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Theory on Ellipses
Questions from Xuan, Drawing Academy student
You mention that the ellipse (representing a circle in perspective) is distorted, with its front portion larger than the back portion, so that it becomes an oval with only one axis of symmetry.
I think, a circle always appears as an ellipse from a linear perspective. While it is true that the closer half of the circle is larger in perspective, the resulting shape that we see is still a perfect ellipse. However, the horizontal line that bisects the circle does not intersect the ellipse at its widest point.
I have made an image to illustrate this:
I modelled a cylinder in perspective using a 3D modelling tool and then overlaid the ellipse outline and its major axis.
As you can see, the darker half of the circular face of the cylinder indeed takes up larger than half the area. However, the 2D projection of the circle in perspective is still an ellipse and the major axis of the ellipse is not the same line as the line that bisects the circular face.
Can you please give your point of view on this subject?
Best regards,
Xuan
Feedback from Vladimir London, Drawing Academy tutor
Dear Xuan,
Many thanks for your question and drawing.
I am very glad that you are thoughtfully reviewing the topics, presented in the drawing course, with an analytical approach.
You are correct, perspective is not a straight-forward subject.
The main challenge is that it is impossible to represent three-dimensional objects, drawing on a flat surface, without distortion. The best method is to use a type of perspective that will minimize distortions.
Sometimes, artists use "incorrect" proportions or geometry to give a viewer a natural feel of objects instead of choosing "geometrically perfect" solutions.
In this painting, Paul Cézanne depicted mountains in the background proportionally larger than in real life. This approach allows the human brain to process the image in a more balanced manner.
You might have such an experience when photographing far-away objects without a zoom function. You may believe that objects in the background would appear larger than they actually do on a photograph.
The same applies to a circle in perspective. Sometimes, you need to step away from a geometrical approach as it would not solve the challenge of drawing without distortion.
It is better to create an "illusion" of a realistic-looking object than to make it geometrically perfect, but looking less natural.
In regard to the 3-D modelling tool, you need to remember that it is done by software calculations. The program has the same difficulties in depicting the real three-dimensional world on the flat surface of a screen.
Developers of the tool used approximations (this is the answer to your question regarding the way 3D and 2D ovals overlap perfectly).
It is great that you ask such questions as it shows you are thinking in a similar manner as great artists of the past.
To your creative success,
Vladimir
Response from Xuan, Drawing Academy student
Dear Vladimir,
Thank you for your kind words and your quick response.
I found a better illustration of what I was trying to say in my previous email.
The widest part of the 2D ellipse does not correspond to the widest part of the circle. In geometric terms, the tangent line where the ellipse intersects the quadrilateral should not be vertical, but should always follow the line that leads to the vanishing point.
You are absolutely correct that there will always be distortion when transferring from 3D to 2D, and it becomes apparent when using a camera.
Anyway, if we draw objects in linear perspective (which is different from how a camera would view them) because it appears to be realistic to our minds, then does it not mean we should follow the rules of linear perspective (e.g., circles becoming ellipses in perspective) to make the objects look believable?
I tested what would happen if I did not follow the rules of linear perspective, but instead used a camera to incorporate some distortion (I am an engineer, after all =]). The result was still an almost perfect ellipse.
This tells me that we should depict circles, in perspective, as ellipses (even if we are approximating) to create the most "natural" look in a picture.
Thank you for helping me think so deep about perspective.
Best regards,
Xuan
Feedback from Vladimir London, Drawing Academy tutor
Hi Xuan,
Many thanks for your response. I really enjoyed you answer and I am very pleased that you are taking questions of perspective seriously. I understand your point and agree with your views.
Regarding this image:
1. The square would be less distorted if a two-point perspective is used;
2. As the result of one-point orthogonal perspective, the oval is tilted, which would not be the case in real life. See the image below.
When drawing ovals, you can rely on your eye (and you should train your eyes to recognize proportions and dimensions) or you can "calculate" it graphically.
If you want to practice this task, you can use this template of a cube with circles (right-click the link and save this file to your computer). You can print it on A3 size paper, cut out the contour, fold along dashed lines, and glue the planes together. Next, place it in various positions and draw it in perspective.
You should use visual measurements. With time and practice, you will learn how to judge proportions by eye, and when in doubt, you can always double check them by measuring with a pencil.
Do not over-complicate this question in your artwork.
There are just a few main rules that will help you draw ovals:
1. Every circle located horizontally will have its main axis parallel to the horizon line.
2. The further away the circle is from the horizon, the fuller the oval will appear; the closer to the horizon, the slimmer it will become.
3. When a horizontal circle coincides with the horizon line, the oval will become a straight line.
4. A non-horizontal circle (like in a vase lying on its sides) will have its main axis perpendicular to the vase's main axis of symmetry (its central vertical line).
5. An oval will never have pointed edges (or a tuna shape); all sides of an oval will be rounded.
The question of how to realistically depict ovals has bothered fine artists for centuries.
I have seen many paintings and drawings by artists of the past where ovals are depicted with obvious mistakes. | 677.169 | 1 |
what are the properties of a triangle | 677.169 | 1 |
What is the relation between line symmetry and point symmetry?
a plane, point symmetry is symmetry on rotation 180º around the originWhich figure has line symmetry and rotational symmetry?
scalene triangle
A scalene triangle – A scalene triangle has no line symmetry because it has all sides unequal in length.
Which shape has line symmetry rotational symmetry?
rectangleWhat is the relationship between the line of symmetry and the line of reflectionHow do you determine whether a figure has line symmetry or rotational symmetry?
Possible answer: To identify line symmetry, look for a line of reflection that divides the figure into mirror-image halves. To identify rotational symmetry, think of the figure rotating around its center. The figure has rotational symmetry if a rotation of at most 180° produces the original figure.
(d) A trapezium, which has equal non-parallel sides, a quadrilateral with line symmetry but not a rotational symmetry of order more than 1.
What has one line of symmetry but no rotational symmetry?
Solution: An isosceles triangle has only line symmetry and no rotational symmetry.
How do you know if a shape has rotational symmetry?
A shape has rotational symmetry when it can be rotated and it still looks the same. The order of rotational symmetry of a shape is the number of times it can be rotated around a full circle and still look the same.
What is point and Line symmetry?
In a plane, point symmetry is symmetry on rotation 180º around the origin. So the letter "S" has point symmetryThere are two basic kinds of symmetry: line or reflection symmetry, where the shape looks the same on both sides of a mirror line, and rotational symmetry, where you can turn the shape through some angle and it will look the same as before you turned it.
What is a group of symmetry?
In group theory, the symmetry group of an object (image, signal, etc.) is the group of all transformations under which the object is invariant with composition as the group operation.
What is symmetry in math?
Symmetry Math definition states that "symmetry is a mirror image". When an image looks identical to the original image after the shape is being turned or flipped, then it is called symmetry. It exists in patterns. You may have often heard of the term 'symmetry' in day to day life | 677.169 | 1 |
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Take exploration of geometry to the next level with Exploragons® 360°! An EAI exclusive, the new Exploragons 360° circle provides a hands-on way to examine properties of a circle, angle relationships and measurement, inscribed angles and arc measurement, inscribed figures, angle intercepts, semi-circles, sectors, secants, segments, and more!
The Exploragons 360° circle has been designed with 24 nubs around the circumference of the circle and one in the center allowing Exploragons pieces to easily snap to it and measure angles in 15° increments.
Included in the Teacher Set are 20 Quick Check Cards that feature definitions, questions, and short activities which can be handed out to individual students or small groups. The set also includes 15 Dry-Erase Task Cards, great for practice with vocabulary, types of angles, and more. Task Cards can be photocopied for use as blackline masters. | 677.169 | 1 |
Given that ( \sec(\theta) = \frac{1}{\cos(\theta)} ), and the cosine of an angle represents the ratio of the adjacent side to the hypotenuse in a right triangle, consider a right triangle where the adjacent side is ( 2\sqrt{3} ) and the hypotenuse is 3.
Since the square of the length of a side of a triangle cannot be negative, this means that there's no real solution for the length of the opposite side. Therefore, the angle whose secant value is ( \frac{2\sqrt{3}}{3} ) is not within the domain of the inverse secant function for real numbers. Thus, ( \sec^{-1}(\frac{2\sqrt{3}}{3}) ) is undefined, and consequently, ( \sec(\sec^{-1}(\frac{2\sqrt{3}}{3})) ) is also undefined.
Since secant is the reciprocal of cosine, and we know that cosine is adjacent over hypotenuse in a right triangle, construct a right triangle where the adjacent side is 2 and the hypotenuse is 3, so that the cosine of one of the acute angles in this triangle is ((2sqrt3)/3).
Now, use the Pythagorean theorem to find the opposite side of this triangle. Then, use the definition of cosine to find the value of cosine of the angle whose secant is ((2sqrt3)/3).
Finally, evaluate 1/cosine of this angle to find the value of sec(sec^-1((2sqrt3)/ | 677.169 | 1 |
SPM Trial Paper 2021 (Selangor) – Paper 1
Question 15:Diagram 6 shows a triangle ABC. Given that the area of triangle ABC is 21 cm2 and ∠BAC is obtuse angle.Diagram 6Find(a) ∠BAC, [3 marks](b) the length of BC, in cm, [2 marks](c) the length of the perpendicular line from A to BC. [3 marks] Solution:(a) Area of Δ= 1 2 absinC 21= 1 2 ( … Read more
Question 12:(a) Find the number of ways to arrange each word SAKU and SAKA if no repetition is allowed. Are the number of arrangement that formed are the same?Explain. [3 marks](b) Two boats are used to cross a river with each boat only able to carry 7 passengers. There are 9 adults and 4 children … Read more
Question 10:(a) Given the gradient of the tangent to the curve y = x2 (3 + px) is -3 when x = -1.Find the value of p. [2 marks](b) Volume, V cm3, of a solid is given by V=8π r 2 + 2 3 π r 3 , r is the radius. Find the approximate … Read more
Question 5:(a) Given a geometric progression with terms a, ar, ar2, ar3, … , arn-2, arn-1 and the sum of first n terms is Sn. Derive the formula for the sum of the first n terms, Sn for the geometric progression when |r| < 1. [3 marks](b) Hence, find the value of n for which the … Read more | 677.169 | 1 |
4th Grade Math - Types of Triangles
apply knowledge of right angles to identify acute, right, and obtuse triangles
Instruction
Students learn that triangles can be classified based on the angles within them.
STAAR Practice
Between 2016 and 2023 (including redesign practice), | 677.169 | 1 |
2-Dimensional Shapes
2-Dimensional Shapes for Class 3 Math
This learning concept will help the students to recall the 2-dimensional shapes of geometry. Also, the students will get to know about open and closed figures. In this learning concept, students will learn to
Identify the open figure and closed figure
Classify circle, square, triangle, rectangle
Define triangle, circle, square, rectangle
The learning concept is explained to class 3 students with examples, illustrations, and a concept map. At the end of the page, two printable worksheets for class 3 with solutions are attached for the students.
Download the worksheets of 2-dimensional shapes and solutions to assess our knowledge of the concept.
What Is a Two-Dimensional Shape?
In geometry, 2-D shapes are completely flat and have only two dimensions – length and width. They do not have any thickness and can be measured only by the two dimensions. There are different types of 2-D shapes, which are open figures and closed figures.
Open and Closed Figures
Open Figure:
Open figures do not begin and end at the same point.
Closed Figure
Closed figures begin and end and the same point.
In our surroundings, we could see some different types of closed figures commonly. Here we discuss some of these closed figures.
These are triangles, squares, rectangles and circles.
What Is Triangle?
A triangle is a 2D shape with three sides and three corners.
Traffic signs, pizza, sandwiches etc are in the shape of a triangle.
What Is Rectangle?
A rectangle is a 2D shape with four sides and four corners.
A blackboard, door, brick etc are in the shape of a rectangle.
What Is Square?
A square is a 2-D shape with four sides and four corners. All sides of squares are equal.
The chessboard, bread etc are in the shape of a square.
What Is Circle?
A circle is a closed 2-D shape made up of a curved line with no sides and no corners. A circle has various parts like centre, radius, diameter, circumference, and so on. | 677.169 | 1 |
Consider a triangle with vertices A, B and C. Call the edge
opposite a given vertex by the same letter, but lower case. So side
a is opposite vertex A etc.
Law of Sines says:
SinA/a= SinB/b=SinC/c
If you prefer, you can split the equation into multiple separate
ones:
SinA/a=SinB/b
Sin A/a=SinC/c etc.
(there is one more part of the law of Sines which most books
leave out. If R is the radius of a circumcircle around triangle
ABC, then SinA/a= SinB/b=SinC/c =2R and in case you forgot a
circumcirlce of a triangle is a unique circle that passes through
each of the triangle 3 vertices.) | 677.169 | 1 |
Pentagram
You are encouraged to solve this task according to the task description, using any language you may know.
A pentagram is a star polygon, consisting of a central pentagon of which each side forms the base of an isosceles triangle. The vertex of each triangle, a point of the star, is 36 degrees.
Task
Draw (or print) a regular pentagram, in any orientation. Use a different color (or token) for stroke and fill, and background. For the fill it should be assumed that all points inside the triangles and the pentagon are inside the pentagram.
This uses the Diagrams library to create an SVG drawing. Compiling, then running it like:
pentagram -w 400 -o pentagram_hs.svg
creates a 400x400 SVG file.
-- Extract the vertices of a pentagon, re-ordering them so that drawing lines-- from one to the next forms a pentagram. Set the line's thickness and its-- colour, as well as the fill and background colours. Make the background a-- bit larger than the pentagram.importDiagrams.PreludeimportDiagrams.Backend.SVG.CmdLinepentagram=let[a,b,c,d,e]=trailVertices$pentagon1in[a,c,e,b,d]#fromVertices#closeTrail#strokeTrail#lwultraThick#fcspringgreen#lcblue#bgFrame0.2bisquemain=mainWith(pentagram::DiagramB)
usingLuxorfunctiondrawpentagram(path::AbstractString,w::Integer=1000,h::Integer=1000)Drawing(h,w,path)origin()setline(16)# To get a different color border from the fill, draw twice, first with fill, then without.sethue("aqua")star(0,0,500,5,0.39,3pi/10,:fill)sethue("navy")verts=star(0,0,500,5,0.5,3pi/10,vertices=true)poly([verts[i]foriin[1,5,9,3,7,1]],:stroke)finish()preview()enddrawpentagram("data/pentagram.png")
// version 1.1.2importjava.awt.*importjava.awt.geom.Path2Dimportjavax.swing.*classPentagram:JPanel(){init{preferredSize=Dimension(640,640)background=Color.white}privatefundrawPentagram(g:Graphics2D,len:Int,x:Int,y:Int,fill:Color,stroke:Color){varx2=x.toDouble()vary2=y.toDouble()varangle=0.0valp=Path2D.Float()p.moveTo(x2,y2)for(iin0..4){x2+=Math.cos(angle)*leny2+=Math.sin(-angle)*lenp.lineTo(x2,y2)angle-=Math.toRadians(144.0)}p.closePath()with(g){color=fillfill(p)color=strokedraw(p)}}overridefunpaintComponent(gg:Graphics){super.paintComponent(gg)valg=ggasGraphics2Dg.setRenderingHint(RenderingHints.KEY_ANTIALIASING,RenderingHints.VALUE_ANTIALIAS_ON)g.stroke=BasicStroke(5.0f,BasicStroke.CAP_ROUND,0)drawPentagram(g,500,70,250,Color(0x6495ED),Color.darkGray)}}funmain(args:Array<String>){SwingUtilities.invokeLater{valf=JFrame()with(f){defaultCloseOperation=JFrame.EXIT_ON_CLOSEtitle="Pentagram"isResizable=falseadd(Pentagram(),BorderLayout.CENTER)pack()setLocationRelativeTo(null)isVisible=true}}}
The following isn't exactly what the task asks for, but it's kind of fun if you have a PS interpreter that progressively updates. The program draws a lot of stars, so it's extremely likely that some of them are pentagrams...
A better more accurate approach utilising equation of a circle using polar coordinates.[1]
5 points are required to draw a pentagram. a circle with centre coordinates x=10 and y=10 with radius 10 was chosen for this example.
In order to find 5 equal points circle needs to be divided by 5 i.e 360/5 = 72
each point on the circumference is 72 degrees apart,
5 points on the circles circumference are calculated and than plotted and line drawn in-between to produce pentagram
#Circle equation#x = r*cos(angle) + centre_x#y = r*sin(angle) + centre_y#centre pointscentre_x=10centre_y=10#radiusr=10deg2rad<-function(d){return((d*pi)/180)}#Converts Degrees to radiansX_coord<-function(r=10,centre_x=10,angle)#Finds Xcoordinate on the circumference {return(r*cos(deg2rad(angle))+centre_x)}Y_coord<-function(r=10,centre_y=10,angle)#Finds Ycoordinate on the circumference {return(r*sin(deg2rad(angle))+centre_x)}# series of angles after dividing the circle in to 5 angles<-list()for(iin1:5){angles[i]<-72*i}angles<-unlist(angles)#flattening the list for(iinseq_along(angles)){print(i)print(angles[i])if(i==1){coordinates<-cbind(c(x=X_coord(angle=angles[i]),y=Y_coord(angle=angles[i])))}else{coordinates<-cbind(coordinates,cbind(c(x=X_coord(angle=angles[i]),y=Y_coord(angle=angles[i]))))}}plot(xlim=c(0,30),ylim=c(0,30),x=coordinates[1,],y=coordinates[2,])polygon(x=coordinates[1,c(1,3,5,2,4,1)],y=coordinates[2,c(1,3,5,2,4,1)],col="#1b98e0",border="red",lwd=5)
packagerequireTk8.6;# lmap is new in Tcl/Tk 8.6setpi[expr4*atan(1)]pack[canvas.c]-expandyes-fillboth;# create the canvasupdate;# draw everything so the dimensions are accuratesetw[winfowidth.c];# calculate appropriate dimensionsseth[winfoheight.c]setr[expr{min($w,$h)*0.45}]setpoints[lmapn{012345}{setn[expr{$n*2}]sety[expr{sin($pi*2*$n/5)*$r+$h/2}]setx[expr{cos($pi*2*$n/5)*$r+$w/2}]list$x$y}]setpoints[concat{*}$points];# flatten the listputs[.ccreateline$points];# a fun reader exercise is to make the shape respond to mouse events,;# or animate it! | 677.169 | 1 |
Course Extras: Jacob's Geometry
This page is designed to help provide support for Kate's Jacob's Geometry eCourse, a video supplement that walk students through Jacob's Geometry in an engaging, visual/auditory way! View samples on MasterBooksAcademy.com.
Helpful Resources
Free Online Graphing Calculators – These calculators can be used instead of a graphing calculator for this course. (It is recommended that college-bound students get a graphing calculator, however, so they are already familiar with using one.)
Desmos (suggested, due to ease of identifying values and zooming in and out) – Note that you have to press the keyboard symbol in the bottom left to get a keyboard. Press y = and the values y equals, using x to stand for the independent variable. Use ^ to show exponents (use your right arrow to get back from an exponent to a regular value). Click on the graph of the curve itself to see the value at any particular point. Zoom in or out using the plus or minus signs. Add a second equation by clicking underneath where the first is typed and typing another one.
Math Is Fun – Use ^ to show exponents; use the first line for inputting the first equation, and the second for the second.
Corrections/Clarifications
As I get questions about problems or discover errors, I will post clarifications or corrections here. If your question isn't here and you're taking the eCourse, please email me so I can get back with you on it (and add it here for others to learn from too).
Students may think the answer is planes, and all the polygons are indeed in different planes. However, in geometry, it's really important to read very carefully and give precise answers, as we then build on those answers. The question was what kind of geometric figure. A plane is not a figure–if you look at page 9, you'll see it make a point about how while it's pictured as having edges, it really has no boundaries. So while the sides are in different planes, the figures they are are polygons. Make sure the understands that a plane doesn't have boundaries.
Chapter 1, Summary and Review, Problem 9
Students may wonder why RO is being counted as an edge; the key here is in reading very carefully. The problem explains that the pattern can be used to form a cube. So in answering the question, it's looking it at not as a flat figure (in which case RO wouldn't be an edge) but as a cube (which you'd get if you folded it up).
Chapter 2, Lesson 4, General Note Solving problems using the thinking of indirect proofs can be tricky at first. Stress understanding the logic of it, not writing it out as a formal proof.
Notice in the example on page 56 that the assumption made is the opposite of a possible conclusion (think answer)–not something that immediately contradicts information given. Think of the conclusions as the possible answers; you need to assume the opposite of what you think the answer is going to be. The whole idea in an indirect proof is to assume a possible answer and see if it is right by seeing if you get a contradiction; the possible answer without a contradiction is the right answer.
Chapter 2, Algebra Review, Problem 26
Students who start by dividing both sides by x may think the problem is unsolvable, getting x + 7 = x. Dividing both sides by x works fine provided the unknown does not equal 0. Since we can't divide by 0, it wouldn't work to divide both sides by x if x = 0. Typically math books operate with the assumption that the unknown is not 0 and allow for dividing by an unknown. Here, though, was one of those rare exceptions where they made x equal to 0. If you ever come across an unsolvable result or a different answer than the Solutions Manual after dividing by an unknown, try solving another way (or plugging in 0 for the unknown) to see if perhaps the unknown is 0.
Chapter 3, Lesson 2, Problems 38-40
You may be tempted to write AC = BD as the answer to 38. Although that is a true statement, it's not true because A-B-C. When the book is asking because A-B-C, it's asking you to think through what you can conclude based on that information alone. A-B-C is a way of writing that B is between A and C. On page 86, we learned If A-B-C, then AB + BC = AC. So that's why we can conclude that AC = AB + BC because A-B-C. The same thing for problem 39–we need to write down what we can conclude because B-C-D. Make sure you're taking notes on what you learn so you can see what you can conclude because of something else. Geometry isn't just about listing a true statement as is the case in other math courses–it's about showing logically how that statement was arrived at, thus teaching critical thinking and deductive reasoning.
On #40, the question is why can we say that AB + BC equals BC + CD. Look closely at what happened and the properties on page 79. In the addition property, we're concluding that if a = b, then a+ c = b + c. In other words, we have 2 equal quantities, and adding an equal quantity to both sides will not change that equality. It's just a fancy way of stating what you've been doing in algebra already. But in the case of AB + BC = BC + CD, we're starting with the answers to 38 and 39: AC = AB + BC and BD = BC + CD. We were told at the beginning that AC = BD. So we'd use substitution to substitute AB + BC for AC and BC + CD for BD.
Note that throughout this course, there are a lot of questions that build on the previous one. The curriculum is trying to break down thinking through logic step by step. But if you struggle on one part, you may end up getting the others wrong too. That's why I don't have the exercises count in grading–it's assumed students will get a lot of the exercises wrong as they get a handle on the material. The goal is to learn so by the tests you've got it.
Chapter 3, Lesson 7, Problem 12
Don't look at the lines that are next to each other, as they are not perpendicular. The picture here shows the 5 sets of perpendicular lines. (The green , pink ones, red ones, penciled ones, and blue ones are all perpendicular.)
Chapter 3, Summary and Review, Problem 19 Notice in the figure where I've marked all the information given (which is the first step to solving the problem). OB's coordinate would be found by taking AOD minus BOD, not minus AOC. That gives 110-80, which equals 30.
Chapter 4, Lesson 6, Problem 35
Note that this is a tough problem, so I'm giving a hint here.
This problem is based on several that come before it. In order to reach these conclusions, as you go through the preceding problems, mark the figure with whatever information you learn at each step. So when you learn certain triangles are congruent, mark their corresponding angles and sides as congruent. Using different colors will help in long problems like this where the tick marks could get to be way too many otherwise. (Rather than using 2 tick marks for the second pair of angles you mark, just use a different color pencil.)
Now look at your notes (which are super important to be taking!) and go through everything you know about triangles, seeing what you can tell about these triangles. You can see in the markings that there are two angles in DEB that are green (so the same) and two in AEC that are red (and thus the same). Looking at your notes, what can you then conclude?
Chapter 4, Lesson 6, Problem 46
The image shows how AB was found using the Pythagorean theorem. The other measurements were obtained the same way.
Chapter 5, Lesson 2, Problem 47
If you are wondering why angle APC is greater than angle AXC, it's because in problem 46 he extended line AP, essentially drawing AX and forming 3 triangles. Angle APC is an exterior angle to triangle PXC, which means it's greater than both the interior angles, one of which is angle PXC. Angle PXC is an exterior angle to triangle AXB, which means that it is greater than both the interior angles in that triangle, one of which is angle B. By the transitive property, then, Angle APC is greater than angle B.
Chapter 5, Lesson 4, Problem 32
This is an alternate valid proof from a student (way to go, Isaac!): After the first three steps, you can say that AB + BC + CD + AC is greater than AC + AD using the Addition Theorem of Inequality. Then use subtraction (of AC) to prove that AB + BC + CD is greater than AD.
Students who list linear pair as the answer are correct; however, that answer doesn't help us prove what we're trying to prove in the next few problems. The questions are building to try to a conclusion in problem 12 that can only be established if we know that they are supplementary–that they add to 180 degrees–because one way we know that lines are parallel is that supplementary interior angles on the same side of a transversal (problem 13).
The Solutions Manual is using whichever definition is used in the next point it's making–supplementary is used as part of the way to know lines are parallel in this case, so it's being used.
