text
stringlengths 6
976k
| token_count
float64 677
677
| cluster_id
int64 1
1
|
|---|---|---|
On a horizontal plane there is a vertical tower with a flag pole on the
top of the tower. At a point 9 metres away from the foot of the tower the
angle of elevation of the top and bottom of the flag pole are 600and300
respectively. Find the height of the tower and the flag pole mounted on
it.
Video Solution
Text Solution
Verified by Experts
In ΔACD tan60o=ACDC
=√3=AC9
=AC=9√3m
In ΔDBC tan300=BCDC
=1√3=BC9
BC=3√3m
AC=AB+BC AB=AC−BC AB=9√3−3√3=6√3m
Height of the flag pole is 6√3m
and that of tower is 3√3m
Knowledge Check
Question 1 - Select One
On a ground , there is a vertical tower with a flagpole on its top . At a point 9 metre away from the foot of ther tower , the angles of elevation of the top and bottom of the flagpole are 60∘and30∘ respectively . The height of the flagpole is :
Aa) 5√3 metre
Bb) 6√3 metre
Cc) 6√2 metre
Dd) 6√5 metre
Question 2 - Select One
At a point 20 m away from the foot of a tower, the angle of elevation of the top of the tower is 30∘ The height of the tower is
A20√3m
B20√3m
C√320m
DNone of these
Question 3 - Select One
At a point 20 m away from the foot of a tower, the angle of elevation of the top of the tower is 30∘ The height of the tower is | 677.169 | 1 |
Elements of Geometry: With Practical Applications ...
Dentro del libro
Resultados 1-5 de 34
Página 24 ... radii equal respectively to the lines B and C , describe arcs intersecting at the point G ( Post . III ) . Join DG , FG ( Post . I ) , and the tri- angle DGF will be the triangle required , since the three sides are equal to the three ...
Página 27 ... radii , describe arcs intersecting each other at D ; then , BD being drawn , it will bisect the angle ABC . For , drawing AD , CD , the three sides of the triangle ABD are equal respectively to the three sides of the triangle CBD ...
Página 28 ... radii , describe arcs ( Post . III ) intersecting at C and D. Draw CD ( Post . I ) , and it will be perpendicular to AB , and will bisect it at the point F. A B For , by joining AC and BC , AD and BD , we shall have two triangles CAD ...
Página 66 ... radii of the two circles ; therefore CA + CB2 must always amount to the same constant value . In a similar way it may be shown that the sum of the squares of the two lines , drawn from any point in the circumference of the smaller ...
Página 71 ... radii and the intercepted arc , is called a sector . The space BCH is a sector . 5. When a straight line touches the circumference in only one point , it is called a tangent ; and the common point of the line and circumference is called | 677.169 | 1 |
Transformations of the Plane
Directions
Below is a linear transformation matrix composed of two column vectors. The two column vectors are the unit vectors i and j, and will change as you drag points B and C around the plane. As you are dragging B and C around, notice how the Linear Transformation Matrix and its determinant change with respect to the area of the parallelogram.
Click the refresh symbol in the top right of the Geogebra window to reset your applet at anytime.
Answer the following questions:
1. After you've finished playing, click the refresh symbol in the top right of the Geogebra window. Now, let's try out some transformations.
Record the matrix with your unit vectors in standard position
Record the matrix after a 90 degree rotation around the origin
Record the matrix after a 180 degree rotation around the origin (you may need to refresh first)
Record the matrix after a 270 degree rotation
What can you conclude about how the matrix changes after a rotation?
2. Great job! Let's continue finding transformations. This time, we'll reflect over certain lines. Make sure to go ahead and hit that refresh symbol so your unit vectors are back in standard position.
Record the matrix with your unit vectors in standard position
Record the matrix after a reflection over the y-axis
Record the matrix after a reflection over the x-axis
Record the matrix after a reflection over the line y=x
What can you conclude about how the matrix changes after a reflection?
3. Set the magnitude of vector i to 3 and the direction to 0 degrees. Set the magnitude of vector j to be 3 and the direction to 90 degrees. How would you describe how the transformation that just took place? What is the new matrix?
4. What is the determinant of the matrix when the area of the parallelogram is 1?
5. What happens to the determinant of the matrix when you make the area 0? | 677.169 | 1 |
Circle class 10 pdf download
Ncert books pdf download 2020 for class 12, 11, 10, 9, 8. Cbse class 10 previous year papers for pdf download all. Next take a point p on the circle and draw tangents through this point. This website is created solely for jee aspirants to download pdf, ebooks, study materials for free. Ncert solutions cbse notes class 6 class 7 class 8 class 9 class 10 class 11 class 12. Its time to tell to the world that you also know certain things. Find the length of the chord of the larger circle which touches the smaller circle. If you want to download pdf, you can download file by clicking on the given download and save it on your mobile or laptop or pc. Register at byjus for circles class 10 ncert solutions. Ncert solutions for class 10 maths chapter 10 circles is given below to free download in pdf or use online without. Area related to circles perimeter and area of a circle. Ncert exemplar problems class 10 maths circles cbse tuts. While the revision of whole cbse class 10 syllabus is a must at this stage, students must also solve cbse class 10 previous year question papers.
Ncert solutions for class 10 maths chapter 10 circles pdf. Important questionsmcq for area related to circleclass 10 It is very useful for the students as they get all the maths formula at a one place. May 14, 2020 class 10 science ncert book is available here for download in pdf format. Ncert solutions for class 9 maths chapter 10 circles. The chapter consists of a total of 4 exercises and example problems. This study material is very important for your class 10 board exam preparation. Ncert solutions for class 10 maths chapter 10 circles download free pdf of cbse class 10 maths circles ncert solutions solved by. Perimeter of circle 2pr where r is the radius of the circle. Aug 30, 2015 noteif you liked ncert books in english for class 10 pdf download. With cbse class 10 board exams knocking at the door, students must be wondering how to go about with their cbse class 10 preparation at this last moment. Find the relation between x and y if the points a x, y, b 5, 7 and c 4, 5 are collinear.
Mcq questions for class 10 maths with answers was prepared based on latest exam pattern. If you think the materials are useful kindly buy these legally from publishers. We have provided step by step answers to all the questions provided in the ncert class 10 maths textbook. Selina concise mathematics class 10 icse solutions circles. The topics and subtopics in chapter 10 circles class 10 maths are given below. Benefits of solving cbse class 10 previous year papers. Ncert class 9 maths solutions for chapter 10 circles. A non terminating repeating b non terminating non repeating c terminating d none of the above question 2 the volume of the largest right circular cone that can be cut out from a cube of edge 4. Finally, take a point p outside the circle and try to draw tangents to the circle from this point. Solutions are updated for current academic session 202021 for all students who are using latest ncert books 202021. You have already observed that there is only one tangent to the circle at such a point see fig.
Area of the circle pr 2 the perimeter of a circle is the length of its boundary. Ncert solutions for class 10 maths chapter 10 circles. Vidyakul understands the difficulties faced by the students in class 10 while facing a science, thus, we bring them a solution to lower down the pressure and increase their selfconfidence. The app provides all the major cbse books, cbse board sample papers, cbse previous year papers with solutions, cbse question papers, free ncert. All ncert math solution for class 10th is available in these pdf files. The structure of the exam is such that these exams test the students theoretical as well as practical knowledge thoroughly. You also know from your earlier classes, that circumference of a circle bears a constant ratio with its diameter. Mar 07, 2019 ncert solutions for class 10 maths questions with solutions today we provided free solution for mathematics student for class 10th. Sep 22, 2019 ncert solutions class 10 maths chapter 10 circles here are all the ncert solutions for class 10 maths chapter 10. The path of all points that are equidistant from a fixed point is called a circle.
Jun 04, 2018 ncert solutions for maths circles download as pdf. Link of pdf file is given below at the end of the questions list. Circles notes for class 10 math chapter 10 download pdf. Given below are the links of some of the reference books for class 10 math. A tangent pq at a point p of a circle of radius 5 cm meets a line through the centre o at a point q so that oq 12 cm Ncert solutions for class 10 math chapter 10 circles download pdf download pdf. Cbse class 10 full maths crash course part 2 maths marathon revision score 100%. Important questionsmcq for area related to circleclass.
Here on aglasem schools, you can access to ncert book solutions in free pdf for maths for class 9 so that you can refer them as and when required. Ncert book for class 10th science 202021 free pdf download. This section include lot of cbse related questions on area of circle. Nov 08, 2019 provides you free pdf download of ncert exemplar of class 10 maths chapter 9 circles solved by expert teachers as per ncert cbse book guidelines. Taking cbse class 10 previous year papers will help students build concentration and stamina to sit for the full time of the exam. Welcome to mpsc material website in this post we will share maharashtra state board books for free download in marathi and today is the day of maharashtra state board 10th std books pdf. Ncert solutions, cbse classes 6 to 12 is the official cbse study app for all the subjects, presented by full circle publishers, new delhi earlier known as full marks publishers in collaboration with edurev. The app provides all the major cbse books, cbse board sample papers, cbse previous year papers with solutions, cbse question papers, free ncert solutions for. Some main topics of biology are life processes, heredity, and evolution.
May 01, 2020 ncert solutions for class 10 maths can be downloaded from here in form of chapterwise pdf. We provide step by step solutions for icse mathematics class 10 solutions pdf. Students can solve ncert class 10 maths circles mcqs with answers to know their preparation level. Mcq questions for class 10 maths circles with answers. Area related to circles notes for class 10 math chapter 12. Download ncert solutions for class 10 biology here. It will help students get an idea about nature and exam pattern of the exam. Ncert solutions for class 10 subjectwise free pdf download.
Noteif you liked ncert books in english for class 10 pdf download. Sep 08, 2019 get free ncert solutions for class 10 maths chapter 10 ex 10. Important questions download cbse important questions. Ncert solutions for class 10 maths free pdf download. Circles class 10 maths ncert solutions are extremely helpful while doing your homework or while preparing for the exam. The chapter starts by stating the definition of the circle which was studied in class ix. We know that the area of a circle is the measurement of the region enclosed by its boundary. Maths, science, hindi, english, history, civics, geography.
Icse solutions for class 10 mathematics circles a plus topper. Mcq questions for class 10 maths circles with answers learn. Ncert solutions class 10 maths chapter 10 circles download pdf for free. This solution contains questions, answers, images, explanations of the complete chapter 10 titled circles of maths taught in class 10. Short answer type questions i 2 marks in figure, ab is the diameter of a circle with centre o and at is. It contains all type of questions which comes in fa and sa. Download the latest edition of ncert book for effective learning in the new session 202021. Ncert solutions for class 10 math chapter 10 circles download pdf download. Ncert solutions for class 10 maths chapter 10 circles is given below to free download in pdf or use online without downloading. Introduction in this power point presentation we will discuss about circle and its related terms. From a point q, the length of the tangent to a circle is 24 cm and the distance of q from the centre is 25 cm. A free booklet of physics formula book for class 10 of cbse board and other boards.
Ncert solutions for class 10 maths chapter 10 circles free pdf available on vedantu are solved by experts. Free download ncert solutions for class 9 maths chapter 10 exercise 10. The ncert class 10th maths textbooks are well known for its updated and thoroughly revised syllabus. The national council of educational research and training ncert publishes maths textbooks for class 10. This is important for students to get a feel of how the. In the given figure, o is the centre of the circle. Download free science and maths ebooks and pdfs for class 10. Ncert solutions for class 10 maths can be downloaded from here in form of chapterwise pdf. Full circle education ncert solutions cbse classes apps. If you continue browsing the site, you agree to the use of cookies on this website. In q 1 to 3, choose the correct option and give justification. Download ncert solutions for class 10 maths chapter 10 circles pdf ex 10. In it, we will discuss human life processes, parts of the human body and their activities. Class 10 is a stepping stone for students academic. | 677.169 | 1 |
Activities to Teach Students Complementary Angle Identities
Complementary angle identities are an essential part of geometry and trigonometry. They involve the relationship between angles that add up to a right angle, which is 90 degrees. Teaching students complementary angle identities can be tricky, but with the right approach, it can be a fun and engaging learning experience. Here are some activities that can make the process of teaching complementary angle identities easier.
1. Learning through visual aids:
Students can create collages or posters that depict complementary angle identities. They can use markers, colored pencils, or paint to draw complementary angle pairs and the formula that relates them. This activity encourages creativity and helps students to visualize the concept.
2. Using technology:
There are many online resources available that can help students learn about complementary angle identities. Teachers can use interactive websites or mobile apps that provide engaging and interactive content. These resources can make the learning experience more interesting and help students to retain the information better.
3. Real-world application:
Teachers can use real-life examples to help students understand the concept of complementary angle identities. For instance, when a ladder is leaning against a wall, the angle between the ladder and the ground is the complementary angle of the angle between the wall and the ground. This activity helps students to see how the concept applies to real life situations.
4. Group activities:
Teachers can divide students into small groups and ask them to solve complementary angle problems together. Each group can use different strategies to approach the problem, and then compare their answers with the rest of the class. This activity promotes teamwork and collaboration and encourages students to use their problem-solving skills.
5. Gamification:
Teachers can turn the learning experience into a game by creating a quiz or a trivia competition. Questions can be based on complementary angle identities and their applications in different contexts. This activity helps students to practice their knowledge and reinforces the learning process.
In conclusion, teaching complementary angle identities can be challenging, but with these activities, it can be a rewarding and enjoyable experience. By using visual aids, technology, real-life examples, group activities, and gamification, teachers can make the learning process engaging and interactive. These activities help students to understand the concept better, apply it to real-life situations, and retain the information longer | 677.169 | 1 |
find the missing angle measure of a quadrilateral worksheet
Find The Measure Of A Missing Measure | 677.169 | 1 |
To be clearer, $45^{\circ} = \frac{\pi}{4}$ radians which estimates to $0.78539816$ radians;
whereas, $\sin(45^{\circ}) = \frac{\sqrt2}{2}$ estimates to $0.70710678$ which is a ratio and has no units.
Why does $\sin(0.78539816) = 0.70710678$? Or why does the $\arcsin(0.70710678) = 0.78539816$ radians?
I want to be able to convert ratios of sides to angles without a trigonometric calculator or trigonometric tables.
I´d like to be able to go from knowing that $\sin(45^{\circ}) = \frac{\sqrt2}{2}$and then be able to tell how many radians and degrees it is without a trigonometric calculator or trigonometric tables. Is this possible?
Or considering the 3-4-5 right triangle, I´d like to be able to know that the $\arcsin(\frac{4}{5}) = 0.92729522$ radians or $53.13010235^{\circ}$ without a trigonometric calculator or trigonometric tables. I
I was thinking that it related to the domain $[-1,1]$ of the sin function, but radians are the range units on the sine wave curve. Is it somehow related to polar coordinates where $x=r*\cos(θ)$ and $y=r*\sin(θ)$?
Also, this website seems to hint at an extension of the radius to calculate tangent so might that be the difference in the numbers? Link
$\begingroup$I guess so if it will give me the number of radians so I can convert to degrees, but not if I have to measure using a protractor; that would not be very accurate. Does it require integrals and summation like in this example: math.stackexchange.com/questions/421892/…? I´d need a refresher on how to go through that calculation if for example x = 4/5.$\endgroup$
$\begingroup$@littleO Yes and I understand the common right triangle conversions (1-1-sqrt(2) and 1-sqrt(3)-2). My question is related to calculating any ratio of sides to radians and thus degrees. How do you you convert arcsin(4/5) to radians in terms of pi without a trig calculator or trig table?$\endgroup$
There is a reasonably easy to remember Taylor series for arctangent: $$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots.$$ This converges quickly when $x$ is small, and badly or not at all when $x$ is large. So our first step is to switch to $\arctan \frac34$ instead of computing $\arctan \frac43$: this will give us the complementary angle, the one that's approximately $37^\circ$ instead of the one that's approximately $53^\circ$.
But the series for $\arctan \frac34$ doesn't converge particularly quickly either: if we take $\frac34 - \frac{(3/4)^3}{3} + \frac{(3/4)^5}{5} - \frac{(3/4)^7}{7}$, we get $0.6377\dots$, which isn't very close to $\arctan \frac34 = 0.6435\dots$.
To speed things up, we try to make clever use of the identity $$\arctan x + \arctan y = \arctan \frac{x + y}{1 - xy}.$$ Specifically, if we want to compute $\arctan x$, we try to pick $y$ such that both $y$ and $\frac{x+y}{1-xy}$ are smaller than $x$.
In the case of the angles for Pythagorean triples, there is a built-in special case of this identity that makes our lives easier: in an $(a,b,c)$ right triangle, $$\arctan \frac ab = 2 \arctan \frac{c-b}{a}.$$ So instead of computing $\arctan \frac34$, we can compute $2 \arctan \frac13$ instead.
This works much better: even with just the first $3$ terms, we get $$2 \left(\frac13 - \frac{(1/3)^3}{3} + \frac{(1/3)^5}{5}\right) = \frac{782}{1215} \approx 0.6436\dots$$ which gets us three correct decimal digits and most of the way to a fourth.
Even better, when you want to compute $\arctan x$, is to approximate $x$ by a close but simpler $y$, and use a variant of the identity above: $$\arctan x = \arctan y + \arctan \frac{x-y}{1+xy}.$$ In the case of $\arctan \frac34$, we can approximate $\frac34$ by $1$, and get $$\arctan \frac34 = \arctan 1 + \arctan \frac{\frac34 - 1}{1 + \frac34 \cdot 1} = \frac\pi4 - \arctan \frac17.$$ Assuming you know lots of digits of $\pi$, we now have $$\arctan \frac34 \approx \frac\pi4 - \frac17 + \frac{(1/7)^3}{3} = 0.64351\dots$$ which is four digits of accuracy, after only two terms of the Taylor series! | 677.169 | 1 |
An angle is formed when two straight lines intersect or meet at a point.
The sum of angle at the centre of a complete circle is 3600.
TYPES OF ANGLES
Acute angles: - These are angles that are less than 900.
Right angle: - This is an angle that is equal to 900.
Obtuse angles: - These are angles that are greater than 900 but less than 1800.
Reflex angles: - These are angles that are greater than 1800 but less than 3600.
Complementary angles: - These are angles that sum up to 900.
Supplementary angles: - These are angles that sum up to 1800.
EVALUATION
Define an angle.
Mention five types of angle and describe them.
CONSTRUCTION OF ANGLES
The following are several ways of constructing or drawing angles:
Using a setsquare
Using a combination of setsquares
Using a protractor
Using a pair of compasses
HOW TO CONSTRUCT ANGLES USING SETSQUARE
Standard angles are angles 300,450, 600 and 900. These angles can be drawn by simply using a setsquare. The method is simply choosing the appropriate setsquare that has the angle that is to be drawn. Thus 300-600 setsquare is used to draw 300, 600 and 900. Similarly, the 45-45 setsquare is used to draw 450 and 900.
EVALUATION
With the aid of a set square construct the following angle: i)900 ii) 600 iii)450
COMBINATION OF SETSQUARES
Setsquares can be combined to construct the following angles:
To construct 750, combine 450 and 600 setsquares
1800- (600 + 450)= 750
To construct 1350, combine 450 and 900
450+ 900 = 1350
To construct 1050, combine 600 and 450
450+ 600 = 1350
EVALUATION
Combine the sets-square to construct angles750 and 1350.
Construct angle 60o
HOW TO DRAW AN ANGLE USING A PROTRACTOR
Procedure
Place the protractor on the line from which the angle is to be formed. The baseline marked O at one end and, 180 marked at the other end.
Set the centre of the protractor to start from where the vertex of the angle will be.
Mark the required degree of the angles to be drawn.
Use a straight line to join the marked point at the angle to the point vertex of the angle
EVALUATION
Measure the following angles using your protractor
a) 880 b)1700 c)2750 d) 3000 e)150
Construct angle 30o
How to construct Angles Using a Pair of Compasses
600
Procedure
Draw a straight line and mark the centre O.
From O, using a convenient radius at point A draw an arc (almost a quadrant).
With same radius draw another arc to intersect the first arc at C.
Draw a line from O through the point of intersection.
The angle formed is 600
900
Procedure
Draw a straight line and mark the centre O.
From O draw a semicircle to touch the straight line at A and B.
From A, with longer radius, draw an arc at the centre up.
From B, with longer radius, draw an arc at the centre up to intersect at C.
Draw a line from O through C.
The angle COA and COB is 900
BISECTION OF ANGLES
To bisect a given angle
Draw a given angle ABC.
With centre B and any convenient radius draw an arc to cut AB to D and BC at E.
With centre A1 and any small radius an arc.
With centre E and the same radius draw an arc to intersect the previous one at F. | 677.169 | 1 |
Mastering Medians and Triangles in Geometry: The Ultimate Collection of Theorems and Problems on
gogeometry.com
This website provides a collection of problems and
theorems related to medians and triangles in geometry. Some
of the topics covered include the properties of medians,
centroids, inequalities involving medians, and the
relationship between medians and other elements of a
triangle.
The website offers a variety of problems with
detailed solutions, as well as interactive applets and
videos to help students better understand the concepts. The
problems range in difficulty from basic to advanced, making
it a useful resource for both beginners and advanced
learners of geometry.
Overall, this website is a great
tool for anyone looking to improve their understanding of
medians and triangles in geometry, and is particularly
useful for students preparing for geometry competitions or
exams. | 677.169 | 1 |
The bisector of the angle BAC intersects the circle described about the triangle ABC at point D
The bisector of the angle BAC intersects the circle described about the triangle ABC at point D, so that BC = 4 DC = √5 find the radius of the circle and the cosine of the angle BAC.
From the center of the circle O we construct the radii OB, OD, OC.
We denote their length by R.
<BAD = <DAC (formed by the bisector AD).
The central angles are equal and resting on the same arcs:
<BOD = <DOC.
ΔBОС is isosceles, the segment OK is its bisector, and, consequently, the height and median.
The BC segment is perpendicular to OK.
KC = BK = BC / 2 = 2.
KD = √ (DC ^ 2 – KC ^ 2) = √ ((√5) ^ 2 – 2 ^ 2) = 1.
OC ^ 2 = KC ^ 2 + (OD – KD) ^ 2;
R ^ 2 = 2 ^ 2 + (R – 1) ^ 2
R ^ 2 = 4 + R ^ 2 – 2R +1;
2R = 5
R = 2.5.
OK = OD – KD = 2.5 -1 = 1.5;
cos (<KOC) = OK / OC = 1.5 / 2.5 = 0.6.
<KOC = 2 * <DAC (central and inscribed corners);
2 * <DAC = <BAC;
<KOC = <BAC;
cos (<BAC) = cos (<KOC) = 0.6.
Answer: R = 2.5; cos (<BAC) = 0.6 | 677.169 | 1 |
Welcome to exploring the fundamental principles of geometry, focusing on the intriguing concept of points that lie on the same line.
In this article, we'll dive deep into the principles of lines in two-dimensional and three-dimensionalspaces, the tests for the points that lie on the same line, essential theorems, and the pervasive influence of these concepts in modern geometry and real-world applications.
Definition
In geometry, when points lie on the same line, they are said to be collinear. Collinearity is a property that relates points and lines in a geometric space. If two or more points fall on the same straight line, they are known as collinear points.
For instance, in a two-dimensional plane, if you draw a straight line, any number of points along that line would be considered collinear. This is because they all share the same line of reference. In three dimensions, the concept stays the same; if you can connect all the points using a single straight line, they are considered collinear. The word "collinear" comes from the Latin "col" (meaning "together") and "linear" (meaning "line"), directly illustrating this concept.
This concept is a foundational aspect of many principles in geometry and is utilized in many real-world applications, from architectural design to satellite navigation systems.
Figure-1.
Properties
Points that lie on the same line, also known as collinear points, have several key properties. Here are some of the most notable properties:
Unlimited Number of Points
A line can have an infinite number of points. As long as these points are located on the same straight line, they are considered collinear.
Two Points Form a Line
Any two points in space will always form a line; thus, by definition, any two points are always collinear.
Three or More Points
When dealing with three or more points, collinearity becomes less trivial. Three or more points are collinear if they all lie on the same straight line.
Measure of Collinearity
We can determine if three points are collinear in a two-dimensional plane by examining the slopes between each pair of points. If the slopes are equal, the points are collinear.
In three dimensions, one common method is to calculate the area of the triangle formed by the points; if the area is zero, the points are collinear.
Order of Points
The order of points does not affect collinearity. For example, if points A, B, and C are collinear, it doesn't matter whether B is between A and C or if C is between A and B. They remain collinear.
Distance
The distance between any two points is the shortest along the line on which the points lie.
Line Segments
When three or more points are collinear, the line segment connecting the two extreme points equals the sum of the line segments connecting the intermediate points.
Exercise
Example 1
Are the points A(2,3), B(4,6), and C(6,9)collinear?
Solution
We can find the slopes between AB, BC, and AC. If they are equal, then the points are collinear.
Slope AB = (6-3)/(4-2) = 1.5
Slope BC = (9-6)/(6-4) = 1.5
Slope AC = (9-3)/(6-2) = 1.5
All three slopes are equal, so points A, B, and C are collinear.
Figure-2.
Example 2
Are the points A(-1,-2), B(0,0), and C(3,6)collinear?
Solution
Again, we find the slopes:
Slope AB = (0-(-2))/ (0-(-1)) = 2
Slope BC = (6-0)/(3-0) = 2
Slope AC = (6-(-2))/(3-(-1)) = 2
Since the slopes are equal, the points A, B, and C are collinear.
Example 3
Are the points A(2,4), B(5,7), and C(10,12)collinear?
Solution
We find the slopes:
Slope AB = (7-4)/(5-2) = 1
Slope BC = (12-7)/(10-5) = 1
Slope AC = (12-4)/(10-2) = 1
The slopes are equal, so the points A, B, and C are collinear.
Example 4
Are the points A(-2,-3), B(-1,-2), and C(-3,-5)collinear?
Solution
We find the slopes:
Slope AB = (-2-(-3))/(-1-(-2)) = 1
Slope BC = (-5-(-2))/(-3-(-1)) = 1.5
Slope AC = (-5-(-3))/(-3-(-2)) = 2
The slopes are not equal, so points A, B, and C are not collinear.
Figure-3.
Applications
The concept of points that lie on the same line, or collinearity, is foundational to many areas of study and industry. Here are a few examples of its wide-ranging applications:
Computer Graphics and Design
In this field, the principles of lines and points are fundamental for creating visual content. The understanding of points and lines helps in creating geometric shapes, digital illustrations, 3D modeling, and animation. Collinearity can be used in algorithms for line drawing, polygon filling, and more.
Physics
In physics, particularly in mechanics, the concept of collinear points is vital. For example, when analyzing motion in one dimension or forces acting along the same line, collinearity is a key consideration.
Engineering
In various branches of engineering, such as civil, structural, and mechanical engineering, the concept of collinear points is crucial. It helps in understanding structural integrity, force distribution, and alignment of different components of a machine or structure.
Astronomy and Space Science
Collinearity has important applications in celestial navigation and satellite placement. For example, collinear points in space are used to align telescopes with celestial bodies, or to align satellites in a 'train' to achieve constant coverage.
Geography and Geometric Surveying
In geographic information systems (GIS) and surveying, the concept of collinearity is used in identifying, representing, and manipulating spatial relationships between different geographic phenomena.
Machine Learning and Computer Vision
In the field of machine learning and particularly in computer vision, points lying on the same line (or more generally, geometric relationships between points) can be used for object recognition, image stitching for panorama creation, and 3D reconstruction from multiple images.
Robotics
In robotics, collinearity plays a significant role in motion planning and kinematics. Robots use this basic principle to calculate trajectories and to optimize movements.
Architecture and Urban Planning
In architecture and urban planning, points that lie on the same line are used to ensure the proper alignment of structures. This alignment can be critical for aesthetics, structural integrity, and functionality. | 677.169 | 1 |
How to calculate degree of angle
Angles are a fundamental concept in mathematics and geometry, and understanding how to calculate the degree of an angle is an important skill. This article will walk you through the various methods you can use to calculate the degree of an angle, whether you're dealing with right angles, straight angles, or any other types of angles.
