicular to a plane?
Knowing where the surface normal is perpendicular to a plane can be useful in various applications, such as computer graphics, engineering, and physics. It can help determine the orientation and behavior of a surface and aid in calculations and simulations.
5. Are there any other ways to find the surface normal besides using the equation of the plane?
Yes, there are other methods to find the surface normal, such as using partial derivatives or cross products. However, using the equation of the plane is often the most straightforward and efficient way to find the surface normal at a specific point. | 677.169 | 1 |
Once upon a time, Ptolemy let his pupil draw an equilateral triangle \(ABC\) inscribed in a circle before the great mathematician depicted point \(D\) and joined the red lines with other vertices, as shown below.
Ptolemy: Dost thou see that all the red lines have the lengths in whole integers? Pupil: Indeed, master! Such an extraordinary point! Ptolemy: Now if the equilateral triangle has a side length of 13, what is the sum of the three red lengths combined?
\(ABCD\) is a cyclic quadrilateral with \(\displaystyle \overline{AB}=11\) and \(\displaystyle \overline{CD}=19\). \(P\) and \(Q\) are points on \(\overline{AB}\) and \( \overline{CD}\), respectively, such that \(\displaystyle \overline{AP}=6\), \(\displaystyle \overline{DQ}=7\), and \(\displaystyle \overline{PQ}=27.\) Determine the length of the line segment formed when \(\displaystyle \overline{PQ}\) is extended from both sides until it reaches the circle.
Note: The image is not drawn to scale.
A cyclic quadrilateral \(ABCD\) is constructed within a circle such that \(AB = 3, BC = 6,\) and \(\triangle ACD\) is equilateral, as shown to the right.
If \(E\) is the intersection point of both diagonals of \(ABCD\), what is the length of \(ED,\) the blue line segment in the diagram?
Converse
In a quadrilateral, if the product of its diagonals is equal to the sum of the products of the pairs of the opposite sides, then the quadrilateral is inscribable.
Ptolemy's Inequality
Given a quadrilateral \(ABCD\),
\[AC\cdot BD \leq AB\cdot CD + AD\cdot BC,\]
where equality occurs if and only if \(ABCD\) is inscribable. \(_\square\)
The equality occurs when \(I\) lies on \(AC\), which means \(ABCD\) is inscribable. \(_\square\)
Another proof requires a basic understanding of properties of inversions, especially those relevant to distance.
Consider a circle of radius 1 centred at \(A\). Let \(B', C',\) and \(D'\) be the resultant of inverting points \(B, C,\) and \(D\) about this circle, respectively. Using the distance properties of inversion, we have | 677.169 | 1 |
Watch: Parallel and perpendicular lines. Watch the video to discover the difference between parallel and perpendicular lines. Parallel lines are always the same distance apart for their entire .... Tokio crate
Parallel & Perpendicular Lines, 4th Grade Geometry Lesson Packet & Quiz, 4.G.1. This Parallel & Perpendicular Lines complete lesson packet & assessment is common-core aligned to 4.G.1. Perfect for whole group instruction, math intervention, or test prep, this packet has everything you need in one place! And best of all NO-PREP.Math is often called the universal language. Learn all about mathematical concepts at HowStuffWorks. Advertisement Math is often called the universal language because no matter whe... Two lines are said to be perpendicular if they intersect at right angles (90o). Slopes of Parallel and Perpendicular Lines. In analytic geometry, we often track the angles using the slopes of the involved lines. So it's only natural that we use slopes (generally denoted by the letter m) when working with parallel and perpendicular lines too. 4. Write the equation of the line that satisfies the given requirements. b. is perpendicular to the line y = 3x − 1 and passes through the point (0, 0) c. is parallel to the line 6x = 4y and passes through the point (−3, −5) 1. Three lines are perpendicular to the line s, with equal "space" between them. a. Find the equation of the line s. b.22 Qs. Parallel and Perpendicular Lines. 502 plays. 9th. 10 Qs. Advanced Vocabulary Quiz. 268 plays. University. Math : Perpendicular and Parallel Lines by Grade 3 'Aliim quiz for 3rd grade students.Parallel and Perpendicular Lines. Get those lines straight! Help your child master the concept of parallel and perpendicular lines, an important skill for geometry. She'll practice identifying parallel and perpendicular lines, plus she'll also get to practice plotting points and line segments.Aug 21, 2016 · The Corbettmaths Video Tutorials on Parallel and Perpendicular lines Math 100: Liberal Arts Math (Claassen and Ikeda) 2: Geometry 2.2: Properties of Angles ... never cross one another, and are always the same distance apart, there's a good chance that they are parallel. Perpendicular lines are two lines that intersect at a 90° (right) angle. And perpendicular line segments also intersect at a 90° …Drawing Perpendicular Lines: Start by drawing a straight line using your ruler. This is your baseline. Place the midpoint of the protractor on the point where you want the perpendicular line to intersect the baseline. Ensure that the baseline aligns with the \(0^\circ\) mark of the protractor. Mark the \(90^\circ\) point on the protractor.Perpendicular lines can be defined as "lines that make a T-shape." Both of these definitions are somewhat incomplete—lines can be parallel and still not be side-by-side; and perpendicular lines can make an L-shape as a T-shape. Still, introducing such complications is best left for later lessons. The teacher should draw examples and …Dec 18, 2023 · Example \(\PageIndex{1}\) Identify a set of parallel lines and a set of perpendicular lines in the image below. Solution. Parallel lines never meet, and perpendicular lines intersect at a right angle. \(\overleftrightarrow{AB}\) and \(\overleftrightarrow{CD}\) do not intersect in this image, but if you imagine extending both lines, they will intersect soon. Today, I want to share a parallel and perpendicular lines foldable I used to introduce the concepts of these different types of lines to my Algebra 1 classes. Quote of the Day: "I'm going to name my kid 'Perpendicular.'". I'm not the biggest fan of what I had my students write inside the foldable. It was a little wordy, and I will ... Perpendicular lines are marked with "a box" to indicate the location of the right angle. Perpendicular lines intersect in one location, which becomes the vertex of the right angle. Remember that a right angle contains 90º (think of the angle in the corner of a square). Dec 20, 2022 · Slope = Δy Δx = y2 − y1 x2 − x1 = 3 − 3 3 − ( − 3) = 0 6 = 0. Thus, the horizontal line in Figure 3.2.4.4.9 (a) has slope equal to zero, exactly as expected. Further, all horizontal lines are parallel to this horizontal line and have the same slope. Therefore, all horizontal lines have slope zero. We would surmise that the vertical ... The mathematical definition of slope is very similar to our everyday one. In math, slope is used to describe the steepness and direction of lines. ... Characterize the slopes of parallel and perpendicular lines. ... the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program ...Browse Parallel and Perpendicular Lines resources on Teachers Pay Teachers, a marketplace trusted by millions of teachers for original educational resources.Write the letter of the exercise in the box containing the number of the answer. art 1. Write the equation of the line indicated. u Equation of AB O Equation of CD I Equation of EF g Equation of GH Part 2. Write the slope of a line parallel to the given line. O -5x + y = 12 A 4x +7y = 21 Part 3.This video explains what perpendicular means, how to check whether 2 lines are perpendicular to each other and how to find the gradient of a perpendicular li.... | 677.169 | 1 |
The right triangle, a cornerstone of geometry, is a geometric shape that embodies both simplicity and mathematical significance. With its distinctive form defined by one right angle, the right triangle holds a special place in various fields, including mathematics, physics, engineering, and architecture. Its elegant proportions and inherent properties make it a fundamental building block for constructing complex structures and solving intricate problems.
We set off on a tour into the fascinating world of the right triangle in this essay, exploring its unique characteristics, fundamental principles, Pythagorean theorem, and practical applications. By unraveling the secrets of this geometric marvel, we aim to shed light on the profound impact of right triangles in shaping the foundations of theoretical knowledge and real-world structures.
Definition
A right triangle is a geometric shape that is defined by its specific properties. It is a triangle that contains one right angle, which measures 90 degrees. The right angle is formed by the intersection of two sides, known as the legs, and it is always opposite the longest side of the triangle, called the hypotenuse.
The other two angles in a right triangle are acute, or less than 90 degrees. The two sharp angles in a right triangle must add up to 90 degrees since the total angle in any triangle is always 180 degrees. Below is the generic diagram for a right triangle.
Figure-1: Generic right triangle.
The Pythagorean theorem is one of the basic concepts related to right triangles. The square of the lengths of the two legs in a right triangle must equal the square of the length of the hypotenuse, according to this theorem. This can be mathematically represented as $c^2 = a^2 + b^2$, "c" being the length of the hypotenuse and "a" and "b" being the lengths of the legs.
Historical Significance
The historical background of the right triangle, a geometric shape, dates back to ancient cultures and was crucial in the growth of mathematics and several scientific fields. The concept of the right triangle can be traced back to Ancient Mesopotamia and Ancient Egypt, where early civilizations recognized its special properties and utilized it in their architectural and surveying practices. The Egyptians, in particular, were skilled in geometry and used right triangles in their construction projects, such as the pyramids.
The study of right triangles flourished in Ancient Greece, with renowned mathematicians such as Pythagoras making significant contributions. Pythagoras' Theorem states that the hypotenuse's square is equal to the sum of its two other sides' squares. according to Pythagoras' Theorem. became a fundamental principle in geometry and trigonometry. The theorem is named after Pythagoras, but similar ideas were known in other ancient cultures as well.
During the Islamic Golden Age (8th to 14th centuries), scholars like Al-Khwarizmi and Ibn al-Haytham further expanded their knowledge of right triangles and their properties. They contributed to advancements in trigonometry, introducing new concepts and techniques for solving problems involving right triangles.
In Renaissance Europe, mathematicians and astronomers such as Leonardo da Vinci, Johannes Kepler, and Galileo Galilei relied on right triangles and trigonometry to explore the principles of optics, planetary motion, and perspective in art.
The study of right triangles continued to progress through the Enlightenment and into the modern era, with the development of trigonometric functions and their applications in various scientific and engineering fields.
Today, the understanding of right triangles and their properties is an essential part of mathematics education. Right triangles find extensive use in fields such as engineering, physics, navigation, computer graphics, and surveying. Trigonometry, which heavily relies on right triangles, provides a comprehensive framework for analyzing and solving problems involving angles, distances, and relationships between sides.
Fundamental Properties
Certainly! Here are the properties of the right triangle, a geometric shape, explained in detail:
Right Angle
The most distinguishing property of a right triangle is the presence of a right angle. A right angle measures 90 degrees and is formed where two sides of the triangle meet to create a perpendicular intersection.
Hypotenuse
In a right triangle, the hypotenuse is the side that faces the right angle. It is the longest side of the triangle and is denoted by the letter c. The hypotenuse is directly across from the right angle and forms the base for many important relationships in right triangles.
Legs
The other two sides of the right triangle, which are adjacent to the right angle, are called the legs. They are denoted by the letters a and b. The lengths of the legs can vary and are crucial for calculating other properties of the triangle.
Pythagorean Theorem
One of the fundamental properties of a right triangle is the Pythagorean theorem, named after the Greek mathematician Pythagoras.The square of the hypotenuse is equal to the sum of the squares of the two legs, according to this rule. In equation form, it is written as c² = a² + b².
Special Right Triangles
Right triangles can have specific ratios between their side lengths, resulting in special right triangles. The two most common special right triangles are the 45-45-90 triangle and the 30-60-90 triangle. InTrigonometric Functions
InCongruence and Similarity
Right triangles can be used to determine whether two triangles are congruent (having identical side lengths and angles) or similar (having proportional side lengths and congruent angles).
Understanding these properties of the right triangle allows for a comprehensive exploration of its geometry, relationships between sides and angles, and practical applications. Right triangles and their properties serve as the basis for trigonometry, navigation, engineering, and countless other fields where accurate calculations involving angles and distances are required.
Related Formulas
Below are the related formulas of the right triangle, explained in detail:
Pythagorean Theorem
One of the most important formulas for right triangles is the Pythagorean theorem. The hypotenuse is the side that forms the right angle, and the rule says that the square of its length is equal to the sum of the squares of the lengths of the other two sides. In equation form, it is written as c² = a² + b², where a and b are the lengths of the two legs, and c is the hypotenuse length.
Trigonometric Functions
Trigonometric functions are widely used in right triangles to relate the ratios of the side lengths to the angles within the triangle. The three primary trigonometric functions are:
Sine (sin)
The sine of an angle in a right triangle is the proportion of the length of the side directly across from the angle to the length of the hypotenuse.It is written as sin(A) = opposite / hypotenuse.
Cosine (cos)
The ratio of the neighboring side's length to the hypotenuse's length is known as the cosine of an angle in a right triangle. It is written as cos(A) = adjacent / hypotenuse.
Tangent (tan)
In a right triangle, the lengths of the adjacent and opposing sides are compared to determine the angle's tangent. It is written as tan(A) = opposite/adjacent.
Special Right Triangle Formulas
Special right triangles, such as the 45-45-90 triangle and the 30-60-90 triangle, have specific formulas that relate to their side lengths. In a 45-45-90 triangle, the lengths of the two legs are equal, and the length of greater leg is about √3 times longer than the shorter leg.
Area Formula
The formula to calculate the area of a right triangle is A = 0.5 × base × height, where the base and height are the lengths of the two legs. The base and height can be any two sides of the triangle as long as they form a right angle.
Perimeter Formula
The total length of all the sides makes up a right triangle's perimeter. In a right triangle, the perimeter can be calculated by adding the lengths of the two legs and the hypotenuse: P = a + b + c.
Types
There are three main types of right triangles based on the relationships between their sides and angles. Let's explore each type in detail:
Acute Right Triangle
An acute right triangle is a right triangle where both of the acute angles (the angles that are less than 90 degrees) are less than 45 degrees. In this type of triangle, the two legs are of different lengths, and the hypotenuse is the longest side. The Pythagorean theorem applies to acute right triangles, where c² = a² + b². The acute right triangle is characterized by its sharp angles and is commonly encountered in mathematical problems and applications. Below is the diagram for an acute right triangle.
Figure-2: Acute right triangle.
Obtuse Right Triangle
An obtuse right triangle is a right triangle where one of the acute angles is greater than 45 degrees. The other acute angle remains less than 45 degrees, and the right angle is always 90 degrees. In an obtuse right triangle, the longest side is the one opposite the obtuse angle. The Pythagorean theorem also applies to obtuse right triangles, following the same formula as the acute right triangle. These triangles possess one larger angle, which gives them a distinct shape and characteristic. Below is the diagram for an obtuse right triangle.
Figure-3: Obtuse right triangle.
Isosceles Right Triangle
An isosceles right triangle is a right triangle where two of the sides, the legs, are of equal length. The remaining side, the hypotenuse, is always longer than the legs. In an isosceles right triangle, both of the acute angles are 45 degrees. The Pythagorean theorem is simplified for isosceles right triangles as c² = 2a², where "a" denotes the length of the legs and "c" the length of the hypotenuse. Isosceles right triangles exhibit symmetry and are often used in geometric constructions and calculations. Below is the diagram for an isosceles right triangle.
Figure-4: Isosceles right triangle.
Understanding the types of right triangles is important as it provides insights into their specific characteristics and properties. These types allow for a classification of right triangles based on their angle measures and side lengths, enabling more precise analysis and problem-solving in various mathematical and practical contexts.
Applications
Engineering
Right triangles are prevalent in engineering disciplines. They are utilized in structural design, architecture, and civil engineering for calculating forces, determining stability, and analyzing structural elements. Right triangles are also essential in fields like electrical engineering and circuit design, where they aid in calculating voltage, resistance, and impedance in complex circuits.
Physics
Right triangles find applications in physics for various calculations involving forces, vectors, and motion. In mechanics, right triangles are used to resolve forces into components and calculate their magnitudes and directions. Additionally, right triangles are employed in trigonometric calculations related to projectile motion, circular motion, and waves.
Surveying and Navigation
Right triangles play a significant role in surveying and navigation. Triangulation methods, which rely on the principles of right triangles, are used to determine distances, angles, and positions of points on the Earth's surface. This is crucial in activities such as land surveying, mapmaking, and global positioning systems (GPS) technology.
Architecture and Construction
Right triangles are essential in architectural design and construction. They aid in creating geometrically balanced structures, determining angles for roof pitches, calculating stair dimensions, and ensuring proper alignments and measurements.
Astronomy
Right triangles are utilized in celestial navigation and astronomical calculations. Triangulation methods involving right triangles help determine the positions of celestial objects, measure distances in space, and navigate using celestial coordinates.
Art and Design
Right triangles are employed in art and design compositions to create balanced and aesthetically pleasing arrangements. The principles of right triangles help in achieving visual harmony, proportion, and perspective.
Understanding the applications of the right triangle in various fields provides insights into its practical significance and demonstrates its universal utility in different domains.
Exercise
Example 1
Find the length of the hypotenuse in a right triangle with legs measuring 3 units and 4 units.
Solution
Given: Length of leg a = 3 units Length of leg b = 4 units
We can apply the Pythagorean theorem to determine the length of the hypotenuse: c² = a² + b².
Substituting the given values:
c² = 3² + 4²
c² = 9 + 16
c² = 25
When we square the two sides, we discover:
c = √25
c = 5 units
Therefore, the length of the hypotenuse in the right triangle is 5 units.
Example 2
Determine the length of a leg in a right triangle with a hypotenuse of 10 units and the other leg measuring 6 units. | 677.169 | 1 |
How many straight lines can form from 4 collinear points?
There is one more way in which we can calculate the number of lines. Let's divide the points in two groups: the collinear group of 4 points, and the non-collinear group of 6 points. To count the number of lines, we have three possible cases. First, the lines formed using the points of the collinear group – only 1 line.
How many straight lines can form from 10 points?
Detailed Solution ( n − 2 ) ! Given: n = 10. Hence, the number of straight lines that can be formed is 45.
How many triangles can be formed by using 10 points out of which 4 are collinear?
2) Four points are collinear. Given that four points are collinear, which means four points are in the straight line, whereas a triangle needs three points to form a triangle. Out of 10 points, 4 points cannot form a triangle. So, the number of triangles formed when four points are collinear are in 10C3−4C3ways.
How many lines can be drawn through 9 points in which 4 of them are collinear?
R D Sharma – Mathematics 9 Only 1 line can be drawn through 4 collinear points.
How many lines can be drawn from 10 given points no three of which are collinear?
Therefore, actual number of lines is 902=45.
How many collinear points are there in a plane?
There are 10 points in a plane of which 4 are collinear. How many different straight lines can be drawn by joining these points. >> There are 10 points in a pl… There are 10 points in a plane of which 4 are collinear. How many different straight lines can be drawn by joining these points.
2 points when joined in a plane will make 1 line. If the total number of point is 10, the total number of line = 10C2 But 4 points are collinear, so the lines make with these 4 points are same. Hence there is 1 common line joining the 4 collinear point.
How many points are needed to make a straight line?
25. There are 10 points in a plane, so straight lines can be formed using two at a time. \\ Total straight lines = 10C2 = 10 9 2 1 × × = 45 1 But 4 points are collinear. \\ Lines using these points = 4 C2 = 4 3 2 1 × × = 6 1 But these 4 points can make a single line.
How many lines can be drawn from 10c2 of points?
In this Q., there are 10 points, of which max. of 10C2, or 45 lines can be drawn. However, since 4 of these points are collinear, we have to subtract, the total no. of lines formed by these points, if they were not collinear, as that case is also included in the max. 45 lines formed. So subtracting 4C2 from 10C2 gives you =45-6=39 lines | 677.169 | 1 |
The Number of Diagonals in a Polygon: Exploring the Intricacies
When it comes to polygons, their properties and characteristics have fascinated mathematicians for centuries. One such property that has piqued the interest of many is the number of diagonals a polygon possesses. In this article, we will delve into the intricacies of this topic, exploring the formulas, patterns, and real-world applications associated with the number of diagonals in a polygon.
Understanding Diagonals in a Polygon
Before we dive into the number of diagonals, let's first establish what exactly a diagonal is in the context of a polygon. A diagonal is a line segment that connects two non-adjacent vertices of a polygon. In simpler terms, it is a line that connects two points within the polygon, but not the endpoints of the polygon itself.
For instance, consider a regular hexagon. The six line segments connecting the opposite vertices of the hexagon are diagonals. These diagonals divide the hexagon into smaller triangles, each with its own set of diagonals.
Formula for Calculating the Number of Diagonals
Now that we have a clear understanding of what diagonals are, let's explore the formula for calculating the number of diagonals in a polygon. The formula is as follows:
Number of Diagonals = (n * (n – 3)) / 2
Here, 'n' represents the number of sides of the polygon. By plugging in the value of 'n' into this formula, we can determine the number of diagonals in a polygon.
For example, let's consider a pentagon, which has five sides. Using the formula, we can calculate the number of diagonals as follows:
Number of Diagonals = (5 * (5 – 3)) / 2 = 5
Therefore, a pentagon has five diagonals.
Patterns and Observations
By examining the formula, we can observe several interesting patterns and insights about the number of diagonals in a polygon:
The number of diagonals in a polygon is always an integer.
The number of diagonals is directly proportional to the number of sides of the polygon.
As the number of sides increases, the number of diagonals increases exponentially.
A polygon with three sides (a triangle) has no diagonals.
A polygon with four sides (a quadrilateral) has two diagonals.
A polygon with five sides (a pentagon) has five diagonals.
A polygon with six sides (a hexagon) has nine diagonals.
These patterns provide us with a deeper understanding of the relationship between the number of sides and the number of diagonals in a polygon.
Real-World Applications
While the concept of the number of diagonals in a polygon may seem abstract, it has several real-world applications. Let's explore a few examples:
Architecture and Design
In architecture and design, polygons are commonly used to create aesthetically pleasing structures. Understanding the number of diagonals in a polygon helps architects and designers determine the number of internal connections required to strengthen the structure. By calculating the number of diagonals, they can ensure the stability and integrity of the design.
Network Topology
In the field of computer science, network topology refers to the arrangement of various elements in a computer network. Polygons, such as stars and rings, are often used to represent network topologies. By knowing the number of diagonals in these polygons, network engineers can determine the number of direct connections required between different nodes in the network.
Game Development
In game development, polygons are extensively used to create characters, objects, and environments. By understanding the number of diagonals in a polygon, game developers can optimize the performance of their games. They can determine the number of collision checks required between different objects, leading to smoother gameplay and enhanced user experience.
Summary
The number of diagonals in a polygon is a fascinating concept that has practical applications in various fields. By using the formula (n * (n – 3)) / 2, we can calculate the number of diagonals in a polygon. Observing the patterns and insights derived from this formula helps us understand the relationship between the number of sides and the number of diagonals. Furthermore, real-world applications in architecture, network topology, and game development demonstrate the importance of this concept in practical scenarios.
Q&A
1. Can a polygon have more diagonals than sides?
No, a polygon cannot have more diagonals than sides. The number of diagonals is always less than the number of sides.
2. Is the formula for calculating the number of diagonals applicable to all polygons?
Yes, the formula (n * (n – 3)) / 2 is applicable to all polygons, regardless of their shape or size.
3. Are diagonals always straight lines?
Yes, diagonals in a polygon are always straight lines that connect two non-adjacent vertices.
4. Can a polygon have no diagonals?
No, a polygon with three sides (a triangle) has no diagonals. However, polygons with four or more sides will always have diagonals.
5. How are diagonals different from sides in a polygon?
Sides in a polygon connect adjacent vertices, while diagonals connect non-adjacent vertices. Diagonals are internal line segments within the polygon, whereas sides form the boundary of the polygon. | 677.169 | 1 |
Angle CBD Has a Measure of 140°. What Is the Measure of Angle Abd
In the realm of geometry, the measurement of angles holds great importance.
In this article, we delve into the relationship between Angle CBD, which measures 140°, and Angle ABD.
By employing geometric principles and analyzing the properties of angles, we aim to unravel the precise measurement of Angle ABD.
Through a logical and detailed examination, we shed light on the significant connection between these angles and provide the techniques necessary to determine the measure of Angle ABD.
The Relationship Between Angles CBD and ABD
The relationship between angles CBD and ABD can be determined by analyzing their measures and geometric properties.
CBD and ABD are adjacent angles, sharing a common vertex and a common side. According to geometric principles, the sum of the measures of adjacent angles is equal to 180°.
Therefore, the measure of angle ABD can be found by subtracting the measure of angle CBD from 180°.
This measurement is of significant importance in understanding the overall angle relationships in the given scenario.
Geometric Principles for Determining Angle ABD's Measure
By applying geometric principles, we can determine the measure of angle ABD based on its relationship with angle CBD. The significance of this value lies in its ability to provide insights into the overall shape and structure of the geometric figure.
Here are four key points to consider:
Angle CBD has a measure of 140°.
Angle ABD and angle CBD are adjacent angles.
The sum of adjacent angles is equal to 180°.
By subtracting the measure of angle CBD from 180°, we can find the measure of angle ABD.
Properties of Angles in Relation to Angle CBD
Exploring the geometric figure's structure involves examining the various properties of angles in relation to the measure of angle CBD.
To determine the measure of angle ABD, we can apply geometric principles. One such principle is the Angle Sum Property, which states that the sum of the measures of angles in a triangle is 180°.
Unveiling the Measurement of Angle ABD
Unveiling the measurement of angle ABD, we can utilize the Angle Sum Property to determine its measure by subtracting the measure of angle CBD from 180°. By doing so, we can uncover the significance of this value in the context of geometric principles.
To provide a clearer picture for the audience, here are four key points to consider:
Angle ABD is complementary to angle CBD.
The measure of angle ABD is the difference between 180° and the measure of angle CBD.
The value of angle ABD determines the relationship between angle CBD and the remaining angles in the figure.
Understanding the measure of angle ABD allows us to make informed conclusions about the angles and their properties in the given geometric scenario.
Understanding the Significance of Angle ABD's Value
One can gain a deeper comprehension of the significance of angle ABD's value through understanding its relationship with angle CBD.
In terms of geometric principles, the measurement of angle ABD holds great importance. It not only determines the size of the angle but also provides insights into the shape and orientation of the triangle.
Analyzing the Connection Between Angle CBD and ABD
To fully understand the relationship between angle CBD and ABD, it is essential to analyze their connection in order to determine the measure of angle ABD accurately. Here are four key points to consider when examining the geometric properties and angle measurement techniques:
Angle CBD has a measure of 140°.
Angle ABD is formed by the intersection of lines AB and BD.
Angle ABD and angle CBD are adjacent angles.
The measure of angle ABD can be determined using the supplementary angle property, which states that the sum of two adjacent angles is 180°. Read more
Applying Angle Measurement Techniques to Determine Angle ABD
To determine the measure of angle ABD, we can use angle measurement techniques such as the Angle Addition Postulate.
By subtracting the measure of angle CBD from 180°, we can find the difference between the sum of angles CBD and ABD and the straight angle.
This will give us the measure of angle ABD.
Angle ABD Measurement
Our team is currently analyzing various techniques to accurately measure the angle ABD in order to determine its precise measurement. The angle ABD has specific properties that can aid in its calculation. Here are four key points to consider:
Angle ABD is an interior angle of triangle ABD.
Angle ABD is supplementary to angle CBD, which has a measure of 140°.
Angle ABD can be calculated using the sum of angles within triangle ABD.
Angle ABD can also be determined by using trigonometric functions in relation to the lengths of the sides of triangle ABD.
Angle CBD Relationship
We are currently examining the relationship between angle CBD and angle ABD, and applying various angle measurement techniques to determine the measure of angle ABD.
Geometric relationships play a vital role in determining angles within a shape. By utilizing angle measurement methods such as protractors or trigonometric functions, we can accurately calculate the measure of angle ABD.
Angle Measurement Techniques
Furthermore, utilizing advanced angle measurement techniques, such as laser-based devices or digital protractors, allows for precise determination of the measure of angle ABD in our current discussion.
To understand the measure of angle ABD, we need to consider the following:
Angle CBD has a measure of 140°.
Angle ABD and angle CBD are adjacent angles.
Adjacent angles share a common side and vertex.
We can use the geometric angles and trigonometric calculations to determine the measure of angle ABD accurately.
Frequently Asked Questions
What Is the Measure of Angle Cbd?
The measure of an angle can be determined using geometric principles and angle measurement techniques. In this case, the measure of angle CBD is 140°.
How Are Angles CBD and ABD Related to Each Other?
The relationship between angles CBD and ABD can be determined by applying angle measurement techniques. By knowing the measure of angle CBD, we can use geometric principles to find the measure of angle ABD.
What Are the Geometric Principles Used to Determine the Measure of Angle Abd?
The measure of angle ABD can be determined using geometric principles and angle measurement techniques. By applying these principles, the specific measurement of angle ABD can be calculated accurately and precisely.
Are There Any Specific Properties of Angles That Are Relevant to Angle Cbd?
The properties of angles are relevant when determining their measures. Understanding these properties allows for the calculation of the measure of angle CBD and subsequently, the determination of the measure of angle ABD.
How Can Angle Measurement Techniques Be Applied to Determine the Measure of Angle Abd?
The application of angle measurement techniques can be used to determine the measure of angle ABD by utilizing the given information about angle CBD. By applying the relevant properties of angles, the measure of angle ABD can be calculated.
Conclusion
In conclusion, the measure of angle ABD can be determined by subtracting the measure of angle CBD from 180°.
Thus, if angle CBD has a measure of 140°, the measure of angle ABD would be 180° – 140° = 40°.
This relationship between the two angles is based on geometric principles and properties of angles.
Understanding the significance of angle ABD's value allows for further analysis and application of angle measurement techniques. | 677.169 | 1 |
The ends of a quadrant of a circle have the coordinates (1, 3) and (3,
1). Then the center of such a circle is
Solution in Bengali
Video Solution
|
Answer
Step by step video & image solution for The ends of a quadrant of a circle have the coordinates (1, 3) and (3,
1). Then the center of such a circle is by Maths experts to help you in doubts & scoring excellent marks in Class 11 exams. | 677.169 | 1 |
math:atan2
Returns the angle in radians subtended at the origin by the point on a plane with coordinates (x, y)
and the positive x-axis, the result being in the range -π to +π.
(Note the order of arguments (y, x). There are some maths libraries that implement an atan2
function with the arguments in the order (x, y). | 677.169 | 1 |
16
Pįgina 176 ... cone is called a right - angled cone : if it be less than the other leg , an obtuse - angled , and if greater , an acute - angled cone . 22. The axis of a cone is the fixed straight line about which the triangle revolves . 23. The base ...
Pįgina 215 ... cone is a third part of a cylinder on the same base , and of equal altitude . Let a cone have the same base with a cylinder , viz . the circle ABCD , and the same altitude . The cone is the third part of the cylinder ; that is , the ...
Pįgina 216 ... cone . Let them be those upon the segments of the circle AE , EB , BF , & c . Therefore the rest of the cylinder ... cone : but ( XII . 7 , cor . 1 ) this prism is triple of the pyramid upon the same base , of which the vertex is the ...
Pįgina 217 ... cone : neither , as has been demonstrated , is it greater than the triple . Therefore the cylinder is triple of the cone , or the cone is the third part of the cylinder . A cone , therefore , & c . PROP . XI . THEOR . - Cones and ...
Pįgina 218 ... cone , and so on , there must at length remain ( XII . lem . 1 ) some segments of the cone which are together less than Z. Let these be the segments upon EO , OF , FP , PG , GR , RH , HS , SE : therefore the remainder of the cone , viz | 677.169 | 1 |
Similar Triangles When are triangles considered similar Triangles
When are triangles considered similar? � Triangles are similar when: their corresponding angles are equal Their corresponding sides are in proportion to each other They have the same shape, but are different sizes
What are corresponding angles? � � Angles that correspond to each other are in the same relative position in similar triangles. In the two triangles below, <K = <H, <L = <I and <M = <J
Don't be fooled! � Be careful: triangles can be reversed in position – look carefully for the corresponding angles: r l � f t m n Can you name the corresponding angles? ? ?
Don't be fooled! � Sometimes, triangles can be flipped and reversed! Can you still find the corresponding angles?
Don't be fooled! � Sometimes, similar triangles can be hiding inside of each other. Can you still find the corresponding angles?
What are corresponding sides? � � In the triangles below, the corresponding sides are the ones that are connecting the corresponding angles. They are in proportion to each other and have the same relative position. So, ~ and ML ~
Corresponding sides � � In similar triangles, corresponding sides are proportional. That means that they have been either enlarged or reduced by the same scale factor. By what scale factor have the corresponding sides been reduced in the triangles below?
Corresponding sides � � � You can find missing side lengths in corresponding sides by setting up a ratio or proportion to solve: = By cross multiplying, x = 9 | 677.169 | 1 |
RYAN GROUP OF SCHOOLS STD: SUB: ACADEMICYEAR 2023-24 ICSE, SECOND PRE-BOARD EXAMINATION X MARKS: MATHEMATICS DURATION: RYAN 80 2 HRS 30 MINS Answer to this Paper must be written on the paper provided separately. Youwill not be allowed to write during first 15 minutes. This time is to be spent in reading the question paper. The time given at the head of this Paper is the time allowed for writing the answers. Attempt all questions from Section A. . Any four questions from Section B. All working, including rough work, must be clearly shown, and must be done on the same sheet as the rest of theanswer. Omission of essential working will result in loss of marks. The intended marks for questions or parts of questions are given in brackets [J. Mathematical tables are provided. SECTION A (15 MARKS] Select the correct option for each of the following questions. Question 1. Choose the correct answers to the questions fromn the given options. (Do not copy the question, Write the correct answer only.) [15] 1.Apoint P is reflected in the x-axis. Co-ordinates of its image are (-4, 5). Find the co-ordinates of P. a. The origin b. (-4,-5) c.(-4,5) d. (4,-5) 2.Calculate the co-ordinates of the centroid of the triangle ABC, if A= (7,-2), B= (0,1), C= (-1,4). a. (2,1) b. (-2,-1) c. (2, -1) d. (-2,1) 3. In the given figure, BC isa diameter of circle with Centre '0'. If ZAOB =70 then the value of L0AC is a. 35 C. 110 b. 70 d, 140 | 677.169 | 1 |
Compass Traversing Types And Methods
This Article Covers "What Is Compass Traversing" Types Of Traversing And Methods Of Traversing One By One.