In a proof, writing a linear pair here would be incorrect, as just knowing it's a linear pair does not automatically lead to concluding the lines are parallel–you'd have to first say that a linear pair forms supplementary angles. The whole idea in geometry is to build step by step. While all linear pairs have supplementary angles, the definition of linear pair emphasizes the idea of a common side and opposite rays–that definition would be used in proofs where that's what we need to show or where we plan on using another theorem or definition that uses linear pair. We'd use supplementary angle when trying to describe the angle sizes or where we plan on using another theorem or definition that uses supplementary angles, as was the case in this problem. While a linear pair answer can be counted as correct here, if you saw it on a test, in the future, when you get to a follow on question (like question 13) and see what the book was trying to prove, go back to your answer and adjust to supplementary.
Test 6, Problem 7c, Clarification
In general, the Solution Manual is looking for the solution that uses the information given in the preceding steps to prove the problem–it's as if you're writing out the steps of a proof, so your answer should be a logical conclusion of the previous steps alone. What was already shown in steps a and b would lead to proving that the triangles congruent based on ASA; now, if we added other steps, we could have used a different way to prove it, but the idea is to view the question as showing the next step in a proof (so it needs to build on the previous statements). Note that you may find it helpful on problems like this to redraw the two triangles seperately, marking them with the information given in the problem.
Chapter 7, Lesson 5, Problems 46-51
To those who answered "ABCD is a trapezoid" for number 50, here's a clarification. In an indirect proof, you assume the opposite of what you're trying to prove (that's your hypothesis) and look for a contradiction that shows that can't be true. Here, in problem 46, we're supposing that AC and DB bisect each other, which is the opposite of what we're trying to prove. That's our hypothesis. So while it is true that the answer to 49 contradicts what we were given that ABCD is a trapezoid, it also means that the hypothesis we assumed in 46 is incorrect, which is what we care about in this case. In an indirect proof, we're assuming the opposite of what we're trying to prove and then looking to prove that can't be the case. We've shown that ABCD is not a trapezoid in problem 49; now, we're stating that means that the hypothesis we had is false…which means that the opposite of it must be true.
Midterm Review, Problem 97 97 should read as follows: "Could this triangle have sides of lengths a^2, b^2, and c^2? Explain." Since we know that a^2 + b^2 = c^2 for this triangle, and we know the sum of any 2 sides has to be greater than the 3rd, we can't have sides that are a^2, b^2 and c^2, as two sides would then equal the third. In other words, we can't make a = a^2, b = b^2, and c = c^2. Chapter 9, Lesson 2, Problem 6The answer is really square units, not just units. Chapter 9, Lesson 4, ScheduleThe schedule should say problems 1-53 on pages 360-364.
Chapter 9, Summary & Review, Problem 45
The problem should really include that the altitudes for triangles BDE and CFD as well and say that they are equal to BH and EG. If we knew that, then the Solution Manual's steps would explain the answer.
Chapter 10, Lesson 1, Problem 47
Seeing the geometric mean is tricky when you're dealing with fractions. However, since we know when simplified that F/G = 9/8 and G/A = 9/8, we know that F/G = G/A…which would mean that G is the geometric mean. We're NOT saying that 9 or 8 is the geometric mean–the geometric mean is G, which is 2/3. We were able to figure out that it's the geometric mean by seeing that the ratios formed a proportion, and since G was in the denominator of one and numerator of the other of two equal ratios, it's the geometric mean. Note that if we were to insert the original values for F/G = G/A, we'd have 3/4/2/3 = 2/3/16/27. You can see then that 2/3 is the geometric mean. But we had to simplify the fractions down to 9/8 to see that they were indeed equal fractions.
Chapter 10, Lesson 2, Number 44
In problem 41, we set up a ratio between the sides of AO and the sides of A1, getting 1188/x = x/594 and solving the proportion in 42 to see that x was 840 mm. We could set up that proportion because we told that the sizes were similar.
We find the area of a rectangle by multiplying length times width, so we multiplied 1188 by 840, the value we found for x. We get 1188 x 840 = 997,920 mm2. That's the answer in square milimeters, but the question asked for it in square meters. 1 meter = 1,000 milimeters, so we have to convert. 997,920 mm2 x 1 m/1,000 mm x 1 m/1,000 mm = approximately 1 m2 when you round. Chapter 10, Lesson 4, Number 50 Alternate Proof for Students Who Already Know About SSS Theorem from Other Material (Thank you, Isaac!)
These particular questions require thinking about a problem a little differently. You don't know actual dimensions of the triangles, but you know they're all congruent. So you that triangle 1 is congruent to triangle 2 and 4. You know that corresponding parts of congruent triangles are congruent, so BC equals the corresponding line segment in triangle 2 and 4…and those segments added together equal DE. So that means the ratio between BC and DE, or BD/DE is going to be 1/2–1 segment divided by 2 of those same segments.
Again, you don't know the actual dimensions. But whatever they are, you know they'd simplify down to 1/2 because you know from congruent triangles that BC is half of DE.
Problem 2 uses your answer to problem 1 and Theorem 48. That theorem says the ratio of the areas of two similar polygons (and we know these triangles are similar polygons since all their angles are the same) is equal to the square of the ratio of the corresponding sides (which is what we found in problem 1). So we square the answer to problem 1.
The other problems follow similar logic to those first two.
Chapter 10, Algebra Review, Problems 21-23
On problem 21, The left hand side of the Solutions Manual shows 1) multiplying both sides by Rr1r2 and distributing the multiplication. When you multiply the left side by that, you end up with r1r2; the right side Rr2 + Rr1. 2) Here they've swapped sides of the equation to put R on the left AND factored out an R at the same time. They skipped showing the step of writing Rr2 + Rr1 = r1r2. 3) Divide both sides by r1 + r2.
Problem 22 follows similar steps to those above. Problem 23, they multiplied both sides by vur and distributed the multiplication, continuing from there.
The general idea is that we multiply both sides by a common denominator, which we find by multiplying each of the denominators together (so RR1r2 in problems 21-22). Then we distributed the multiplication and simplify.
This skill was covered on pages 614-617 in Elementary Algebra. Do not stress over this if you used a different course that did not cover this. It is covered in Algebra 2 in more depth. Feel free to skip these problems if you don't have Elementary Algebra to look back at and the above explanation is not sufficient.
Chapter 11, Lesson 6, Problem 63
The answer can be found algebraically. You can take the reciprocal of both sides of an equation without changing the value, as shown in pic 1 below.
If you're wondering why you can take the reciprocal of both sides, it's really the same as multiplying both sides by the denominators and then dividing them both by the numerators, as shown in pic 2 below.
Chapter 12, Lesson 4, Problem 189 This is an explanation of how this case could be proved in a similar way for those curious.
Chapter 12, Lesson 6, Problem 29 (Number 39 Is Similar) This is an explanation of how the answer is obtained. Note that problem 39 follows similar reasoning. Look at the graph drawn in the Solution manual for problem 28. The distance formula is to find the distance between two points–if you look it up in the Index, you'll find it is explained on page 134. Here, we know that the distance between the center point (which is 0,0) and any point on the circle is 5, as we're told the radius is 5. Thus, d in the formula is 5, and we can plug in 0,0 as x1 and y1 (or as x2 and y2–it really wouldn't matter) and x and y as the others since those are the symbols we're using in this case. That leaves us with √[(x – 0)2 + (y – 0)2] = 5 OR √[(0 – x)2 + (0 – y)2] = 5. In either case, if we square both sides, we get that x2 + y2 = 25.
Chapter 13, Lesson 3, Number 14 You could also use SAS to prove the triangles congruent. Note on page 244 that SAS is part of the proof of HL.
Chapter 13, Lesson 3, Number 19 The answer is based on the answer to problem 16, where we saw that the sums of the lengths of the quadrilateral's opposite sides must be equal. The bases are two sides of a trapezoid and the legs the other two sides. We're basically just restating problem 16's conclusion for a trapezoid. Chapter 13, Lesson 3, Numbers 34-36The answer to 35 can also be found using the Reflexive Property with PC and PB and still applying AAS. The answer to 36 is based on the answer to 35. If you look at the triangles shown congruent in 35, PF and PG are corresponding parts of one of the pair of triangles, and PG and PH are corresponding parts of the other pair. Corresponding parts of congruent triangles are congruent. And since both PF and PH equal PG, they must equal each other as well.
Chapter 13, Lesson 4, Number 51 The answer mentions that the altitude is the third altitude. The idea of it being the third altitude is not important to have in the answer. The other two altitudes were already identified in 49, and each triangle has 3 altitudes, so that's why it said the third. It would have been the first if it was the first one drawn. The important part of the answer is that AF is the altitude and an understanding as to how we know that.
Chapter 13, Lesson 5, Set IV (Optional Exercise; Explanation) For this problem, we are told the area of the frame equals the area of the picture, which is 4^2, or 16. We can think of the frame as 4 rectangles with width x and the same length as the length of the picture, which we know is 4 cm, plus the 4 corners, which are going to be squares with sides x. That gives us this: 4(4x + x^2) = 16 We have 4 cm times x, so 4x plus the corner area of x^2, and there are four sides and four corners, so we're multiplying all that by 4. We can then distribute the multiplication to get this: 16x + 4x^2 = 16 Rearrange to be in quadratic form: 4x^2 + 16x – 16 = 0 Factor out a 4: 4(x^2 + 4x – 4) = 0 Solve the quadratic using the quadratic formula to get -2 plus or minus 2 times the square root of 2. The solution left off the minus sign as with geometry, we're looking for the positive answer (we're not going to have a negative length).
Chapter 13, Lesson 6, Problems 1-3
A key to understanding these problems is to remember that on a coordinate graph, unlike in regular geometry, we take into account direction and thus have positive and negative. We consider down and left as negative, and right and up as positive.
The slope of DA is -3/2 because if you start at D and look at the rise/run, you're going down 3 (rise) and to the right 2 (run), getting -3/2 (the 3 is negative since it is in the downward direction). For CB, if you start at C, you're going down 3 and to the right 2, so -3/2 as well.
The product of the slopes of two consecutive sides would be 2/3 x -3/2, which simplifies to -1. We saw in problem 1 the two sides that have a slope of 2/3; the other two (that are consecutive to those) have -3/2, as seen in problem 2.
Problem 3 is asking why it equals -1. It's referring back to a theorem covered on page 463.
Chapter 14, Lesson 3, Problems 34 and 35 Explanation of Answer: We know from 31 that we're dealing with a 30-60 right triangle. We know based on Theorem 51 that PB is r/2 and a is r/2(SQRT3). (Answers to 32 + 33.) 35 asks for the area of the larger triangle (AOB) in terms of r. We find the area by taking 1/2 times the base times the height, as A = 1/2(bh). The height is a, which we saw in 33 that we can write as r/2(SQRT3). The base is 2 times PB, which we can saw in 32 that we can write PB as r/2. Multiplying that by 2 would just yield r, as r/2(2) = r. So plugging in r for the base and r/2(SQRT3) for the height in the area equation gives us A = 1/2(SRT3))r, which equals r^2/4(\sqrt3), which could also be written the way it is in the Solutions Manual as \sqrt3/4(r^2). (Algebraically, they mean the same thing.) 35 is asking for the area of the hexagon, which we find by multiplying the area of triangle ABO by 6. So we'd multiply \qsrt3/4(r^2) times 6 to get the answer given in 35.
Chapter 15, Lesson 7, Number 45 In the Solution Manual, there should be an A instead of a V = and the answer should be mi^2 not mi^3.
Schedule
Day 125 should be Take Test 11 on pages 81-84 (TG). Review test results.
Help! I'm Struggling.
If you're struggling through the first several chapters, take heart! It should get easier. Geometry requires training the brain in a lot of logic. Keep in mind that the problems are designed to be exercises and to stretch you! It is okay if you get some wrong–use the tests as your assessment. Here are few things to make sure you're doing:
Make sure you're taking notes in a way that you can easily find everything you know about a specific topic! More info on this is in the eCourse Downloads under Topical Notes.
Make flashcards. (Whether on Index cards or using a site like Quizlet.com.) Becoming familiar with the definitions/postulates/theorems as you go is really important so you don't get overwhelmed.
Use the answers given in the back of the textbook to help you. Check those problems as you go and use them to help you on the other problems.
Parents, you can give hints on some problems if needed. Avoid having the student look at the Solution Manual when trying to solve unless absolutely necessary. Hints would be more helpful in building their critical thinking skills.
Use colored pencils to mark every piece of information you learn as you learn it. That way, when you're asked to draw a conclusion, you can see visually all you know already. An example is given below from Chapter 4, Lesson 6, problem 35. (Rather than using 2 tick marks for the second pair of angles you mark, just use a different color pencil.)
On true and false questions, remember you're trying to decide if the statement is always true. If it's not always true, then it's a false statement, as you can't make that conclusion all of the time. I know these questions are tough. Don't give up and try to remember you're trying to really think through the definitions and the theorems/postulates related to see if you can make the statements definitively or not. Often we're using geometry to measure things (like objects in outer space) we can't physically measure, so we need to know if we can safely make conclusions or not.
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Chapter 9 Areas of Parallelograms and Triangles Class 9 Maths NCERT Solutions is available on this page that will help you in completing your homework on time and obtain maximum marks in the exams. You can also Download PDF of Chapter 9 Areas of Parallelograms and Triangles NCERT Solutions Class 9 Maths to practice in a better manner. There are variety of concepts given in this chapter 6 Class 9 Maths textbook that will develop your necessary skill to solve more and more questions.
These NCERT Solutions for Class 9 will prove useful guide in making a student confident. These solutions are updated according to the latest NCERT Class 9 Maths textbook. Chapter 9 NCERT Solutions will develop you understanding of the chapter and make a student confident.
Page No: 155 Exercise 9.1 1. Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels. (i) Trapezium ABCD and ΔPDC lie on the same DC and between the same parallel lines AB and DC. (ii) Parallelogram PQRS and trapezium SMNR lie on the same base SR but not between the same parallel lines. (iii) Parallelogram PQRS and ΔRTQ lie on the same base QR and between the same parallel lines QR and PS. (iv) Parallelogram ABCD and ΔPQR do not lie on the same base but between the same parallel lines BC and AD. (v) Quadrilateral ABQD and trapezium APCD lie on the same base AD and between the same parallel lines AD and BQ. (vi) Parallelogram PQRS and parallelogram ABCD do not lie on the same base SR but between the same parallel lines SR and PQ. Page No: 159 Exercise 9.2 1. In Fig. 9.15, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD. Given, AB = CD = 16 cm (Opposite sides of a parallelogram) CF = 10 cm and AE = 8 cm Now, Area of parallelogram = Base × Altitude = CD × AE = AD × CF ⇒ 16 × 8 = AD × 10 ⇒ AD = 128/10 cm ⇒ AD = 12.8 cm 2. If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = 1/2 ar(ABCD). Answer E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD. To Prove, ar (EFGH) = 1/2 ar(ABCD) Construction, H and F are joined. Proof, AD || BC and AD = BC (Opposite sides of a parallelogram) ⇒ 1/2 AD = 1/2 BC Also, AH || BF and and DH || CF ⇒ AH = BF and DH = CF (H and F are mid points) Thus, ABFH and HFCD are parallelograms. Now, ΔEFH and ||gm ABFH lie on the same base FH and between the same parallel lines AB and HF. ∴ area of EFH = 1/2 area of ABFH — (i) also, area of GHF = 1/2 area of HFCD — (ii) Adding (i) and (ii), area of ΔEFH + area of ΔGHF = 1/2 area of ABFH + 1/2 area of HFCD ⇒ area of EFGH = area of ABFH ⇒ ar (EFGH) = 1/2 ar(ABCD) 3. P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar(APB) = ar(BQC). Answer ΔAPB and ||gm ABCD are on the same base AB and between same parallel AB and DC. Therefore, ar(ΔAPB) = 1/2 ar(||gm ABCD) — (i) Similarly, ar(ΔBQC) = 1/2 ar(||gm ABCD) — (ii) From (i) and (ii), we have ar(ΔAPB) = ar(ΔBQC) 4. In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that (i) ar(APB) + ar(PCD) = 1/2 ar(ABCD) (ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD) [Hint : Through P, draw a line parallel to AB.] In a parallelogram, AB || GH (by construction) — (i) Thus, AD || BC ⇒ AG || BH — (ii) From equations (i) and (ii), ABHG is a parallelogram. Now, In ΔAPB and parallelogram ABHG are lying on the same base AB and between the same parallel lines AB and GH. ∴ ar(ΔAPB) = 1/2 ar(ABHG) — (iii) also, In ΔPCD and parallelogram CDGH are lying on the same base CD and between the same parallel lines CD and GH. ∴ ar(ΔPCD) = 1/2 ar(CDGH) — (iv) Adding equations (iii) and (iv), ar(ΔAPB) + ar(ΔPCD) = 1/2 {ar(ABHG) + ar(CDGH)} ⇒ ar(APB) + ar(PCD) = 1/2 ar(ABCD) (ii) A line EF is drawn parallel to AD passing through P. In a parallelogram, AD || EF (by construction) — (i) Thus, AB || CD ⇒ AE || DF — (ii) From equations (i) and (ii), AEDF is a parallelogram. Now, In ΔAPD and parallelogram AEFD are lying on the same base AD and between the same parallel lines AD and EF. ∴ ar(ΔAPD) = 1/2 ar(AEFD) — (iii) also, In ΔPBC and parallelogram BCFE are lying on the same base BC and between the same parallel lines BC and EF. ∴ ar(ΔPBC) = 1/2 ar(BCFE) — (iv) Adding equations (iii) and (iv), ar(ΔAPD) + ar(ΔPBC) = 1/2 {ar(AEFD) + ar(BCFE)} ⇒ ar(APD) + ar(PBC) = ar(APB) + ar(PCD) 5. In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that (i) ar (PQRS) = ar (ABRS) (ii) ar (AXS) = 1/2 ar (PQRS) (ii) In ΔAXS and parallelogram ABRS are lying on the same base AS and between the same parallel lines AS and BR. ∴ ar(ΔAXS) = 1/2 ar(ABRS) — (ii) From (i) and (ii), ar(ΔAXS) = 1/2 ar(PQRS) Page No: 106 Answer Area of ΔPSA + ΔPAQ + ΔQAR = Area of PQRS — (i) Area of ΔPAQ = 1/2 area of PQRS — (ii) Triangle and parallelogram on the same base and between the same parallel lines. From (i) and (ii), Area of ΔPSA + Area of ΔQAR = 1/2 area of PQRS — (iii) Clearly from (ii) and (iii), Farmer must sow wheat or pulses in ΔPAQ or either in both ΔPSA and ΔQAR. Page No: 162 Exercise 9.3 1. In Fig.9.23, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar(ACE). Adding (i) and (ii) we get, ar(AOC) + ar(BOC) = ar(AOD) + ar(BOD) ⇒ ar(ABC) = ar(ABD)
Exercise 9.1
1. Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.
Answer
(i) Trapezium ABCD and ΔPDC lie on the same DC and between the same parallel lines AB and DC.
(ii) Parallelogram PQRS and trapezium SMNR lie on the same base SR but not between the same parallel lines.
(iii) Parallelogram PQRS and ΔRTQ lie on the same base QR and between the same parallel lines QR and PS.
(iv) Parallelogram ABCD and ΔPQR do not lie on the same base but between the same parallel lines BC and AD.
(v) Quadrilateral ABQD and trapezium APCD lie on the same base AD and between the same parallel lines AD and BQ.
(vi) Parallelogram PQRS and parallelogram ABCD do not lie on the same base SR but between the same parallel lines SR and PQ.
2. If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that
ar (EFGH) = 1/2 ar(ABCD).
Answer
Given,
E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD.
To Prove,
ar (EFGH) = 1/2 ar(ABCD)
Construction,
H and F are joined.
Proof,
AD || BC and AD = BC (Opposite sides of a parallelogram)
⇒ 1/2 AD = 1/2 BC
Also,
AH || BF and and DH || CF
⇒ AH = BF and DH = CF (H and F are mid points)
Thus, ABFH and HFCD are parallelograms.
Now,
ΔEFH and ||gm ABFH lie on the same base FH and between the same parallel lines AB and HF.
∴ area of EFH = 1/2 area of ABFH — (i)
also, area of GHF = 1/2 area of HFCD — (ii)
Adding (i) and (ii),
area of ΔEFH + area of ΔGHF = 1/2 area of ABFH + 1/2 area of HFCD
⇒ area of EFGH = area of ABFH
⇒ ar (EFGH) = 1/2 ar(ABCD)
3. P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar(APB) = ar(BQC).
Answer
ΔAPB and ||gm ABCD are on the same base AB and between same parallel AB and DC.
Therefore,
ar(ΔAPB) = 1/2 ar(||gm ABCD) — (i)
Similarly,
ar(ΔBQC) = 1/2 ar(||gm ABCD) — (ii)
From (i) and (ii),
we have ar(ΔAPB) = ar(ΔBQC)
4. In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar(APB) + ar(PCD) = 1/2 ar(ABCD)
(ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD)
[Hint : Through P, draw a line parallel to AB.]
Answer
In a parallelogram,
AB || GH (by construction) — (i)
Thus,
AD || BC ⇒ AG || BH — (ii)
From equations (i) and (ii),
ABHG is a parallelogram.
Now,
In ΔAPB and parallelogram ABHG are lying on the same base AB and between the same parallel lines AB and GH.
∴ ar(ΔAPB) = 1/2 ar(ABHG) — (iii)
also,
In ΔPCD and parallelogram CDGH are lying on the same base CD and between the same parallel lines CD and GH.
∴ ar(ΔPCD) = 1/2 ar(CDGH) — (iv)
Adding equations (iii) and (iv),
ar(ΔAPB) + ar(ΔPCD) = 1/2 {ar(ABHG) + ar(CDGH)}
⇒ ar(APB) + ar(PCD) = 1/2 ar(ABCD)
(ii) A line EF is drawn parallel to AD passing through P.
In a parallelogram,
AD || EF (by construction) — (i)
Thus,
AB || CD ⇒ AE || DF — (ii)
From equations (i) and (ii),
AEDF is a parallelogram.
Now,
In ΔAPD and parallelogram AEFD are lying on the same base AD and between the same parallel lines AD and EF.
∴ ar(ΔAPD) = 1/2 ar(AEFD) — (iii)
also,
In ΔPBC and parallelogram BCFE are lying on the same base BC and between the same parallel lines BC and EF.
∴ ar(ΔPBC) = 1/2 ar(BCFE) — (iv)
Adding equations (iii) and (iv),
ar(ΔAPD) + ar(ΔPBC) = 1/2 {ar(AEFD) + ar(BCFE)}
⇒ ar(APD) + ar(PBC) = ar(APB) + ar(PCD)
5. In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = 1/2 ar (PQRS)
Answer
(i) Parallelogram PQRS and ABRS lie on the same base SR and between the same parallel lines SR and PB.
∴ ar(PQRS) = ar(ABRS) — (i)
(ii) In ΔAXS and parallelogram ABRS are lying on the same base AS and between the same parallel lines AS and BR.
∴ ar(ΔAXS) = 1/2 ar(ABRS) — (ii)
From (i) and (ii),
ar(ΔAXS) = 1/2 ar(PQRS)
Page No: 106
6
Answer
The field is divided into three parts. The three parts are in the shape of triangle. ΔPSA, ΔPAQ and ΔQAR.
Area of ΔPSA + ΔPAQ + ΔQAR = Area of PQRS — (i)
Area of ΔPAQ = 1/2 area of PQRS — (ii)
Triangle and parallelogram on the same base and between the same parallel lines.
From (i) and (ii),
Area of ΔPSA + Area of ΔQAR = 1/2 area of PQRS — (iii)
Clearly from (ii) and (iii),
Farmer must sow wheat or pulses in ΔPAQ or either in both ΔPSA and ΔQAR.
Page No: 162
Exercise 9.3
1. In Fig.9.23, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar(ACE).
Answer
Given,
AD is median of ΔABC. Thus, it will divide ΔABC into two triangles of equal area.
∴ ar(ABD) = ar(ACD) — (i)
also,
ED is the median of ΔABC.
∴ ar(EBD) = ar(ECD) — (ii)
Subtracting (ii) from (i),
ar(ABD) – ar(EBD) = ar(ACD) – ar(ECD)
⇒ ar(ABE) = ar(ACE)
2. In a triangle ABC, E is the mid-point of median AD. Show that ar(BED) = 1/4 ar(ABC).
Answer
ar(BED) = (1/2) × BD × DE As E is the mid-point of AD,
As E is the mid-point of AD,
Thus, AE = DE As AD is the median on side BC of triangle ABC, Thus, BD = DC Therefore,
6. In Fig. 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD.
If AB = CD, then show that:
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint : From D and B, draw perpendiculars to AC.]
Answer
Given,
OB = OD and AB = CD
Construction,
DE ⊥ AC and BF ⊥ AC are drawn.
Proof:
(i) In ΔDOE and ΔBOF,
∠DEO = ∠BFO (Perpendiculars)
∠DOE = ∠BOF (Vertically opposite angles)
OD = OB (Given)
Therefore, ΔDOE ≅ ΔBOF by AAS congruence condition.
Thus, DE = BF (By CPCT) — (i)
also, ar(ΔDOE) = ar(ΔBOF) (Congruent triangles) — (ii)
Now,
In ΔDEC and ΔBFA,
∠DEC = ∠BFA (Perpendiculars)
CD = AB (Given)
DE = BF (From i)
Therefore,ΔDEC ≅ ΔBFA by RHS congruence condition.