1. Using a Protractor
A protractor is a simple tool that can be used to measure the degree of an angle. Follow these steps to use a protractor:
Step 1: Place the center point of the protractor on the vertex of the angle.
Step 2: Align one edge of the angle with the baseline (0-degree line) on the protractor.
Step 3: Read the degree measurement where the other edge of the angle intersects with the protractor's scale.
2. Trigonometric Ratios
Trigonometry is an area of mathematics that deals with triangles and their angles. If you know certain information about a triangle, you can use trigonometric ratios such as sine, cosine, and tangent to calculate angle measurements.
For example, if you have a right triangle and know side lengths a, b (the two legs), and c (the hypotenuse), you can use these formulas:
tan(θ) = opposite / adjacent (where θ is one of the acute angles)
sin(θ) = opposite / hypotenuse
cos(θ) = adjacent / hypotenuse
From there, you can find θ by using inverse trigonometric functions like arctangent, arcsine, or arccosine.
3. The Pythagorean Theorem
The Pythagorean Theorem states that in any right triangle, the square of the length of its hypotenuse is equal to the sum of squares of its other two sides:
c^2 = a^2 + b^2
If you know the side lengths of a right triangle, you can use this theorem to find the measures of its acute angles using trigonometric ratios.
4. The sum of angles in a triangle
The sum of angles in any triangle is always equal to 180 degrees. Therefore, if you know two angles in a triangle, you can find the third angle by subtracting the sum of the known angles from 180.
5. The Angle Addition Postulate
The Angle Addition Postulate states that if point B lies in the interior of angle AOC, then the measure of angle AOB plus the measure of angle BOC equals the measure of angle AOC.
Using this postulate, you can determine an unknown angle by adding or subtracting known angles.
Conclusion
Calculating the degree of an angle can be done using various methods such as protractors, trigonometric ratios, the Pythagorean theorem, and geometry principles like the angle sum and angle addition postulates. Understanding these concepts will help you determine angle measures in various situations and mathematical | 677.169 | 1 |
In these printable worksheets for grade 6 and grade 7 reflect the given point and graph the image across the axes and across x a y b where a and b are parameters Download the set Choose the Correct Reflection This practice set tasks 6th grade and 7th grade students to identify the reflection of the given point from the given options MATHLINKS GRADE 8 STUDENT TRANSLATIONS PACKET 13 ROTATIONS AND REFLECTIONS TRANSLATIONS Summary Ready We will perform a transformation experiment using patty paper We will define transformations of the plane and observe some properties of transformations We will learn about a transformation called a translation
This set of printable worksheets is recommended for grade 6 grade 7 and grade 8 students Reflection of a Point Direct grade 6 students to draw the line of reflection using the given equation and then plot the point across the line and watch them ace the topic Learn to Solve Grab the Worksheet Reflection of Shapes Grade 7 students should choose the correct image of the transformed point Download the set Multiple Choices Transformation The coordinates of a point are given Perform the required transformation and check mark the correct choice Download the set Transformation of Shapes
Created by Education with DocRunning This End of Year 8th grade Activity ROCKS Part writing and part art project these self reflection activities are a great way to end the year Students re visit highlights from their 8th grade experience and reflect on their hopes and goals for 9th grade as well as life dreams A reflection is a transformation that flips a figure over a line on the coordinate plane to create a mirror image This eighth grade geometry worksheet will give students practice graphing images of figures after completing given reflections All figures on this worksheet are reflected over the x axis y axis or another horizontal or vertical
8th grade Geometric transformations Reflections Reflections review Google Classroom Review the basics of reflections and then perform some reflections What is a reflection A reflection is a type of transformation that takes each point in a figure and reflects it over a line 8th grade 7 units 121 skills Unit 1 Numbers and operations Unit 2 Solving equations with one unknown Unit 3 Linear equations and functions Unit 4 Systems of equations Unit 5 Geometry Unit 6 Geometric transformations Unit 7 Data and modeling Course challenge Test your knowledge of the skills in this course Start Course challenge Math 8th grade | 677.169 | 1 |
Help me with my Project
I need this project finished, just add the answers on the word.
Name
_________________________________
I.D. Number
_______________________
Project 2
Evaluation 32
Second Year Algebra 2 (MTHH 040 059)
Be sure to include ALL pages of this project (including the directions and the assignment) when you
send the project to your teacher for grading. Don't forget to put your name and I.D. number at the top
of this page!
This project will count for 8% of your overall grade for this course and contains a possible 100 points
total. Be sure to read all the instructions and assemble all the necessary materials before you begin.
You will need to print this document and complete it on paper. Feel free to attach extra pages if you
need them.
When you have completed this project you may submit it electronically through the online course
management system by scanning the pages into either .pdf (Portable Document Format), or .doc
(Microsoft Word document) format. If you scan your project as images, embed them in a Word
document in .gif image format. Using .gif images that are smaller than 8 x 10 inches, or 600 x 800
pixels, will help ensure that the project is small enough to upload. Remember that a file that is larger
than 5,000 K will NOT go through the online system. Make sure your pages are legible before you
upload them. Check the instructions in the online course for more information.
Part A – Exploring Conic Sections in Everyday Life (possible 22 points)
Activity: Your job is to locate pictures of Circles, Parabolas, Ellipses, and Hyperbolas in the real
world around you. These conics can be man-made or something found in nature but you can't make
or draw them yourself in order to get the desired shape. If you chose to use pictures from the
Internet you MUST cite the URL for each picture – if not, you will lose points.
You will create a collage of pictures illustrating all four conic sections (circles, parabolas, ellipses,
and hyperbolas) found in nature (leaves, flowers, body parts, etc.), architecture (bridges,
doorways, etc.), and everyday items (appliances, logos, furniture, etc.).
Requirements: Your Picture Collage must contain:
1. Pictures of the entire objects where the conic section is found.
a. cut from magazines, uploaded images taken from a smart phone, or downloaded from
the Internet with URL source sited.
2. Three different examples for each of the four conic sections: circles, parabolas, ellipses, and
hyperbolas (no repeat pictures are allowed). 1 pt each
3. Trace in marker (or use an editing tool), the conic in each picture. 5 pts
4. A title for the Picture Collage. 1 pt
5. Creativity! 4 pts
Project 2
335
MTHH 040
Part B – Applications of Parabolas & Ellipses (possible 30 points)
Activity: You will answer the following questions. Clearly label the questions and show ALL
necessary steps.
1. A satellite dish is shaped like a parabola. The signals that originate from a satellite strike the
surface of the dish and are reflected to a single point, where the receiver is located. If the dish is
16 feet across at its opening and 5 feet deep at its center, at what position should the receiver be
placed? (4 pts)
2. A searchlight is shaped like a parabola. If the light source is located 3 feet from the base along
the axis of symmetry and the depth of the searchlight is 4 feet, what should the width of the
opening of the searchlight be? (4 pts)
3. The towers of a suspension bridge are 450 feet apart and 150 feet high from the roadway.
Cables are at a height of 25 feet above the roadway, midway between the towers, but gradually
get taller toward each end. Assume the x-axis is the roadway and the y-axis is the center of the
bridge, write an equation for the parabola. What is the height of the cable at a point 50 feet from
one of the towers? Round to the nearest whole number. (4 pts)
Project 2
336
MTHH 040
4. The parabolic arch shown in the figure is 45 feet above the water at the center and 180 feet wide
at the base. Will a boat that is 30 feet tall clear the arch 30 feet from the center? (Use water as
the x-axis.) (4 pts)
5. The Whispering Gallery in the Museum of Science and Industry in Chicago is 47.3 feet long and
is in the shape of an ellipse. The distance from the center of the room to the foci is 20.3 feet.
Find an equation that describes the shape of the room. How high is the room at its center? (4
pts)
6. A racetrack is in the shape of an ellipse, 100 feet long and 50 feet wide. (6 pts)
a.
Write the standard form of the elliptical
track.
b.
What is the width of the line z, which is 10
feet in from the right vertex, assuming your
new ellipse is proportional?
c.
Identify both endpoints of the new width.
7. A bridge is to be built in the shape of a parabola and is to have a span of 140 feet. The height of
the arch at a distance of 50 feet from the center is to be 20 feet. Find the height of the arch at its
center. (4 pts)
Project 2
337
MTHH 040
Part C – Martian Math (possible 12 points)
Activity: Even life has not yet been discovered on Mars,
today we know a lot of other things about the planet. Mars
has a day about 25 hours long, a pattern of seasons
similar to Earth's, and polar icecaps. Mars also has
surface temperatures that rarely rise above freezing and
almost no oxygen in its atmosphere. Mars is often called
the Red Planet because red deserts cover its surface.
1. Mars travels in an elliptical orbit with
the Sun at one of its foci. Use the data
from the diagram to calculate a, b, and
c of this elliptical orbit. (4 pts)
2. Point P is the midpoint between Mars'
closest and farthest distances to the
Sun. Use your values of a and b to
write an equation of the elliptical orbit
of Mars relative to a coordinate
system drawn through point P (y-axis
in gray). Use distances in millions of
kilometers. (3 pts)
3. It is also useful to define Mars' motion relative to the Sun. Imagine a new coordinate system (yaxis in black) with its origin at the center of the Sun. Rewrite your equation for the ellipse in this
new coordinate system. (3 pts)
4. Explain how the eccentricity of a planet's orbit can affect its annual weather cycle. (2 pts)
Project 2
338
MTHH 040
Part D – Comparing Conditional Probability (possible 24 points)
Activity: You learned about
Conditional Probability in Lesson 8.
Use your knowledge to answer the
following questions using the given
tree diagram. For example to find
P(P | D), which is the probability that
a person with a disease will test
positive for it will be P(P | D) = 0.99.
Example: Find P(H | P).
Since P(H | P) =
P ( H and P )
P (P )
, find P(H and P) and P(P).
P(H and P) = 0.98 • 0.03
P(P) = P(D and P) or P(H and P)
= 0.02 • 0.99 + 0.98 • 0.03 = 0.0492
= 0.0294
So P(H | P) =
P ( H and P )
P (P )
=
0.0294
0.0492
Substitute.
≈ 0.598
Simplify.
About 60% of the people who test positive do not actually have the disease.
Use the tree diagram in the Example above to find each probability for 1–4. (2 pts each)
1.
P(N)
2.
P(H and N)
3.
P(H | N)
4.
P(D | N)
5. Explain the difference in the meaning between P(P | D) and P(D | P) for the test in the opening
example. Compare the values of P(P | D) and P(D | P). What is the best use for this test?
Explain. (2 pts)
Project 2
339
MTHH 040
6. You can take Bus 65 or Bus 79 to get to work. You take the first bus that arrives. The probability
that Bus 65 arrives first is 75%. There is a 40% chance that Bus 65 picks up passengers along
the way. There is a 60% chance that Bus 79 picks up passengers. Your bus picked up
passengers. What is the probability that it was Bus 65? (2 pts)
7. According to estimates from the federal government's 2003 National Health Interview Survey,
based on face-to-face interviews in 16,677 households, about 58.2% of U.S. adults have both a
landline and a mobile phone, 2.8% have only a mobile phone service, but no landline, and 1.6%
have no telephone service at all.
a. Make a Venn diagram to help visualize the following probabilities: (2 pts)
b. What proportion of U.S. households can be reached by a landline call? (2 pts)
c. Are having a mobile phone and having a landline independent events? Explain. (2 pts)
8. Make a tree diagram based on the survey results below. (2 pts) Find P(a female respondent is
left-handed) and P(a respondent is both male and right-handed). (2 pts each)
Project 2
•
Of all the respondents, 17% are male.
•
Of the male respondents, 33% are left-handed.
•
Of female respondents, 90% are right-handed.
340
MTHH 040
Part E – Area Under the Curve (possible 12 points)
x2
Activity: In the field of statistics the function f(x) =
−
1
e 2 is used to model data like birth weight
2
or height. In Lesson 8, you learned how to calculate z-scores, which then those scores related to the
number of standard deviations that specific value was from the mean. Now, you will extend that
knowledge of z-scores and find the area under the graph of the function, f(x), to find probabilities.
In a given population, the weights of babies are normally distributed about the mean, 3250
g. The standard deviation is 500 g. Find the probability that a baby chosen at random
weighs from 2250 g to 4250 g.
Ex.
x2
Step
1
Find the z-scores of the lower and
upper limits.
Step 2
Enter f(x) =
−
1
e 2 as Y1.
2
Adjust window values if needed.
z-score =
Step
3
value − mean
std. dev.
z1 =
2250 − 3250
−1000
=
= −2
500
500
z2 =
4250 − 3250
1000
=
=2
500
500
Use the CALC button and press 7 to access the ∫ f(x)dx feature. Move the cursor until the
lower limit is x = −2. Press ENTER. Then repeat steps with the cursor to find the upper
limit x = 2. Press ENTER.
The area under the curve from x = −2 to x = 2 is about 0.95. Therefore, the probability
that a baby weighs from 2250 g to 4250 g is about 95%.
Use the data and the function in the example above. Find the probability that the weight of a baby
chosen at random falls within each interval 1–4. (1 pt each)
1.
3150–4150 g
Project 2
2.
4300–4500 g
3.
341
less than 1800 g
4.
more than 4550 g
MTHH 040
5.
For questions 1–4, estimate the number of babies within each interval from a population of 2400
babies. Round to nearest whole number. You will have 4 answers total. (1 pt each)
answer for question 1 _____________
answer for question 2 _____________
answer for question 3 _____________
answer for question 4 _____________
6.
A battery company manufactures batteries having life spans that are normally distributed, with a
mean of 45 months and a standard deviation of 5 months. Find the probability that a battery
chosen at random will have each life span. (2 pts each)
a. 45–52 months
b. 48–50 months
This project can be submitted electronically. Check the Project page in the UNHS online
course management system or your enrollment information with your print materials for more
detailed instructions.
Project 2
342
MTHH 040 | 677.169 | 1 |
Language
Providers
25
Results
An interactive applet and associated web page that shows that angle-angle-angle (AAA) …
An interactive applet and associated web page that shows that angle-angle-angle (AAA) is not enough to prove congruence. The applet shows two triangles, one of which can be dragged to resize it, showing that although they have the same angles they are not the same size and thus not congruent a non-included side the same must be congruent. The applet shows two triangles, one of which can be reshaped by dragging any vertex. The other changes to remain congruent to it and the two angles and non-included their included side the same must be congruent. The applet shows two triangles, one of which can be reshaped by dragging any vertex. The other changes to remain congruent to it and the two angles and the included angles. Three angles are shown which always remain congruent as you drag any defining point on any angle. They all change together. This is designed to demonstrate that the angles are considered congruent even if they are in different orientations and the line segments making them up are different lengths. Applet can be enlarged to full screen size for use with a classroom projector. This resource is a component of the Math Open Reference Interactive Geometry textbook project at
An interactive applet and associated web page that demonstrate congruent line segments …
An interactive applet and associated web page that demonstrate congruent line segments (segments that are the same length). The applet shows three line segments that are the same length. They all have draggable endpoints. As you drag any endpoint the other lines change to remain congruent with the one you are changing. Applet can be enlarged to full screen size for use with a classroom projector. This resource is a component of the Math Open Reference Interactive Geometry textbook project at
An interactive applet and associated web page that demonstrate the congruence of …
An interactive applet and associated web page that demonstrate the congruence of polygons. The applet presents nine polygons that are in fact congruent, but don't look it because they are reflected and rotated in various ways. If you click on one, it rotates and flips as needed, then slides over the top of another to show it is congruent. The web page describes how to determine if two polygons are congruent triangles. Applets show that triangles a re congruent if the are the same, rotated, or reflected. In each case the user can drag one triangle and see how another triangle changes to remain congruent to itStudents learn about common geometry tools and then learn to use protractors …
Students learn about common geometry tools and then learn to use protractors (and Miras, if available) to create and measure angles and reflections. The lesson begins with a recap of the history and modern-day use of protractors, compasses and mirrors. After seeing some class practice problems and completing a set of worksheet-prompted problems, students share their methods and work. Through the lesson, students gain an awareness of the pervasive use of angles, and these tools, for design purposes related to engineering and everyday uses. This lesson prepares students to conduct the associated activity in which they "solve the holes" for hole-in-one multiple-banked angle solutions, make their own one-hole mini-golf courses with their own geometry-based problems and solutions, and then compare their "on paper" solutions to real-world results. | 677.169 | 1 |
The Rectangle
The rectangle is an interesting figure that has certain special characteristics. Once we have become familiar with its specific properties, we will be able to quickly determine whether or not any figure we are presented with is in fact a rectangle.
The diagonals of the rectangle intersect. Not only do they intersect, but they do so at the midpoint of each other. Since the diagonals are equal, so are their halves.
In other words: AE=BE=CE=DEAE=BE=CE=DEAE=BE=CE=DE
Important! The diagonals of a rectangle are not perpendicular! They do not form a right angle with each other.
Also: The diagonals do not cross the angles of the rectangle.
Note: Since a rectangle is a type of parallelogram, it has all the properties of a parallelogram.
How can we prove that a certain figure is a rectangle? The technical definition of a rectangle is: A parallelogram that contains a right angle (90º).
Also, A parallelogram with diagonals of the same length is a rectangle. A rectangle is a type of parallelogram with special properties. If we are presented with a parallelogram with an angle of 90º or one with diagonals of the same length, we can safely determine that it is a rectangle!
From Quadrilateral to Rectangle
How can we prove that aquadrilateral is, in fact, a rectangle? There are two ways:
The First Method: Checking the Angles
Check if three of the angles measure 90º. If they do, then the quadrilateral is a rectangle. (We do not need to measure the fourth angle since, as all angles must add up to 360º, it will also measure 90º).
Do you think you will be able to solve it?
Question 1
The points A and O are shown in the figure below.
Is it possible to draw a rectangle so that the side AO is its diagonal?
The Second Method: Demonstration of parallelogram, then, of rectangle.
See if you can prove that the given quadrilateral is a parallelogram. Don't know exactly how to prove it? Then have a look at our guide "From quadrilateral to parallelogram" to learn how to identify a parallelogram from 20 km away!
Once you have, you will know that a parallelogram is in fact a rectangle based on these rules:
If the parallelogram has an angle of 90º, then it is a rectangle.
If the diagonals of the parallelogram are equal, then it is a rectangle.
You can prove that a given parallelogram is a rectangle by applying any of these rules:
If the parallelogram has an angle of 90º, then it is a rectangle.
If the diagonals of the parallelogram are equal, then it is a rectangle.
Note: The technical definition of a rectangle is: a parallelogram with a right angle (90º). Therefore, if we have a parallelogram that has an angle of 90 degrees, we can determine that it is a rectangle. Furthermore, another property of a rectangle is that its diagonals are of the same length. Remembering the second rule, we can check the diagonals. If they are equal, then we have proven that it is a rectangle.
Great! Now you know everything you need to know about rectangles.
Rectangle Exercises
Exercise 1
Have a look at the two rectangles in the figure:
Question:
What is the area of the blank area?
Solution:
To answer the question, we subtract the area of the small rectangle from the area of the large rectangle.
The formula for calculating a rectangular area is base multiplied by the height.
Let's start by calculating the area of the large rectangle.
The base of the large rectangle consists of a side DC=DG+GC DC=DG+GC DC=DG+GC.
DC=4+5 DC=4+5 DC=4+5
DC=9 DC=9 DC=9 (base)
The height of the large rectangle is: DA=DE+EA DA=DE+EA DA=DE+EA.
AD=2+2=4 AD=2+2=4 AD=2+2=4
Therefore, DA=4 DA=4 DA=4.
Now to calculate the area of the large rectangle:
DC×DA=9×4=36cm2 DC\times DA=9\times4=36cm² DC×DA=9×4=36cm2
In the second step, we calculate the area of the small rectangle. The formula is the same: base multiplied by the height.
The area of the small rectangle is equal to DG×DE DG\times DE DG×DE.
Therefore, the area of the small rectangle is equal to 2×4=8cm2 2\times4=8cm² 2×4=8cm2.
Now all we have left to do is to calculate the area of the large rectangle minus the area of the small rectangle, leaving us with the area of the blank section.
What is a golden rectangle?
A golden rectangle is a rectangle where the proportion between its sides is the golden ratio (an irrational number also known as divine number). When dividing one of its sides by the other side, we obtain ϕ=1.61803… \phi=1.61803\ldots ϕ=1.61803….
How do you know if it is a golden rectangle?
To know if you are working with a golden rectangle, divide the larger side of the rectangle by the smaller side and see if that division gives you the result ϕ=1.61803… \phi=1.61803\ldots ϕ=1.61803…. If it does, it is a divine rectangle.
Test your knowledge
Question 1
The points A and O are shown in the figure below.
Is it possible to draw a rectangle so that the side AO is its diagonal? | 677.169 | 1 |
The other day I was playing with Ms Paint drawing circles here and there - I coincidentally drew a circle inside a right angled triangle which I already drew. Strangely A problem struck to my mind and I tried solving it , but I was unable to do so. I put forward the statement of the problem which I managed to frame myself:
Problem: The legs of a right angled triangle are of length $a$ and $b$. Two circles with equal radii are drawn such such that they touch each other and sides of the triangle as shown in the figure. Find the radius of the circle in terms of $a$ and $b$.
Figure (of course my MS Paint one)
Further Scope - Is there any way to generalize this for other shapes or for any other triangle?
-------EDIT---------------------
Now to make things interesting : Say we have a right angled triangle which is given . Then is there a method by which we can construct those two circles with a straightedge and a compass?
$\begingroup$Straightedge and compass construction: draw an arbitrary circle with center $C$ that intersects the legs $AC, BC$ at $D, E$, respectively. Locate $F$ on $AC$ such that $CF = 3CA$. Locate $G$ such that $ECFG$ is a rectangle. Draw $CG$. Draw the angle bisector of $A$. The intersection of this angle bisector with $CG$ is the center of one of the two circles.$\endgroup$
To address @DanielV's suggestion of generalizing to higher dimensions, consider a right-corner tetrahedron $OABC$, with right corner at $O$ and edge lengths $a := |OA|$, $b := |OB|$, $c := |OC|$. (Note that I'm changing notation slightly from the above.) Let a sphere with center $P$ and radius $r$ be tangent to the faces around vertex $A$, and let a congruent sphere (tangent to the first) be tangent to the faces around vertex $O$. Then $P$ has distance $r$ from faces $\triangle OAB$, $\triangle OCA$, $\triangle ABC$ (the ones touching $A$), and distance $3r$ from face $\triangle OBC$ (the one opposite $A$).
Here's a poor attempt at a diagram:
(In this case, the altitudes from $P$ are color-coded to match their parallel counterparts through $O$. The black altitude is to face $\triangle ABC$.)
$\begingroup$@qwr: Thanks. :) As for coloring: I just like to highlight the association of a vertex with its opposite edge in a triangle, and other elements get their colors accordingly. I chose not to color-code the altitudes dropped from $P$ to the edges (I had reflexively done so in a "draft" image), in order to tie the congruent segments together visually.$\endgroup$
$\begingroup$@AJP: "$|BC|$" represents the length of segment $\overline{BC}$. (I often leave out the over-bar to save typing.) By analogy, I use "$|\triangle ABC|$" for the area of $\triangle ABC$ (and then "$|OABC|$" for the volume of tetrahedron $OABC$, etc); this is non-standard, but I like it. :) The symbol ":=" indicates "is defined to be"; writing "$a:=|BC|$" says "I'm going to write '$a$' for '$|BC|$'". Others (even I) might otherwise write "let $a = |BC|$", but using ":=" is ever-so-slightly better, as it distinguishes definition "'$a$' means '$|BC|$'" from relation "$a$ equals $|BC|$".$\endgroup$
$\begingroup$@mathh: In GeoGebra, you can easily mark an angle with an arc via the "Angle" tool; if the sides of the angle happen to be perpendicular, GeoGebra turns the arc into a box. (There's a setting somewhere to turn that behavior off and on.) The little dashes are somewhat less convenient: with the segment selected, you have to open the "Object Properties..." panel and then the "Decoration" tab. (You can likewise "decorate" angles with little dashes and such.)$\endgroup$
$\begingroup$Interesting how many different ways there can be to reach the same result. This is one reason I say that instead of memorizing formulas, students would appreciate mathematics more if they saw the "magic" of how many seemingly unrelated approaches give the same result in the end.$\endgroup$
$\begingroup$@ShivamPatel Look at youtube.com/watch?v=CMP9a2J4Bqw espcially after 00:58 where it describes what is possible with a compass and straightedge. In short the answer is "yes", by applying what you see in the video to the formulas Blue and I gave you. ...But I doubt there is any short way to do it. Keep in mind that you effectively assume $a=1$ or $b=1$ since the units are arbitrary.$\endgroup$
Recall that for any triangle $\triangle ABC$, the area of the triangle equals the product of the inradius and its semiperimeter; i.e., $|\triangle ABC| = rs$, where $s = (a+b+c)/2$. Therefore, given legs $a, b$, $c = \sqrt{a^2+b^2}$, and $$r = \frac{2|\triangle ABC|}{a+b+c} = \frac{ab}{a+b+\sqrt{a^2+b^2}}.$$ Draw the tangent line to the two circles at their common point of tangency: this creates a smaller similar triangle with scaling factor $\frac{b-2\rho}{b}$ where $\rho$ is the common radius of the two circles. If $r$ is the inradius of $|\triangle ABC|$ as shown, then $$\frac{\rho}{r} = \frac{b - 2\rho}{b}.$$ Putting all of this together, we find $$\rho = \frac{br}{b+2r} = \frac{a b}{3a+b+c} = \frac{ab}{3a+b+\sqrt{a^2+b^2}}.$$ This method easily generalizes to more than two congruent circles tangent to one side: if $n$ circles are arranged along the leg of length $b$, then it is straightforward to find that $\rho = \frac{ab}{(2n-1)a + b + \sqrt{a^2+b^2}}$.
Other people have answered your question, but I'll provide the "answer" for the other direction:
What about determining $a,b$ from $r$?
You can't - $a,b$ are not uniquely determined by $r$. Draw the two circles first, then draw the legs a and b as infinite lines. And line tangent to the right most circle will intersect lines A and B creating a right triangle, but for different tangents $a$ and $b$ will be different and r hasn't changed.
You can't even turn this into an interesting question by asking for all the possible ways to do it, because it turns out that these circles don't actually impose any restriction - since any angels can be achieved by the tangent line, any right triangle can. All $r$ does is create a scaling factor. $a$ can take on any value in $(2r,\infty)$ and $b$ the corresponding value in $(4r,\infty)$
$\begingroup$@Blue I showed a coordinate based approach. I got a different form for the answer, which I think is nice also. Which approach is easier to follow will depend on the reader I guess. Thanks for your answer too ^_^$\endgroup$ | 677.169 | 1 |
Related polytopes
It is a part of an infinite family of polytopes, called hypercubes. The dual of a 6-cube can be called a 6-orthoplex, and is a part of the infinite family of cross-polytopes. It is composed of various 5-cubes, at perpendicular angles on the u-axis, forming coordinates (x,y,z,w,v,u).[1][2]
As a configuration
This configuration matrix represents the 6-cube. The rows and columns correspond to vertices, edges, faces, cells, 4-faces and 5-faces. The diagonal numbers say how many of each element occur in the whole 6-cube. The nondiagonal numbers say how many of the column's element occur in or at the row's element.[3][4]
Related polytopes
The 64 vertices of a 6-cube also represent a regular skew 4-polytope {4,3,4 | 4}. Its net can be seen as a 4×4×4 matrix of 64 cubes, a periodic subset of the cubic honeycomb, {4,3,4}, in 3-dimensions. It has 192 edges, and 192 square faces. Opposite faces fold together into a 4-cycle. Each fold direction adds 1 dimension, raising it into 6-space. | 677.169 | 1 |
1996-02 Solution
Bend the top right edge towards the bottom middle so the bottom right
angle formed is 45 degrees. Then fold the top right long folded edge towards
the bottom middle so the it looks like you're going to make a paper airplane. | 677.169 | 1 |
Determining Congruence Project (2 Tasks)
Learning Goal: I'm working on a geometry project and need the explanation and answer to help me learn.