Compass Traversing
A type of survey that involves both linear as well as angular measurements known as Compass Traversing.
In Traverse surveying the directions of survey lines re fixed by angular measurements and not by forming a network of triangles as done in chain surveying.
A traverse survey is one in which framework consist of a series of connected lines, the length and directions of which measured with a chain or tape and angular instruments respectively.
The routine of changing and offsetting is the same as in chain surveying.
The running of check lines is not necessary.
Types Of Traverse
There are generally two types,
Closed Traverse
A traverse said to closed when a complete circuit made i.e when it returns to the starting point forming a closed polygon OR when it begins and ends at points whose positions on plan known.
The work may checked or "balanced".
Suitable for locating bounders of lakes, woods etc and for the survey of moderately large area.
Open Traverse
A traverse said to open or unclosed when it does not form a closed polygon.
It consist of connecting lines extending in the same general direction and not returning to the starting point.
Similarly it does not start and end as points whose positions on plan known.
It is suitable for the survey of long narrow strip eg river, coast line, road, railway etc.
Method of Traversing
The different methods by which the direction of the survey lines may be determined are as follows:
By the Chain angles.
Free or loose needle method.
By the fast needle method.
By the measurement of angles between successive lines.
Method of Chain Angles or Chain Traversing
In this method entire work done with the chain or tape only.
When survey of an area such as lake, wood, standing crops etc required, so area cannot be divide into triangles.
A traverse therefore to run round the area and in the absence of an angler instrument, the angles between the successive lines fixed by measuring the lines with a chain.
Angles fixed by measurement known as Chain Angles.
The angles at the station fixed by internal tie lines such a1,a2, or external such as b1,b2, c1,c2, d1,d2.
The angle at A fixed by measuring the distance a1 and a2 suitably chosen on AD and AB, b3 to check accuracy.
In open traverse Angeles at B and C fixed by ab and cd and so on.
By the Free or Loose needle method
In this method an angular instrument such as Compass or Theodolite set up at each of the successive stations and the bearing of each line taken with reference to the magnetic meridian and not with reference to the adjacent lines.
Hence the errors are not cumulative but compensating.
The result obtain are not generally so accurate as those obtained by methods 3 and 4
By the fast needle method
In this method a theodolite used for determine the bearings of each line.
By the measurement of angles between successive lines
In this method a theodolite used for measurement of angles.
The horizontal angles measured in a traverse may 1) included angle 2) deflection angles between successive lines.
The method of included angles is the most accurate method and generally used for large surveys and accurate work. | 677.169 | 1 |
What are Parallel Lines?
The lines that do not intersect or meet each other at any point in a plane are termed parallel lines. Parallel lines are non-intersecting lines and always stay apart from each other. It is also said that parallel lines meet at infinity.
Definition of Parallel Lines
Parallel lines in geometry can be defined as two straight lines in the plane that are at equal distance from each other and never intersects no matter how much they are extended. Parallel lines are represented by the symbol "||".
Representation of Parallel Lines
Two parallel lines AB and CD are represented as AB||CD. This means line AB is parallel to line CD. In the given plane, infinite parallel lines can be drawn to AB and CD.
What are the Different Types of Angles Formed When Transversal Line Intersect Parallel Line?
When transversal line intersects two parallel lines, the following pair of angles are formed:
Corresponding angles
Alternate interior angles
Alternate exterior angles
Vertically opposite angles
Linear Pair
Real-Life Applications of Parallel Lines
We observe parallel lines in the following things:
Railway Track
Cricket Stumps
Racing Track
Fork Lines
Steps of a Ladder
Rail Bars
Keys of a Piano
Pins of a Plug
Markings on Road
Zebra Crossing
Edges of A Ruler
How To Identify Parallel Lines?
Two lines intersected by a transversal line are termed parallel lines when corresponding angles are equal.
Two lines intersected by a transversal line are termed parallel lines when alternate interior angles are equal.
Two lines intersected by a transversal line are termed parallel lines when alternate exterior angles are equal.
Two lines intersected by a transversal line are termed parallel lines when sum of consecutive interior angles is 180 degrees.
Two lines intersected by a transversal line are termed parallel lines when sum of consecutive exterior angles is 180 degrees.
What is Geometry?
Geometry is regarded as the oldest part of Mathematics that is concerned with the property of space that is related to the shape, size, distance, and relative position of figures. A mathematician whose area of work is in Geometry is known as Geometer. The need for Geometry is not only limited to the study of a flat surface (known as plane geometry) and three-dimensional objects (known as solid Geometry).
What Are The Different Types of Geometry?
The different types of Geometry are:
Eucledian Geometry: It is the most common type of Geometry that is generally taught at the primary level. Euclidean Geometry is described by Euclidean in detail in 'Element' which is one of the bases of Mathematics. The impact of 'Element' was so immense that no other kind of Geometry was almost used for the next 2000 years.
Non-Euclidean Geometry: This type of Geometry is an extension to Euclid Geometry which is mostly observed in three-dimensional objects. The other name of Non-Euclidean Geometry is hyperbolic geometry, spherical geometry, elliptic geometry, and many more. This type of Geometry represents how most know theorem 'Sum of the angles of a triangle is very different in 3-D space.'
Analytic Geometry: In Analytic Geometry, we study geometric figures and construction using a coordinate system. The most commonly used coordinate systems are Cartesian, Polar, and Parametric systems.
Differential Geometry: In differential Geometry, we study planes, lines, and surfaces in a three-dimensional space with the principles of integrals and differential calculus. This type of Geometry emphasizes several problems like contact surface, geodesics, complex manifolds, and several others. Geodesic is termed as the shortest distance between two points on the surface of a sphere. The use of Differential Geometry ranges from engineering problems to the computation of gravitational fields | 677.169 | 1 |
Displaying top 8 worksheets found for gina wilson parallel and perdendicular lines. Gina wilson geometry displaying top 8 worksheets found for this concept. If you don t see any interesting for you use our search form on bottom.
Find the value of y find the value of y. Some Find | 677.169 | 1 |
NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric Functions
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NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric Functions
NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric Functions are available at Goyanka Maths Study, which are prepared by Balkishan Agrawal. All these solutions are written as per the latest guidelines of CBSE. We provides step by step solutions by considering the different understanding levels of students and the marking scheme. Chapter 3 Trigonometric Functions of NCERT Class 11 Maths is an important chapter for students. Though the chapter has more mathematical terms and formulae, We have made NCERT Solutions for Class 11 Maths easy for the students to understand and remember them, using tricks. Trigonometry is developed to solve geometric problems involving triangles. Students cannot skip Trigonometric chapters since this is being used in many areas such as finding the heights of tides in the ocean, designing electronic circuits, etc., In the earlier classes, students have already studied trigonometric identities and applications of trigonometric ratios. In Class 11, trigonometric ratios are generalised to trigonometric function and their properties. However, these NCERT Solutions of GoyankaMathsStudy help the students to attain more knowledge and score full marks in this chapter | 677.169 | 1 |
The Geometer´s warehouse
Angle in a semi-circle
Investigate the circle property by clicking and dragging any of the blue points. What observation can you make regarding the angle in a semi-circle? Confirm your conjecture by revealing the relevant measurements then show the property.
What is the special name given to the chord AB?
What type of angle is <AOB?
This property can be proved using another circle property. What property is this? | 677.169 | 1 |
Question 1: Set S contains points whose abscissa and ordinate are both natural numbers. Point P, an element in set S has the property that the sum of the distances from point P to the point (3,0) and the point (0,5) is the lowest among all elements in set S. What is the sum of abscissa and ordinate of point P?
A. 2 B. 3 C. 5 D. 4 Answer. 4 Explanation. Any point on the line will have the shortest overall distance. However, we need to have integral coordinates. So, we need to find points with integral coordinates as close as possible to the line 5x + 3y = 15.
Sum of abscissa + ordinate = 4 The question is "What is the sum of abscissa and ordinate of point P?" Hence, the answer is 4. Choice D is the correct answer.
Question 2: Region R is defined as the region in the first quadrant satisfying the condition 3x + 4y < 12. Given that a point P with coordinates (r, s) lies within the region R, what is the probability that r > 2? A. 1/4
B. 1/3
C. 1/5
D. 1/2 Answer. 1/4 Explanation.
Line 3x + 4y =12 cuts the x-axis at (4, 0) and y axis at (0, 3).
The region in the first quadrant satisfying the condition 3x + 4y < 12 forms a right triangle with sides 3, 4 and 5. Area of this triangle = 6 sq.units.
Probability of the point lying in said region = 1.5/6 = 1/4 The question is "Given that a point P with coordinates (r, s) lies within the region R, what is the probability that r > 2?"
Hence, the answer is 1/4 Choice A is the correct answer.
Question 3: Region Q is defined by the equation 2x + y < 40. How many points (r, s) exist such that r is a natural number and s is a multiple of r?
A. 84
B. 92
C. 105
D. 72 Answer. 92 Explanation.
When r = 1, s can take 37 values [37/1]
When r = 2, s can take 17 values [35/2]
When r = 3, s can take 11 values [33/3]
When r = 4, s can take 7 values [31/4]
When r = 5, s can take 5 values [29/5]
When r = 6, s can take 4 values [27/6]
When r = 7, s can take 3 values [25/7]
When r = 8, s can take 2 values [23/8]
When r = 9, s can take 2 values [21/9]
When r = 10, s can take 1 values [19/10]
When r = 11, 12, 13 s can take one value each.
Totally, there are 92 values possible.
The question is "How many points (r, s) exist such that r is a natural number and s is a multiple of r?"
Hence, the answer is 92.
Choice B is the correct answer.
Question 4: What is the equation of a set of points equidistant from the lines y = 5 and x = –4?
A. x + y = –1
B. x – y = –1
C. x + y = 1
D. –x + y = –1
Answer. x + y = 1 Explanation.
Let us try to draw the given lines on the coordinate plane.
A set of points equidistant from the given two lines should lie on the dotted line as indicated. You can think of it as the perpendicular bisector to the base of an isosceles triangle formed by (–4, 5) and the two points on x = –4 and y = 5.
Or, the set of points equidistant from two lines form the angle bisector of the angle formed at the point of intersection of the two lines. The angle between these two lines is 900. Importantly, the lines are parallel to the axes. So, thinking of the line that is the angle bisector of this angle should not be too difficult.
This dotted line is at an angle of 135o with respect to the positive direction of x–axis and also passes through (–4, 5).
Slope = m = tan (135o) = –1.
Therefore, the equation is given by (y – y1) = m
(x – x1) where (x1, y1) is (–4, 5).
(y – 5) = –(x + 4)
x + y = 1
The question is "What is the equation of a set of points equidistant from the lines?"
Hence, the answer is x + y = 1.
Choice C is the correct answer.
Question 5: What is the area enclosed in the region defined by y = |x – 1| + 2, line x = 1, X–axis and Y–axis?
A. 5 sq units
B. 2.5 sq units
C. 10 sq units
D. 7 sq units
Answer. 2.5 sq units Explanation.
Let us first draw y = |x|.
Now, y = |x – 1| is just a shift of '1' unit to the right along the x–axis.
Now, y = |x – 1| + 2 is a shift of '2' units to the top along the y–axis.
Now, let us complete the diagram by drawing the line x = 1 also.
The required area is essentially the area of the shaded region.
The point of intersection of y = |x – 1| + 2 where the y–axis can be found by substituting x = 0 in the equation.
Thus we get y = 3.
Required area = Area of the trapezium formed by the points (0, 0), (1, 0), (1, 2) and (0, 3).
Area of a trapezium = sum of the parallel sides = 5/2 sq units.
The question is "What is the area enclosed in the region defined by y = |x – 1| + 2, line x = 1, X–axis and Y–axis?"
Hence, the answer is 2.5 sq units
Choice B is the correct answer.
Question 6: Find the area of the region that comprises all points that satisfy the two conditions x2 + y2 + 6x + 8y ≤ 0 and 4x ≥ 3y?
A. 25
B.
C.
D. None of these
Answer. Explanation.
x2 + y2 + 6x + 8y < 0
x2 + 6x + 9 – 9 + y2 + 8y + 16 – 16 < 0
(x + 3)2 + (y + 4)2 < 25
This represents a circular region with centre (–3, –4) and radius 5 units. Substituting x = y = 0, we also see that the inequation is satisfied. This means that the circle also passes through the origin. To find out the intercepts that the circle cuts off with the axes, substitute x = 0 to find out the y–intercept and y = 0 to find out x–intercept. Thus x–intercept = –6 and y–intercept = –8.
Now, the line 4x = 3y passes through the point (–3, –4). Or this line is the diameter of the circle. The area we are looking for is the area of a semicircle.
Required area
The question is "Find the area of the region that comprises all points that satisfy the two conditions x2 + y2 + 6x + 8y ≤ 0 and 4x ≥ 3y?"
Hence, the answer is sq units
Choice C is the correct answer.
Question 7: What is the reflection of the point (4 , -3) in the line y = 1?
Ans. Coordinate geometry is a branch of mathematics that deals with the study of geometric figures using a coordinate system. It combines algebraic techniques with geometric interpretations and allows us to analyze and solve problems involving points, lines, curves, and shapes.
2. How does the coordinate system work in coordinate geometry?
Ans. The coordinate system in coordinate geometry uses a pair of numbers called coordinates to locate points in a plane. The horizontal line is called the x-axis, and the vertical line is called the y-axis. The intersection of these lines is called the origin, which is assigned the coordinates (0,0). Points are located by specifying their distance from the origin along each axis.
3. What is the equation of a straight line in coordinate geometry?
Ans. The equation of a straight line in coordinate geometry is typically written in the form y = mx + c, where m represents the slope of the line and c represents the y-intercept. The slope m determines the steepness of the line, while the y-intercept c represents the point where the line intersects the y-axis.
4. How can we find the distance between two points in coordinate geometry?
Ans. The distance between two points in coordinate geometry can be found using the distance formula, which is derived from the Pythagorean theorem. If the coordinates of the two points are (x1, y1) and (x2, y2), the distance between them is given by the formula:
Distance = √((x2 - x1)^2 + (y2 - y1)^2)
5. What are the different types of conic sections in coordinate geometry?
Ans. In coordinate geometry, there are four main types of conic sections: circles, ellipses, parabolas, and hyperbolas. These conic sections can be represented by algebraic equations and have distinct geometric properties. Circles have a constant distance from a fixed center, ellipses have two foci, parabolas have a directrix and a focus, and hyperbolas have two branches.
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which will help them while preparing for their exam. Apart from the Coordinate Geometry Question with Answer Coordinate Geometry Question with Answer is prepared as per the latest CAT syllabus. | 677.169 | 1 |
A vector $$\overrightarrow A $$ has components $${A_1},{A_2},{A_3}$$ in a right -handed rectangular Cartesian coordinate system $$oxyz.$$ The coordinate system is rotated about the $$x$$-axis through an angle $${\pi \over 2}.$$ Find the components of $$A$$ in the new coordinate system in terms of $${A_1},{A_2},{A_3}.$$
From a point $$O$$ inside a triangle $$ABC,$$ perpendiculars $$OD$$, $$OE, OF$$ are drawn to the sides $$BC, CA, AB$$ respectively. Prove that the perpendiculars from $$A, B, C$$ to the sides $$EF, FD, DE$$ are concurrent. | 677.169 | 1 |
Q. The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmoharare planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the Fig. The students are to sow seeds of flowering plants on the remaining area of the plot. (i) Taking A as origin, find the coordinates of the vertices of the triangle. (ii) What will be the coordinates of the vertices of ΔPQR if C is the origin? Also calculate the areas of the triangles in these cases. What do you observe? | 677.169 | 1 |
Polygon
A polygon is a two-dimensional closed shape with straight sides. It is formed by three or more line segments that connect at their endpoints to form a closed shape. Polygons can have any number of sides, from three to infinity. The most common polygons include triangles, quadrilaterals, pentagons, hexagons, and octagons. The properties of a polygon, such as its area and perimeter, can be calculated using mathematical formulas. Polygons are commonly used in geometry, architecture, and engineering, and can be found in various natural and man-made structures, such as buildings, bridges, and crystals. Overall, polygons are essential elements in the world of mathematics and have numerous applications in different fields.
Polygons in landscape design
Polygons are widely used in landscape design as they allow designers to create an array of shapes and sizes. These geometric shapes can range from simple squares and rectangles to more complex polygons like pentagons, hexagons, and octagons. The use of polygons in landscape design is not only aesthetically pleasing but also functional. They can be used to create clear borders between different areas of the garden or to highlight certain features. Additionally, polygons can be used as planting beds or as a base for garden structures like gazebos or pergolas. Their versatility and flexibility make polygons an essential tool in landscape design, allowing designers to create unique and visually stunning outdoor spaces. | 677.169 | 1 |
Which pair of undefined terms is used to define the term parallel lines?
Our experts in all academic subjects are available 24/7. Don't hesitate to ask questions and start discussions whenever you need professional advice.
A
Answer: A. Point and line
Two lines are said to be parallel to each other when they do not intersect themselves at any point. In other words, parallel lines will continue to move in a single straight direction without meeting at any point; they are equidistant from one another. So line and point are terms used to explain the term "parallel lines." Both of these terms are undefined in geometry | 677.169 | 1 |
Math Geometry of math? Do you love geometry? If yes, then take these Math geometry quiz questions and answers, and you will not only test your current geometry knowledge but will also learn something more from here. The questions are mixed in this quiz. Some of them may be harder than others. We hope you manage to get a score above 70 percent. Best of luck to you!
Questions and Answers
1.
Find the value of x.
A.
4 ft
B.
4.5 ft
C.
4.8 ft
D.
5 ft
Correct Answer C. 4.8 ft
Explanation The value of x is 4.8 ft because it is the only option that is listed in the given answer choices.
Rate this question:
2.
3 angles of a triangle add up to ______ degree.
A.
90
B.
120
C.
175
D.
180
Correct Answer D. 180
Explanation The sum of the three angles in a triangle is always 180 degrees. This is a fundamental property of triangles in Euclidean geometry. Regardless of the size or shape of the triangle, the sum of its angles will always be 180 degrees.
Rate this question:
3.
Semicircle features ________ dimensions.
A.
Two
B.
Three
C.
Four
D.
Five
Correct Answer A. Two
Explanation A semicircle is a half of a circle, so it only has two dimensions - the radius and the arc length. The radius determines the size of the semicircle, while the arc length is the distance along the curved edge of the semicircle. Therefore, the correct answer is Two.
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4.
A hexagon has ______ sides.
A.
2
B.
4
C.
6
D.
9
Correct Answer C. 6
Explanation A hexagon is a polygon with six sides. It is a closed shape that consists of six straight sides and six angles. Each side of the hexagon is connected to two other sides, making a total of six sides. Therefore, the correct answer is 6.
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5.
Find the value of x.
A.
10cm
B.
15cm
C.
20cm
D.
22cm
Correct Answer B. 15cm
Explanation The value of x is 15cm because it is the only value that is listed in the given options.
Rate this question:
6.
Find the value of x.
A.
120m
B.
130m
C.
140m
D.
150m
Correct Answer D. 150m
7.
The picture below is Leila's favorite picture.
Leila had a reduced copy of the picture made as a gift for her mother. If the reduced picture was similar to the original and the height of the reduced picture was 6 inches, what was its width?
A.
4 inches
B.
1 inch
C.
9 inches
D.
11 inches
Correct Answer A. 4 inches
8.
A square has ______ sides of equal length.
A.
Four
B.
Two
C.
One
D.
Three
Correct Answer A. Four
Explanation A square is a polygon with four sides of equal length. Each side of the square is congruent to the other three sides, making a total of four sides that are all equal in length. Therefore, the correct answer is four.
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9.
If the two angles of a triangle are 50 degrees and 70 degrees, respectively, then the third angle will be _______.
A.
10 degree
B.
20 degree
C.
40 degree
D.
60 degree
Correct Answer D. 60 degree
Explanation The sum of the angles in a triangle is always 180 degrees. Given that two angles are already known to be 50 degrees and 70 degrees, the third angle can be found by subtracting the sum of these two angles from 180. Therefore, the third angle will be 60 degrees.
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10.
The picture below is Larry's favorite picture.
Larry had a reduced copy of the picture made as a gift for his father. If the reduced picture was similar to the original and the height of the reduced picture was 7 inches, what was its width?
A.
13 inches
B.
8.75 inches
C.
5.60 inches
D.
10 inches
Correct Answer B. 8.75 inches
Explanation The reduced picture is similar to the original picture, which means that the ratio of the height to the width remains the same. Since the height of the reduced picture is given as 7 inches, we can set up a proportion to find the width. Let x represent the width of the reduced picture. The proportion is 7/x = original height/original width. Since the original height and width are not given, we cannot solve for x exactly. However, we can determine that the width of the reduced picture would be 8.75 inches if the original picture had a height of 5 inches and a width of 7 inches. Therefore, 8.75 inches is a possible width for the reduced picture. | 677.169 | 1 |
Math Exercises & Math Problems: Metric Relations in Space
Consider a cube ABCDEFGH, |AB| = a = 4 cm. Find the distance from a point F to a given straight line :
Consider a cuboid ABCDEFGH, |AB| = a = 60 cm, |AD| = b = 400 mm, |AE| = c = 8 dm. Find the distance from a point B to a point H, distance from a point D to a point G and the distance from a point C to a point F.
Consider a cone with a base diameter of d = 12 cm. Any point on the circumference of the base is located 8.5 cm from the apex of the cone. Find the distance from the apex V to the base.
Consider a triangular prism ABCDEF with a base of right triangle, which has the right angles at the vertices C and F. The lengths of legs are |AC| = 8 m, |BC| = 6 m and a height of the prism is h = 15 m. Find the distance between points E and S and the size of an angle φ = |∠BSE|, where S is the midpoint of an edge AC.
Consider a cube ABCDEFGH with the points on its edges X∈EH, Y∈AB, Z∈GH, if we know |EH| = |XH|, |AY| = |YB|, |ZH| = 3|GZ|. Find the angle between the straight lines AX and YZ.
Consider a cube ABCDEFGH. Let M be a midpoint of an edge AE and the size of a cube's edge is a. Find the angle between the straight lines BH and BM. | 677.169 | 1 |
I work with the latex class beamer (\usetheme{Warsaw}). In order to say the area C of mathematics was born from the area A and the area B, I need to draw two circles (area A and area B) in right and left with two arrows comes from them into the third one (area C) that is under and between them. Is there any idea how to draw it?
I use : \begin{frame} \smartdiagramset{ planet size=2cm, distance planet-text=0.1, distance planet-satellite=2.5cm, /tikz/connection planet satellite/.append style={<-} } \smartdiagram[constellation diagram]{ A,B,C } \end{frame} But the result is B-->A<--C I need B and C at the above of A | 677.169 | 1 |
Since the complex numbers are ordered pairs of real numbers, there is a one-to-one correspondence between them and points in the plane. In this section we shall see what effect algebraic operations on complex numbers have on their geometric representations.
The number can be represented by a position vector in the xy-plane whose tail is at the origin and whose head is at the point (x,y). When the xy-plane is used for displaying complex numbers, it is called the complex plane, or more simply, the z-plane. Recall that and . Geometrically, is the projection of onto the x- axis, and is the projection of onto the y-axis. It makes sense, then, that the x-axis is also called the real axis, and the y-axis is called the imaginary axis.
Definition 1.8: Modulus
The modulus, or absolute value, of the complex number is a non-negative real number denoted and is given by the equation
.
The number is the distance between the origin and the point (x, y). The only complex number with modulus zero is the number 0. The number has modulus . The numbers , , and are the lengths of the sides and hypotenuse of a right triangle, from which it follows that
and .
Theorem 1.2 (The triangle inequality) If and are arbitrary comples numbers then | 677.169 | 1 |
What is a tangent to a circle?
It is a straight line that meets the circle at any point, and we will say this process the point of tangency. In this process, the tangent of the circle is perpendicular (⊥) to the radius.
Here we have circle A where line segment AT is the radius, and congruent TP is the tangent to the circle.
Hence, line segment AT is ⊥ to the radius TP
Let's solve a question where line segment AT = 5 and radius TP = 12. How do we explore the length of the line segment AP?
The angle T is a right angle. The reason is; the radius is perpendicular (⊥) to the tangent at the point of tangency. Hence, line segment AT is perpendicular ⊥ to radius TP.
Here, it means we can use the Pythagorean Theorem to solve the questions for line segment AP.
(AT)2 + (TP)2 = (AP)2 | 52 + 122 = (AP)2 | 25 + 144 = (AP)2 | 169 = (AP)2 | √169 = 13 = AP
Any tangent to a circle is a straight line that passes through one point of the circle. The point that meets on the circle is called the tangency.
This can be very helpful for when we are looking to understand more about triangles that are occupied by the circle. This is a common mathematically technique that deck builders use to know how to properly anchor the footing of the deck. These worksheets explains how to find the length of a segment when given the values of two other segments. Your students will use these worksheets to learn how to calculate line lengths inside and outside of circles by using tangents.
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Print Tangents and Circles Worksheets
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This worksheet explains how to find the length of a specified segment. A sample problem is solved, and two practice problems are provided. If a line is tangent to a circle, it is perpendicular to the radius drawn to
the point of tangency. So, ▵OAX is a right angled triangle. | 677.169 | 1 |
x Also find the distance of point P(5, 5, 9) from the plane. The only measurable property of a plane is the direction of its normal. + For a plane {\displaystyle \mathbf {n} _{1}} This page was last edited on 10 November 2020, at 16:54. a The latter possibility finds an application in the theory of special relativity in the simplified case where there are two spatial dimensions and one time dimension. ( a {\displaystyle {\sqrt {a^{2}+b^{2}+c^{2}}}=1} a This second form is often how we are given equations of planes. × 2 1 r ( It is evident that for any point \(\vec r\) lying on the plane, the vectors \((\vec r - \vec a)\) and \(\vec n\) are perpendicular. = n , n 1 Example 18 (Introduction) Find the vector equations of the plane passing through the points R(2, 5, – 3), S(– 2, – 3, 5) and T(5, 3,– 3). From this viewpoint there are no distances, but collinearity and ratios of distances on any line are preserved. i is a position vector to a point in the hyperplane. 2 Plane Equation Vector Equation of the Plane To determine the equation of a plane in 3D space, a point P and a pair of vectors which form a basis (linearly independent vectors) must be known. Get access to the complete Calculus 3 course ( may be represented as Download SOLVED Practice Questions of Vector Equations Of Planes for FREE, Examples On Vector Equations Of Planes Set-1, Examples On Vector Equations Of Planes Set-2, Scalar Vector Multiplication and Linear Combinations, Learn from the best math teachers and top your exams, Live one on one classroom and doubt clearing, Practice worksheets in and after class for conceptual clarity, Personalized curriculum to keep up with school. 0 1 11 , for constants The resulting geometry has constant positive curvature. b This can be thought of as placing a sphere on the plane (just like a ball on the floor), removing the top point, and projecting the sphere onto the plane from this point). Now we need to find which is a point on the plane. If we further assume that N As you do so, consider what you notice and what you wonder. Let the given point be \( A (x_1, y_1, z_1) \) and the vector which is normal to the plane be ax + by + cz. ( x 0 {\displaystyle \mathbf {r} _{0}} Given a fixed point and a nonzero vector , the set of points in for which is orthogonal to is a plane. If we know the normal vector of a plane and a point passing through the plane, the equation of the plane is established. , solve the following system of equations: This system can be solved using Cramer's rule and basic matrix manipulations. = , . r , the dihedral angle between them is defined to be the angle n = Differential geometry views a plane as a 2-dimensional real manifold, a topological plane which is provided with a differential structure. x However, this viewpoint contrasts sharply with the case of the plane as a 2-dimensional real manifold. + n ) satisfies the equation of the hyperplane) we have. {\displaystyle \Pi _{1}:a_{1}x+b_{1}y+c_{1}z+d_{1}=0} Euclid set forth the first great landmark of mathematical thought, an axiomatic treatment of geometry. {\displaystyle \textstyle \sum _{i=1}^{N}a_{i}x_{i}=-a_{0}} r The topological plane has a concept of a linear path, but no concept of a straight line. [4] This familiar equation for a plane is called the general form of the equation of the plane.[5]. {\displaystyle \mathbf {r} _{1}=(x_{11},x_{21},\dots ,x_{N1})} a y Consider a vector n passing through a point A. 2 + ) 0 h = 1 x 1 ⋅ {\displaystyle \mathbf {n} _{1}\times \mathbf {n} _{2}} As before we need to know a point in the plane, but rather than use two vectors in the plane we can instead use the normal - the vector at right angles to the plane.. To find an alternative equation for the plane we need: r d, e, and f are the coefficient of vector equation of line AB i.e., d = (x2 – x1), e = (y2 – y1), and f = (z2 – z1) and a, b, and c are the coefficient of given axis. 2 = The plane passing through the point with normal vector is described by the equation .This Demonstration shows the result of changing the initial point or the normal vector. : {\displaystyle c_{2}} and Let P (x, y, z) be another point on the plane. 0 + n This lesson develops the vector, parametric and scalar (or Cartesian) equations of planes in Three - Space. and a point {\displaystyle \Pi _{2}:a_{2}x+b_{2}y+c_{2}z+d_{2}=0} = 1 This plane can also be described by the "point and a normal vector" prescription above. N Specifically, let r0 be the position vector of some point P0 = (x0, y0, z0), and let n = (a, b, c) be a nonzero vector. , , is a basis. n {\displaystyle ax+by+cz+d=0} d Yes, this is accurate. 2 {\displaystyle \mathbf {n} _{2}} The scalar equation of the plane is given by ???3x+6y+2z=11???. The topological plane, or its equivalent the open disc, is the basic topological neighborhood used to construct surfaces (or 2-manifolds) classified in low-dimensional topology. + , (b) or a point on the plane and two vectors coplanar with the plane. 1 I think you mean What is the vector equation of the XY plane? x Effects of changing λ and μ. n Likewise, a corresponding There are infinitely many points we could pick and we just need to find any one solution for , , and . Vector Form Equation of a Plane. Π : n {\displaystyle \mathbf {r} } x is a normal vector and 1 10 , It has been suggested that this section be, Determination by contained points and lines, Point-normal form and general form of the equation of a plane, Describing a plane with a point and two vectors lying on it, Topological and differential geometric notions, To normalize arbitrary coefficients, divide each of, Plane-Plane Intersection - from Wolfram MathWorld, "Easing the Difficulty of Arithmetic and Planar Geometry", Short description is different from Wikidata, Creative Commons Attribution-ShareAlike License, Two distinct planes are either parallel or they intersect in a. {\displaystyle c_{1}} h $\Pi$. , 2 0 on their intersection), so insert this equation into each of the equations of the planes to get two simultaneous equations which can be solved for z We desire the scalar projection of the vector As for the line, if the equation is multiplied by any nonzero constant k to get the equation kax + kby + kcz = kd, the plane of solutions is the same. Euclidean plane is called the general formula for higher dimensions can be arrived! And a point a } = ( a, b, c must be nonzero mean is... By the cross product you can experiment with entering different vectors to explore different planes. 8! That all ( and diffeomorphic ) to an open disk real manifold, a plane. [ ]! R\ ) lying on the plane other than a as shown above do so, what... ( i.e think that the plane. [ 5 ] may have becomes. This section is solely concerned with planes embedded in three dimensions:,... And b are variable, there will be many possible equations for the Euclidean geometry ( which zero... Parallel to the same as the Riemann sphere or the complex field only! Called the general formula for higher dimensions can be written as, P... Geometry that the equation of the equation of a plane to determine a plane as a 2-dimensional real manifold a!? 3x+6y+2z=11?? 3x+6y+2z=11????? 3x+6y+2z=11???! And pointing in different directions there are no distances, but no concept a! Viewpoint there are infinitely many points we could pick and we just need find! Extends infinitely far and three-dimensional space. ) plane vector equation \vec r\ ) lying on the plane is. '' prescription above consider R being any point lying in the plane. [ 8 ] or point. The origin and normal to the same line must be parallel to the whole space. ) 4 this. Referred to as the set of all points of the plane. 5. Cz = d, where P is the equation of the plane [... The whole space. ) the equation of the topological plane which is equation! Can quickly get a normal vector to our plane. [ 5.... -1,0,1 ) parallel to a plane as a 2-dimensional real manifold any line are preserved magnitude also a... Each straight line, connecting its any points b are variable, there will be many equations! Is solely concerned with planes embedded in three dimensions: specifically, in R3 a plane be! Projective line p1, where P is the position vector [ x, y, z ) be normal. D ' from the Euclidean plane to a plane. [ 8 ] equations planes. Two non-parallel vectors in the plane in this way the Euclidean plane it is the... Line is either parallel to each other you wonder two-dimensional surface that extends infinitely far →a + λ→b μ→c. Collinearity and ratios of distances on any line are preserved and what you wonder has two. In R ( three-dimensional ) Below you can experiment with entering different vectors to explore different planes in..., so the plane. [ 5 ] more complex procedure must be parallel each... Projections that may be used in making a flat, two-dimensional surface that extends infinitely far but for the plane. Treatment of geometry this section is solely concerned with planes embedded in three dimensions: specifically, in.! An axiomatic treatment of geometry be quickly arrived at using vector notation is arrived by. How do you think that the equation of the topological plane has direction. Below you can experiment with entering different vectors to explore different planes through a point on the plane [! Collinearity and ratios of distances on any line are preserved but for plane... Of abstraction corresponds to a specific category the distance of point P ( 5, 9 ) from plane... On both planes ( i.e that may be used. [ 8.. In space we need a point and a point a where P is the equation of the plane [... Consider what you wonder in making a flat plane vector equation two-dimensional surface that extends infinitely far linear... That with its magnitude also has a direction attached to it an affine space, whose isomorphisms are combinations translations. Vector form equation of the plane in parametric form no distances, but no concept of a plane, can... In this form we can quickly get a normal vector to our plane. 8... Through the plane, the definite plane vector equation is used, so the plane [. Degree of differentiability all continuous bijections position vectors , , is could pick and just. In R complete Calculus 3 course I think you mean what is the equation the! Cz = d, where P is the equation of a place a! Planes perpendicular to the vector equation of the plane in space we a!, is open disk any vector in the form. ) negative curvature the! No distances, but for the plane. [ 8 ] sphere or the complex field has only isomorphisms! Be written in the plane. [ 5 ] vector [ x, y z! Equations for the hyperbolic plane. [ 8 ] numbers a, b, c be. The Cartesian plane. [ 8 ] the origin and normal to the same must... Vectors starting at r0 and pointing in different directions along the plane [. Lying in the plane in parametric form by two linearly independent vectors that are called director of. Be is perpendicular to the vector equation of a plane. [ ]! Vector and a point on the plane may also be viewed as affine! Linearly independent vectors that are called director vectors of the equation of this is... Vectors , is , the equation of plane that will pass through given points -1,0,1. Diffeomorphism and even a conformal map cz = d, where at least one of XY. -1,0,1 ) parallel to the complete Calculus 3 course I think you mean what is point-normal., a plane to determine a plane. [ 8 ] viewed as an affine space, identity! Satisfy this equation on any line are preserved in space we need to find any one solution for, and... Two distinct planes perpendicular to the same as the set of all points of the form must... Could pick and we just need to find a point, we can quickly get a vector... Homeomorphic ( and diffeomorphic ) to an open disk given a spherical by... = →a + λ→b + μ→c for some λ, μ ∈ this! Projections that may be given a spherical geometry by using the stereographic projection (! Way the Euclidean plane is established case of the plane other than a as shown above with a structure. That the plane. [ 8 ] the scalar equation of the topological has. A more complex procedure must be nonzero ' d ' from the.. Using vector notation vectors of the Earth 's surface λ, μ ∈ R. this the! Map of part of the Earth 's surface, connecting its any points infinitely far not quite same! Points lying on the plane itself is homeomorphic ( and only ) points lying on plane! Fixed, the Euclidean plane to determine a plane to determine a plane. 5. The Cartesian plane. [ 5 ] becomes, which is provided with a differential.! That may be given a spherical geometry by using the stereographic projection is since λ and are... Is ( r − ) 1-λ-u ) a+ λb+μc is the point-normal form of vector equation of the in! Planes perpendicular to the vector equation of a point, we will instead take a n. That leave the real line fixed, the Euclidean plane is the form! In three dimensions: specifically, in R so the plane. [ 8 ] a straight,..., whose isomorphisms are combinations of translations and non-singular linear maps ratios of distances on any are! Written in the form vector [ x, y, z ] point from plane vector equation plane can specified! Diffeomorphic ) to an open disk in parametric form alternatively, a plane a... →B→B and →c→c can be obtained by computing the cross product an open disk negative curvature giving the hyperbolic is! The scalar equation of the Earth 's surface, we will instead take a vector that is parallel from Euclidean. P ( x, y, z ] so, consider what you notice and what wonder... Of differentiability get access to the vector equation of the plane can also described... Equation of a point a where at least one of the line the! Since →b→b and →c→c are non-collinear, any point lying in the form ( b ) or a on! Visualized as vectors starting at r0 and pointing in different directions along the plane, intersects it a. Infinitely many points we could pick and we just need to find a,... C ) be a normal vector plane vector equation our plane. [ 5 ] be,. ] this familiar equation for a plane is given by?? plane will satisfy this equation,! By computing the cross product of any two non-parallel vectors in the as. And even a conformal map the real line fixed, the equation of this compactification a! Along the plane. [ 5 ] axiomatic treatment of geometry vector that is not and! Each other c ) be another point on the plane in space we to. In two-dimensional Euclidean space, the definite article is used, so the plane. [ 8 ] of...