Thus, ar(ΔDEC) = ar(ΔBFA) (Congruent triangles) — (iii)
Adding (ii) and (iii),
ar(ΔDOE) + ar(ΔDEC) = ar(ΔBOF) + ar(ΔBFA)
⇒ ar (DOC) = ar (AOB)
(ii) ar(ΔDOC) = ar(ΔAOB)
⇒ ar(ΔDOC) + ar(ΔOCB) = ar(ΔAOB) + ar(ΔOCB) (Adding ar(ΔOCB) to both sides) ⇒ ar(ΔDCB) = ar(ΔACB) (iii) ar(ΔDCB) = ar(ΔACB) If two triangles are having same base and equal areas, these will be between same parallels DA || BC — (iv) For quadrilateral ABCD, one pair of opposite sides are equal (AB = CD) and other pair of opposite sides are parallel. Therefore, ABCD is parallelogram.
7. D and E are points on sides AB and AC respectively of ΔABC such that ar(DBC) = ar(EBC). Prove that DE || BC.
Answer
ΔDBC and ΔEBC are on the same base BC and also having equal areas. Therefore, they will lie between the same parallel lines. Thus, DE || BC.
8. XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that
ar(ΔABE) = ar(ΔAC)
Answer
Given,
XY || BC, BE || AC and CF || AB
To show,
ar(ΔABE) = ar(ΔAC)
Proof:
EY || BC (XY || BC) — (i)
also,
BE∥ CY (BE || AC) — (ii)
From (i) and (ii),
BEYC is a parallelogram. (Both the pairs of opposite sides are parallel.)
Similarly,
BXFC is a parallelogram.
Parallelograms on the same base BC and between the same parallels EF and BC.
⇒ ar(BEYC) = ar(BXFC) (Parallelograms on the same base BC and between the same parallels EF and BC) — (iii)
Also,
△AEB and parallelogram BEYC are on the same base BE and between the same parallels BE and AC.
⇒ ar(△AEB) = 1/2ar(BEYC) — (iv)
Similarly,
△ACF and parallelogram BXFC on the same base CF and between the same parallels CF and AB.
⇒ ar(△ ACF) = 1/2ar(BXFC) — (v)
From (iii), (iv) and (v),
ar(△AEB) = ar(△ACF)
9. The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Fig. 9.26). Show that
ar(ABCD) = ar(PBQR).
[Hint : Join AC and PQ. Now compare ar(ACQ) and ar(APQ).]
Answer
AC and PQ are joined.
ar(△ACQ) = ar(△APQ) (On the same base AQ and between the same parallel lines AQ and CP)
⇒ ar(△ACQ) – ar(△ABQ) = ar(△APQ) – ar(△ABQ)
⇒ ar(△ABC) = ar(△QBP) — (i)
AC and QP are diagonals ABCD and PBQR. Thus,
ar(ABC) = 1/2 ar(ABCD) — (ii)
ar(QBP) = 1/2 ar(PBQR) — (iii)
From (ii) and (ii),
1/2 ar(ABCD) = 1/2 ar(PBQR)
⇒ ar(ABCD) = ar(PBQR)
10. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
Answer
△DAC and △DBC lie on the same base DC and between the same parallels AB and CD.
∴ ar(△DAC) = ar(△DBC)
⇒ ar(△DAC) − ar(△DOC) = ar(△DBC) − ar(△DOC)
⇒ ar(△AOD) = ar(△BOC)
11. In Fig. 9.27, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.
Show that
(i) ar(ACB) = ar(ACF)
(ii) ar(AEDF) = ar(ABCDE)
Answer
(i) △ACB and △ACF lie on the same base AC and between the same parallels AC and BF.
∴ ar(△ACB) = ar(△ ACF)
(ii) ar(△ACB) = ar(△ACF)
⇒ ar(△ACB) + ar(△ACDE) = ar(△ACF) + ar(△ACDE)
⇒ ar(ABCDE) = ar(△AEDF)
Page No: 164
12. A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Answer
Let ABCD be the plot of the land of the shape of a quadrilateral.
Construction,
Diagonal BD is joined. AE is drawn parallel BD. BE is joined which intersected AD at O. △BCE is the shape of the original field and △AOB is the area for constructing health centre. Also, △DEO land joined to the plot.
To prove:
ar(△DEO) = ar(△AOB)
Proof:
△DEB and △DAB lie on the same base BD and between the same parallel lines BD and AE.
ar(△DEB) = ar(△DAB)
⇒ ar(△DEB) – ar△DOB) = ar(△DAB) – ar(△DOB)
⇒ ar(△DEO) = ar(△AOB)
13. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
[Hint : Join CX.]
Answer
Given,
ABCD is a trapezium with AB || DC.
XY || AC
Construction,
CX is joined.
To Prove,
ar(ADX) = ar(ACY)
Proof:
ar(△ADX) = ar(△AXC) — (i) (On the same base AX and between the same parallels AB and CD)
also,
ar(△ AXC)=ar(△ ACY) — (ii) (On the same base AC and between the same parallels XY and AC.)
From (i) and (ii),
ar(△ADX)=ar(△ACY)
14. In Fig.9.28, AP || BQ || CR. Prove that ar(AQC) = ar(PBR).
ar(AQC) = ar(PBR).
Answer
Given,
AP || BQ || CR
To Prove,
ar(AQC) = ar(PBR)
Proof:
ar(△AQB) = ar(△PBQ) — (i) (On the same base BQ and between the same parallels AP and BQ.)
also,
ar(△BQC) = ar(△BQR) — (ii) (On the same base BQ and between the same parallels BQ and CR.)
Adding (i) and (ii),
ar(△AQB) + ar(△BQC) = ar(△PBQ) + ar(△BQR)
⇒ ar(△ AQC) = ar(△ PBR)
15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(AOD) = ar(BOC). Prove that ABCD is a trapezium.
Answer
Given,
ar(△AOD) = ar(△BOC)
To Prove,
ABCD is a trapezium.
Proof:
ar(△AOD) = ar(△BOC)
⇒ ar(△AOD) + ar(△AOB) = ar(△BOC) + ar(△AOB)
⇒ ar(△ADB) = ar(△ACB)
Areas of △ADB and △ACB are equal. Therefore, they must lying between the same parallel lines.
Thus, AB ∥ CD
Therefore, ABCD is a trapezium.
16. In Fig.9.29, ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
Answer
Given,
ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC)
To Prove,
ABCD and DCPR are trapeziums.
Proof:
ar(△BDP) = ar(△ARC)
⇒ ar(△BDP) – ar(△DPC) = ar(△DRC)
⇒ ar(△BDC) = ar(△ADC)
ar(△BDC) = ar(△ADC). Therefore, they must lying between the same parallel lines.
Thus, AB ∥ CD
Therefore, ABCD is a trapezium.
also,
ar(DRC) = ar(DPC). Therefore, they must lying between the same parallel lines.
In the Chapter 9 Areas of Parallelograms and Triangles, we are focused on the relationship between the areas of these geometric figures under the condition when they lie on the same base and between the same parallels.
• Figures on the Same Base and Between the Same Parallels: Two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices (or the vertex) opposite to the common base of each figure lie on a line parallel to the base.
• Parallelograms on the same Base and Between the same Parallels: Parallelograms on the same base and between the same parallels are equal in area.
• Triangles on the same Base and between the same Parallels: Two triangles on the same base (or equal bases) and between the same parallels are equal in area. Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.
There are total 4 exercises in the Chapter 9 Class 9 Maths NCERT Solutions which are very important for examinations purpose. We have prepared exercisewise NCERT Solutions of every questions which can find from the given links.
Exercise 9.1 Chapter Class 9 Maths NCERT Solutions
Exercise 9.2 Chapter Class 9 Maths NCERT Solutions
Exercise 9.3 Chapter Class 9 Maths NCERT Solutions
Exercise 9.4 Chapter Class 9 Maths NCERT Solutions
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You can easily download Chapter 2 Polynomials NCERT Solutions just by visiting this page as it will help you in clearing your doubts just at a click. These NCERT Solutions are updated as per the latest marking pattern of CBSE.
How many exercises in Chapter 9 Areas of Parallelograms and Triangles?
How many exercises in Chapter 9 Areas of Parallelograms and Triangles?
There are total four exercises in the Chapter 9 Class 9 Maths in which the last one ins optional. We have provided accurate and also detailed every step through which one can always clear their doubts.
How can I understand the topics given in Chapter 9 Class 9 Maths?
NCERT Solutions is one of the best way through which one can understand all the topics given in the Chapter 9 Class 9 Maths. Through the help of these solutions, one will be able to solve supplementary books and exemplar questions as well. | 677.169 | 1 |
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Want to find the area of a trapezoid? If you have the length of each base and the height, you can use them to find the area. In this tutorial, you'll see how to identify those values and plug them into the formula for the area of a trapezoid. Then see how to simplify to get your answer!
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Trying to figure out the formula for the area of a trapezoid? You could start by creating a parallelogram out of two trapezoids. Then, use the formula for the area of a parallelogram to figure out the formula for the area of one trapezoid. This tutorial shows you how!
Did you know that you can figure out the formula for the area of a circle by first turning the circle into a parallelogram? It seems a little weird, but it really works! Watch this tutorial to see how it's done! | 677.169 | 1 |
Which statement best shows the difference between a line and a point?
Which statement best shows the difference between a line and a point?
@Mathematics
12 years ago
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OpenStudy (anonymous):
A line and a point cannot be collinear.
A point has no location but a line can lie on a plane.
A point has no dimension but a line has length as a dimension to measure.
A line has width and depth but a point has only length as a dimension to measure.
12 years ago
OpenStudy (hoblos):
i think the 3rd
12 years ago
OpenStudy (anonymous):
A point has no dimension but a line has length as a dimension to measure.
12 years ago
OpenStudy (anonymous):
but technically a line is infinite .. only a segment or ray has a dimension length to measure. | 677.169 | 1 |
This problem has been solved! You'll receive a detailed solution to help you master the concepts.
Solution 1
#### Solution By Steps***Step 1: Understanding the Geometry*** In Euclidean geometry, the sum of the internal angles of a triangle is always 180 degrees.#### Final AnswerThe sum of angles in a triangle is 180 degrees.
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Its volume can be obtained by slicing it into a regular triangular prism and an equilateral square pyramid, and adding their volume subsequently:[2]22+3312a3≈0.669a3.{\displaystyle {\frac {2{\sqrt {2}}+3{\sqrt {3}}}{12}}a^{3}\approx 0.669a^{3}.}
It has three-dimensional symmetry group of the cyclic group C2v{\displaystyle C_{2\mathrm {v} }} of order 4. Its dihedral angle can be calculated by adding the angle of an equilateral square pyramid and a regular triangular prism in the following:[4]
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The dihedral angle of an augmented triangular prism between two adjacent squares is that of a triangular prism between two lateral faces, the interior angle of a triangular prism π/3=60∘{\displaystyle \pi /3=60^{\circ }}.
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Worksheets Class 11 Mathematics Trigonometric Functions
Students should refer to Worksheets Class 11 Mathematics Trigonometric Functions Chapter 3 provided below with important questions and answers. These important questions with solutions for Chapter 3 Trigonometric Functions have been prepared by expert teachers for Class 11 Mathematics based on the expected pattern of questions in the class 11 exams. We have provided Worksheets for Class 11 Mathematics for all chapters on our website. You should carefully learn all the important examinations questions provided below as they will help you to get better marks in your class tests and exams.
Question. A circular wire of radius 3 cm is cut and bent so as to lie along the circumference of a hoop whose radius is 48 cm. The angle in degrees which is subtended at the centre of hoop is (a) 21.5° (b) 23.5° (c) 22.5° (d) 24.5° | 677.169 | 1 |
Elements of Geometry: With Practical Applications ...
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Página 8 ... perimeter of the polygon . ( 17. ) The surfaces of level fields , bounded by straight fences , are polygonal figures . Floors of buildings are polygons , usually having four sides . XIV . The simplest kind of polygon is one having only ...
Página 92 ... figures are those which have the angles of the one equal to the angles of the other , each to each , and the sides about the equal angles proportional . 4. The perimeter , or contour of a figure , 92 ELEMENTS OF GEOMETRY .
Página 93 With Practical Applications ... George Roberts Perkins. 4. The perimeter , or contour of a figure , is the sum of all its sides , or the length of the bounding line . 5. Two magnitudes are said to be identical , when they : are equal in ...
Página 113 ... perimeters of similar polygons are to each other as their homologous sides ; and their areas are to each other ... perimeter of the second polygon , as any one antecedent is to its corresponding consequent , and there- fore as AB | 677.169 | 1 |
TANCET 2014 DS 67: Geometry
DirectionsWhat is the distance from point X to point Z?"
The answer to the question will the measure of the distance between point X and point Z. It is essentially a number followed by a unit of distance - which in this question is cm.
When is the data sufficient?
If we are able to come up with a unique value for the distance between X and Z, the data is sufficient.
If we are either not able to find an answer or if we are not able to find a unique value, the data is NOT sufficient.
What additional information do we have from the question stem?
If point X is directly north of point Y and directly west of point Z. The adjacent figure shows how the 3 points lie in the x-y plane.
Because the point X is directly north of Y and directly west of Z, the three points form the vertices of a right triangle that is right angled at point X.
Statement (1) ALONE
The distance from Y to Z is 20 cm.
The distance from Y to Z provides the length of the hypotenuse of the right triangle. We will not be able to find the distance from X to Z distance from X to Y is equal to half the distance from Y to Z.
Remember: When you are evaluating statement (2) ALONE, please do not recall information that you read in statement (1). Any information from statement (1) should not be used while evaluating statement (2).
We know the ratio of the lengths of the two sides of the right triangle.
We can infer that the distance from X to Z will be \\sqrt { 3 } \\) times the distance from X to Y.
This information is not enough to find the exact distance form X to Z.
We could NOT find the answer to the question with the information given in statement (2).
Statement (2) ALONE is NOT sufficient.
Hence, we can eliminate choice (2).
Answer narrrows down to choice 3 or choice 5.
Combine the Statements
The distance from Y to Z is 20 cm.
+
The distance from X to Y is equal to half the distance from Y to Z.
We can deduce that the distance from X to Y = 10 cm.
We need to find the distance from X to Z. It happens to be the length of one of the perpendicular sides of the right triangle.
We know the measure of the remaining two sides. The hypotenuse measures 20. Side YZ measures 10. | 677.169 | 1 |
Can you explain this in English? I don't understand how the equation defines a segment. (A segment of what?)Correct me if I'm wrong, but the notation is just standard set notation. For example, {x in R | x > 2} would be read as "all x beloning to the set of real numbers such that x is greater than 2."
Similarly, {P | d[sub]T[/sub](P, A) = d[sub]T[/sub](P, B)} would be the set off all points equidistant from two fixed points (A and B, in this case). Similarly, the set {P | dT(P, A) + dT(P, B) = dT(A, B)} would be the set of all points that are colinear with fixed points A and B (ie the sum of the distances from a point to each of the fixed points is the same as the total distance between the two fixed points).
Further examples:
{P | d[sub]T[/sub](P, A) = r} is the set of all points of fixed distance from A (ie a circle of radius r centered at A.)
{P | d[sub]T[/sub](P, A) + d[sub]T[/sub](P, B) = c} is an elipse
As for taxicab geometry, if this kind of thing interests you, there are entire branches of mathematics devoted to the study of non-Euclidean gemetry. I took half a semester of this stuff last year, so if you have any questions, please post and I'll try to answer. Also, if you want to do more research on your own, the formal mathematical name for this kind of thing is a Metric Space.
Hello, It's been 13 years since I learned Taxi. Here is a better distance equation.
D = abs(P1x - p2x) + abs(P1y - p2y)
Or if you had two points (2,1) and (3,4) the distance would be 4. abs(2-3) + abs(1-4). There are four paths from one point to the other.
I just pulled out my "text" book on it. Its inventor is Eugene Krunse. He is (was maybe) a professor at the U of Michigan. (I have his e-mail if you need it).
The purpose of it is really to introduce people to Non-Euclidean Geometry, and was used for me as a warm up to a really hard Sr. Level College Geometry Course.
The circle (or the set of all points equal distance from a single point) turns out to be a square. However the perpendicular bisector (all points on a line perpendicular to another line and equal didtance to two points and the line) turns out really weird. Even weirder if the points on the first line are an odd length away from each other. If memory serves all of the conic shapes are represented. | 677.169 | 1 |
A Treatise on Surveying, Containing the Theory and Practice: To which is ...
When two sides of a right-angled triangle are given, the other side may be found by the following rules, without first finding the angles.
1. When the hypothenuse and one leg are given, to find the other leg.
RULE.
Subtract the square of the given leg from the square of the hypothenuse; the square root of the remainder will be the leg required.* Or by logarithms thus,
To the logarithm of the sum of the hypothenuse and given side, add the logarithm of their difference; half this sum will be the logarithm of the leg required.
2. When the two legs are given to find the hypothenuse.
RULE.
Add together the squares of the two given legs; the square root of the sum will be the hypothenuse.* Or by logarithms thus,
* DEMONSTRATION. The square of the hypothenuse of a right-angled tri angle is equal to the sum of the squares of the sides (47.1). Therefore the truth of the first part of each of the rules, is evident.
h+bxh-b; whence from the nature of logarithms, the latter part of the first rule is evident.
which solved by logarithms will correspond with the latter part of
From twice the logarithm of the perpendicular, subtract the logarithm of the base, and add the corresponding natural number to the base; then, half the sum of the logarithms of this sum, and of the base, will be the logarithm of the hypothenuse.
EXAMPLES.
1. The hypothenuse of a right angled triangle is 272, and the base 232; required the perpendicular.
2. Given the base 186, and the perpendicular 152, to
3. The hypothenuse being given equal 403, and one leg 321; required the other leg. Ans. 243.7.
4. What is the hypothenuse of a right-angled triangle, the base of which is 31.04, and perpendicular 27.2. Ans. 41.27.
The following examples, in which trigonometry is applied to the mensuration of inaccessible distances and heights, will serve to render the student expert in solving the different cases, and also to elucidate its use.
The Application of Plane Trigonometry to the Mensuration of Distances and Heights.
EXAMPLE 1. Fig. 54.
Being on one side of a river and wanting to know the distance to a house on the other side, I measured 500 yards along the side of the river in a right line AB, and found the two angles* between this line and the object to be CAB = 74° 14', and CBA Required the distance between each station and the object.
49° 23'.
Calculation.
The sum of the angles CAB and CBA is 123° 37', which subtracted from 180° leaves the angle ACB = 56° 23. Then by case 1;
The angles may be taken with a surveyor's compass or any other similar
Suppose I want to know the distance between two places A and B, accessible at both ends of the line AB, and that I measured AC 735 yards, and BC= 840; also the angle ACB 55° 40'. What is the distance between A and B?
=
Calculation.
The angle ACB = 55° 40', being subtracted from 180°, leaves 124° 20′; the half of which is 62° 10′. Then by case 3.
= 7° 12′ and we shall have CAB = 69° 22′, and CBA≈ 54° 58′. Then,
Wanting to know the distance between two inaccessible objects A and B, I measured a base line CD = 300 yards: at C the angle BCD was 58° 20′ and ACD 95° 20'; at D the angle CDA was 53° 30′ and CDB 98° 45'.
Calculation.
1. In the triangle ACD, are given the angle ACD → 95° 20′, ADC = 53° 30′, and the side CD
=
300, to find
300, to find BC
2. In the triangle BCD, are given the angle BCD 58° 20′, BDC = 98° 45', and side CD = 761.47.
=
3. In the triangle ACB we have now given the angle ACB ACD - BCD = 37°, the side AC465.98 and BC = 761.47, to find AB
tance required.
479.8 yards, the dis
EXAMPLE 4. Fig. 57.
Being on one side of a river and observing three objects A, B and C stand on the other side, whose distances apart 1 knew to be, AB = 3 miles, AC = 2, and BC= 1.8, I took a station D, in a straight line with the objects A and C, being nearer the former, and found the angle ADB = 17° 47'. Required my distance from each
of the objects.
Construction.
With the three given distances, describe the triangle ABC; bisect BC in F and draw FE perpendicular to it; draw CE making the angle BCE 72° 13′ the complement of the given angle ADB; with E as a centre and distance EC, describe the circle BCD, meeting CA produced in D: then AD, CD and BD will be the distances required.* | 677.169 | 1 |
A right triangle is formed by connecting the 3 coordinates A, B and C. Find the area of the ΔABC, if AB and BC are parallel to the coordinate axes as shown in the figure.
A
10 sq. units
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B
20 sq. units
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C
6 sq. units
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D
12 sq. units
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Open in App
Solution
The correct option is C
6 sq. units
Since BC is parallel to X-axis, ordinate of point B is 1.
AB is parallel to Y-axis, so its abscissa is 1.
The point B is (1, 1).
Length of AB is difference of ordinates since it is parallel to Y-axis = 4 - 1 = 3 units
Length of BC is difference of abscissa since it is parallel to X-axis = 5 - 1 = 4 units AreaofΔABCis12×AB×BC=12×3×4sq.units=6sq.units. | 677.169 | 1 |
point O is the center of the circle and OC = AC =
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14 Dec 2013, 17:41
4
Kudos
3
Bookmarks
In the figure attached, point O is the center of the circle and OC = AC = AB. What is the value of x (in degrees)?
OC = AC = AB tells us that triangle OAC and BAC are isosceles. Also, as with any triangle on a straight line with an exterior angle, the exterior angle (or in this case, angle ACB) can be found by subtracting the interior angle from 180.
We know that in triangle OAC (because it is an isosceles triangle) that angle o and angle a are equal to one another. Therefore, angle c = 180 - x - x --> angle c = 180-2x. Angle C (of the smaller isosceles triangle) also happens to share the same angle measurement as angle B. Furthermore, because OA and OB are radii, we know that they equal one another and that angle A = angle B.
We know that angle c in the smaller isosceles triangle = 180 - the measure of the obtuse angle C. As shown above the obtuse angle C = (180-2x) So, the small angle C = 180 - (180-2x) --> angle C = 2x. So, angle C = B = 2x. If angle A = B then A = 2x and becaause the obtuse triangle has two x measurements, we know that the measure of A in the small isosceles triangle = x. Therefore, in the small isosceles triangle we have 2x+2x+x = 180. 5x = 180. x = 36.Re: In the figure above, point O is the center of the circle and OC = AC =
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27 May 2014, 16:15
VeritasPrepKarishma wrote:
russ9 wrote:
Hi Karishma,Hi Karishma,
This is news to me -- you can't assume that because length AC is shared, that implies that the corresponding angle in Triangle ACO will equal the corresponding angle in ACB?
Re: In the figure above, point O is the center of the circle and OC = AC =
[#permalink]
27 May 2014, 21:18
Expert Reply
russ9 wrote:
Hi Karishma,
This is news to me -- you can't assume that because length AC is shared, that implies that the corresponding angle in Triangle ACO will equal the corresponding angle in ACB?
Hmm -- learn something new everyday! Thanks!
Corresponding angles are equal only when you have parallel lines with a common transversal. Make two triangles with a common side. Can you make the angles made on the common side such that the angles have very different measures? Sure!Re: In the figure above, point O is the center of the circle and OC = AC =
[#permalink]
29 Dec 2014, 20:07
Expert Reply center of the circle and OC = AC =
[#permalink]
29 Dec 2014, 20:38 center of the circle and OC = AC =
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29 Dec 2014, 20:41
Expert Reply
GMAT01 wrote:Doing it your way: x+ x + (180 - 2y)+ y = 180 2x = y
Thank you.
I must have overlooked something because I get:
x+x+(180-2y)+y= 180 2x-y=0 2x-2x= 0
Re: In the figure above, point O is the center of the circle and OC = AC =
[#permalink]
29 Dec 2014, 21:14
I must have overlooked something because I get:
x+x+(180-2y)+y= 180 2x-y=0 2x-2x= 0[/quote][/quote] IRe: In the figure above, point O is the center of the circle and OC = AC =
[#permalink]
29 Dec 2014, 21:20
1
Kudos
Expert Reply
bankerboy30 wrote:
How are we getting different variables x and y when the sides are equal. Can you explain Krishna. Cause three sides are equal shouldn't their angles be noted with the same variable?
Note the sides that are equal OC = AC = AB
OC and AC are sides if a triangle and the angles opposite to them are marked as x each (i.e. they are equal)
AC and AB are equal sides of another triangle and angles opposite to them are marked as y each. Note that you cannot mark them as x too because they are equal angles in a different triangle. Their measure could be different from x. I have explained this in detail in a post given below. Giving the explanation here:
"."
Re: In the figure above, point O is the center of the circle and OC = AC =
[#permalink]
29 Dec 2014, 21:25
2
Kudos
Expert Reply
GMAT01 wrote:
IFrom this equation: x+x+(180-2y)+y= 180, you derive that 2x = y. If you try to substitute 2x = y in this equation itself, you will just get 2x - 2x = 0 which implies 0 = 0. This equation has 2 variables and you need two distinct equations to get the value of the two variables. If both equations are just 2x = y, you cannot get the value of x. You need another equation to get the value of x. | 677.169 | 1 |
0 users composing answers..
To find the slope \( m \) of the angle bisector of the lines \( y = 3x \) and \( y = 2x \), we can use the formula for the slope of the angle bisector between two lines given their slopes \( m_1 \) and \( m_2 \):
Unfortunately, I don't think it's used to calculate the angle bisector of a line.
It is mainly used to calculate if two lines are parrallel to eachother.