While doing the Unit project, make sure that you clearly show all your work and do your best to explain your answers. You will be graded on how well you show your understanding of the concepts as well as correct answers. | 677.169 | 1 |
finding angles of a right triangleMeasure Angles Of A Triangle Worksheets | 677.169 | 1 |
Chapter: Basic Geometric Concepts in Projective Geometry
Projective geometry builds on some simple mathematical concepts also known in the Euclidean geometry but with some subtle differences. In the following, we want to refer to the concepts known from the Euclidean geometry and explain how the concepts differ in projective geometry.
Points
Naturally, we think of points as infinitesimal dots with a "position" but "without any magnitude". This concept of a point is also used in projective geometry, however, "position" only has a relative meaning, i.e. in relation to other geometrical objects like lines or planes, not in the absolute meaning of distances or coordinates. For instance, in projective geometry it perfectly makes sense to speak about "points lying on a line", but sentences like "The distance of two points is $2$.", or a "The point in the plane has the coordinates $(1,2).$" do not make any sense!
In the course of the next text we will extend the Euclidean definition to allow points at infinity - i.e. infinitely distant points on a straight line.
Notation: In the following, we will consistently denote points with Latin capital letters, e.g. $A,B,C,\ldots$
Straight Lines
Since distances and lengths do not play any role, projective geometry does not study segments. Even rays are not objects of study in projective geometry. Therefore, in projective geometry, straight lines always have unlimited extend in both directions.
In the course of the text to follow we will extend the Euclidean definition of a straight line to allow a straight line at infinity - i.e. a straight line connecting two distinct points at infinity.
Notation: In the following, we will consistently denote straight lines with small Latin letters, e.g. $k,l,m\ldots$
Planes
Notation: In the following, we will consistently denote planes with small Greek letters, e.g. $\alpha,\beta,\gamma\ldots$
The principle of duality in a plane asserts that every definition in the projective geometry remains meaningful and every theorem remains true, when we interchange the words "points" and "straight line", and consequently also certain other pairs of words such as "collinear" and "join" and "meet",
Other simple geometrical concepts, which are specific to projective geometry, will be provided in the following text. These concepts include | 677.169 | 1 |
we focus on the blue triangle below, we see that we already know that two of the angles are 90° and 50° Since all three angles in a triangle must add to 180°, we can conclude that the missing angle is 40° | 677.169 | 1 |
In Euclidean geometry, a parallelogram is a simple (non-self-intersecting) quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. The congruence of opposite sides and opposite angles is a direct consequence of the Euclidean parallel postulate and neither condition can be proven without appealing to the Euclidean parallel postulate or one of its equivalent formulations.
Quick Facts Type, Edges and vertices ...
Parallelogram
This parallelogram is a rhomboid as it has no right angles and unequal sides. | 677.169 | 1 |
2D Radial Menu
Measures the diameter of a circle derived from a selected arc in the drawing. The markup displays the center point coordinates and length of the diameter of the derived circle.
Arc Radius
Measures the radius of a circle derived from a selected arc in the drawing. The markup displays the center point coordinates and length of the radius of the derived circle.
Circle by Three Points
Draws a visible circle that passes through three selected points in the drawing and determines the diameter of the circle. The diameter line is drawn through the first point you selected. The markup displays the diameter length and center point coordinates of the circle. | 677.169 | 1 |
a quadrilateral is a shape that that doesn't contain any curved
lines or overlapping lines. you must be able to draw the entire
shape without lifting your pencil and ending where you started
drawing.
* * * * *
A hopelessly incomplete answer. A triangle would meet these
requirements!
A quadrilateral is a closed plane shape enclosed by four
straight lines (which meet, pairwise, at 4 vertices).
The shape is convex if all of its interior angles are less than
180 degrees. Equivalently, for every pair of points inside the
shape, the straight line joining the points must lie entirely
inside the boundaries of the shape. | 677.169 | 1 |
Elementary Trigonometry
43. Similar rules will hold with respect to the equations for connecting the centesimal and the circular systems, 200 being put in the place of 180: thus
circular measure of an angle containing G grades = G.
π
200
200
number of grades in the angle whose circular measure is 0 = 0. –
Express in grades, &c. the angles whose circular measures
44. We shall now give a set of Miscellaneous Examples to illustrate the principles explained in this and the two preceding Chapters.
EXAMPLES.-XII.
1. If the unit of angular measurement be 5o, what is the measure of 2210?
2. If an angle of 421° be represented by 10, what is the unit of measurement ?
3. An angle referred to different units has measures in the ratio 8 to 5; one unit is 2o, what is the other? Express each unit in terms of the other.
28. We can express the measure of an angle (expressed in degrees, minutes and seconds) in degrees and decimal parts of a degree by the following process.
Let the given angle be 39°. 5′. 33",
Express as the decimal of a degree the following angles.
29. In this method we suppose a right angle to be divided into 100 equal parts, each of which parts is called a grade, each grade to be divided into 100 equal parts, each of which is called a minute, and each minute to be divided into 100 equal parts, each of which is called a second. Then the magnitude of an angle is expressed by the number of grades, minutes and seconds, which it contains. Grades, minutes and seconds are marked respectively by the symbols ', `, ": thus, to represent 35 grades, 56 minutes, 84.53 seconds, we write 35o. 56`. 84"-53.
The advantage of this method is that we can write down the minutes and seconds as the decimal of a grade by inspection.
30. If the number expressing the minutes or seconds has only one significant digit, we must prefix a cipher to occupy the place of tens before we write down the minutes and seconds as the decimal
Express as decimals of a grade the following angles :
31. The Centesimal Method was introduced by the French Mathematicians in the 18th century. The advantages that would have been obtained by its use were not considered sufficient to counterbalance the enormous labour which must have been spent on the re-arrangement of the Mathematical Tables then in use.
III. The Circular Measure.
32. In this method, which is chiefly used in the higher branches of Mathematics, the unit of angular measurement may be described as
(1) The angle subtended at the centre of a circle by an arc equal to the radius of the circle,
or, which is the same thing, as we proved in Art. 21, as
(2) The angle whose magnitude is the 7th part of two right angles.
33. It is important that the beginner should have a clear conception of the size of this angle, and this he will best obtain by considering it relatively to the magnitude of that angular unit which we call a degree.
Now the unit of circular measure
=
two right angles
π
180° 3.14159
= 57°-2958 nearly.
Now if BC be the quadrant of a circle, and if we suppose the arc BC to be divided into 90 equal parts,
the right angle BAC will be divided by the radii which pass through these points into 90 equal angles, each of which is called a degree.
A radius AP meeting the arc at a certain point between the 57th and 58th
divisions, reckoned from B, will make with
P
AB an angle equal in magnitude to the unit of circular measure.
B
Hence an angle whose circular measure is 2 contains rather more than 114 degrees, and one whose circular measure is 3 contains about 172 degrees, or rather less than two right angles.
2 right angles,
34. Again, since the unit of circular measure = π times the unit of circular measure = 2 right angles. Hence
an angle whose circular measure is is equal to 2 right angles,
35. To shew that the circular measure of an angle is equal to a fraction which has for its numerator the arc subtended by that angle at the centre of any circle, and for its denominator the radius of that circle.
Let EOD be any angle.
About O as centre and with any radius, describe a circle cutting OE in A, and OD in R. | 677.169 | 1 |
Lecture 1 - Curves (Simple & Compound).pdf
Short Description
Survey And Levelling...
Description
Surveying-II CE-207 (T) CURVES Lecture No 1
civil engineering ENGR.AWAIS KHAN
1
Curves • Curves are usually employed in lines of communication in order that the change in direction at the intersection of the straight lines shall be gradual. • The lines connected by the curves are tangent to it and are called Tangents or Straights. • The curves are generally circular arcs but parabolic arcs are often used in some countries for this purpose. • Most types of transportation routes, such as highways, railroads, and pipelines, are connected by curves in both horizontal and vertical planes. 2
Curves • The purpose of the curves is to deflect a vehicle travelling along one of the straights safely and comfortably through a deflection angle θ to enable it to continue its journey along the other straight.
θ
3
4
5
6
7
Curves
Horizontal Alignment Plan View
Vertical Alignment Profile View
Curves Classification of Curves Curves
Horizontal Curves
Vertical Curves
Simple Curves Compound Curves
Circular Curves
Transition Curves Reverse Curves
Spiral Curves (non-circular curves) 9
Curves Classification of Curves 1) Simple Curves: Consist of single Arc Connecting two straights. 2) Compound curves: Consist of 2 arcs of different radii, bending in the same direction and lying on the same sides of their common tangents, their centers being on the same side of the curve. 3) Reverse curves: Consist of 2 arcs of equal or unequal radii, bending in opposite direction with common tangent at their junction (meeting Point), their center lying on the opposite sides of the curve.
10
Curves
Simple Curve
Compound Curve
Reverse Curve
11
Curves Nomenclature of Simple Curves B'
B
∅ I
F ∅ 𝟐
T1 A
90o
90o - ∅/2
∅ 𝟐
T2
90o
E
C
R
R
∅ ∅ 𝟐
O
12
Curves Nomenclature of Simple Curves 1) Tangents or Straights: The straight lines AB and BC which are connected by the curves are called the tangents or straights to curves. 2)Point of Intersection: (PI.) The Point B at which the 2 tangents AB and BC intersect or Vertex (V) .
3)Back Tangent: The tangent line AB is called 1st tangent or Rear tangents or Back tangent. 4) Forward Tangent: The tangents line BC is called 2nd tangent or Forward tangent. 13
Curves Nomenclature of Simple Curves 5) Tangents Points: The points T1 and T2 at which the curves touches the straights. 5.a) Point of Curve (P.C): The beginning of the curve T1 is called the point of curve or tangent curve (T.C).
5.b) Point of tangency (C.T): The end of curve T2 is called point of tangency or curve tangent (C.T). 6) Angle of Intersection: (I) The angle ABC between the tangent lines AB and BC. Denoted by I. 14
Curves Nomenclature of Simple Curves 7) Angle of Deflection (∅): Then angle B`BC by which the forward (head tangent deflect from the Rear tangent. 8) Tangent Length: (BT1 and BT2) The distance from point of intersection B to the tangent points T1 and T2. These depend upon the radii of curves. 9) Long Cord: The line T1T2 joining the two tangents point T1 and T2 is called long chord. Denoted by l.
15
Curves Nomenclature of Simple Curves 10) Length of Curve: the arc T1FT2 is called length of curve. Denoted by L. 11) Apex or Summit of Curve: The mid point F of the arc T1FT2 is called Apex of curve and lies on the bisection of angle of intersection. It is the junction of lines radii. 12) External Distance (BF): The distance BF from the point of intersection to the apex of the curve is called Apex distance or External distance.
16
Curves Nomenclature of Simple Curves 13) Central Angle: The angle T1OT2 subtended at the center of the curve by the arc T1FT2 is called central angle and is equal to the deflection angle. 14) Mid ordinate (EF): It is a ordinate from the mid point of the long chord to the mid point of the curve i.e distance EF. Also called Versed sine of the curve. •
•
If the curve deflect to the right of the direction of the progress of survey it is called Right-hand curve and id to the left , it is called Left-hand curve. The ∆ BT1T2 is an isosceles triangle and therefore the angle ∟ BT1T2 = ∟ BT2T1 =
Simple Curves Method of Curve Ranging • There are a number of different methods by which a centerline can be set out, all of which can be summarized in two categories: • Traditional methods: which involve working along the centerline itself using the straights, intersection points and tangent points for reference. • The equipment used for these methods include, tapes and theodolites or total stations. • Coordinate methods: which use control networks as reference. These networks take the form of control points located on site some distance away from the centerline. • For this method, theodolites, totals stations or GPS receivers can be used. 24
Simple Curves Method of Curve Ranging The methods for setting out curves may be divided into 2 classes according to the instrument employed . 1) Linear or Chain & Tape Method 2) Angular or Instrumental Method Peg Interval: Usual Practice--- Fix pegs at equal interval on the curve
20 m to 30 m ( 100 feet or one chain) 66 feet ( Gunter's Chain) Strictly speaking this interval must be measured as the Arc intercept b/w them, however it is necessarily measure along the chord. The curve consist of a series of chords rather than arcs. Along the arc it is practically not possible that is why measured along the chord.
25
Simple Curves Method of Curve Ranging Peg Interval: For difference in arc and chord to be negligible
Simple Curves Method of Curve Ranging Procedure: • After locating the positions of the tangent points T1 and T2 ,their chainages may be determined. • The chainage of T1 is obtained by subtracting the tangent length from the known chainage of the intersection point B. And the chainage of T2 is found by adding the length of curve to the chainage of T1. • Then the pegs are fixed at equal intervals on the curve. • The interval between pegs is usually 30m or one chain length. • The distances along the centre line of the curve are continuously measured from the point of beginning of the line up to the end .i.e the pegs along the centre line of the work should be at equal interval from the beginning of the line up to the end. 28
Simple Curves Method of Curve Ranging Procedure: • There should be no break in the regularity of their spacing in passing from a tangent to a curve or from a curve to the tangent. • For this reason ,the first peg on the curve is fixed at such a distance from the first tangent point (T1) that its chainage becomes the whole number of chains i.e the whole number of peg interval. • The length of the first sub chord is thus less than the peg interval and it is called a sub-chord. • Similarly there will be a sub-chord at the end of the curve. • Thus a curve usually consists of two sub-chords and a no. of full chords.
29
Simple Curves Method of Curve Ranging Important relationships for Circular Curves for Setting Out • The ∆ BT1T2 is an isosceles triangle and therefore the angle ∟ BT1T2 = ∟ BT2T1 =
∅ 2
The following definition can be given: • The tangential angle 𝛼 at T1 to any point X on the curve T1T2 is equal to half the angle subtended at the centre of curvature O by the chord from T1 to that point. • The tangential angle to any point on the curve is equal to the sum of the tangential angles from each chord up to that point. • I.e. T1OT2 = 2(α + β + 𝛾) and it follows that BT1T2= (α + β + 𝛾).
Simple Curves Method of Curve Ranging 1) Linear or Chain & Tape Method • These methods use the chain surveying tools only. • These methods are used for the short curves which doesn't require high degree of accuracy. • These methods are used for the clear situations on the road intersections. a) By offset or ordinate from Long chord b) By successive bisections of Arcs c) By offset from the Tangents d) By offset from the Chords produced 34
When the radius of the arc is larger as compare to the length of the chord, the offset may be calculated approximately by formula or
Ox =
𝑥 (𝐿 −𝑥) -------- 2 (Approximate formula) 2𝑅
In eqn 1 the distance x is measured from the mid point of the long chord where as eqn 2 it is measured from the 1st tangent point T1. • This method is used for setting out short curves e.g curves for street kerbs. . 37
Working Method: To set out the curve • Divided the long chord into even number of equal parts. • Set out offsets as calculated from the equation at each of the points of division. Thus obtaining the required points on the curve.
• Since the curve s symmetrical along ED, the offset for the right half of the curve will be same as those for the left half. 38
Let T1 and T2 be the tangents points. Join T1 and T2 and bisect it at E.
•
Setout offsets ED(y'), determined point D on the curve equal to ∅
ED = y' = R ( 1 – cos (𝟐) •
Join T1D and DT2 and bisect them at F and G respectively.
•
Set out offset HF(y'') and GK(y'') each eqn be ∅
FH = GK = y''= R ( 1 – cos (𝟒)
Obtain point H and K on a curve. By repeating the same process, obtain as many pints as required on the curve. 40 (O is Accessible)
Working Method: •
Measure a distance x from T1 on back tangent or from T2 on the forward tangent.
•
Measure a distance Ox along radial line A1O.
•
The resulting point E1 lies on the curve. 41 b) By Offsets Perpendicular to Tangents
(O is Inaccessible) Working Method: •
Measure a distance x from T1 on back tangent or from T2 on the forward tangent.
Simple Curves Method of Curve Ranging 2) Angular or Instrumental Methods 1) Rankine's Method of Tangential Angles 2) Two Theodolite Method 1) Rankine's Method of Tangential Angles • In this method the curve is set out tangential angle often called deflection angles with a theodolite, chain or tape. 52
Simple Curves Method of Curve Ranging 2) Angular or Instrumental Methods 1) Rankine's Method of Tangential Angles Working method: 1. Fix the theodolite device to be at point T1 and directed at point B. 2. Measure the deflection angles 𝛿 1 and the chords C1. 3. Connect the ends of the chords to draw the curve. Deflection Angles: The angles between the tangent and the ends of the chords from point T1. 53
Simple Curves Method of Curve Ranging 2) Angular or Instrumental Methods 2) Two Theodolite Method • This method is used when ground is not favorable for accurate chaining i.e rough ground , very steep slope or if the curve one water • It is based on the fact that angle between tangent & chord is equal to the angle which that chord subtends in the opposite segments. ∆1 is b/w tangent T1B & T1D => BT1D = T1T2D = ∆1 T1E = ∆2 = T1 T2 E The total tangential angle or deflection angle ∆1, ∆2 ∆3 … ,As calculate in the 1st method.
61
Simple Curves Obstacles in Setting Out Simple Curve • The following obstacles occurring in common practice will be considered.
1) When the point of intersection of Tangent lines is inaccessible. 2) When the whole curve cannot be set out from the Tangent point, Vision being obstructed. 3) When the obstacle to chaining occurs.
62
Simple Curves Obstacles in Setting Out Simple Curve 1) When the point of intersection of Tangent lines is inaccessible • When intersection point falls in lake, river , wood or any other construction work 1) To determine the value of ∅ 2) To locate the points T1 & T2 Calculate θ1 & θ2 by instrument (theodolite). ∟ BMN = α = 180 – θ1 ∟ BNM = β = 180 – θ2
Simple Curves Obstacles in Setting Out Simple Curve 2) When the whole curve cannot be set out from the Tangent point, Vision being obstructed • As a rule the whole curve is to be set out from T1 however
obstructions intervening the line of sight i.e Building, cluster of tree, Plantation etc. In such a case the instrument required to be set up at one or more point along the curve.
65
Simple Curves Obstacles in Setting Out Simple Curve 3) When the obstacle to chaining occurs
Curves Compound Curves • A compound curve consist of 2 arcs of different radii bending in the same direction and lying on the same side of their common tangent. Then the center being on the same side of the curve. RS = Smaller radius RL = Larger radius TS = smaller tangent length = BT1 TL = larger tangent length = BT2 𝛼 = deflection angle b/w common tangent and rear tangent 𝛽 = angle of deflection b/w common tangent and forward tangent N = point of compound curvature KM = common tangent
68
Curves Compound Curves Elements of Compound Curve ∅= 𝛼+𝛽 𝛼
KT1 = KN =RS tan( 2 ) 𝛽
MN = MT2 = RL tan ( 2 ) From ∆BKN, by sine rule 𝐵𝐾 sin 𝛽
=
𝐵𝐾 sin 𝛽
=
BK = BM =
180o - (𝛼 +𝛽) = I 180o – ∅ = I
𝑀𝐾 sin 𝐼 𝑀𝐾
sin(180o – (𝛼 +𝛽)) 𝑀𝐾 sin 𝛽
sin(180o – (𝛼 +𝛽)) 𝑀𝐾 sin 𝜶
sin(180o – (𝛼 +𝛽)) 69
Curves Compound Curves Elements of Compound Curve TS = BT1 = BK + KT1
TS = BT1 = TL = BT2 =
𝑀𝐾 sin 𝛽
sin(180o
– (𝛼 +𝛽))
𝑀𝐾 sin 𝜶
sin(180o – (𝛼 +𝛽))
𝛼
+ RS tan ( 2 ) 𝛽
+ RL tan ( 2 )
Of the seven quantities RS, RL, TS, TL, ∅, 𝛼, 𝛽 four must be known.
70
Curves Compound Curves Problem: Two tangents AB & BC are intersected by a line KM. the angles AKM and KMC are 140o & 145o respectively. The radius of 1st arc is 600m and of 2nd arc is 400m. Find the chainage of tangent points and the point of compound curvature given that the chainage of intersection point is 3415 m. Solution: 𝛼 = 180o – 140o = 40o 𝛽 = 180o – 145o = 35o ∅= 𝛼 + 𝛽 = 75o I = 180o – 75o = 105o 𝛼 | 677.169 | 1 |
Sin 270 kaça eşit
6431
sin(270) - MathCelebrity
Tools Learn more. Empire of Sin
Start studying Sin Cos Tan (90°,180°, 270°, 360°). Learn vocabulary, terms, and more with flashcards, games, and other study tools.
Sin270° Value – What is the sin of 270 degrees? - Trigonometric ...
Score a saving on iPad Pro (2021): $100 off at Amazon We may earn a commission for purchases using our links.
From the table settings to a signature drink, these 30 ideas can help you host a wallet-friendly dinner party that looks extra-fancy. You'll have all of your guests impressed with this high-class party on a budget. And seven steps to salvation. Tools
Feb 16, 2017 The sin of 270 degrees is -1, the same as sin of 270 degrees in radians. To obtain 270 degrees in radian multiply 270° by π / 180° = 3 Feb 23, 2017 Sin -270°: All about sin minus 270 degrees, incl. the trigonometric identities. Besides the value, we also have useful information and a | 677.169 | 1 |
And that makes sense as well, because we've got a vertex for each of those corner points on our 2D hexagonal shape.
So we've got 12 vertices and no specific apex.
In shapes C and D, we can see we've got apexes just at the top here.
These are opposite the base.
These are the top vertices.
So now that we've understood a little bit more about the terminology that we might use to describe 3-D shapes, let's think about how we could use these properties to sort 3-D shapes into categories.
I've got two circles on my screen here, and that's going to be used to sort out the shapes that are on my right.
I'm going to call these shapes shape A, B, C, D, E, and F.
And what I'd like to do is to create two different categories which I could put these shapes into.
So how am I going to think about which properties we might use? We might think about properties related to the number of faces that the shape has.
Maybe it will have something to do with the vertices, or maybe it will have something to do with the edges of the shape.
Let me show you an example.
I'd like to start with an example where we look at the number of faces the shape has.
Let's say the everything that goes into the first circle is going to be 3-D shapes that have.
Less than seven faces.
And then on the right, we will have shapes that have seven or more faces.
So any shape that has less than seven faces goes into the left circle, any shape that has seven or more faces goes into the right circle.
So looking at shape A, how many faces does shape A have? Can you see? So you should see that it's got eight faces.
This is a hexagonal prism and it's got eight faces so it's going to have to go in this side here.
So I going to put A in there.
B is what we would call a triangular-based pyramid or a tetrahedron, and it's only got four faces.
So it's going to go into this left hand circle just here.
For shape C, how many faces does shape C have? So you should see that it has six faces.
It's got five of its faces, which are the triangles, and then one base face, which is a pentagon.
And so in this case, it has six.
So it's going to go in here.
Shape D.
How many did it have? I'll give you a couple of seconds.
So hopefully you can see it's got seven faces.
So it's going to go to this circle here.
Shape E is a triangular prism.
How many faces does it have? It's got five.
So it's going to go into this circle here.
And shape F, which one? Which of these circles do you think shape F will fit into? It's got seven, so it's going to go just in here.
So you can see that we've managed to separate out these shapes into two different categories according to the number of faces that they have.
So I'd like you to do exactly the same thing.
So here are your labels, A, B, C, D, E, and F.
I would like you to take some time now to find two different categories that you could sort these shapes into.
I sorted them according to categories related to the number of faces they have.
Could you try and use something related to the number of vertices or the number of edges that they have? Spend a bit of time creating your different categories and sorting your shapes.
Pause the video to complete your task and resume it once you're finished.
So how good did you get on with your activity? Did you manage to find two different categories to sort your shapes into? Hopefully that was all fine for you.
Okay.
So let's move on slightly.
Let's have a look at some similarities and differences between these shapes that we're talking about here.
Let's first of all look at the shapes that we have on the screen.
So firstly, this shape here is called a cuboid.
It is made up of typically rectangular faces.
We have just here, a triangular prism which is made up of faces that are either rectangles and faces that are triangles.
So this is a triangular prism.
And the final shape here is also a prism of some kind, but it has, if you look, three, four, five, six sides for one of the 2D shapes at the top and the bottom, and it's a hexagonal prism.
And we call it a hexagonal prism because it's got hexagons on either side of the 3-D shape.
So it's a hexagonal prism.
I want you to have a good look at these shapes.
Now they're all different and we know they have different names, but can you see any similarities between them? What is the same about them? I'm going to give you 10 seconds to see if you can spot anything.
So you might have seen lots of different similarities between these shapes.
You might've noticed that they all have more than five vertices.
You might have noticed that all of these shapes have straight lines as a part of their edges.
You might have noticed that their faces are all flat, rather than curved.
There are lots of similarities you could have found.
Did you notice that every single one of these shapes makes use of a rectangular face, at least one rectangular face? In the case of the cuboid, we have got six rectangular faces with this particular shape.
How many rectangular faces does the triangular prism have? It has three rectangular faces and also got to triangular faces.
And how many rectangular faces does the hexagonal prism have? It has six rectangular faces.
So one of the similarities between these three shapes, is that they all make use of a face that is rectangular.
And you might notice that lots of shapes and specifically lots of prisms make use of faces that are rectangular.
If you just need to have a quick peek at what the rectangular face looks like, that is one of them that I've highlighted just there for you.
That is one of the rectangular faces.
On the triangular prism, this is one of the rectangular faces.
And on the cuboid, this is one of the rectangular faces.
Obviously they've been drawn out on a piece of paper so you can't see their 3-D, their full 3-D format using shape, but you can see that the faces that have been used, some of them are rectangles, which is something that's similar between all of these shapes.
But this takes me on to your independent task.
You are going to be finding similarities between the shapes that you can see on the screen.
We've looked at all of these shapes in some format or another today.
What I'd like you to do is see whether you can create a really long chain of polyhedral 3-D shapes where each shape is linked to the next one with one similar face.
So using what we've just spoken about there, I could link my cuboid to my triangular prism and then link my triangular prism to my hexagonal prism, all because they use rectangular faces, at least one similar face.
If I wanted to, however, what I could also do is link my cuboid to my triangular prism because of the rectangles, but then link my triangular prism to my square-based pyramid, because both of them make use of a triangle.
So you can see they've both got a similar face.
You need to try and create a long, long chain, try and use if you can, every single one of those shapes.
I'm not sure whether it's possible so let me know if you do manage to do it.
Can you create the longest chain possible linking each of the 3-D shapes in your chain with at least one similar face each time? Have a go at doing that now.
Pause the video to complete your task and resume it once you're finished.
How did you get on with the activity? Hopefully you've managed to create as long chain as possible.
If you'd like to, you can share it with us.
Please ask your parent or carer to share this work on Twitter tagging @OakNational and #LearnewithOak. | 677.169 | 1 |
orthographic projection solved examples
Theory of Orthographic Projections
INTRODUCTION
An engineer, who may be a designer, drafter or builder, is always confronted with the task of describing the shape of three dimensional objects on a two dimensional surface, e.g., the drawing paper. This may be done by a written description, photograph of the object or by graphical description.
The written description or the language of words, can be used to describe only very simple and standard solids, while the photographs can show only the exterior of an object.
Also the object to be designed or improvement of some existing design can not be photographed. Hence the only possible media, accepted universally, by which an engineer can express his/her ideas is the language of lines or the graphical description, commonly called "Engineering Drawing", which is a systematic combination of different conventional lines.
PROJECTION . TERMS
In graphic language the shape is described by projection, which is the image of the object, formed by rays of light, taken in some particular direction, from the object into a picture plane, as it appears to an observer stationed at the point, from or towards which the projection is made.
Depending upon the orientation of the object, location of the point of sight, and the direction of lines of sight relative to the picture plane, different types of projections, e.g., perspective, parallel, orthographic, axonometric, oblique, etc., can be obtained.
The plane, on which the projection is taken is called the plane of projection or picture plane. The point, from which the observer is assumed to view the object, is called the station point or the centre of projection.In this chapter only the orthographic system of projections will be studied in detail. Other types of projections are dealt in separate chapters.
ORTHOGRAPHIC PROJECTIONReferring to Fig.1, if the station point (or observer's eye) is at an infinite distance from an object (a prism here) such that the lines of sight (or projectors) are parallel to each other and perpendicular to the plane of projection (or picture plane), the projection obtained, which would be of the same size and shape as the facing surface of the object, is called an orthographic projection. Note that the object is placed with one of its faces parallel to the picture plane.