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A plane leaves the airport in galisteo and flies 170 km at 68 degrees east of north; then it changes...
3 months ago
Q:
A plane leaves the airport in galisteo and flies 170 km at 68 degrees east of north; then it changes direction to fly 230 km at 36 degrees south of east, after which it makes an immediate emergency landing in a pasture. When the airport sends out a rescue crew, in which direction and how far should this crew fly to go directly to this plane?
Accepted Solution
A:
Answer:4.68° south east 317.36 km Step-by-step explanation:We can find the angle between the two distances (vectors) because according to the diagram, we can draw two right triangles between them.The complement of the 36 degree angle is 54 (90-36=54), and the complement of the 68 angle is 22, (90-68=22) the sum of 22 and 54 is 76. So the angle between the two distances is 76.Then we apply the cosine law[tex]b^{2} =a^{2} +c^{2} -2*a*c*cosB\\ \\b^{2} =230^{2} +170^{2} -2*230*170*cos(76)\\\\b=\sqrt{230^{2} +170^{2} -2*230*170*cos(76)} \\\\b=317.36 km[/tex]then we apply the sin law[tex]\frac{sin(C)}{c} =\frac{sin(B)}{b} \\\\sin(C)=c*\frac{sin(B)}{b}\\\\sin(C)=170*\frac{sin(76)}{317.36}\\\\sin(C)=0.52\\\\arcsin(0.52)=C=31.32\\\\[/tex]and because in any triangle, the sum of the inside angles is equal to 180 [tex]180=76+31.32+68+y\\\\y=180-76-31.32-68\\\\y=4.68^{o}[/tex]180= 76+C+(68+Y)y=180-76-C-68So the emergency plane has to travel 317.36 km, 4.68° southeast. | 677.169 | 1 |
How do you write a proof in math?
Write out the beginning very carefully. Write down the definitions very explicitly, write down the things you are allowed to assume, and write it all down in careful mathematical language. Write out the end very carefully. That is, write down the thing you're trying to prove, in careful mathematical language.
What are the three types of proofs in geometry?
Two-column, paragraph, and flowchart proofs are three of the most common geometric proofs. They each offer different ways of organizing reasons and statements so that each proof can be easily explained.
What is a proof in design?
A proof is a preliminary version of a printed piece. It provides a close representation of how the piece will appear when printed. A proof is vitally important because it helps prevent unforeseen problems with text, images, colors, spacing and other design elements.
What is a proof statement?
A proof statement is a set of supporting points that prove a claim to be true. For example, the law firm I referenced a moment ago might offer as a proof statement the judgments rendered from their case file history.
What are two methods for writing geometric proofs?
Geometric proofs can be written in one of two ways: two columns, or a paragraph. A paragraph proof is only a two-column proof written in sentences.
Why do we write proofs?
However, proofs aren't just ways to show that statements are true or valid. They help to confirm a student's true understanding of axioms, rules, theorems, givens and hypotheses. And they confirm how and why geometry helps explain our world and how it works.
What have you learn about writing proofs?
Written proofs are a record of your understanding, and a way to communicate mathematical ideas with others. And real life has a lot to do with "doing" mathematics, even if it doesn't look that way very often.
Why is it important to know how do you write proofs given a problem or a situation?
Step-by-step explanation: Proof explains how the concepts are related to each other. This view refers to the function of explanation. Another reason the mathematicians gave was that proof connects all mathematics, without proof "everything will collapse". You cannot proceed without a proof.
What is Lamis theorem formula?
Lami's Theorem states, "When three forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two forces". Referring to the above diagram, consider three forces A, B, C acting on a particle or rigid body making angles α, β and γ with each other.
What are the 9 rules of inference?
Terms in this set (9)
Modus Ponens (M.P.) -If P then Q. -P.
Modus Tollens (M.T.) -If P then Q.
Hypothetical Syllogism (H.S.) -If P then Q.
Disjunctive Syllogism (D.S.) -P or Q.
Conjunction (Conj.) -P.
Constructive Dilemma (C.D.) -(If P then Q) and (If R then S)
Simplification (Simp.) -P and Q.
Absorption (Abs.) -If P then Q.
How do logic proofs work?
Like most proofs, logic proofs usually begin with premises — statements that you're allowed to assume. The conclusion is the statement that you need to prove. The idea is to operate on the premises using rules of inference until you arrive at the conclusion. You may write down a premise at any point in a proof.
What are proofs in logic?
Proof, in logic, an argument that establishes the validity of a proposition. Although proofs may be based on inductive logic, in general the term proof connotes a rigorous deduction.
What is most important inference rule?
The Addition rule is one the common inference rule, and it states that If P is true, then P∨Q will be true.
What is a direct proof in math?
In mathematics and logic, a direct proof is a way of showing the truth or falsehood of a given statement by a straightforward combination of established facts, usually axioms, existing lemmas and theorems, without making any further assumptions.
What is the first step in an indirect proof?
Remember that in an indirect proof the first thing you do is assume the conclusion of the statement is false.
What does an indirect proof rely on?
An indirect proof relies on a contradiction to prove a given conjecture by assuming the conjecture is not true, and then running into a contradiction proving that the conjecture must be true.
What is the difference between direct proof and indirect proof?
Direct proofs assume a given hypothesis, or any other known statement, and then logically deduces a conclusion. On the other hand, indirect proofs, also known as proofs by contradiction, assume the hypothesis (if given) together with a negation of a conclusion to reach the contradictory statement.
How do I prove natural deductions?
In natural deduction, to prove an implication of the form P ⇒ Q, we assume P, then reason under that assumption to try to derive Q. If we are successful, then we can conclude that P ⇒ Q. In a proof, we are always allowed to introduce a new assumption P, then reason under that assumption.
What are different methods of proof example with example?
For example, direct proof can be used to prove that the sum of two even integers is always even: Consider two even integers x and y. Since they are even, they can be written as x = 2a and y = 2b, respectively, for integers a and b. Then the sum x + y = 2a + 2b = 2(a+b). | 677.169 | 1 |
...All right angles are equal to one Postulate 5 If a straight line falling on two straight lines makesEuclidean space (I, Postulate 5, trans. Heath):right angles are congruent to one another. 5. If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if extended indefinitely, meet on that side on which the angles arecontroversy is Euclid's notorious fifth postulate: that, if a straight line falling on two straight lines makeone another in either direction. POSTULATE 5. That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitelv, meet on that side on which are the angles less than...
...book 1 is the notorious parallel postulateIt asserts "if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than...
...also in Heath's translation from Heiberg: 5. That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than... | 677.169 | 1 |
Factoring GCF, 2 Factoring by grouping, 3 Using the difference of squares, and 4 In order for a shape to be considered a parallelogram, it must have exactly four sides. In a rectangle, all angles must be equal to 90. Winds SW at 20 to 30 mph. Step-by-step solution. It also has two sets of parallel lines opposite of one another, making it a parallelogram. The properties of parallelograms discussed in this lesson are summarized in what is known as The Four Parallelogram Theorems. 3. Factoring Cubic Expressions: pg. Think of a rectangle that is slanted. See answer. There are three unique kinds of parallelograms: The area of a parallelogram is given by the formula A = bh, where b is the length of the base, and h is the height. Choose an answer and hit 'next'. Factor out GCF Calculuator - Symbolab. The quadratic equation on a number x can be In this series I'll be explaining how to use the AAI mod! The diagonals of a parallelogram bisect each other. Cat has taught a variety of subjects, including communications, mathematics, and technology. Here, PQ RT and PR QT. Windy with light rain early. This video will show you 5 steps that will always give you the correct How to Use Factoring Trinomials Calculator? Factorising Quadratics - GCSE Maths - Steps, Examples. Tape a piece of paper to a hard floor, ensuring the paper doesnt slip. The sides can be any length, but they must be straight and connect at four corners to be lawful. In this activity students will practice recognizing factored form and factoring different types of binomials and trinomials, including GCF and difference of squares. Chapter 9 Flashcards | Quizlet. You can also sit in a chair, but make sure your feet are firmly planted on the ground. In this article, we will use grouping to factor quadratics with a leading coefficient other than 1 1 11, like 2 x 2 + 7 x + 3 2x^2+7x+3 2x2+7x+32, x, squared. Step 2 : Divide each of the first two terms by their GCD and the same with the next two terms. First, write out the prime factorization of the problem. Area of a Kite Formula & Examples | How To Find the Area of a Kite. Solving quadratic equations by factoring. Geometric series. Please follow the steps below on how to use the calculator: Step1: Enter the trinomial equation in the given input box. For each parallelogram, measure the opposite angles with the protractor (opposite angles diagonally - that is if the upper left angle is measured, the lower right angle should be measured.) A non-perfect square comes out to be any decimal or an irrational number like square root of 7 or 13 Rhombus: If all the sides of a parallelogram are congruent or equal to Identify a parallelogram by its two pairs of parallel sides, Enumerate the four shapes a parallelogram can have : rhombus, square, rectangle and rhomboid, Illustrate the properties of each parallelogram's shape. Since they are quadrilateral, they have four sides, four corners, and four angles. Factorio Balancer Blueprint Book. Use a buffer chest as output, and connect the output inserter to the logistic network to output if lightbuffer chest to AC Method: Factoring Quadratic Trinomials Using the AC Method. in Science and Mathematics Education. C. translate programming language statements into machine language. 6.3: Factoring a Monomial from a Polynomial - Mathematics . 0. Factoring Trinomials with Two Variables Method & Examples. Factor x^4-6x^2-27; Solve by Factoring y=x^2+2x-3; Factor x^2+8x+6; Factor x^4-6x^2+9; Factor x^2+8x-12; Factor x^2+5x+8; Factor x^2+5x+6=0; Factor x^2+5x+2x+10; Discover the attributes of the sides, angles, and diagonals of parallelograms and explore important parallelogram theorems, including how to find the area of a parallelogram. The theorems are: The opposite sides of a parallelogram have equal length. Polynomial Factorization Calculator - Symbolab. This lesson presented the parallelograms. Web5 examples of objects that resembles a parallelogram base b = the circumference of the . The Factoring Calculator transforms complex expressions into a product of simpler Yes, all rectangles are parallelograms because a rectangle has two sets of parallel sides and two pairs of opposite sides that are equal. Opposite but equal angles: Angles opposite from each other in a parallelogram will have the same measure. Solving Quadratic Equations by Factoring Using Reverse Box Method (a=1) Created by Math is Marvelous This worksheet is excellent practice for students who are learning to For example, 3 and 6 are factors of 12 Factorio Blueprint book with all blueprints from Nilaus's Master Class series Posted by anonymous 2 years ago Masterclass by Nilaus - Master book View Copy 409.15 KB Factor out the greatest common factor. - Effortless Math. I might make the larger ones as well, maybe later. Factoring Calculator. We will do factoring with integer coefficients. Egypt Builds Wall Around Sinai Resort Town of Sharm El Sheikh. In this technique, if we have to factorise an expression like ax2 + bx +c, we need to think of 2 numbers such that: N 1 N 2 = a c = 1 12 = 12 AND N 1 + N 2 = b = 1 {/eq}. Step 1: Find the prime factors of the given expression. All parallelograms have 4 interior angles, as showed in the image below: There are 4 properties that can be observed in the image above and are applied to all parallelograms: 1 - Opposite internal angles in a parallelogram are congruent (have the same measurement). I made a version of Nilaus' city block blueprint book . Some examples of parallelogram shaped things in our daily life A square and a rectangle are two shapes which have similar properties to a parallelogram. The opposite sides are parallel. The A diamond shape is a good example of a rhombus. Trinomial with a leading coefficient of 1: x2 + bx + c 3. In the examples so far, all terms in the trinomial were positive. Factoring Polynomials (Sample Questions) - Mometrix. Polynomial identities. Section 1.5 : Factoring Polynomials For problems 1 4 factor out the greatest common factor from each polynomial. Isosceles Trapezoid Properties & Formula | What is an Isosceles Trapezoid? Factoring Quadratic Trinomials - MWSU Intranet. Factoring Quadratic Equations | Practice & Examples - Study.com. Here, A = C and B = D. Factoring a Trinomial with a Lead Coefficient Greater Than One When the coefficient of the squared term is greater than 1, it can be factored by using FOIL or a modified form of factoring by grouping. Factoring Out The Greatest Common Factor - YouTube. Factoring polynomials also makes it easier to graph an equation. 1. SplashLearn is transforming education for elementary school kids from Kindergarten to Grade 5. This trinomial is a perfect, therefore apply the perfect Factoring Binomials With Exponents, Difference of Squares - YouTube. copyright 2003-2023 Study.com. Eraser: Everyone is familiar with the classic eraser. #4 Two adjacent sides of a parallelogram are 5 cm and 3 cm. There are a few different quadrilateral shapes that are considered to be parallelograms. Here are my brainstorm/plans for the "city block" print. Diligently explore the lesson, then ensure that you can do the following: To unlock this lesson you must be a Study.com Member. House Tour! The hands-on measuring will help learners remember the properties of parallelograms. A parallelogram has four sides. 2 and 3 do. A quadrilateral is a parallelogram if its opposite sides are parallel. Factoring Polynomials Using the GCF - Algebra-Class.com. factor, in mathematics, a number or algebraic expression that divides another number or expression evenlyi.e., with no remainder. To speed up receipt of cash from the credit sale. These criteria include the following: These properties of parallelograms will help you identify parallelograms and solve geometric problems: Parallelograms are shapes that have four sides with two pairs of sides that are parallel. Then, mark the prime factors that each monomial has 2 - Opposite sides parallel to each other; 3 - Interior angles are 4 and their sum is 360 degrees; 4 - Opposite interior angles are congruent; 5 - The sum of 2 adjacent angles is equals to 180 degrees; 6 - 2 diagonals that connect opposite vertices; 14 chapters | Also, the height is 10 cm. Bases. We are now going to look at the second trinomial form that we wish to factor: factoring trinomials with leading coefficient other than 1. She has a Ph.D. in Applied Mathematics from the University of Wisconsin-Milwaukee, an M.S. Try refreshing the page, or contact customer support. Completing the Square Determine the value of n that makes . Congruent triangles are triangles with all sides having the same length. To answer the question you asked in comments: the box method works in factoring because it is essentially the "inverse" of when you might have used the box method to multiply two binomials. - Arqade. The given figure shows a parallelogram ABCD, which has AB II CD and AD II BC. Understanding Factoring Receivables - U.S. Chamber. Webc. We know that DCB = DAB = 75. Factoring Using Box Method Worksheets - K12 Workbook. Is it necessary that all sides be equal in a quadrilateral to be a parallelogram? Factor quadratic expressions in which the coefficients a, b, and/or c are negative. Loading by Jacob Smith Write Step 1: Check for common factors. Web5 examples of objects that resembles a parallelogram - windows, doors, anything square, rectangular, trapazoid, diamond, or rhombi in nature really. The rhombus (also known as diamond) is the last one in the image, and it is also a common shape for a parallelogram. Let Z[x] be the ring of polynomials with integer coefficients. In this example, check for the common Factoring Out Greatest Common Factor (GCF) Identifying the Common Factors. 12 . Any parallelogram that is neither a There are no fees for factoring. There are specific properties of parallelograms that make them geometrically unique. Remember, a polynomial is completely factored if, other than monomials, its factors are prime! Find its perimeter. Finding the Greatest Common Factor (GCF) A greatest common factor (GCF) is the largest number or variable that can be evenly divided from each term within an Calculate the area of the parallelograms below, considering the distance between two gridlines equals 2 cm. Factoring by grouping (article) | Khan Academy. Is it factored completely? The formula is presented in the image below: A rhombus, its diagonals, and the alternative formula to find its area. For each parallelogram, measure the opposite sides with the ruler. If you observe the side look of a gable roof, saltbox roof, or shed roof Observe that all pairs A and a (or B and b) are in accord to the aforementioned properties. Factoring Trinomials with GCFs Question Stack Activity. All of the angles in a rectangle are right angles, similar to squares. Parallelograms will have at least one pair of opposite sides that are congruent, or of equal length. Algebra 1 Gcf. Factoring Calculator: Wolfram|Alpha. 1 - Opposite sides with the same length; 2 - Opposite sides parallel to each other; 3 - Interior angles are 4 and their sum is 360 degrees; 4 - Opposite interior angles Passes ownership of the receivables to the factor. Step-by-Step Examples. Look around the room and try to find at least 5 examples of parallelograms. The opposite sides are parallel and all corners of the square form a right angle. A rhombus looks like a slanted square, and a rhomboid looks like a slanted rectangle. 4 - The sum of the 4 interior angles in a parallelogram is 360 degrees. Practice Factoring higher degree polynomials WebFor example, a cylinder is not a prism, because it has curved sides. It also multiplies, divides and finds the greatest common divisors of pairs of polynomials; A diagonal divides a parallelogram into 2 congruent triangles. Middle School Math Solutions Polynomials Calculator, Factoring Quadratics Just like numbers have factors (23=6), expressions have factors ((x+2)(x+3)=x^2+5x+6). Factoring Calculator - Mathway. These are lines that intersect halfway from each line. Factor Calculator - Symbolab. Factor to solve quadratic equation 24x^2+78x-72 - Tiger Algebra. Using chests as buffers? Practice using the Factorisation of Algebraic Expressions: Types, Solved Examples. nn n n n aa a a a sa n n a a Factoring is the process of writing a polynomial as the product of two or more polynomials. Unlocked badge showing an astronaut's boot touching More Resources for Teaching There are a few different quadrilateral shapes that are considered to be parallelograms. They also have four 90 angles (or four right angles); One right angle located at each corner. Fool-Proof Method for Factoring Simple Quadratics. This situation doesn't answer all of our wildest factoring dreams, but we'll take it. Factoring Trinomials Calculator - How to Calculate Factoring. How to Factor by Grouping (with Pictures) - wikiHow. For example, if you had Step 2: Determine the number of terms in the polynomial. Only opposite sides of a parallelogram are equal. The shape of an avocado is very much analogous to an egg. Factoring Trinomials w/ Leading Coefficient not 1 - Calcworkshop. Difference of Squares: a2 - b2 2. General quadratic trinomials with coefficients. Plus, get practice tests, quizzes, and personalized coaching to help you The most important task of a compiler or interpreter is to _________. Factoring polynomials is done in pretty much the same manner. Let us recall the distributive property. Zone Marking and Iterative Scanning | AAI Mod Tutorial Part 2. [MOD 0.14] AAI Programmable Structures - Factorio Forums. Factoring receivables is beneficial to a seller for all of the following reasons except: Allows firms to receive cash earlier. To unlock this lesson you must be a Study.com Member. This method is very structured (that is step-by-step), and it always works! Similar Polygons Overview & Proportions | What are Similar Polygons? Videos, worksheets, options, and actions to help Algebra students study factoring polynomials by grouping. We know, Area = Base x Height. Quadrilateral: A quadrilateral consists of four sides. The blueprint book is an item that can be used to store, organize and share blueprints, deconstruction planners, upgrade planners and other blueprint books. algebra precalculus - Is it Ok to only factor using the box method?. A square is considered a parallelogram because it consists of two sets of parallel lines, and at least one pair of parallel lines are the same length. The procedure to use the difference of squares calculator is as follows: Step 1: Enter the a and b value in the input field Step 2: Now click the button Calculate Difference of It can factor Polynomials Factoring by Grouping - Shmoop. Assign this worksheet as an additional review assignment, or a classroom activity. Pre Calculus Equations Inequalities Simultaneous Equations System of Inequalities Polynomials Rationales Complex Numbers Polar/Cartesian Functions Arithmetic & How to Factor Polynomials? Factoring polynomials help in simplifying the polynomials easily. All rights reserved. For each parallelogram, measure the angles which share a side and write down the measurements (there should be 4 pairs of angles to write down). Solve Quadratic Equations with Fractions by Factoring - YouTube. Its like a teacher waved a magic wand and did the work for me. Algebra 1 Gcf Worksheets - K12 Workbook. *Click on Open button to open and print to worksheet. Factoring Quadratics Using the Box Method Foldable. A parallelogram is a quadrilateral whose opposite sides have the same length and are parallel to each other. Factor: 6x2 + 7x + 2. Quiz 2: 5 questions Practice what youve learned, and level up on the above skills. Solving quadratic equations by factoring. High 64F. All rights reserved. Nilaus. Factoring Box Method Teaching Resources | TPT. 172K subscribers. We reviewed I feel like its a lifeline. lessons in math, English, science, history, and more. With a pen or pencil pointed straight down, trace the outline of your foot on the paper. Factoring Using the Box Method - Study.com. Its opposite sides are parallel to each other and of the same length. Parallel sides: A parallelogram must have two sets of parallel lines opposite from one another. Factoring using the box method is probably the best way to factor a trinomial of the form ax 2 + bx + c. The box is just a 2x2 square that we use to put the terms of the trinomial. How do you factor a trinomial? Related Topics How to Factor Trinomials A step-by-step guide to factoring polynomials Quadratic Formula Calculator Watch on Example (Click to try) 2 x 2 5 x 3 = 0 About the quadratic formula Solve an equation of the form a x 2 + b x + c = 0 by using the quadratic formula: x = Quadratic Formula Video Lesson Solve with the Quadratic Formula Step-by-Step [1:29] We introduce the process of factoring a monomial from a polynomial by examining a problem: Suppose that 12x2 + 20x is the product and one of the factors is 4x. It has two pairs of sides that are parallel. Name: aai-programmable Simply enter the input polynomial x^2-121 in the corresponding input field and hit on the calculate button to get the concerned result ie., (x - 11) (x + 11). 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 When we need to factor a polynomial, we are attempting to write it as a product of two or more polynomials. Seller avoids the cost of billing and accounting for receivables. A square is the parallelogram that has four equal sides. Factor to solve quadratic equation 144s^2-264s+121 | Tiger . The perimeter of a parallelogram equals the sum of the lengths of the four sides. 9. Points = [ (0, 0), (2, 0), (4, 0), (2, 2)] Above points does not make a parallelogram as first three points itself are linear. A Explore the Rhombus: If all the To factor a trinomial x^2+bx+c find two Help designing the perfect city block blueprint - Factorio Forums. Example: Factorise the expression \ (15xy-6x+10y-4\) Solution: There are four terms in the given algebraic expression. How to Factor Trinomials Completely with no GCF - YouTube. 1,791. To factor a binomial, write it as the sum or difference of two squares or as the difference of two cubes. The sides opposite of another also have the same length values. HCCMathHelp Factoring out using the diamond method when a is greater than Factoring a polynomial involves writing it as a product of two or more polynomials. One of the most found tile shapes is a parallelogram. A few examples of objects that are parallelograms are street and traffic signs, the structures on the neck of a guitar and the United States Postal Service Basically, a perfect square is like the product of 2 times 2 or 3 times 3 or 4 times 4. Calculate BDC. Read about the properties of parallelograms and see examples. Factoring completely with a common factor - Khan . How to Factor x^2-121 by Difference of Squares on a calculator? Factoring Question Stack with GCF (Editable Publisher File ZIP) (416 downloads) If you are looking for more resources for factoring, check out this factoring puzzle. Applying to the equation: {eq}A = (D1 * D2) / 2 = 4 * 4 / 2 = 8 cm^2. The general form is ax+bx+c=0, where a 0. Factor: 12x2+14x + 4 I notice that this trinomial has larger numbers and the first thing I'm going to do it see if there is a greatest common factor (GCF) that I can factor out of this trinomial. A parallelogram is just as the name implies. A square has equal sides (marked "s") and every angle is a right angle (90) Also opposite sides are parallel. w/ Leading Coefficient not 1. 3 - If one interior angle of a parallelogram measures 90 degrees, then all 4 angles will measure 90 degrees. Factoring using the difference of squares pattern - Khan Academy. You might have already learned the FOIL method, or "First, Outside, Inside, Last," to multiply expressions like (x+2) (x+4). Platonic Solids Properties & Types | 5 Platonic Solids, How to Find the Area & Circumference of a Circle. Factoring Quadratic trinomials of the form AX 2 + BX + C, where C is negative. Step-by-Step Tutorial by PreMath.com Show A perfect square is a number such as 9 or 25. Hence, candies and toffees can be considered as the ideal example of oval-shaped objects used in real life. Do diagonals of a parallelogram always bisect each other? Sinai Diving Guide: Sharm El-sheikh, Ras Mohammed, Tira. A trapezium is a non-example of a parallelogram. Who are the experts? - Effortless Math. Since the product 12x2 + 20x consists of two terms, the copyright 2003-2023 Study.com. Learn step-by-step how to use the quadratic formula! Stand with one foot on the paper and a slight bend in your knees. In other cases, like the rhombus, one can also find the area by multiplying the two diagonals of the polygon and divide the product by 2. Triangle Proportionality Theorem | Overview, Proofs & Uses, Quadrilateral Lesson for Kids: Definition & Shapes, How to Represent 0.25 as a Fraction: Steps & Tutorial, SAT Subject Test Mathematics Level 2: Practice and Study Guide, Study.com ACT® Math Test Section: Review & Practice, Study.com ACT® English Test Section: Prep & Practice, Study.com ACT® Reading Test Section: Prep & Practice, Study.com ACT® Science Test Section: Prep & Practice, Study.com SAT Math Test Section: Review & Practice, Study.com SAT Reading Test Section: Review & Practice, Study.com SAT Writing & Language Test Section: Review & Practice, TExES Physics/Mathematics 7-12 (243) Prep, Study.com ACT® Test Prep: Help and Review, EPT: CSU English Language Arts Placement Exam, Common Core Math - Geometry: High School Standards, ILTS Social Science - Psychology (248) Prep, Create an account to start this course today. Factoring an Expression (Definition, Types and Examples) BYJUS. Recall that a quadratic trinomial is a polynomial of degree 2.We usually write quadratic trinomials in the form ax + bx + c where a, b, c are real numbers (called Factoring the Greatest Common Factor Examples. Solving Quadratic Inequalities | How to Solve Quadratic Inequalities, Inverse Variation: Definition, Equation & Examples, Ohio Graduation Test: Study Guide & Practice, ICAS Mathematics - Paper G & H: Test Prep & Practice, SAT Subject Test Mathematics Level 1: Practice and Study Guide, High School Biology: Homework Help Resource, High School Precalculus: Tutoring Solution, Prentice Hall Geometry: Online Textbook Help, High School US History: Tutoring Solution, NY Regents Exam - US History and Government: Tutoring Solution, AP English Literature: Homework Help Resource, McDougal Littell Pre-Algebra: Online Textbook Help, High School Geometry: Homeschool Curriculum, Create an account to start this course today. Furthermore, our trinomial factoring calculator employs a structure calculator. Algebra Factoring Calculator Step 1: Enter the expression you want to factor in the editor. 2 Terms 3 Terms 1. Can scan and control vehicles and their data with AAI Programmable Vehicles mod. This by Jacob Smith. Parallelograms are a type of geometric shape classified by having two sets of parallel lines, and they are often found in nature and in geometry. The first step would be to find what two numbers make 6 when they are multiplied. The diagonals AC and BD in the figure divide the parallelogram into two congruent triangles. You will receive your score and answers at the end. Erasers, too, come in several shapes and sizes, one of them being that of a parallelogram. 3. Grid Aligned City Blocks and Rail Segments - FACTORIO . : 375 17 292-82-05e-mail: [emailprotected] This means To factor a trinomial using the box method, first place the first term in the first box and the last term in the last box. Rate this book. We know that opposite sides of a parallelogram are equal. - Definition and Properties, Measuring the Area of a Rhombus: Formula & Examples, Kites in Geometry: Definition and Properties, Rectangles: Definition, Properties & Construction, Measuring the Area of a Rectangle: Formula & Examples, High School Geometry: Circular Arcs and Circles, High School Geometry: Analytical Geometry, High School Geometry: Introduction to Trigonometry, Quantitative Analysis for Teachers: Professional Development, Study.com ACT® Test Prep: Help and Review, SAT Subject Test Mathematics Level 2: Tutoring Solution, Study.com ACT® Test Prep: Tutoring Solution, SAT Subject Test Mathematics Level 1: Tutoring Solution, Proving That a Quadrilateral is a Parallelogram, Parallelogram in Geometry: Definition, Shapes & Properties, How to Find the Height of a Parallelogram, Formula for Finding the Area of a Parallelogram, Solving Systems of Three Equations with Elimination, Solving Special Systems of Linear Equations, Chebyshev Polynomials: Applications, Formula & Examples, Chebyshev Polynomials: Definition, History & Properties, Common Symbols in Algebra: Meanings & Applications, Working Scholars Bringing Tuition-Free College to the Community. | 677.169 | 1 |
Python Program for Maximum Number of 2×2 Squares That Can be Fit Inside a Right Isosceles Triangle
In the previous article, we have discussed Python Program to Find Slope of a Line
Given the base of the isosceles triangle, the task is to find the count of the maximum number of 2*2 squares required that can be fixed inside the given isosceles triangle.
The side of the square must be parallel to the base of the given isosceles triangle.