Since \(\frac{Ax+By+C}{\sqrt{A^2+B^2}} \) calculates the distance from the origin for the line \(Ax + By + C =0\) and the same applies for the left hand side, if the two are equal, they are parrallel lines.
However, I do believe I know what formula was used.
If we set the slopes of the two lines given in the question to \(m_1\) and \(m_2\), the slope of the angle bisector is | 677.169 | 1 |
The axes are at right angles to each other so that a point in
the plane, unless it is on an axis, forms a rectangle with the
origin and the perpendiculars to the axes. The feet of these
perpendiculars are the points from that determine the coordinates
of the point. | 677.169 | 1 |
Interactive Problem 70: Euclidian Normal Form of a Quadric
Find the eigenvalues of
and give them in descending order into the diagonal of the
following matrix
:
0
0
0
0
0
0
.
Find (using eigenvectors of
) an orthogonal matrix
with
.
The first row of
shall consist of nonnegative entries. Bring the entries of
to
the common denominator
. Give the numerator values of
:
.
The next intention is to find the Euclidean normal form of
. First you have to find a
representation of
with respect to a coordinate system adapted to the symmetry (principle axes transformation), i.e. make the transformation
. In the new
coordinates
is specified as:
.
Eliminate the linear terms by translation: The transformation
,
,
leads to the normal form
.
The total transformation is:
.
The quadric
is the following geometric figure:
n/a
ellipsoid
one-sheeted hyperboloid
hyperbolic paraboloid
zeppelin
cone
The signature of the quadric
is:
n/a
equal to
the determinant of
The length of the principle axes of a quadric determines:
n/a
the intersection of the quadric and the coordinate axes
the intersection of the quadric and the unit sphere
the expansion of the quadric
Find two intersecting lines
and
, which are contained in
and are
orthogonal to each other. Thereby the point
shall be element of
. The
intersection
of
and
is:
, ,
.
Complete the following two normalized direction vectors of the lines
and
: | 677.169 | 1 |
Picture This
Let's imagine that you are walking down the middle of a long street. You look ahead and see a tall building just up ahead. Suddenly, you have an idea! You want to find the exact center point of the road in front of the building. But the street is so long, you can't see the end of it! What can you do?
Enter the Midpoint Theorem
This is where the Midpoint Theorem comes in. This concept is all about finding the midpoint, or halfway point, of a line segment. In other words, it can help you mark the exact center point of any line, even if you can't see the ends of it.
Why is it Important?
The Midpoint Theorem may seem like a small thing, but it's actually incredibly important in many fields, including math, engineering, and architecture. Knowing how to find the midpoint of a line segment can help you make precise measurements, create accurate blueprints, and design the perfect structure.
Let's Get Started
Now that you know a bit about the Midpoint Theorem and why it's important, it's time to dive in! We'll start by exploring the basics of this concept and how it works, and then we'll move on to some more advanced applications. By the end of this lesson, you'll be a pro at finding midpoints and using them to solve problems. So buckle up, and let's get started! | 677.169 | 1 |
Which Angles Are Corresponding Angles Check All That Apply
Which angles are corresponding angles check all that apply – Delving into the concept of corresponding angles, this exploration unravels their significance in the realm of geometry. Corresponding angles, as their name suggests, are pairs of angles that occupy specific positions relative to each other, forming a fundamental concept in geometric relationships.
This article aims to shed light on the intricacies of corresponding angles, exploring their properties, applications, and implications for understanding geometric figures.
Corresponding angles arise when two parallel lines are intersected by a transversal, creating four distinct angles at the points of intersection. These angles exhibit unique properties that govern their relationships, providing valuable insights into the behavior of lines and angles in geometric constructions.
Corresponding Angles: Which Angles Are Corresponding Angles Check All That Apply
Corresponding angles are angles that are formed when two lines are intersected by a transversal. They are located in the same relative position on opposite sides of the transversal.
Properties of Corresponding Angles, Which angles are corresponding angles check all that apply
The theorem relating to corresponding angles states that if two lines are intersected by a transversal, then the corresponding angles are congruent.
This theorem has several implications for angle relationships:
If two angles are corresponding angles, then they have the same measure.
If two angles are congruent, then they are corresponding angles.
Applications of Corresponding Angles
Corresponding angles are used in solving geometry problems involving parallel lines. For example, if you know that two lines are parallel and you know the measure of one angle, then you can use corresponding angles to find the measure of the other angles.
Corresponding angles are also used in real-life applications, such as:
Architecture: Architects use corresponding angles to ensure that buildings are symmetrical.
Surveying: Surveyors use corresponding angles to measure distances and angles.
Engineering: Engineers use corresponding angles to design bridges and other structures.
Constructing Parallel Lines Using Corresponding Angles
You can use corresponding angles to construct parallel lines. To do this, follow these steps:
Draw a line.
Draw a transversal that intersects the line.
Mark the corresponding angles on opposite sides of the transversal.
Use a protractor to measure one of the corresponding angles.
Use the protractor to draw an angle congruent to the measured angle on the other side of the transversal.
Draw a line through the point of intersection of the transversal and the angle you just drew.
The line you just drew is parallel to the first line.
Answers to Common Questions
What are corresponding angles?
Corresponding angles are pairs of angles that occupy specific positions relative to each other when two parallel lines are intersected by a transversal.
How can corresponding angles be used to construct parallel lines?
Corresponding angles can be used to construct parallel lines by ensuring that the corresponding angles formed by the transversal and the two lines are congruent.
What are the properties of corresponding angles?
Corresponding angles are congruent, meaning they have the same measure. | 677.169 | 1 |
Geometrical Problems Deducible from the First Six Books of Euclid, Arranged ...
(5.) If the base of any triangle be bisected by the diameter of its circumscribing circle, and from the extremity of that diameter a perpendicular be let fall upon the longer side; it will divide that side into segments, one of which will be equal to half the sum, and the other to half the difference of the sides.
Let the base BC of the triangle ABC be bisected in E, by the diameter of the circumscribing circle ACD; and from D draw DF perpendicular to AB the longer side; BF will be equal to half the sum, and AF to half the difference of AB, BC.
=
Join DA, DB, DC; and make BG BC; join DG. Since BG= BC, and BD is common, and the angle GBD=CBD, since the arc AD=DC; .. DG= DC=DA, and DF is at right angles to AG, :. AF= FG. Whence the sum of AB and BC is equal to AG and 2 BG, i. e. to 2 BF; and the difference of AB and BC is equal to the difference of AB and BG, i. e. to 2 AF
(6.) The same supposition being made, as in the last proposition; if from the point, where the perpendicular meets the longer side, another perpendicular be let fall on the line bisecting the vertical angle; it will pass through the middle of the base.
The same construction being made as before; (see last Fig.) let FH be drawn perpendicular to BD, which bisects the vertical angle; FH will pass through E.
Because CB=BG, and the angle CBD=GBD,
.. BD is perpendicular to CG; and .. FH is parallel to CG. But since AF=FG, .. (Eucl. vi. 2.) AC is bisected by FH; which ... passes through E.
(7.) If a point be taken without a circle, and from it tangents be drawn to the circle, and another point be taken in the circumference between the two tangents, and a tangent be drawn to it; the sum of the sides of the triangle thus formed is equal to the sum of the two tangents.
From a given point D let two tangents DA, DB be drawn; and to C any point in the circumference between them, let a tangent ECF be drawn. The sum of the sides of the triangle is equal to the two tangents DA and DB.
E
F
B
FB, .. DE, EF, FD DB together. In the
Since AE-EC, and FC together are equal to AD and same manner, if through any other point in the arc ACB a tangent be drawn, it will be equal to the two segments of DA, DB intercepted between it, and the points of contact A and B ; and the three sides of the triangle so formed will be equal to DA, and DB together.
(8.) Of all triangles on the same base and between the same parallels, the isosceles has the greatest vertical angle.
Let ABC be an isosceles triangle on the base AC, and between the parallels AC, BD. It has a greater vertical angle
E
than any other triangle ADC on the same base, and between the same parallels.
About ABC describe a segment of a circle ABC; and since B is the middle point of the arc, and BD is parallel to AC, BD is a tangent at B. Let the arc cut AD in E; join EC. Then the angle ABC= AEC, and .. is greater than ADC.
COR. Of all triangles on the same base and having the same vertical angle, the isosceles is the greatest. For the triangle AEC has the same vertical angle with ABC, and ABC= ADC on the same base and between the same parallels; but ADC is greater than AEC, :. ABC is greater than AEC.
(9.) If through the vertex of an equilateral triangle a perpendicular be drawn to the side, meeting a perpendicular to the base drawn from its extremity; the line intercepted between the vertex and the latter perpendicular is equal to the radius of the circumscribing circle.
Let BE perpendicular to AB meet AE, which is perpendicular to the base AC, in E; BE is equal to the radius of the circle described about ABC.
E
H G
Draw BF, CG perpendiculars to the sides; and produce CG to H. Then CI is equal to the radius of the circle described about ABC; and EBIH is a parallelogram. And since CF is equal to FA, (Eucl. vi. 2.) CI is equal to IH, i. e. to the opposite side BE; and .. BE is equal to the radius of the circumscribing circle.
(10.) If a triangle be inscribed in a semicircle, and a perpendicular drawn from any point in the diameter, meeting one side, the circumference, and the other side produced; the segments cut off will be in continued proportion.
Let ABC be a triangle in the semicircle ABC; and from any point D in the diameter, let DF be drawn perpendicular to AD, meeting BC, the circumference, and AB produced, in E, G, F;
DE: DG :: DG : DF.
A
F
B
G
D
For the angles at E being equal, and the angles at Bright angles, .. the angle ECD is equal to BFD; and the angles at D are right angles; .. the triangles EDC, ADF are similar, and therefore
(11.) If a triangle be inscribed in a semicircle, and one side be equal to the semi-diameter; the other side will be a mean proportional between that side and a line equal to that side and the diameter together.
Let ABC be a triangle inscribed
in the semicircle, and let BC be equal to the semi-diameter; then will
A
BC: BA :: BA: BC+CA.
Produce AC to D, making CD equal to the semidiameter. Take O the centre. Join BD, BO.
Since
BO=BC, the angle BCO is equal to BOC, i. e. to OAB and OBA together, or to 2 BAC. But BCA is equal to CBD and CDB together, i. e. to 2 CDB, since CB= CD; hence the angle BAC BDC, and BA=BD; also the triangles BAD, BCD are similar;
=
.. BC : (BD=) BA :: BA : AD, which is equal to BC and CA together.
(12.) If a circle be inscribed in a right-angled triangle; to determine the least angle that can be formed by two lines drawn from the extremity of the hypothenuse to the circumference of the circle.
Let ABC be a right-angled triangle, in which a circle DEG is inscribed. On AC describe a segment of a circle ADC, which may touch the inscribed circle in
some point, as D. The lines AD, DC, drawn to this point from A and C, contain an angle less than the lines drawn to any other point in the circumference of the circle DEG.
For take any other point E, and join AE, EC; produce CE to F, and join AF. The exterior angle AEC is greater than AFC, i. e. than ADC, which is in the same segment. And the same may be proved of lines drawn to every other point in the circumference of the circle DEG. | 677.169 | 1 |
Your chance of acceptance
Your chancing factors
Extracurriculars
Understanding the 30-60-90 triangle
Hey everyone! So, in my geometry class, we've started ramping up on trigonometry and there's this one triangle that keeps confusing me - the 30-60-90. Can anyone explain what it is and share the formula? Is it something that I have to memorize for the SAT or ACT?
7 days ago
Sure, I can help explain the 30-60-90 triangle. It's a special type of right triangle where the angles measure exactly 30 degrees, 60 degrees, and 90 degrees (the right angle).
Now, here's the fun part: the sides of a 30-60-90 triangle have a specific ratio that will always hold true. If we call the shortest side (which is opposite the 30-degree angle) "x," then the length of the side opposite the 60-degree angle (often called the longer leg) follows as: x√3. The hypotenuse (the side opposite the 90-degree angle), will be 2x.
So, to summarize: if the shortest side is x, the longer side is x√3, and the hypotenuse is 2x.
As for the SAT and ACT, yes, you should have a good grasp of the 30-60-90 triangle rules. Keep in mind that starting from spring 2024, the SAT is fully digital and there isn't a no-calculator section, which means you'll be able to use a calculator for all math problems. However, knowing this ratio rule by heart will save you a lot of time even if you have your calculator handy.
In ACT, however, knowing these rules becomes particularly important because the math section does not provide you with geometry formulas. Again, the better you've got these rules down, the quicker you'll be able to answer questions related to 30-60-90 triangles.
My suggestion would be to familiarize yourself with these rules and practice applying them to various types of problems. Good luck | 677.169 | 1 |
each point, we need to check two things:
1) Is the point close enough to the center that it could be within the circle at all?
This is as simple as computing the distance between the point and the center of the circle: ~\sqrt{(X-50)^2 + (Y-50)^2}~. This distance must be no greater than ~50~, the radius of the circle.
2) Is the point between the two lines that define the bounds of the current circle sector?
We'd like to know the angle of the given point in the reference frame where ~0~ degrees is up and ~90~ degrees is to the right, since that's the way the boundary of the circle sector moves as the progress increases.
Most languages have an atan2() function which can give you the angle from ~(0, 0)~ to a given point. However, this function puts ~0~ degrees to the right and ~90~ degrees upwards, so it's necessary to translate the output to the desired reference frame. | 677.169 | 1 |
The angle of elevation of the top of an unfinished tower at a point distant 120 m from its base is 45°. If the elevation of the top at the same point is to be 60°, the tower must be raised to a height :
120 ( √3 + 1 ) m
120 ( √3 - 1 ) m
10 ( √3 + 1 )
None of these
Correct Option: B
Let us draw the figure from the given question. Given :- OB = h + x In figure , we have | 677.169 | 1 |
finding missing sides of congruent triangles worksheet pdf
Finding Missing Angles Of Triangles Worksheet | 677.169 | 1 |
Cotangent
Feeling:
Dumb
Language:
Arabic
Prompt:
Cotangent
Cotangent (cot) is a trigonometric function that represents the ratio of the adjacent side to the opposite side in a right triangle. It is calculated by dividing the length of the adjacent side by the length of the opposite side.
The cotangent function can be used to calculate angles in trigonometry. For example, if you have a right triangle with an adjacent side of 4 and an opposite side of 3, you can use the cotangent function to find the angle θ:
cot(θ) = adjacent/opposite
cot(θ) = 4/3
To find the angle θ, you can take the arctangent (or inverse cotangent) of the result:
θ = arctan(4/3)
One interesting fact about cotangent is that it is the reciprocal of the tangent function. In other words, cotangent is equal to 1 divided by tangent:
cot(θ) = 1/tan(θ)
This relationship can be useful in trigonometric calculations and proofs.
Overall, cotangent is a useful trigonometric function for calculating angles in right triangles and can be used in various applications in mathematics and physics | 677.169 | 1 |
...sides of it, make the adjacent angles together equal to two right angles, these two straight lines are in one and the same straight line. At the point B, in the straight line AB, let the two straight linns BC, BD, upon the opposite sides of AB, make the adjacent angles, ABC, ABD, together equal to...
...sides of it, make the adjacent angles together equal to two right angles, these two straight lines are in one and the same straight line. At the point B, in the straight line AB, let the two straight linos BC, BD, upon the opposite sides of AB, make the adjacent angles, ABC, ABD, together equal to...
...straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line. The 15th, Book I. If two straight lines cut one another, the vertical or opposite angles shall be equal....
...sides of it, make the adjacent angles together equal to two right angles, these two straight lines are in one and the same straight line. At the point B...lines BC, BD upon the opposite sides of AB, make the adjacnnt angles ABC, ABD equal togethe. to two right angles. BD is in the same straight line with CB....
...straight line, two other straight lines, upon tlie opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line. Given that at the point B in the straight line AB, the two straight lines BC and BD, upon the opposite...
...line, two other straight lines upon the opposite sides of it make the adjacent angles together .jequal to two right angles, these two straight lines shall be in one and the same straight line. 4. Make a triangle of which the sides shall be equal to three given straight lines. Can this be always...
...point ma straight line two other straight line» malte the two angles on opposite sides of it, together equal to two right angles, these two straight lines shall be in the same straight line ; but if they make two right angles on the same side of it, they shall coincide...
...straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line. (References — Prop. I. 13; ax. 1, 3.) Hypothesis. — At the point B in the straight line AB, let...
...lines, Upon the opposite sides of it, make the adjacent angles together equal to two right angles, then these two straight lines shall be in one and the same straight line. (References — Prop. i. 13; ax. 1, 3.) At the point B in the straight line AB, let BC, BD, upon the...
...straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line. 4. At a given point in a given straight line, make a rectilineal angle equal to a given rectilineal... | 677.169 | 1 |
Parallel Lines & Related Angles Activity
In this activity, you will explore the different angle relationships formed by parallel lines cut by a transversal.
In the applet below, the dashed brown line is a transversal that intersects 2 parallel lines.
Take a few minutes to explore the special relationships among the various types of angle pairs formed when a transversal intersects PARALLEL LINES, by following the steps below:
STEPS:
1. DRAG the green points around to change the red and/or blue angle measures that appear.
2. BE SURE TO CLICK ON EACH BOX TO DETERMINE EACH RELATIONSHIP. MAKE NOTE OF EACH RELATIONSHIP.
Use your observations to complete the statements below. Be sure to check your answer
When a transversal intersects PARALLEL LINES, all pairs of corresponding angles are | 677.169 | 1 |
Example Question #1 : Line And Angle Understanding And Applications
Complementary angles are any two angles in a triangle that sum to be 90
Complementary angles are any two angles in a triangle that sum to be 180
Correct answer:
Complementary angles are any two angles that sum to be 90
Explanation:
The definition of complementary angles is: any two angles that sum to 90. We most often see these angles as the two angles in a right triangle that are not the right angle. These two angles do not have to only be in right triangles, however. Complementary triangles are any pair of angles that add up to be 90.
Example Question #1 : Line And Angle Understanding And Applications
Solve for angle 1.
Possible Answers:
Correct answer:
Explanation:
Even though it may not be obvious at first, the given angle is actually a supplementary angle to angle 1. This is because the given angle is corresponding (and therefore congruent) angles to the angle adjacent to angle 1. Since they are supplementary we can set up the following equation.
Example Question #2 : Line And Angle Understanding And Applications
Lines and are parallel. Using this information, find the values for angles 1,2,3, and 4.
Possible Answers:
Correct answer:
Explanation:
We must use the fact that lines and are parallel lines to solve for the missing angles. We will break it down to solve for each angle one at a time.
Angle 1:
We know that angle 1's supplementary angle. Supplementary angles are two angles that add up to 180 degrees. These two are supplementary angles because they form a straight line and straight lines are always 180 degrees. So to solve for angle 1 we simply subtract its supplementary angle from 180.
Angle 2:
We now know that angle 1 is 130 degrees. We can either use the fact that angles 1 and 2 are opposite vertical angles to find the value of angle 2 or we can use the fact that angle 2's supplementary angle is the given angle of 50 degrees. If we use the latter, we would use the same procedure as last time to solve for angle 2. If we use the fact that angles 1 and 2 are opposite vertical angles, we know that they are congruent. Since angle then angle .
Angle 3:
To find angle 3 we can use the fact that angles 1 and 3 are corresponding angles and therefore are congruent or we can use the fact that angles 2 and 3 are alternate interior angles and therefore are congruent. Either method that we use will show that .
Angle 4:
To find angle 4 we can use the fact that the given angle of 50 degrees and angle 4 are alternate exterior angles and therefore are congruent, or we can use the fact that angle 3 is angle 4's supplementary angle. We know that the given angle and angle 4 are alternate exterior angles so .
Example Question #3 : Line And Angle Understanding And Applications
What are the values of angles and ?
Possible Answers:
Correct answer:
Explanation:
We are able to use the relationship of opposite vertical angles to solve this problem. The given angle and angle are opposite vertical angles and therefore must be congruent. So . is supplementary to both angles and so they must be congruent. and are also opposite vertical angles so they must be congruent in that respect as well. So
To answer this question, we must understand the definition of alternate exterior angles. When a transversal line intersects two parallel lines, 4 exterior angles are formed. Alternate exterior angles are angles on the outsides of these two parallel lines and opposite of each other.
Example Question #124 : Congruence
These are vertically opposite angles, there is not enough information to determine any further relation
These are vertically opposite angles, therefore they are equal
These are corresponding angles, there is not enough information to determine any further relation
Correct answer:
These are vertically opposite angles, therefore they are equal
Explanation:
To answer this question, we must understand the definition of vertically opposite angles. Vertically opposite angles are angles that are formed opposite of each other when two lines intersect. Vertically opposite angles are always congruent to each | 677.169 | 1 |
... , at the ...
Page 8 ... straight line AL is equal to BC . ( ax . 1. ) Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC . Which was to be done . PROPOSITION III . PROBLEM . From the greater of two given ...
Page 13 Euclides Robert Potts. PROPOSITION IX . PROBLEM . To bisect a given rectilineal angle , that is , to divide it into two equal angles . Let BAC be the given rectilineal angle . It is required ... a straight line BOOK I. PROP . IX , X. 13.
Page 14 Euclides Robert Potts. PROPOSITION XI . PROBLEM . To draw a straight line at right angles to a given straight line , from a given point in the same . Let AB be the given straight line , and C a given point in it . It is required to ...
Page 15 Euclides Robert Potts. It is required to draw a straight line perpendicular to AB from the point C. E AF GB D Take ... given point C shall be perpendicular to the given straight line AB . Join CF , and CG . And because FH is equal | 677.169 | 1 |
A similar figure to the Webb toroid can be made with ordinary pentagonal rotundae instead of tunnelled ones. It would have genus 29. Similar toroids can be made with a mixture of ordinary and tunnelled pentagonal rotundae, to achieve quasi-convex Stewart toroids in the genera 30-40, although these have lower symmetry than either the Webb toroid or the version with 12 ordinary pentagonal rotundae.
Its pentagonal prisms can also be removed (while bringing the other components together), and the toroid would maintain its quasi-convexity and genus. The convex hull would remain equilateral, but would no longer be a near-miss Johnson solid. Instead, it would be the Minkowski sum of a rectified chamfered dodecahedron and a regular dodecahedron. It would maintain the 12 decagons and 80 triangles, while the dodecagons would "contract" into rectangular-symmetricoctagons (which would appear somewhat squashed). If the perpendicular edges of these octagons were further "contracted" into points, it would produce a rhombus with one diagonal double the length of the other.
The Webb toroid is also used in the construction of a quasi-convex Stewart toroid of even greater genus than the holey monster. The new toroid is an outer-blend of a Webb toroid, a holey monster, and a triangular cupola that connects the two larger toroids. (The holey monster most readily admits outer blends on its hexagonal faces, and the Webb toroid has no hexagons but plenty of triangles and squares.) The empty space in the middle of the Webb toroid is large enough to contain the holey monster, allowing the blend to share the convex hull of the Webb toroid and thus be quasi-convex. In the resulting figure, the holey monster is off-center, so there is no obvious way to add more supports between the figures to increase the genus further. | 677.169 | 1 |
Moving a vector involved in a linear transformation increases the scale of the linear transformation by the number of vectors moved. If you move it parallel to the floor, it will increase by 0. If you move it non-parallel to the floor, it will have a non-zero value. | 677.169 | 1 |
Synopsis
Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It is a fundamental concept in geometry and is used in various fields such as physics, engineering, and astronomy.
The study of trigonometry involves understanding trigonometric functions such as sine, cosine, and tangent, which relate the angles of a triangle to the lengths of its sides. These functions are essential for solving problems involving triangles and can be used to calculate distances, heights, and angles.
Trigonometry is also used to analyze periodic phenomena, such as sound waves and electromagnetic waves, as well as to model complex systems in mathematics and physics. It provides a powerful tool for understanding the world around us and is essential for many scientific and technical disciplines.
Download and play the Soundtrack list
Play
Title
Artist
Trigonometry
Bootie Call
Friday Sky
I Get Lifted
Rock Me More and More
La Panorama
Who Do You Think You Are
Pra Fazer um Samba
Toast
Manteca
I'd Like To Walk Around In Your Mind
Shade
Magnifique
We Be Comin In Hot
Dancing On My Own - Radio Edit
Witness (1 Hope)
May You Never
Stay Low
Before
Guess Who?
You Make Me Feel (Mighty Real)
Ambersand
Alexis Grapsas:
Performer
Three Mornings
Alexis Grapsas:
Performer
Different Kinds Of Cry
Alexis Grapsas:
Performer
Unicorn
Alexis Grapsas:
Performer
Study Of Triangles
Alexis Grapsas:
Performer
Northern Lights
Alexis Grapsas:
Performer
Six Months Later
Alexis Grapsas:
Performer
About Last Night
Alexis Grapsas:
Performer
Bad Luck
Alexis Grapsas:
Performer
Unconventional Love
Alexis Grapsas:
Performer
Mother's Consent
Alexis Grapsas:
Performer
Gemma's Grief
Alexis Grapsas:
Performer
Leave Behind
Alexis Grapsas:
Performer
Take A Walk In The Flat
Alexis Grapsas:
Performer
Synchronized
Alexis Grapsas:
Performer
We Are Not Bad People
Alexis Grapsas:
Performer
Panic Attacks
Alexis Grapsas:
Performer
You Are My Person
Alexis Grapsas:
Performer
Out Of Options
Alexis Grapsas:
Performer
Back In The Water
Alexis Grapsas:
Performer
Trigonometry Impromptu (Bonus Track)
Alexis Grapsas:
Performer
Gemma's Grief Reworked (Bonus Track)
Alexis Grapsas:
Performer
User reviews
Timothy Parker
7/10
The soundtrack of Trigonometry is a masterful example of how music can elevate a visual medium, bringing an extra layer of emotion and depth to the characters and their relationships.