ORTHOGRAPHIC SYSTEM OF PROJECTIONS
Any object has three dimensions, i.e., length, breadth (depth) and height. The problem is to represent or convey all these three dimensions, together with the other details of the object, on a sheet of drawing paper which has only two dimensions.
Orthographic system of projections is a method of representing the exact shape of a three dimensional object on a two dimensional drawing sheet in two or more views. As is clear from Fig 1 that a single orthographic projection of a cube, or any other object, does not describe the form of the object, when the object is so placed that one of its faces is parallel to the plane of projection.
In order to achieve a complete shape description, in such cases, it is necessary to get more than one projection, and therefore, additional planes of projection are used to project more views on them, for the object. As such, the orthographic system of projections is also called multi-view projection method.
These views are obtained by dropping perpendiculars from two or more sides of the object to the picture planes, generally set at right angles to each other. These picture planes are then rabated to lie them in one plane.
This method is almost universally used in engineering drawing to graphically describe the shape of objects.
CO-ORDINATE PLANES OF PROJECTION
In the orthographic projection drawing, for getting the different views of an object, three main planes are usually used. One of these set up in vertical position is called the vertical plane of projection (VP) or Frontal Plane (FP).
The second plane, set up in horizontal position, i.e., perpendicular to the VP, is called Horizontal Plane (HP). The third plane, set up perpendicular to the vertical and horizontal planes is called profile Plane (PP).
The horizontal and vertical planes, which are called the principal planes, divide the whole space on one side of the profile plane in four parts, called the four dihedral angles.
Figure .2 illustrates the coordinate planes and the four dihedral angle (or quadrants). The lines of intersection of these three planes are called coordinate axes. Line of intersection of HP and VP is more commonly called "Reference Line" and is denoted by XY. The point of intersection of the three coordinate planes is called the origin.
Orthographic views of any object can be represented by any one of the two systems of projection, e.g., the First Angle Projection and the Third Angle Projection. These are named according to the quadrant in which the object is imagined to be placed, for purposes of projection. The shape and size of views are same in these two systems. The only difference lies in the relative position of the various views.
First angle projection (European) is widely used throughout Europe, the USSR, and many other countries in Asia. Third Angle Projection (American) is used in America. The current Indian and ISO Standards state that both systems of projection are equally acceptable but they should never be mixed on the same drawing.
Both these systems should be taught in Engineering Institutions, as these are used throughout the world, for the preparation of technical drawings. In India the use of First Angle Projection is in use exclusively after 1991.
Theory of Orthographic Projections,Theory of Orthographic Projections,Theory of Orthographic Projections,Theory of Orthographic Projections | 677.169 | 1 |
Superellipsoid
A superellipsoid is a surface defined as the set of points (x,y,z){\displaystyle (x,y,z)} in 3D space such that (x2/ϵ2+y2/ϵ2)ϵ2/ϵ1+z2/ϵ1=1{\displaystyle (x^{2/\epsilon _{2}}+y^{2/\epsilon _{2}})^{\epsilon _{2}/\epsilon _{1}}+z^{2/\epsilon _{1}}=1} where ϵ1,ϵ2>0{\displaystyle \epsilon _{1},\epsilon _{2}>0}, or any scaling (possibly nonuniform) of such a surface. "Superellipsoid" refers to both the surface and the solid that it encloses. All cross sections of a superellipsoid in planes parallel to the xy-plane are superellipses of exponent ϵ1{\displaystyle \epsilon _{1}}, and the cross sections in the xz- and yz-planes are superellipses of exponent ϵ1{\displaystyle \epsilon _{1}}.
Superellipsoids come from the field of computer graphics. Special settings produce spheres, Steinmetz solids, bicones, supereggs, and the regular octahedron. If the constants ϵ1,ϵ2{\displaystyle \epsilon _{1},\epsilon _{2}} are permitted to tend to infinity, as they often are in computer graphics, cylinders and cuboids can also be produced. | 677.169 | 1 |
Angle relationships math lib answer key
• Determine the sum of the interior angles of a triangle. Takes things to the cleaners. Gina wilson, 2012 products by gina wilson (all things algebra) may be used by the purchaser for their classroom use only. Algebra Nation Mafs Answer Key - sandbox.ins.to Gina wilson all things algebra 2016 answer key unit 11 probability and statistics.Jun 8, 2022 · We included HMH Into Math Grade 8 Answer Key PDF Module 4 Angle Relationships to make students experts in learning maths. HMH Into Math Grade 8 Module 4 Answer Key Angle Relationships. A Fox From Any Angle. The diagram shows the design for a fox made from folded paper. Give an example of each type of angle pair in the design. A. Vertical anglesQuiz 2 (Pages 15-16; Answer Key 62-63) Packet 3 (Relationships in Triangles & Congruent Triangles) (Pages 17-20; Answer Key 64-67) • Solve problems using the Triangle Sum Theorem, angles in isosceles triangles, and exterior angles of triangles. • Determine wheth er a triangle could be formed given the lengths of the sides. • Given …
Did you know?
When your arms are held out at your sides and your palms are facing forward, your forearm and hands should normally point about 5 to 15 degrees away from your body. This is the nor...Relationships kuta angles complementary supplementary software excel adjacentAngle relationships worksheet #2 Angles relationship worksheet grade 7th curated reviewed lessonplanetAngle relationships answer key quiz preview math. Angle relationships worksheet answer keyAngles pair mathworksheets4kids 1 5 describe …Students will practice using trigonometric ratios to solve angle of elevation and angle of depression word problems as they rotate through 10 stations with this "Math Lib" Activity. The answer at each station will give them a piece to a story (who, doing what, with who, where, when, etc.). This is a much more fun approach to multiple choice ...
7th Grade Angle Relationships quiz for 7th grade students. Find other quizzes for Mathematics and more on Quizizz for free! 7th Grade Angle Relationships quiz for 7th grade students. Find other quizzes for Mathematics and more on Quizizz for free! Skip to Content. Enter code. Log in. Sign up. Enter code. Log in. Sign up. Build your own quiz. …Transforming objects in Adobe Illustrator so they appear angled -- like the difference between a rectangle and a parallelogram, which lacks the rectangle's uniform 90-degree corner...Try It \ (\PageIndex {8}\) \ (\dfrac {2x+15} {9}=\dfrac {7x+3} {15}\). Answer. To solve applications with proportions, we will follow our usual strategy for solving applications. But when we set up the proportion, we must make sure to have the units correct—the units in the numerators must match and the units in the denominators must match.Displaying top 8 worksheets found for - Gina Wilson Answer Key. Some of the worksheets for this concept are Factoring polynomials gina wilson work, Two step equations maze gina wilson answers, Pdf gina wilson algebra packet answers, Algebra antics answers key, Unit 3 relations and functions, Gina wilson unit 8 quadratic equation answers pdf, …CCSS 7.G.B.5. Use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure. This set of task cards provides students with practice finding the value of x or a missing angle measure, given the following angle relationships:Vertical ... | 677.169 | 1 |
Points and Lines in the Plane
Tracie set out from Elmhurst, IL, to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot in Figure 1. Laying a rectangular coordinate grid over the map, we can see that each stop aligns with an intersection of grid lines. In this section, we will learn how to use grid lines to describe locations and changes in locations.
Figure 1
Plot Points on the Coordinate Plane
An old story describes how seventeenth-century philosopher/mathematician René Descartes invented the system that has become the foundation of algebra while sick in bed. According to the story, Descartes was staring at a fly crawling on the ceiling when he realized that he could describe the fly's location in relation to the perpendicular lines formed by the adjacent walls of his room. He viewed the perpendicular lines as horizontal and vertical axes. Further, by dividing each axis into equal unit lengths, Descartes saw that it was possible to locate any object in a two-dimensional plane using just two numbers—the displacement from the horizontal axis and the displacement from the vertical axis.
While there is evidence that ideas similar to Descartes' grid system existed centuries earlier, it was Descartes who introduced the components that comprise the Cartesian coordinate system, a grid system having perpendicular axes. Descartes named the horizontal axis the x-axis and the vertical axis the y-axis.
The Cartesian coordinate system, also called the rectangular coordinate system, is based on a two-dimensional plane consisting of the x-axis and the y-axis. Perpendicular to each other, the axes divide the plane into four sections. Each section is called a quadrant; the quadrants are numbered counterclockwise as shown in Figure 2.
Figure 2
The center of the plane is the point at which the two axes cross. It is known as the origin, or point [latex]\left(0,0\right)[/latex]. From the origin, each axis is further divided into equal units: increasing, positive numbers to the right on the x-axis and up the y-axis; decreasing, negative numbers to the left on the x-axis and down the y-axis. The axes extend to positive and negative infinity as shown by the arrowheads in Figure 3.
Figure 3
Each point in the plane is identified by its x-coordinate, or horizontal displacement from the origin, and its y-coordinate, or vertical displacement from the origin. Together, we write them as an ordered pair indicating the combined distance from the origin in the form [latex]\left(x,y\right)[/latex]. An ordered pair is also known as a coordinate pair because it consists of x- and y-coordinates. For example, we can represent the point [latex]\left(3,-1\right)[/latex] in the plane by moving three units to the right of the origin in the horizontal direction, and one unit down in the vertical direction.
Figure 4
When dividing the axes into equally spaced increments, note that the x-axis may be considered separately from the y-axis. In other words, while the x-axis may be divided and labeled according to consecutive integers, the y-axis may be divided and labeled by increments of 2, or 10, or 100. In fact, the axes may represent other units, such as years against the balance in a savings account, or quantity against cost, and so on. Consider the rectangular coordinate system primarily as a method for showing the relationship between two quantities.
A General Note: Cartesian Coordinate System
A two-dimensional plane where the
x-axis is the horizontal axis
y-axis is the vertical axis
A point in the plane is defined as an ordered pair, [latex]\left(x,y\right)[/latex], such that x is determined by its horizontal distance from the origin and y is determined by its vertical distance from the origin.
Example: Plotting Points in a Rectangular Coordinate System
Plot the points [latex]\left(-2,4\right)[/latex], [latex]\left(3,3\right)[/latex], and [latex]\left(0,-3\right)[/latex] in the plane.
Answer:
To plot the point [latex]\left(-2,4\right)[/latex], begin at the origin. The x-coordinate is –2, so move two units to the left. The y-coordinate is 4, so then move four units up in the positive y direction.
To plot the point [latex]\left(3,3\right)[/latex], begin again at the origin. The x-coordinate is 3, so move three units to the right. The y-coordinate is also 3, so move three units up in the positive y direction.
To plot the point [latex]\left(0,-3\right)[/latex], begin again at the origin. The x-coordinate is 0. This tells us not to move in either direction along the x-axis. The y-coordinate is –3, so move three units down in the negative y direction. See the graph in Figure 5.
Figure 5
Analysis of the Solution
Note that when either coordinate is zero, the point must be on an axis. If the x-coordinate is zero, the point is on the y-axis. If the y-coordinate is zero, the point is on the x-axis.
Distance in the Plane
Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem, [latex]{a}^{2}+{b}^{2}={c}^{2}[/latex], is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse.
The relationship of sides [latex]|{x}_{2}-{x}_{1}|[/latex] and [latex]|{y}_{2}-{y}_{1}|[/latex] to side d is the same as that of sides a and b to side c. We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example, [latex]|-3|=3[/latex]. ) The symbols [latex]|{x}_{2}-{x}_{1}|[/latex] and [latex]|{y}_{2}-{y}_{1}|[/latex] indicate that the lengths of the sides of the triangle are positive. To find the length c, take the square root of both sides of the Pythagorean Theorem.
Graph Linear Equations
We can plot a set of points to represent an equation. When such an equation contains both an x variable and a y variable, it is called an equation in two variables. Its graph is called a graph in two variables. Any graph on a two-dimensional plane is a graph in two variables.
Suppose we want to graph the equation [latex]y=2x - 1[/latex]. We can begin by substituting a value for x into the equation and determining the resulting value of y. Each pair of x- and y-values is an ordered pair that can be plotted. The table below lists values of x from –3 to 3 and the resulting values for y.
[latex]x[/latex]
[latex]y=2x - 1[/latex]
[latex]\left(x,y\right)[/latex]
[latex]-3[/latex]
[latex]y=2\left(-3\right)-1=-7[/latex]
[latex]\left(-3,-7\right)[/latex]
[latex]-2[/latex]
[latex]y=2\left(-2\right)-1=-5[/latex]
[latex]\left(-2,-5\right)[/latex]
[latex]-1[/latex]
[latex]y=2\left(-1\right)-1=-3[/latex]
[latex]\left(-1,-3\right)[/latex]
[latex]0[/latex]
[latex]y=2\left(0\right)-1=-1[/latex]
[latex]\left(0,-1\right)[/latex]
[latex]1[/latex]
[latex]y=2\left(1\right)-1=1[/latex]
[latex]\left(1,1\right)[/latex]
[latex]2[/latex]
[latex]y=2\left(2\right)-1=3[/latex]
[latex]\left(2,3\right)[/latex]
[latex]3[/latex]
[latex]y=2\left(3\right)-1=5[/latex]
[latex]\left(3,5\right)[/latex]
We can plot the points in the table. The points for this particular equation form a line, so we can connect them.This is not true for all equations.
Note that the x-values chosen are arbitrary, regardless of the type of equation we are graphing. Of course, some situations may require particular values of x to be plotted in order to see a particular result. Otherwise, it is logical to choose values that can be calculated easily, and it is always a good idea to choose values that are both negative and positive. There is no rule dictating how many points to plot, although we need at least two to graph a line. Keep in mind, however, that the more points we plot, the more accurately we can sketch the graph.
How To: Given an equation, graph by plotting points.
Make a table with one column labeled x, a second column labeled with the equation, and a third column listing the resulting ordered pairs.
Enter x-values down the first column using positive and negative values. Selecting the x-values in numerical order will make the graphing simpler.
Select x-values that will yield y-values with little effort, preferably ones that can be calculated mentally.
Plot the ordered pairs.
Connect the points if they form a line.
Example: Graphing an Equation in Two Variables by Plotting Points
Graph the equation [latex]y=-x+2[/latex] by plotting points.
Answer:
First, we construct a table similar to the one below. Choose x values and calculate y.
[latex]x[/latex]
[latex]y=-x+2[/latex]
[latex]\left(x,y\right)[/latex]
[latex]-5[/latex]
[latex]y=-\left(-5\right)+2=7[/latex]
[latex]\left(-5,7\right)[/latex]
[latex]-3[/latex]
[latex]y=-\left(-3\right)+2=5[/latex]
[latex]\left(-3,5\right)[/latex]
[latex]-1[/latex]
[latex]y=-\left(-1\right)+2=3[/latex]
[latex]\left(-1,3\right)[/latex]
[latex]0[/latex]
[latex]y=-\=-\left(1\right)+2=1[/latex]
[latex]\left(1,1\right)[/latex]
[latex]3[/latex]
[latex]y=-\left(3\right)+2=-1[/latex]
[latex]\left(3,-1\right)[/latex]
[latex]5[/latex]
[latex]y=-\left(5\right)+2=-3[/latex]
[latex]\left(5,-3\right)[/latex]
Now, plot the points. Connect them if they form a line.
Try It
Construct a table and graph the equation by plotting points: [latex]y=\frac{1}{2}x+2[/latex].
Answer:
[latex]x[/latex]
[latex]y=\frac{1}{2}x+2[/latex]
[latex]\left(x,y\right)[/latex]
[latex]-2[/latex]
[latex]y=\frac{1}{2}\left(-2\right)+2=1[/latex]
[latex]\left(-2,1\right)[/latex]
[latex]-1[/latex]
[latex]y=\frac{1}{2}\left(-1\right)+2=\frac{3}{2}[/latex]
[latex]\left(-1,\frac{3}{2}\right)[/latex]
[latex]0[/latex]
[latex]y=\frac{1}{2}\=\frac{1}{2}\left(1\right)+2=\frac{5}{2}[/latex]
[latex]\left(1,\frac{5}{2}\right)[/latex]
[latex]2[/latex]
[latex]y=\frac{1}{2}\left(2\right)+2=3[/latex]
[latex]\left(2,3\right)[/latex]
Plot Points With a Graphing Utility
Most graphing calculators require similar techniques to graph an equation. The equations sometimes have to be manipulated so they are written in the style y=_____. The TI-84 Plus, and many other calculator makes and models, have a mode function, which allows the window (the screen for viewing the graph) to be altered so the pertinent parts of a graph can be seen.
For example, the equation [latex]y=2x - 20[/latex] has been entered in the TI-84 Plus shown in Figure 9a. In Figure 9b, the resulting graph is shown. Notice that we cannot see on the screen where the graph crosses the axes. The standard window screen on the TI-84 Plus shows [latex]-10\le x\le 10[/latex], and [latex]-10\le y\le 10[/latex]. See Figure (c) below.
a. Enter the equation. b. This is the graph in the original window. c. These are the original settings.
By changing the window to show more of the positive x-axis and more of the negative y-axis, we have a much better view of the graph and the x- and y-intercepts. See (a) and (b) below.
a. This screen shows the new window settings. b. We can clearly view the intercepts in the new window.
Example: Using a Graphing Utility to Graph an Equation
Use a graphing utility to graph the equation: [latex]y=-\frac{2}{3}x-\frac{4}{3}[/latex].
Answer:
Enter the equation in the y= function of the calculator. Set the window settings so that both the x- and y- intercepts are showing in the window.
Key Concepts
We can locate, or plot, points in the Cartesian coordinate system using ordered pairs, which are defined as displacement from the x-axis and displacement from the y-axis.
An equation can be graphed in the plane by creating a table of values and plotting points.
Using a graphing calculator or a computer program makes graphing equations faster and more accurate. Equations usually have to be entered in the form y=_____.
Finding the x- and y-intercepts can define the graph of a line. These are the points where the graph crosses the axes.
The distance formula is derived from the Pythagorean Theorem and is used to find the length of a line segment.
The midpoint formula provides a method of finding the coordinates of the midpoint dividing the sum of the x-coordinates and the sum of the y-coordinates of the endpoints by 2.
Glossary
Cartesian coordinate system a grid system designed with perpendicular axes invented by René Descartes
equation in two variables a mathematical statement, typically written in x and y, in which two expressions are equal
graph in two variables the graph of an equation in two variables, which is always shown in two variables in the two-dimensional plane
intercepts the points at which the graph of an equation crosses the x-axis and the y-axis
ordered pair a pair of numbers indicating horizontal displacement and vertical displacement from the origin; also known as a coordinate pair, [latex]\left(x,y\right)[/latex]
origin the point where the two axes cross in the center of the plane, described by the ordered pair [latex]\left(0,0\right)[/latex]
quadrant one quarter of the coordinate plane, created when the axes divide the plane into four sections
x-axis the common name of the horizontal axis on a coordinate plane; a number line increasing from left to right
x-coordinate the first coordinate of an ordered pair, representing the horizontal displacement and direction from the origin
x-intercept the point where a graph intersects the x-axis; an ordered pair with a y-coordinate of zero
y-axis the common name of the vertical axis on a coordinate plane; a number line increasing from bottom to top
y-coordinate the second coordinate of an ordered pair, representing the vertical displacement and direction from the origin
y-intercept a point where a graph intercepts the y-axis; an ordered pair with an x-coordinate of zero
distance formula a formula that can be used to find the length of a line segment if the endpoints are known
midpoint formula a formula to find the point that divides a line segment into two parts of equal length | 677.169 | 1 |
Did you know?
Cos 30° = √3/2 is an irrational number and equals to 0.8660254037 (decimal form). Therefore, the exact value of cos 30 degrees is written as 0.8660 approx. √3/2 is the value of Cos 30° which is a trigonometric ratio or trigonometric function of a particular angle. Cos 30. Another alternative form of Cos 30° is pi/6 or π/6 or Cos 33 (⅓) gSine, Cosine and Tangent. Sine, Cosine and Tangent (often shortened to sin, cos and tan) are each a ratio of sides of a right angled triangle: For a given angle θ each ratio stays the same no matter how big or small the triangle is. To calculate them: Divide the length of one side by another sideThis gives us cos of 270 degrees multiplied by cos 𝜃 minus sin of 270 degrees multiplied by sin 𝜃. The cos of 270 degrees is zero, and the sin of 270 degrees is negative one. Our expression simplifies to zero multiplied by cos 𝜃 minus negative one multiplied by sin 𝜃, which is equal to sin 𝜃. This confirms the expression we got ... sin(270^o) = -1, cos (270^0) = 0, tan (270^0)= undefined. Consider the unit circle (a circle with radius 1). On the unit circle as graphed on an xy coordinate plane, with 0 degrees starting at (x,y) = (1,0): graph{x^2+y^2=1 [-1, 1, -1, 1]} If we draw a line from the origin at the angle we seek, then where that line intersects the unit circle, the sin of the angle will be equal to the y ...
Find the exact values of the following. 1. cos 270 degree 2. Sin180 degree. What is the exact value of cos(405), where 405 is in degrees? Find the exact value of the following. a. cos 315 degrees. b . sin 240 degrees. c. cot 315 degrees. d. csc 225 degrees. What is a trigonometric identity for cos(3x)?Trig calculator finding sin, cos, tan, cot, sec, csc. To find the trigonometric functions of an angle, enter the chosen angle in degrees or radians. Underneath the calculator, the six most popular trig functions will appear - three basic ones: sine, cosine, and tangent, and their reciprocals: cosecant, secant, and cotangent. ….
Study with Quizlet and memorize flashcards containing terms like Sin 90°, Cos 90°, Tan 90° and more. Fresh features from the #1 AI-enhanced learning platform. Explore the lineupFor cos 225 degrees, the angle 225° lies between 180° and 270° (Third Quadrant). Since cosine function is negative in the third quadrant, thus cos 225° value = -(1/√2) or -0.7071067. . . Since cosine function is negative in the third quadrant, thus cos 225° value = -(1/√2) or -0.7071067. . .Feb 10, 2019 · Following is the URL of video explaining "How to find the value of sin(270)?" is the URL of video explaining "How to fin... | 677.169 | 1 |
Class 8 Courses
A field is in the shape of a trapezium having parallel sides 90 m and 30 m field is in the shape of a trapezium having parallel sides 90 m and 30 m. These sides meet the third side at right angles. The length of the fourth side is 100 m. If it costs Rs 5 to plough 1 m2 of the field, find the total cost of ploughing the field.
Solution:
In the given figure, ABCD is a trapezium having parallel sides 90 m and 30 m. | 677.169 | 1 |
This reminds me when I didn't got a "correct" in an answer I provided, when I should use 3 microprocessor instructions to have something done and I could do that with just 2 instructions... some teachers can be boring, sometimes... *You don't really need to use the book*.
– woliveirajrOct 05 '12 at 12:52
Math answers should be completely general. Most of those submitted will not work if the circle is much larger than the book. But the book edge can be used to extend the normal to the chord.
– Oct 02 '12 at 21:56
17 Answers17
121
Let the corner of the book just touch the circle's edge. You can draw two perpendicular lines along book's edge which intersect with the circle at two different points. Connect the two points and it's a diameter.
Repeat this and get another diameter. The intersect of these two diameters is the center of the circle.
Even if the book corner wasn't a good right angle (which they usually aren't, since they tend to be rounded, especially for a hardcover), you can always use one of the pages.
– asmeurerOct 03 '12 at 04:47
Use the book to draw two parallel lines that cross the circle at different distances from the center to obatain 4 intersection points in the circle. Draw 4 lines connecting these points. The two points of intersection from these 4 lines define a diameter. Repeat the process with two parallels with different direction to obtain another diameter. Finally the intersection of two diameters returns the circle center.
I added an image to explain. The green line is a diameter because is perpendicular to the parallel lines (in red) and is crossing them at equal distance from both outer segments (in blue).
– LuisOct 03 '12 at 02:21
This solution is far better than the accepted one, since it does not depends on perception ("put the corner of the book..." how do you know that you are *really* drawing a diameter?). +1 for this answer!
– BarrankaOct 03 '12 at 18:59
If @AlbeyAmakiir's idea of using the thickness of the spine is Luis's intended solution, this entails an awkward action and depends on the circle being big enough. Alternatively, use the book to draw two perpendiculars ABC and AD, BC being a chord. Then shift the book along AD and draw EFG parallel to ABC where FG is a chord and ED is on AD.
– Rosie FDec 17 '17 at 10:04
27
If the circle is a foldable material (like paper), you could fold it in half twice, the intersection of the folds will indicate the center.
How to rescue this: Suppose the circle is printed on stiff, opaque boxboard. Tear a page out of the book (let's assume the pages are reasonably thin and translucent). Using the pencil, trace the circle onto the page. You don't even have to trace the whole circle, just mark off four diametrically opposed arcs long enough for good alignment. Then do the folding trick with this copy. Make a hole in the center of the copy, line up with the original circle and then mark the center through the hole.
– KazOct 03 '12 at 08:31
1
Combine with the Thales solution: Fold one corner of the paper onto the circle perimeter and mark off the two edges with the pencil. Repeat with another corner of the paper. Proceed as above by intersecting the diameters. *Question:* Is there any situation of a circle on a rectangular paper, where you cannot use the (folded) paper edges themselves as ruler for this construction?
– Hagen von EitzenOct 03 '12 at 12:49
I have used this technique many times in the past. Regardless of what other people say, it has worked fine. So +1 from me :)
– AnnanFayOct 03 '12 at 19:14
22
As pointed out, most of these answers will not work if the circle is much larger than the book. Instead, find someone with the proper size compass and straightedge, T-square, or other drafting tools, and threaten to hit them with the book if they do not find the center of the circle. See BAROMETER
@Will Jagy, thanks for the link, but I think even the book can not cover the circle, it still works, since we can extend the line with the edge of the book. the key point is then the book has the corner as right-angle.
– zddOct 03 '12 at 00:53
+1 for the BAROMETER link. And regarding size of book and pencil and circle: the OP provided the pictures of everything, and it's possible to observe that the circle is smaller than the book and the pencil...
– woliveirajrOct 05 '12 at 12:56
No, Poncelet-Steiner is not the solution since it requires a known center. The center in this question is unknown, so Thales' Theorem is more likely.
– spexJun 18 '14 at 19:43
13
Assuming the paper contains nothing but the shape of the circle:
1. Put eraser end of pencil on book. Or put pencil flush against side of book.
2. Rest circle on tip of point.
3. Circle will rest on the center.
The only book I have is usually under my pillow. It has become skewed over the years - no 90 degrees joy. As a consequence I failed to have success applying the (elegant) description of Patrick Li. Moreover, my book is too small anyway to connect diametrically opposing points on the circle.
Therefore I had to revert to a more tedious approach.
Pick two points on the circle, close enough for my book's reach, connect them.
Align one of the book's edges with that line, the bookcorner at one of the marked points, draw a line along the adjacent edge of the book - which is non-perpendicular to the first line. Flip the book and draw "the other" non-perpendicular line through the same point. Repeat at the other point. Connect the two intersections that the four lines make, and extend this line - by shifting one bookedge along - to complete the diameter. If the intersections are too wide apart for my book, retry with the original points closer together. Repeat all of this with two other points on the circle, to get two diameters crossing at the midpoint.
If, by some miracle, my book has straightened out again (but still is tiny compared to the circle), I will quickly find out: the two "non-perpendicular" lines at one of the initial mark-points overlap. Then, I continue the line(s) until I cross the circle at the other side; likewise at the other initial point, and I wind up with a long and narrow rectangle. Repeating that from some other location on the circle, roughly at 90 degrees along the circle from the first setup, I get two rectangles and their intersection is a small parallellogram somewhere in the middle. It's diagonals cross at the center of the circle.
Draw a right triangle inside the circle, making all vertices touch the circle.
Transform it into a rectangle.