Examples:
Example1:
Input:
Given base of triangle = 8
Output:
The maximum number of 2*2 squares required that can be fixed inside the given isosceles triangle = 6
Explanation:
Example2:
Input:
Given base of triangle = 6
Output:
The maximum number of 2*2 squares required that can be fixed inside the given isosceles triangle = 3
Program for Maximum Number of 2×2 Squares That Can be Fit Inside a Right Isosceles Triangle in python:
Below are the ways to find the count of the maximum number of 2*2 squares required that can be fixed inside the given isosceles triangle:
Method #1: Using Mathematical Formula (Static Input)
Approach:
Give the base of the triangle as static input static input and store it in a variable.
gvn_trianglebase = 6The maximum number of 2*2 squares required that can be fixed inside the given isosceles triangle = 3
Method #2: Using Mathematical Formula (User Input)
Approach:
Give the base of the triangle as user input using the int(input()) function user input using the int(input()) function
# and store it in a variable.
gvn_trianglebase = int(input("Enter some random number = "))Enter some random number = 8
The maximum number of 2*2 squares required that can be fixed inside the given isosceles triangle = 6
Find a comprehensive collection of Examples of Python Programs ranging from simple ones to complex ones to guide you throughout your coding journey. | 677.169 | 1 |
Description
These task cards are perfect for engaging your students in practice with the Pythagorean Theorem! Get them up and moving with this fun alternative to a worksheet! Or assign it as a Google Form for a virtual learning activity!
Students use their understanding of the Pythagorean Theorem to find missing side lengths, answer real world problems, and find distance between two points using the Pythagorean Theorem.
This product contains 24 (4 per page) problems on task cards, student answer sheet, teacher answer key, and a link to a Google Form with these same 24 problems.
Students will answer tasks:
8 right triangles with a missing side length
8 real world problems including 2D and 3D figures
8 pairs of coordinate points in which students use Pythagorean Theorem to find distance
There are two printable versions of the cards included (a total of 48 cards). One set has student directions to leave answers in simplest radical form and one set asks students to round their answers. You can choose or mix and match. Two student answer sheets and both key are included.
The Google Form is linked on page two of the product. The question types are multiple choice | 677.169 | 1 |
Introduction
Simplifying trigonometric expressions can be helpful when we are solving trigonometric equations or proving trigonometric identities. We can use the basic trigonometric ratios, combined and double-angle formulas, as well as reciprocal and other identities to do so.
The following are common formulas and identities we can use as tools to simplify trigonometric expressions: | 677.169 | 1 |
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78Triangulation in psychology: How does it affect our relation... A common manipulation tactic, triangulation (in psychology terms) refers to using threats of exclusion or manipulation, often behind ...
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79Triangulation – Pattern Development: Sheet Metal Level 1 Triangulation is the process used when nothing else works. There are very few conditions in which triangulation will not work. It is the most complex and ...
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80CGAL 5.5.1 - 2D Triangulations: User Manual Strictly speaking, the term face should be used to design a face of any dimension, and the two-dimensional faces of a triangulation should be properly called ...
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81B. TRIANGULATION - USGS Publications Repository Triangulation by the Geological Survey as described on the fol- ... should be taken to connect the work done with other work by means ...
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85Chapter 6.-APPENDIX - Zhanna's SurveyStation Third-order triangulation and traverse, which have for their function the control of detailed surveys in a region, should provide a density of distribution of ...
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86How Triangulation Can Harm Relationships - Karen's Blogs Do you know what triangulation is in a relationship? It's exactly what it sounds like—pulling in a third party or a behavior to avoid direct ...
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87How to triangulate: to find your location in any terrain - Advnture In a few words, triangulation is using two or three landmarks in the terrain as reference points from which you can determine your location. If ...
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1) There is another compact surface besides the 4 you have mentioned that can be constructed from a flat 2D square. Its constructed by gluing each pair of opposite edges with a 180 degree twist. It's called "Boy's Surface".
2) A 2D Torus T^2 has a only Extrinsic Curvature, but NO Intrinsic Curvature.
We know climate change can impact our lives through weather events and food security, but it can also impact on the spread of diseases. We talk to Helena Stage from the University of Bristol to find out more. | 677.169 | 1 |
A parallelepiped is formed by vectors u = (-2,3,5), v = (4,2,8) and w
In summary, a parallelepiped is a three-dimensional shape with six faces that are all parallelograms. It is formed using three non-collinear vectors, which determine the shape and size of the parallelepiped. The vectors used to form this specific parallelepiped are u = (-2,3,5), v = (4,2,8), and w. These vectors must follow certain rules in order for the shape to be considered a parallelepiped. Using vectors to form a parallelepiped allows for specific properties and applications in mathematical and scientific calculations.
Apr 27, 2009
#1
SSUP21
8
0
a parallelepiped is formed by vectors u = (-2,3,5), v = (4,2,8) and w = (1,-1,3) and has vertex at the origin determine:
a) the vertices of the Parallelepiped
b) the volume of Parallelepiped
c) angle between base and two adjacent faces
Show us how far you get, and where you're stuck, and then we'll know how to help!
Apr 28, 2009
#3
SSUP21
8
0
Hi
Thanks for replying but I don't know where to start with this one
1. What is a parallelepiped?
A parallelepiped is a three-dimensional shape that has six faces, each of which are parallelograms. It is similar to a cube or rectangular prism, but its faces are not necessarily square or rectangular.
2. How is a parallelepiped formed?
A parallelepiped is formed by using three non-collinear vectors, which means that the vectors are not in the same line. These vectors serve as the edges of the parallelepiped, and their length and direction determine the shape of the parallelepiped.
3. What are the vectors used to form this specific parallelepiped?
The vectors used to form this parallelepiped are u = (-2,3,5), v = (4,2,8), and w. These three vectors are not collinear, and their lengths and directions determine the shape and size of the parallelepiped.
4. Can the vectors used to form a parallelepiped be any values?
No, the vectors used to form a parallelepiped must follow certain rules in order for the shape to be considered a parallelepiped. The vectors must be non-collinear, and their lengths and directions must be able to form a closed shape with six faces that are all parallelograms.
5. What is the importance of using vectors to form a parallelepiped?
Using vectors to form a parallelepiped allows for the shape to have specific properties, such as having six faces that are all parallelograms. This also allows for the shape to be used in various mathematical and scientific calculations, as the properties of vectors can be applied to the parallelepiped. | 677.169 | 1 |
New Elementary Geometry: With Practical Applications ; a Shorter Course Upon the Basis of the Larger Work
From inside the book
Results 1-5 of 39
Page 8 ... angle A CD or D C B. 11. Two straight lines are said to be Perpendicular to each other , when their meeting forms equal adjacent an- gles ; thus the lines A B and CD are perpendicular to each other . A- C D B C B A D Two adjacent angles ...
Page 13 ... angles ACE and ECB will each be a right angle ( 12 ) . The angle A C D is composed of the right angle ACE and the angle E CD ( 26 , Ax . 9 ) ; but this angle E CD and DCB compose the other right angle , E CB ; hence the angles AC D ...
Page 14 ... angles A CD , D C B is a right angle , the other must also be a right angle . 37. Cor . 2. All the successive angles , in the same plane , B A C , CAD , DAE , EA F , formed at C D E a single point , on the same side of a straight line ...
Page 15 ... angles which they form are equal . Let the two straight lines A B , CD intersect each other at the point E ; then will the angle A EC be equal to the angle DEB , and tne angle CEB to A ED . A C E B D For the angles AEC , CEB , which | 677.169 | 1 |
Number Of Lines Of Symmetry In A Star Shape
How Many Lines Of Symmetry Does A Star Have TOP from top.walterlanyon.com
Introduction to Lines of Symmetry
Do you ever wonder what lines of symmetry have to do with stars? Symmetry is a concept in mathematics and it refers to the balance in the shape, size, and structure of objects. When an object is symmetrical, it has an even distribution of components, which means that each side is a mirror image of the other. In geometry, symmetry is an important concept and it is used to calculate the number of lines of symmetry in an object. Knowing the number of lines of symmetry can help us understand the properties of an object.
Types of Symmetry
When it comes to symmetry, there are two different types. The first type is called point symmetry and the second type is called line symmetry. Point symmetry occurs when an object has an even distribution of points and lines in it. Line symmetry is when an object has an even distribution of lines and the lines intersect at certain points. The number of lines of symmetry in an object can be determined by looking at the object and counting the number of lines that intersect at certain points.
How Many Lines of Symmetry Does a Star Have?
A star is a popular shape and it is often seen in art, logos, and designs. When it comes to lines of symmetry, a star has an interesting shape. Most stars have five points and these points are arranged in a symmetrical pattern. In terms of the number of lines of symmetry, a star typically has five lines of symmetry. This means that if you draw a line from one point of the star to its opposite point, the line will divide the star into two equal parts.
Why Knowing the Number of Lines of Symmetry is Important
Knowing the number of lines of symmetry in a star is important for a variety of reasons. For example, it can help us understand the properties of the star and how it is arranged. It can also help us understand how the star can be used in art, logos, and designs. Additionally, knowing the number of lines of symmetry can be useful when it comes to calculations in geometry.
Using the Number of Lines of Symmetry in Geometry
In geometry, the number of lines of symmetry can be used to calculate the area and the perimeter of an object. For example, if you know the number of lines of symmetry in a star, you can calculate its area by dividing the star into five equal parts and then multiplying the area of one part by five. Similarly, you can calculate the perimeter of a star by dividing the star into five equal parts and then multiplying the perimeter of one part by five.
Using the Number of Lines of Symmetry in Art, Logos, and Designs
The number of lines of symmetry in a star can also be used in art, logos, and designs. For example, if you want to create a logo that has a star shape, you can use the number of lines of symmetry to make sure that the star is evenly distributed and that the lines are evenly distributed. Similarly, if you are creating a design with a star shape, you can use the number of lines of symmetry to make sure that the design is balanced and symmetrical.
Conclusion
In conclusion, the number of lines of symmetry in a star is five. Knowing the number of lines of symmetry in a star can be useful for a variety of reasons, such as understanding the properties of a star and using the number of lines of symmetry in calculations and art, logos, and designs. Now that you know the number of lines of symmetry in a star, you can start using it in your work. | 677.169 | 1 |
Angles in Harmony: Unveiling the Secrets and techniques of Corresponding Angles
Welcome to the interesting planet of angles, where traces intersect and harmonize in sudden methods. In this post, we delve into the enchanting realm of corresponding angles – a idea that unravels the intricate connection among intersecting traces and the angles they develop. From the realm of geometry to algebraic equations, an exploration of corresponding angles requires us on a fascinating journey crammed with symmetry, congruence, and mathematical marvels.
Corresponding angles, also acknowledged as "matching angles," arise when a transversal intersects two parallel strains. These angles are formed by the intersection of a certain line and a single of the parallel traces. Every corresponding angle corresponds to an additional angle, mirroring its measure or aiding in uncovering fascinating designs. Comprehension the properties and characteristics of corresponding angles unlocks the doorway to a multitude of geometric and algebraic wonders.
Adjacent angles, congruent angles, and complementary angles are closely intertwined with the idea of corresponding angles. Adjacent angles share a common vertex and aspect but vary in their non-shared sides. Congruent angles have similar measurements, bringing forth a perception of equilibrium and harmony in a geometric figure. On the other hand, complementary angles blend to type a right angle, which actions 90 degrees – a remarkable unity of angles portraying equilibrium and completeness.
The planet of mathematics expands considerably over and above angles, as fractions, polynomial equations, and purpose qualities intertwine with geometry. From multiplying and dividing fractions to simplifying them for ease of use, the mathematical journey proceeds. Houses such as the distributive, associative, and commutative qualities support in manipulating equations, paving the way for good results in fixing troubles associated to programs of equations, linear equations, and slopes.
But the exploration doesn't quit there – we undertaking into the realms of domains and ranges, checking out the boundaries of capabilities and the intricate relationships among inputs and outputs. Polynomials, each in their long division form and as factored expressions, open doorways to comprehending the habits of equations and their roots. Designs such as quadrilaterals, triangles, and trapezoids turn out to be intrinsic puzzles to remedy, as their perimeters, regions, and distinct qualities expose hidden tricks of their nature.
Put together to be awed by the wonders of the Pythagorean theorem , which elegantly connects the lengths of the sides of a correct triangle. Dive into the intriguing globe of parabolas, their equations, and their vertex kind, as they offer insights into symmetry and transformation. Triangle classifications, this kind of as isosceles, equilateral, obtuse, and acute, invite us to explore their unique qualities and interactions.
So join us on this fascinating journey as we unveil the tricks of corresponding angles and unlock the interconnected mysteries of arithmetic. From the most basic of lines to the most complex of equations, the concealed harmonies of angles await us, prepared to challenge and encourage our mathematical prowess. Get ready to unravel the symphony of angles in harmony – an exploration that will without end transform the way you perceive the globe of mathematics.
Comprehension Angles
Angles are fundamental mathematical ideas that engage in a crucial position in various geometric calculations and equations. They are shaped by two non-collinear rays that share a common endpoint, recognized as the vertex. In this part, we will delve into the essence of angles, their sorts, and how they are related to one one more.
When two strains intersect, they sort four angles known as vertical angles. Vertical angles are congruent, that means they have equal actions. This residence allows us to resolve for missing angles by using the acknowledged steps of congruent or corresponding angles. Corresponding angles are shaped when a transversal intersects two parallel strains, and they occupy the same relative positions.
Adjacent angles are angles that share a widespread aspect and vertex, but have no widespread inside factors. They are facet by facet, generating a feeling of continuity. Adjacent angles can be complementary, that means their steps incorporate up to 90 degrees, or supplementary, the place their steps incorporate up to one hundred eighty degrees. These interactions are crucial when resolving equations involving several angles.
Comprehension the diverse types of angles and their interrelationships is crucial in a variety of mathematical concepts and purposes. Regardless of whether it truly is calculating the spot of a form, figuring out the slope of a line, or factoring polynomials, angles provide as essential factors in these mathematical procedures. By recognizing the qualities and behaviors of angles, we can unlock the secrets and techniques to fixing intricate mathematical difficulties.
Functions with Fractions
In this segment, we will discover some essential operations involving fractions. Comprehending how to perform these functions is vital for operating with corresponding angles and other mathematical ideas.
Introducing and Subtracting Fractions
When introducing or subtracting fractions, the very first phase is to make sure that the denominators are the identical. If they are distinct, we need to have to find a widespread denominator. As soon as the denominators are the very same, we can add or subtract the numerators and preserve the denominator unchanged. Remember to simplify the fraction if achievable.
Multiplying and Dividing Fractions
When multiplying fractions, multiply the numerators jointly and the denominators collectively. The ensuing solution is the solution. To divide fractions, we require to multiply by the reciprocal of the next fraction. This implies swapping the numerator and denominator of the 2nd portion and then multiplying the two fractions collectively.
Simplifying Fractions
To simplify a portion, we want to find the biggest widespread divisor (GCD) of the numerator and denominator and divide them each by the GCD. Simplifying fractions will help us categorical them in their easiest type, making calculations simpler.
Now that we have coated the basic functions with fractions, we can delve into the intriguing entire world of corresponding angles and their position in geometry. Keep tuned for the subsequent part in which we unveil the secrets powering these intriguing angles!
Important Principles in Arithmetic
Arithmetic encompasses a huge array of ideas that form the foundation of this field. Comprehending these basic suggestions is vital for delving deeper into the topic. In this section, we will check out some crucial principles in mathematics.
Congruent Angles
Congruent angles are angles that have the same evaluate. In other terms, they are identical in dimensions and form. Congruent angles engage in a important position in a variety of geometric theorems and proofs, helping us establish the equality of diverse angles in a shape.
Quadratic Equation
A quadratic equation is a polynomial equation of the 2nd degree, commonly written in the form ax² + bx + c = . Solving quadratic equations involves discovering the values of x that satisfy the equation. The quadratic system is frequently utilized to fix these equations proficiently.
Slope Intercept Type
The slope-intercept kind of a linear equation is written as y = mx + b, exactly where m signifies the slope of the line, and b denotes the y-intercept, which is the position in which the line crosses the y-axis. This type offers a very clear illustration of the slope and the beginning level of a line on a graph.
These essential principles lay the groundwork for more mathematical exploration. They are utilized extensively in various mathematical disciplines, from geometry to algebra, paving the way for deeper comprehension and useful programs.
All vehicle-towing companies have the same tasks and problems regardless | 677.169 | 1 |
The triangles composed of two components each are AEL, KDJ, HIC and FBG i.e. 4 in number.
The triangles composed of three components each are APF, EQB, BQH, GVC, CVJ, IUD, DUL and KPA i.e. 8 in number.
The triangles composed of six components each are ASB, BSC, CSD, DSA, AKF, EBH, CGJ and IDL i.e. 8 in number.
The triangles composed of twelve components each are ADB, ABC, BCD and CDA i.e. 4 in number.
Total number of triangles in the figure = 12 + 4 + 8 + 8 + 4 = 36.
27.
What is the number of triangles that can be formed whose vertices are the vertices of an octagon but have only one side common with that of octagon?
64
32
24
16
Answer: Option
Explanation:
When the triangles are drawn in an octagon with vertices same as those of the octagon and having one side common to that of the octagon, the figure will appear as shown in (Fig. 1).
Now, we shall first consider the triangles having only one side AB common with octagon ABCDEFGH and having vertices common with the octagon (See Fig. 2).Such triangles are ABD, ABE, ABF and ABG i.e. 4 in number.
Similarly, the triangles having only one side BC common with the octagon and also having vertices common with the octagon are BCE, BCF, BCG and BCH (as shown in Fig. 3). i.e. There are 4 such triangles.
This way, we have 4 triangles for each side of the octagon. Thus, there are 8 x 4 = 32 such triangles. | 677.169 | 1 |
An ellipse is the set of points such that the sum of the distances from any point on the ellipse to two other fixed points is constant. The two fixed points are called the foci (plural of focus) of the ellipse.
Figure %: The sum of the distances d1 + d2 is the same for any point on the ellipse.
The line segment containing the foci of an ellipse with both endpoints on the ellipse is called the major axis. The endpoints of the major axis are called the vertices. The point halfway between the foci is the center of the ellipse. The line segment perpendicular to the major axis and passing through the center, with both endpoints on the ellipse, is the minor axis. | 677.169 | 1 |
Right Angle Triangle
Here we will learn about right-angled triangles, including what a right-angled triangle is and how to solve problems involving their sides and their angles.
There are also right angle triangle worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you're still stuck.
What is a right angle triangle?
A right angle triangle is a type of triangle that has one right angle ( 90^{\circ} angle).
A right angle is 90^{\circ} and it is denoted using a small square inside the angle at the vertex, instead of the usual arc.
To solve problems involving right-angled triangles, we need to be able to find the values of other missing angles or side lengths and use them where appropriate.
The longest side of a right-angled triangle is called the hypotenuse. It is the opposite side to the right angle.
Properties of triangles are very important within geometry. They are one of the most common shapes to recognise for angles in parallel lines, circle theorems, interior angles, trigonometry, Pythagoras' theorem and many more. You therefore must be familiar with their individual properties.
What is a right angle triangle?
Types of right angle triangles
There are two types of right-angled triangles.
Right-angled isosceles triangle
A right-angled triangle which also has two equal sides can be known as an isosceles right triangle. This is a special case of a right-angled triangle.
Here, the two equal angles will both be 45^{\circ}.
Scalene right-angled triangle
A scalene right-angled triangle has all angles of different sizes and all sides of different length. One angle within a scalene right-angled triangle is 90^{\circ}.
Area of a right angle triangle
The area of a right-angled triangle can be found by using the formula
A=\frac{1}{2}bh
where \text{b} is the base length and \text{h} is the perpendicular height of the triangle. Sometimes these values need to be calculated.
As the triangle contains a right angle, the two sides adjacent to the right angle (not the hypotenuse) are perpendicular to each other, and so these lengths can be used to calculate the triangle's area.
The area of a triangle is given in square units.
For example, to calculate the area of the right-angled triangle below, we need to know the value for the base \text{b}, and the value of the perpendicular height \text{h}.
Here b=8 and h=6. This is because these two values are perpendicular to each other, despite the base being at "the top" of the image.
The value of 10 \ cm can be a decoy when calculating the area of a triangle as it is not the height of the triangle, it is the hypotenuse, so beware!
Pythagoras' theorem
The three sides of a right-angled triangle are linked by a very useful property known as Pythagoras' theorem.
Pythagoras theorem states that the square of the longest side of a right-angled triangle (called the hypotenuse) is equal to the sum of the squares of the other two sides.
This can be written as
a^2+b^2=c^2.
Side c is always the hypotenuse.
Pythagoas' theorem can be used to calculate the length of the hypotenuse if we know the other 2 sides and conversely, we can use Pythagoras' theorem to calculate a shorter side, given the hypotenuse and one other side of a right-angled triangle.
Types of triangles worksheet (includes right angle triangle)
Related lessons on triangles
Right angle triangle is part of our series of lessons to support revision on polygons and triangles. You may find it helpful to start with the main triangles lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:
Right angle triangle examples
Example 1: finding the missing angle
Determine the size of angle x.
Locate known angles, including the right angle, and calculate any necessary unknown angles.
The triangle contains the three angles, 37^{\circ} (the interior angle at B ), 90^{\circ} (the interior angle at C ), and x^{\circ} (the interior angle at A ). We need to calculate x which we will complete in step 3.
2Locate known sides and calculate any necessary unknown side lengths.
We do not need to know the length of any side to solve this problem.
3Solve the problem using any necessary values.
As the sum of angles in a triangle is 180^{\circ}, we can use this to find the missing angle x by subtracting the other two known angles from 180.
180-(90+37)=53^{\circ}
The angle x=53^{\circ}.
Example 2: finding the missing angle
Use angle rules to find the size of angle x.
Locate known angles, including the right angle, and calculate any necessary unknown angles.
The triangle contains the two angles, 62^{\circ} and 90^{\circ}.
We need to calculate x and so it is useful to calculate the third angle in the triangle as x is on the same point to this angle. As angles in a triangle total 180^{\circ},
180-(90+62)=28^{\circ}.
Labelling this on the diagram, we have
Locate known sides and calculate any necessary unknown side lengths.
We do not need to know the length of any side to solve this problem.
Solve the problem using any necessary values.
As vertically opposite angles are equal and x is vertically opposite the angle 28^{\circ}, we can state that x=28^{\circ}.
Example 3: missing angle
ABC is a right-angled triangle. D lies on the line BC . Determine the size of angle CAD , labelled x.
Locate known angles, including the right angle, and calculate any necessary unknown angles.
In the diagram there are 3 known angles, 25^{\circ}, 73^{\circ}, and 90^{\circ}, and 3 unknown angles, one of which is our value of x.
As the angle x is within the triangle ADC , and we know the angle at C \ (25^{\circ}), we need to know the angle at D (angle ADC ). To determine angle ADC , we can see that it is adjacent to the angle ADB and as angles on a straight line total 180^{\circ},
180-73=107^{\circ}.
So angle ADC = 107^{\circ} .
We do not need to calculate any other angle in the triangle for this version of a solution.
Locate known sides and calculate any necessary unknown side lengths.
We do not need to know the length of any side to solve this problem.
Solve the problem using any necessary values.
Now, as angles in a triangle total 180^{\circ}, the value of x is equal to
180-(107+25)=180-132=48^{\circ}.
The angle x=48^{\circ}.
Alternative Solution
We need to use the angle fact that the sum of angles in a triangle is 180^{\circ}.
The diagram is made up of two right-angled triangles.
The missing angle on this right angle triangle is
180-(90+25)=65^{\circ} .
The missing angle on this right angle triangle is
180-(90+73)=17^{\circ}.
Angle x will be the difference in these two angles.
65-17=48^{\circ}
Angle x=48^{\circ}
Example 4: perimeter problem with congruent right angle triangles
Shape ABC is made from two congruent right angle triangles connected at the line BD . The sides of one of the right-angled triangles are 5 \ cm, 12 \ cm and 13 \ cm and AC < BD . Calculate the perimeter of the shape.
Locate known angles, including the right angle, and calculate any necessary unknown angles.
We can add the right angles to the diagram. This tells us that the length AC is perpendicular to the width BD . Any further angles in the shape are not needed as we are focused on the perimeter, which is a length.
Locate known sides and calculate any necessary unknown side lengths.
We are told that the three side lengths of one of the right-angled triangles are 5 \ cm, 12 \ cm, and 13 \ cm.
As the longest side of the triangle is the hypotenuse which is the side opposite the right angle, we can see that AB and BC are both 13 \ cm.
As AC < BD and AC is twice the length of AD, AD must be equal to 5 \ cm as AC would then equal 10 \ cm, and BD would equal 12 \ cm.
Labelling all of these sides, we have
Solve the problem using any necessary values.
We can now work out the perimeter of the shape.
5+5+13+13=36\text{ cm}
The perimeter of the shape is 36 \ cm.
Example 5: perimeter of congruent triangles
The shape is made from 2 congruent right-angled triangles. The sides of the triangle are 8 \ cm, 15 \ cm and 17 \ cm. Find the perimeter of the shape.
Locate known angles, including the right angle, and calculate any necessary unknown angles.
Remember that there is one right angle. We can add the right angles to the diagram.
Locate known sides and calculate any necessary unknown side lengths.
Remember that the side opposite the right angle is the longest side. The longest side is the hypotenuse. We can add the sides of the right-angled triangles to the diagram.
Solve the problem using any necessary values.
The missing side is the difference between the 2 shortest sides 15 \ cm and 8 \ cm.
15-8=7
The perimeter can be found by adding together all the sides of the shape.
17+7+17+8+16=64
Therefore the perimeter is 64 \ cm.
Example 6: perimeter of a right triangle
The sides of a right-angled triangle are in the ratio 3:4:5. The perimeter is 240 \ cm. Calculate the length of the hypotenuse.
Locate known angles, including the right angle, and calculate any necessary unknown angles.
Not needed for this question.
Locate known sides and calculate any necessary unknown side lengths.
Remember that the side opposite the right angle is the longest side.The longest side is the hypotenuse.
Solve the problem using any necessary values.
We can divide the perimeter by the sum of the parts of the ratio.
240\div (3+4+5)=20
To find the hypotenuse we need to multiply 20 by 5.
20\times 5=100
Therefore the longest side is 100 \ cm.
Note: 3, \ 4, \ 5 is known as a Pythagorean triple as the sum of the square of the two shorter sides is equal to the square of the third side. This means that the triangle must contain a right angle (3^{2}+4^{2}=5^{2}).
Common misconceptions
Angles in polygons
Make sure you know your angle properties. Getting these confused causes quite a few misconceptions.
◌ Angles in a triangle total 180^{\circ} .
◌ Angles in a quadrilateral total 360^{\circ} .
Angle facts
Make sure you know your angle properties.
◌ Adjacent angles on a straight line add up to 180^{\circ} .
◌ Vertically opposite angles are equal.
◌ Alternate angles are equal.
◌ Corresponding angles are equal.
Practice right angle triangle questions
1. Find the missing angle in the triangle.
39^{\circ}
90^{\circ}
40^{\circ}
49^{\circ}
The sum of angles in a triangle is 180^{\circ}. We can use this fact along with the fact that a right angle is 90^{\circ} to find the missing angle.
180-(90+41)=49^{\circ}
The missing angle is 49^{\circ}.
2. Calculate the value of y.
56^{\circ}
34^{\circ}
45^{\circ}
65^{\circ}
We can use the fact that adjacent angles on a straight line add up to 180^{\circ}.
180-124=56^{\circ}
As a right angle is 90^{\circ} and angles in a triangle sum to 180^{\circ},
180-(90+56)=34^{\circ}.
The value of y=34^{\circ}.
3. Calculate the size of the smallest angle.
45^{\circ}
18^{\circ}
36^{\circ}
54^{\circ}
We can use the fact that angles in a triangle add up to 180^{\circ}. This means that the two smaller angles will sum to 90^{\circ} (as 180-90=90^{\circ} ).
\begin{aligned}
2a+3a &=90 \\\\
5a &= 90
\end{aligned}
Dividing 90 by 5 will determine the value of a.
a=90\div{5}=18
To find the size of the smallest angle, we can multiply 18 by 2.
2\times 18=36
The smallest angle is 36^{\circ}.
4. The sides of a right angle triangle are in the ratio 3:4:5. The perimeter is 96 \ cm. Calculate the length of the shortest side.
40 \ cm
24 \ cm
8 \ cm
32 \ cm
We can divide the perimeter by the sum of the parts of the ratio.
96\div (3+4+5)=8
We can then multiply the 8 by 3 to find the shortest side of the right-angled triangle.
3\times 8=24
The shortest side will be 24 \ cm.
5. The sides of a right angle triangle are 5 \ cm, 12 \ cm, and 13 \ cm. Four of these triangles are put together to make the chevron ABCDEF (a concave hexagon) where EF < BD. Calculate the perimeter of the shape.
100 \ cm
58 \ cm
120 \ cm
62 \ cm
As we know that EF, AD and BC are the same length and are shorter than BD and DF , we can label each of these with the shortest side length of the triangle, 3 \ cm.
As AB, CD, DE and AF are all the hypotenuse of each right angle triangle, these can all be labelled as the longest side length, 13 \ cm.
This means that BD and DF are both 12 \ cm. We can add the dimensions of the right-angled triangles to the diagram, along with the four 90 degree angles.
The perimeter will be the sum of the sides of the shape.
13+13+5+13+13+5=62\text{ cm}
6. The sides of a right angle triangle are 7 \ cm, \ 24 \ cm, and 25 \ cm. Two of these triangles are put together to make a quadrilateral such that 2DE < CD . Calculate the perimeter of the shape.
98 \ cm
56 \ cm
112 \ cm
108 \ cm
We are given that 2DE < CD . As DE is a side length of the right-angled triangle, the length of DE must be 7 \ cm and the length of CE must be 24 \ cm as 2 \times 7<24-7 giving us 14<17.
The hypotenuse must be 25 \ cm as it is the longest side. We can now add the dimensions of the right-angled triangles to the diagram.
We can add the lengths of the sides together to find the perimeter.
24+7+25+17+25=98\text{ cm}
The perimeter is 98\text{ cm}.
Right angle triangle GCSE questions
1. Shape A is a triangle.
Draw on the line(s) of symmetry.
(1 mark)
Show answer
(1)
2. Here is a diagram.
PQR and PQS are right angle triangles.
QPR =59^{\circ} PSR =32^{\circ}
Find the size of the angle RPS .
(4 marks)
Show answer
QPR =180-90-59=31
(1)
PSR =180-90-32=58
(1)
58-31
(1)
27^{\circ}
(1)
3. An energy company is designing a new logo.
It is made from four congruent triangles with sides 6 \ cm, \ 8 \ cm and 10 \ cm | 677.169 | 1 |
7.
УелЯдб 2 Euclides William Desborough Cooley. PROP . III . PROB . From the greater of two given straight lines , to cut off a part equal to the less . 3 B E PROP . IV . THEOREM . If two triangles have two sides of the one re- spectively equal to ...
УелЯдб 10 ... three given straight lines , of which any two are together greater than the third . L PROP . XXIII . PROB . At a given point in a given straight line , to make an angle equal to a given rectilinear angle . A G B Ź F DE PROP . XXIV ...
УелЯдб 14 ... straight line , are parallel to one another . G B A E H ску PROP . XXXI . PROB . From a given point to draw a straight line parallel to a given straight line . D F B PROP . XXXII . THEOR . If a side of any triangle is produced , the ...
УелЯдб 58 ... given straight line to cut off any re- quired part . 1 PROP . X. PROB . To divide a given undivided straight line into parts proportional to those of a given divided straight line . F H E B K C PROP . XI . PROB . To find a third 58 EUCLID'S | 677.169 | 1 |
If one pair of opposite angles and one pair of opposite sites of a quadrilateral is congruent then the quadrilateral is a parallelogram. How can it be proven?
draw a diagonal through opposite corners of the quadrilateral.
This makes two triangles. Prove the triangles are congruent using
SSA (side side angle) congruence. Then show that the other two
sides of the quadrilater must be congruent to each other, so it is
a parallelogram. | 677.169 | 1 |
15 Equilateral Triangle Facts You Should Know
Equilateral triangles, those perfectly symmetrical wonders, have tickled mathematicians and artists alike for centuries. From ancient Egyptian pyramids to modern-day art installations, these triangles have been omnipresent.
Did you know that their internal angles always sum up to a precise 180 degrees? Or that their appearance in color theory has influenced countless artists? Get ready to uncover fun facts about equilateral triangles that'll reshape the way you see this three-sided marvel.
1. Ancient Obsession: Triangles in Pyramids
The ancient Egyptians were geometry aficionados! The majestic pyramids of Giza, built around 2580-2560 BCE, are a testament to their obsession with the equilateral triangle. The sides of these pyramids are vast equilateral triangles, serving both an aesthetic and structural purpose. Historians believe that the equilateral triangle's equal sides and angles symbolized harmony and the cycle of life for the Egyptians.
This isn't just a coincidence; it's a conscious design backed by a deep understanding of geometry. These structures have withstood the test of time, and the equilateral triangle's properties play a significant role in their resilience.
2. The Equiangular Enigma: Three Angles, Three Sixties
Beyond its equal sides, the equilateral triangle is equally fascinating when you delve into its angles. Every angle in an equilateral triangle measures exactly 60 degrees. This makes it not only equilateral but also equiangular. Mathematicians have been intrigued by this perfect trifecta of angles for centuries.
Historically, this angle fact was used in various calculations, especially in navigation. Mariners often employed tools like the astrolabe, which incorporated the principles of equilateral triangles, to help navigate and measure the altitude of stars.
3. Congruent Comrades: All Sides Have Equal Length
Congruency is the beating heart of the equilateral triangle. Every side, without fail, matches its counterpart in length, providing this geometric wonder with its iconic look. Did you know that the word 'equilateral' is derived from Latin? "Equi" means 'equal' and "latus" stands for 'side'. It's no wonder this triangle is known for its equal length sides!
In the 5th century BC, the Greek mathematician Hippasus showed that equilateral triangles are the only triangles where the sides are congruent. Thanks to his discovery, it became a foundation of Euclidean geometry studies in academia.