Kenneth Williams
6/10
Overall, the soundtrack of Trigonometry is a thought-provoking and immersive musical journey that celebrates the beauty and complexity of trigonometry, highlighting its importance in shaping our understanding of the world.
Richard Thomas
4/10
The use of generic and cliché musical motifs in the soundtrack of Trigonometry was disappointing. It missed the opportunity to explore innovative and unconventional sounds that could have reflected the dynamic and versatile nature of trigonometry as a mathematical discipline.
Laura Anderson
4/10
The soundtrack of Trigonometry failed to capture the complexity and beauty of the mathematical concepts it represents. It lacked depth and creativity, making it feel disconnected from the profound nature of trigonometry.
Andrew White
6/10
The use of electronic beats and subtle melodies in the soundtrack of Trigonometry mirrors the complexity and elegance of trigonometric functions, creating a dynamic and engaging musical experience.
Linda Jones
8/10
The soundtrack of Trigonometry effectively conveys the mathematical precision and elegance of trigonometric concepts through its carefully crafted compositions and arrangements.
Ronald Martin
3/10
The music in Trigonometry was repetitive and uninspiring, failing to evoke any emotions or enhance the viewing experience. It felt like a missed opportunity to create a soundtrack that could have elevated the themes of triangles, angles, and relationships in a more engaging way.
Brian Phillips
5/10
The soundtrack of Trigonometry effectively conveys the interconnectedness of trigonometry with various scientific disciplines, showcasing the versatility and applicability of trigonometric concepts in real-world scenarios.
Brian Turner
7/10
The use of subtle melodies and ambient sounds in the soundtrack creates a soothing and immersive listening experience that enhances the viewing experience of the series.
Donna Martin
8/10
The memorable themes and motifs in the music of Trigonometry stay with you long after watching the show, making it a soundtrack worth revisiting and enjoying on its own merits.
Joseph Campbell
5/10
The soundtrack of Trigonometry skillfully blends modern electronic elements with classical influences, reflecting the timeless nature of trigonometry as a fundamental branch of mathematics.
Brian Davis
8/10
The integration of electronic beats and orchestral arrangements in the music adds depth and complexity to the overall atmosphere of Trigonometry, making it a standout feature of the show.
Lisa Taylor
10/10
The instrumental choices and arrangements in the Trigonometry soundtrack are truly exceptional, blending elements of classical music with modern electronic sounds to represent the intersection of tradition and innovation in the study of trigonometry. Each track evokes a sense of mathematical exploration and discovery, making it a captivating and enriching musical journey.
Kimberly Adams
7/10
The soundtrack of Trigonometry is a mesmerizing blend of electronic and classical music that perfectly captures the essence of the show's themes and emotions.
Steven White
5/10
The soundtrack of Trigonometry perfectly captures the essence of mathematical precision and beauty, creating a unique atmosphere that immerses the listener in the world of geometry and trigonometric concepts.
Brian Taylor
10/10
The soundtrack of Trigonometry captures the essence of mathematical precision and beauty with its intricate melodies and harmonies. The music perfectly mirrors the elegance and complexity of trigonometric functions, creating a mesmerizing listening experience that immerses you in the world of triangles and angles.
Charles Jones
5/10
The use of atmospheric soundscapes and rhythmic patterns in the soundtrack of Trigonometry enhances the listener's understanding of the mathematical principles at play, making the subject more accessible and engaging.
David Williams
5/10
The music in Trigonometry evokes a sense of exploration and discovery, much like the process of delving into trigonometric functions to uncover hidden relationships and patterns in triangles.
Carol Clark
7/10
The music in Trigonometry enhances the storytelling by creating a sense of tension, drama, and beauty that perfectly complements the unfolding narrative of the series. | 677.169 | 1 |
Construction Of Rhombus - Definition, Examples, Properties (Class 8)
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Steps for the construction of rhombus, where the measurement of its two diagonals is given. Learn how to construct a rhombus when a side, a diagonal and measure of angle are given with simple steps at BYJU Steps for the construction of rhombus, where the measurement of its two diagonals is given. Learn how to construct a rhombus when a side, a diagonal and measure of angle are given with simple steps at BYJU'S. | 677.169 | 1 |
Congruency, Symmetry for Grade 4 (examples, solutions, videos)
In these lessons, we will learn congruence of 2-D shapes and the symmetry of 2-D shapes. This lesson is suitable for Grade 3 and Grade 4 kids.
Related Topics: More Math Lessons for Grade 4 More Lessons on Geometry Congruent Math Games
Definition: Two 2-D shapes are congruent if they are identical in shape and size.
It is important to recognize that the term congruent applies only to size and shape. The figures can be different colors, or oriented in different ways, and they will still be congruent as long as they are the same shape and the same size.
Characteristics: Congruent shapes must have • corresponding sides congruent • corresponding vertices congruent • the same area • the same shape
The following Frayer Model gives a summary of congruency for 2-D shapes.
Congruent – Grade 4 Common Core Standards This video examine the meaning of congruence. Example: Denise drew the figure below. Which figure is congruent to the figure Denise draw?
Definition: A 2-D figure has line symmetry when it can be divided or folded so that the two parts match exactly. The fold line is called the line of symmetry. Any given line of symmetry divides a figure into equal halves. It may also be said that each of the halves are mirror images of each other. Line symmetry is also called reflective symmetry or mirror symmetry. Characteristics: Symmetrical shapes must have: • two congruent parts separated by a line of symmetry • corresponding vertices and sides matching when the shape is folded along the axis of symmetry
The following Frayer Model gives a summary of symmetry for 2-D shapes.
Shapes may have multiple lines of symmetry and the lines of symmetry can be vertical, horizontal, or diagonal. The more sides that a regular polygon has, the greater the number of lines of symmetry there are. A circle has an infinite number of lines of symmetry.
Lines of symmetry – Grade 3 Common Core Standards In this video, we find lines of symmetry in a 2D shape. Example: Which figure shows a line of symmetry?
Multiple Lines of Symmetry Looking at a few problems with more than one line of reflective symmetry.
Line Symmetry This video will review with you the basics of line symmetry and how to find line symmetry. A figure has line symmetry if a line can be drawn through the figure so that each half is a mirror image of the other | 677.169 | 1 |
area and perimeter of quadrilaterals and triangles worksheets
Area And Perimeter Of Quadrilaterals And Triangles Worksheets – Triangles are among the most fundamental designs in geometry. Understanding triangles is crucial for learning more advanced geometric terms. In this blog post We will review the various kinds of triangles including triangle angles and the methods to calculate the perimeter and area of a triangle and will provide specific examples on each. Types of Triangles There are three kinds in triangles, namely equilateral isosceles, and scalene. Equilateral … Read more | 677.169 | 1 |
vertex
In mathematics, a vertex is a point where two or more lines, curves, or edges intersect
In mathematics, a vertex is a point where two or more lines, curves, or edges intersect. It is commonly used to refer to the meeting point of edges in a geometric figure, such as the corners of a polygon or the intersection of edges in a three-dimensional shape.
For example, in a triangle, each of the three corners is a vertex. Similarly, in a cube, each of the eight corners is a vertex. In graph theory, a vertex represents a point or node in a network, and the connections between vertices are represented by edges.
In algebraic terms, a vertex can also refer to a maximum or minimum point on a curve or a parabola. In the context of a quadratic function, the vertex is the highest or lowest point on the graph, depending on whether the parabola opens upwards or downwards.
To find the vertex of a quadratic function in standard form (f(x) = ax^2 + bx + c), you can use the formula x = -b / (2a) to find the x-coordinate of the vertex, and then substitute this value back into the function to find the corresponding y-coordinate.
Overall, a vertex represents a significant point of intersection or extremum in various mathematical contexts, serving to provide valuable information about the characteristics and properties of the given shape or | 677.169 | 1 |
Parallel Lines
If distance between two lines is the same at each and every point on two lines, then two lines are said to be parallel. If lines l and m do not intersect each other at any point then l || m.
Transversal line
A line is said to be transversal which intersect two or more lines at distinct points.
1. Corresponding angles
Pair of angles having different vertex but lying on same side of the transversal are called corresponding angles. Note that in each pair one is interior and other is exterior angle.
∠1 and ∠2
∠3 and ∠4
∠5 and ∠6
∠1 and ∠8
These angles are pair of corresponding angles.
2. Alternate interior angles
Pair of angles having distinct vertices and lying can either side of the transversal are called alternate interior angles.
∠1 and ∠2
∠3 and ∠4
These angles are alternate interior angles
3. Consecutive interior angles
Pair of interior angles of same side of transversal line.
∠1 and ∠2
∠2 and ∠4
These angles are consecutive interior angles or co-interior angles
Axiom 6.3: If two parallel lines are intersected by a transversal then each pair of corresponding angles are equal.
If AB || CD, then
∠PEB = ∠EFD
∠PEA = ∠EFC
∠BEF = ∠DFQ
∠AEF = ∠CFQ
Theorem 6.2: If two parallel lines are intersected by a transversal then pair of alternate interior angles are equal.
If AB || CD, then ?
∠AEF = ∠EFD
∠BEF = ∠CFE
Theorem 6.3: If two parallel lines are intersected by a transversal then the ! sum of consecutive interior angles of same side of transversal is equal to 180°. If AB || CD then
(i) ∠BEF + ∠DFE = 180° (ii) ∠AEF + ∠CFE = 180°
Axiom 6.4: If two lines are intersected by a transversal and a pair of corresponding angles are equal, then two lines are parallel. (i) If ∠PEB = ∠EFD (corresponding angles), then AB || CD
Theorem 6.4: If two lines intersected by a transversal and a pair of alternate interior angles are equal, then two lines are parallel. If ∠AEF = ∠EFD (alternate interior angles), then AB || CD.
Theorem 6.5: If two lines are intersected by a transversal and the sum of consecutive interior angles of same side of transversal is equal to 180°, the lines are parallel. If ∠AEF + ∠CFE = 180°, then AB || CD.
Theorem 6.6: Lines which are parallel to the same line are parallel to each other. If AB || EF and CD || EF then AB || CD | 677.169 | 1 |
Hint: First we have to define what the terms we need to solve the problem are. Since from the given set of questions we need to construct an arrangement of the square using a diagonal given point is the only known value; also, if we need to draw a square first, we need to draw a diagonal point and then only we can able to construct the given set of arrangements
Complete step by step answer: First, we need to draw anything so then only we can able to proceed further and in the five steps given above has two basics are draw at first and in that draw a diagonal is the given and know value Hence first draw a diagonal AC = \[5cm\].(step 2 is the first arrangement) Similarly, now after the diagonal is drawn; we now draw a PQ the perpendicular with the bisector of AC. So that only using the perpendicular and diagonal points we can able to draw a square; (step four is the second arrangement) now from the given PQ cut AC at O. (internally cutting the axis) (step one is at the arrayments of third) and then forth part of arrangement is With O as center and OA radius drawing the circle. And let the circle cut QP at points B and D. (fourth arrangement) so that we are able to cut QP points. Hence finally balance one is the step three also the fifth arrangement joins all the parts by joining AB, BC, CD and DA. Then ABCD is the required square.
So, the correct answer is "Option D".
Note: since if any of the arrangement process done with the mistakes like options; $A)2,1,4,5,3$,$B)2,5,4,1,3$,$C)2,4,5,1,3$ we cannot able to obtain a square; like if the first arrangement if one then without diagonal we cannot start drawing. | 677.169 | 1 |
JAC Class 9 Maths Solutions Chapter 10 Circles Ex 10.6
JAC Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.6
Page-186
Question 1.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Answer:
Given: Two intersecting circles, in which OO' is the line of centres and P and Q are two points of intersection.
To prove: ∠OPO' = ∠OQO'
Construction: Join PO, QO, PO' and QO'.
Proof: In APOO' and AQOO,'
we have PO = QO [Radii of the same circle]
PO' = QO'[Radii of the same circle]
OO' = OO' [Common]
APOO' = AQOO' [SSS axiom]
⇒ ∠OPO' ≅ ∠OQO' [CPCT]
Hence, the line of centres of two intersecting circles subtends equal angles at the two points of intersection. Proved.
Question 5.
Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
Answer:
Given: A rhombus ABCD whose diagonals intersect each other at O.
To prove: A circle with AB as diameter passes through O.
Proof: ∠AOB = 90°
[Diagonals of a rhombus bisect each other at 90°]
⇒ ∆AOB is a right triangle right angled at O.
⇒ AB is the hypotenuse of right ∆AOB.
⇒ If we draw a circle with AB as diameter, then it will pass through O because angle in semicircle is 90° and ∠AOB = 90°.
Question 8.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are
90° – \(\frac{1}{2}\) A, 90° – \(\frac{1}{2}\) B and 90° – \(\frac{1}{2}\) C.
Answer:
Given: AABC and its circumcircle. AD, BE, CF are bisectors of ∠A, ∠B, ∠C respectively.
Construction: Join DE, EF and FD.
Proof: We know that angles in the same segment are equal.
Page-187
Question 9.
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
Answer:
Given: Two congruent circles which intersect at A and B. PAQ is a line through A.
To Prove: BP = BQ.
Construction: Join AB.
Proof: AB is a common chord of both the circles.
As the circles are congruent,
arc ADB = arc AEB
⇒ ∠APB = ∠AQB [Angles subtended by equal arcs]
So, in APBQ, BP = BQ [Sides opposite to equal angles are equal]
Question 10.
If any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Answer:
(i) Let bisector of ∠A meet the circumcircle of ∆ABC at M.
Join BM and CM.
∴ ∠MBC = ∠MAC [Angles in same segment]
and ∠BCM = ∠BAM [Angles in the same segment]
But ∠BAM = ∠CAM [∴ AM is bisector of ∠A]
∴ ∠MBC = ∠BCM So, MB = MC [Sides opposite to equal angles are equal].
⇒ M lies on the perpendicular bisector of BC
Hence, angle bisector of ∠A and perpendicular bisector of BC intersect on the circumcircle of ∆ABC.
(ii) Let M be a point on the perpendicular bisector of BC which lie on circumcircle of ∆ABC
Join AM,
∴ M lies on perpendicular bisector of BC.
∴ BM = CM
∠MBC = ∠MCB [Angle opposite to equal sides are equal]
But ∠MBC = ∠MAC [Angles in the same segment]
and ∠MCB = ∠BAM [Angles in the same segment]
So, from (i)
∠BAM = ∠CAM AM is bisector of ∠A
⇒ bisector of ∠A and perpendicular bisector of BC intersect at M which lies on circumcircle of ∆ABC. | 677.169 | 1 |
149 Geometry Quizzes, Questions, Answers & Trivia - ProProfs (2024)
Quiz: How Well Do You Know Inscribed Angles?
Quiz: How Well Do You Know Inscribed Angles?
Welcome to the intriguing world of inscribed angles within circles! Prepare to embark on a journey of geometric discovery with the "How Well Do You Know Inscribed Angles?" quiz. This quiz is designed to test your...
Welcome to the "Test Your Knowledge of Scale Drawing!" quiz! This quiz will assess your understanding of scale drawing concepts, its applications, and the mathematical principles involved. Scale drawings are essential...
Questions: 10|Attempts: 205 |Last updated: Nov 17, 2023
Sample Question
What is a scale drawing used for?
Measuring length
Creating 3D models
Representing data
Enlarging images
Test Your Expertise On Geometry Dash!
Test Your Expertise On Geometry Dash!
Welcome to the "Test Your Expertise on Geometry Dash!" quiz. If you are a fan of this addictive rhythm-based platformer game, this quiz is designed to challenge your knowledge of its levels, icons, characters, and...
Questions: 10|Attempts: 668 |Last updated: Apr 9, 2024
Sample Question
In Geometry Dash, what is the primary objective of the game?
Collect coins
Complete levels
Defeat bosses
Solve puzzles
A Quiz On Geometry Formulas
A Quiz On Geometry Formulas
As a student of math, you must have studied geometry. Haven't you? Would you like to try a quiz on geometry formulas right now? Geometry is the branch of mathematics that is concerned with the shape of individual objects,...
Questions: 10|Attempts: 1602 |Last updated: Aug 16, 2023
Sample Question
The formula for the area of a triangle
A = 1/2 × b × h
A = 2 × b × h
A = 1/4 × b × h
A = b × h
Geometry Quiz For Grade 6
Geometry Quiz for Grade 6
Check out this geometry quiz if you're studying in grade 6 and have started learning about geometrical shapes. Also, it will test your knowledge of important geometry concepts if you know about them already. Geometry is a...
Questions: 10|Attempts: 4866 |Last updated: Jan 30, 2024
Sample Question
Identify this polygon.
Triangle
Convex
Concave
Hexagon
Quadrilateral And Its Types Quiz
Quadrilateral and its Types Quiz
Aquadrilateralis a polygon that has exactly four sides. Here is an informative quiz about Quadrilaterals, that will help in the betterment of your geometric understanding. The quiz will help you learn more about all...
The notion of line or straight line is to represent straight objects in geometry. Lines are an idealization of such objects, often described in terms of two points. Take this quiz to test your knowledge and clear your concepts...
Questions: 16|Attempts: 1333 |Last updated: Mar 22, 2023
Sample Question
Find the slope through the points (4, -2) and (-1, 6).
-5/8
8/5
-8/5
-3
Geometrical Figures: 3rd Grade Quiz!
Geometrical Figures: 3rd Grade Quiz!
.
Questions: 17|Attempts: 922 |Last updated: Mar 21, 2023
Sample Question
How many edges do a kite have?
1
2
4
3
Intermediate Level Geometry Test: Quiz!
Intermediate Level Geometry Test: Quiz!
Do you think you can master intermediate geometry? With this quiz, you will need to understand diagonal parallelogram division, whether altitudes drawn from every vertex of a triangle are equal, and what is the correct...
Questions: 20|Attempts: 265 |Last updated: Mar 21, 2023
Sample Question
The diagonal of a parallelogram divides into two congruent triangles.
True
False
Math Quiz: Geometry Exam!
Math Quiz: Geometry Exam!
Questions: 11|Attempts: 490 |Last updated: Mar 22, 2023
Sample Question
How many degrees are in a triangle?
90
360
180
Cyclic Quadrilateral Quiz: Test!
Cyclic Quadrilateral Quiz: Test!
Questions: 8|Attempts: 770 |Last updated: Mar 21, 2023
Sample Question
How many degrees are in a triangle?
90
360
180
MS Power Point Presentation Quiz #4
MS Power Point Presentation Quiz #4
Questions: 25|Attempts: 419 |Last updated: Mar 20, 2023
Sample Question
Which tab is not available on left panel when you open a presentation?
Outline
Slides
Notes
All of above are available
Assessment 1 - Coordinate Geometry - Ix
Assessment 1 - Coordinate geometry - IX
Answer these geometry questions. Find out your score at the end. You have only 12 minutes to complete the quiz.
Questions: 10|Attempts: 54 |Last updated: Mar 15, 2023
Sample Question
The ordinate of all the points in the x-axis is:
1
-1
Any natural number
Geometry Trivia: Trivia Test On Shapes And Lines! Quiz
Geometry Trivia: Trivia Test on Shapes and Lines! Quiz
Questions: 9|Attempts: 248 |Last updated: Mar 22, 2023
Sample Question
Which among the following is not a three-dimensional shape?
Circle
Cube
Cylinder
Cuboid
A Quiz On Coordinate Plane
A Quiz on Coordinate Plane
In coordinate geometry, planes that can be folded into different directions equally. This property is because of the presence of ordered pairs in coordinate planes. Operations in the geometry of different shapes involve finding...
Questions: 10|Attempts: 395 |Last updated: Mar 22, 2023
Sample Question
Which of these does not require the application of the principles of coordinate planes?
Differential geometry
Linear algebra
Multivariate calculus
Probability theory
Quadrilaterals And Polygons Properties Quiz!
Quadrilaterals and Polygons Properties Quiz!
Questions: 29|Attempts: 357 |Last updated: Aug 29, 2023
Sample Question
All angles are right angles.
Parallelogram, Rectangle, Rhombus, Square
Rhombus, Square
Rectangle, Parallelogram
Rhombus, Rectangle
Rectangle, Square
ACT Prep Geometry Quiz #1
ACT Prep Geometry Quiz #1
Questions: 10|Attempts: 421 |Last updated: Mar 22, 2023
Sample Question
What is the next term in the geometric sequence 16, - 4, 1, - 1/4, .....?
- 1/8
1/16
1/2
What Do You Know About Pseudo-polyomino?
What do you know about Pseudo-polyomino?
The term "Pseudo-polyomino" is a form of recreation mathematics that deals with plane figures and side polygons featuring square cells. Are you looking to improve your knowledge of Pseudo-Polyomino? This quiz has been...
Questions: 10|Attempts: 151 |Last updated: Mar 20, 2023
Sample Question
Pseudo polyamino is basically aplane geometric figure formed by joining one or more equal squares edge to edge or to corner at which degree?
60
90
120
360
What Do You Know About Polyknights?
What do you know about polyknights?
A polyknight is described as a plane geometric structure formed through the selection of the cells present in the square lattice that clearly shows the path of the chess knight. If you are willing to learn more, take this short,...
Questions: 10|Attempts: 117 |Last updated: Mar 18, 2023
Sample Question
In how many ways can polyominoes be distinguished?
3
4
5
6
Differential Geometry Test
Differential Geometry Test
Do you understand differential geometry well? Take this quiz on differential geometry and test how much you know. Differential geometry employs the principles of calculus, both differential and integral, as well as multilinear...
Questions: 10|Attempts: 1768 |Last updated: Feb 19, 2024
Sample Question
These theories are principles on which differential geometry is based, except
Theory of surfaces
Theory of plane curves
Theory of space curves
Theory of dimensional shapes
How Good Are You In Algebraic Geometry
How good are you in Algebraic Geometry
Questions: 10|Attempts: 244 |Last updated: Mar 18, 2023
Sample Question
The fundamental objects of study in Algebraic Geometry are
Algebraic Varieties
Algebraic Functions
Polynomial Equation
Quadratic Equation
How Well Do You Know Coordinate Plane?
How well do you know coordinate plane?
Acoordinate planeis an intersection of two number lines. Just like a regular number line, the ones on a coordinate plane could stretch on infinitely.
Questions: 10|Attempts: 157 |Last updated: Mar 21, 2023
Sample Question
Acoordinate plane, which is an intersection of how many number lines?
One
Two
Three
Four
How Much Do You Know About Areas Of Rectangles?
How much do you know about areas of rectangles?
The area of apolygonis the number of square units inside the polygon. To understand the difference betweenperimeterand area, think of perimeter as the length of fence needed to enclose the yard, whereas...
Questions: 10|Attempts: 177 |Last updated: Mar 21, 2023
Sample Question
To find the area of a rectangle, multiply the .....?
Length by the width
Breath by the length
Width by the breath
Length by the breath
A Quick Coordinate Plane Assessment Test
A Quick Coordinate Plane Assessment Test
Step into the fascinating world of mathematics with our "Quick Coordinate Plane Assessment Test". This engaging quiz is designed to test your knowledge and understanding of the coordinate plane. A coordinate plane is a...
Questions: 10|Attempts: 2673 |Last updated: Sep 4, 2023
Sample Question
The horizontal number line on a coordinate plane is known as the...
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Do You Know Geometric Transformations?
Do you know Geometric Transformations?
A geometric transformation is any bijection of a set having some geometric structure to itself or another such set. take the quiz and have fun. | 677.169 | 1 |
Elements of geometry and mensuration
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Página 14 ... given ' line means a line ' given ' sometimes in position , sometimes in magnitude , sometimes in both , ac- cording to circumstances ; and the word ' given ' means fixed or known . ( 3 ) A proposition ' is something proposed to be done ...
Página 16 Thomas Lund. STRAIGHT LINES AND RECTILINEAL PLANE FIGURES . * 23. PROPOSITION I. To describe an equilateral tri- angle upon a given straight line ↑ . Let AB be the given straight line , which is to be one side of the triangle ; with ...
Página 17 ... line AB upon DE ; then the point B will fall B upon E , because AB = DE . Again , since AB falls upon DE , AC will ... given angle , that is , to divide it into two equal angles . Let BAC be the given angle ; it is required to ...
Página 18 ... line AF . For , since D and E are points in the circumference of the same circle whose centre is A , AD = AE ; and since DEF is an equilateral triangle , DF = EF . There ... given straight line . Upon AB 18 GEOMETRY AS A SCIENCE .
Página 20 Thomas Lund. cular to a given straight line of unlimited length from a given point without it . Let AB be the given straight line , and C a given point without it , from which it is required to draw a perpendicular to AB . Take a point D | 677.169 | 1 |
Key to Geometry, Book 7: Perpendiculars and Parallels, Chords and Tangents, Circles
(From Amazon): Key they do sophisticated constructions involving over a dozen steps and are prompted to form their own generalizations. When they finish, students have been introduced to 134 geometric terms and are ready to tackle formal proofs. Book 7 includes three comprehensive chapters which cover perpendiculars and parallels, chords and tangents and circles. Student workbook, consumable student will also need a compass, a straight edge and a sharp pencil.Pages: 154, PaperbackPublisher: Key Curriculum Press | 677.169 | 1 |
4 2 study guide and intervention angles of triangles
4-2-study-guide-and-intervention-angles-of-triangles 2 Downloaded from cdn.ajw.com on 2021-05-06 by guest SAT with confidence—very few questions will surprise you, and even fewer will be able to withstand your withering attacks. Stand tall, intrepid student. Destiny awaits. Updated for the New SAT This new edition of the Math Guide has been ...We would like to show you a description here but the site won't allow us.