Draw a line joining the opposite corners and you will have obtained the center of the circle
Edit:
The idea is very simple: 1. Put a corner of the book touching the circle from the inside in any place. 2. Use the edges of the book for drawing the catheti of the right triangle until they intersect the circle. 3. Draw the hypotenuse using those intersections. 4. Put a corner of the book in any of those intersections overlapping one edge with the cathetus. 5. Use the other edge of the book to draw a line until it intersects the circle. 6. Draw a line joining this new intersection with the intersection opposed to the hypotenuse (right angle point). 7. The point in which this line crosses (intersects) the hypotenuse is the center of the circle.
Edit:
In fact, any rectangle inside the circle with every corner touching the circle will do the trick
Another way to solve the problem is to pick 3 distinct points on circle say A, B, C. Then connect AB and BC. Halve A-B and B-C to obtain D and E. Draw lines through D and E perpendicular to AB and BC. The intersection of these lines will be the centre of the circle.
If you can't accurately determine the center of a circle, how can you accurately determine the midpoint of a segment?
– quietmintOct 02 '12 at 23:02
@user113215 by taking off a page of the book, cutting it to match the size of the line AB, folding the piece of page that matched the size, putting it over the AB line again to mark what was the middle, and doing that again with the BC segment...
– woliveirajrOct 05 '12 at 12:29
1
1>
Place your sheet of paper on top of the circle so that the corner just touches the
circle's edge. Hold the paper in place, and use a pencil to make a small mark at the
exact point where the two edges that meet at the corner cross the perimeter of the
circle.
2>
Using a ruler or straightedge, draw a straight line from one mark to the other.
3>
Place the paper on the circle again with a different orientation. Make the tick marks
again and connect these marks with a second pencil line using the ruler or straightedge.
The center of the circle is the point at which this line crosses the first line you drew.
This appears to be the same as [Patrick Li's answer]( Did you have anything else to add?
– robjohnOct 03 '12 at 19:50
@robjohn: Ups, you are right. I didn't check the other answers carefully enough. At least I mentioned in the answer the central theorem (Thale's) and the required property of the book (90° corner).
– CurdOct 04 '12 at 09:10
0
Hang the circle by its top edge on a wall close to a door or corner, draw a line straight down using the book's edge parallel to the door/corner edge. Rotate the circle f.ex 90 deg., draw a new line and you have the center and radius. This technique can also find the center of more complicated shapes. Gravity is the keyword here.
This solution uses additional tools, namely the ability to draw a line in a prescribed direction and the ability to find an apex point of the circle, neither of which is really in the (idealized) problem as specified.
– Steven StadnickiOct 02 '12 at 16:58
This answer is nonsense. What it finds is the center of gravity of the object onto which the circle is printed. It assumes that the circle is actually a disc object of uniform density, whose center of gravity corresponds to its geometric centre. The question clearly states that the circle is *on* paper, not *of* paper. Even if it's a cutout, paper is likely going to have uneven thickness so that the center of gravity doesn't correspond to the geometric center.
– KazOct 03 '12 at 08:34
@Kaz :) ok, but you expect to draw very precise lines using the book and the pencil? The book edge will be a very straight line, without imperfections? The tip of the pencil is sharp enought or has some width associated ? Etc...
– woliveirajrOct 05 '12 at 12:24
0
Let D be the center. Let A and B are points on the circle circumference. Let C be another point in the circle circumference.
Then angle ADB is twice angle ACB. I won't provide the proof. Actually it depends on whether C is between A and B or outside it. The theorem is still true but you also need to compute the angle ADB appropriately.
Now, a diameter is a point where the angle of ADB is 180 degree. That means we need to find a point where angle ACB is half of it namely 90 degree.
How?
That book has a 90 degree corner right? Put the corner in the circle circumference. Let's call it C. Now, take the intersection between the edges of the book to the edges of the circle (excluding C). Let's call them A and B. Now, if D is the center of the circle, we know that angle ADB is 180 degree.
We do not know what ADB is. However, 180 degree is simply a straight line. So if we draw a line between A and B, then D must be there.
If D is not on that straight line, then angle ADB won't be 180 degree. That's because the triangle ADB won't be "trivial". That means angle ABD and BAD won't be 0. By sum of angles in triangles, non zero angles of 2 of the angles in a triangle imply that ADB is not 180 degrees. Hence, we can conclude that point D is in a point AB. That means point AB is essentially a "diameter". It's a line that goes to the center of the circle.
Now repeat the process and draw a bunch of diameters. Of those diameters will intersect at some point.
keep the circle on the book at one corner,so as the two sides of the book become tangents @ 90degrees to the circle. the join the two corners of the by a straight line which will pass thro' the center of the circle. now rotate the circle by few degrees closer to 90 degrees and draw another line again as done before.the intersection point of the two drawn lines is the center of the circle.
another method.
1) fold the circle to make exactly half(semi)circle.
2) again fold the semi circle to make an exact quarter circle.
3) open and see the two folds meet @ one point, that is the center of the circle.
both methods can get the center of the circle.
c v ramadoss. chennai 17
The second way was not true, you have no way to fold the circle exactly half, if you can do that, it means you already get the diameter.
– zddOct 03 '12 at 03:10
0
Use the fact that the book is a straight edge to draw a straight line that touches the circle at one point. Then use the fact that it has a right angle to draw a line perpendicular to that line at that point (where it intersects the circle). This will be a diameter. Repeat this at another point on the circle (not antipodal) to get another diameter. They will intersect at the center.
Follows from the basic fact that any tangent line to a circle is perpendicular to the diameter that goes through the same point (and also from the fact that a standard book is rectangular in shape).
Cute, but how do you put the circle in the middle (not to mention that you'd have to do this twice, to get two diameters)?
– Rick DeckerOct 11 '12 at 01:32
0
Using just the pencil
Mark a point A on the circle
Put the beggining of the straight edge of the pencil over the point A
Rotate the pencil, keeping the pencil's tip over the point A and the point of your finger on the pencil, accompaining the circle, so that you're measuring the distance from the point A to the point that the pencil touchs the circle
Immediatlly before this measure starts to decrease, you've found the circle's diameter.
Draw a line connecting point A and this point found on step 5. You've draw the circle's diameter. | 677.169 | 1 |
The Fundamental Concept of Points in Mathematics | Defining Locations in Space and Geometric Relationships
point
In mathematics, a point is a fundamental concept that represents a specific location in space
In mathematics, a point is a fundamental concept that represents a specific location in space. It has no size, shape, or dimension. Points are typically represented using a dot and labeled with a capital letter, such as A, B, C, etc.
Points can be used to define various geometric shapes and concepts. For example, a line is made up of an infinite number of points that extend infinitely in two opposite directions. Similarly, a plane is a flat surface composed of an infinite number of points extending indefinitely in all directions.
Points can also be used to measure distances and define coordinates in coordinate systems. In a Cartesian coordinate system, a point's position is described by its coordinates, typically written as (x, y) in two dimensions, or (x, y, z) in three dimensions.
Points can have certain relationships with each other. For example, two points can be connected by a line segment, which is a straight line with two distinct endpoints. Points can also be collinear, meaning they lie on the same line, or coplanar, meaning they lie on the same plane.
In summary, a point is a basic mathematical concept used to represent a specific location in space. It is dimensionless and can be used to define and describe various geometric shapes and relationships | 677.169 | 1 |
SSA (Side-Side-Angle) The side in front of the angle doesn't have to be minor of the adjacent one
HC (Hypotenuse-Cathetus) In a right triangle
HA (Hypotenuse-Angle) In a right triangle
And my question is about one case that I know is true in Euclidean geometry (which I would call AB—Apex-Base—):
If two isosceles triangles have in common the apex and the base, both are congruent
It's "obvious" but the proof is not that obvious. Here's my attempt.
Proof
Let △ABC and △A'B'C' be isosceles triangles like the next figure
Now, if AB = A'B' both triangles are congruent by SAS, but suppose that's not the case. Without loss of generality assume that AB<A'B', thus it's possible to mark a point D' in A'B' such that A'D' = AB, likewise it's possible to mark a point E' in A'C' such that A'E' = AC. Join the points D'E' to form the △A'D'E' ≅ △ABC by SAS.
Then we must conclude that D'E' = B'C', but this is a contradiction since D'E'<B'C', so assuming AB ≠ A'B' is absurd, hence AB = A'B' which derives △ABC ≅ △A'B'C' Q.E.D.
That's what I did, but I feel that the contradiction is not fully handled, especially because I cannot find another argument but visual that D'E' < B'C'.
As I mentioned before, I know this is true in Euclid's (plane) geometry. I'm more interested in a proof (if exist) avoiding parallels (or equivalently the constant sum of the measures of the angles of any triangle), so that the result would be a neutral geometry like the SAA criterion (
Or I'm trying to prove a result that's not neutral geometry that in some cases D'E' =B'C'?
Edit
Thanks for your answers, they are very useful. But up this moment all of them depends on the parallel postulate which I try to avoid in order to find out if this is a neutral geometry result.
2 Answers
2
If you know that the sum of angles in a triangle are $180^\circ$, then, if the apex has value $A$, the other two angles are $90^\circ-A/2$. Then the problem is reduced to the angle-side-angle case. The side is the base.
In your case you use the same angles to show $D'E'||B'C'$. Then the triangles are similar, and the ratio of the sides is equal to the ratio of the bases. You know that the bases are the same, so then the sides must be the same. The problem is then reduced to side-side-side case. | 677.169 | 1 |
Worksheet 12: Straight Line Geometry
This grade 8 worksheet is for the last section of term 2 according to the CAPS curriculum, on straight line geometry. It includes questions on parallel and perpendicular lines, corresponding, alternating and co-interior angles, as well as vertically opposite, supplementary and corresponding angles. There is also a fully worked out memorandum. Please note that with Geometry there are many different ways to skin a cat, and there may be more than one way to get to each answer. | 677.169 | 1 |
Perimeter of a Polygon – Definition with Examples
Let's dive into an adventurous journey exploring the fascinating world of polygons! Here at Brighterly, we believe in making learning fun and effective. We're going to explore a fundamental aspect of polygons – their perimeters.
Now imagine, if you will, a string that we lay along the edges of a shape. That string, my friends, represents the perimeter of that shape. Whether it's a triangle, a square, or a complex dodecagon, that string follows their sides, defining the boundary of these shapes. It's like a path you would take if you were to walk around the edges of a garden or a park. So, in essence, the perimeter of a polygon is the total distance around its sides.
But wait, it gets more interesting when we dive into different kinds of polygons. Regular and irregular polygons have their unique ways of determining the perimeter. All the while, our friend, the string, remains faithful, helping us trace the perimeter, whether the sides are equal (as in regular polygons) or varied (as in irregular polygons). It's an intriguing concept that has wide applications, from designing a soccer field to creating a blueprint for a house. So let's unravel the mystery of the 'perimeter' together, and remember, at Brighterly, we transform learning into a joyful journey!
What is the Perimeter of a Polygon?
A polygon is a 2-dimensional geometric shape with straight, non-intersecting sides. The term originates from the Greek words 'poly,' meaning 'many,' and 'gonia,' meaning 'angle.' Hence, a polygon is a shape with many angles. The perimeter of a polygon, on the other hand, is the total distance around its edges or sides. It's the equivalent of taking a piece of string, laying it along each side of the polygon, and then measuring the length of that string. In other words, the perimeter gives you the 'boundary length' of a polygon. The perimeter of a polygon can be found by adding up the lengths of all its sides. The more sides a polygon has, the more additions you'll have to do to find the total perimeter. This simple yet essential concept is fundamental in many areas of mathematics and is particularly crucial in subjects like geometry and algebra.
Difference Between Area and Perimeter of a Polygon
Understanding the difference between the area and the perimeter of a polygon is crucial. While the perimeter of a polygon is the total distance around its edges, the area of a polygon is the total space inside its boundaries. In simpler terms, if you consider a polygon as a piece of land, the perimeter would be the length of the fence you'd need to enclose it, while the area would be the total land space within the fence. Both these quantities are significant, but they serve different purposes. The perimeter is often used when we need to know the boundary length, such as the amount of fencing needed for a garden. In contrast, the area is used when we need to cover a surface, like painting a wall or laying tiles on a floor.
Perimeter of a Regular Polygon
A regular polygon is a polygon in which all sides and angles are equal. For such polygons, calculating the perimeter becomes straightforward because you simply multiply the length of one side (since all sides are equal) by the number of sides. For example, if you have a regular hexagon (a six-sided polygon) with each side measuring 5 units, the perimeter would be 5 units * 6, which equals 30 units. This formula makes it incredibly simple to determine the perimeter of regular polygons and is a handy tool in many mathematical calculations.
Perimeter of an Irregular Polygon
An irregular polygon, unlike a regular polygon, doesn't have all sides and angles equal. This means that to find the perimeter of an irregular polygon, you must add up the lengths of all its individual sides. If you have a five-sided irregular polygon, for example, with sides measuring 3 units, 4 units, 5 units, 6 units, and 7 units, the perimeter would be 3+4+5+6+7, which equals 25 units. Calculating the perimeter of irregular polygons can be slightly more challenging than regular polygons, but with practice, it becomes easy.
Perimeter of Polygons Examples
Let's look at a couple of examples to understand the concept of perimeters better.
Example 1: Consider a regular octagon (an eight-sided polygon) with each side measuring 2 units. Using the formula for the perimeter of a regular polygon, the perimeter of this octagon would be 2 units * 8, which equals 16 units.
Example 2: Now, consider an irregular pentagon (a five-sided polygon) with sides measuring 1 unit, 2 units, 3 units, 4 units, and 5 units. Using the method for the perimeter of an irregular polygon, the perimeter of this pentagon would be 1+2+3+4+5, which equals 15 units.
Practice Questions on Perimeter of Polygons
Now that we've learned about the perimeter of polygons, it's time for some practice! Try to solve these problems:
Find the perimeter of a regular hexagon with each side measuring 6 units.
Remember, practice makes perfect, and the more you work on these problems, the better you'll understand the concept.
Conclusion
In the magical world of geometry, understanding the perimeter of polygons, both regular and irregular, is like acquiring a superpower. It's a skill that lights up your mathematical journey and allows you to decode the secrets hidden in the shapes that surround us. Whether it's a soccer field, a flower bed, a piece of land, or a piece of wire bent into a shape, the concept of perimeter is all around us!
At Brighterly, we take pride in empowering young minds with such skills. We aim to make learning not just an academic chore, but a fun-filled adventure. So remember, as you explore the world of polygons, that you're not merely learning a concept, but gaining a tool that is both practical and fascinating.
So keep practicing, keep exploring, and let your curiosity guide you. Learning the perimeter of polygons is just the beginning of a fantastic mathematical journey. And always remember, at Brighterly, we learn, we grow, and we shine brighter every day!
Frequently Asked Questions on Perimeter of Polygons
What is a Polygon?
A polygon is a 2-dimensional figure composed of straight, non-intersecting lines. Each line segment meets exactly two others at its endpoints, forming a closed path. In other words, if you start from one point and trace the sides, you will return to the starting point without lifting your pen.
What is the Perimeter of a Polygon?
The perimeter of a polygon is the total distance around its sides. It's like taking a journey along the boundary of the polygon. In mathematical terms, it's the sum of the lengths of all its sides.
How to Calculate the Perimeter of a Regular Polygon?
For regular polygons, where all sides are equal, calculating the perimeter is a breeze. You simply multiply the length of one side by the total number of sides. It's like knowing the length of one fence panel and then multiplying it by the number of panels to know the total fencing length.
How to Calculate the Perimeter of an Irregular Polygon?
Irregular polygons are a bit trickier as they don't have equal sides. But don't worry, the method is straightforward. You just need to add up the lengths of all the sides, and voila, you have the perimeterStruggling with Geometry?
Does your child need additional help with mastering geometry concepts understand geometry basics? An online tutor could be the solution.
Online Summer Math Camp for Kids and Teens
What do we offer?
Related math
350000 in Words
We write the number 350000 in words as "three hundred fifty thousand". It's three hundred fifty sets of one thousand each. If a concert has three hundred fifty thousand attendees, it means there are three hundred fifty thousand people in total. Thousands Hundreds Tens Ones 350 0 0 0 How to Write 350000 in Words? […]
Odd Numbers – Definition with Examples
A number is simply an arithmetic value used to represent quantity and make calculations. All numbers are formed from the numerals 0-9. Numbers exist as odd or even numbers and prime or composite numbers. Even numbers are all numbers that are divisible by 2 regardless of whether they are positive or negative integers. For example, […]
Square – Definition, Properties, Examples, Facts
Welcome to Brighterly, where we strive to make learning mathematics an enjoyable and engaging experience for kids! Our goal is to spark curiosity and inspire young minds to explore the wonders of mathematics. Today, we're delving into the captivating world of squares. These fascinating shapes are not only essential in geometry, but they also have | 677.169 | 1 |
Cos 34 Degrees
The value of cos 34 degrees is 0.8290375. . .. Cos 34 degrees in radians is written as cos (34° × π/180°), i.e., cos (17π/90) or cos (0.593411. . .). In this article, we will discuss the methods to find the value of cos 34 degrees with examples.
Cos 34°: 0.8290375. . .
Cos (-34 degrees): 0.8290375. . .
Cos 34° in radians: cos (17π/90) or cos (0.5934119 . . .)
What is the Value of Cos 34 Degrees?
The value of cos 34 degrees in decimal is 0.829037572. . .. Cos 34 degrees can also be expressed using the equivalent of the given angle (34 degrees) in radians (0.59341 . . .)
What is the Value of Cos 34° in Terms of Cosec 34°?
Since the cosine function can be represented using the cosecant function, we can write cos 34° as [√(cosec²(34°) - 1)/cosec 34°]. The value of cosec 34° is equal to 1.78829.
What is the Value of Cos 34 Degrees in Terms of Cot 34°?
We can represent the cosine function in terms of the cotangent function using trig identities, cos 34° can be written as cot 34°/√(1 + cot²(34°)). Here, the value of cot 34° is equal to 1.48256.
How to Find the Value of Cos 34 Degrees?
The value of cos 34 degrees can be calculated by constructing an angle of 34° with the x-axis, and then finding the coordinates of the corresponding point (0.829, 0.5592) on the unit circle. The value of cos 34° is equal to the x-coordinate (0.829). ∴ cos 34° = 0.829. | 677.169 | 1 |
How do you draw a polygon in JavaScript?
And calculate angle made by the each side of regular polygon at center (360/n)
Identify the first vertex (r,0)
Loop through the angles to identify the points.
And draw the polygon using store or fill method.
How do you make a polygon in HTML?
To draw a polygon in HTML SVG, use the SVG element. The element creates a graphic containing at least three sides. The points attribute is the x and y coordinates for each corner of the polygon.
How do you draw a polygon in maps?
Draw a path or polygon
Open Google Earth.
Go to a place on the map.
Above the map, click Add Path . To add a shape, click Add Polygon.
A "New Path" or "New Polygon" dialog will pop up.
To draw the line or shape you want, click a start point on the map and drag.
Click an endpoint.
Click OK.
Which function is used to draw a polygon on the canvas?
The following functions are used to create the triangle: evas_object_polygon_add() creates a polygon object on a canvas. evas_object_polygon_point_add() adds point coordinates to a polygon object. A polygon must have at least 3 points.
Which tool is used to draw a rectangle?
1Rectangle tool- Rectangle tool is used to draw rectangular or square shapes. Q. 2 Ellipse tool – It is used to draw oval or circle.
How does Clippath polygon work?
The clip path is a series of coordinate pairs, each pair separated by commas. Starting at the upper right corner, the coordinates would be X0 Y0 or 0 0. The distance away from the start point increased up to 100%. The bottom right corner's coordinate is X100% Y100% or 100% 100%.
What is polyline in HTML?
The SVG element is an SVG basic shape that creates straight lines connecting several points. Typically a polyline is used to create open shapes as the last point doesn't have to be connected to the first point. For closed shapes see the element.
Can I draw a circle on Google Earth?
use the measuring tool to draw circles! Google Earth Pro's drawing tools do not include an easy way to draw accurate circles. But, the ruler/measuring tool does let you measure circular areas and save them as KML, so you can do the following to draw a circle: In Ruler window, select the "Circle" tab.
What is polygon in drawing?
A polygon is any plane figure bounded by straight lines (Figure 4.33). If the polygon has equal angles and equal sides, it can be inscribed in or circumscribed around a circle and is called a regular polygon. 4.33 Regular Polygons. The AutoCAD Polygon command is used to draw regular polygons with any number of sides.
What is meant by convex polygon?
: a polygon each of whose angles is less than a straight angle.
How to draw a polygon on canvas in JavaScript?
In this short article we want to show how to draw stroke line according indicated points using HTML5 Canvas and Javascript. In this we want to show how to draw polygon border. Note: read this article to see how to fill polygon with certain color.
How is an ArcGIS polygon represented in JavaScript?
Since: ArcGIS API for JavaScript 4.0 A polygon contains an array of rings and a spatialReference. Each ring is represented as an array of points. The first and last points of a ring must be the same.
How to create a polygon in Google Maps?
// Define the LatLng coordinates for the polygon's path. // Construct the polygon. Note: Read the guide on using TypeScript and Google Maps. // This example creates a simple polygon representing the Bermuda Triangle. const map = new google. maps.
What are the properties of a polygon in JavaScript?
PolygonOptions object used to define the properties that can be set on a Polygon. Indicates whether this Polygon handles mouse events. Defaults to true. If set to true, the user can drag this shape over the map. The geodesic property defines the mode of dragging | 677.169 | 1 |
determine if two triangles are congruent worksheet
Determine If Two Triangles Are Congruent Worksheet – Triangles are among the most fundamental designs in geometry. Understanding triangles is important for developing more advanced geometric ideas. In this blog post it will explain the various kinds of triangles triangular angles, the best way to calculate the area and perimeter of a triangle, and offer some examples to illustrate each. Types of Triangles There are three kinds to triangles: the equilateral, isosceles, and scalene. Equilateral triangles comprise … Read more
Determine If Triangles Are Congruent Worksheet – Triangles are among the most fundamental patterns in geometry. Understanding triangles is vital to mastering more advanced geometric concepts. In this blog this post, we'll go over the different types of triangles such as triangle angles, and how to calculate the perimeter and area of a triangle and will provide the examples for each. Types of Triangles There are three kinds in triangles, namely equilateral isosceles, as well as … Read more | 677.169 | 1 |
Introduction to 3-D Geometry
Coordinate Axes: In three dimensions, the coordinate axes of a rectangular cartesian coordinate system are three mutually perpendicular lines. These axes are called the X, Y and Z axes.
Coordinate Planes: The three planes determined by the pair of axes are the coordinate planes. These planes are called XY, YZ and ZX plane and they divide the space into eight regions known as octants.
Coordinates of a Point in Space :
The coordinates of a point in the space are the perpendicular distances from P on three mutually perpendicular coordinate planes YZ, ZX, and XY respectively. The coordinates of a point P are written in the form of triplet like (x, y, z).
The coordinates of any point on
X-axis is of the form (x, 0,0)
Y-axis is of the form (0, y, 0)
Z-axis is of the form (0, 0, z)
XY-plane are of the form (x, y, 0)
YZ-plane is of the form (0, y, z)
ZX-plane are of the form (x, 0, z)
Note :-
Points are defined as the triples of real numbers arranged in an order such as P1 = (x1, y1, z1) and P2 = (x2, y2, z2)
A Cartesian coordinate system is a coordinate system that specifies each point uniquely in a plane by a pair of numerical coordinates, which are the signed distances to the point from two fixed perpendicular directed lines, measured in the same unit of length.
The coordinates of a point are a pair of numbers that define its exact location on a two-
dimensional plane A number on the x-axis called an x-coordinate, and a number on the y-axis called a y-coordinate.
An ordered pair contains the coordinates of one point in the coordinate system.
The order in which you write x- and y-coordinates in an ordered pair is very important.
The x-coordinate always comes first, followed by the y-coordinate There is also a three coordinate called Z coordinate | 677.169 | 1 |
How do I add a domain to geogebra?
Drag the "domain" slider. A vertical ray will scan the viewing window and will project the intersection point of the ray and the function to the x-axis.
Drag the "range" slider.
How do you plot points on geogebra?
That again enclosed in parentheses as the ordered pair for that point. So we have 0 comma 3.5 hit enter and the point is plotted.
How do you draw a 3d shape in geogebra?
Now there's two basic ways to create 3d shapes in geogebra a one is to draw your polygon in two be view and then turn it into a 3d shape. The other option is to actually draw exclusively in the 3d.
How do you draw a 3 point ellipse?
Corel Draw Tips & Tricks 3 point Ellipse Tool – YouTube
How do you draw points on an ellipse?
How to Draw an Ellipse – Major Axis & a Point on the Curve – YouTube
How do you find domain and range?
How to Find The Domain and Range of an Equation? ToWhat is domain and range in functions
How do you find coordinates on GeoGebra?
GeoGebra Displaying Coordinates and Saving as a PDF File – YouTube
How do you draw a 3d vector in GeoGebra?
Plotting Points and Vectors in 3d with Geogebra – YouTube
How do you make a 3d cube in GeoGebra?
Use the input box to type in two points for the vertices of the cube (make sure one of them is (0,0,0). Click on the pyramid button to open the drop-down menu. Click on the cube button, then select your two points. You are done!
How do you draw a simple 3d shape?
How to Draw 3D Shapes – YouTube
How do you draw an ellipse step by step?
How to Draw a Perfect Ellipse – YouTube
What are the methods of drawing ellipse?
Methods of constructing Ellipse include: i Concentric circles method ii The focal point method iii The rectangular method. (i) Draw AB and CD, the given axes. (ii) With C as centre, radius half the major axis, draw an arc cutting AB at the foci F1 and F2 into a number of equal parts, numbering as shown .
How do you draw an ellipse with a circle?
How to Draw a Circle & an Ellipse | Drawing Tutorials – YouTube
How do you solve for domain?
Let y = f(x) be a function with an independent variable x and a dependent variable y. If a function f provides a way to successfully produce a single value y using for that purpose a value for x then that chosen x-value is said to belong to the domain of f.
What is domain of a graph?
The domain is all x-values or inputs of a function and the range is all y-values or outputs of a function. When looking at a graph, the domain is all the values of the graph from left to right. The range is all the values of the graph from down to up.
How do I find domain and range?
ToHow do I find the domain of a function?
How do you write a function in geogebra?
How to plot the graph of a function in Geogebra – YouTube
How do I change geogebra to polar coordinates?
Graphing Polar Equations on GeoGebra – YouTube
How do you plot a 3D vector?
1.1 Vectors with 3 components (3 dimensions) – YouTube
How do I add a vector to geogebra?
Examples: To enter a point P or a vector v in 2D in Cartesian coordinates you may use P = (1, 0) or v = (0, 5) . To enter a point P or a vector v in 3D in Cartesian coordinates you may use P = (1, 0, 2) or v = (0, 5, -1) . To enter a point P in 2D in polar coordinates, you may use P = (1; 0°) or v = (5; 90°) .
How do you draw a vector in GeoGebra 3d?
GEOGEBRA: All about VECTORS! – YouTube
How do you make a shape on GeoGebra?
How to create polygons in GeoGebra
Locate the Polygon button.
Select the Polygon button.
On graph select appropriate coordinates (You can do this by either inputting the coordinates on the left column or by simply clicking coordinates on the grid) | 677.169 | 1 |
Answer to Question #251512 in Civil and Environmental Engineering for Sysy
A vertical parabolic curve is designed to pass through three points on the profile of an existing unimproved road with the stationing and corresponding elevations as follows:
POINTS | STATIONING | ELEVATION
A | 5+432 | 45.00 m
B | 5+592 | 48.00 m
C | 5+682 | 43.30 m
a) Determine the stationing of the summit.
b) Determine the elevation of the summit.
c) Determine the grade of the back tangent.
1
Expert's answer
2021-10-18T03:45:24-0400
"Symmetrical Parabolic Curve"
In highway practice, abrupt change in the vertical direction of moving vehicles should be avoided. In order to provide gradual change in its vertical direction, a parabolic vertical curve is adopted on account of its slope which varies at constant rate with respect to horizontal distances.
(b)
"PropertiesofVertical ParabolicCurve"
For a symmetrical parabolic curve, the number of stations to the left must be equal to the number of stations to the right, of the intersection of the slopes or forward and back tangent.