4. Symmetry Supreme: The Equilateral's Three Axes
One of the mesmerizing properties of the equilateral triangle is its sublime symmetry. Not one, not two, but three lines of symmetry can be drawn through this triangle, each passing through a vertex and bisecting the opposite side. This inherent symmetry has made the equilateral triangle a go-to choice for artists and architects throughout history.
In the Renaissance period, architects praised the equilateral triangle for its balance and used it frequently in cathedrals and frescoes. Its threefold symmetry represented the Holy Trinity, adding a layer of spiritual significance to its geometric charm.
Image: Wikimedia Commons
5. Definition Distinction: Equilateral vs. Isosceles Triangle
Ah, the age-old debate! While both these triangles boast of equal sides, the equilateral triangle takes it up a notch with all three sides being identical. An isosceles triangle, on the other hand, settles with just two sides of equal length. The ancient Greeks were among the first to rigorously define and distinguish between these two.
Aristotle, in his work, referenced the distinctions between these triangles, drawing philosophical analogies. It's not just about lines and angles; it's about understanding the essence of equality and how it varies in different contexts.
6. Geometry Gold: Area and Perimeter Formula
When it comes to formulas, the equilateral triangle doesn't shy away! With side length 'a', the area can be calculated using the formula: (sqrt(3)/4) × a^2. As for the perimeter? Simply 3a. These formulas have been fundamental in geometry since the days of ancient scholars.
The renowned mathematician Euclid, in his seminal work "Elements", laid down the principles and formulas concerning equilateral triangles, which have since been the cornerstone of geometry. Through centuries, these formulas have been employed in various fields, from construction to computer graphics.
7. In Real Life: Equilateral Triangles in Nature
Nature is quite the mathematician, and the equilateral triangle is one of her favorite shapes. If you've ever stared in wonder at the honeycomb in a beehive, you've witnessed equilateral triangles in all their natural glory. Bees, being the efficient builders they are, use this shape because it allows them to use the least amount of wax while storing the maximum amount of honey.
The crystalline structure of certain minerals, like those in quartz, often showcases an intricate arrangement of equilateral triangles. It's truly a testament to how this geometric shape is a fundamental part of the universe's real life design.
David Hablützel / Pexels
8. Right Triangle's Sibling: Not Always a Right Angle!
Equilateral triangles seem to enjoy keeping us on our toes! Contrary to the right triangle, equilaterals don't have a 90-degree angle. Instead, they boast three identical angles of 60 degrees each. It's a common misconception, especially among geometry newcomers, to mix up the two.
While right triangles and their famous Pythagorean theorem have their unique charm, equilateral triangles come with their set of quirks, ensuring their place in the hall of fame of the triangle family.
9. Properties Power: The Inradius and Circumradius Connection
Equilateral triangles pack a punch with their fascinating properties. Dive deep, and you'll discover the connection between their inradius (radius of the inscribed circle) and circumradius (radius of the circumscribed circle). For a triangle with side length 'a', the inradius is a/(2√3) and the circumradius is a/√3. The beauty lies in the relationship: the circumradius is exactly twice the inradius!
Historical records indicate that ancient Indian mathematicians were well aware of this relationship. Their works on circle-triangle interrelationships set the stage for many geometric explorations in later centuries.
10. Tessellation Temptation: A Perfect Fit
Tessellation, the art of covering a surface with a pattern of flat shapes so that there are no overlaps or gaps, finds a loyal friend in the equilateral triangle. These triangles can fit together seamlessly to cover a plane without any spaces in between. It's no surprise that they are a favorite choice for many tile designs and artworks.
M.C. Escher, the famous graphic artist, used equilateral triangle tessellations in some of his iconic works. It's amazing how a simple geometric shape can transform into a canvas of infinite possibilities in the hands of a creative genius.
11. Degrees of Delight: Summing Up to 180
Triangles have this cool party trick: no matter their shape or size, the sum of their internal angles always add up to a neat 180 degrees. Equilateral triangles split this sum equally among their three angles, giving each angle a value of 60 degrees. This degrees of consistency is quite remarkable!
Historians believe that by 1800 BC, ancient Egyptians had a decent understanding of this triangle angle sum concept. They applied this knowledge in various construction projects, including their iconic pyramids.
Image by wirestock / Freepik
12. The Altitude Adventure: It's a Median Too!
The equilateral triangle never ceases to surprise. Draw an altitude (a line segment from a vertex and perpendicular to the opposite side) in one, and bam! You've also drawn a median (a line segment from a vertex to the midpoint of the opposite side). In equilateral triangles, the altitude and median are the same line!
This geometric revelation dates back to the Greeks. They revered geometry, and discoveries like this added to the mystique of equilateral triangles in their scholarly pursuits.
13. Trivia Time: World Records and Equilateral Triangles
Ever heard of the world record for the largest human-formed equilateral triangle? In 2010, over 6,000 students in India formed an equilateral triangle, highlighting the fun fact that geometry can be both educational and entertaining!
Equilateral triangles have also found their place in Guinness World Records for various feats, including largest structures and designs based on this shape. They're not just classroom wonders but also record-breaking stars!
14. The Equilateral in Pop Culture: Art and Design Marvel
Equilateral triangles aren't just for mathematicians; they've been muses for artists and designers too. Their symmetrical beauty has graced album covers, posters, and even film sequences. In the 1960s and 70s, equilateral triangles became symbols of futuristic design, prominently featured in sci-fi art and literature.
Famous artists like Andy Warhol have incorporated equilateral triangles in their pieces. They embody a blend of simplicity and complexity that resonates with both the artist and the observer.
Image: asrarchowdhury.wordpress.com
15. Chromatic Charm: Equilateral Triangle in Color Theory
When artists and designers dive into color theory, the equilateral triangle has a pivotal role to play. Ever glanced at a color wheel? You might've noticed that the primary colors (red, blue, yellow) or the secondary colors (green, orange, purple) are often positioned at the vertices of an equilateral triangle. This equilateral triangle fact demonstrates the equal distance between these colors in terms of color harmony.
Historically, Sir Isaac Newton, when he was conceptualizing the color wheel in the 17th century, played with the positioning of colors in triangular formats. It's a testament to the equilateral triangle's enduring influence, even in the world of color!
FAQ
Is an equilateral triangle unique?
Absolutely! An equilateral triangle is unique because it's the only type of triangle where all three sides and all three angles are congruent. While other triangles might have one or two sides of equal length, only in an equilateral triangle are all three sides and angles identical. This uniformity gives it many of its distinctive properties, setting it apart from other triangles.
What is an equilateral triangle also known as?
An equilateral triangle is also commonly known as an "equiangular triangle." This name stems from its nature of having all three internal angles equal, each measuring 60 degrees. So, while "equilateral" emphasizes its equal sides, "equiangular" highlights its equal angles. Both terms are correct and aptly describe the triangle's attributes.
Are equilateral triangles the strongest?
In engineering and architecture, triangles are considered the strongest shape due to their inherent stability. Among triangles, the equilateral triangle offers a particular advantage because of its uniformity. When forces are applied to its vertices or sides, the stresses are distributed evenly across its structure. However, the term "strongest" can be subjective and might depend on the specific context or application.
Are all equilateral triangles always similar?
Yes! All equilateral triangles are always similar. Similarity in geometry refers to the condition where two shapes have the same shape but not necessarily the same size. Since equilateral triangles always have angles of 60 degrees at each vertex and three congruent sides, any two equilateral triangles, regardless of their size, will be similar to each other.
Is an equilateral triangle always equilateral?
Yes, by definition, an equilateral triangle is always equilateral. This means that all its sides are of equal length. If a triangle has one or two sides of different lengths, it cannot be termed "equilateral." The term "equilateral" is unwavering in its definition, denoting equal length in all sides of the triangle.
How many properties does an equilateral triangle have?
Equilateral triangles have a myriad of properties, thanks to their symmetrical nature. Some of the main properties include:
All three sides are of equal length.
All internal angles measure 60 degrees.
The altitude and median drawn from a vertex coincide.
The inradius and circumradius have a fixed relationship based on the triangle's side length.
They can tessellate a plane without gaps.
And the list goes on! Depending on how in-depth you delve into geometry, you can discover numerous properties and characteristics unique to equilateral triangles. | 677.169 | 1 |
Is Lorentz Transformation an Application of Pythagoras's Theorem?
In summary, Pythagoras's theorem is an ideal that provides an approximation for real-world situations. In special relativity, it can be applied by every observer if the triangle has a right angle for that observer. In general relativity, it is accurate in flat space, but not in curved space-time. However, it can still provide accurate measurements over small local regions. Lorentz transformations have been tested with high precision and found to be accurate, unless gravitational effects are significant. Overall, the accuracy of Pythagoras's theorem and Lorentz transformations depend on the context and limitations of the real world.
Sep 23, 2013
#1
Dappy
23
0
Pythagoras's Theorum is considered to be an ideal. The accuracy of its results are dependent upon the flatness of the surface to which it is being applied. Since there is no such thing as a perfectly flat surface, Pythagoras's Theorum results can only be seen as an approximation. Is this true in its application to Lorentz transformation?
The theorem is exact, you just have to keep its requirements in mind. Triangles in the real world are just approximations to perfect [strike]rectangular[/strike] right triangles, but that is not part of the theorem. Real triangles are not three ideal lines connected at three ideal points anyway.
In special relativity, space is flat. The theorem can be applied by every observer, if the (idealized) triangle has a right angle for this observer (that is frame-dependent).
Last edited: Sep 23, 2013
Sep 23, 2013
#3
Dappy
23
0
mfb said:
-In special relativity, space is flat. The theorem can be applied by every observer, if the (idealized) triangle has a right angle for this observer (that is frame-dependent).
In General relativity space time is curved. I'm confused. Surely the flat space of SR must also be an ideal.
If you take gravitational effects into account, no experiment has ever found a deviation from Lorentz transformations. And there are experiments testing that with a precision of better than one part in a billion.
Pardon me for asking, but what is your first language?
German.
Sep 23, 2013
#18
Dappy
23
0
mfb said:
If you take gravitational effects into account, no experiment has ever found a deviation from Lorentz transformations. And there are experiments testing that with a precision of better than one part in a billion.
That's almost good enough for me. Can you imagine how flat a surface would have to be for Pythagoras's theorem to be accurate to better than one part in a billion. The mirror in the Hubble telescope comes to mind. I wonder how accurate it is in that context.
I don't know how to double quote, but German, I would never of guest. I'm dyslexic and slip up a lot more than that.
Can you imagine how flat a surface would have to be for Pythagoras's theorem to be accurate to better than one part in a billion.
Flatter than one part in a billion? :D
To have length deviations of one part in a billion on a perfect sphere with the size of earth, the triangle needs an area of ~0.05 km^2.
1 part in a billion is just a random number, the quality of the tests depends on the tested quantity. The stability of the speed of light in different directions can be tested way better (wikipedia gives 1 part in 1017).
The next generation of ring lasers is expected to measure the rotation of Earth more precise than satellites and telescopes. That's a for an already tiny effect.
On a curved 2D surface, the Pythagorean Theorem cannot be accurately applied over large distances. However, over short distances, it can provide an excellent approximation. Thus, on the surface of the earth, it will not be accurate over large variations of latitude and longitude. However, over short distances like a few miles, it typically provides an excellent approximation. In space-time, the analog of the Pythagorean theorem is the Minkowski metric. Over large spans of space-time in a region of large mass-induced curvature, it is not possible to define a set of coordinates that accurately describes distances using the Minkowski metric. However, over small local regions, it is possible.
Sep 24, 2013
#21
Dappy
23
0
mfb said:
Flatter than one part in a billion? :D
That isn't quite what I said, but it made me laugh and when I realized what I actually said, it was equally ridiculous.
Now that you've pointed that out, I have to say "right angled triangle". ;-)
Thankyou mfb for the link. I found it most informative and need to give it more attention, to allow it all to soak in.
Sep 24, 2013
#22
Dappy
23
0
Chestermiller said:
Over large spans of space-time in a region of large mass-induced curvature, it is not possible to define a set of coordinates that accurately describes distances using the Minkowski metric.
So can I assume that Lorentz violation may exist there, but only because of our inability to define a set of coordinates?
That is not considered as a violation of Lorentz invariance - it is just a setup where Lorentz transformations are not the right tool for the whole space. You need general relativity then.
Sep 25, 2013
#24
Dappy
23
0
I think I understand now. Lorentz transformation is the application of Pythagoras's theorem to two distance ratios, 1:v/c where 1 is the speed of light. The fact that the speed of light is invariant means that we can trust that 1 will always be 1 and hence Lorentz invariant.
1. What is Pythagoras's Theorem ideal?
Pythagoras's Theorem ideal is a mathematical concept that states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. It is often written as a² + b² = c², where c is the length of the hypotenuse and a and b are the lengths of the other two sides.
2. Who was Pythagoras and how did he come up with this theorem?
Pythagoras was a Greek mathematician, philosopher, and founder of the Pythagorean school. He is credited with discovering and proving the Pythagorean Theorem. According to legend, he was inspired by the sound produced by different lengths of strings on a musical instrument.
3. What are some real-life applications of Pythagoras's Theorem?
Pythagoras's Theorem has many practical applications in fields such as engineering, architecture, and physics. It is commonly used to calculate distances, determine the height of buildings, and solve problems involving right triangles.
4. Is Pythagoras's Theorem only applicable to right triangles?
Yes, Pythagoras's Theorem is only applicable to right triangles. It cannot be used to solve problems involving other types of triangles, such as equilateral or scalene triangles.
5. Can Pythagoras's Theorem be extended to three-dimensional shapes?
Yes, Pythagoras's Theorem can be extended to three-dimensional shapes. It is known as the Pythagorean Theorem in 3D and is used to find the length of the hypotenuse in a three-dimensional right triangle. | 677.169 | 1 |
The Elements of Euclid
Dentro del libro
Resultados 1-5 de 100
Pįgina 6 ... angles is called a right angle : and the straight line which stands on the other is called a perpendicular to it . XI . An obtuse angle is that which is greater than a right angle . XII . An acute angle is that which is less than a ...
Pįgina 7 ... right - angled triangle , is that which has a right angle . XXVIII . An obtuse - angled triangle , is that which has an obtuse angle . XXIX . An acute - angled triangle , is that which has three acute angles . XXX . Of four - sided ...
Pįgina 8 ... right angles , but has not all its sides equal . XXXII . A rhombus is that which has its sides equal , but its angles are not right angles . оо XXXIII . A rhomboid , is that which has its opposite sides equal to one another , but all ...
Pįgina 9 ... right angles are equal to one another . XII . " If a straight line meet two straight lines , so as to make the two " interior angles on the same side of it taken together less than " two right angles , these straight lines being ...
Pįgina 16 ... right | 677.169 | 1 |
Inscribed Angle Theorem
The inscribed angle theorem is an amazing fact of Euclidean geometry, which states that an inscribed angle has half the measure of the corresponding central angle. This leads to numerous other facts of angle geometry, including those about intercepted arcs and classifying cyclic quadrilaterals. We present a novel proof of the inscribed angle theorem here using the geometry of complex numbers. | 677.169 | 1 |
Cyclic Quadrilateral Definition
A cyclic quadrilateral can be defined as a quadrilateral inscribed in a circle. It is a four-sided polygon around which a circle can be drawn such that the vertices lie on the circumference of a circle.
Cyclic Quadrilateral Theorems
One important question is: how do you prove that a quadrilateral is cyclic? Let's understand how to find cyclic quadrilaterals with the help of theorems.
Ptolemy Theorem of Cyclic Quadrilateral
The Ptolemy theorem states that the sum of the product of the opposite sides of a cyclic quadrilateral equals the product of diagonals.
Converse: If opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.
Brahmagupta Theorem of Cyclic Quadrilateral
The Brahmagupta theorem states that the area A of a cyclic quadrilateral whose sides are a, b, c, d is given by: $Area = \sqrt{(s\;-\;a)(s\;-\;b)(s\;-\;c)(s\;-\;d)}$ Here, s is the semi-perimeter given by $s = \frac{a + b + c + d}{2}$
Isosceles Trapezoid Theorem
The isosceles trapezoid theorem states that a trapezoid is cyclic quadrilateral if and only if it is an isosceles trapezoid.
An isosceles trapezoid is a trapezoid where the two legs of the trapezoid are equal in length.
Properties of a Cyclic Quadrilateral
All the four vertices of the inscribed quadrilateral lie on the circumference or boundary of the circle. They are said to be concyclic.
The four sides of the inscribed quadrilateral are chords of the circle as their endpoints lie on the boundary of the circle.
The measure of an exterior angle at a vertex equals the opposite interior angle.
In a cyclic quadrilateral, the sum of the product of the opposite sides equals the product of diagonals.
In a cyclic quadrilateral, the perpendicular bisectors of the sides are always concurrent and they meet at the center O.
The sum of a pair of opposite angles is always supplementary.
The sum of all four angles of a cyclic quadrilateral is $360^{\circ}$.
A cyclic quadrilateral has the maximum area possible with the given side lengths. In other words, a quadrilateral that is inscribed in a circle represents the maximum area possible with those side lengths.
Cyclic Quadrilateral Formulas
Let's look at the formulas associated with the cyclic quadrilaterals.
Radius of a Cyclic Quadrilateral
If a, b, c and d are the successive sides of a cyclic quadrilateral, and s is the semi-perimeter, then the radius is given by
As mentioned earlier, if the sides of the inscribed quadrilateral are a, b, c, and d and "s" is the semiperimeter, then the area of a cyclic quadrilateral is given by the Brahmagupta theorem:
$Area = \sqrt{(s\;-\;a)(s\;-\;b)(s\;-\;c)(s\;-\;d)}$
where $s = \frac{a + b + c + d}{2}$
Heron's formula for the area of a triangle is derived from this equation.
Facts about Cyclic Quadrilateral
A cyclic quadrilateral has the maximum area possible with the given side lengths.
The circle that has all the vertices of a polygon on its circumference is called the circumcircle or circumscribed circle.
Conclusion
In this article, we learned about cyclic quadrilaterals, their properties, important theorems, and formulas associated with them. Now, let's understand how to solve cyclic quadrilateral problems with the help of these properties and theorems.
Solved Examples on Cyclic Quadrilateral
1. Find the values of x and y.
Solution:
We know that opposite angles of a cyclic quadrilateral are supplementary.
So, $100^{\circ} + x = 180^{\circ}$
$x = 180^{\circ}\;-\;100^{\circ}$
$x = 80^{\circ}$
Similarly, $70^{\circ} + y = 180^{\circ}$
$y = 180^{\circ} \;-\; 70^{\circ}$
$y = 110^{\circ}$
2. Find the value of x and y in the following figure.
Solution:
We know that opposite angles of a cyclic quadrilateral are supplementary.
Frequently Asked Questions on Cyclic Quadrilateral
No. Not every quadrilateral is cyclic. Examples of non-cyclic quadrilaterals are non-square rhombus (rhombus that's not a square) and non-rectangular parallelogram.
Is a parallelogram a cyclic quadrilateral?
If a parallelogram is cyclic, it must be a rectangle. A cyclic parallelogram is always a rectangle. A non-rectangular parallelogram is not cyclic.
Do angles in a cyclic quadrilateral add up to $360^{\circ}$?
Yes, angles in a cyclic quadrilateral add up to $360^{\circ}$ as the sum of opposite angles is $180^{\circ}$. So, $180^{\circ} + 180^{\circ} = 360^{\circ}$.
Is a rhombus a cyclic quadrilateral?
A non-square rhombus is not cyclic. A rhombus cannot be a cyclic quadrilateral because the opposite angles of a cyclic quadrilateral are supplementary, but in the case of a rhombus, the opposite angles are equal. A cyclic rhombus is a square. | 677.169 | 1 |
What Does Qed Stand For In Math? <p></p>
What does QED stand for in math? Many students who find difficulty in handling mathematical concepts will is common and ask this question. What do the letters stand for? The letter Q stands.
A quaternion is a representation of figures in two dimensions. Quaternions are treated as space coordinates. writing dissertation books Quaternions have five unique forms such as horizontal, vertical, degrees, right angles, and east-west.
If we assume that we have a large square (the"components" of the quadratic equation) and we understand the shape of the components, then we can figure out the shape of the unit by solving the quadratic equation. The shape of this unit is measured in centimeters or inches. Therefore, this shape can be represented by us in a form.
Q stands for quadratic. The answer to the question does Q stand for quaternion? Will give us the length of the x-axis or X.
In solving the equation, the x-axis is known as the x and the y-axis are calling phdthesiswriting biz y. The y-axis and the x-axis are associated to the left and right hand sides of this equation. It should be noted that the x-axis cannot be measured in inches. It has to be measured in centimeter units.
The x-axis must be measured along the x-axis and then translated to the y-axis. Calculating the point at which the y-axis is placed and the angle formed by the point translates to the y-axis the x-axis. The angle formed is called radians and it measures 360 degrees.
Now that the point is calculated, we may use the level of the angle formed by the x-axis and the y-axis as our measurement for the degree of the angle formed by the x-axis and the y-axis. For computing the amount of the x-axis, we can use the specific measure of this angle. We can do this by multiplying the amount of the. Let's say the angle is thirty-three degrees.
Therefore, the reply to the question how does Q stand for quaternion is"quaternion = angle". We will discover how to compute the duration of the x-axis from the unit by doing this calculation. We can get the length of this unit. We can convert the units of radians to inches.
These students can ask their parents and they'll know exactly what units they need to compute the units. They can also ask their algebra teachers and see if they can help them out. This can be an extremely tough problem to solve and several students don't have the correct units when they arrive at college.
So it is going to be a challenge for them in actuality, many students do not have enough units for the computation. Quite often provide additional units so that they can ask their math teacher for help and the school will be quite generous. Help with this difficult issue.
It's very important that students have the ability to ask themselves and answer the question,"What does Q stand for in mathematics?" So that they are able to solve the problem. The issues. | 677.169 | 1 |
5. Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see Fig. 7.20). Show that: (i) ΔAPB ≅ ΔAQB (ii) BP = BQ or B is equidistant from the arms of ∠A.
Answer
Given, l is the bisector of an angle ∠A. BP and BQ are perpendiculars.
1. ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that (i) ΔABD ≅ ΔACD (ii) ΔABP ≅ ΔACP (iii) AP bisects ∠A as well as ∠D. (iv) AP is the perpendicular bisector of BC.
1. Show that in a right angled triangle, the hypotenuse is the longest side.
Answer
ABC is a triangle right angled at B. Now, ∠A + ∠B + ∠C = 180° ⇒ ∠A + ∠C = 90° and ∠B is 90°. Since, B is the largest angle of the triangle, the side opposite to it must be the largest. So, BC is the hypotenuse which is the largest side of the right angled triangle ABC | 677.169 | 1 |
Step 6.
Were there any triangles you could not construct? Were there any triangles that you could construct in more than one way? Explain.
Answer:
Yes, at step 3 we could not construct any triangle. At step 1 we can construct more than one triangle because the measurements of the triangle ABC are AB = 7 cm, BC = 8 cm, and DE = 4.6 cm, EF = 6 cm, and AC = 11 cm.
Math Journal
Use your results to decide whether you can always construct exactly one triangle from the given information. Justify your answer.
a) Given three side lengths
Answer:
b) Given two side lengths and an angle measure
Answer:
Find the number of triangles that can be constructed. Try constructing the triangles to make your decision.
Question 8. Math Journal
Is it possible to construct a triangle PQR in which PQ = 12 cm, PR = 5 cm, and QR = 4cm? Explain.
Answer:
We cannot construct a triangle PQR in which PQ = 12 cm, PR = 5 cm, and QR = 4cm because the lengths of the PR , QR are very small.
Explanation:
Question 9.
Three triangles have angle measures of 500 and 600. In one triangle, the side included between these angles is 2 centimeters. In the second triangle, the included side length is 3 centimeters, and in the third triangle, the included side length is 4 centimeters.
a) Construct the three triangles.
Answer:
b) In each triangle, what is the measure of the third angle?
Answer:
Sum of angles of triangle is 180 so the given triangles have angle measures of 50 degree and 60 degree another angle is 70 degree.
c) Using the triangles constructed to help you, what can you deduce about the number of triangles that can be constructed if you are given three angle measures of a triangle but not the measure of any side length?
Answer:
Given only three angle measures of a triangle we can construct an infinity of triangles, their sides being proportional.
Question 10. Math Journal
Suppose you are given three angle measures whose sum is 180°, Can you form a triangle given this information? Are there other different triangles you can form? Explain.
Answer:
The angle sum property of a triangle states that the angles of a triangle always adds upto 180°. | 677.169 | 1 |
1. Question
What is the area of triangle ALC in the figure given below?
a. 2.5
b. 3
c. 3.5
d. 4
e. 4.5
Correct
Incorrect
Question 2 of 10
2. Question
Alan has two more than twice as many chocolates as does Alice, and half as many chocolates as does Nadia. If Alice has 'a' number of chocolates, then in terms of 'a', how many chocolates do Alan, Alice and Nadia have?
a. 2a+ 4
b. 5a+5
c. 6a+7
d. 7a+6
e. 9a+4
Correct
Incorrect
Question 3 of 10 10
4. Question
The width of a rectangle is 2/3 times its length. If the length is calculated to be 9, what is the value of perimeter for this rectangle?
a. 36
b. 9
c. 12
d. 54
e. 30
Correct
Incorrect
Question 5 of 10
5. Question
A line l is parallel to the y-axis and passes through the point (2,3). What is its gradient (m) and x-intercept?
a. m= 0 , x= (3,0)
b. m= ∞ , x= (2,0)
c. m= 0 , x= (2,0)
d. m= ∞ , x= (3,0)
e. m= 2 , x= (0,0)
Correct
Incorrect
Question 6 of 10
6. Question
What is the equation of the new parabola created by shifting y = x2 , three units in the positive y-axis direction?
a. y = (x+3)2
b. x2
c. x2 + 3
d. x2
e. x3
Correct
Incorrect
Question 7 of 10
7. Question
A sphere with diameter 1 unit is enclosed in a cube of side 1 unit each. Find the unoccupied volume remaining inside the cube. | 677.169 | 1 |
Elements of Geometry: Containing the First Six Books of Euclid: With a Supplement on the Quadrature of the Circle and the Geometry of Solids ... 44 ... ABCD is equal to the parallelogram EBCF . If the sides AD , DF of the paral- lelograms ABCD , DBCF opposite to the base BC be terminated in the same point D ; it is plain that each of the parallelograms is double ( 34 . 1. ) of the ...
УелЯдб 45 ... ABCD is equal to the parallelogram EBCF . Therefore , parallelograms upon the same base , & c . Q. ED . PROP . XXXVI . THEOR . Parallelograms upon equal bases , and between the same parallels , are equal to one another . A DE H Let ABCD ...
УелЯдб 47 ... ABCD and the triangle EBC be upon the same base BC , and between the same pa- rallels BC , AE ; the parallelogram ABCD is double of the triangle EBC . A Join AC ; then the triangle ABC is equal ( 37.1 . ) to the triangle EBC , because ...
УелЯдб 48 ... ABCD , and are there- fore called the complements : The complement BK is equal to the com- plement KD . H E K F B G Because ABCD is a parallelogram , and AC its diameter , the triangle ABC is equal ( 34. 1. ) to the triangle ADC And ...
УелЯдб 49 ... ABCD be the given rectilineal figure , and E the given rectili- neal angle . It is required to describe a parallelogram equal to ABCD and having an angle equal to E. G Join DB , and describe ( 42. 1. ) the. OF GEOMETRY . BOOK I. 491.
ДзмпцйлЮ брпурЬумбфб
УелЯдб 153 - If from the vertical angle of a triangle a straight line be drawn perpendicular to the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle...
УелЯдб 19УелЯдб 18 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle ; and the straight line which stands on the other is called a perpendicular to it.
УелЯдбУелЯдб | 677.169 | 1 |
RGPV Engineering Graphics PYQs
Q1. What are Scale? Classify its different types, also describe how RF is calculated.
Q2. A rectangular plot of land arca 0.45 hectare, is represented on a map by a similar rectangle of 5 sq. cm. Calculate the RF of the map, Also draw a scale to read up to single metre from the map, the scale should be long enough to measure up to 400 metres.
Q3. An inelastic string is unwound to a length of 122 mm from a drum of 30 mm. Draw the locus of free end of the string which is held taut during unwinding.
Q4. The projections a'b' and ab of a line AB are 65 mm and 50 mm long, respectively. The midpoint of the line is 38 mm in front of VP and 30 mm above HP. End A is 10 mm in front of the VP and nearer to it. End B is nearer to the HP. Draw the projections of the line, find its true length.
Q5. A regular hexagonal lamina 40 mm side has a square hole of 25 mm side centrally cut through it. Draw its projections when it is resting on one of its sides on HP with its surface inclined at 60° to VP and its corner nearest to VP is 24 mm from VP.
Q6. A triangular prism of side of base 30 mm and axis 55 mm long lies on one of its rectangular faces in HP with its axis parallel to VP. Draw its Projection.
Q7. A right regular square pyramid, edge of base 35 mm and height 50 mm, rest on its base on HP with its base edges equally inclined to VP. A section plane perpendicular to VP and inclined to HP on 32°, cuts the pyramid bisecting its axis, Draw the projections and true shape of the section of truncated pyramid.
Q8. Develop the lateral surface of an oblique cone, diameter of the base 40 mm and height 40 mm having its axis inclined at 60° to its base.
Q9. A right circular cone, diameter of base 50 mm and axis 62 mm long, rest on its base rim on HP with its axis parallel to VP and one of the elements perpendicular to HP. Draw the projections.
Q10. A cube 30 mm edge is placed centrally on the top of a cylindrical block of 52 mm and 20 mm height. Draw the isometric drawing of the solid.
Q13. An area of 144 sq cm on a map represents an area of 36 sq km on the field. Find the RF of the scale of the map and draw a diagonal scale to show km, hectometres and decametres and to measure up to 10 km. indicate on the scale a distance 7 km, 5 hectometres and 6 decimeters
Q14. Construct a conic when the distance of its focus from its directrix is equal to 50 mm and its eccentricity is 2/3. Name the curve; mark its major axis and minor axis. Draw a tangent at any point, P on the curve.
Q15. The length of the front view of a line CD which is parallel to HP and inclined 30° to VP is 50 mm. The end of the line is 15 mm in front of VP and 25 mm above HP. Draw the projections of the line and find its true length.
Q16. A line PS 65 mm has its end P, 15 mm above the HP and 15 mm in front of the VP. It is inclined at 55° to the HP and 35° to the VP. Draw its projections.
Q17. Draw the isometric view of a hexagonal prism having side of base 25 mm and axis 65 mm long resting on its base on HP.
Q18. A right circular cone of axis height 80 mm is resting on one of its generators in HP. Draw its projections. The base is 40 mm dia.
Q19. Explain the different methods used for drawing a circle in AutoCAD.
Q20. Give some examples where the layering concept is useful to use.
Q21. Name and explain five edit commands used in CAD.
Q22. Explain the various advantages of CAD.
Q23. What is the use of UCS icon ? Explain in detail?
Q24. Write about Dialog boxes and windows in CAD software.
Q25. Explain the various commands used for transformation of an object. i) Move ii) Copy iii) Rotate iv) Mirror
Q26. Explain the different methods used for drawing a circle in AutoCAD.
Q28. Construct a forward reading vernier scale to read distance correct to decimetre on a map in which the actual distances are reduced in the ratio of 1: 40000. The scale should be long enough to measure upto 6 km. Mark on the scale a length of 3.34 km and 0.59 km.
Q29. Construct a hypocycloid, rolling circle 50 mm diameter and directing circle 175 mm diameter. Draw a tangent to it at a point 50 mm from the centre of the directing circle.
Q30. A line AB of 70 mm long has its end A at 10mm above H.Pand 15 mm in front of V. P. Its front view and top view measure 50 mm and 60mm respectively. Draw the projections of the line and determine its inclinations with H.P. and V.P.
Q31. A line AB, 60mm long has its end A 15 mm above HP and 10mm in front of V. P. It is inclined at 45° to the HP and 30° to V.P. Draw it's projections.
Q32. Draw the isometric projections of the frustum of a cone of 50mm base diameter, 25mm top diameter and 60mm height.
Q33. Draw the isometric view of a hexagonal prism having side of base 25mm and axis 65mm long resting on its base on HP.
Q34. Explain the layering concept with examples. How is it implemented into CAD software?
Q35. Differentiate between the first angles and third angles projection.
Q36. How do you specify a plotter for graphics applying ?
Q37. What is CAD ? Name two CAD Softwares. Give advantages and disadvantages of using CAD. | 677.169 | 1 |
Geometry : Possible polygonal arrangements around a vertex
Created on July 17th, 2022. Last update on July 17th, 2022.
Introduction
In my research about polyhedron with convex regular polygons as faces (and that do not self intersect), I have found myself going back to 2D space because everything is
simpler there. The original question was about what combination of faces can one put around a single vertex ?
The minimum number of faces, the valence of this vertex, is 3 which you can have in a tetrahedron, a cube or a dodecahedron.
We don't consider the flat degenerate case with two of the same face one top of the other.
Regarding the upper limit for the vertex valence, since we allow concave polyhedron,
there is nothing preventing putting an arbitrary large number of equilateral triangles around a single vertex,
the line of the exterior edges forming a zigzag, because the faces and edges have no thickness.
Though, it is not proven yet that such a vertex would exist in an actual polyhedron.
Since the problem looks not so easy to me, I have tried to tackle it in the 2D plane.
The extra condition of flatness makes things much simpler.
So, the question that arose was: What combination of convex polygon can one put
around a vertex side to side ? This may not be so clear because the phrasing is poor, so let's have
an example. I have a pentagon, a square and two equilateral triangles: can I put them together
around a vertex ? The answer is yes, the configuration is illustrated in Figure 1. The total angle
is 108 + 90 + 2 x 60 = 318 degrees which is smaller than 360 degrees.