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angles of those n 2 triangles. Interior Angle If a convex polygon has n sides, and S is the sum of the measures of its interior angles, Sum Theorem then S 180(n 2). A convex polygon has 13 sides. Find the sum of the measures of the interior angles. S 180(n 2) 180(13 2) 180(11) 1980 The measure of an interior angle of a regular polygon is 120 ...Exterior Angle Theorem At each vertex of a triangle, the angle formed by one side and an extension of the other side is called an exterior angle of the triangle. For each exterior angle of a triangle, the remote interior angles are the interior angles that are not adjacent to that exterior angle. In the diagram below, /B and /A are the remote ...Course: common core geometry (8947589357) 3 Documents. University: Studocu University - USA. Info. Download. AI Quiz. View full document. A worksheet of Bisectors of Triangles name date period study guide and intervention bisectors of triangles perpendicular bisectors perpendicular bisector is.
Step 1 Draw a right triangle and label one acute angle B. Label the adjacent side 3 and the hypotenuse 7. Step 2 Use the Pythagorean Theorem to find b. ... View ANSwER_OF_8-4_Study_Guide_and_Intervention.pdf from MATH 245 at San Francisco Stat... 5_1practice_solutions.pdf.12.3 1BS2 Study Guide and Intervention Congruent Triangles Congruence and Corresponding Parts Triangles that have the same size and same shape are congruent triangles. Two triangles are congruent if and only if all three pairs of corresponding angles are congruent and all three pairs of corresponding sides are congruent. In the figure, ABC ≅ RSTQuestion: NAME DATE PERIOD 4-1 Study Guide and Intervention Classifying Triangles Classify Triangles by Angles One way to classify a triangle is by the measures of its angles. . If all three of the angles of a triangle are acute angles, then the triangle is an acute triangle. If all three angles of an acute triangle are congruent, then the ...5x+2 Subtract from each side. Divide each side by LL. Substitute for and b. You know that ZE So, mLE — 1110 mLL. (5y + Subtract from each side. Divide each side by ANGLES THEOREM THEOREM 4.3: THIRD Chapter 4 Geometry Notetaking Guide 76 If two angles of one triangle are congruent to two angles of another triangle, then theView Geometry 2-7.pdf from AA 1NAME _ DATE _ PERIOD _ 2-7 Study Guide and Intervention Parallel Lines and Transversals Transversal Angle Pair Relationships A line that intersects two or more other ... B4.07_Sides and Angles of Congruent Triangles_CA (1).doc. ... View 3-2_Study_Guide.pdf from MATH MISC at Braden River High School. NAME _ DATE ...
4-4 Study Guide and Intervention (continued) Proving Triangles Congruent—SSS, SAS SAS Postulate Another way to show that two triangles are congruent is to use the Side-Angle-Side (SAS) Postulate. SAS Postulate If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent.Lesson 4-1 Classify Triangles by AnglesOne way to classify a triangle is by the measures of its angles. • If oneof the angles of a triangle is an obtuse angle, then the triangle is an obtuse triangle. • If oneof the angles of a triangle is a right angle, then the triangle is a right triangle. • If all threeof the angles of a triangle are ...6. If cot A = 8, find the exact values of the remaining trigonometric functions for the acute angle A. Find the measu of angle e. Round to the nearest degree, if necessary. S 50 Solve each triangle. Round side measures to the nearest tenth and angle measures to the nearest degree. 10. 14 cos Il. ….
version 2 0 8 2 study guide and intervention st louis public schools guidance on how to develop complex interventions to ... angles study guide and intervention 4. 2. Overcoming Reading Challenges Dealing with Digital Eye Strain Minimizing Distractions Managing Screen Time 3. Promoting Lifelong Learning Utilizing eBooks for SkillLesson 4-5. PERIOD. 4-4 Study Guide and Interventiongraph the point (0, -2). From … Web4-3 Study Guide and Intervention Congruent Triangles Third Angles Theorem If two angles of one triangle are congruent to two angles of a second triangle, then the third angles of … WebSep 18, 2014 · b2 - 4ac < 0 2 complex roots Example Find the value of the discriminant for each equation.
shape - triangles are congruent if their corresponding angles are equal, and theirtriangle as long as you know the other two lengths. However, now we can relate angle measures to the right triangle side lengths. This allows us to find both missing side lengths when we only know one length and an acute angle measure. The acute angle measures in a right triangle can be found based on any two side lengths.4 2 Study Guide And Intervention Angles Of Triangles 4-2-study-guide-and-intervention-angles-of-triangles 3 Downloaded from cdn.ajw.com on 2019-04-06 by guest recommended actions for parents and caregivers, teachers, administrators, and policy makers, stressing the importance that everyone work together to ensure a mathematically literate society.
sks 15 After the news of Ryanair's diverted flight to Belarus, we explain how airline pilots handle military interventions. As I write this, the aviation world is still trying to get its ...10 4 Study Guide And Intervention Inscribed Angles - [desc-2] Skip to content. ... 10 4 Study Guide And Intervention Inscribed Angles [desc-7] [desc-6] [desc-8] 10 4 Study Guide And Intervention Inscribed Angles. 10 4 Study Guide And Intervention Inscribed Angles ... Solved Study Guide And Intervention Similar Triangles The Chegg [title-1] [url ... craigslist madison cars by ownermeggie b Section 4-1: Angles of Triangles. Section 4-2: Congruent Triangles. Section 4-3: Proving Triangles Congruent - SSS, SAS. Page 310: ... Section 4-7: Triangles and Coordinate Proof. Page 341: Chapter 4 Study Guide and Review. Page 345: Chapter 4 Practice Test. Page 348: Chapter 4 Preparing for Assessment. Exercise 1. Exercise 2. Exercise 3 ...The other two angles are called remote interior angles. 4.3 Exterior Angle Theorem: The measure an exterior angle is equal to the sum of the measures of its remote interior angles. m m m4 1 2 4.3 mJ 4.3 Third Angle Theorem: If two angles of one triangle are congruent to two angles of another cheap sam Study Guide And Intervention Postulates And Paragraph Proofs Answers. Dec 3, 2020 — 5-1 study guide and intervention bisectors of triangles find the measure of fm. ... guide and intervention, isosceles triangles skills practice 4 6 answer key ... in exercises 28-30, write a paragraph proof for these other theorems .... woodypercent27s north eastsks amrkayaverage man Two lines have the same slope if and only if they are parallel. Two lines are perpendicular if and only if the product of their slopes is 21. Example 1 Find the slope of a line parallel to the line containing A(23, 4) and B(2, 5). Find the slope of @#$. AB Use (23, 4) for (x 1, y 1) and use (2, 5) for (x 2, y 2). m. legend zelda link Practice set 1: Solving for a side. Trigonometry can be used to find a missing side length in a right triangle. Let's find, for example, the measure of A C in this triangle: We are given the measure of angle ∠ B and the length of the hypotenuse , and we are asked to find the side opposite to ∠ B . The trigonometric ratio that contains both ...Lesson 4-5. PERIOD. 4-4 Study Guide and Intervention sks rjal ma badhkhwrdn kyrayranywhy do you look 50 and you 11.03.2022 · About 5 Key Unit Answer Geometry 1 Basics Homework . Some of the worksheets displayed are Geometry unit 1 workbook, Unit 1 tools of geometry reasoning and proof, Geometry unit answer key, Unit 1 points lines and planes homework, Coordinate geometry mathematics 1, Practice test unit 1 name, 2 the angle addition postulate, The ... | 677.169 | 1 |
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Warm upsupplementary Properties of parallelograms • Opposite sides of a parallelogram are parallel • Opposite sides are congruent • Opposite angles of a parallelograms are congruent. • Diagonals of a parallelogram bisect each other • Consecutive angles of a parallelogram are supplementary
Properties of a Rhombus (Rhombi) • A rhombus is a parallelogram (this means it has ALL of the characteristics of a parallelogram) In addition: • A rhombus has four congruent sides • The diagonals of a rhombus are perpendicular • The diagonals bisect opposite angles
Properties of Rectangles • A rectangle is a parallelogram (this means it has ALL the characteristics of a parallelogram) IN ADDITION: • Four right angles • The diagonals of a rectangle are congruent and they bisect each other
Properties of Squares • A square is a parallelogram, a rectangle, and a rhombus (It has ALL those characteristics!!!) • Has four congruent sides • Has four right angles • The diagonals of a square: • bisect each other • are congruent • are perpendicular. • Bisect opposite angles
Cut and sort • Go to the back of the room and get worksheets 2-15A and 2-15B and a pair of scissors • Your instructions are on 2-15B • Have fun Ü • Then complete 2-16 | 677.169 | 1 |
You need to construct a regular polygon. When you draw two sides, the interior angle created between them is 120°. What will be the sum, in degrees, of the measures of the interior angles of this polygon when it is completed?
We can now calculate the number of exterior angles the shape has and, since an exterior angle is formed by the extension of one side of the shape, this will also equate to the number of sides the shape has:
#"number of exterior angles" = "number of sides" = 360^@/60^@ = 6#
Our calculation would suggest that our shape is a regular hexagon, which does indeed have interior angles of #120^@#.
We can now find the total of all of our interior angles: we can use this equation to do so:
#"sum of interior angles" = 180(n-2)#
Where #n =# the number of sides the shape has. Therefore:
#"sum of interior angles" = 180(6-2) = 180xx4 = 720^@#
To verify, dividing our answer by #6# should result in the size of one of the shape's interior angles, as dictated by the question:
#720/6 = 120^@ color(green)(" TRUE")#
Observe that both of our equations to find the total of the shape's interior angles require us to know the number of sides the shape has. This was a piece of information with which we were not provided: this is why we had to work this out using exterior angles.
The sum of the interior angles of a polygon can be calculated using the formula: Sum = (n - 2) * 180°, where 'n' is the number of sides of the polygon. Since the interior angle of the regular polygon is 120°, the exterior angle is 180° - 120° = 60°. For a regular polygon, the exterior angle is equal to the interior angle. Therefore, each exterior angle is 120°. The number of sides, 'n', can be found by dividing 360° (the total degrees in a circle) by the measure of each exterior angle, which is 120°. Hence, n = 360° / 120° = 3. Substituting 'n' into the formula, Sum = (3 - 2) * 180° = 1 * 180° = 180°. Thus, the sum of the measures of the interior angles of the completed regular polygon is 180° | 677.169 | 1 |
Pythagorean theorem gina wilson
Notes adapted from gina wilson, all things algebra. His backyard is a 24 meter by 45 meter rectangle. Webconsecutive interior angles. Real Estate | How To WRITTEN BY: Gina Baker Pu. 4 Pythagorean theorem, sine, cosine, tangent, and other triginoimetric identities and formulas c2>a2+b2 c^2 = a^2 + b^2 angle formed by a horizontal line and a line of sight to a point above the line. angle of depression. Unit 4 in high school mathematics usually focuses on topics such as quadratic equations, functions or geometry depending on the curriculum Pythagorean theorem, or properties of quadrilaterals.
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In the world of mathematics education, Gina Wilson's "All Things Algebra" is a renowned name, and her answer keys are highly sought after by teachers, students, and parents alike. 2) Classify as either acute, right or obtuse.
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I have two rays on a 2D plane that extend to infinity, but both have a starting point. They are both described by a starting point and a vector in the direction of the ray extending to infinity. I want to find out if the two rays intersect, but I don't need to know where they intersect (it's part of a collision detection algorithm).
Everything I have looked at so far describes finding the intersection point of two lines or line segments. Is there a fast algorithm to solve this?
It is tempting, at a first glance, to look for some fancy use of vector products and comparing angles, but think about calculations needed to get those products and look at Adam's or Peter's solution. Calculating the determinants for the equation set is almost the same as calculating vector products
If you only need to know if the rays intersect, the signs of u and v are enough, and these two divisons can be replaced by num*denom<0 or (sign(num) != sign(denom)), depending on what is more efficient on your target machine.
Please note that the rare case of det==0 means that the rays do not intersect (one additional comparison).
Small nit: for the case of det==0, it could mean there is no intersection, or it could also mean there is intersection everywhere (the input lines overlap). Perhaps it could more accurately be phrased "the rays do not have a unique intersection".
I understand this post is old, and I am not sure why or how, but this solution causes different solutions when the vector a and be are switched, notably when the solution would have been v = 2 but it equals 1 instead. The points I used were: for the first vector (7.64475393f, 5.59931898f), (6.30824f, 4.91833f) for the second vector (6.43122959f, 7.98099709f),(6.43122864f, 6.48099709f)
A ray can be represented by the set of points A + Vt, where A is the starting point, V is a vector indicating the direction of the ray, and t >= 0 is the parameter. Thus, to determine if two rays intersect, do this:
I found this post while trying to find the intersection point between two rays, based on other answers here. Just in case someone else has arrived here looking for the same answer, here's an answer in TypeScript / JavaScript.
I want only to check if the two rays intersect. I will go about it by calculating the direction of rotation of two "triangles" created from the two rays. They aren't really triangles, but from a mathematical standpoint, if I only wanted to calculate the rotation of the triangle, I only need two vectors with a common starting point, and the rest doesn't matter.
The first triangle will be formed by two vectors and a starting point. The starting point will be the first ray's starting point. The first vector will be the first ray's direction vector. The second vector will be the vector form the first ray's starting point to the second ray's starting point. From here we take the cross product of the two vectors and note the sign.
We do this again for the second triangle. Again, the starting point is the second ray's starting point. The first vector is the second ray's direction and the second vector is from the second ray's starting point to the first ray's starting point. We take the cross product again of the vectors and note the sign.
Now we simply take the two signs and check if they are the same. If they are the same, we have no intersection. If they are different we have an intersection. That's it!
An edge case in my application is very unlikely and even a false positive doesn't really hurt me. I need raw speed since I'm iterating over possibly billions of such cases. I'll just simply say if anything collinear we will just call it an intersection rather than add a few extra statements to nail down exactly what happened.
If the lines are of infinite length, then they will always intersect, unless they are parallel. To check if they are parallel, find the slope of each line and compare them. The slope will just be (y2-y1)/(x2-x1).
Sorry i edited my post to reflect what i was really asking. They lines are actually rays with a starting point but no end. Checking the slope will not give me the answer as it is possible for the "lines" to intersect on the "wrong" side of the point which counts as a miss.
I believe the question implies that they are not infinite in both directions: they have a starting vector and a direction vector. Thus a line starting at (1,2) and going off on (0,1) will not intersect with a line starting at (2,1) and going off on (1,0). If they were lines rather than 'rays', then of course your answer would be correct. | 677.169 | 1 |
Nine Geometricall Exercises: For Young Sea-men, and Others that are Studious ...
F Triangles there are Two Kinds; viz. Plain, (or Rightlined) and Spherical, (or Circular.) Either of which do confift of Six Parts; namely, of Three Sides, and as many Angles; but in this Place we. fhall only treat of the Plain.
I. A Plain (or Right Lined) Triangle confifteth of Three Sides, Fig. I. and as many Angles: And fuch are the Two Figures CBA and CDB; in which, in the firft Figure A B, BC and C A, are the Three Sides of the Triangle CBA, and CA B,. A CB and CBA, are the Three Angles of the fame Triang CB A.
And Note here] That an Angle (in any Cafe) is always noted with Three Letters; the middlemoft whereof reprefents the Angular Point. So in the Triangle ABC, if I would exprefs the Angle at C, I would fay, The Angle. ACB; or, The Angle BC A.
Alfo in the fecond Figure D B C, the Lines DC, D B and CB, are the Three Sides of the Triangle DB C, and the Angles BDC, CBD and B C D, are the Angles of the fame Triangle DBC.
II. Any
Fig. 1.
Fig. II.
II. Any Two Sides of a Triangle, are called the Sides of that Angle contained by (or comprehended between) them: So the Sides B C and CA are the Sides containing the Angle BC A.
III. Every Side of a Triangle, is the fubtending Side of that Angle which is oppofite unto it. As in the Triangle ABC, BC is the fubtending Side of the Angle CAB; the Side B A fubtends the Angle BCA, and the Side C A fubtends the Angle CBA.
And here Note again,] That the greateft Side, always, Subtends the greatest Angle, the leffer Side the leffer Angle, and equal Sides fubtend equal Angles.
IV. The Meafure of an Angle is the Arch of a Circle defcribed upon the Angular Point, and is intercepted between the Two Sides containing the Angle, (increafing the Sides, if Need be). So in the Triangle A B C, the Measure of the Angle CBA, is Arch C F.
V. Every Circle is divided into 360 Degrees, and every Degree into 60 Minutes, (or rather into 100 or 1000 Parts, &c.) Which Degrees are fo much the Greater, by how much the Circle is Greater; and thofe Arches which contain the fame Number of Degrees in equal Circles, are Equal: But in unequal Circles they are termed Like Arches. So the Arches C F and DE are Equal Arches, they being equal Parts of the fame Circle DHFK. But the Arches CF and O'P are Like-Arches: For, as CF is 40 Parts (or Degrees) of the Greater Circle DE FC, fois OP 40 Degrees of the Leffer Circle POG.
VI. A Quadrant (or Quarter) of a Circle, is an Arch of 90 Degrees. As is the Quadrant (or Arch) H F.
VII. The Complement of an Arch less than a Quadrant, is fo much as that Arch wanteth of 90 Degrees. So the Complement of the Arch C F 40 Degrees, is the Arch HC 50 Degrees.
VIII. The Excefs of an Arch Greater than a Quadrant, is fo many Degrees as that Arch exceedeth 90 Degrees. So the Arch DHC being 140 Degrees, is the Arch H C, 50 Degrees more than the Quadrant DH.
IX. A Semicircle is an Arch of 180 Degrees. As is the Semicirele DHF.
X. The Complement of an Arch less than a Semicircle, to a Semicircle, is fo much as that Arch wants of 180 Degrees. So the Complement of the Arch D HC 140 Degrees, to the Semicircle D HE 180 Degrees, is the Arch C F 40 Degrees.
XI. The
XI. The oppofite Angles made by the croffing of Two Diameters Fig. II. in a Circle, (or any Two other Right Lines croffing each other) are equal. So the Angles C B F and CB E, (made by the Interfection of the Two Diameters DF and CE in the Center B) are equal.
XII. An Angle is either Right or Oblique.
XIII. A Right Angle is that whofe Measure is a Quadrant or 90 Deg. So the Angles H BD, and HBF, are Right Angles, their Measures being the Quadrants DH and H F.
XIV. All Oblique Angles are either Acute or Obtufe.
XV. An Acute Angle, is that whofe Meafure is lefs than 90 Deg. So the Angles HBC 50 Deg. and C B F 40, are Oblique Acute Angles.
XVI. An Obtufe Angle, is that whofe Meafure is more than a Quadrant or 90 Deg. So the Angle DB C (confifting of 90 and 50 Deg. viz. of 140 Deg ) is an Oblique Obtufe Angle.
XVII. The Complements of Angles are the fame, as are the Complements of Arches.
XVIII. All Angles concurring (or meeting) together upon One Right Line, all of them being taken together, are equal to a Semicircle, or 180 Deg. So the Angle DBH 90 Deg. HBC 50 Deg. and CBF 40 Deg. (made by the concurring, or meeting, of the Three Lines DB, H B and C B, upon the Diameter DF in the Center B) are all of them Equal to the Semicircle D HCF, or 180 Deg.
XIX. A Triangle hath fome of its Sides Equal, or elfe they be all Unequal.
XX. A Triangle of fome Equal Sides is either Equicrural or Equilateral.
XXI. An Equicrural Triangle is that which hath only Two Equal Sides. And fuch is the Triangle D B E, whofe Sides BD and BE are Equal.
XXII. An Equicrural Triangle is Equi-angled at the Bafe. So in the Equicrural Triangle BDE, the Angles B D E and BED, at the Bafe D E, are Equal, viz. each of them 70 Deg. for the Angle DBE being 40 Deg. that taken from 180 Deg. leaves140 Deg. the half, whereof 70 Deg. is equal to the Angle BDE or DEB, and all the Three Angles equal to 180 Deg. or Two Right Angles.
This is Demonftrated in the XIth Theorem hereof.
XXIII, An
Fig. II.
XXIII. An Equilateral Triangle, is that whofe Sides are all Equal, and whofe Angles contain (each of them) 60 Deg. So the Triangle E BK hath its Sides BE, B K and E K, all of them equal; and the Angles E BK, EK B and KE B equal alfo, and each of them equal to 60 Deg. and confequently all of them equal to 180 Deg.
XXIV. A Triangle is either Right Angled or Oblique Angled. XXV. A Right-angled Triangle, is that which hath one Right Angle. And fuch is the Triangle C A B, Right-angled at A.
XXVI. An Oblique-angled Triangle is that which hath all its Angles Oblique. And fuch is the Triangle B CD.
XXVII. An Oblique-angled Triangle, is either Acute-angled, or Obtufe-angled.
XXVIII. An Oblique Acute-angled Triangle is that which hath all its Three Angles Acute. And fuch are the Triangles D B E, and EB K.
XXIX. An Oblique Obtufe-angled Triangle is that which hath One Obtufe, and Two Acute Angles. And fuch is the Triangle DBC, whofe Angle DBC is Obtufe, and the Angles BDC and BCD Acute.
Fig. III.
CHA P. II.
Of Right Lines, applied to a Circle.
Orafmuch as the Ratio or Proportion of an Arch Line to a Right Line, is as yet unknown, yet it is abfolutely neceffary that Right Lines be applied to a Circle, for the Calculation of Triangles, wherein Arch Lines come in Competition: For the Angles of Plain (or Right-lined) Triangles are measured by Arches of Circles.
Now, the Right Lines applied (or relating) to a Circle, are Chords, Sines, Tangents, Secants and Verfed Sines.
1. A Chord, or Subtenfe, is a Right Line, joining the Extremities of an Ark, as A C is the Chord of the Arks ABC and ADC.
2. A Right Sine, which is fingly called a Sine, is a Right Line, drawn from one end of an Ark, perpendicular to the Diameter drawn through to the other End: Or, it is half the Chord of twice the Ark; fo AE is the Right Sine of the Arks A B and A D. The Radius (or Sine of 90 Deg.) is called the Whole
Sine, and is the greatest of all Sines: For the Sine of an Ark Fig. III. greater than a Quadrant, is lefs than the Radius; fo F G is the
whole Sine or Radius.
3. A Verfed Sine is the Segment of the Radius between the Ark and its Right Sine; fo E B is the Verfed Sine of the Ark A B, and of the Ark AG D.
4. The Secant of an Ark, is a Right Line, drawn from the Center through one end of an Ark, till it meet with the Tangent: That is, a Right Line touching the Circle at the neareft end of that Diameter which cuts the other end of the Ark. FM is the Secant, and BM the Tangent, of the Ark A B, or of A D.
5. The Difference of an Ark from a Quadrant, (or 90 Deg.) whether it be Greater or Lefs, is called the Complement of that Ark, foGA is the Complement of the Arks A B and A G D, and HA is the Sine of that Complement: GI the Tangent of that Complement and F I the Secant of that Complement. All which (for Brevity) we write Co-Sine, Co-Tangent, Co-Secant of the Ark.
6. The Difference of an Ark from a Semicircle (or 180 Deg.) is called its Supplement; fo the Ark A B is the Supplement of the Ark DG A, to a Semicircle.
7. That Part of the Radius which is between the Centre and its Right Sine, is equal to the Co-Sine. As FE is equal to H A, and FO is equal to the Co-Sine of the Ark DS.
8. If an Ark be Greater or Lefs than a Quadrant, the Sum or Difference, accordingly, of the Radius and Co-Sine, is equal to the Verfed Sine For F D and H A together, are equal to D E, the Verfed Sine of the Ark DGA; and F B lefs by HA (or E F) is equal to E B, which is the Verfed Sine of the Ark A B. | 677.169 | 1 |
Line-segment Properties
Line-segment properties: Several of the computational-geometry algorithms in this chapter will require answers to questions about the properties of line segments. A convex combination of two distinct points p1 = (x1, y1) and p2 = (x2, y2) is any point p3 = (x3, y3) such that for some α in the range 0 ≤α≤ 1, we have x3 = αx1 (1 - α)x2 and y3 = αy1 (1 - α)y2. We also write that p3 = αp1 (1 - α)p2. Intuitively, p3 is any point that is on the line passing through p1 and p2 and is on or between p1 and p2 on the line. Given two distinct points p1 and p2, the line segment is the set of convex combinations of p1 and p2. We call p1 and p2 the endpoints of segment . Sometimes the ordering of p1 and p2 matters, and we speak of the directed segment. If p1 is the origin (0, 0), then we can treat the directed segment as the vectorp2.
In this section, we shall explore the following questions:
Given two directed segments and , is clockwise from with respect to their common endpoint p0?
Given two line segments and , if we traverse and then , do we make a left turn at point p1?
Do line segments and intersect?
There are no restrictions on the given points.
We can answer each question in O(1) time, which should come as no surprise since the input size of each question is O(1). Moreover, our methods will use only additions, subtractions, multiplications, and comparisons. We need neither division nor trigonometric functions, both of which can be computationally expensive and prone to problems with round-off error. For example, the "straightforward" method of determining whether two segments intersect-compute the line equation of the form y = mxb for each segment (m is the slope and b is the y-intercept), find the point of intersection of the lines, and check whether this point is on both segments-uses division to find the point of intersection. When the segments are nearly parallel, this method is very sensitive to the precision of the division operation on real computers. The method in this section, which avoids division, is much more accurate.