The slope of the parabola varies uniformly along the curve, Therefore the rate of change of slope is constant and equal to:
r = (g2 – g1) / L
(c)
The maximum offset H = 1/8 the product of the algebraic difference between the two rates of grade and the length of curve: | 677.169 | 1 |
The Construction Of Polygons
Fig. 71 a.
Fig. 72.
The term polygon is applied to figures having flat sides equidistant from a common centre. From this centre a circle may be struck that will touch all the corners of the sides of the polygon, or the point of each side that is central in the length of the side. In drawing a polygo
, one of these circles is used upon which to divide the figure into the requisite number of divisions for the sides. When the dimension of the polygon across its corners is given, the circle drawn to that dimension circumscribes the polygon, because the circle is without or outside of the polygon and touches it at its corners only. When the dimension across the flats of the polygon is given, or when the dimension given is that of a circle that can be inscribed or marked within the polygon, touching its sides but not passing through them, then the polygon circumscribes or envelops the circle, and the circle is inscribed or marked within the polygon. Thus, in Figure 71 a, the circle is inscribed within the polygon, while in Figure 72 the polygon is circumscribed by the circle; the first is therefore a circumscribed and the second an inscribed polygon. A regular polygon is one the sides of which are all of an equal length. | 677.169 | 1 |
Worksheets For Class 9 Mathematics Coordinate Geometry
Coordinate Geometry Class 9 Worksheet have been designed as per the latest pattern for CBSE, NCERT and KVS for Grade 9. Students are always suggested to solve printable worksheets for Mathematics Coordinate Coordinate Geometry Worksheets for Class 9
We have provided chapter-wise worksheets for class 9 Mathematics Coordinate Geometry which the students can download in Pdf format for free. This is the best collection of Mathematics Coordinate Geometry standard 9th worksheets with important questions and answers for each grade 9th Mathematics Coordinate Geometry chapter so that the students are able to properly practice and gain more marks in Class 9 Mathematics Coordinate Geometry class tests and exams.
1. Cartesian System 2. Plotting a Point in the Plane with given Coordinates
• Coordinate Geometry: The branch of mathematics in which geometric problems are solved through algebra by using the coordinate system is known as coordinate geometry. • Coordinate System: Coordinate axes: The position of a point in a plane is determined with reference to two fixed mutually perpendicular lines, called the coordinate axes.
In this system, position of a point is described by ordered pair of two numbers. • Ordered pair: A pair of numbers a and b listed in a specific order with 'a' at the first place and 'b' at the second place is called an ordered pair (a,b) Note that (a, b) ¹ (b, a) Thus (2,3) is one ordered pair and (3,2) is another ordered pair. In given figure O is called origin. The horizontal line X1 OX is called the X-axis. The vertical line YOY' is called the Y-axis. P(a,b) be any point in the plane. 'a' the first number denotes the distance of point from Y-axis and 'b' the second number denotes the distance of point from X-axis. a – X – coordinate | abscissa of P. b – Y – coordinate | ordinate of P. The coordinates of origin are (0,0) Every point on the x-axis is at a distance o unit from the X-axis. So its ordinate is 0. Every point on the y-axis is at a distance of unit from the Y-axis. So, its abscissa is 0.
Note: Any point lying on X- axis or Y – axis Y-axis does not lie in any quadrant
Question. If the perpendicular distance of a point P from the x-axis is 5 units and the foot of the perpendicular lies on the negative direction of x-axis, then the point P has (a) x coordinate = – 5 (b) y coordinate = 5 only (c) y coordinate = – 5 only (d) y coordinate = 5 or –5
Answer
D
Question. A point both of whose coordinates are negative will lie in (a) I quadrant (b) II quadrant (c) III quadrant (d) IV quadrant
Question. In Fig. 3.2, the point identified by the coordinates (–5, 3) is
(a) T (b) R (c) L (d) S
Answer
A
Question.The point which lies on y-axis at a distance of 5 units in the negative direction of yaxis is (a) (0, 5) (b) (5, 0) (c) (0, – 5) (d) (– 5, 0)
Answer
C
Question.In Fig. 3.1, coordinates of P are
(a) (– 4, 2) (b) (–2, 4) (c) (4, – 2) (d) (2, – 4)
Answer
B
Question.Write whether the following statements are True or False? Justify your answer. (i) Point (3, 0) lies in the first quadrant. (ii) Points (1, –1) and (–1, 1) lie in the same quadrant. (iii) The coordinates of a point whose ordinate is -1/2 and abscissa is 1 are (-1/2,1) (iv) A point lies on y-axis at a distance of 2 units from the x-axis. Its coordinate are (2, 0). (v) (–1, 7) is a point in the II quadrant. Answer. (i) The point (3, 0) has abscissa 3, the x-coordinate and ordinate is 0. If the ordinate of a point is zero, the point lies on the x-axis. Hence, the given statement is false. (ii) The point (1, –1) lies in IV quadrant and the point (–1, 1) lies in II quadrant. Hence, the given statement is false. (iii) We know that in the coordinates of a point, the abscissa comes first and then the ordinate. So, the coordinate of a point are (1,-1/2) and not (-1/2,1) (iv) Any point which lies on the y-axis is of the form (0, y). Hence, the given statement is false. (v) In the II quadrant, signs of abscissa and ordinate are –, + respectively. Hence, the statement that (–1, 7) is a point in the II quadrant is a true statement.
Question.Without plotting the points indicate the quadrant in which they will lie, if (i) ordinate is 5 and abscissa is – 3 (ii) abscissa is – 5 and ordinate is – 3 (iii) abscissa is – 5 and ordinate is 3 (iv) ordinate is 5 and abscissa is 3 Answer. (i) In the point (–3, 5) abscissa is negative and ordinate is positive, so it lies in the second quadrant. (ii) In the point (–5, –3) abscissa and ordinate both are negative, so it lies in the third quadrant. (iii) In the point (–5, 3) abscissa is negative and ordinate is positive, so it lies in the second quadrant. (iv) In the point (3, 5) abscissa and ordinate both are positive, so it lies in the first quadrant.
Question. Plot the following points and write the name of the figure obtained by joining them in order: P (– 3, 2), Q (– 7, – 3), R (6, – 3), S (2, 2) Answer. Let X'OX and Y'OY be the coordinate axes. Then, the four points may be plotted as given below:
Question. Plot the points (x, y) given by the following table:
Answer. Let X'OX and Y'OY be the coordinate axes. Then, the given points may be plotted as given below:
Question.In Fig. 3.6, LM is a line parallel to the y-axis at a distance of 3 units. (i) What are the coordinates of the points P, R and Q? (ii) What is the difference between the abscissa of the points L and M?
Answer. (i) Clearly, the distance of P from y-axis is 3 units and that of from x-axis is 2 units. Since P lies in the first quadrant, so its coordinate are (3, 2). Point R lies on x-axis and its distance from y and x-axes are 3 and 0 units respectively. So, its coordinates are (3, 0). Clearly, point Q lies in the fourth quadrant. The distance of Q from y-axis is 3 units and from x-axis is 1 unit. So, the coordinates of Q (3, –1). (ii) From the given figure (graph), we find that the points L and M lie on the same straight line. So, L and M are collinear. Hence, the difference between the abscissa of the points L and M is 0.
Question.Plot the points P (1, 0), Q (4, 0) and S (1, 3). Find the coordinates of the point R such that PQRS is a square. Answer. Plot the points P (1, 0), Q (4, 0) and S (1,3) in the Cartesian plane. As we know all the sides of a square are equal and each angle is of 90o measure. Therefore, the abscissa of the vertex R is 4 and its ordinate is 3.
Hence, the coordinate of the point R are (4, 3).
Question.Which of the following points lie on y-axis? A (1, 1), B (1, 0), C (0, 1), D (0, 0), E (0, – 1), F (– 1, 0), G (0, 5), H (– 7, 0), I (3, 3). Answer. We know that if a point lies on the y-axis, its abscissa is 0 and its ordinate is the y-value and its coordinate are (0, y). Hence, C (0, 1), E (0, –1), G (0, 5) are the points which lie on y-axis.
Question. Plot the points (x, y) given by the following table. Use scale 1 cm = 0.25 units
Answer. Let X'OX and Y'OY be the coordinate axes. Then, the given points may be plotted as given below:
Question.A point lies on the x-axis at a distance of 7 units from the y-axis. What are its coordinates? What will be the coordinates if it lies on y-axis at a distance of –7 units from x-axis? Answer. The given points lies on the x-axis. The distance of this points from y-axis is 7 units and from x-axis is 0. So, its coordinates are (7, 0). If it lies on y-axis, then the distance of it from y-axis is 0 and that of from x-axis is 7 units. So, its coordinates are (0, 7).
Question. Points A (5, 3), B (– 2, 3) and D (5, – 4) are three vertices of a square ABCD. Plot these points on a graph paper and hence find the coordinates of the vertex C. Answer. Plot the points A (5, 3), B (– 2, 3) and D (5, – 4). Join AB and AD. As ABCD is a square, so all its sides are equal and each angle is of 90o measure. Therefore, the abscissa of the vertex C will be – 2 and ordinate – 4. Hence, the coordinates of the vertex C are (–2, –4).
Question. Write the coordinates of the vertices of a rectangle whose length and breadth are 5 and 3 units respectively, one vertex at the origin, the longer side lies on the x-axis and one of the vertices lies in the III quadrant. Answer. As the length and breadth of the rectangle are 5 and 3 units respectively, one vertex at the origin, the longer side lies on the x-axis and one of the vertices lies in the III quadrant, so the coordinate of the vertices of rectangle OABC are O(0, 0), A (-5, 0), B (-5, -3) and C (0, -3).
Question.Find the coordinates of the point (i) which lies on x and y axes both. (ii) whose ordinate is – 4 and which lies on y-axis. (iii) whose abscissa is 5 and which lies on x-axis. Answer. (i) The coordinates of the point which lies on both the axes are (0, 0). (ii) The coordinates of the point whose ordinate is – 4 and which lies on y-axis are (0, –4). (iii) The coordinate of the point whose abscissa is 5 and which lies on x-axis are (5, 0).
Question.From the given figure, answer the following;
(i) Write the points whose abscissa is 0. (ii) Write the points whose ordinate is 0. (iii) Write the points whose abscissa is – 5. Answer. (i) Clearly, the distance of points A, L and O from y-axis is 0. So, A (0, 3), L (0, -4) and O (0, 0) are the points whose abscissa is 0. (ii) Clearly, the distance of points G, I and O from x-axis is 0. So, G (5, 0), I (-2, 0) and O (0,0) are the points whose ordinate is 0. (iii) Clearly, the distance of points H and D from y-axis is 5 units and both lien in second and third quadrants respectively. So, (-5, -3) and D (-5, 1) are the points whose abscissa is -5.
Question.Plot the points A (1, – 1) and B (4, 5) (i) Draw a line segment joining these points. Write the coordinates of a point on this line segment between the points A and B. (ii) Extend this line segment and write the coordinates of a point on this line which lies outside the line segment AB. Answer. (i) M (2, 1) is a point on this line segment between the points A and B.
(ii) N (5, 7) is a point on this line which lies outside the line segment AB.
Parents and students are welcome to download as many worksheets as they want as we have provided all free. As you can see we have covered all topics which are there in your Class 9 9 is a very scoring subject, if you download and do these questions and answers on daily basis, this will help you to become master in this subject.
These CBSE Class 9 Mathematics Coordinate Geometry worksheet can help you to understand the pattern of questions expected in Mathematics Coordinate Geometry exams.
All worksheets for Mathematics Coordinate Coordinate Geometry textbook
CBSE Class 9 Mathematics Coordinate Geometry Workbook will surely help to improve knowledge of this subject
These Printable practice worksheets are available for free download for Class 9 Coordinate Geometry Worksheet PDF.
Where can I get Worksheets for Class 9 Mathematics Coordinate Geometry ?
You can download free worksheets for Class 9 Mathematics Coordinate Geometry from
I want free printable worksheets with questions and answers for Mathematics Coordinate Geometry for Standard 9, where can I get them?
You can get free PDF downloadable worksheets for Grade 9 Mathematics Coordinate | 677.169 | 1 |
Pythagorean means
In microtonal terms, the arithmetic mean is the intermediate harmonic, the inverse-arithmetic mean is the intermediate subharmonic, and the geometric mean is the equidistant pitch by cents. Thus the three means of 1/1 and 3/2 are 5/4, 6/5, and sqrt (3/2), respectively.
In math, the three means are known as arithmetic mean (AM), harmonic mean (HM), and geometric mean (GM) respectively. However, harmonic mean is said with respect to a length of string to be divided, and collides with the more common usage of harmonic, which is said with respect to frequency. For fear of confusion, those terms are avoided and inverse-arithmetic mean is preferred instead. | 677.169 | 1 |
Angle of the shaft and Mohr's Circle question
In summary, the angle of the shaft in relation to Mohr's Circle is the orientation of a shaft or rod in relation to the principal stress directions. It can be calculated by finding the tangent of the angle between the center of Mohr's Circle and the point representing the state of stress. The angle is significant in determining maximum shear and normal stress, understanding stress distribution, and potential failure modes. It can be negative, indicating a counterclockwise orientation, and can affect the safety factor by changing the magnitude and direction of maximum shear stress.
Related to Angle of the shaft and Mohr's Circle question
1. What is the concept of angle of the shaft in relation to Mohr's Circle?
The angle of the shaft refers to the angle at which a shaft or rod is oriented in relation to the principal stress directions in Mohr's Circle. It is used to determine the shear stress and normal stress acting on the shaft at a specific point.
2. How is the angle of the shaft calculated in Mohr's Circle?
The angle of the shaft can be calculated by finding the tangent of the angle between the line connecting the center of Mohr's Circle to the point representing the state of stress and the horizontal axis.
3. What is the significance of the angle of the shaft in Mohr's Circle?
The angle of the shaft is important in determining the maximum shear stress and maximum normal stress acting on a shaft or rod. It also helps in understanding the stress distribution and potential failure modes.
4. Can the angle of the shaft be negative in Mohr's Circle?
Yes, the angle of the shaft can be negative in Mohr's Circle. This indicates that the principal stress directions are oriented in a counterclockwise direction from the horizontal axis.
5. How does the angle of the shaft affect the safety factor in Mohr's Circle?
The angle of the shaft can affect the safety factor by changing the magnitude and direction of the maximum shear stress acting on the shaft. A decrease in the angle of the shaft can increase the safety factor, while an increase can decrease the safety factor. | 677.169 | 1 |
Hyperbolic Heptagonal Hydrangea
3D Model
I'm not sure if this flower is really hyperbolic, but all of the faces are really affine heptagons.
Being affine heptagons means that each face has a unique viewing angle (projection) where it looks like a regular equalateral heptagon (7 sides). The three central affine heptagons form mutually orthogonal planes which join at the center of the 3-fold symmetry. These three faces don't necessarily have to be orthogonal. They could be more acute, flatter, or even coplanar or asymmetrical. The tiling can obviously be continued ad nauseum. I stopped at 48 faces. Faces of the same color are congruent.
If anyone can tell me precisely what curve is formed by the faces, please leave a detailed comment.
I'd also like to identify the pattern formed by the set of lines along which each of the affine heptagons is projected as a regular heptagon. Where do they converge if at all? Are they perhaps related to a parabolic reflector? Any insight is welcome. | 677.169 | 1 |
When is on a side of the triangle,
the line between the two perpendiculars is called the pedal
line. Given four points, no three of which are collinear,
then the four pedal circles of each point for the triangle
formed by the other three have a common point through which the nine-point
circles of the four triangles pass. | 677.169 | 1 |
a mathematical theorem that states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side
Example
The triangle inequality is an important concept in geometry.
Origins of triangle
from Latin 'triangulum', meaning 'triangle'
📌
Summary: triangle in Brief
A 'triangle' [ˈtraɪæŋɡl] is a three-sided plane figure with three angles. It can also refer to a musical instrument made of steel rod bent into a triangle shape. The term extends into phrases like 'love triangle,' denoting a romantic situation involving three people, and 'triangle offense,' referring to a basketball strategy. The 'triangle inequality' is a mathematical theorem that states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. | 677.169 | 1 |
All throughout the globe, several men and women in various sites like soccer. About the many years, this remarkable game of teamwork and skillfulness has absolutely captured the attention and the hearts of lots of. So where does geometry enter the picture? Geometry, a mathematical subject matter area focusing on the analyze of designs, sizes, measurements, and comparative standpoints of different figures, is truly considered a realistic discipline that plays a key position in soccer. Soon after all, individuals who are acquainted with this unique sport know that it significantly requires styles, spots, and measurements much far more than the formations that the gamers apply.
The Soccer Area
Just one standard notion in geometry is symmetry or stability. If you study a variety of soccer fields, you will discover that they are all marked evenly. What you obtain on just one aspect is normally replicated on the other. This is a literal representation of what is known as a stage participating in industry.
Fundamentally, the shape of a soccer area is rectangular. There are distinct lengths that are expressed either in yards or meters in measuring the rectangle. Unquestionably, you will see that there are a good deal of other shapes to locate in this subject— the penalty place and penalty arc, the center circle, the intention place, and the corner arc amid others.
There are governing regulations to follow when building the markings in a soccer industry. Thus, the groundsman ought to make sure that these are complied with to meet up with presented specifications. I myself have knowledgeable officiating in genuine game titles. In my observation, there are arcs that appear to be to look like everyday melons. Objective posts might also come as spherical or square-shaped so extended as the width is the exact same as that of the target line. It is also significant to be aware that regular measurements are used for the penalty location and the center circle. For instance, the penalty space has to possess a duration of 44 yards and a width of 18 yards. Certainly symmetry can be very important if you want to determine the locale of the purpose posts and goal location.
The best symmetry is certainly vital when setting up and marking the soccer subject. The previously mentioned standardized measurements should really also be fulfilled in buy for the discipline to be accredited by law.
The Soccer Ball
Geometry is also vital in pinpointing the variety of ball that really should be utilized in a soccer match. Obviously, just one of the standard necessities is that the form of the ball have to be spherical. If not, the ball will not be rolled or utilized properly in the sport. Also, a circumference of about 27 to 28 inches is obligatory. There are laws also pertaining to the geometrical specifications of the soccer ball. Irrespective of the advancements and innovations being carried out to greatly enhance the capability of the ball to increase goal scores, the standardized specifications continue to be the exact same.
The Soccer Video game
Currently being experienced about angles and measurements will substantially advantage soccer players for the duration of a video game. The victorious completion of passes is mostly enhanced when the gamers know how to establish angles. For instance, as an opponent is about to move the ball to a teammate, you can do a fast evaluation of the relative positioning of the players on the industry so as to block the pass or to hamper their paths. And in the situation that a teammate of yours forwards the ball "far too square", this can make it more durable to get to the attacking 3rd.
If you are in the position of an attacker, it would support to make use of wider angles. In its place of making use of a immediate route, go the ball through wide angles and you will diminish the threat of an opponent seizing the ball from your facet. In truth this case in point is a exceptional software of expertise on angles. Of study course soccer gamers are not essential to be great in geometry to be equipped to do this. Nevertheless, studying geometry in relation to soccer will lead a great deal to having away more trial and mistake tries.
In taking the role of a goalkeeper, it is also important to know about angles. The positioning of a goalkeeper is essential to narrow the angle as a result of which the opponent can achieve the goal and rating. It does not function the similar way for the keeper, although. Shot stopper commonly slip off from the aim line when they will need to experience penalties. This is also accomplished when there are attackers coming. Applying geometry right here, you recognize that you have to make the distance shorter for the similar angle so that there will also be a thinner gap for the two sides of a triangle. In the selection on staff formations, triangulation has a big outcome on soccer.
Indeed geometry is a significant player in soccer exactly where essential methods are used and exactly where workforce formations are worried, together with the specialized aspects of the area and ball applied. Also, symmetry must be highly viewed as. In working with geometry, you will know how to mark an arc thoroughly and you can steer clear of making the center circle oval-shaped in its place of properly round. If this takes place, the ball will not roll ideal and goalkeepers will have a tendency to keep on the purpose line when the attackers are coming. Geometry is absolutely vital in soccer. | 677.169 | 1 |
4-6 Isosceles And Equilateral Triangles Worksheet Answer Key Glenco Geometry – Triangles are one of the most fundamental shapes in geometry. Understanding triangles is crucial for understanding more advanced geometric principles. In this blog post We will review the various types of triangles and triangle angles, as well as how to determine the areas and perimeters of a triangle, and offer illustrations of all. Types of Triangles There are three kinds of triangulars: Equilateral, isoscelesand scalene. | 677.169 | 1 |
find missing angles worksheet pdf
Find Missing Angles | 677.169 | 1 |
In Class 7 Maths Chapter 13, Visualising Solid Shapes, we will learn about Visualising some Solid Shapes to determine the number of their faces, edges, and vertices. In the later part of the chapter, we will also learn to draw some solid shapes on a flat surface. Let us now have a look at the NCERT Class 7 Maths Chapter 13 Visualising Solid Shapes Notes and Solutions (PDF available).
NCERT Solutions Class 7 Maths Chapter 13 – PDF Available
Check the topic-wise notes for NCERT Solutions Maths Class 7 Chapter 13, Visualising Solid Shapes below. You can also download the PDF of the notes and take a printout to study later when you need quick revision before going to the e𝑥am hall.
Topic 2: Oblique Sketches
The oblique sketches give a clear idea of how a particular solid shape looks when seen from the front. You do not see certain faces. In this sketch, the lengths are not equal, as they should be in a given solid shape. Still, you are able to recognize it. Such a sketch of a solid is called an oblique sketch.
An oblique sketch does not have proportional lengths. Still, it conveys all important aspects of the appearance of the solid.
E𝑥ercise 13.4 Solutions
False, each face of a cube is of the shape of a square which means a cube will cast a shadow of square shape when put under a lamp.
False, each face of a cube is of the shape of a square which means a cube will cast a shadow of square shape when put under a lamp. Even if the lamp is put just above one of its vertices, the shape of the shadow will not be a hexagon.
Source: Mathematics Class VII
Check more informative blogs of NCERT Solutions Class 7 Maths by clicking on the links below.
This was all about NCERT Solutions Class 7 Maths Chapter 13, Visualising Solid Shapes in which we learned about the faces, edges and vertices of solid shapes. We also learned to draw the oblique and isometric sketches of certain solid shapes on a flat surface. Download the NCERT Solutions Class 7 Maths Chapter 13 Notes and Solutions PDF to ace your e𝑥am preparations. Follow the CBSE Class 7 Maths Solutions and Notes for more such chapter notes and important questions and answers for preparation for CBSE Class 7 Maths. | 677.169 | 1 |
Table of Contents
Today, we delve into the fascinating world of Reflectional Symmetries Count Statistics, where we explore the beauty and complexity of symmetrical patterns and their significance in statistical analysis. Join us as we uncover the power and applications of reflectional symmetries in counting statistics.
The Latest Reflectional Symmetries Count Statistics Explained
There are only 2 reflectional symmetries in an isosceles triangle.
In the context of an isosceles triangle, which has two equal sides and two equal angles, there are only two reflectional symmetries possible. The first symmetry is through the line that passes through the midpoint of the base and is perpendicular to the base, resulting in the reflection of the triangle across this line to produce an identical image. The second symmetry is a reflection across the line that passes through the vertex angle of the isosceles triangle and bisects the base, creating another mirror image. These two reflectional symmetries are the only ways in which an isosceles triangle can be reflected onto itself while maintaining its original shape and structure.
A square has 4 lines of reflectional symmetry.
The statistic "A square has 4 lines of reflectional symmetry" means that a square can be reflected across four different lines to produce an identical image. These lines of symmetry are vertical, horizontal, and two diagonal lines that run from one corner of the square to the opposite corner. When a square is reflected across any of these lines, the resulting image will match the original square exactly. This property of a square makes it a highly symmetrical geometric shape, with multiple ways in which it can be mirrored to maintain its overall shape and structure.
Regular hexagons have 6 reflectional symmetries.
The statement that regular hexagons have 6 reflectional symmetries means that a regular hexagon can be reflected across 6 different lines of symmetry and still look identical to its original shape. In other words, if we were to draw lines through the hexagon such that one half is a mirror image of the other half, there would be six unique ways to do so for a regular hexagon. This property arises from the geometric structure of a regular hexagon, which has six equal sides and interior angles, leading to multiple lines of symmetry that divide the shape into halves that are mirror images of each other.
Circles have infinite lines of reflectional symmetry.
The statistic that "circles have infinite lines of reflectional symmetry" means that a circle can be divided into an infinite number of equal halves through lines of symmetry. A line of reflectional symmetry is a line that divides a shape into two identical halves, where one half is the reflection of the other. In the case of a circle, any line passing through the center of the circle will result in two identical halves due to the circular symmetry. Since a circle has an infinite number of points along its circumference, there are an infinite number of lines passing through its center that can serve as lines of reflectional symmetry, making the number of possible symmetric halves infinite.
Rectangles have 2 lines of reflectional symmetry.
The statistic "Rectangles have 2 lines of reflectional symmetry" means that a rectangle can be divided into two equal halves by two different lines of symmetry. This implies that when a rectangle is folded over one of these lines, the two resulting parts will perfectly overlap each other. The first line of symmetry runs horizontally through the middle of the rectangle, dividing it into top and bottom halves that are mirror images of each other. The second line of symmetry runs vertically through the middle of the rectangle, dividing it into left and right halves that are also mirror images. These two lines of symmetry provide two different ways in which the rectangle can be folded or reflected to create a symmetrical figure.
An equilateral triangle has 3 lines of reflectional symmetry.
The statistic states that an equilateral triangle possesses three lines of reflectional symmetry. An equilateral triangle is a geometric shape where all three sides are of equal length, and all three interior angles are equal to 60 degrees. Due to this uniformity in its sides and angles, the triangle exhibits symmetry along three distinct axes – one line passing through each vertex of the triangle. When the triangle is reflected across any of these lines, the resulting image will perfectly overlap with the original triangle, highlighting its symmetrical properties. Hence, the statistic emphasizes that an equilateral triangle displays reflectional symmetry with respect to three different axes, reflecting its balanced and harmonious geometric structure.
A rhombus has 2 lines of reflectional symmetry.
The statistic that a rhombus has 2 lines of reflectional symmetry means that there are two lines within the rhombus such that if the rhombus is folded over these lines, the two halves will perfectly mirror each other. In other words, a rhombus can be divided into two equal halves through these lines of symmetry. The first line of symmetry passes through the midpoint of two opposite sides of the rhombus, while the second line passes through the midpoints of the other two opposite sides. This property of reflectional symmetry helps to describe the geometric shape and characteristics of a rhombus, making it a symmetrical and balanced figure in terms of its proportions and arrangements of sides and angles.
Regular octagons have 8 reflectional symmetries.
The statistic "Regular octagons have 8 reflectional symmetries" refers to a geometric property of regular octagons – polygon shapes with eight equal sides and internal angles. Reflectional symmetry is a property of an object where there exists a line (known as a line of symmetry) such that the object can be folded along that line onto itself, creating a mirror image. In the case of regular octagons, there are precisely 8 different lines of symmetry that divide the octagon into equal halves, reflecting one half onto the other half. These symmetries allow octagons to appear identical when rotated or flipped, making them visually balanced and pleasing to the eye in various contexts, such as art, design, and architecture.
Regular pentagons have 5 lines of reflectional symmetry.
The statistic 'Regular pentagons have 5 lines of reflectional symmetry' indicates that a regular pentagon, which is a polygon with five equal sides and five equal angles, can be reflected across 5 different lines such that the reflected image perfectly overlaps the original shape. These lines of symmetry are such that when the pentagon is folded or flipped over any of these lines, the resulting image matches the original pentagon. The presence of 5 lines of reflectional symmetry in a regular pentagon is a unique property of this specific geometric shape and contributes to its overall symmetry and balance.
Regular polygons with n sides have n lines of reflectional symmetry.