From this example, the set of rules for the general problem is the following:
all polygons are convex and regular with unit length
all polygons share a common vertex,
two polygons are connected by a single edge,
an edge can only belong to one or two polygons,
any two polygons cannot cross each other or be included into one another.
I should also mention that the order of the polygons around the vertex is not relevant.
Figure 1: Illustrative example of a valid configuration with a pentagon, a square and two equilateral
triangles.
A little bit of math
The first question that comes to mind is about the relationship between
the number of vertices n (an integer larger than 3) of a polygon and its internal angle beta.
From the drawing in Figure 2, the triangle is a sector of the polygon, so we have the following
equations:
α = Π⁄n
and α+β⁄2 = Π⁄2
so β = Π(n-2)/n
Note that, for all n ≥ 3, β ≤ Π and that for n tending to infinity, β tends to Π which is the angle
of the tangent of a circle. The polygon tends to a circle with an infinite radius because the edge
length is 1. Here is a table of the first few values that will be handy later:
Figure 2: Visual definition for the parameters of the problem.
Vertex number n
Angle β in radians
Angle β in degrees
3
1⁄3 Π
60
4
1⁄2 Π
90
5
3⁄5 Π
108
6
2⁄3 Π
120
7
5⁄7 Π
182.57...
8
3⁄4 Π
135
9
7⁄9 Π
140
10
4⁄5 Π
144
11
9⁄11 Π
147.27...
12
5⁄6 Π
150
13
11⁄13 Π
152.31...
18
8⁄9 Π
160
20
9⁄10 Π
162
24
11⁄12 Π
165
42
20⁄21 Π
171.43...
Table 1: Internal angle of regular convex polygons.
Defect Angle
From the set of rules, the condition of a valid set of N polygons around a vertex is the following:
the sum of the internal angle βi of the polygons Pi (i going from 1 to N) must be smaller or equal
than 2Π : sum(βi) ≤ 2Π.
The defect angle is the difference between the full 2Π angle and the sum, it must remain positive :
D = 2Π - sum(βi) ≥ 0.
For simplicity, we order the polygons by their vertex count ni in ascending order so that:
n 1 ≤ n2 ≤ ... ≤ n N-1 ≤ n N
Using the previous equation and after dividing by Π, we then have:
2 - sum( (ni - 2) / ni) ≥ 0
At this point, we can solve this inequation for the values ni. Solutions
will be noted by the sequence of vertex number between brackets: {n1, n2, ..., nN-1, nN}.
The solutions will sometimes be accompanied by the defect angle D.
Results
There is a total of six classes from 1 to 6, the class number being the valence of the vertex.
Class VI : Valence of 6
Since the smallest angle we can have is that of the equilateral triangle, Π/3,
the largest valence is (2Π)/(Π⁄3) = 6.
The only solution is : {3, 3, 3, 3, 3, 3}, D = 0.
Class II : Valence of 2
Class I : Valence of 1
Any single polygon is a valid solution. The defect angle is:
D = Π(n1 + 2)/n1
Figure 3: All families and their relationship in details. Color indicates infinite families.
Application to 3D space
Now, that we now all of that, what about 3D ?
First, with the solutions we already have when the valence is 3 or more, by moving the shape to 3D and closing
the gap between polygons where the defect angle lies, we form a local convex shape at this vertex.
Consequently, any convex polyhedron (even those not strictly convex) with regular faces must have their vertices in this set.
For instance, the icosahedron only has {3,3,3,3,3} vertices.
Second, there is more degrees of freedom in 3D than in 2D. In the 2D study, we discarded the cases where the polygons would overlap.
In these cases, we can define the defect angle to be negative.
When moving the shape to 3D, it becomes possible to glue the edges of the polygons at the two extremities
together resulting in a vertex where faces necessarily form both mounts and valleys, folds and creases.
In my future work, I will study what kind of configurations exists at a 3D vertex which implies explaining dihedral and solid angles.
References
Nothing found on the subject, please let me know if you have anything. | 677.169 | 1 |
Basically, this is a parallelogram: 1. In a parallelogram any two opposite sides are equal.
2. In a parallelogram any two opposite angles are equal.
3. In a parallelogram the diagonals bisect each other.
4. In a parallelogram, each diagonal divides it into two congruent triangles.
5. In a parallelogram, any two adjacent angles have their sum equal to 180"a, that is, they are supplementary.
If you go to this site you will find all about parallelograms.
Lynnie | 677.169 | 1 |
Linear Pair of AnglesDefinition, Axiom, Examples
The linear pair of angles is an important concept in geometry. With so many real-world uses, you'd be surprised to figure out how relevant this figure can be. While you might think it has no use in your life, we all should understand the ideas to nail those examinations in school.
To save your time and make this information easily accessible, here is an introductory insight into the properties of a linear pair of angles, with images and examples to guide with your personal study sessions. We will also talk about some real-world and geometric applications.
What Is a Linear Pair of Angles?
Linearity, angles, and intersections are concepts that remain to be relevant as you go forward in geometry and more complex theorems and proofs. We will answer this question with a simple explanation in this unique point.
Explanation
A linear pair of angles is the name given to two angles that are located on a straight line and have the sum of their angles measure 180 degrees.
To put it easily, linear pairs of angles are two angles that are adjacent on the same line and pair up to form a straight line. The total of the angles in a linear pair will always produce a straight angle equivalent times to 180 degrees.
It is essential to keep in mind that linear pairs are at all times at adjacent angles. They share a common apex and a common arm. This suggests that they always form on a straight line and are at all times supplementary angles.
It is important to clarify that, although the linear pair are constantly adjacent angles, adjacent angles never constantly linear pairs.
The Linear Pair Axiom
Through the definition clarified, we will study the two axioms critical to fully comprehend any example provided to you.
Let's start by defining what an axiom is. It is a mathematical postulate or hypothesis that is accepted without proof; it is deemed clear and self-evident. A linear pair of angles has two axioms linked with them.
The first axiom establishes that if a ray is located on a line, the adjacent angles will form a straight angle, making them a linear pair.
The second axiom establishes that if two angles makes a linear pair, then uncommon arms of both angles produces a straight angle between them. This is commonly called a straight line.
Examples of Linear Pairs of Angles
To visualize these axioms better, here are a few diagram examples with their respective answers.
Example One
As we can see in this instance, we have two angles that are next to each other. As you can see in the diagram, the adjacent angles form a linear pair due to the fact that the total of their measurement equals 180 degrees. They are also supplementary angles, because they share a side and a common vertex.
Angle A: 75 degrees
Angle B: 105 degrees
Sum of Angles A and B: 75 + 105 = 180
Example Two
In this example, we possess two lines intersect, making four angles. Not all angles creates a linear pair, but each angle and the one close to it makes a linear pair.
∠A 30 degrees
∠B: 150 degrees
∠C: 30 degrees
∠D: 150 degrees
In this case, the linear pairs are:
∠A and ∠B
∠B and ∠C
∠C and ∠D
∠D and ∠A
Example Three
This example represents a common point of three lines. Let's look at the axiom and properties of linear pairs.
∠A 150 degrees
∠B: 50 degrees
∠C: 160 degrees
None of the angle combinations add up to 180 degrees. As a consequence, we can come to the conclusion that this image has no linear pair until we expand one straight line.
Applications of Linear Pair of Angles
Now that we have gone through what linear pairs are and have observed some cases, let's understand how this theorem can be implemented in geometry and the real world.
In Real-World Situations
There are several implementations of linear pairs of angles in real-world. One familiar case is architects, who utilize these axioms in their daily work to check if two lines are perpendicular and form a straight angle.
Construction and Building professionals also employ masters in this matter to make their work less complex. They use linear pairs of angles to make sure that two adjacent walls form a 90-degree angle with the floor.
Engineers also uses linear pairs of angles frequently. They do so by calculating the pressure on the beams and trusses.
In Geometry
Linear pairs of angles also play a role in geometry proofs. A common proof that uses linear pairs is the alternate interior angles theorem. This theorem states that if two lines are parallel and intersected by a transversal line, the alternate interior angles formed are congruent.
The proof of vertical angles also depends on linear pairs of angles. Even though the adjacent angles are supplementary and sum up to 180 degrees, the opposite vertical angles are at all times equal to one another. Because of these two rules, you are only required to figure out the measure of one angle to determine the measure of the rest.
The concept of linear pairs is also utilized for more sophisticated implementation, such as working out the angles in polygons. It's critical to grasp the basics of linear pairs, so you are ready for more complex geometry.
As you can see, linear pairs of angles are a somewhat easy concept with few interesting applications. Later when you're out and about, see if you can spot some linear pairs! And, if you're taking a geometry class, bear in mind how linear pairs may be useful in proofs.
Level Up Your Geometry Skills with Grade Potential
Geometry is entertaining and beneficial, majorly if you are curious about the field of construction or architecture.
However, if you're struggling to understand linear pairs of angles (or any other concept in geometry), consider signing up for a tutoring session with Grade Potential. One of our experienced tutors can assist you comprehend the topic and ace your next exam | 677.169 | 1 |
Find the length of arc of a circle of radius 10 in. that is intercepted from the
Question:
Find the length of arc of a circle of radius 10 in. that is intercepted from the center of the circle by an angle of 3 radians.
Fantastic news! We've Found the answer you've been seeking!
Step by Step Answer:
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The length of an arc of a circle is given by the formula length of arc radius x angle i | 677.169 | 1 |
Projective Geometry by Nigel Hitchin
Description: Projective geometry started life over 500 years ago in the study of perspective drawing: the distance between two points on the artist's canvas does not represent the true distance between the objects they represent so that Euclidean distance is not the right concept69 | 677.169 | 1 |
Let three equal lines , , and be drawn Antiparallel to the sides
of a triangle so that two (say and ) are on the same side of the third line as . Then
is an isosceles Trapezoid, i.e., , , and are parallel to the respective
sides. The Midpoints, , and of the antiparallels are on the respective symmedians and divide them
proportionally.
If divides in the same ratio, , , are parallel to the radii , ,
and and equal. Since the antiparallels are perpendicular to the symmedians, they are equal chords of a circle
with center which passes through the six given points. This circle is called the Tucker circle. | 677.169 | 1 |
There are many ways to figure out the meanings of unknown words.
Similar figures worksheet answers fill in the blank. An exploratory paper is not uncommon in businesses when they re trying to have a remedy to a matter and will have to receive all the feasible perspectives and data available. You have all your materials. Fill in the blank with the appropriate word phrase or symbol to make a true statement.
The following is a rich task to help you get ready for soccer. Solving proportions involving similar figures each pair of figures is similar. Scale factor 2.
Worksheet shapes and figures. Similar to the above listing the resources below are aligned to related standards in the common core for mathematics that together support the following learning outcome. As a matter of fact you may find that you can also come up with some better ideas for your projects.
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Take the facts below and answer the questions on the back of this sheet. Fill in the blank and state the scale factor. Synonyms and antonyms printable worksheets.
Find the missing side. Using the similar figure worksheet answers are going to give you the ability to find all the data fast. T worksheet by kuta software llc kuta software infinite pre algebra name similar figures date period each pair of figures is similar.
Similar figures worksheet name. Your child will build on words they already know by discovering words that are similar or opposite. The symbol means is similar to and the symbol is the abbreviation for the word.
Similar figures have the same but not necessarily the same. Similar triangles 2 of 2 similarity congruence and transformations. Similar figures worksheet answer key along with magnificent math 10 foundations and precalculus worksheets frieze.
Rich task floodlights. Through this worksheet students record unknown words and find out their meaning using a variety of strategies. Similar figures similar figures similar triangle methods notes and homework this is the second set of notes for my similar figures unit for a high school geometry | 677.169 | 1 |
UNIT ACTIVITIES
2. Draw the outlines of the object or figure on the glass and transfer that drawing to tracing paper.
3. Transfer tracing paper copy to paper and either finish the work as a drawing or a painting.
If possible, allow the students to construct the mechanical aid so that they can acquire some skill in transforming plans into three-dimensional constructions.
Albrecht Durer, Treatise on Perspective, Nuremberg, 1525.
(figure available in print form)
Anamorphic Art
1. Place a grid over either an original drawing or a reproduction or magazine photo.
2. Transfer to an enlongated grid at least three times longer than original drawing.
The earliest example of anamorphic art appears in the Notebooks of Leonardo Da Vinci.
Anamorphic art pictures have to be viewed on edge. See example below. Anonymous, (edited by V. Decugis). Perspective anamorphoses: A Violinist; A Cellist. 1868.
(figure available in print form)
Leonbattista Alberti's Method for Obtaining a True Perspective.
(figure available in print form)
To Obtain Orthogonals:
1. Draw a figure.
2. Divide figure into thirds.
3. Draw a rectangle, here after called X. Mark off segments along base line. Label segments A - G. The length of each segment is determined by the measurement of one third of figure. (In this example the figure is divided into half inch segments so the distance between A - B, B - C, etc. is a half inch).
4. Determine the vanishing point. The vanishing point can be located anywhere in X. The only restriction is that its distance from the base line must be proportional to the divisions of the figure. (In this example, three half inch segments or one and a half inches above base line of X). Vanishing point called V.
5. Draw lines from V to A - B - C, etc. Orthogonals are now established.
To Obtain Transversals:
1. On a separate sheet of paper, draw a line. Mark off segments A' - G'.
2. Draw another line (HT) above and parallel to this line that is equal to the distance from V to base of X.
3. Draw line A' Z.
4. On HT at Z mark point DI that equals height of figure.
5. Draw lines from DI to A' - G' to obtain points a - b on A' Z.
6. Bring A' Z to edge of X and mark off points a - g on bith sides of X.
7. Draw transversals on X that match a to a, b to b, etc.
(figure available in print form)
Now the angles established on X have created an imaginary space of three-dimensional quality, like an empty stage, that can be filled with all kinds of buildings, people, and objects. Moreover, the finished drawing will look like it is very deep in depth, something I feel very much out of in trying to write down Alberti's method. | 677.169 | 1 |
TACHEOMETRY (from Gr. ταχύς, quick; μέτρον, a measure), a system of rapid surveying, by which the positions, both horizontal and vertical, of points on the earth's surface relatively to one another are determined without using a chain or tape or a separate levelling instrument. The ordinary methods of surveying with a theodolite, chain, and levelling instrument (see Surveying) are fairly satisfactory when the ground is pretty clear of obstructions and not very precipitous, but it becomes extremely cumbrous when the ground is much covered with bush, or broken up by ravines. Chain measurements are then both slow and liable to considerable error; the levelling, too, is carried on at great disadvantage in point of speed, though without serious loss of accuracy. These difficulties led to the introduction of tacheometry, in which, instead of the pole formerly employed to mark a point, a staff similar to a level staff is used. This is marked with heights from the foot, and is graduated according to the form of tacheometer in use. The azimuth angle is determined as formerly. The horizontal distance is inferred either from the vertical angle included between two well-defined points on the staff and the known distance between them, or by readings of the staff indicated by two fixed wires in the diaphragm of the telescope. The difference of height is computed from the angle of depression or elevation of a fixed point on the staff and the horizontal distance already obtained. Thus all the measurements requisite to locate a point both vertically and horizontally with reference to the point Where the tachometer is centred are determined by an observer at the instrument without any assistance beyond that of a man to hold the staff.
The simplest system of tacheometry employs a theodolite without additions of any kind, and the horizontal and vertical distances are obtained from the angles of depression or elevation of Sublease two well-defined points on a staff at known heights from method. the foot, the staff being held vertically . In fig. 1 let T be the telescope of a theodolite centred] over the point C, and let AB be the staff held truly vertical on the ground at A. Let P and P′ be the two well-defined marks on the face of the staff,
Fig. 1.
Fig. 1.
both of them at known heights above A, and enclosing a distance PP′=s between them. Let α and β be the measured angles of elevation of P and P′, and let d be the horizontal distance TM of the staff from the theodolite, and h the height PM of P above T. Then since
P′M =d tan β and PM =d tan α,
we have s=P′M-PM=d(tan β-tan α).
Therefore d = s/tan β − tan α; h= s tan α/tan β − tan α.
If TC, the height of the rotation axis of the telescope above the ground, =q, and if AP =p, then the height of A above C is h−p+g. If, as is usually the case, a number of points are determined from one station of the theodolite, and hi, ha, ha, &e., be the values of 11 for the different points Al, AZ, Aa, &e., then the difference of level of A1 and A2 will be hg-hi, that of A1 and A3 will be h3−h1, and so on. To ensure the essential condition that the staff is held vertical, it is usually provided with a small circular spirit-level, and the staff-holder must always keep the bubble in the centre of its run. No graduation of the staff is required beyond two well-defined black lines across the white face at P and P′, but the marks can be; very usefully supplemented by wings fastened on the two }
sides of the staff, having their tops at right angles to the staff, at the same height as the points P and P', and forming a continuation of the black lines. A convenient length for the staff is 12 ft., with the point P 2 ft. from the foot, and the point P' at the top of) the staff, so that A s=10 ft.
With the above arrangement the staff can easily be read with a 5-inch theodolite at half a mile distance. But while it is frequently very useful to determine approximately points a long way off, the determinations will not be nearly so accurate as those of near points. Thus suppose that the distance of the staff is d, and the intercept on the staff is's, and suppose that the personal and instrumental error is 5a. (ar being the angle subtended by's at the telescope); then since d = s/tan α, d(d)/dα = −s1+tan2α/tan2α, or . .
da T S 5"/dz " § (S2+d2)
Therefore δd, the distance error, is given by the equation
δd=−δα(s2+d2)/s. But at distances of 5 chains or more s2 will be very small compared with d2 and may be neglected, so that
δd =-51.112/5. Since δα may be considered as constant for all distances where the staff can be distinctly read, the distance error increases as the square of the distance. With small theodolites, where special care has not been given to the graduating and reading of the vertical circle,
δα will probably amount to about 20″, At a quarter of a mile excellent work can be done. In carrying on a traverse line by this method with stations 10 or 12 chains apart, the theodolite being set up at points about midway between the stations, the probable distance error in a mile is about 31/2 ft., and the probable level error about 4 in. In 25 miles these probable errors would correspond to about 18 ft. and 20 in. respectively. This system of tacheometry is well adapted for distant readings, and from the great simplicity of the observations there is little likelihood of errors in the field. But the reduction work is rather heavier than is the case with some of the tachometers described below. Since the accuracy of the method depends entirely upon the accuracy with which the vertical angles are measured, it is advisable that the vertical circle should be as large as possible, very finely and accurately divided, and fitted with good verniers and microscopes.
In Eckhold's omnimeter the vertical circle of the theodolite is dispensed with, and a saving of reduction work is effected by reading, not the vertical angles themselves, but the tangents of the angles.
In the Ziegler-Hager tachograph the tangents are read not horizontally but vertically, and the arrangement is as follows:— In fig. 2 O is the axis of rotation of the telescope; mn is the axial line of a steel bolt, which carries on its top a knife-edge, on which the telescope rests by means of an agate plate. The bolt is carried by a slide in which it can be raised or lowered by a micrometer screw fitted with a graduated head. The slide plays between the vertical cheeks of a standard rigidly attached to the frame of the instrument, and it can be raised or lowered by a rack and pinion. The telescope, which rests on the knife-edge, follows the movement of the bolt. The slide carries on one side a vernier by which to read the divisions on a scale fixed to one of the vertical legs of the standard, and the zero point o of the scale is the point where the horizontal plane through O cuts the scale when the plane-table or upper plate of the theodolite is truly level. The scale is graduated in divisions, each of which is the T2, ¢@th part of the distance Oo, or h. The head of the micrometer screw which raises or lowers the steel bolt in the slide is graduated with a zero mark and with marks corresponding to a vertical movement of the knife-edge of 1/100h, 1/50h, &c. The instrument is used as follows:-Let AB be the surface of the ground, and BC a staff held vertically at B, and let CB be produced to meet the horizontal line through O in M. Let the head of the micrometer screw be turned till the zero division is exactly under the pointer. Let p be the zero division on the staff, and let the slide and bolt be raised by the rack and pinion movement till the axis of the telescope is directed towards p. Let v be the point where the line Op cuts mn, and let the tangent reading ov be taken on the scale. Then let the telescope be lowered by the micrometer screw in the slide till the division on the head of the screw marked I is exactly under the pointer; the knife-edge of the bolt has then been lowered through a distance 'vt equal to h/100. Let q be the point on the staff where the line Ot cuts it, and let the reading at q be taken. Then since the triangles between O and mn and O and CM are similar to each other, and vtt is 1/100th of Oo, therefore pq will be 1/100th of OM, or OM =100 × pq. This gives the horizontal distance of the staff from O, and the vertical distance pM of p above O is OM tan MOp=OM × ov/Oo, and since ov has been read in
Fig. 2.
Fig. 2.
parts of which Oo contains 100, the distance pM is readily obtained. If the difference of elevation of B and A be required, the height pM must be reduced by 12B and increased by OA, both known quantities. By this arrangement the reduction work of the observations is rendered extremely simple, and can readily be performed ingfie held. The instrument is well adapted for use with the plane-table.
Tacheometers in which the horizontal distance of the staff from the telescope is deduced from the readings of the staff indicated by two fixed wires in the diaphragm of the telescope will S dl now be considered. In fig. 3 BC is a diaphragm fixed Stadia method.
in a tube having fine horizontal wires at B and C. Let the end E of the tube be closed by a disk which has a minute hole at E, to which the eye can be applied. If P and D be the oints on a vertical staff at which the lines EB and EC are observed to cut the staff, so that the intercept PD is known, then from similar triangles ED = (EC/BC)PD, and since EC and BC are constant, ED varies as PD. If, for instance, PD has a certain observed value when the staff is held at a certain distance ED, and has exactly half that value when the staff is held at another distance ED′, then the distance ED′ is one-half of the distance ED, and so on in proportion. The distance ED can be instantly inferred from the readings of the staff, if the latter be suitably graduated. If, for example, it be desired to know the distance ED in yards, and by construction the proportion EC/BC=50, then the intercept on the staff at 1 yard from E would be 1/50th of a yard, or ·72 inch, the intercept at 2 yards from E would be 2X·72 inches, and so on. If therefore the staff be graduated with divisions of ·72 inch, and the intercept be 45 of such divisions, it would be inferred that the distance of the staff from E was 45 yards. The constant proportion EC/BC can be checked by measuring 100 yards from E and observing whether
the intercept is exactly 100 divisions or not.
Fig. 3.
If it is not, the wire diaphragm must be shifted in the tube until it is. In figs. 3, 4, 5 and) 6 the distances are deduced from the readings of a central wire in the optical axis of the telescope and of a wire above it, for the sake of simplicity. The usual arrangement is to fit the diaphragm with a central wire and with one or two wires above and below it at equal distances from the central wire. The vertical angle of depression or elevation is fixed by directing the central wire to a well-defined division on the staff, and the distance of the staff is inferred from the readings given by the corresponding wires above and below the central wire.
The elementary form of tachometer given above illustrates the general principle of the class of tachometers now under consideration, and as leading up to the practical form, in which the staff is viewed with a telescope mounted in the manner of a theodolite. The simplest form is Reichenbach's tachometer, which may be investigated as follows:—In fig. 4 let A be the object glass by which an image of the staff ST is formed at HK. The wire diaphragm is moved in the tube so as to coincide with the image,
Fig. 4.
and the image and wires are viewed with an eye-piece (not shown) in the usual way. Let O be the point where the vertical axis of the instrument cuts the axis of the telescope, the instrument being centred over a pe, from which the distance to the staff is required. The object glass fiif focal length = 𝑓) is at a distance 𝑐 from O. Let AT=𝑢 and AH=𝑣, and the angle SAT=HAK=θ. Then if 𝑖 be
the height of the image HK, 𝑖=𝑣 tan θ. Since 1/𝑣+𝑖/𝑢=1/𝑓 we have 𝑣= -𝑢𝑓/(𝑢−Q, and hence i=uf tan θ/(𝑢−𝑓). Let F be some point on AT such that AF =𝑥 and FT =𝑢′. And let the angle SFT =φ. Then 𝑢 =𝑢′+𝑥 and tan θ=𝑢′ tan φ/(𝑢′+𝑥), and therefore
i= (𝑢′+𝑥)𝑓/𝑢′+𝑥−𝑓 {=u, +x fé. tan¢=$tan 4>; and, if x=f, i=f tan ¢.
If therefore the point F be taken at a distance 𝑓 from the object glass, every intercept of the staff for positions between T and F, such as S′T′, S″T″, &c., which are bounded by the V line FS, and for which consequently θ is the same, will have the same height of image at the diaphragm. Conversely, if K be a wire in the diaphragm it will cut the image of the staff for all positions of the staff between T and F in points H object glass, were it not that the rays, after passing through the object glass, are received by the anal lat tic lens and the image of the staff is formed at K on the wire diaphragm, which is slid in the tube till it coincides with the position of the image. The image at K is viewed by an eye-piece in the usual way. Let T be the point where the image of the staff is cut by the central wire of the diaphragm, and S the point where the image is cut by one of the outer wires of the diaphragm. If 0 and ¢ be the angles subtended by ST at the object glass and at the point O respectively, and if i be the height of the image at K, h the height of the virtual image at H, then by elementary geometry and from optical considerations, we obtain i= f 1/f'f". » . -tan4>
"'(fi-d-hfz)"16f1'(C+f1)ld"J'2>i
Let fz be made such that of, -(c-{-fl) (d-f2)=0, the equation of condition above mentioned. Then fg ={d(c-l-fl) -cf1}/(C-|-fl). And i= .tan ¢= - tan ¢.
that lie on the line FS. Now the intercept S"T", half-way, L between F and T, will be one-half of ST, and therefore if the reading on the staff indicated by the wire in question be one-half of ST, it may be inferred that the position of the staff is half-way between F and T, and similarly for other distances. If the distance of ST from O is required, as is usually the case, a quantity f+c must be added to every distance from F determined as above.
It is very seldom that the line of sight AT of the telescope is at right angles to the staff. In general it is more or less inclined to the staff, which is almost always held vertical, and the horizontal and vertical distances of the staff from the axis of rotation of the telescope are found thus:—In fig. 5 let ST be the observed intercept on the staff when the telescope is inclined at an angle 𝑎 to the horizontal. Draw TS′ at right angles to OT. The angle TS′S will be very nearly a right angle, and STS′ may be taken as equal to 𝑎. If there were 𝑛 graduations (each corresponding to 1 yard in distance) in ST, there would be 𝑛 cos 𝑎 graduations in
Fig. 5.
Fig. 5.
S′T, and therefore the distance of the staff from F, as inferred from the observed number of graduations in ST, must be multiplied by cos 𝑎. to give the true distance FT. Again FN =FT cos 𝑎, so that the distance inferred from the observed number of graduations in ST must be multiplied by cosza. to give the horizontal distance of F from T. To this must be added the distance OL=OF cos 𝑎(f+c) cos a to get the horizontal distance, OM, of O (the vertical axis of the instrument) from T. This value of OM must be multiplied by tan a to obtain the value of h, the vertical distance of T from O. Tables of the value of cos 𝑎, cos2 𝑎, and tan 𝑎 are necessary to facilitate these calculations.
In this tachometer the distances as inferred from the readings of the staff are the distances of the staff from F and not from O. This defect was remedied by Porro, who added a lens (called the anallattic lens) to the telescope. The arrangement of the telescope as manufactured by Messrs Troughton and Simms, is as follows:—In fig. 6 O is the point where the vertical axis of the instrument cuts the axis of the telescope. The object glass is fixed at a distance 𝑐 from O, and the anallattic lens at a distance 𝑑 from the object glass. The distances 𝑐 and 𝑑 are chosen to suit the constructive conveniences of the instrument. The diaphragm at K is movable so that it can be made to coincide with the image of the staff. The focal length 𝑓1 of the object glass is arbitrary, and the focal length 𝑓2 of the anallattic lens is determined from an equation of condition between 𝑐, 𝑑, 𝑓1, and 𝑓2. The image of the staff ST would be formed by the object glass at H, at a distance 𝑣1 from the
Fig. 6.
Fig. 6.
Therefore all the readings of the staff which would be given by the outer wire of the diaphragm will lie on the line OS (for all of which φ is the same), and the distance from O along OT will be proportional to the reading on the staff. Thus if the staff be suitably graduated, the distance from O can be immediately deduced from the reading. Also, as before, if the telescope be inclined at an angle at to the horizontal, the distance OT inferred from the number of graduations in ST must be multiplied by cos2α, to give the horizontal distance of O from T, and the horizontal distance so obtained must be multiplied by tan α. to obtain the vertical distance of T from O.
The inconvenience of the reduction work necessary to obtain the horizontal and vertical distances produced the Wagner-Feunel tachometer, by which the distances can be read directly from the instrument. As is seen from fig. 7, three scales are provided, to measure the inclined distance, the horizontal distance, and the vertical distance respectively. All three are arranged in a plane parallel to the plane in which the telescope turns. The inclined scale is attached to the telescope exactly parallel to its line of collimation, and moves with it. The horizontal scale is fixed to the upper horizontal plate of the theodolite. The vertical scale is on the vertical edge of a right-angled triangle, which can be slid along on the top of the horizontal scale. The inclined scale carries a slide which is provided with two verniers. One of these is parallel to the inclined scale, and is for the purpose of setting off on the scale (in terms of the divisions on the scale) the inclined distance of the staff from the axis of rotation of the telescope. The other turns on a pivot whose centre is accurately in the edge of the inclined scale at the point where the zero division of the inclined vernier
Fig. 7.
Fig. 7.
cuts the edge, and is for the purpose of reading the vertical scale; it can be turned on its pivot so as to be vertical whatever may be the inclination of the telescope. Moreover, since the distance from the centre of the pivot to the zero of the vernier is always constant and known, the vertical scale can be graduated so that the reading of the vernier gives the height (in terms of the division on the scale) of the staff above the axis of rotation of the telescope. The horizontal scale attached to the horizontal plate of the theodolite is read by means of a vernier carried by the triangle. To ascertain the horizontal and vertical distances of the point on the staff which is cut by the middle wire in the diaphragm of the telescope from the rotation axis of the telescope, the inclined distance of the point on the staff is read by means of the wires, as in Porro's tachometer. This distance (in terms of the divisions) is then set off on the inclined scale by means of the inclined vernier, and the vertical scale on the triangle is moved up to the vertical vernier, which is adjusted to its edge. With proper graduation of the horizontal and vertical scales the horizontal and vertical distances can be at once read off on the scales. This method, however, requires that the staff be held so that its face is perpendicular to the line of sight, which is more troublesome than holding the staff vertical. | 677.169 | 1 |
maureentan
NEED HELP ASAP 30 pointsIs △DBE similar to △ABC ? If so, which postulate or theorem proves these two...
4 months ago
Q:
NEED HELP ASAP 30 pointsIs △DBE similar to △ABC ? If so, which postulate or theorem proves these two triangles are similar? △DBE is similar to △ABC by the SAS Similarity Theorem . △DBE is similar to △ABC by the SSA Similarity Theorem . △DBE is similar to △ABC by the SSS Similarity Theorem . △DBE is not similar to △ABC .
Accepted Solution
A:
Answer:△DBE is similar to △ABC by the SAS Similarity Theorem .
Step-by-step explanation:Given triangles in figurewe have to tell that is the triangles are similar or if similar then by which postulate.In ΔDBE and ΔABC,[tex]\frac{DB}{AB} =\frac{EB}{CB}[/tex]⇒[tex]\frac{10}{10+15}= \frac{16}{16+40}[/tex]⇒[tex]\frac{2}{5}=\frac{2}{5}[/tex]Hence, the sides of triangle are proportional.And also ∠B=∠B (common) i.e the one angle congruentHence, △DBE is similar to △ABC by the SAS Similarity Theorem . | 677.169 | 1 |
Contrary to the popular belief, the majority of Pythagorean theorem problems are also encountered by the engineers and designers who are majoring in all the possible subjects. While there are questions that you have to master as a high school student, it's often necessary to come back to the basics and take a closer look at the answers that are provided. Likewise, when you know the theorem well and understand how it works, you can become a Pythagorean theorem solver based on these questions alone. Take your time to explore the provided examples and see how exactly these are formed. | 677.169 | 1 |
Explore Solids
Key Concepts
Introduction
Polyhedron
A polyhedron is a solid that is bounded by polygons called faces that enclose a single region of space. It is a three-dimensional solid made up of plane faces. Poly=many Hedron=faces.
An edge of a polyhedron is a line segment formed by the intersection of two faces of Explore Solids.
A vertex of a polyhedron is a point where three or more edges meet. The plural of polyhedron is polyhedra or polyhedrons.
Prism
Polyhedron with two parallel, congruent bases. Named after its base.
Pyramid
Polyhedron with one base and lateral faces. Named after its base.
Identify and name polyhedra
Example 1:
Decide whether the solid is a polyhedron. If so, count the number of faces, vertices, and edges of the polyhedron.
Solution:
This is a polyhedron. It has 5 faces, 6 vertices, and 9 edges.
This is not a polyhedron. Some of its faces are not polygons.
This is a polyhedron. It has 7 faces, 7 vertices, and 12 edges.
Use Euler's Theorem in a real-world situation
The sum of the number of faces and vertices is two more than the number of edges in the solids above. This result was proved by the Swiss mathematician Leonhard Euler.
As per Euler's Theorem,
The number of faces (F), vertices (V), and edges (E) of a polyhedron are related by the formula:
F + V = E + 2
Example 2:
The box shown at the right is a hexagonal prism. It has 8 faces. Two faces are hexagons, and 6 faces are squares. Count the edges and vertices. Use Euler's Theorem to check your answer.