Cross products: Computing cross products is at the heart of our line-segment methods. Consider vectors p1 and p2, shown in Figure 33.1(a). The cross productp1×p2 can be interpreted as the signed area of the parallelogram formed by the points (0, 0), p1, p2, and p1p2 = (x1x2, y1y2). An equivalent, but more useful, definition gives the cross product as the determinant of a matrix:
Figure 33.1: (a) The cross product of vectors p1 and p2 is the signed area of the parallelogram. (b) The lightly shaded region contains vectors that are clockwise from p. The darkly shaded region contains vectors that are counterclockwise from p.
If p1×p2 is positive, then p1 is clockwise from p2 with respect to the origin (0, 0); if this cross product is negative, then p1 is counterclockwise from p2. Figure 33.1(b) shows the clockwise and counterclockwise regions relative to a vector p. A boundary condition arises if the cross product is 0; in this case, the vectors are collinear, pointing in either the same or opposite directions.
To determine whether a directed segment is clockwise from a directed segment with respect to their common endpoint p0, we simply translate to use p0 as the origin. That is, we let p1 - p0 denote the vector , where and , and we define p2 - p0 similarly. We then compute the cross product
(p1 - p0) × (p2 - p0) = (x1 - x0)(y2 - y0) - (x2 - x0)(y1 - y0).
If this cross product is positive, then is clockwise from ; if negative, it is counterclockwise.
Determining whether consecutive segments turn left or right: Our next question is whether two consecutive line segments and turn left or right at point p1. Equivalently, we want a method to determine which way a given angle ∠p0p1p2 turns. Cross products allow us to answer this question without computing the angle. As shown in Figure 33.2, we simply check whether directed segment is clockwise or counterclockwise relative to directed segment . To do this, we compute the cross product (p2 - p0) × (p1 - p0). If the sign of this cross product is negative, then is counterclockwise with respect to , and thus we make a left turn at p1. A positive cross product indicates a clockwise orientation and a right turn. A cross product of 0 means that points p0, p1, and p2 are collinear.
Figure 33.2: Using the cross product to determine how consecutive line segments and turn at point p1. We check whether the directed segment is clockwise or counterclockwise relative to the directed segment . (a) If counterclockwise, the points make a left turn. (b) If clockwise, they make a right turn. | 677.169 | 1 |
What are some common misconceptions about W.D. Gann Arcs and Circles?
What are some common misconceptions about W.D. Gann Arcs and Circles? A. They are symbols to represent a particular moment (time, day of week, event, seasons etc.) B. They are used to show a period of days or weeks in a new calendar. C. Arcs and Circles are used in symbols and are derived from mathematics. D. Different ways of circles and arcs are used (Circles: Rectangles, Rhombiform and Ellipses Circles: Sq's, Cross, Tri, Pentagons Arcs: Straight, Half-Right, Half, Right, Full, Quarter, Half quarter, 1/4) The definition of arcs and arches are: Formed by approximating the shape of a circle. Figure of art (Arch) A word derived from Greek for bridge. An object or device that spans a gap, click to read or entrance. see this here word derived from Greek for bridge.
Harmonic Vibrations
An object or device that spans a gap, opening, or entrance. Figure of art (Arch) link word derived from Greek for bridge. An object not used for actual purposes of life. Arch as it relates to arches It can describe: A ridge or ledge that divides open land on a valley floor. Bridge A main road crossing a river valley or a town. The arch style window. A window made of overlapping horizontal spandrels (the spaces above the arch's horizontal panels) It can describe: A ridge or ledge that divides open land on a valley floor. Bridge A main road crossing a river valley or Homepage town. The arch style window. It describes: A window made of 2 or more arc panels. For instance, the upper and lower edges of the arch. How are Arcs and Circles Used? A common place to see Arcs and Circles are on clocks, wrist watches, mirrors, wall clock, etc. In geography they are a common symbol used for mountain ranges, countries, cities.
Hexagon Analysis
Arcs and circles sometimes used in books as symbols for chapters, partsWhat are some common misconceptions about W.D. Gann Arcs and Circles? 1/4/13 10:30 AM – 4/28/13 11:30 AM The next time you are reading an ad copy in your paper, take a moment to consider it. Consider every line and every color. Then think about the way advertising works. It's not the line, or color of the ad, or the pictures that make someone draw a check. All these things are important. But there's another step to consider. It's called psychology. Try to get into the mind of the consumer and consider what makes someone run, or not run, your ad. The mental picture someone makes of an ad in order to make that cash trip to the bank will always be influenced by the shape of the ad itself. You can't control that process. So what happens if you gave someone a circle instead of a line? If you gave 20% color to a line on a black and white ad, would they be more willing to invest in that product or service? More likely you'll give the customer a line today, because advertisers already know that the line is all they really need to create a brand image.
Time Cycles
If you hand them a circle, the ad will still have a line in it, but you'll be creating an image in the mind of the consumer that This Site there. They're not completely safe. They still know you're just as honest as the line, in the worst sense of the word honesty. What I'm getting at is that there are several things they know about the W.D. Gann & Co. ad company that any competitor would understand. But it's those very things about the company that has helped the W.D. Gann & Co. advertising company grow so well and so consistently in fifty year. Gann may have started his career as a painter, but he became an advertising manWhat are some common misconceptions about W.D.
Celestial Resonance
Gann Arcs and Circles? Please respect our community guidelines when posting comments on this site. Any profanity, malicious or vulgar remarks, lewd or more than friendly suggestions will be removed. Users will not choose to publish your comments. You will have 180 seconds to edit your comments before it is published. This is cool blog thanks a lot for take time to share this with us and I Love some things in this blog Thanks a lot for sharing with us. God gave us an awesome life with Him…. God Bless and God Keep you,.. This is a good website, thanks to share, i am impressed. Good site, good luck!!!!!! GOD ALMIGHTY LETS HAVE MORE AND MORE BLESSINGS IN OUR LIVES!!! Thank you for your articles on archery.
Swing Charts
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Forecasting Methods
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Two Young Black Girls Find Trigonometric Proof Of The Pythagorean Theorem
Two young Black high school students, Calcea Johnson and Ne'Kiya Jackson from St. Mary's Academy in New Orleans, have achieved a groundbreaking mathematical feat by finding a trigonometric proof of the Pythagorean Theorem—a task considered impossible for over 2,000 years. This theorem, fundamental in geometry, states that in a right triangle, the square of the hypotenuse (c²) is equal to the sum of the squares of the other two sides (a² + b²). The challenge to prove this theorem using trigonometry was posed by their math teacher, Michelle Blouin Williams, who offered a $500 reward for a new proof, seeking to inspire creativity and ingenuity among her students.
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After two months of dedicated work, Johnson and Jackson succeeded where many believed no solution existed. They developed their proof, which they whimsically named the "Waffle Cone," utilizing a unique approach involving congruent triangles and a pattern of progressively smaller right triangles forming a larger "waffle cone" shape. This achievement not only demonstrates their exceptional skill but also places them in a rare group of individuals who have contributed a new proof to this ancient mathematical principle. Their success underscores the potential every student holds when encouraged and supported in an educational environment like St. Mary's Academy, which promotes high standards and opportunities for all its students. | 677.169 | 1 |
Define the trigonometric function for sine of angle $\alpha$ in a right triangle.
The sine of angle $\alpha$ is defined as the ratio of the side opposite angle $\alpha$ to the hypotenuse.
What is the reciprocal function of the cosine of an angle $\alpha$ in a right triangle?
The reciprocal function of the cosine of angle $\alpha$ is the secant of angle $\alpha$.
Express the tangent function of an angle $\alpha$ in terms of the side opposite and the side adjacent to $\alpha$.
The tangent of angle $\alpha$ is equal to the ratio of the side opposite angle $\alpha$ to the side adjacent to $\alpha$.
What is the definition of the cosecant of an angle $\alpha$ in trigonometry?
The cosecant of angle $\alpha$ is defined as the reciprocal of the sine of angle $\alpha$.
How is the cotangent of an angle $\alpha$ related to the tangent of the same angle?
The cotangent of angle $\alpha$ is the reciprocal of the tangent of the same angle.
Study Notes
Trigonometric Identities: Unraveling the Relationships in Angles
Trigonometry, a discipline that studies the relationships between the sides and angles of triangles, is a fundamental tool in mathematics with countless applications in physics, engineering, and other fields. At the heart of trigonometry lie the trigonometric identities, which establish connections between the six primary trigonometric functions: sine, cosine, tangent, cosecant, secant, and cotangent.
Defining Trigonometric Functions
Let's imagine a right triangle with angles (\alpha), (\beta), and (\gamma), where (\gamma = 90^\circ). The side opposite (\alpha) is denoted as (opposite)(\alpha) (or (opp_\alpha)), the side adjacent to (\alpha) is (adjacent)(\alpha) (or (adj_\alpha)), and the hypotenuse, (hyp), is the longest side. The trigonometric functions are then defined as follows:
[
\sin\alpha = \frac{opp_\alpha}{hyp}
]
[
\cos\alpha = \frac{adj_\alpha}{hyp}
]
[
\tan\alpha = \frac{opp_\alpha}{adj_\alpha}
]
The cosecant, secant, and cotangent functions are the reciprocals of the sine, cosine, and tangent functions, respectively:
[
\csc\alpha = \frac{1}{\sin\alpha}
]
[
\sec\alpha = \frac{1}{\cos\alpha}
]
[
\cot\alpha = \frac{1}{\tan\alpha}
]
Trigonometric Identities
Trigonometric identities are mathematical equations that connect different trigonometric functions or the same function evaluated at different angles. They are essential for simplifying and solving trigonometric problems. Some of the most commonly used identities include:
Uses of Trigonometric Identities
Converting expressions with one trigonometric function into expressions with another trigonometric function.
Simplifying and rearranging trigonometric expressions.
Deriving relationships between sides and angles in triangles.
Solving problems in physics, engineering, and other fields.
Conclusion
Trigonometric identities are the cornerstone of trigonometry, providing the tools needed for working with angles and their associated trigonometric functions. By appropriately applying these identities, we can simplify and solve problems in a variety of fields, from math and physics to engineering and beyond. These identities not only serve as a foundation for the subject but also demonstrate the beautiful connections within mathematics.
Delve into the world of trigonometry and discover the intricate relationships between angles through trigonometric identities. Learn how these identities connect various trigonometric functions and how they are crucial for solving complex mathematical problems. Explore Pythagorean identities, co-function identities, angle addition formulas, and more. | 677.169 | 1 |
By now, you are familiar with sine, cosine, and tangent, and how to graph these curves. However, there is so much more to trigonometry than just these three functions. We'll start by exploring the inverses of these functions -- , , and -- and how they relate to the more familiar . Be sure to pay special attention to the restrictions placed on the original functions to get to the inverse functions.
But first, let's start with a quick review of what trigonometric functions are. Trigonometric functions are a set of functions that take an angle as an input and output a value that represents a ratio of the sides of a right triangle. The most commonly used trigonometric functions are sine, cosine, and tangent.
Inverses of Sine, Cosine, and Tangent
Now, let's move on to the . These functions are the inverses of the trigonometric functions we just discussed. In other words, they take a ratio of the sides of a right triangle as an input and output an angle. The inverse trigonometric functions are denoted with an "arc" in front of the function name, such as arcsine, arccosine, and arctangent.
The inverse trigonometric functions allow us to find the missing angle in a right triangle when we know the ratios of the sides. For example, if we know the ratio of the opposite side to the hypotenuse in a right triangle, we can use the arcsine function to find the angle opposite to that side. Similarly, if we know the ratio of the adjacent side to the hypotenuse, we can use the arccosine function to find the angle adjacent to that side. And if we know the ratio of the opposite side to the adjacent side, we can use the arctangent function to find the angle opposite to that side.
The shorthand forms of these functions are
The graphs of the functions are shown below:
Graph of arcsine. Image courtesy of Math LibreTexts.
Graph of arccosine. Image courtesy of Math.net.
Image courtesy of Wiktionary.
As you can see, the graphs of arcsine and arccosine are finite, but the graph of arctangent goes on forever. This is due to the domain and range restrictions placed on the original sine, cosine, and tangent functions.
Restricting the Domain of the Sine Function
The sine function is periodic, which means that it repeats its values after a certain interval, known as the . The period of the sine function is 360 degrees or 2𝛑 radians. This means that for any angle, there are an infinite number of angles that will give the same sine value. For example, the sine of 30 degrees is the same as the sine of 390 degrees, because 30 + 360 = 390. Similarly, the sine of 45 degrees is the same as the sine of 405 degrees, and so on.
When finding the inverse sine, we need to take the sine of an angle and find the angle that produced it. But since the sine function has a period of 2𝛑 radians, there are an infinite number of angles that can produce the same sine value, making it impossible to determine a unique angle. To ensure that we get a unique solution, we need to restrict the domain of the input value of sine to a specific range of angles.
The arcsine function, which is the inverse of the sine function, is defined over the range [-90,90] degrees or [-𝛑/2,𝛑/2] radians. This means that the output value for the arcsine function must be in this domain, so that we get a unique solution for the angle. To achieve this restricted range, we need to restrict the domain of the sine function to [-𝛑/2, 𝛑/2] radians since the domain of the original function becomes the range of the inverse function. The reason we denote the range in angle brackets is this is a , meaning -𝛑/2 and 𝛑/2 are included in the range and are valid outputs for the arcsine function.
Restricting the Domain of the Cosine Function
The cosine function, like the sine function, is periodic and has a period of 360 degrees or 2π radians. This means that for any angle, there are an infinite number of angles that will give the same cosine value. For example, the cosine of 60 degrees is the same as the cosine of 420 degrees (60+360=420), and the cosine of 75 degrees is the same as the cosine of 435 degrees, and so on.
Also like the sine function, there are an infinite number of angles that can produce the same cosine value, making it impossible to determine a unique angle. To ensure that we get a unique solution, we need to restrict the domain of the input value to a specific range of angles for the cosine function.
The arccosine function, which is the inverse of the cosine function, is defined over the range [0,180] degrees or [0,𝛑] radians. This means that the input value for the arccosine function must be in this range, so that we get a unique solution for the angle. Again, notice the square brackets denoting the closed interval for this function.
The range of the cosine function is [-1,1], which means that the output values of the cosine function are within this range. However, the cosine is non-negative for values in the range of [0,180] degrees or [0,𝛑] radians. Therefore, when we use the arccosine function, which is the inverse of the cosine function, we can restrict the domain of the input values to [0,180] degrees or [0,𝛑] radians to ensure that we get a unique solution.
Restricting the Range of the Tangent Function
The tangent function is defined over all real numbers except for multiples of 𝛑/2, and it is periodic with a period of 180 degrees or 𝛑 radians, which means that it repeats itself every 180 degrees or 𝛑 radians. The tangent function takes an angle as input and returns a value that represents the ratio of the opposite side to the adjacent side of a right triangle.
The range of the tangent function is all real numbers, which means that the output values of the tangent function can be any real number. However, the tangent function is not defined at certain angles, such as 90, 270, 450 and so on, which are known as the "singularities" or "undefined" values of the tangent function. Therefore, when we use the arctangent function, which is the inverse of the tangent function, we can restrict the domain of the input values to the tangent function to (-90,90) degrees or (-𝛑/2,𝛑/2) radians to ensure that we get a unique solution. This is because the arctangent function takes a real number as an input, and returns the angle that produced that value. Notice the open parentheses on this interval that show the arctangent function is defined on an and will not output a value that is equal to -𝛑/2 or 𝛑/2, because the tangent value for these angles is undefined.
For example, if we want to find the angle that produced the tangent value of 1, we can use the arctangent function. The arctangent of 1 is 45 degrees, which is the angle that produced the tangent value of 1. If the input value for tangent is not in the range of (-90, 90) degrees or (-𝛑/2, 𝛑/2) radians, it might return an angle that is not unique, for example, 315 degrees, which also produces the same tangent value of 1. By restricting the domain, we can ensure that the output angle is unique and corresponds to a real-world angle.
Using the Unit Circle to Evaluate Inverse Trig Functions
The , as mentioned previously, is a very powerful tool that has a variety of uses. One of these uses is solving an inverse trigonometric function. The input of an inverse trig function (the value inside the parentheses) is the sine/cosine/tangent of the angle, so the question you are trying to answer is: what angle has a sine/cosine/tangent value of x?
Let's make this idea more concrete with an example: What does arctangent(1) evaluate to?
The question we are trying to answer here is: what angle has a tangent value of 1?
Image courtesy of Remind.
Looking at the unit circle, and trying out different angles within (-𝛑/2, 𝛑/2), we can see that the tangent value at 𝛑/4 is 1, so our answer is 𝛑/4 because that is the angle.
Practice Problems
1. What is the value of arcsin(-1/2)?
a) -30 degrees
b) -60 degrees
c) 30 degrees
d) 60 degrees
Answer: b) -60 degrees
2. What is the value of arccos(√2/2)?
a) 30 degrees
b) 45 degrees
c) 60 degrees
d) 90 degrees
Answer: d) 45 degrees
3. What is the value of arctan(√3)?
a) 45 degrees
b) 60 degrees
c) 90 degrees
d) 120 degrees
Answer: a) 60 degrees
Key Terms to Review (11)
Arccosine: The arccosine of a number is the angle whose cosine is that number. It gives us the measure of an angle in radians or degrees.
Arcsine: The arcsine of a number is the angle whose sine is that number. It gives the measure of an angle in radians or degrees.
Arctangent: The arctangent of a number is the angle whose tangent is that number. It gives us the measure of an angle in radians or degrees.
Closed Interval: A closed interval is a set that includes both endpoints. It represents a range of values that starts at one point and ends at another, including those points.
Domain and Range: The domain of a function refers to all the possible input values or x-values for which the function is defined. The range of a function refers to all the possible output values or y-values that the function can produce.
Inverse Trigonometric Functions: Inverse trigonometric functions are functions that undo or reverse the effects of trigonometric functions. They allow us to find angles based on ratios, rather than finding ratios based on angles.
Open Interval: An open interval is a set of real numbers between two endpoints, where the endpoints are not included in the interval.
Period: The period of a function is the distance between two consecutive points on the graph that have the same value. It represents the length of one complete cycle of the function.
Periodic Function: A periodic function is a type of mathematical function that repeats its values in regular intervals. The interval at which the function repeats is called the period.
Trigonometric Functions: Trigonometric functions are mathematical functions that relate angles in a right triangle to ratios of side lengths. Common trigonometric functions include sine, cosine, and tangent.
Unit Circle: The unit circle is a circle with a radius of 1 unit centered at the origin (0,0) on a coordinate plane. It is used to understand trigonometric functions and their relationships. | 677.169 | 1 |
Circle A has a radius of #1 # and a center of #(5 ,2 )#. Circle B has a radius of #2 # and a center of #(4 ,5 )#. If circle B is translated by #<-3 ,4 >#, does it overlap circle A? If not, what is the minimum distance between points on both circles?
To determine if circle B overlaps circle A after being translated by <-3, 4>, we need to calculate the distance between the centers of the circles after the translation and compare it to the sum of their radii.
The distance between the centers of the circles can be found using the distance formula:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
For circle A, the center is at (5, 2), and for circle B after translation, the center becomes (4 - 3, 5 + 4) = (1, 9).
\[ \text{Distance} = \sqrt{(1 - 5)^2 + (9 - 2)^2} = \sqrt{16 + 49} = \sqrt{65} \]
The sum of the radii of circle A and circle B is \( 1 + 2 = 3 \).
Since the distance between the centers of the circles (\( \sqrt{65} \)) is greater than the sum of their radii (3), the circles do not overlap.
To find the minimum distance between points on both circles, we subtract the sum of the radii from the distance between their centers:
\[ \text{Minimum distance} = \sqrt{65} - 3 \ | 677.169 | 1 |
72
Page 7 ... greater than a quadrant , the angle is called obtuse . The magnitude of an angle may be estimated or measured by means of any particular angle , taken as the unit angle . The right angle is generally the angle chosen as the unit angle ...
Page 10 ... greater than a right angle . XV . When the sum of two angles is equal to a right angle , they are called Complementary angles ; each being the comple- ment of the other . XVI . When the sum of two angles is equal to two right angles ...
Page 13 ... greater than any of them . IX . Things which coincide , or fill the same space , are iden- tical . X. All right angles are equal to one another . XI . A straight line is the shortest distance between two points . XII . Through the same ...
Page 18 ... greater than the angle ACB , opposite this side . A D C B For , the sum of the two angles DAB , DBA is obviously less than the sum of the two angles CAB , CBA . And since the sum of the three angles of every triangle is equal to two ...
Page 19 ... greater than their difference . C First . The straight line AB is the shortest distance between A and B ( A. XI . ) , and there- fore shorter than the broken line AC + CB . The same reasoning applies to each of the sides . Hence , we ...
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Page 80 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. A D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw the altitudes BD and B'D'.
Page 28 - If two triangles have two sides of the one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first is greater than the third side of the second. Given A ABC and A'B'C ' with Proof STATEMENTS Apply A A'B'C ' to A ABC so that A'B | 677.169 | 1 |
Unit 8 polygons and quadrilaterals answer key
Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere. Saxon 1992-09 Geometry and Billiards Serge Tabachnikov 2005 This book is devoted to billiards in their relation with differential geometry, classical mechanics, and geometrical optics. Check Details Unit 7 polygons and quadrilaterals answer key pdf contents 1. Final answer: A trapezoid is a type of quadrilateral with at least one pair of parallel sides, called bases. Proof: Opposite angles of a parallelogram. In today's competitive job market, it is crucial to be well-prepared for interviews. | 677.169 | 1 |
What angle magnitude of cross product and dot product of two vectors are equal?
Ans: When angle between two vectors is 45 degree, cross product and dot product of two vectors are equal.
Is magnitude of cross product equal to dot product?
Cross product will results in a vector, with a magnitude and direction – even if the magnitude may be 0, and dot product will results in a scalar. Adding to that, the magnitude of the cross product can be the same as the result of the dot product when , assuming that is the angle between the two operand vectors.
What does it mean when the dot product equals the cross product?
A dot product is the product of the magnitude of the vectors and the cos of the angle between them. A cross product is the product of the magnitude of the vectors and the sine of the angle that they subtend on each other. The dot product is zero when the vectors are orthogonal ( θ = 90°).
How is the magnitude of the cross product of two vectors related to the magnitude of the individual vectors and the angle between them?
The equation to calculate a cross product is pretty simple. The cross product between vectors A and B is equal to the magnitude of vector A multiplied by the magnitude of vector B multiplied by sine of the angle between them. That will give you the magnitude of your answer.
What is the cross product of two equal vectors?
When two vectors are multiplied with each other and the product of the vectors is also a vector quantity, then the resultant vector is called the cross product of two vectors or the vector product. The resultant vector is perpendicular to the plane containing the two given vectors.
What is the magnitude of a cross product?
The magnitude of the resulting vector from a cross product is equal to the product of the magnitudes of the two vectors and the sine of the angle between them.
What is the dot product of two cross products?
Given two unit vectors, their cross product has a magnitude of 1 if the two are perpendicular and a magnitude of zero if the two are parallel. The dot product of two unit vectors behaves just oppositely: it is zero when the unit vectors are perpendicular and 1 if the unit vectors are parallel.
How do you find the magnitude of the cross product of two magnitudes?
Can cross product and dot product of two vectors be the same?
Dot product and Cross product of two vectors cannot be same because resultant of dot product is a scalar quantity and that of vector product is a vector. What should be the angle between two vectors of equal magnitude so that the magnitude of the resultant is equal to the magnitude of two vectors? The required angle is 120 degree.
Using the known formulas for dot and cross product relating the magnitudes and the angle we get for vectors a,b The ratio of the magnitude of cross and dot products of two vectors is tan θ . If they were equal , then tan θ = 1 which implies θ = 45 degree.
How do you find the angle between two vectors using cross product?
Angle Between Two Vectors Using Cross Product. The formula of the angle between two vectors using the cross product is as follows: a → × b → = | a → | | b → | s i n θ n ^. , where, n ^. denotes the unit vector that shows the direction of the multiplication of two vectors.
What is the difference between dot product and cos product?
Both the definitions are equivalent when working with Cartesian coordinates. However, the dot product of two vectors is the product of the magnitude of the two vectors and the cos of the angle between them. To recall, vectors are multiplied using two methods | 677.169 | 1 |
In construction of similar triangles, if scale factor is more than 1, then the new triangle is the given one.
A
larger than
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
smaller than
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C
congruent to
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Solution
The correct option is A larger than Scale factor basically defines the ratio between the sides of the constructed triangle to that of the original triangle. So when we see the scale factor is greater than 1, it means the sides of the constructed triangle is larger than the original triangle i.e. the triangle constructed is larger than the original triangle. Similarly, if scale factor is less than 1 then the sides of the constructed triangle is smaller than that of the original triangle i.e. the constructed triangle is smaller than the original triangle. And when we have scale factor equal to 1, then the sides of both the constructed triangle and that of the original triangle is equal. | 677.169 | 1 |
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Syed Misbahuddin Husain,
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Take x and y in the polar form of the ellipse. Assume the vertiecs of the triangle on the elliipse. If I have the apex of the isoceles triangle in the x axis, the two equal sides of the triangle will extend to the either Ist and 4thOR 2ndand 3rd quadrant. Accordingly, name the coordinates polarity and find the area of the triangle using cartesian coordinates as you have coordinates of all the three vertices. Then substitute the x and y with their polar eqns as done in the first step. Then using the application of derivatives, find the maxima value using 2ndorder differentiation test. | 677.169 | 1 |
Geometry Essentials For Dummies
Geometry Essentials For Dummies(9781119590446) was previously published asGeometry Essentials For Dummies (9781118068755). While this version features a newDummiescover and design, the content is the same as the prior release and should not be considered a new or updated product.