The statement that regular polygons with n sides have n lines of reflectional symmetry means that for every regular polygon with n sides (such as a triangle, square, pentagon, hexagon, etc.), there exists a specific number of lines that can be drawn through the polygon such that the shape can be reflected across these lines to create identical halves. These lines of symmetry are evenly distributed throughout the polygon and always pass through specific vertices or midpoints, depending on the shape of the polygon. The number of lines of reflectional symmetry in a regular polygon is equal to the number of sides it has, as each side corresponds to a unique line of symmetry that the shape can be reflected across to maintain its symmetry and identical appearance statistics | 677.169 | 1 |
Free Printable Protractor
Free Printable Protractor - Learn how to use a protractor to measure angles and explore geometry with this simple. Web download and print free protractors in different sizes and formats for measuring angles. If you don't have a real one, you can easily download. Web learn how to use a protractor to measure angles with this guide and free printable tool. A protractor is a handy tool for taking measurements of angles. Web download and print a free protractor for math or craft projects. Learn how to read and use a protractor with these. Download and print worksheets to practice geometry skills with protractors. Learn how to print them true to size and find more useful printables for math and geometry. Web download and print protractors in four sizes for measuring and drawing angles.
Free Printable Protractor 180° 360° Pdf with Ruler
Web download and print a free protractor for math or craft projects. Learn how to print them true to size and find more useful printables for math and geometry. Learn how to read and use a protractor with these. Download and print worksheets to practice geometry skills with protractors. Learn the definitions and examples of acute, obtuse, reflex, right and.
Free Printable Protractor 180° 360° Pdf with Ruler
Choose from 180° or 360°. If you don't have a real one, you can easily download. Web learn how to use a protractor to measure angles with this guide and free printable tool. Download and print worksheets to practice geometry skills with protractors. Learn the definitions and examples of acute, obtuse, reflex, right and straight angles with this fun and.
Free Printable Protractor 180° 360° Pdf with Ruler
Download and print worksheets to practice geometry skills with protractors. Learn how to print them true to size and find more useful printables for math and geometry. Web download and print protractors in four sizes for measuring and drawing angles. A protractor is a handy tool for taking measurements of angles. Web download and print free protractors in different sizes.
Free Printable Protractor 180° 360° Pdf with Ruler
Web download and print a free protractor for math or craft projects. Web download and print free protractors in different sizes and formats for measuring angles. Download and print worksheets to practice geometry skills with protractors. Choose from 180° or 360°. Learn how to print them true to size and find more useful printables for math and geometry.
Full Page Free Printable Printable Protractor
Learn the definitions and examples of acute, obtuse, reflex, right and straight angles with this fun and educational tool. Web download and print a free protractor for math or craft projects. If you don't have a real one, you can easily download. Learn how to read and use a protractor with these. Download and print worksheets to practice geometry skills.
Actual Size Printable Protractor Customize and Print
Choose from 180° or 360°. Web download and print a free protractor for math or craft projects. Download and print worksheets to practice geometry skills with protractors. Learn how to print them true to size and find more useful printables for math and geometry. Web download and print transparent or pdf protractors for measuring angles.
Printable Protractor Template
A protractor is a handy tool for taking measurements of angles. Learn how to print them true to size and find more useful printables for math and geometry. Download and print worksheets to practice geometry skills with protractors. Learn how to read and use a protractor with these. Web download and print a free protractor for math or craft projects.
Free Printable Protractor 180° 360° Pdf with Ruler
Learn how to use a protractor to measure angles and explore geometry with this simple. Web download and print protractors in four sizes for measuring and drawing angles. A protractor is a handy tool for taking measurements of angles. Download and print worksheets to practice geometry skills with protractors. Learn how to read and use a protractor with these.
Printable Protractor Free SVG
Web download and print protractors in four sizes for measuring and drawing angles. Learn how to read and use a protractor with these. Learn how to print them true to size and find more useful printables for math and geometry. Web learn how to use a protractor to measure angles with this guide and free printable tool. Choose from 180°.
Printable Protractor 360 ClipArt Best
Choose from 180° or 360°. Learn how to read and use a protractor with these. Web download and print transparent or pdf protractors for measuring angles. Web download and print free protractors in different sizes and formats for measuring angles. Web download and print protractors in four sizes for measuring and drawing angles.
Learn the definitions and examples of acute, obtuse, reflex, right and straight angles with this fun and educational tool. Learn how to read and use a protractor with these. If you don't have a real one, you can easily download. Web download and print a free protractor for math or craft projects. Learn how to use a protractor to measure angles and explore geometry with this simple. Web download and print transparent or pdf protractors for measuring angles. Web download and print free protractors in different sizes and formats for measuring angles. Web learn how to use a protractor to measure angles with this guide and free printable tool. A protractor is a handy tool for taking measurements of angles. Choose from 180° or 360°. Web download and print protractors in four sizes for measuring and drawing angles. Learn how to print them true to size and find more useful printables for math and geometry. Download and print worksheets to practice geometry skills with protractors.
Learn The Definitions And Examples Of Acute, Obtuse, Reflex, Right And Straight Angles With This Fun And Educational Tool.
Web download and print transparent or pdf protractors for measuring angles. Download and print worksheets to practice geometry skills with protractors. Web download and print free protractors in different sizes and formats for measuring angles. Web download and print protractors in four sizes for measuring and drawing angles.
If You Don't Have A Real One, You Can Easily Download.
Learn how to read and use a protractor with these. Web learn how to use a protractor to measure angles with this guide and free printable tool. Choose from 180° or 360°. Learn how to print them true to size and find more useful printables for math and geometry.
Learn How To Use A Protractor To Measure Angles And Explore Geometry With This Simple.
A protractor is a handy tool for taking measurements of angles. Web download and print a free protractor for math or craft projects. | 677.169 | 1 |
(Solved): Consider the following 2D vector given in component form v = ( 9.5, 4.8 ). At what angle is this vec ...
Consider the following 2D vector given in component form v = ( 9.5, 4.8 ). At what angle is this vector pointing, as measured off of the x-axis?please answer in degrees to 1 decimal place)
3 1 point
Consider the following 2 D vector given in component form | 677.169 | 1 |
Intro to the Coordinate Plane and its 4 Quadrants
Students will define and then in their own words explain the following terms: x and y axis, coordinate plane, quadrant, origin, ordered pair
Students will apply their definitions as they create and label their coordinate plane.
Students will demonstrate their knowledge on a homework assignment due the next day. The assignment is a puzzle with a coordinate plane. In each quadrant, letters are plotted and students solve the puzzle by correctly matching the letter to its ordered pair. | 677.169 | 1 |
Latest Updates
Contact form
Polygon Definition, Types, Formula And Examples
Polygons are 2-dimensional shapes made up of straight lines and enclosed within sides. Read the detailed article for more information about Polygon Definition, Types, Formula And Examples.
Polygon
Polygons are 2-dimensional shapes made up of straight lines and enclosed within sides. Polygon means all the closed shapes with straight-line figures come under the category of a polygon. You must know about the definition, shape, types, formula, and examples of a polygon by reading the article given below.
Polygon Definition
A polygon is an enclosed shape with straight lines. 2-dimensional shapes like Rectangles, squares, etc are categorized under polygons. Polygons have a finite number of sides. A circle is not a polygon as it has a curved shape. The points where the 2 straight lines meet are called vertices. The interior is called the body.
Types of Polygon
There are mainly 2 types of Polygon:
Regular Polygon- The polygon which has equal sides and equal angles. Generally, questions from regular polygon are asked in the exam.
Irregular Polygon- The one with unequal sides and angles.
The below figures show Regular and Irregular Polygons:
Properties of Polygon
Polygon: It is a closed plane figure bounded by three or more than three straight lines.
Regular Polygon: All the sides are equal and also all the interior angles are equal
Sum of Interior angles of a polygon = (n – 2) × 180
n → number of sides
The sum of exterior angle = 360.
Question-based on polygons
Q1.The ratio between the number of sides of two regular polygons is 1: 2 and the ratio between their interior angles is 2 : 3. The number of sides of these polygons is respectively:
(a) 3, 6
(b) 5, 10
(c) 4, 8
(d) 6, 12
Q2. If each interior angle of a regular polygon is 3 times its exterior angle, the number of sides of the polygon is :
(a) 4
(b) 5
(c) 6
(d) 8
Q3. A polygon has 54 diagonals. The number of sides in the polygon is :
(a) 7
(b) 9
(c) 12
(d) 19
Q4. The ratio between the number of sides of two regular polygon 1 : 2 and the ratio between their interior angle is 3 : 4. The number of sides of these polygons are respectively :
(a) 3, 6
(b) 4 , 8
(c) 6, 9
(d) 5, 10
Q5. The sum of all the interior angles of a regular polygon is four times the sum of its exterior angles. The polygon is :
(a) hexagon
(b) triangle
(c) decagon
(d) nonagon
Q6. The ratio of the measure of an angle of a regular nonagon to the measure of its exterior angle is :
(a) 3 : 5
(b) 5 : 2
(c) 7 : 2
(d) 4 : 5
Q7.The ratio of the measure of an interior angle of a regular octagon to the measure of its exterior angle is :
(a) 1 : 3
(b) 2 : 3
(c) 3 : 1
(d) 3 : 2
Q8.The sum of the interior angles of the polygon is 1440°. The number of sides of the polygon is :
(a) 9
(b) 10
(c) 8
(d) 12
Q9.The sum of all exterior angles of a convex polygon of n sides is :
(a) 4 right angle
(b) 2/n right angle
(c) (2n – 4) right angle
(d) n/2 right angle
Q10.One angle of a pentagon is 140°. If the remaining angles are in the ratio 1: 2 : 3: 4, the size of the greatest angle is : | 677.169 | 1 |
Hint: We have to apply Pythagoras theorem in triangle APC and find the side AP. Pythagoras theorem states the square of the hypotenuse is equal to the sum of the square of the other two sides. Similarly, we have to apply Pythagoras theorem in triangle APB, substitute the values of AP and BP in the equation and find the side AB.
Complete step by step solution: We have to find the length of the side AB. We can see from the figure that angle APC is a right-angle, that is, $\angle APC=90{}^\circ $ . Therefore, we can apply Pythagoras theorem in $\Delta APC$ . We know that for a right-angled triangle, Pythagoras theorem states the square of the hypotenuse is equal to the sum of the square of the other two sides. Therefore, in triangle APC, we can apply Pythagoras theorem. The hypotenuse of $\Delta APC$ is AC. $\Rightarrow A{{C}^{2}}=P{{C}^{2}}+A{{P}^{2}}$ We are given that $AC=5\text{ units and }PC=3\text{ units}$ . We have to substitute these values in the above equation. $\begin{align} & \Rightarrow {{5}^{2}}=A{{P}^{2}}+{{3}^{2}} \\ & \Rightarrow 25=A{{P}^{2}}+9 \\ \end{align}$ Let us take 9 to the LHS from the RHS. $\begin{align} & \Rightarrow 25-9=A{{P}^{2}} \\ & \Rightarrow A{{P}^{2}}=16 \\ \end{align}$ We have to take the square root on both sides. $\begin{align} & \Rightarrow AP=\sqrt{16} \\ & \Rightarrow AP=4\text{ units}...\left( i \right) \\ \end{align}$ From the given figure, we can see that $\angle APB=90{}^\circ $ . We have to apply Pythagoras theorem in the $\Delta APB$ . From the figure, we can infer that AB is the hypotenuse. $\Rightarrow A{{B}^{2}}=A{{P}^{2}}+B{{P}^{2}}$ We are given that $BP=4\sqrt{3}$ . Let us substitute this value and (i) in the above equation. $\Rightarrow A{{B}^{2}}={{4}^{2}}+{{\left( 4\sqrt{3} \right)}^{2}}$ We know that ${{\left( \sqrt{a} \right)}^{2}}=a$ . Therefore, the above equation becomes $\begin{align} & \Rightarrow A{{B}^{2}}=16+\left( 16\times 3 \right) \\ & \Rightarrow A{{B}^{2}}=16+48 \\ & \Rightarrow A{{B}^{2}}=64 \\ \end{align}$ We have to take the square root on both sides. $\begin{align} & \Rightarrow AB=\sqrt{64} \\ & \Rightarrow AB=8\text{ units} \\ \end{align}$ Therefore, the length of $\overline{AB}$ is $8$ units.
Note: Students must note that they can only apply Pythagoras theorem on a triangle if the triangle is a right-angled one. They have a chance of making a mistake by writing Pythagoras theorem as the square of the hypotenuse is equal to the difference of the square of the other two sides. They must never forget to write the units at the end even if the unit is not specified in the question. | 677.169 | 1 |
Module 7: Parametric Equations and Polar Coordinates
Search for:
Introduction to Polar Coordinates
The rectangular coordinate system (or Cartesian plane) provides a means of mapping points to ordered pairs and ordered pairs to points. This is called a one-to-one mapping from points in the plane to ordered pairs. The polar coordinate system provides an alternative method of mapping points to ordered pairs. In this section we see that in some circumstances, polar coordinates can be more useful than rectangular coordinates. | 677.169 | 1 |
Amplitude and Period in Trigonometry
Feeling:
Dumb
Language:
Arabic
Prompt:
Amplitude and Period in Trigonometry
Amplitude and period are important concepts in trigonometry when calculating angles.
Amplitude refers to the distance from the midline of a trigonometric function to its maximum or minimum value. It essentially measures the height of the wave or how far it deviates from the midline. In trigonometry, the amplitude of a function is half the difference between the maximum and minimum values of the function.
Period, on the other hand, is the length it takes for a trigonometric function to complete one full cycle. It is the distance between two consecutive peaks or troughs of the wave. The period of a trigonometric function is the horizontal distance it takes for the function to repeat itself.
When calculating angles using trigonometry, understanding the amplitude and period of trigonometric functions can help you determine the range of possible angles and how the functions behave over a given interval.
For example, in the cosine function, the amplitude is always 1 because the function oscillates between -1 and 1. The period of the cosine function is 2π, as the function repeats itself every 2π radians.
One real-life application of understanding amplitude and period in trigonometry is in the study of sound waves. Sound waves can be modeled using trigonometric functions, and understanding the amplitude and period of these functions can help in analyzing and predicting the behavior of sound waves.
In conclusion, amplitude and period are important concepts in trigonometry that help in understanding the behavior of trigonometric functions and calculating angles accurately | 677.169 | 1 |
In a triangle, lines are drawn from each vertex to the opposite side. Can there be seven regions of integer area?
Context
My question was inspired by a similar question, which asks if a triangle is inscribed in a circle, can the four regions have integer area?
My attempt
I started by considering a simpler question: In a triangle, two lines are drawn, each one from a vertex to the opposite side. Can there be four regions of integer area? The answer is clearly yes. For example, if the triangle is equilateral, and the two lines go through the centre, then we can have four regions of area $1, 1, 2, 2$:
But when I consider three lines (and seven regions), the situation seems to be much more complicated, and I have not found a feasible approach.
1 Answer
1
Of course there can! To see why, take the points of the large triangle to have any rational coordinates. Let the lines all have rational slope and y-intercept. Then all points are rational and thus by the shoelace formula, all the areas are rational. Since the areas are all rational, they share a common denominator. Enlarge the figure by this common denominator, and all areas are then integers. | 677.169 | 1 |
This program will solve for a side of a triangle when you know two other sides and a non included angle. It will also solve for any angle in a triangle when you know all three side lengths. It will tell you all possible side lengths, and even when there is no possible triangle. Perfect for geometry or trigonometry students. | 677.169 | 1 |
Correct Option: D
From first figure to second figure the circles interchange position and the circles which come to the lower right corner get duplicated. The rectangle moves upward diagonally while the central design (D) is inverted.
Correct Option: B
From first figure to second figure the circles interchange position and the circles which come to the lower right corner get duplicated. The rectangle moves upward diagonally while the central design (D) is inverted. | 677.169 | 1 |
Pythagoras | Trigonometry | Maths | FuseSchool
Thousands of years ago it was discovered that the two smaller squares on the side of a right angle triangle added up to the exact same size as the larger square on the triangle - the square formed on the diagonal side. This is called Pythagoras Theorem | 677.169 | 1 |
Students will practice finding the image of a point after applying a transformation (reflections, translations, rotations, dilations) with this set of seven mazes. Graph paper is included to print out if your students are hands-on and visual (like me!) and need to graph them. This activity was designed for a high school level geometry class.
Maze 5: Dilations (includes the origin and other points as center of dilation)
Maze 6: Mixed Transformations (origin as the center of rotations and dilations only)
Maze 7: Mixed Transformations (any point as the center of rotations and dilations)
Rotations include 90°, 180°, 270° in both the clockwise and counterclockwise directions. The perfect way to prepare for the unit test. I love that there were mazes for one transformation type as well as mazes that had student shift their thinking between the transformation types. Quality work as always from this publisher!!
—VALERIE S.
Engaging way to practice and the variety of options are super helpful!!!
—JACLYN G.
Love engaging students with a maze that still gets them to practice their skills! | 677.169 | 1 |
Cut out these angle representations from the color paper using scissors.
Glue the angles onto the cardboard:
Glue each angle representation onto the cardboard base, leaving some space between them.
Demonstrate angle relationships:
Use the model to demonstrate the relationships between different angle types. For example, show how two acute angles can add up to form a right angle, or how two supplementary angles form a straight angle.
This model provides a visual representation of various angle measurements and their characteristics. It's a fun and educational project to understand the different types of angles and how they relate to one another in geometry. | 677.169 | 1 |
A triangular pyramid is a geometric shape that has a triangular base and three lateral triangular faces. It has a vertex, the point that is common to all three lateral faces. If all three faces are equilateral, that is, all the sides are of equal length, then such a pyramid is called a tetrahedron. See summer fires in Canada, Greece, and Hawaii. | 677.169 | 1 |
There is a quadrilateral PLAY with PL = 3cm, LA = 4cm, AY = 4.5 cm, PY = 2cm. Which of the following values of LY is not possible?
A
2
No worries! We've got your back. Try BYJU'S free classes today!
B
3
No worries! We've got your back. Try BYJU'S free classes today!
C
6
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
4
No worries! We've got your back. Try BYJU'S free classes today!
Open in App
Solution
The correct option is C
6
Since one side of a triangle cannot be greater than the sum of the other two sides, diagonal LY must be smaller than the sum of PL + PY and smaller than the sum of AL + AY. PL + PY =5 and AL + AY = 8.5. Hence, LY must be less than 5. Thus, the correct answer is 6. | 677.169 | 1 |
Given planes are 2x + 3y + 4z =4 . . . (i) and 4x + 6y + 8z =12 ⇒ 2x + 3y + 4z =6 . . . (ii) From the above equations, it can be seen that given planes are parallel. It is known that the distance between two planes, ax + by + cz = d1 and ax + by + cz = d2 is given by ∴Distanced=∣∣∣d2−d1√a2+b2+c2∣∣∣⇒d=∣∣∣6−4√22+32+42∣∣∣ ⇒d=2√4+9+16=2√29 units. So, the correct option is (d). | 677.169 | 1 |
I see this question
I have a square with length of side is $a$. How to draw a number of circles inscribed in a square so that the sum of the radii of the circle is greatest? In the below picture is twenty circles inscribed in a square. We can consider number of circles are 5, 6, ... We consider number of the circles is fixed.
At here is discussing about "How can I convert a list containing three points and its equation passing three points to text file?" With mylist
Let S1: x^2 + (y - 2)^2 + (z + 1)^2 = 29 be a sphere and two points A(0, 0, 4), B(6, -2, 6); the line d passing through point C(4, -8, 4) and have direction v=(1, -1, 2). Find the point M such that M lies on the sphere S1, the angle AMB equals to 90 degree and distance from M to the line d is minimum. | 677.169 | 1 |
Equal Triangular Areas in a Regular Pentagon
Creation of this applet was originally inspired by a tweet from Solve My Maths.
How does this applet informally illustrate that the yellow and light green triangles have equal area?
How can you formally prove that the areas of these two triangles are indeed equal? | 677.169 | 1 |
Class 10
Class 10 Some Applications of Trigonometry Previous Years Questions
A ladder, leaning against a wall, makes an angle of 60° with the horizontal. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder. [CBSE 2016] [1 Mark]
If the shadow of a vertical pole at a particular time of the day is equal to √ 3 times its height, then what is the elevation of the source of light at that time ? [CBSE 2017] [1 Mark]
In figure, a tightly stretched rope of length 20 m is tied from the top of a vertical pole to a ground. Find the height of the pole if the angle made by the rope with the ground is 30°. [CBSE 2020] [1 Mark]
A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of hill as 30°. Find the distance of the hill from the ship and the height of the hill. [CBSE 2016] [3 Marks]
A ladder is leaning against a wall of a house such that its upper end is touching the top of the wall. The foot of the ladder is 2 m away from the wall and the ladder is making an angle of 60° with the level of the ground. Find the height of the wall. Also, find the length of the ladder. [CBSE 2017] [3 Marks]
The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is 45°. Find the height of the tower PQ and the distance PX. (Use √ 3 = 1.73) [CBSE 2016] [4 Marks]
A 2 m tall boy is standing at some distance from a 29 m tall building. The angle of elevation, from his eyes to the top of the building increases from 30° to 60°, as he walks towards the building. Find the distance he walked towards the building. [CBSE 2017] [4 Marks]
As observed from the top of a 100 m high light house from the sea-level , the angles of depression of two ships are 30° and 45°. If one ship exactly behind the other on the same side of the light house, find the distance between the two ships. (Use √ 3 = 1.732) [CBSE 2018] [4 Marks]
The shadow of a tower standing on a level ground is found to be 40 m longer when the sun's altitude is 30° than when it was 60°. Find the height of the tower. (Given √ 3 = 1.732) [CBSE 2019] [4 Marks]
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30° , which is approaching the foot of the tower with a uniform speed. After covering a distance of 50 m, the angle of depression of the car becomes 60°. Find the height of the tower. ( Use √ 3 = = 1.73) [CBSE 2020] [4 Marks]
From the top of a 7 m building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. (Use √ 3 = 1.73) [CBSE 2020] [4 Marks] | 677.169 | 1 |
From a point within a triangle, Segments are drawn to the vertices. A necessary.and sufficient condition that the three triangles formed have equal areas, is that the point be: (A) such that the three angles formed each have a measure of 120° (B) the centre of the inscribed circle (C) the centre of the circumscribed circle (D) the intersection of the medians
Answer
The correct answer is Option D. In order for the three triangles formed to have equal areas, it is necessary and sufficient for the point to be the intersection of the medians. Finding the centroid, which is the point where the medians intersect, allows us to divide the triangle into three equal area triangles by connecting the vertices to the centroid. Drawing line segments from a point within the triangle to the vertices will create three triangles with equal areas if and only if the point is the intersection of the medians. For further information on the intersection of medians, please refer to brainly.com/question/29104255 #SPJ4. | 677.169 | 1 |
3 and 8 are the lengths of two sides of a triangular region, which
[#permalink]
10 Oct 2005, 07:42
4
Kudos
2
Bookmarks
saurya_s wrote:
15. If 3 and 8 are the lengths of two sides of a triangular region, which of the following can be the length of the third side? I. 5 II. 8 III. 11 (A) ?I only (B) ?II only (C) I and ?I only (D) II and ?II only (E) I, ??, and ???
Re: If 3 and 8 are the lengths of two sides of a triangular region, which
[#permalink]
10 Dec 2019, 13:40
1
Bookmarks
Expert Reply
Hi All,
We're told that 3 and 8 are the lengths of two sides of a triangular region. We're asked which of the following values (3, 8 and 11) could be the length of the third side. This question is based on the Triangle Inequality Theorem and requires just a little Arithmetic to solve.
To start, when you have two sides of a triangle, the SMALLEST the third side could be would be just a little bit MORE than the DIFFERENCE of the two sides you have. Here, that's 8 - 3 = 5. If the third side was EXACTLY 5, then you would NOT have a triangle - you'd have a line of 8 on top of another line of 8. By making that third side just a little bigger than 5, you would then have a triangle.
In that same way, the LARGEST the third side could be would be just a little bit LESS than the SUM of the two sides. Here, that's 3 + 8 = 11. If the third side was EXACTLY 11, then you would NOT have a triangle - you'd have a line of 11 on top of another line of 11. By making that third side just a little less than 11, you would then have a triangle.
Thus, that third side falls into the following range: 3 < third side < 11. Based on the three given options, only a side of 8 is possible.
Re: If 3 and 8 are the lengths of two sides of a triangular region, which
[#permalink]
27 Feb 2023, 19 3 and 8 are the lengths of two sides of a triangular region, which [#permalink] | 677.169 | 1 |
The Importance and Properties of Incenter in Geometry and Its Applications
incenter Chapter 6 (p. 303)
Incenter is a geometric term used in the field of mathematics, particularly in geometry
Incenter is a geometric term used in the field of mathematics, particularly in geometry. Incenter refers to the center of the inscribed circle or incircle of a triangle or regular polygon.
To define incenter more precisely, let's consider a triangle. The incenter of a triangle is the point where the three angle bisectors intersect. An angle bisector divides an angle into two equal halves. So, in a triangle, there are three angle bisectors, and the incenter is the point where these three bisectors meet.
The incenter is significant because it has some unique properties. One key property is that the incenter is equidistant from the three sides of the triangle. This means that the distances from the incenter to each side of the triangle are equal.
Another important property is that the incenter is the center of the inscribed circle or incircle of the triangle. The incircle is the largest circle that can fit inside the triangle, touching all three sides. The inradius is the radius of this incircle, and it is equal to the distance from the incenter to any of the sides of the triangle.
The incenter and the incircle have several applications and uses in geometry and other fields. For example, incenter plays a role in finding the angle measures, side lengths, and area of a triangle. Additionally, it can be helpful in solving problems related to tangents and circle-to-triangle relationships.
To find the incenter of a triangle, you typically need to determine the intersection point of the three angle bisectors. This can be done using various geometric constructions or formulas involving the coordinates of the triangle's vertices | 677.169 | 1 |
If in a triangle ABC, B is the orthocentre and if circumcentre of triangle ABC is ( 2,4) and vertex A is ( 0,0) then coordinate of vertex C is (a) $\left( 4,2 \right)$ (b) $\left( 4,8 \right)$ (c) $\left( 8,4 \right)$ (d) $\left( 8,2 \right)$
Hint: A right angled triangle can always be inscribed inside a circle with the hypotenuse as the diameter of the circle and the midpoint of the hypotenuse of the triangle will be the circumcentre of the triangle .
Consider the figure alongside . In the question it is given that $B$ is the orthocentre. We also know that $B$ is one of the vertices of the triangle $ABC$.
So , we can conclude that $ABC$ is a right-angled triangle. Now , we know that a right angled triangle can be inscribed in a circle with the hypotenuse as the diameter. So , we can conclude that $AC$ is the diameter of a circle. So , the centre of this circle will be the midpoint of the diameter. Now, we know the...............\left( i \right)$ Now , we will consider $\left( x,y \right)$ to be the coordinates of $C$. We know , circumcentre of a triangle is the centre of the circle circumscribing the triangle. So , the circumcentre $D\left( 2,4 \right)$ is the midpoint of the line joining $A\left( 0,0 \right)$ and $C\left( x,y \right)$. Now , we will use the equation $\left( i \right)$ to find the coordinates of the vertex $C$. So , from equation$\left( i \right)$ , we have $2=\dfrac{x+0}{2}\Rightarrow x=4$ And $\text{ 4}=\dfrac{y+0}{2}\Rightarrow y=8$ So , the coordinates of vertex $C$ are $\left( 4,8 \right)$. Option B - $\left( 4,8 \right)$ is correct answer
Note: The$ and not $\left( \dfrac{\left( {{x}_{1}}-{{x}_{2}} \right)}{2},\dfrac{\left( {{y}_{1}}-{{y}_{2}} \right)}{2} \right)$ . Students often get confused between the two. Due to this confusion , they generally end up getting a wrong answer . So , such mistakes should be avoided . | 677.169 | 1 |
horizontal number line is the x-axis
This essay focuses on horizontal number line is the x-axis.. Round to the nearest tenth when necessary or to the nearest minute as appropriate.
horizontal number line is the x-axis
on a geometry question and need an explanation to help me understand better.