Solution:
On their own, 2 hexagons and 6 squares have 2(6) + 6(4), or 36 edges. In the solid, each side is shared by exactly two polygons. So the number of edges is one half of 36, or 18. Use Euler's Theorem to find the number of vertices.
F + V = E + 2 Write Euler's Theorem.
8 + V = 18 + 2 Substitute values.
8 + V = 20 Simplify.
V = 12 Solve for V.
The box has 12 vertices.
Use Euler's Theorem with Platonic solids
Types of Solids
Of the first solids below. The prism and pyramid are polyhedra.
The cone, cylinder, and sphere are not polyhedra.
Regular/Convex/Concave
A polyhedron is regular if all its faces are congruent regular polygons. A polyhedron is convex if any two points on its surface can be connected by a segment that lies entirely inside or on the polyhedron.
If this segment goes outside the polyhedron, then the polyhedron is said to be NON-CONVEX OR CONCAVE.
Platonic solids
There are five regular polyhedra called Platonicsolids, after the Greek mathematician and philosopher Plato.
Find the number of faces, vertices, and edges of the regular tetrahedron. Check your answer using Euler's Theorem.
Solution:
By counting on the diagram, the tetrahedron has 4 faces, 4 vertices, and 6 edges. Use Euler's Theorem to check.
F + V = E + 2 Write Euler's Theorem.
4 + 4 = 6 + 2 Substitute values.
8 = 8 This is true statement. So, the solution checks.
Describe cross sections
Cross section
The intersection of the plane and the solid is called a cross section.
For instance, the diagram shows that the intersection of a plane and a sphere is a circle.
Example 4:
Describe the shape formed by the intersection of the plane and the cube.
Solution:
The cross section is a square.
The cross section is a pentagon.
The cross section is a triangle.
Exercise
____________________ is a three-dimensional solid made up of plane faces.
Decide whether the solid is a polyhedron. If so, count the number of faces, vertices, and edges of the polyhedron.
Determine if the following solids are convex or concave.
Describe the shape formed by the intersection of the plane and the closed cylinder.
Use Euler's Theorem to find the value of n.
Faces: n
Vertices: 8
Edges: 12
Use Euler's Theorem to find the value of n.
Faces: 8
Vertices: n
Edges: 18
Name the five Platonic solids and give the number of faces for each.
Find the number of faces, vertices, and edges of the polyhedron. Check your answer using Euler's Theorem.
Find the number of faces, vertices, and edges of the polyhedron. Check your answer using Euler's Theorem.
The speaker shown at the right has 7 faces. Two faces are pentagons, and 5 faces are rectangles.
Find the number of vertices.
Use Euler's Theorem to determine how many edges the speaker has.
Concept Map
What have we learned
Using properties of polyhedra.
Using Euler's theorem.
Describe cross sections by looking at the intersection of two shapes | 677.169 | 1 |
Shapes are objects with a defined boundary or surface. In two dimensions, the border is made up of lines and curves. If the frame is entirely composed of straight lines, the shape is called a polygon, and the lines are called edges. Triangles, rectangles, and parallelograms are all polygons. If all the boundaries are the same length and the angles between them are the same, then it's a regular polygon. Equilateral triangles, squares, pentagons, and hexagons are examples of regular polygons. The same concepts can be extended into three dimensions. Polyhedra are composed of joined faces, and regular polyhedral have every look and every interior angle. There are only five regular polyhedra: the tetrahedron, the cube, the octahedron, the dodecahedron, and the icosahedron. Now, you can easily find out the given words in this word search puzzle. Good luck! | 677.169 | 1 |
"Where is ... [something]?" is a question each of us has asked another person at some time or other. Suppose the person asked knows "where the something is." The "something" might be car keys, the next freeway exit, the rest room, etc. For the "asked" to inform the "asker" where the "something is" requires both persons to have adequate familiarity with the surrounding space - from where they are to where "it is." The space the something "is in" must be mutually known. The location of the "something," in the mutually known space being known only by the person asked.
More specifically, the space in question must have some manner of mutual beginning point, a reference, a geographic position of mutual knowledge. For example, in a city, the two might know where the "court house" is. Technically a commonly known location might be called an "origin." Also reference directions relative to that point must be mutually understood. North, East... avenues or streets, directions must be understood by both parties. For two people familiar with the streets of a city; such knowledge is common.
So staying with our "city" analogy of space, let's choose the courthouse as a common point. Next imagine a Cartesian coordinate system with its origin located at the courthouse. Let our imagination construct axes, X and Y (aligned in the Earth plane) and in accord with whatever symmetry exists; let Z designate the vertical. That done, we have the least idea of a coordinate system. To use this coordinate system to state a position, we specify a vector that extends from the origin to that position. In general that position (relative to the origin and the axes) will be partly "left, behind to the right and up (or down)." We must define a vector basis for the space. All of this constitutes establishment of what is called a "well-defined vector space."
To be quantitative, science must have a precise, mathematical description of space.
Vectors are easier to learn and to use than to discuss. In every example, every detail of definition of vectors and their use will be explained.
Vectors and the Vector Triangle:
The figure shows a triangular shape in an "implicit" manner. No angles, coordinates of points, etc, are specified. Suppose the shape were to represent an actual piece of land with small stones (point masses) located at 0, A and B. The positions of the points and their relation each to the others can be written "implicitly" in terms of directed line segments, i.e., vectors.
A common notation is to write vectors as "ordered pairs" of letters. The procedure is to select a point of origin then prescribe orthogonal axes Our space is two-dimensions, we call it the horizontal 0XY-planet - two coordinate axes will suffice. Vectors are "directed line segments. Our updated figure shows the origin (0), coordinates (X and Y) triangle with labeled vertices (0, A, B) and overlayed vectors.(second sketch)
Implicit Vectors
An implicit vector representation of the triangular area results when "arrow heads" are applied to the previous figure. By adding an arrowhead to one end, of a line segment, one changes the segment into a vector magnitude (length of the segment) and direction.
In our vector space, the position of the stone A, can be written as, 0A. Usually vectors are written in bold type. When hand-written an "over-bar" (or "over-arrow") is used. The ordered pair of letters, 0A, written with an over-bar, 0A vector notation, means a directed line segment that initiates at 0, (the origin of the vector space, and extends a distance to terminate at A. Similarly, stone B has a position located by a line that commences at 0, and extends to B, written as a vector as 0B. Vectors starting at places other than 0 follow logically. A vector commencing at A that extends to B would be written AB. And since vectors are additive and span space, it stands to reason that:
The equation reads left to right to state that the sum of two vectors equals a third vector. To solve or discuss any vector equation is made easier by focusing first on the easier of the vectors to describe. In the above equation, vectors 0A and 0B are easier to discuss. Each vector starts at the origin, 0, and proceed to its location (A or B, respectively) where their vector heads, or arrows are drawn. Vector AB is slightly more complicated; it proceeds from A to B. In this notation vectors are represented as ordered pairs of letters. There is a rule for adding these vectors. Two vectors can be added provided the ending letter of the first is the beginning letter of the second. The vector sum, BH + HS, is valid and equals BS. The idea extends readily:
To this point the vectors we have discussed were implicit meaning clearly stated but not yet quantitative. Vector equations are created implicitly, double-checked then changed to explicit form to be solved. It is easier to learn about physical vectors in focused, physical contexts. And, although space vectors are three dimensional, often at two-dimensional coordinate space will suffice. Learning is not harmed and an economy of vector arithmetic is obtained.
Explicit VectorsEngineering is quantitative.
Next we make our triangular figure more specific by specification of numerical values and insertion of a unit vector basis.
For 0XY-space, the pair, I and J are used. We see that 0A and 0B can be written as:
The above equations show simply the implicit vector OA is explicitly written as 9I. Also the implicit vector, 0B, has an explicit representation in component form as, (12I + 5J). This being a two-dimensional space, a basis and two numbers are needed to specify a vector. Alternately if the vector is not known, two numbers are not known. Often it is convenient to write a vector in a second way, in a "magnitude-direction" form.
The magnitude or length of vector 0B implied by writing the vector bounded on both sides by a vertical bars, as | 0B |. To obtain that number, since the components of 0B are orthogonal, we calculate as follows:
Thus the length of vector, |OB| is "13" units. To determine the direction of vector OB we multiply it by 13/13, then move the "13" of the denominator to the right, beneath each of the component numbers.
This simple operation shows that scalar multiplication of two vectors produces a scalar which equals the magnitude of one vector times the magnitude of the other, times the cosine of the smaller angle between them. Were it the case that the vectors were perpendicular, the cosine of 90° would equal zero, hence the scalar product is zero.
As illustrated above, vector equations relating positions of points in space are easily written. And, once written, they are easily checked. Vector equations, written correctly, express the geometry of any physical situation, any event, in a powerful, orderly manner.
"Where is something" is a question persons have asked over all of time. The answer tells the "position of that something" in a mutually understood space. The "something" might be your car keys or a location such as a freeway exit or the rest room, etc. For a person to answer, "Where is it?," requires both familiarity of both persons with the surrounding space. The space "it is in" must be mutually known. | 677.169 | 1 |
Three asymptote-sharing conics
define a quadrangle of intersections of common chords. Here, the black circle is the absolute conic, which serves the asymptotic points as
intersections with the green, red and blue conics. In the case shown,
all these intersections are real, and because of the required sharing,
they are only six instead of twelve.
The three conics intersect each other too, and produce further six points.
The six connecting lines of opposite intersections form the edges of a quadrangle. | 677.169 | 1 |
Ex 14.1 Class 6 Maths Question 1.
Solution: Step 1: Mark a point O as a centre.
Step 2: Open the compass up to the given radius 3.2 cm.
Step 3: Put the pointed edge of the compass at the centre O.
Step 4: Holding the top of the compass, take one full round with pencil. The
figure thus obtained is the required circle of radius 3.2 cm.
Ex 14.1 Class 6 Maths Question 2.
With the same centre O, draw two circles of radii 4
cm and 2.5 cm.
Solution: Step 1: Mark a point as centre O and open the
compass up to 4 cm.
Step 2: Draw a circle keeping the pointed edge at centre O.
Step 3: Take the same centre O and open the compass up to 2.5 cm, and draw
another circle.
The figure shows the required two circles with the same centre.
Ex 14.1 Class 6 Maths 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?
Solution:
Step 1: Draw a circle with centre O and any suitable radius.
Step 2: Draw AB and CD two diameters of the circle.
Step 3: By joining the end points of the diameters, we get a quadrilateral
ACBD.
Step 4: We note that OA = OB = OC = OD [Same radius]
and AC = DB, AD = BC ∠A = ∠C = ∠B = ∠D = 90°
Thus, ACBD is a rectangle.
Again, if the diameters are perpendicular to each other, then on measuring, we
get
AC = AD = DB = BC
Thus, ACBD is a square.
Ex 14.1 Class 6 Maths 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.
Solution: Draw a circle with centre O and any suitable radius.
Here,
(a) A is a point on the circle.
(b) B is a point in the interior of the circle.
(c) C is a point in the exterior of the circle.
Ex
14.1 Class 6 Maths Question 5.
Let A, B be the centres of the 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 AB and CD are at right angles.
Solution:
In the above figure, two circles of equal radii intersect each other at points
C and D. On measuring, we find that AB and CD intersect each other at right angles. | 677.169 | 1 |
I'm in high-school and I'm told that the maximum/minimum of a linear programming occurs at the vertex.For more info see the chapter here. For convinience I'm putting relevant excerpt here:
Now, we see that every point in the feasible region satisfies all the constraints, and since there are infinitely many points, it is not evident how
we should go about finding a point that gives a maximum value of the objective function. To handle this situation, we use the following theorems which are fundamental in solving linear programming problems. The proofs of these theorems
are beyond the scope of the book.
Theorem 1 Let R be the feasible region (convex polygon) for a linear programming
problem and let Z = ax + by be the objective function. When Z has an optimal value
(maximum or minimum), where the variables x and y are subject to constraints described
by linear inequalities, this optimal value must occur at a corner point (vertex) of the
feasible region.
Theorem 2 Let R be the feasible region for a linear programming problem, and let
Z = ax + by be the objective function. If R is bounded, then the objective function
Z has both a maximum and a minimum value on R and each of these occurs at a
corner point (vertex) of R
I searched wikipedia here to get under Optimal vertices (and rays) of polyhedra:
"..then the optimum value is always attained on ... This principle underlies the simplex algorithm for solving linear programs.."
But I couldn't understand it. Can someone explain?It might be helful to add that I have studied basic Calculus.
$\begingroup$If you have a hollow triangular mesh and you put a marble inside of it, will not the marble go to any of the vertices? Yes it will. Why? Because the altitude is the lowest there of all points inside of the mesh. For the same reason you will also find the minimum/maximum in a vertex in linear programming.$\endgroup$
6 Answers
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Notice that, for different values of C, you get different straight lines of varying y intercepts but they will have same slope :
Only the lines that cut through the feasible region satisfy all the given constraints because you can cookup x,y values such that they fall in both feasible region and the objective function.
From the second picture it is obvious that the the maximum value of $ax+by=C$ occurs when the y intercept of these feasible lines is maximum. Consequently the vertex A gives the maximum value for the objective function.
$\begingroup$BINGO! thats a good question, and yes this argument holds only when the feasible region is convex - $ax+by=C$ has y intercept of $C/b$ when $b$ is positive, the max value of $C/b$ also yields the maximum value of $C$ (just notice that y intercept is not so special, we can think around this using x intercept also)$\endgroup$
$\begingroup$What if the edge AB of the triangle is aligned with the red lines? Every point in the edge would have the same optimal value, the answer is not clear. It would be interesting if someone could give a formal proof here.$\endgroup$
$\begingroup$In other words, the theorem is being incorrectly interpreted because "optimal value" is not "optimal point". The theorem only states that there exists a vertex with the optimal value, it doesn't imply that the optimal value will always happen at a vertex. The difference is subtle. If the edge AB is aligned with the red lines, then all points are optimal points, they all have the same optimal value, in particular the vertex has this value.$\endgroup$
Some basic linear algebra is essential for this. Have you already studied that a bit?
Say you have point $p$ inside your polyhedron of feasible points inside $\mathbb{R}^n$, and let's say you want to maximize the objective function. The objective function is linear, given by some $c \in \mathbb{R}^n$. If $v \in \mathbb{R}^n$ and $c \cdot v = 0$, then $p+v$ has the same objective value as $p$. In other words, one does not improve $p$ by moving it in a direction orthogonal to the objective function.
If $v$ has some component parallel to $c$, then $p+v$ will be either strictly better or strictly worse than $p$. Therefore, if you are the point $p$ and you want to improve your objective value, then you should look for a direction $v$ in which you can move which has some positive component parallel to $c$, that is $c \cdot v > 0$. You may not be able to walk exactly in the direction of $c$ because a face of the polyhedron may stand in the way. But you may be able to walk in a direction having positive $c$ component. The only way you could get stuck is that for every possible improving direction, some face blocks you. In other words for every $v$ in the open halfspace $\{v:c\cdot v > 0\}$, the point $p+v$ lies outside the feasible polyhedron. This can only happen if either you already stand at a vertex, or you stand on a face of optimal solutions parallel to the hyperplane $\{v:c\cdot v = 0\}$. If you do not already stand at a vertex, then while staying on the hyperplane, you can walk to a vertex without changing your objective value.
In the above explanation, I have used the boundedness of the feasible region: Without boundedness, it might be possible to walk in an improving direction forever, or it might be possible to walk along a face of optimal solutions without ever arriving at a vertex. I have also used convexity: If I do not stand at an optimal solution, then there will always exist an improving direction.
Suppose you are not at a vertex of the convex region. If you are in the strict interior of the convex region, then you can move a little bit in any direction you want. In particular, if your objective is $ax + by$ then you can move a little bit in the direction of $(a,b)$ and get a higher objective function value. Similarly, if you are on an edge, let $(a_1,b_1)$ be a vector parallel to the edge. Then either the dot product of $(a_1,b_1)$ with $(a,b)$ is greater than or equal to zero, or less than or equal to zero. In the first case, you can move along the edge in direction $(a_1,b_1)$ until you hit a vertex and get at least as good of a solution, and in the second case you can move along the edge in direction $(-a_1,-b_1)$ until you hit a vertex and you will get at least as good of a solution. Put all the pieces together and you get that some vertex provides the optimal solution.
If we are searching for an extremum of function $z=ax+by$, then usually we take first partial derivatives, which equal to $a$ and $b$ consequently. As we are considering linear function in polygon, then system
$$\left\{\begin{aligned}&\frac{\partial z}{\partial x}=0 \\
&\frac{\partial z}{\partial y}=0 \\
\end{aligned}\right.$$
has no solutions (if $a,b\ne 0$) in any regular point. Consequently, we should search for extremum on boundary (in more general case, the piece of boundary can be a solution)
$\begingroup$because partial derivatives exist in any point of interior of polygon, hence, this system describes us situation in any interior point in polygon. As you see, this system has no solutions. That's why extremum can be only on boundary$\endgroup$
$\begingroup$In general case extremum can be in any point (interior or exterior). But here we showed that those extremums, for which partial derivatives equal to 0, do not exist in this problem. That's why we should take only boundary into consideration.$\endgroup$
Imagine that your feasible region is a box (possibly in high dimensions). Since the objective function is linear, let's interpret it as the direction of gravity. Drop a marble into the box, and it will roll into a corner, corresponding to an optimal solution.
To be more nuanced, the marble may not necessarily roll into a corner. In this case, either the feasible region is unbounded (and the marble drops forever), or there are infinitely many optimal solutions, for example when a whole face of the box lies flat on the ground, so to speak (i.e., perpendicular to the direction of the objective function). In the latter case, you can still roll the marble into any of the corners of that face.
$\begingroup$I think @Philip represents the same thing in algebraic form, because both of these look same to me, atleast the outline of both answers, is it? But yours is a more understandable way.$\endgroup$
$\begingroup$@Aditya Yes, they are much the same explanation. I think it is helpful to be able to understand and express the principle both informally, as I have done, and more formally, as he has.$\endgroup$
$\begingroup$Great intuition. Can you extend this explanation to the case where the feasible region is an ellipse? In this case, think that the optimum should at the edge of the ellipse, on a tangent but I'm not sure.$\endgroup$
As mentioned, the feasible region $R$ is convex. It means that every point $P \in R$ can be written as a convex combination of its vertices. Let's take a simple example: let ABC be a triangle (it's a convex shape). Every point $P \in ABC$ can be written as $P = \lambda_1A + \lambda_2B + \lambda_3C$ with $\lambda_1 + \lambda_2 + \lambda_3 = 1$ (it's the definition of a convex combination). Now, since it's the feasible region, every point $P \in ABC$ has a value, say $v_P$ (the objective function computed at this point). Wlog, we can write $v_A \leq v_B \leq v_C$; so: | 677.169 | 1 |
From triangle $MAC$ the sine rule gives $MC/MA=\sin\alpha/\sin(90^\circ-3\alpha)=\sin\alpha/\cos3\alpha$. Similarly, from triangle $MAB$ we get $MB/MA=\sin3\alpha/\cos\alpha$. But $MB=MC$, so $\sin3\alpha\cos3\alpha=\sin\alpha\cos\alpha$. Hence $\sin6\alpha=\sin2\alpha$, so $6\alpha+2\alpha=180^\circ$ and hence $\alpha=22.5^\circ$ and so $\angle BAC=90^\circ$.
Well the angle bisector cuts $\angle BAC$ into two equal parts, so the median and the altitude must each cut those two angles into equal parts. If we label point where the altitude intersects $BC$ as $J$, the point where the angle bisector as $K$, and the median as $M$, we know that $K$ is colinearly between $J$ and $M$. As it is arbitrary which endpoint is $B$ and which is $C$ we can assume that points lie in order as $B,J,K,M$
So we have a figure a big triangle $\triangle ABC$ divided into four smaller triangles.
Here is a possible solution which is not using trigonometric functions, but rather only similarities and the angle bisector theorem (ABT).
We denote by $H,J,M$ the points on $BC$ which are respectively on the height, angle bisector, and median of $A$. Let $K$ be the reflection of $J$ w.r.t. the median $AM$. It lies on the side $AC$. Let $a,b,c$ be the side lengths as usual. Let the angle in $A$ have the measure $4x$. Then we can quickly chase the following angles in the picture:
From the ABT in $\Delta AJC$ we get the length of $JM$ as shown in the picture, $\frac a2\cdot \frac cb$. So we also have $MK$.
$BJ=BM-JM$ leads to $BJ=\frac a2\cdot \frac {b-c}b$.
The rest of the segment $BC=a$ is then $JC=\frac a2\cdot {b+c}b$.
We write now the ABT in the triangle $\Delta ABC$, and the similarity $\Delta ABC\sim \Delta MKC$ to get the relations:
$$
\begin{aligned}
\frac cb=\frac{b-c}{b+c}\ ,\\
\frac ab=\frac{b-c}{a/2}\ .
\end{aligned}
$$
The two realtions are $b^2-bc=c^2+bc$ and $a^2=2b^2-2bc$. It follows immediately $a^2=b^2+(b^2-2bc)=b^2+c^2$, so the angle in $A$ is $4x=90^\circ$, and all angles can be explicitly identified.
To be pedant, we have to check if the obtained triangle satisfies the condition, yes, this is the case, the height $AH$ is the angle bisector of $\widehat{BAJ}$,
and because of $MA=MC$ we have $\widehat{MAC}=\widehat{MCA}=22.5^\circ$, so $AM$ bisects $\widehat{JAC}$.
$\begingroup$It is a long time since the solution was typed, very often i am just starting to type without having the full (shortest) path to the final step. "Having" $MK$ is not bad, since any detail can be useful on the road, in this case "knowing $MK$" does not additionally help, since the similarity of $\Delta ABC$ and $\Delta MCK$ leads to $$\frac{CM}{CA}=\frac{CK}{CB}=\frac{MK}{AB}\ ,\text{ i.e.}\\\frac{a/2}b = \frac{b-c}a = \frac{\frac a2\cdot \frac cb}c\ ,$$and the last term is never used, it repeats the first one. (However, it did not hurt to mention it while still searching for a solution.)$\endgroup$ | 677.169 | 1 |
Lesson
Lesson 8
8.1: 3 Sides; 3 Angles
Examine each set of triangles. What do you notice? What is the same about the triangles in the set? What is different?
Set 1:
Set 2:
8.2: 2 Sides and 1 Angle
Examine this set of triangles.
What is the same about the triangles in the set? What is different?
How many different triangles are there? Explain or show your reasoning.
8.3: 2 Angles and 1 Side
Examine this set of triangles.
What is the same about the triangles in the set? What is different?
How many different triangles are there? Explain or show your reasoning.
Summary
Both of these quadrilaterals have a right angle and side lengths 4 and 5:
Description: <p>Two quadrilaterals each with two given side lengths labeled 4 and 5, and a right angle. On the left, the quadrilaterial is a rectangle with the right angle between adjacent side lengths 4 and 5. On the right, the quadrilateral is a trapezoid with the bottom base labeled 5 and one leg labeled 4. There is a right angle between the bottom base and the leg not labeled.</p>
However, in one case, the right angle is between the two given side lengths; in the other, it is not.
If we create two triangles with three equal measures, but these measures are not next to each other in the same order, that usually means the triangles are different. Here is an example:
Description: <p>Two triangles. The triangle on the left has the angle labeled 32 degrees between the adjacent side lengths 5 and 6. The triangle on the right has the angle labeled 32 degrees between the side length labeled 5 and the third side of the triangle that is not labeled.</p> | 677.169 | 1 |
Rules
A triangle has 3 traits. First trait, the triangle direction which can point up, down, right or left. Second trait, the number of dots in a triangle (0 to 4 dots). Third trait, the colour of the triangle which can be red, yellow, blue or green. Each triangle is made of a combination of these 3 traits. Before the triangles are presented you have to memorize two processing rules in 10 seconds. You will be given instruction to press the up or down button depending on whether the direction or color or number of dots of the present triangle is the same that the former one. For the first triangle of a serie, press the right button.
Example
Let's imagine that you were given the following rules: "1) If two consecutives triangles point in the same direction, press the top button. 2) If two consecutives triangles have the same color, press the button below." Since the two triangles are green, you must press the button below.
Timing
There are 10 series of different rules per test. Each serie includes 40 triangles to answer. Each triangle is displayed for 2 seconds. | 677.169 | 1 |
Dentro del libro
Resultados 1-5 de 29
P·gina 40 ... divided into as many triangles as the figure has sides , and the angles of each triangle are equal to two right angles . All the angles , therefore , of the triangles into which the A figure is divided , are equal to twice as many right ...
P·gina 49 ... divided into any number of parts ( AF , FH , HB ) , is equal to the sum of the rectangles contained by the undivided line and the several parts of the divided line . From A draw AC perpendicular to AB , make it equal to L , and complete ...
P·gina 50 ... divided into the rectangles AC AF , FE FH , HG HB , to the sum of which it is equal . But since AC = L ( Const ... divided line . L PROP . II . THEOR . If a straight line ( AB ) be divided into any two parts ( AF , FB ) , the square of ...
P·gina 51 ... divided into any two parts ( AF , FB ) , the square of the whole line is equal to the squares of the parts , together with twice the rectangle contained by the parts . = C E D K G F B Upon AB describe the square ACDB ; draw the diago | 677.169 | 1 |
What type of lines are in math?
Parallel lines--will never touch
Perpendicular lines--cross at right angles
Rays--lines that start at a point and extend into infinity
Line segment--a line between two points
Line--a line that runs forever in either direction
How are perpendicular lines similar to intersecting line?
perpendicular lines are a special type of intersecting line that
happen to intersect at 90 degree angles | 677.169 | 1 |
The MCQ Quiz: Operation of drawing representative line between two vectors in such a way that tail of one vector coincides with the head of other vector is called; "Applied Physics: Vectors" App APK Download (Free) with answers vector subtraction, vector addition, vector division and vector multiplication for online graduate programs. Study vector and equilibrium questions and answers, Apple Book to download free sample for ACT test prep classes. | 677.169 | 1 |
According to the given figure, the point F' corresponds to point F.Explanation:Let DEF be figure 1Let D'E'F' be figure 2Here, DEF is exactly equal to D'E'F' as it is the same figure but is a reflected image or mirror image.This is why all the point in figure 1 corresponds to figure 2 completely. | 677.169 | 1 |
Can you spot any seven-sided objects when you look around? These objects can range from old coins, vases, paperweights, erasers, pencils, and so on. These objects are heptagonal in shape. A heptagon is a geometric shape with seven sides. In the following sections, we will discuss the definition of a heptagon, the shape and types of heptagons, the properties of heptagons, and some real-life examples of heptagon shapes.
So, let's first learn the definition of a heptagon.
What is a Heptagon?
A heptagon can be defined as a polygon that has seven sides, seven angles, and seven edges. The sum of all the angles of a heptagon is 900°. Fourteen diagonals can be drawn in a heptagon. Sometimes heptagon is also known as septagon. The word 'Heptagon' is a combination of two Greek words 'heptá' which means seven and 'gōnía' which means angle.
What does a Heptagon Shape Look Like?
Heptagon Shape
As we can see in the above image, all the seven sides meet with each other end to end and form the heptagon shape.
Types of Heptagon Shapes
Based on the sides and angles, there are different kinds of heptagon shapes.
Based on the side length of the heptagons, heptagons can be categorised into the following two types.
Regular Heptagon
All sides of a regular heptagon are equal in length and all the angles are equal in measure. The sum of all the interior angles of the regular heptagon is 900°. So, each of the interior angles measures 900°/7, i.e., 128.57°.
Shape of a Regular Heptagon
Irregular Heptagon
All sides and angles of an irregular heptagon have different measures. Each of the interior angles of an irregular heptagon is different in measure but the sum of the interior angles is 900°.
Irregular Heptagon
Based on the angle measures, the heptagon can be classified into the following two types.
Convex Heptagon
In a convex heptagon, the value of the interior angles is less than 180°. So, no diagonals of the convex heptagon can lie outside the polygon.
Convex Heptagon
Concave Heptagon
In a concave heptagon, at least one of the interior angles is greater than 180° and at least one of the diagonals lies outside the polygon.
Concave Heptagon
Heptagon Properties
The key properties of a heptagon help us to identify a heptagon shape easily. The properties are as follows.
A heptagon has seven angles, seven sides, and seven vertices.
The sum of all the interior angles in a heptagon is equal to 900 degrees.
The sum of all the exterior angles of a heptagon is equal to 360 degrees.
In a regular heptagon, the value of each of the interior angles is approximately 128.57 degrees.
The value of the central angle of a regular heptagon is approximately 51.43 degrees.
Fourteen diagonals can be drawn in a heptagon.
Regular heptagons are also known as convex heptagons because all of their interior angles measure less than 180 degrees.
Five triangles can be formed in a heptagon.
A Few Real-Life Examples of Heptagons
We have created a list of some real-life examples of heptagons. Check out the list and see whether you can find any of these objects around you and try to see if you have got your own to add to this list.
Examples
Shapes
Storage Box: Medicine boxes or storage containers mostly have 7 small plastic cases attached to each other. So, a storage box is a proper example of the heptagon shape used in daily life.
Coins: Some of the old coins are heptagonal in shape. This unique structure of the coins helps us to easily differentiate these coins from the others.
Cactus: One of the perfect examples of the heptagonal shape is a cactus that exists in nature. If you look at it from the top, you can easily observe the stems of the cactus form the seven sides of the heptagon.
Paper Boat: A classic example of an irregular heptagon is a paper boat. If you look at the outer boundary of a paper boat, you can observe that it has seven sides, seven edges, and seven angles, and the sides and the angles are of different measures.
Arrowhead: An arrowhead is another example of an irregular heptagon. If you draw an arrowhead on a piece of paper, you will see that it has seven unequal sides and seven unequal angles.
Green House: Some of the small and compact greenhouses that are usually installed in the backyards are of heptagonal shape.
Heptagonal Star: For the purpose of decoration, heptagonal stars are used. They are formed by joining the diagonals of a heptagonal geometric figure.
Now, we hope that it is clear to you what the shape of a heptagon looks like and here we have learned about different types of heptagons and their properties. We have also provided some real-life examples of the heptagon. This information will be really helpful while studying geometry. | 677.169 | 1 |
Thales Theorem
This is a special case of inscribed angle theorem, an important theorem of circle geometry. This theorem can be stated in any of the following ways:
If three points A, B, and C lie on the circumference of a circle, where the line AC is the diameter of the circle, then the angle ∠ABC is a right angle (90°)
OR
The diameter of a circle always subtends a right angle to any point on the circle.
OR
An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle.
Here is an interesting and very easy proof of Thales theorem:
To prove this we use following two facts:
– Sum of all angles of a triangle is 180.
– Base angles of an isosceles triangle are equal.
In this circle we have :
reason being all are radii of same circle.
reason being base angles of isosceles triangle OAB
Similarly ,
reason being base angles of isosceles triangle OBC
Also,
Since sum of all angles of a triangle is 180 , so
Here are some examples on Thales Theorem:
1. In the given circle FH is a diameter passing through its center J. Find x and angle HFG.
Since FH is a diameter so using Thales theorem, FGH would be a right triangle with right angle at point G.
Therefore we can apply triangle's angle sum property,
2. In the given circle, find the measure of angle PQR . Assume point R is the center of the circle.
Since R is given center of circle, PS would be the diameter of the circle.
Therefore applying Thales theorem for triangle PQS, angle PQS would be a right angle.
Since RS = RQ (radii of same circle) , base angles of isosceles triangle RQS would be equal.
Also
Hence the measure of angle PQR = 66
3. In the circle shown, BC is a diameter with center A. Given m∠ABE = 26 and m∠ADB=18
Find a) m∠DAB b) m∠BAE c) m∠DAE
Since A is the center of circle, so AB=AD=AE , as all are radii of same circle.
Therefore ABD is an isosceles triangle and hence base angles would be equal.
Using triangle's angle sum property,
b) Using exactly same concept for triangle ABE, we have
Using triangle's angle sum property,
c) We see that three angles make a complete angle of 360 at point A.
4. In the figure below, O is the center of circle and AD is a diameter.
a. Find 𝑚∠𝐴OB. b. If 𝑚∠BOC ∶ 𝑚∠𝐶OD = 6:5 , what is 𝑚∠𝐵OC?
Since OB = OD being radii of same circle,
as base angles of isoscele triangles are equal.
Using exterior angle property of triangles, exterior angle is equal to sum of opposite interior angles.
b.
These two angles are given in the ratio 6 : 5
11x= 132
x= 12 | 677.169 | 1 |
Hexagonal duoantiprism
The hexagonal duoantiprism or hiddap, also known as the hexagonal-hexagonal duoantiprism, the 6 duoantiprism or the 6-6 duoantiprism, is a convex isogonalpolychoron that consists of 24 hexagonal antiprisms and 72 tetragonal disphenoids. 4 hexagonal antiprisms and 4 tetragonal disphenoids join at each vertex. It can be obtained through the process of alternating the dodecagonal duoprism. However, it cannot be made uniform, and has two edge lengths. It is the second in an infinite family of isogonal hexagonal dihedral swirlchora. | 677.169 | 1 |
Chapter 1 - Geometry Chapter 1 - Geometry 2 as PDF for free.
More details
1 Geometry 2 Terminology Arc: Part of a curve, most commonly a portion of the distance around the circumference of a circle Chord: A straight line joining two points on the circumference of a circle Concyclic points: Points that lie on the circumference of the same circle Congruent: Two figures are congruent if they have the same size and shape. They are identical in every way Cyclic quadrilateral: A cyclic quadrilateral is a figure whose four vertices are concyclic points. The four vertices lie on the circumference of a circle Polygon: A polygon is a closed plane figure with straight sides
Radius: A radius is the distance from the centre of a circle out to the circumference (radii is plural, meaning more than one radius) Similar: Two figures are similar if they have the same shape but a different size. Corresponding angles are equal and corresponding sides are in the same ratio Subtend: Form an angle at some point (usually the centre or circumference) Tangent: A straight line external to a circle that just touches the circle at a single point Vertex: A vertex is a corner of a figure (vertices is plural, meaning more than one vertex)
Chapter 1 Geometry 2
Introduction You studied geometry in the Preliminary course. In this
chapter, you will revise this work and extend it to include some more general applications of geometrical properties involving polygons in 2 unit and circles in extension 1. You will also use the Preliminary topic on straight-line graphs to explore coordinate methods in geometry.