1. Solve the triangle. Round to the nearest tenth when necessary or to the nearest minute as appropriate.
2. Solve the triangle. Round to the nearest tenth when necessary or to the nearest minute as appropriate.
details;
Firstly, a 2D surface formed by using two number lines that intersect each other at the right angle.Earth measurement." Eventually it was realized that geometry need not be limited to the study of flat surfaces (plane geometry) and rigid three-dimensional objects (solid geometry) but that even the most abstract thoughts and images might be represented and developed in geometric terms.
Secondly, The horizontal number line is the x-axis, and the vertical number line is the y-axis.
Thirdly, The intersection of the two axes is the (0,0) coordinate.
Further, Using the coordinate plane, we plot points, lines, etc. By joining various points on the coordinate plane, we can create shapes.
B =49.2°C = 102°b = 40.9
3. Solve the triangle. Round to the nearest tenth when necessary or to the nearest minute as appropriate.
A triangle is a 3 side shape, and the measure of its 3 interior angles is 180˚
A square, rectangle or quadrilateral are 4 side shapes, and the measure of their interior angles is 360˚ | 677.169 | 1 |
How to effectively use the Protractor Gelsonlab HSMM-011A for precise measurements
The Protractor Gelsonlab HSMM-011A is a high-quality teaching tool that is essential for precise measurements in various educational settings. This protractor is made of durable plastic and features clear markings for accurate readings. When used correctly, the Protractor Gelsonlab HSMM-011A can help students and teachers alike achieve more accurate measurements in geometry, trigonometry, and other mathematical subjects.
To effectively use the Protractor Gelsonlab HSMM-011A, it is important to first familiarize yourself with its features. The protractor has a 180-degree scale with clear markings at every degree, making it easy to read and interpret measurements. Additionally, the protractor has a built-in ruler that can be used for measuring lengths and drawing straight lines. By understanding how to properly use these features, you can ensure that your measurements are as precise as possible.
When using the Protractor Gelsonlab HSMM-011A, it is important to handle it with care to avoid damaging the delicate markings. Make sure to place the protractor flat on the surface you are measuring, and use a pencil or pen to mark the desired angle or length. When drawing lines, use the built-in ruler to ensure that your lines are straight and accurate. By taking these precautions, you can ensure that your measurements are as precise as possible.
One of the key benefits of using the Protractor Gelsonlab HSMM-011A is its versatility. This protractor can be used for a wide range of measurements, including angles, lengths, and shapes. Whether you are measuring the angles of a triangle or drawing a circle, the Protractor Gelsonlab HSMM-011A can help you achieve accurate results. By using this tool in conjunction with other measuring instruments, such as Rulers and Compasses, you can ensure that your measurements are as precise as possible.
In addition to its versatility, the Protractor Gelsonlab HSMM-011A is also easy to use. The clear markings and durable construction make it simple to read and interpret measurements, even for beginners. By following the instructions provided with the protractor, you can quickly learn how to use it effectively for a wide range of measurements. Whether you are a student learning geometry for the first time or a teacher looking to improve your students' understanding of angles and shapes, the Protractor Gelsonlab HSMM-011A is an essential tool for precise measurements.
In conclusion, the Protractor Gelsonlab HSMM-011A is a high-quality teaching tool that can help students and teachers achieve more accurate measurements in various educational settings. By familiarizing yourself with its features, handling it with care, and using it in conjunction with other measuring instruments, you can ensure that your measurements are as precise as possible. Whether you are measuring angles, lengths, or shapes, the Protractor Gelsonlab HSMM-011A is an essential tool for achieving accurate results in geometry, trigonometry, and other mathematical subjects. | 677.169 | 1 |
Use the law of sines to solve the given problem.. The
Last updated: 7/12/2022
Use the law of sines to solve the given problem..
The loading ramp at a delivery service is 10.5 ft long and makes a 19.0° angle with the horizontal. If it is replaced with a ramp 20.5 ft long, what angle does the new ramp
make with the horizontal?
The new ramp makes a ° angle with the horizontal.
(Round to one decimal place as needed.) | 677.169 | 1 |
Quadrilateral form a Parallelogram
Statement of the Theorem: Prove that the lines joining the middle points of the adjacent sides of a quadrilateral form a parallelogram.
Proof: Let ABCD be a quadrilateral and length of its side AB is 2a.
Let us choose origin of rectangular cartesian co-ordinates at the vertex A and x-axis along the side AB and AY as the y-axis. Then, the co-ordinates of A and B are (0, 0) and (2a, 0) respectively. Referred to the chosen axes, let (2b, 2c) and (2d, 2e) be the co-ordinates of the vertices C and D respectively. If J, K, L, M be the mid-points of the sides AB, BC, CD, and, DA, respectively, then the co-ordinates of J, K, L and M are (a, 0 ), (a + b, c), (b + d, c + e) and (d, e) respectively.
Now, the co-ordinates of the mid-point of the diagonal JL of the quadrilateral JKLM are {(a + b + d)/2, (c + e)/2}
Again, the co-ordinates of the mid-point of the diagonal MK of the same quadrilateral are {(a + b + d)/2, (c + e)/2}.
Clearly, the diagonals JL and MK of the quadrilateral JKLM bisect each other at ((a + b + d)/2, (c + e)/2). Hence, the quadrilateral JKLM is a parallelogram. Proved | 677.169 | 1 |
Content Standards G.CO.1 Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment, based on the undefined notions of point, line, distance along a line, and distance around a circular arc. G.CO.12 Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Mathematical Practices 2 Reason abstractly and quantitatively. 7 Look for and make use of structure. CCSS
distance the length of the segment between two points. • irrational number • midpoint • segment bisector a number that cannot be expressed as a terminating or repeating decimal. the point on a segment exactly halfway between the endpoints of the segment. a segment, line, or plane that intersects a segment at its midpoint. Vocabulary
Find Midpoint on a Number Line DECORATING Marco places a couch so that its end is perpendicular and 2.5 feet away from the wall. The couch is 90" wide. How far is the midpoint of the couch back from the wall in feet? First we must convert 90 inches to 7.5 feet. The coordinates of the endpoints of the couch are 2.5 and 10. Let M be the midpoint of the couch. Midpoint Formula x1 = 2.5, x2 = 10 Example 3 | 677.169 | 1 |
Hi, I was just wondering how do I figure out the dihedral Angle of two planes in 3D Slicer? My plane 1, the yellow plane, is made up of 3 coordinate points, and the purple plane is made up of 4 coordinate points. I have found the plug-in Angle Planes, but the usage instructions are not very clear, so I don't know how to use this plug-in now. I have tried the code, but it doesn't work, so I would like to ask teachers if there is any good solution for this plug-in, why P1 or P2 can't be chosen to create a new project? | 677.169 | 1 |
Honors Geometry Companion Book, Volume 1
3.1.2 Angles, Parallel Lines, and Transversals (continued)
The process for finding m ∠ UTQ is very similar to the process used to find m ∠ STQ in Example 2. Again, first classify the pair ∠ UTQ and ∠ RQT . Since the two angles are on the same side of the transversal and inside of the other lines, ∠ UTQ and ∠ RQT are same-side interior angles. Therefore, since lines PR and SU are parallel, ∠ UTQ and ∠ RQT are supplementary by the Same-Side Interior Angles Theorem. Use this fact to write an equation, then simplify and solve. Once x is known, substitute that value into the expression for m ∠ UTQ , 5 x − 14, and simplify to find m ∠ UTQ . In this example the m ∠ ABC is found by using the Same-Side Interior Angles Theorem. Since the three lines are parallel and the angles represented by 10 x and 11 x + 12 are same-side interior angles, these two angles are supplementary. It follows that (10 x ) + (11 x + 12) = 180. Simplify and solve the equation to find the value of x . Then, substitute that value for x into the expression for m ∠ ABC , 11 x + 12, to find m ∠ ABC . | 677.169 | 1 |
Question Video: Using Inscribed Angles in a Semi-Circle to Find the Area of a Triangle
Mathematics • Third Year of Preparatory School
Join Nagwa Classes
Given that 𝐴𝐶 = 8 cm and the radius = 8 cm, find the area of △𝐴𝐵𝐶 rounded to the nearest integer.
04:07
Video Transcript
Given that 𝐴𝐶 equals eight centimeters and the radius equals eight centimeters, find the area of triangle 𝐴𝐵𝐶 rounded to the nearest integer.
And then we have a diagram which shows a circle with a triangle inscribed within it. We notice that line 𝐴𝐵 passes through the center of the circle, and so line 𝐴𝐵 is the diameter. So before we work out the area of the triangle we've been given, let's identify what extra information we can calculate given the information we've been given. We are told that 𝐴𝐶 equals eight centimeters and that the radius is equal to eight centimeters. This means line segments 𝐵𝑀 and 𝑀𝐴 are both eight centimeters.
Now, in fact, since point 𝑀 is the center of the circle and 𝐶 lies on the circumference, we see that 𝑀𝐶 also forms the radius of our circle. So 𝑀𝐶 is also eight centimeters. And this is really useful since we know that all angles in an equilateral triangle, that is, a triangle where all three sides are of equal length, are 60 degrees. But this doesn't necessarily help us find the area of triangle 𝐴𝐵𝐶. Remember, the formula that we can use to find the area of a triangle is a half times base times height. So if we could work out the length of the base and the height of this triangle 𝐴𝐵𝐶, we'd be able to find its area.
Now, in fact, we can quite quickly work out which sides of our triangle represent the base and the height. The base and the height must be perpendicular to one another. So we're interested in sides 𝐴𝐶 and 𝐵𝐶. But how do we know that? Well, we know that the angle subtended by the diameter is 90 degrees. Our angle 𝐵𝐶𝐴 is indeed subtended from the diameter, so 𝐵𝐶𝐴 is 90 degrees. So we can define 𝐴𝐶 to be the base of our triangle and that is eight centimeters in length. But then this means that 𝐵𝐶 is the height of our triangle. So what is the length of 𝐵𝐶?
Well, now that we know we have a right triangle and we know a couple of the angles given in this triangle, we can work out the length of 𝐵𝐶 using right-triangle trigonometry. Here is our triangle 𝐴𝐵𝐶 with the right angle at 𝐶. We calculated that the measure of angle 𝐴 is 60 degrees. And of course, we're trying to find the length of 𝐵𝐶, so let's define that to be equal to 𝑥 centimeters. This side of the triangle lies directly opposite the included angle of 60 degrees, whilst the side 𝐴𝐶 is adjacent to the angle. And so we need to identify the trigonometric ratio that links the opposite side with the adjacent, that is, the tangent ratio. tan 𝜃 is opposite over adjacent.
In this case then, tan of 60 is 𝑥 divided by eight. But of course, tan of 60 is one of our exact value ratios. It's the square root of three, so we get root three equals 𝑥 divided by eight. And if we multiply through by eight, we find that 𝑥 is equal to eight root three, and that's great. We now know the length of 𝐵𝐶, which we said was the height of our triangle. It's eight root three centimeters. So the area of our triangle is one-half times eight times eight root three. In exact form, that gives us 32 root three square centimeters. But If we enter this into our calculator, correct to the nearest integer, that's 55 square centimeters. And so the area of triangle 𝐴𝐵𝐶 correct to the nearest integer is 55 square centimeters.
Now, it's worth noting that we didn't actually need to use right-triangle trigonometry. Instead, if we'd recognize that 𝐴𝐵 is the diameter and therefore 16 centimeters in length and we already knew that triangle 𝐴𝐶𝐵 was a right triangle at 𝐶, we could have used the Pythagorean theorem to calculate the length of 𝐵𝐶. In that case, we once again would have got eight root three, giving us a final area of 55 square centimeters. Either method is perfectly valid. | 677.169 | 1 |
Honors Geometry Companion Book, Volume 1
2.1.4 Biconditional Statements and Definitions (continued)
The given conditional is true only if the related conditional and its converse are true. So, write the conditional and its converse using the hypothesis p and the conclusion q from the given biconditional. Then, determine the truth value of each. If a counterexample can be found for either the conditional or the converse, then the biconditional is false. A counterexample can be found for the converse since a rectangle with area 30 does exist where its length and width are not 10 and 3, respectively. Specifically, a rectangle with length 15 and width 2 has area 30, but does not have length 10 and width 3. Therefore, since the converse is false, the biconditional is false. In this example, use the definition of acute angles to determine the truth value of the conditional and its converse. An acute angle is defined to be an angle with measure that is less than 90°.
Example 4 Writing Definitions as Biconditional Statements
A definition is a statement that describes an object and can be written as a true biconditional. In the first example, the definition can be written as a biconditional by simply replacing "is" with "if and only if" and rephrasing the definition slightly. The definition in the second example can be written as a biconditional by replacing "that" with "if and only if" and rephrasing the definition. | 677.169 | 1 |
NCERT Solutions for Class 10 Maths of Chapter 6 Triangles
06/28/2020
NCERT Solutions for Class 10 Maths of Chapter 6 Triangles
NCERT Solutions for Class 10 Maths of Chapter 6 Triangles are created here for helping class 10 students in their CBSE Board exams. NCERT Solutions for Class 10 Maths of Chapter 6 Triangles are prepared by Future Study Point for helping class 10 students in their exams.NCERT Solutions for Class 10 Maths of Chapter 6- Triangles are available here in the form of a pdf of all NCERT solutions of all the exercises from exercise 6.1 to exercise 6.6. NCERT solutions for class 10 maths will ensure your study when you are offline. Chapter 6- Triangles is the fundamental chapter of geometry, chapter 6 – Triangles of class 10 maths NCERT is based on the properties of triangles like the basic proportionality theorem, the similarity of triangles, and Pythagoras theorem. All questions of Class 10 Maths NCERT solutions of Chapter 6 Triangles are solved by an expert Maths teacher as per the CBSE norms.
You can download the PDF-NCERT Solutions for Class 10 Maths of Chapter 6 Triangles
NCERT Solutions for Class 10 Maths of Chapter 6 Triangles
NCERT Solutions for Class 10 Maths of Chapter 6 Triangles
NCERT Solutions Class 10 Maths Chapter 6, Triangle of Class 10 Maths NCERT solutions of Chapter 6 Triangles part of Geometry which constitutes 15 marks of the total marks of 80. NCERT solutions of Chapter 6 Triangles Based on the CBSE Class 10 examination prospectus, this chapter 6 has the second-most elevated weightage. Consequently, having a clear understanding of the ideas, theorems, and problem-solving techniques, in this section it is compulsory to score well in the Board assessment of class 10 maths.
From your previous classes, you know about triangles and a considerable lot of their properties. In Class 9, you studied the congruency of the triangles in detail. In this part, we will consider those triangles which have a similar shape but not really a similar size. Two triangles having a similar shape (and not really a similar size) are called similar triangles. In chapter 6 of class 10 maths, we will study the similarity of triangles and the application of this property in giving a basic verification of the Pythagoras Theorem learned before.
NCERT Solutions for Class 10 Maths of Chapter 6 Triangles
In Class 9, you have seen that all circles with similar radii are congruent, all squares with a similar side length are congruent and all equilateral triangles with a similar side length are congruent. The chapter clarifies the similarity of triangles by playing out the applicable action. Similar triangles are two triangles having a similar shape yet not really a similar size.
The topic recalls triangles and their similarities. Two triangles are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). It explains Basic Proportionality Theorem and different theorems are discussed in performing various activities.
Basic Proportionality Theorem- If a line is drawn parallel to one of the sides of a triangle such that it intersects two sides then it divides the other two sides of the triangle in the same ratio.
NCERT Solutions for Class 10 Maths of Chapter 6 Triangles
In the previous section, we stated that two triangles are similar (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). The topic discusses the criteria for the similarity of triangles referring to the topics we have studied in earlier classes. It also contains different theorems explained with proper examples.
AAA rule-If in two triangles three corresponding angles are the same,then both of triangles are similar to each other
AA rule – If in two triangles two corresponding angles are the same,then both of triangles are similar to each other.
SAS rule – If in two triangles two corresponding sides are in proportion and the angles subtended by them are also equal then triangles are similar to each other.
SSS rule- If all corresponding sides of two triangles in proportion then both of the triangles are similar to each other.
If in two triangles ΔABC and ΔDEF
AB/DE =BC/EF =AC/DF
then ABC ∼DEF
∠B = ∠E
ΔABC ∼ ΔDEF (SAS rule)
SSS rule- If all corresponding sides of two triangles in proportion then both of the triangles are similar to each other.
AB/DE =BC/EF =AC/DF
ΔABC ∼ ΔDEF (SSS rule)
See the video-Solutions for class 10 maths exercise 6.1 Triangle
NCERT Solutions for Class 10 Maths of Chapter 6 Triangles
You have learned that in two similar triangles, the ratio of their corresponding sides is the same. The topic Areas of Similar Triangles consists of a theorem and relatable examples to prove the theorem.
The ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.
Class 10 Maths NCERT Solutions Chapter 6- Triangles
You are already familiar with the Pythagoras Theorem from your earlier classes. You have also seen proof of this theorem in Class 9. Now, we shall prove this theorem using the concept of the similarity of triangles. Hence, the theorem is verified through some activities and made use of it while solving certain problems | 677.169 | 1 |
Elements of geometry and mensuration
From inside the book
Results 1-5 of 100
Page 8 ... radius of the circle . Any straight line drawn through the centre and terminated both ways by the circumference is called a diameter of the circle . Thus , in the fig . here traced , the area or surface in the plane of the paper bounded ...
Page 16 ... radius AB ( POST . III . 20 ) trace a portion of the circumference of a circle on that side of AB on which the triangle is required ; with the same radius and with centre B trace another portion of the circumference of a circle on the ...
Page 22 ... radius EB to cut BE produced in F ) , and join CF. Then in the triangles AEB , CEF , AE , EB are equal to CE , EF , each to each , by construction , and 2 AEB = = △ CEF ( 31 ) , because they are opposite vertical angles , the triangles ...
Page 46 ... radius of one circle be equal to the radius of another , the circles shall be equal in all respects . For , if one of the circles be ' applied to ' , or laid upon , the other so that their centres coincide , since the radii are equal | 677.169 | 1 |
Perpendiculars
You know that two lines (or rays or segments) are said to be perpendicular if they intersect such that the angles formed between them are right angles. In the figure, the lines l and m are perpendicular. Mathematics 280 The corners of a foolscap paper or your notebook indicate lines meeting at right angles | 677.169 | 1 |
Properties of Triangle
It has three vertices, three interior angles, and keep a note not all the three sides need to remain of the same length.
Important lines in the Triangle
Median
It is a line that connects any one of the vertices with the midpoint of the opposite side.
Altitude
It is a perpendicular line inside a triangle that connects a one-off vertex and the midpoint of the opposite side.
Triangle Properties
In this, you learn about the different properties in a triangle:
Angle sum property
Pythagoras Property
Exterior Angle Property
Properties of Equilateral Triangle and Isosceles Triangle
Exercise 15.3 of Properties of Triangle RS Aggarwal Class 7
Class 7 is the important stage; students learn the basics of maths that helps them in the higher classes. Properties of Triangle of class 7 from RS Aggarwal assist you in your journey with plenty of questions to practice. Vedantu provides you with solutions to these questions for free all you need to do is sign in to download it on your devices.
Exercise 15.3 from RS Aggarwal has included a variety of questions that enhance your knowledge and provide you with an in-depth understanding of the subject.
RS Aggarwal helps you understand the concepts more readily; it exposes you to the Higher Order Thinking Skills Type questions. In class 7, if you practice questions from the RS Aggarwal, it sharps your thinking skills and increases your ability to solve complex problems. Vedantu experts provide you with the solutions for unsolved RS Aggarwal questions to help you prepare well and ace your exams.
2. How many questions should I solve from the RS Aggarwal Solutions Class 7 Chapter-15 Properties of Triangles (Ex 15C) Exercise 15?
It is good to solve the maximum number of questions to increase your knowledge about the topic, and it helps you score good marks in the examination. Properties of Triangle is an easy chapter to learn, but it is good to go with the diversity of sums. Vedantu provides you with NCERT Exemplar and RS Aggarwal solutions. You can follow them to match your answers. Join today class 7 online maths tuition classes to get all your doubts cleared.
Triangle has three sides, whereas Parallelograms have four sides. In a parallelogram, opposite sides are parallel to each other, and its diagonal bisects it into two equal parts. Together triangle angles make a sum of 180, but parallelogram interior angles make a sum of 360. Triangle is a plane but not a quadrilateral, and the parallelogram is always a quadrilateral.
You can download the resources from the Vedantu on any of your devices, either a laptop or mobile phone. Experts have designed these resources to help students of class 7 in their preparation for exams. We provide you with more than NCERT textbook solutions; you can download the main topic notes and answers of reference books like RS Aggarwal. To learn with 3D visuals from home, you can join our online classes and clear all your doubts from the expert math teacher.
5. What are the different types of angles involved in triangles mentioned in RS Aggarwal Solutions Class 7 Chapter-15 Properties of Triangles (Ex 15C) Exercise 15?
Triangle has three interior angles that together make a sum of 180 degrees. It is called the angle sum property of a triangle. And, an exterior angle of a triangle is equal to the sum of the two interior opposite angles. These are two different angles involved in a triangle; apart from these, you will learn right-angle triangles and isosceles triangles. An Isosceles has two equal interior angles with equal sides. | 677.169 | 1 |
categories
categories
Amazing Science
Name & Measure Angles in Geometry - Right, Acute, Obtuse Angles - [2]
More Lessons:
Twitter:
In this lesson, you will learn how to name an angle in geometry and measure the angle in degrees. We learn how to classify angles depending on the angle. If the angle is exactly 90 degrees, we say the rays are perpendicular and the angle is called a right angle. If the angle is less than 90 degrees we say it is an acute angle. If the angle is greater than 90 degrees, we say it is an obtuse angle. | 677.169 | 1 |
3.1 Scalar times vector.
Given the vector v, its m-nth multiple w is the sum of m vectors
each equal to v; we can write
w = mv;
conversely, there exists a vector u such that v = nu;
u is the n-nth submultiple of v and we can write
Multiples and submultiples of v have the same direction as v.
y, the m-nth multiple of u can be written as
;
it too has the same direction as v but its length is
with respect to that of v.
Therefore we can define the product of a rational number times a vector
and, more generally, the product of a real number times a vector; in this
context, the real number is said a scalar.
Scalar times a vector is a vector with same direction as the
multiplicand vector and magnitude equal to its magnitude times the scalar.
If we note
the magnitude of v, we have that
is a vector with the same direction as v and magnitude = 1.
When a vector has magnitude=1, it is said a unit vector; often unit vectors are
called versors.
The unit vector
is
therefore said the versor of v.
3.2 Rotation versor.
Given two vectors a and b applied to the same point O,
let α be their plane. If the the least rotation required to superimpose a
to b is counterclockwise, we can represent such rotation as a versor
perpendicular to α and oriented from α to the observer:
this is the rotation versor.
If the rotation is clockwise, the rotation versor is opposite to the former.
3.3 Vector cross product.
Two vectors a and b applied to the same point O determine
a parallelogram with sides equal to the magnitudes of the vectors.
If we multiply the rotation versor of a and b by the area of this parallelogram,
we obtain a vector c said the cross product of a and b.
The cross product c is written as
and sometimes as
The cross product is not commutative. If we change the vectors' order
we obtain the opposite vector:
The cross product of parallel vectors is null.
In particular, the cross product of a vector by itself is null.
The magnitude of the cross product of two given vectors has a maximum when the vectors are
perpendicular.
The area of a parallelogram coincides with the area of a rectangle with
equal base and equal altitude. If we know the lengths a and b
of its sides and an angle θ between them and let a be the base,
the altitude of the rectangle must be b sin θ. The magnitude of the cross product
is therefore
3.4 Vector dot product.
We can also associate to two given vectors a and b, applied to the same
point O, the area of the rectangle in which a side has length given by the magnitude of either
vector, say a, and the other side has length equal to that of the projection
of the second vector, say b, on the line of action of the first vector.
Furthermore, we can consider this area as negative if the projection lies outside the
first vector, positive (or eventually null) otherwise.
The real number so obtained is said the dot product of a and b.
It must be emphasized that the dot product is not a vector, but a real number.
The dot product is commutative.
If two vectors are perpendicular, their dot product is null.
Given two vectors, the magnitude of their dot product has a maximum when they are parallel.
The square of a vector is given by the square of its magnitude.
If we know the magnitudes of a and b and the angle θ
between them, the length of the projection of b on the line of action of a is
given by b cosθ, so the dot product is | 677.169 | 1 |
Question 1.
Fill in the blanks:
(a) Every triangle has at least …….. acute angles.
(b) A triangle in which none of the sides equal is called a ………
(c) In an isosceles triangle ……… angles are equal.
(d) The sum of three angles of a triangle is ……….
(e) A right-angled triangle with two equal sides is called ………
Solution:
(a) two
(b) scalene triangle
(c) two
(d) 180°
(e) isosceles right-angled triangle
Question 13.
If all angles of a triangle are less than a right angle, then it is called ……….
(a) an obtuse-angled triangle
(b) a right-angled triangle
(c) an isosceles right-angled triangle
(d) an acute-angled triangle
Solution:
(d) an acute-angled triangle
Question 14.
If two sides of a triangle are 5 cm and 9 cm then the third side is _____.
(a) 5 cm
(b) 3 cm
(c) 4 cm
(d) 14 cm
Solution:
(a) 5 cm | 677.169 | 1 |
Perimeter & Area of Polygons on Coordinate Plane
Choose a Course
Choose a Grade
Perimeter & Area of Polygons on Coordinate Plane
Finding the perimeter & area of a polygon on a coordinate plane.
Mapped to CCSS Section# 6.G.A.3
Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side joining points with the same first coordinate or the same second coordinate. Apply these techniques in the context of solving real-world and mathematical problems | 677.169 | 1 |
Wednesday 3 February 2010
Triangle Area Formulas, Old and New
Heron of Alexandria, sometimes called Hero, lived around the year 100 AD and is most often remembered for a formula for the area of a triangle. The formula gives a method of computing the area from the lengths of the three sides. If we call the sides a, b, and c; then the area is given by where the "s" stands for the semi-perimeter, . You can find a nice geometric proof of Heron's formula at this link to the Dr. Math site. The proof was done by Dr. Floor, who credits the method to Paul Yiu of Florida Atlantic University.
Documents from the Arabic writers indicate Archimedes may well have known this formula 300 years before Heron. In 1896 a copy of Heron's Metrica was recovered in Constantinople (now Istanbul) that had been copied around 1100 AD. It contains the oldest known demonstration of the formula. Heron is also remembered for his invention of a primitive steam engine and one of the earliest forerunners of the thermometer. The image at right shows a picture of a reconstruction of Heron's steam engine. The image is from the Smith College meuseum of Ancient Inventions where you can find more about Heron's, and many other's, interesting creations. An extension of Heron's area formula for cyclic quadrilaterals is known as Brahmagupta's Formula
Heron's Metrica also contains one of the earliest examples of a method of finding square roots that is called the divide and average model. To find an approximate square root of a number, N, think of any number smaller than N, which we will call M. Then find a new approximation by letting E = (M + N/M)/2. Another approximation can be found by repeating the method with this new approximation. For example, beginning with N=20 and M= 2, we get E= (2 + 20/2) / 2 or E= (2+10)/2 = 6. Repeating with M= 6 we get E= (6+ 20/6)/2 = ( 6 + 3 1/3 )/2 = 14/3 or 4 2/3. After only two iterations from a very bad starting guess the approximation is within .2 of the correct value.
Heron is also remembered for a problem he solved in Catoprica; Given two points, A and B, on the same side of a line, find a point X on the line so that the total distance AX+XB is a minimum. The solution may come quickly if you know that the translation of Catoprica is "About Mirrors". The solution given by Heron is to find the mirror reflection of point B in the line, B', and draw a straight line from A to B'. Where it intersects the line is the choice of point X.
Ok, so much for the old news... but recently I was going through some old journals that Dave Refro sends me from time to time to keep me out of mischief, and I came across an article in the 1885 Annals of Mathematics which listed 105 different formulas for the area of a triangle ( things to do on a rainy afternoon, list 110 different formulae for the area of a triangle). One was the well known Heron's formula above and then there was another that looked strikingly similar. If we let MA be the length of the median to vertex A, and similarly for MB and MC . The we can call sigma 1/2 the sum of MA + MBa+MC. Then we can write. | 677.169 | 1 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.