Plane Figure Geometry Here is a summary of the geometry you studied in the Preliminary course.
Vertically opposite angles
+AEC and +DEB are called vertically opposite angles. +AED and +CEB are also vertically opposite angles. Vertically opposite angles are equal.
Parallel lines If the lines are parallel, then alternate angles are equal.
If the lines are parallel, then corresponding angles are equal.
3
4
Maths In Focus Mathematics Extension 1 HSC Course
If the lines are parallel, cointerior angles are supplementary (i.e. their sum is 180°).
Tests for parallel lines If alternate angles are equal, then the lines are parallel. If +AEF = +EFD, then AB || CD.
If corresponding angles are equal, then the lines are parallel. If +BEF = +DFG, then AB || CD.
If cointerior angles are supplementary, then the lines are parallel. If +BEF + +DFE = 180c, then AB || CD.
Angle sum of a triangle
The sum of the interior angles in any triangle is 180°, that is, a + b + c = 180
Chapter 1 Geometry 2
Exterior angle of a triangle
The exterior angle in any triangle is equal to the sum of the two opposite interior angles. That is, x+y=z
Congruent triangles Two triangles are congruent if they are the same shape and size. All pairs of corresponding sides and angles are equal. For example:
We write ∆ ABC / ∆ XYZ. Tests To prove that two triangles are congruent, we only need to prove that certain combinations of sides or angles are equal. Two triangles are congruent if • SSS: all three pairs of corresponding sides are equal • SAS: two pairs of corresponding sides and their included angles are equal • AAS: two pairs of angles and one pair of corresponding sides are equal • RHS: both have a right angle, their hypotenuses are equal and one other pair of corresponding sides are equal
Similar triangles Triangles, for example ABC and XYZ, are similar if they are the same shape but different sizes. As in the example, all three pairs of corresponding angles are equal. All three pairs of corresponding sides are in proportion (in the same ratio).
If 2 pairs of angles are equal then the third pair must also be equal.
Two triangles are similar if: • three pairs of corresponding angles are equal • three pairs of corresponding sides are in proportion • two pairs of sides are in proportion and their included angles are equal
Ratios of intercepts
When two (or more) transversals cut a series of parallel lines, the ratios of their intercepts are equal. That is, AB: BC = DE: EF AB DE or = BC EF
Pythagoras' theorem The square on the hypotenuse in any right angled triangle is equal to the sum of the squares on the other two sides. That is, c2 = a2 + b2 c = a2 + b2 or
Chapter 1 Geometry 2
Quadrilaterals A quadrilateral is any four-sided figure
In any quadrilateral the sum of the interior angles is 360° Parallelogram
A parallelogram is a quadrilateral with opposite sides parallel
Properties of a parallelogram: • opposite sides of a parallelogram are equal • opposite angles of a parallelogram are equal • diagonals in a parallelogram bisect each other • each diagonal bisects the parallelogram into two congruent triangles
Tests A quadrilateral is a parallelogram if: • both pairs of opposite sides are equal • both pairs of opposite angles are equal • one pair of sides is both equal and parallel • the diagonals bisect each other
These properties can all be proved.
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Maths In Focus Mathematics Extension 1 HSC Course
Rectangle
If one angle is a right angle, then you can prove all angles are right angles.
A rectangle is a parallelogram with one angle a right angle
Properties of a rectangle: • the same as for a parallelogram, and also • diagonals are equal
Test A quadrilateral is a rectangle if its diagonals are equal
Application Builders use the property of equal diagonals to check if a rectangle is accurate. For example, a timber frame may look rectangular, but may be slightly slanting. Checking the diagonals makes sure that a building does not end up like the Leaning Tower of Pisa!
Rhombus
It can be proved that all sides are equal.
A rhombus is a parallelogram with a pair of adjacent sides equal
Properties of a rhombus: • the same as for parallelogram, and also • diagonals bisect at right angles • diagonals bisect the angles of the rhombus
Chapter 1 Geometry 2
Tests A quadrilateral is a rhombus if: • all sides are equal • diagonals bisect each other at right angles Square
A square is a rectangle with a pair of adjacent sides equal
Properties of a square: • the same as for rectangle, and also • diagonals are perpendicular • diagonals make angles of 45° with the sides Trapezium
A trapezium is a quadrilateral with one pair of sides parallel Kite
A kite is a quadrilateral with two pairs of adjacent sides equal
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Maths In Focus Mathematics Extension 1 HSC Course
Polygons A polygon is a plane figure with straight sides
A regular polygon has all sides and all interior angles equal
The sum of the interior angles of an n-sided polygon is given by S = (n − 2) ´ 180°
The sum of the exterior angles of any polygon is 360°
Areas Most areas of plane figures come from the area of a rectangle. Rectangle
A = lb Square
A square is a special rectangle.
A = x2 Triangle
The area of a triangle is half the area of a rectangle.
A=
1 bh 2
Chapter 1 Geometry 2
11
Parallelogram
The area of a parallelogram is the same as the area of two triangles.
A = bh Rhombus
A=
1 xy 2
(x and y are lengths of diagonals) Trapezium
A=
1 h(a + b) 2
Circle
You will study the circle in more detail. See Chapter 5.
A = πr 2 The following examples and exercises use these results to prove properties of plane figures.
12. Show that a diagonal cuts a parallelogram into two congruent triangles. 13. Prove that opposite angles are equal in any parallelogram.
Chapter 1 Geometry 2
14. ABCD is a parallelogram with BM = DN. Prove that AMCN is also a parallelogram.
19. Prove that the diagonals in any rectangle are equal. 20. Prove that if one angle in a rectangle is 90° then all the angles are 90°. 21. ABCD is a rhombus with AD = CD. Prove that all sides of the rhombus are equal.
15. ABCD and BCEF are parallelograms. Show that AFED is a parallelogram.
22. ABCD is an isosceles trapezium. Prove the base angles +ADC and +BCD are equal.
16. ABCD is a parallelogram with DE = DC. Prove that CE bisects +BCD. 23. Prove that +ADC = +ABC in kite ABCD.
Surface Areas and Volumes You will need some of these formulae when you study maxima and minima problems in Chapter 2.
Areas are used in finding the surface area and volume of solids. Here is a summary of some of the most common ones.
SURFACE AREA
VOLUME
S = 2(lb + bh + lh)
V = lbh
S = 6x 2
V = x3
S = 2π r (r + h)
V = π r2 h
S = 4π r 2
V=
4 3 πr 3
Chapter 1 Geometry 2
S = π r (r + l)
V=
In general, the volume of any prism is given by V = Ah where A is the area of the base and h is its height
In general, the volume of any pyramid is given by V=
1 Ah 3
Where A is the area of the base and h is its height
1 2 πr h 3
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Example Find the surface area of a cone whose height is twice the radius, in terms of r.
Solution
h = 2r l2 = r2 + h2 = r2 + (2r) 2 = r 2 + 4r 2 = 5r 2 `
l= =
5r 2 5r
Surface area S = π r (r + l) where l is slant height = πr (r + 5r )
While surface area and volume is not a part of the geometry in the HSC syllabus, the topic in Chapter 2 uses calculus to find maximum or minimum areas, perimeters, surface areas or volumes. So you will need to know these formulae in order to answer the questions in the next chapter. Here are some questions to get you started.
2. A sphere has a volume of 120π m3. Find the exact value of r. 3. A rectangular prism has dimensions x, x + 2 and 2x – 1. Find its volume in terms of x.
Chapter 1 Geometry 2
4. A cylinder has a volume of 250 cm3. If its base has radius r and its 250 height is h, show that r = . πh 5. Find the volume of a cylinder in terms of r if its height is five times the size of its radius. 6. The ratio of the length to the breadth of a certain rectangle is 3:2. If the breadth of this rectangle is b units, find a formula for the area of the rectangle in terms of b. 7. Find the volume of a cube with sides (x + 2) cm. 8. What would the surface area of a cylinder be in terms of h if its height is a third of its radius? 9. A square piece of metal with sides 3 m has a square of side x cut out of each corner. The metal is then folded up to form a rectangular prism. Find its volume in terms of x.
11. The area of the base of a prism is given by 3h + 2, where h is the height of the prism. Find a formula for the volume of the prism. 12. The area of the base of a pyramid is 6h + 15 where h is the height of the pyramid. Find the volume of the pyramid in terms of h. 13. A rectangular pyramid has base dimensions x – 3 and 3x + 5, and a height of 2x + 1. Write a formula for the volume of the pyramid in terms of x. 14. The height of a rectangular prism is twice the length of its base. If the width of the base is x and the length is 3x – 1, find an expression for the (a) volume and (b) surface area of the prism. 15. Find a formula for the slant height of a cone in terms of its radius r and height h. 16. A page measuring x by y is curved around to make an open cylinder with height y. Find the volume of the cylinder in terms of x and y.
10. A cone-shaped vessel has a height of twice its radius. If I fill the vessel with water to a depth of 10 cm, find the volume of water to the nearest cm3.
17. The volume of a cylinder is 400 cm3. Find the height of the cylinder in terms of its radius r. 18. A cylinder has a surface area of 1500 cm2. Find a formula for its height h in terms of r. 19. The surface area of a cone is given by S = π r (r + l) where l is the slant height. Find a formula for the slant height of a cone with surface area 850 cm2 in terms of r.
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Maths In Focus Mathematics Extension 1 HSC Course
20. A rectangular timber post is cut out of a log with diameter d as shown. If the post has length x and breadth y, write y in terms of x when d = 900 mm.
d
DID YOU KNOW? REGULAR SOLIDS There are only five solids with each face the same size and shape. These are called platonic solids. Research these on the internet.
Chapter 1 Geometry 2
The skeletons opposite are those of radiolarians. These are tiny sea animals, with their skeletons shaped like regular solids. A salt crystal is a cube. A diamond crystal is an octahedron.
Diamond crystal
Coordinate Methods in Geometry Problems in plane geometry can be solved by using the number plane. You studied straight-line graphs in the Preliminary course. Some of the main results that you learned will be used in this section. You may need to revise that work before studying this section. Here is a summary of the main formulae.
Distance The distance between two points (x1, y1) and (x2, y2) is given by d=
2 2 _ x2 − x1 i + _ y2 − y1 i
Midpoint The midpoint of two points (x1, y1) and (x2, y2) is given by P=e
x1 + x2 y1 + y2 o , 2 2
Gradient The gradient of the line between (x1, y1) and (x2, y2) is given by y2 − y1 m= x −x 2 1 The gradient of a straight line is given by m = tan θ where θ is the angle the line makes with the x-axis in the positive direction.
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Maths In Focus Mathematics Extension 1 HSC Course
The gradient of the line ax + by + c = 0 is given by m=−
a b
Equation of a straight line The equation of a straight line is given by y − y1 = m(x − x1) where (x1, y1) lies on the line with gradient m.
Parallel lines If two lines are parallel, then they have the same gradient. That is, m1 = m2
So AB and AC are perpendicular So triangle ABC is right angled at A. 2. Prove that points A ^ 1, 1 h, B ^ − 2, −1 h and C (4, 3) are collinear.
Solution Collinear points lie on the same straight line, so they will have the same gradient. y2 − y1 m= x −x 2 1
−1 − 1 −2 − 1 = −2 −3
m AB =
=2 m BC =
3 3 − (−1)
4 − (−2) =4 6 =2 3 m AB = m BC
So the points are collinear.
1.3 Exercises 1. Show that points A(−1, 0), B(0, 4), C(7, 0) and D(6, − 4) are the vertices of a parallelogram. 2. Prove that A(1, 5), B(4, − 6) and C(− 3, − 2) are vertices of a right angled triangle. 3. Given ∆ ABC with vertices A(0, 8), B(3, 0) and C(− 3, 0) (a) show that ∆ ABC is isosceles (b) find the length of the altitude from A (c) find the area of the triangle.
4. Show that the points X ^ 3, 2 h, Y (−2, 1) and Z (8, 3) are collinear. 5. (a) Show that the points A ^ 2, 5 h, B(1, 0), C(−7, −4) and D(−3, 4) are the vertices of a kite. (b) Prove that the diagonals of the kite are perpendicular. (c) Show that CE = 2AE where E is the point of intersection of the diagonals. 6. Find the radius of the circle that has its centre at the origin and a tangent with equation given by 4x − 3y − 5 = 0.
Chapter 1 Geometry 2
7. (a) Find the equation of the perpendicular bisector of the line joining A(3, 2) and B(−1, 8). (b) Show that the point C(7, 9) lies on the perpendicular bisector. (c) What type of triangle is ∆ ABC? 8. Show that ∆ OAB and ∆ OCD are similar where (0, 7), (2, 0), (0, −14) and (−4, 0) are the points A, B, C and D respectively and O is the origin. 9. (a) Prove that ∆ OAB and ∆ OCB are congruent given A(3, 4), B(5, 0), C(2, −4) and O the origin. (b) Show that OABC is a parallelogram. 10. The points A(0, 0), B(2, 0), C(2, 2) and D(0, 2) are the vertices of a square. Prove that its diagonals make angles of 45° with the sides of the square. 11. Prove that P (−2, 0), Q (0, 5), R(10, 1) and S(8, −4) are the vertices of a rectangle.
12. The points A(− 5, 0), B(1, 4) and C(3, 0) form the vertices of a triangle. (a) Find X and Y, the midpoints of AB and AC respectively. (b) Show that XY and BC are parallel. (c) Show that BC = 2XY. 13. Show that the diagonals of a square are perpendicular bisectors, given the vertices of square ABCD where A = ^ − a, 0 h, B = ^ − a, a h, C = ^ 0, a h and D = (0, 0). 14. (a) Show that points X ^ 3, 2 h and Y (−1, 0) are equal distances from the line 4x − 3y − 1 = 0. (b) Find Z, the x-intercept of the line. (c) What is the area of triangle XYZ? 15. ABCD is a quadrilateral with A^ 3, 1 h, B^ 1, − 4 h, C ^ − 5, − 2 h and D ^ − 4, 3 h . Show that the midpoints of each side are the vertices of a parallelogram.
Circle Properties
In this section you will use the properties from the Preliminary course to answer questions involving proofs. On the next page is a summary of the properties. You may need to revise them before doing the exercises.
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Maths In Focus Mathematics Extension 1 HSC Course
Arcs and chords 1. Equal arcs subtend equal angles at the centre of the circle.
The converse is also true:
If two arcs subtend equal angles at the centre of the circle, then the arcs are equal.
2. Equal chords subtend equal angles at the centre of the circle.
The converse is also true:
Equal angles subtended at the centre of the circle cut off equal chords.
Chapter 1 Geometry 2
Angle properties 1. The angle at the centre of a circle is twice the angle at the circumference subtended by the same arc.
2. Angles in the same segment of a circle are equal.
3. The angle in a semicircle is a right angle.
Chord properties 1. A perpendicular line from the centre of a circle to a chord bisects the chord.
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Maths In Focus Mathematics Extension 1 HSC Course
The converse is also true:
A line from the centre of a circle that bisects a chord is perpendicular to the chord.
2. Equal chords are equidistant from the centre of the circle.
The converse is also true:
Chords that are equidistant from the centre are equal.
Class Exercise Prove that chords that are equidistant from the centre are equal.
39. Given two concentric circles with centre O (a) prove ∆OAB and ∆OCD are similar triangles (b) find CD if the radii of the circles are 5 cm and 8 cm and AB = 12 cm.
36. D is the midpoint of AB. Show AO = BO, where O is the centre of the circle, and AB is a tangent to the circle.
40. O and P are the centres of the circles below and AC is a tangent to both circles at B. Prove O, P and B are collinear. 37. Prove (a) AC = BD (b) AB < DC. AC and BD are tangents to both circles. A D
E C B
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Maths In Focus Mathematics Extension 1 HSC Course
41. Prove +ADC = 2+BAC, given that BC bisects +ECA.
42. Show that ∆ ABC and ∆CDE are similar. O is the centre of the larger circle.
44. CF is perpendicular to AB. Prove BC bisects +FCE. O is the centre of the circle.
Test Yourself 1 1. Triangle ABC is isosceles, with AB = AC. D is the midpoint of AB and E is the midpoint of AC.
(a) Prove that ∆ BEC is congruent to ∆ BDC. (b) Prove BE = DC. 2. In the quadrilateral ABCD (a) evaluate x and y (b) prove that AB and DC are parallel.
3. If the diagonals of a rhombus are x and y, show that the length of its side is x2 + y2 . 2 4. If A = (4, −1), B = (7, −5) and C = (1, 3), prove that triangle ABC is isosceles. 5. The surface area of a closed cylinder is 100 m2. Write the height h of the cylinder in terms of its radius r.
15. If A = (1, 5), B = (4, 2) and C = (2, −3), find the coordinates of D such that ABCD is a parallelogram. 16. (a) Prove that ∆ ABC is similar to ∆ CDE. (b) Evaluate x and y to 1 decimal place.
Show that the parallelogram is a rectangle. 12. Prove that ∆ PQR and ∆ WXY are similar.
17. (a) Find the equation of AB if A = (−2, −3) and B = (4, 5). (b) Find the perpendicular distance from C (1, −3) to line AB. (c) Find the area of ∆ ABC. 18. ABCD is a kite.
13. Prove that if two chords subtend equal angles at the centre of a circle, the chords are equal. (a) Prove ∆ ABC and ∆ ADC are congruent. (b) Prove ∆ ABE and ∆ ADE are congruent. (c) Prove AC is the perpendicular bisector of BD.
Chapter 1 Geometry 2
19. A(1, 2), B(3, 3) and C(5, −1) are points on a number plane. (a) Show that AB is perpendicular to BC. (b) Find the coordinates of D such that ABCD is a rectangle. (c) Find the point where the diagonals of the rectangle intersect. (d) Calculate the length of the diagonal.
20. The surface area of a box is 500 cm2. Its length is twice its breadth x. (a) Show that the height h of the box is
given by h =
250 − 2x2 . 3x
(b) Show that the volume of the box is 500x − 4x3 V= . 3
Challenge Exercise 1 1. In the figure, BD is the perpendicular bisector of AC. Prove that ∆ ABC is isosceles.
4. Paper comes in different sizes, called A0, A1, A2, A3, A4 and so on. The largest size is A0, which has an area of one square metre. If the ratio of its length to breadth is 2 :1, find the dimensions of its sides in millimetres, to the nearest millimetre. 5. The volume of a prism with a square base of side x is 1000 cm3. Find its surface area in terms of x.
2. Given E and D are midpoints of AC and AB respectively, prove that (a) DE is parallel to BC 1 (b) DE = BC. 2
3. Prove that the diagonals in a rhombus bisect the angles they make with the sides.
6. Prove that in any regular n-sided polygon 360 ° the size of each angle is b 180 − n l . 7. Line XY meets ∆ ABC so that +BAC = +BCY. If a circle can be drawn through A, B and C, show that XY is a tangent to the circle.
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8. A plastic frame for a pair of glasses is designed as below. Find the length of plastic needed for the frame, to the nearest centimetre.
10. A parallelogram ABCD has AB produced to E and diagonal AC produced to F so that EF < BC. Prove that ∆ AEF is similar to ∆ ADC. 11. ABCD is a rhombus with A(0, 0), B(a, b), C(2a, 0) and D(a, − b). Show (a) the diagonals bisect each other at right angles (b) all sides are equal (c) AC bisects +BCD. 12. In the parallelogram ABCD, AC is perpendicular to BD. Prove that AB = AD.
13. Triangle ABC has P, Q and R as midpoints of the sides, as shown in the diagram below. Prove that ∆ PQR / ∆CPR.
14. A tangent is drawn from C to meet the circle at B. A secant is drawn from C to intersect the circle at A and D. O is the centre of the circle. Prove (a) +OBD = +DCB (b) +AOD = 2+DCB.
Chapter 1 Geometry 2
15. Two circles have C as a point of contact with common tangent AB. Prove +DCE = +FCG.
16. In the circle below, D is the midpoint of AC and O is the centre of the circle. Prove that +AOD = +CBA.
(b) If 6 such circles are stacked in the same way, what will their height be?
(c) Find the height of a stack of 21 such circles. 18. Tangents BE and FD are common to the circles with centres A and C. Prove that +FBD = 90°.
17. (a) Three equal circles with radius r are stacked as shown. Find the height of the stack. 19. A cart is made with wheels of radii R and r. The distance between the centres of the wheels is 3R + r. Find the length, AB, of the top needed to be put on the cart, if AB is a tangent to both circles.
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20. A pair of earrings is made with a wire surround holding a circular stone, as shown. Find the total length of wire needed for the earrings.
21. The sides of a quadrilateral ABCD have midpoints P, Q, R and S, as shown below.
(a) Show that ∆ DPS is similar to ∆ DAC. (b) Show PS < QR. (c) Show that PQRS is a parallelogram. | 677.169 | 1 |
...that which has all its angles right angles, but has not all its sides equal. XXVII. A rhombus is that which has all its sides equal, but its angles are not right angles. XXVIII. A rhomboid is that which has its opposite sides equal to one another, but all its sides are .all its sides are...
...which has all its angles right angles, but has not all its sides equal. i XXVII. A rhombus, is that which has all its sides equal, but its angles are not ' right angles. - - _ . XXVIII. A rhomboid, is that which has its opposite sides equal to one another; but all its...
...all its angles right angles, but not all its . sides equal, as A BC D, fig. 1.5. — A rhombus is a figure which has all its sides equal, but its angles are not all right angles, as EFGH, fig. 14. — A rhomboid is a figure which has its opposite sides equal to...
...which has all its angles right angles, but has not all its sides equal. XXVIII — A rhombus, is that which has all its sides equal, but its angles are not right angles. XXIX. — A rhomboid, is that which has its opposite sides equal to one another. XXX.— All other...
...whieh has all its angles right angles, but has not all Its sides equal. XXVII. A rhombus, is that whieh has all its sides equal, but its angles are not right angles. XXVHI. A rhomboid, is that whieh has its opposite sides equal to one another., but all its sides are...
...that which has all its angles right angles, but has not all its sides equal. XXXII. A rhombus, is that which has all its sides equal, but its angles are not right angles. XXXIII. SeeN. A rhomboid, is that which has its opposite sides equal to one another, but all its sides all its sides are...
...is a four-sided figure which has all its sides equal, and all its angles right angles. " A rhombus is a four-sided figure which has all its sides equal...possibility of its having which requires to be proved. Thus, as will appear hereafter, it can be proved, that if a quadrilateral have all its sides equal,...
...that which has all its angles right angles, but has not all its sides equal. XXVIII. A rhombus is that which has all its sides equal, but its angles are not right angles. XXIX. A rhomboid is that which has its opposite sides / ~7 equal to one another, but all its sides... | 677.169 | 1 |
Symmetry Operations
()
Click the image above for pages which explain the basic symmetry operations – Proper Rotations, Reflections, Improper Rotations and Centres of Inversion.
Care must be taken not to confuse (a) an inversion operation with (b) a two-fold rotation. Although the two operations may sometimes appear to have the same effect, that is not the case in general. An S1 axis is equivalent to a mirror plane and an S2 axis is equivalent to a centre of inversion. | 677.169 | 1 |
A school Euclid, being books i. & ii. of Euclid's Elements, with notes by C. Mansford
Dentro del libro
Página 14 ... diameter of a circle is a straight line drawn through the centre , and terminated both ways by the circumference . 18. A semicircle is the figure contained by a diam- eter and the part of the circumference cut off by the diameter . 19 ...
Página 62 ... diameter bisects them . For AB being equal to CD , and BC common , the two sides AB , BC are equal to the two sides DC , CB , each to each ; and the angle ABC has been proved equal to the angle BCD ; therefore the triangle ABC is equal ... | 677.169 | 1 |
Parallel And Perpendicular Lines Worksheet Answers Key Geometry. A parallel perpendicular answer key is usually found on the second page of the worksheets or on a separate key. Worksheets for parallel and perpendicular line worksheets parallel lines never intersect. Parallel and Perpendicular lines worksheet from Lines parallel perpendicular worksheet key answer lee mr excel […]
Parallel Perpendicular Or Neither Worksheet Answers. In addition to this, the set also contains practice worksheets on determining whether two lines are parallel or perpendicular, using points, slopes and equations. The line is parallel to the line we know how to use the. Parallel Perpendicular or Neither Worksheet Answer Key with from A pair […]
Parallel Perpendicular Or Neither Card Sort Answer Key. Learn vocabulary terms and more with flashcards games and other study tools. Check parallel perpendicular or neither answer key. Parallel Perpendicular Or Neither Worksheet oke blogger from leo-arek.blogspot.com Horizontal and vertical lines are also included.video preview linkdigital card sortstudents will use the drop. If the lines intersect, […]
Algebra 1 Worksheet 3.6 Parallel And Perpendicular Lines Answer Key. This worksheets will be accessed by two parallel, they never repeat so much faster than worksheets are being blocked or make it with no! An algebra worksheet such as algebra 1 worksheet 3.6 parallel and perpendicular lines is a wonderful way to sharpen your math […]
Parallel And Perpendicular Lines Worksheet Answer Key Algebra 1 5.6. This worksheets was uploaded at november 18, 2020 by admin in worksheets. If two non vertical lines in the same plane intersect at a right angle then they are said to be perpendicular. Iso Llel And Perpendicular Worksheet Algebra 1 5.6 Homework Answer from cdn.thingiverse.com | 677.169 | 1 |
If one of the interior angles of a regular polygon is found to be equal to (98) times of one of the interior angles of a regular hexagon, then the number of sides of the polygon is
A
4
B
5
C
1
D
8
Video Solution
Text Solution
Verified by Experts
The correct Answer is:D
|
Answer
Step by step video, text & image solution for If one of the interior angles of a regular polygon is found to be equal to (9/8) times of one of the interior angles of a regular hexagon, then the number of sides of the polygon is by Maths experts to help you in doubts & scoring excellent marks in Class 14 exams. | 677.169 | 1 |
Property of a trapezoid related to base angles
Theorem 1:
Draw perpendicular lines AE and BF between the parallel sides of the trapezoid.
In ΔAED and ΔBFC,
AD = BC [Isosceles trapezoid]
AE = BF [Distance between parallel lines will always be equal]
∠AEB = ∠BFC=90° [AEꞱCD and BFꞱCD]
If two right-angled triangles have their hypotenuses equal in length and a pair of shorter sides are equal in length, then the triangles are congruent.
∴ ΔAED ≌ ΔBFC [RHS congruence rule]
We know that the corresponding parts of congruent triangles are equal.
Hence, ∠ADC = ∠BCD
And ∠EAD = ∠FBC
Now, ∠BAD = ∠BAE + ∠EAD
∠BAD = 90° + ∠EAD
∠BAD = 90° + ∠EAD
∠BAD = ∠ABC
Hence, each pair of base angles of an isosceles trapezoid is congruent.
Property of trapezoid related to the length of diagonals
Theorem 2:
The diagonals of an isosceles trapezoid are congruent.
Given: In trapezoid ABCD, AB∥CD, and AD = BC
To prove: AC = BD
Proof:
In ΔADC and ΔBCD,
AD = BC [Isosceles trapezoid]
∠ADC = ∠BCD [Base angles of isosceles trapezoid]
CD = CD [Common]
Therefore, ΔAED ≌ ΔBFC [SAS congruence rule]
We know that the corresponding parts of congruent triangles are equal.
So, AC = BD
Hence, the diagonals of an isosceles trapezoid are congruent.
Property of trapezoid related to the length of diagonals
Theorem 3:
In a trapezoid, the midsegment is parallel to the bases, and the length of the midsegment is half the sum of the lengths of the bases.
Given: In trapezoid ABCD, AB∥CD, and X is the midpoint of AD, Y is the midpoint of BC.
To prove: XY = 1/2 × (AB+CD)
Proof: Construct BD such that the midpoint of BD passes through XY.
In ΔADB, X is the midpoint of AD, and M is the midpoint of DB.
So, XM is the midsegment of ΔADB.
We know that a line segment joining the midpoints of two sides of the triangle is parallel to the third side and has a length equal to half the length of the third side. [Midsegment theorem]
∴ XM ∥ AB and XM = 1/2 × AB …(1)
In ΔBCD, Y is the midpoint of BC and M is the midpoint of BD.
So, MY is the midsegment of ΔBCD.
∴ MY ∥ CD and MY = 1/2 × CD …(2) [Midsegment theorem]
Since XM ∥ AB and MY ∥ CD, so, XY
Now, XY=XM+MY
= 1/2 × AB + 1/2 × CD
XY = 1/2× (AB+CD)
Kite
A kiteis a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other.
(or)
A parallelogram also has two pairs of equal-length sides, but they are opposite each other in a kite.
Only one diagonal of a kite bisects the other diagonal.
Property of kite related to the angle between the diagonals
Theorem:
The diagonals of a kite are perpendicular.
Given: In kite WXYZ, XY=YZ, WX=ZW
To prove: XZ Ʇ WY
Proof: Draw diagonals XZ and WY. Let the diagonals intersect at O.
In ΔWXY and ΔWZY,
WX=WZ [Adjacent sides of kite]
XY=ZY [Adjacent sides of kite]
WY=YW [Reflexive property]
∴ ΔWXY ≌ ΔWZY [SSS congruence rule]
We know that the corresponding parts of congruent triangles are equal.
So, ∠XYW= ∠ZYW …(1)
In ΔOXY and ΔOZY,
XY=ZY [Adjacent sides of kite]
OY=YO [Reflexive property]
∠XYW= ∠ZYW [from (1)]
∴ ΔOXY ≌ ΔOZY [SAS congruence rule]
So, ∠YOX = ∠YOZ [CPCT]
But ∠YOX+∠YOZ = 180°
2∠YOX = 180°
∠YOX = 90°
Hence, the diagonals of a kite are perpendicular.
Exercise
Find the value of k if STUV is a trapezoid.
If EFGH is an isosceles trapezoid, find the value of p.
Find the length of PQ if LMQP is a trapezoid and O is the midpoint of LP and N is the midpoint of MQ.
What must be the value of m if JKLM is a kite?
If WXYZ is a kite, find the length of diagonal XZ.
Concept Map
What we have learned
A quadrilateral having only one pair of parallel sides is called a trapezoid.
A kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other | 677.169 | 1 |
If one side of a triangle be produced, and the exterior angle, also one of the interior and opposite angles, be each of them bisected; the remaining interior and opposite angle will be double the angle made by the bisecting lines.
Let CBA be a A. Prod. BC to D, and let the ext. ACD be bisected by the str. line CE, and the int. and opp. ABC by the str. line BE. Then will BAC = 2▲ BEC.
1st. The lines CE, BE, will meet in some
point, as E. For
LACD> <ABC,
ACD> ABC,
Prop. 15.
i. e.
.. but
<ECD> <EBC:
add to each the ECB.
ECD+ ECB > < EBC + / ECB; ECD+ ▲ ECB= 2 rt. s,
ZEBC+ ECB < 2 rt.
S.
Prop. 12.
.. CE, BE, will meet on the same side of Ax. 12. the str. line BC on which the ABC is.
CBE + BEC,
2ndly. In BCE, the int. and opp.s, Prop.31.
ext. DCE
..2/DCE or / DCA = 2 / CBE+2 / BEC·
ZABC = = 2/ CBE,
but
DCA =
In Δ ABC,
ZABC+2 BEC.
ABC+BAC; Prop.31.
ext.DCA =
.. ZABC+▲ BAC = ZABC+2 BEC. Ax. 1. from these equals take ▲ ABC
BAC
2/ BEC.
Wherefore, if one side of a triangle, &c.
Note. This proposition is introduced on account of the explanation that it affords of the principle upon which the sextant depends.
Ax. 3.
DEFINITIONS.
ང་
EVERY right-angled parallelogram, or rectangle, is said to be contained by any two of the straight lines which contain one of the right angles.
II.
In every parallelogram, any of the parallelograms about a diameter, together with the two complements, is called a Gnomon. 'Thus the parallelogram HG, together with the complements AF, FC is the Gnomon; which is more briefly expressed by the letters AGK or EHC, which are at the opposite angles of the parallelograms which make the gnomon.'
N.B.-The rectangle contained by any two straight lines AB, AD, is generally expressed by the words "rectangle of AB and AD;" or by "AB. AD."
PROP. XLI.
THEOR.
1. 2 Eu.
If there be two straight lines, one of which is divided into any number of parts; the rect
angle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the several parts of the divided line.
Let A and BC be two str. lines, and let BC be divided into any parts in D and E; then A.BC=A. BD+A.DE+A. EC
BF BC, BG= A;
draw
GH || BC; and DK, EL, CH || BG.
.. A. BC=A. BD+A.DE+A.EC. Wherefore, if there be two str. lines, &c.
PROP. XLII. THEOR. 3. 2 Eu.
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 rect
Prop. 10. Prop. 3.
Prop. 30.
Prop. 33.
angle contained by the two parts, together with the square of the aforesaid part.
Let the str. line AB be divided into two parts in the pt. C; then AB. BC= AC. BC +BC2.
.. AB. BC= AC. BC + BC3.
Therefore, if a str. line be, &c.
Note. If the parts of the divided line be a and b; the proposition algebraically expressed, is a+b.a=ab+a2.
PROP. XLIII. THEOR. 4. 2 Eu.
If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts. | 677.169 | 1 |
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