Just the critical concepts you need to score high in geometry
This practical, friendly guide focuses on critical concepts taught in a typical geometry course, from the properties of triangles, parallelograms, circles, and cylinders, to the skills and strategies you need to write geometry proofs. Geometry Essentials For Dummies is perfect for cramming or doing homework, or as a reference for parents helping kids study for exams.
Get down to the basics — get a handle on the basics of geometry, from lines, segments, and angles, to vertices, altitudes, and diagonals | 677.169 | 1 |
Exploring a Sequence of Transformations Level 3
Move a figure with a sequence of transformations while avoiding the barrier.
Putting It All Together
Answer these open ended questions on your own or with others to form deeper math connections.
Open-ended question
Is there a set of transformations that will always give the same image as reflecting a preimage across the -axis and then the -axis? If so, describe an example. If not, describe how you would change the question so the answer is yes. | 677.169 | 1 |
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.
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Solution
Take a point C on the graph such that ABCD is a square i.e all sides AB, BC, CD and AD are equal. So, the abscissa of C should be equal to the abscissa of B i.e -2 and ordinate of C should be equal to the ordinate of D. i.e -4.
Hence, the coordinates of C are (-2, - 4).
The graph obtained by plotting the points A, B and C and D is given below. | 677.169 | 1 |
Triangles
Do you know the features of a triangle? In this article, we are going to complete your information about triangles.
Triangles are closed shapes having three \(3\) angles, three sides, as well as three vertices.
Triangles with \(3\) vertices say \(P, Q,\) and \(R\) are characterized as \(△PQR\). It's additionally called a \(3\)-sided polygon or a trigon.
Crucial properties of triangles are shown here:
Triangles have\(3\) sides, angles, and vertices.
The angle total property of a triangle says the amount of the \(3\) inner triangles is constantly \(180°\). Like with any particular triangle \(PQR\), the angle \(P +\) angle \(Q +\) angle \(R = 180°\).
Triangle's inequality property says the amount of the two sides' length for triangles is bigger than the \(3\)rd side.
Like in the Pythagorean theorem, with a right triangle, the square of the hypotenuse equates to the quantity of the squares of the additional \(2\) sides such as \((Hypotenuse² = Base² + Altitude²)\).
The side which is opposite the larger angle is the one that is the longest.
The outer angle's property of a triangle says the outer triangle angle always is equivalent to the total of the inside opposite angles.
Related Topics
Right Triangles
A right angle's definition says if one of the triangle's angles is a right angle – \(90º\), it's known as a right-angled triangle or merely, a right triangle.
Several critical properties characterize and assist in identifying right triangles.
The biggest angle is constantly \(90º\).
The biggest side is known as a hypotenuse, and constantly it's the side opposite of a right angle.
The Pythagoras rule governs the dimensions of the sides.
It can't contain an obtuse angle.
Acute triangles:
Acute triangles are those classified based on the angles' measurements. Should every inner angle in the triangle be lower than \(90°\), it's an acute triangle.
Acute Angle Triangles' properties are:
Based on the angle's sum property, all \(3\) inner angles of the acute triangle combine to form \(180°\).
Triangles can't be both right-angled triangles and acute-angled triangles all at once.
Triangles can't be acute-angled triangles as well as obtuse-angled triangles all at once.
The angle's property of an acute triangle declares the inner angles of acute triangles are constantly fewer than \(90°\) or are in-between (\(0°\) to \(90°\)).
The side that is opposite to the tiniest angle is the tiniest triangle side.
Obtuse Triangles:
Within geometry, obtuse scalene triangles are defined as triangles with one angle measuring over \(90\) degrees, yet lower than \(180\) degrees, plus the additional \(2\) angles are a smaller amount than \(90\) degrees. All \(3\) sides, as well as the angles, vary in length. | 677.169 | 1 |
Parts Of A Circle Worksheet
Parts Of A Circle Worksheet - Parts of a circle worksheet description. Each worksheet has 9 problems identifying basic parts of a circle such as center, radius and diameter. What are the parts of a circle? Know the definitions for the different parts of a circle Identify and name the different parts of a circle; In section a, pupils will mark the centre point on five different circles.
These worksheets ( with solutions) help students take the first steps and then strengthen their knowledge of the names of the parts of a circle. A simple activity to name the parts of a circle pdf, with a powerpoint for sharing the answers, plus an answer sheet for printing. After completing this worksheet, the aim is to be able to name the. Web radius, sector, arc, tangent, chord, radii, diameter, circumference. Web what does this parts of a circle worksheet teach?
Web there are also parts of a circle worksheets based on edexcel, aqa and ocr exam questions, along with further guidance on where to go next if you're still stuck. You could print off any of it for display. Learners will illustrate and name parts of circles throughout this worksheet. Web we have a set of 4 different worksheets to help you learn the parts of a circle. Web what does this parts of a circle worksheet teach?
Area of Circles Worksheet Printable PDF Worksheets
Identify and name the different parts of a circle; A simple activity to name the parts of a circle pdf, with a powerpoint for sharing the answers, plus an answer sheet for printing. Remember, we call name of a circle by its center. Know the definitions for the different parts of a circle Web radius, sector, arc, tangent, chord, radii,.
Parts of a Circle (Worksheets with Solutions) Teaching Resources
After completing this worksheet, the aim is to be able to name the. Each worksheet has 9 problems identifying basic parts of a circle such as center, radius and diameter. Draw a diagram to show each of these parts of the circle. You could print off any of it for display. Web parts of a circle worksheet.
Parts of a Circle Poster Math methods, Math posters high school, Gcse
Know the definitions for the different parts of a circle (b) a radius of 6cm. There is also a printable cheat sheet which includes a diagram and definitions for you to print. A simple activity to name the parts of a circle pdf, with a powerpoint for sharing the answers, plus an answer sheet for printing. This section contains identifying.
Parts Of A Circle Worksheet
There is also a printable cheat sheet which includes a diagram and definitions for you to print. You could print off any of it for display. The corbettmaths practice questions on the parts of the circle. Web radius, sector, arc, tangent, chord, radii, diameter, circumference. Identify and name the different parts of a circle;
Parts Of A Circle Worksheet - In section a, pupils will mark the centre point on five different circles. Web help your students prepare for their maths gcse with this free parts of a circle worksheet of 33 questions and answers. This section contains identifying parts of a circle such as tangent, chord, secant, and other basic parts. What are the parts of a circle? Web what does this parts of a circle worksheet teach? Web parts of a circle worksheet. There is also a printable cheat sheet which includes a diagram and definitions for you to print. Know the definitions for the different parts of a circle Illustrate and name parts of circles. After completing this worksheet, the aim is to be able to name the.
Web radius, sector, arc, tangent, chord, radii, diameter, circumference. Name the parts of the circle shown in each diagram. Draw a diagram to show each of these parts of the circle. Using these sheets will help your child to: (a) a radius of 4cm.
These worksheets ( with solutions) help students take the first steps and then strengthen their knowledge of the names of the parts of a circle. Web there are also parts of a circle worksheets based on edexcel, aqa and ocr exam questions, along with further guidance on where to go next if you're still stuck. Web what does this parts of a circle worksheet teach? Web help your students prepare for their maths gcse with this free parts of a circle worksheet of 33 questions and answers.
In section a, pupils will mark the centre point on five different circles. The parts of a circle are the radius, diameter, circumference, arc, chord, secant, tangent, sector and segment. Web help your students prepare for their maths gcse with this free parts of a circle worksheet of 33 questions and answers.
Web there are also parts of a circle worksheets based on edexcel, aqa and ocr exam questions, along with further guidance on where to go next if you're still stuck. What are the parts of a circle? Identify and name the different parts of a circle;
After Completing This Worksheet, The Aim Is To Be Able To Name The.
Web what does this parts of a circle worksheet teach? Illustrate and name parts of circles. Using these sheets will help your child to: Identify and name the different parts of a circle;
Web Help Your Students Prepare For Their Maths Gcse With This Free Parts Of A Circle Worksheet Of 33 Questions And Answers.
The parts of a circle are the radius, diameter, circumference, arc, chord, secant, tangent, sector and segment. Parts of a circle worksheet description. Web there are also parts of a circle worksheets based on edexcel, aqa and ocr exam questions, along with further guidance on where to go next if you're still stuck. A simple activity to name the parts of a circle pdf, with a powerpoint for sharing the answers, plus an answer sheet for printing.
The Corbettmaths Practice Questions On The Parts Of The Circle.
Learners will illustrate and name parts of circles throughout this worksheet. Name the parts of the circle shown in each diagram. Remember, we call name of a circle by its center. There is also a printable cheat sheet which includes a diagram and definitions for you to print.
You could print off any of it for display. Web parts of a circle worksheet. What are the parts of a circle? This section contains identifying parts of a circle such as tangent, chord, secant, and other basic parts. | 677.169 | 1 |
If true then enter 1 and if false then enter 0
Can two right angles be complement to each other?
We know, right angle is the angle equal to 90o.
Two right angles can never be a complement to each other. Since the right angle means 90 degrees, so the sum of two right angles will always be 180 degrees.
E.g.: Let us suppose two obtuse angles are 90o & 90o. Here, the sum of these two obtuse angles will give us 180o. Since, for being complementary, sum should be equal to 90o, we can say that no pairs of right angles are complementary. Hence, the statement is false and 0 is the answer. | 677.169 | 1 |
How do you convert the Cartesian coordinates (3,2) to polar coordinates?
1 Answer
Explanation:
To express the position of your point #P# in polar coordinates you need to supply the length #r# of the segment joining your point to the origin and the angle #theta# formed by this segment with the positive side of the #x# axis:
From Pythagoras Theorem you get: #r=sqrt(3^2+2^2)=sqrt(9+4)=sqrt(13)=3.6#
From trigonometry you get that: #theta=arctan(2/3)=tan^-1(2/3)=33.7°# | 677.169 | 1 |
A 2D Rotation Demo using SIN() and COS()
This VB code graphically shows the relationship between Sine and Cosine when drawing a cirlce. In my humble opinion, I think drawing a circle is the first thing all graphics programmers need to learn. Although it may look like I'm just plotting a single dot in a cirlce, there is another way of looking at this. You could say I am rotating the dot around it's origin. This is exactly the sort of code you will need to rotate a space ship around it's origin, ala 'Asteroids' style. In all my years of programming, I've never once seen an application such as this one. Hopefully, it will clear up what SIN and COS actually look like when plotted. | 677.169 | 1 |
Improve student understanding of angle measurement with this set of 24 task cards.
Practice Measuring Angles with a Protractor
How do you teach students to measure angles? Angles can be measured in degrees. Degrees are the most common unit of measurement for angles and are represented by the symbol °. A full rotation is equal to 360°.
If you are looking for an activity for students to practise using a protractor to measure and draw angles, you have come to the right place! Teach Starter has created a set of 24 task cards that will give them the practice they need to feel confident with this skill.
Through this activity, students will practise drawing angles with a given measurement as well as determining the measurement of an angle with their protractor.
Tips for Differentiation + Scaffolding
A team of dedicated, experienced educators created this resource to support your maths lessons.
In addition to individual student work time, use this set of task cards to enhance learning through guided maths groups, whole class lessons or remote learning assignments.
If you have a mixture of above- and below-level learners, check out these suggestions for keeping students on track with the concepts:
Support Struggling Students
Help students who need support understanding the concepts by inviting them to reference previous assignments or posters and reminding them how to use a protractor properly. Additionally, this activity can be completed in a small group or 1-on-1 setting.
Challenge Fast Finishers
Encourage students who need a bit of a challenge to create additional angles and trade with a classmate. Students can then practise measuring these angles with their protractors.
Sc 'SCOOT!' Continue in this manner until students return to their starting point.
Exit Ticket
Use these cards as a formative assessment after your lesson. Pick a random assortment of cards and project them on the board for the whole class to see. Students can draw angles with the given measurement on a sheet of paper, sticky note or their workbook.
Easily Prepare This Resource for Your Students
Use the dropdown icon on the Download button to choose between the PDF or editable Google Slides version of this resource. A recording sheet and answer key are also included with this download.
Print on cardbaord for added durability and longevity. Place all pieces in a folder or large envelope for easy access.
This resource was created by Kaylyn Chupp, a Teach Starter collaborator.
Don't stop there! We've got more activities and resources that cut down on lesson planning time: | 677.169 | 1 |
Videos in this series
Please select a video from the same chapter
This is the second video in the series for the Further Maths Units 3 and 4 course. It continues looking at Measurement and Geometry by reviewing the work which has previously been covered on triangles. Looking at the different types of triangles as well and interior and exterior angles. A quick look at bisecting angles leads into a number of worked examples which complements the theory from the start of the video | 677.169 | 1 |
Geometry – Drawing Extreme Diagrams
December 25, 2021 per se; instead, they are logical puzzles. If you can prove why some things will not work, it means whatever is left will work.
Let me explain with the help of an official Data Sufficiency question.
Question:
In the figure above, is the area of the triangle equal to the area of the triangle ?
Statement 1:
Statement 2: Triangle is isosceles.
Solution:
When presented with this question, people see right triangles and jump to Pythagorean theorem, isosceles triangles and then wage a war on , , and relations. Well, that is our traditional approach. But what do we do if making equations and solving for relations isn't our style?
We make diagrams and figure out the relations! One thing that is apparent the moment we read statement 1 is that the figure is not to scale. From the figure it looks as if is greater than or at best, equal to . That itself is an indication that if you draw the figure on your own, you could see something that will make this question very simple. The question setter doesn't want to show you that and hence he made the distorted figure.
Anyway, let's first analyze the question. Then we will look at the statements.
We need to compare areas of and . Both are right angled triangles.
We need to figure out whether these two are the same.
Think about it this way – we are given a triangle with a particular area. So the length of must be defined. If is very small, (shown by the dotted lines in the diagram given below) the area of will be very close to 0. If is very large, the area will be much larger than the area of . So for only one value of , the area of will be equal to the area of .
We need to figure out whether for the given relations, the triangles have equal area.
Statement 1:
This gives us . Let's draw and such that is somewhat shorter than . Now can we say that the areas of the two triangles are the same? No. The area of is decided by and both not just . We can vary the length of to see that the relation between and is not enough to say whether the areas will be the same (see the diagrams given below).
So this statement alone is not sufficient.
Statement 2: Triangle is isosceles.
This means that . Notice that the triangle is right angled so the hypotenuse must be the largest side. If is isosceles, it means that the two legs of the triangle must be equal. Hence sides of must be in the ratio .
Since we only need to consider relative length of the sides, let's say that , and or some multiple thereof.
We have no idea about the length of so this statement alone is also not sufficient.
Let's consider both statements together now:
(Since )
Both triangles have the same area. Sufficient!
Answer (C).
Now compare this approach with your Pythagorean approach. Is this simpler? | 677.169 | 1 |
Geometry
If you are aware of elementary facts of geometry, then you might know that the area of a disk with radius $ R$ is $ \pi R^2$ . Premise The radius is actually the measure(length) of a line joining the center of disk and any point on the circumference of the disk or any otherIn 1904, the french Mathematician Henri Poincaré (en-US: Henri Poincare) posed an epoch-making question, which later came to be termed as Poincare Conjecture, in one of his papers, which asked: If a three-dimensional shape is simply connected, is it homeomorphic to the three-dimensional sphere? Henri Poincare - 1904 So what does it really mean? How | 677.169 | 1 |
Convex
In computer graphics, a convex polygon is a polygon in which all of the interior angles are less than or equal to 180 degrees. This means that a convex polygon can be drawn without crossing itself.
Here are some examples of convex polygons:
A triangle: A triangle is always a convex polygon. This is because the sum of the angles in a triangle must be 180 degrees, and if any angle is greater than 180 degrees, then the sum of the angles would be greater than 180 degrees.
A square: A square is also a convex polygon. This is because the sum of the angles in a square is 360 degrees, and no angle in a square is greater than 180 degrees.
A pentagon: A pentagon is a convex polygon if and only if none of its interior angles measures more than 180 degrees.
Here are some of the benefits of using convex polygons in computer graphics:
They can be more easily rendered: Convex polygons can be more easily rendered than concave polygons because they have fewer edges and vertices. This is because each edge and vertex must be calculated individually when rendering a convex polygon.
They can be more easily tesselated: Tessellation is the process of dividing a polygon into smaller polygons, and convex polygons can be more easily tesselated than concave polygons. This is because the smaller polygons can be created in a way that preserves the convex shape of the original polygon.
They can be more easily used in collision detection: Collision detection is the process of determining whether two objects have collided. Convex polygons are easier to use in collision detection than concave polygons because the boundaries of convex polygons are simpler to define.
Here are some of the drawbacks of using convex polygons in computer graphics:
They can be less efficient for rendering some types of objects: Convex polygons can be less efficient for rendering some types of objects, such as objects with curved surfaces. This is because convex polygons cannot represent curved surfaces as accurately as concave polygons.
They can be less efficient for collision detection: Convex polygons can be less efficient for collision detection for objects with many convex polygons. This is because the collision detection algorithm must check for collisions between all of the convex polygons in the object.
They can be less efficient for tessellation: Convex polygons can be less efficient for tessellation for objects with many convex polygons. This is because the tessellation algorithm must divide each convex polygon into smaller polygons.
Overall, convex polygons offer a number of benefits, but they can also have some drawbacks. It is important to consider the needs of your application when deciding whether or not to use convex polygons.
Feedback
Please be sure to submit issues or feature requests through the embedded feedback form. In the event it is a major issue please contact us directly through Discord. | 677.169 | 1 |
Geometry
Geometry is one of the oldest and main branches of mathematics.
Measurement of the earth is the exact meaning of the word 'Geometry'.
The geometry began when men felt the need to measure their lands while
buying and selling. Various shapes and figures with which we deal in
geometry are called geometrical figures. In 5th grade geometry, we learn
about the construction of geometrical figures and study their basic
properties. Thus, one can say that geometry is the science of properties
and relations of figures.
We will learn about some basic concepts and terms in geometry.
In geometry terms namely point, line and plane form the foundation of geometry. These terms point, line and plane are cannot be precisely defined. However we give examples to illustrate the meaning of these terms. Geometry is all about describing shapes and their properties.
We have already learned about points, line segments, rays, straight lines, angles etc. Hence, we will revise these terms in brief.
Geometry is a science in which we study some properties and
relations of points, lines, planes and solids in space.
Let us recall and review some of the concepts we developed
earlier.
Point, Line, Line Segment, Ray and Straight Line:
Point: A circle of zero radius is known as point. Point is
simply a dot (.) A point has no length, no breadth or no height (thickness).
A point is represented by a dot (.)
Points are named with capital letters A, B, C, D, etc.
Collinear Points: The points lying on the same straight line are called collinear points. Points A, B, C, D, E are collinear points because all of them are lying on the same straight line.
Non-Collinear Points: The points which are not on the same straight line are called non Collinear Points. Points A, B, C and D are non collinear points because all of them are not on the same straight line.
Line: Points join together to form a line.
A line has no end points.
Given below is line MN.
A line represented by two points on it.
Line MN = \(\overleftrightarrow{MN}\)
The two arrows show that it extends indefinitely to both
directions.
Line Segment:
A line segment is a part of a line.
It has two end points.
The given figure shows a line segment AB. It has two
end points A and B. Every line segment has a definite measure which is equal to
its length.
A line segment is represented with the two end points.
Line segment AB = AB
Ray:
A ray has an end point on one side and it extends
indefinitely on the other side. Given below is a ray ST.
A ray is represented by the end point and another point on
it.
The given figure shows ray AB. The symbol for the ray AB
is \(\overrightarrow{AB}\). It has one end point. It can be extended to
any length in the direction of B
from A.
Straight Line: The given figure shows a straight line AB. The
symbol for the straight line AB is \(\overleftrightarrow{AB}\). It has no end points.
Plane: A plane is a flat surface. A
plane extends in all the four directions infinitely endlessly in length and
breadth.
A plane cannot be drawn on a piece of paper. It has no
boundary and what we draw on a paper is a part of a plane only but not the
plane itself. Table top, wall, roof of the room etc. are the examples of the
part of the plane.
In short, a plane is a flat surface like a table top, a book etc. The
plane can be extended in all the directions. So, a part of plane can only be
represented on a sheet of paper. Points and lines lie on a plane. It is named
by marking three points on it.
Hence, flat shapes like lines; circles and triangles that can be drawn on a flat surface for example on a piece of paper are called plane geometry.
The three dimensional objects like cubes, cuboids, prisms, cylinder and pyramids are called solid geometry | 677.169 | 1 |
Measuring Angles Worksheets
When students get into Grade 4, teachers introduce symmetry, points, segments, lines, and other aspects of geometry to them. With each following grade, kids continue to learn how to classify and measure objects based on their angles. Continue reading to see more about a measuring angles worksheet.
Measuring Angles Worksheet PDF
Measuring Angles Worksheet PDF
Measuring Angles Worksheet PDF
How Brighterly's Tutors Can Help Your Child Using Angle Measuring Worksheets
At Brighterly, an online math learning platform, the tutors help students with their elementary math problems. Tutors use angle measuring worksheets to show that an angle is right, acute, or obtuse. Exercises in angles measurement worksheets move on to broad classifications depending on the angle and the type of parallel or perpendicular lines. Kids can follow in-depth instructions on how to measure angles precisely.
1:1 Math Lessons
Want to raise a genius? Start learning Math with Brighterly
While students learn how to identify and measure angles, they should complete additional activities to better understand the concept. Students will get a newfound knowledge of angles by using the worksheet. They will conveniently identify and sketch triangles and quadrilaterals based on side length or angle type (acute, obtuse, or right angles). If you want, tutors can even recommend a measuring angles worksheet in PDF, which kids can use for personal practices.
Angles And Angle Measure Worksheet Answers
Angles And Angle Measure Worksheet Answer Key
Measuring Angles Worksheet Answer Key
Measuring And Constructing Angles Worksheets
Interior And Exterior Angle Measures Worksheet
Measure Angles Worksheet
Using Measuring Angles Worksheet in a Fun Way
Kids can get involved in fun, lovely, and exciting activities with angle measure worksheets. Students can use masking tape to make intersecting lines and other angles on the board. Worksheets allow students to practice what they learn in their geometry classes. All these activities encourage learning-by-doing, so children become familiar with the concept.
Typically, a measuring angles worksheet comes with answers at the end of the book so that kids can confirm their calculations. A child can keep trying and calculating using angles or angle measure worksheet answers. Worksheets prove that math is not a subject that is impossible to learn by offering tasks that start from the basics and move towards more complex concepts. With colorful pictures, kids are going to love measuring worksheets completely struggle with understanding geometry basicsSimilar Triangles Worksheets
Ever heard of similar triangles? As kids climb up the academic ladder, they're introduced to a wide range of mathematical concepts, including basic geometry. To a large extent, geometry is fun. However, keeping up with the different types of shapes can be a tad overwhelming for any learner. Similar triangles are an essential part of […]
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2 Digit by 1 Digit Multiplication Worksheets
Multiplication can be complex for kids when it gets to two-digit numbers. Their young minds may get things mixed up and multiply from the right instead of the left. To give your child a better shot at understanding this math concept, get 2 digit by 1 digit multiplication worksheets. If you want to know more | 677.169 | 1 |
Solution 9: (i) For acute angles, remember what sine means: opposite over hypotenuse. If we increase the angle, then the opposite side gets larger. That means "opposite/hypotenuse" gets larger or increases. (ii) For acute angles, remember what cosine means: base over hypotenuse. If we increase the angle, then the hypotenuse side gets larger. That means "base/hypotenuse" gets smaller or decreases. (iii) For acute angles, remember what tangent means: opposite over base. If we decrease the angle, then the opposite side gets smaller. That means "opposite /base" gets decreases. | 677.169 | 1 |
Class 10 Maths Chapter 7 Coordinate Geometry Important Questions
Updated by Tiwari Academy
on January 25, 2024, 11:16 AM
Class 10 Maths Chapter 7 Coordinate Geometry Important Questions in Hindi Medium with Solutions prepared for CBSE and state board exam 2024-25. Chapter 7 of the Class 10 Mathematics named Coordinate Geometry, introduces students to the study of geometry using a coordinate system. This chapter is pivotal as it bridges algebra with geometry, allowing the representation of geometric shapes in a numerical format and solving geometric problems algebraically.
Class 10 Maths Chapter 7 Coordinate Geometry Important Questions
The beginning of the chapter introduces the Cartesian coordinate system, which is fundamental to coordinate geometry. It describes how this system uses two perpendicular lines, known as axes (the x-axis and the y-axis), to define a plane. Every point in this plane is represented by a pair of numerical coordinates: (x, y).
These coordinates indicate the point's position relative to the two axes. The chapter 7 explains the concept of quadrants in the Cartesian plane and how the sign of the coordinates changes according to the quadrant in which a point lies. This understanding is crucial for locating points and navigating the coordinate plane.
Extra Questions of 10th Maths Chapter 7
We also learn here the concept of the distance formula is introduced. This formula is used to calculate the distance between any two points in the coordinate plane. The chapter 7 of 10th Maths derives the distance formula, from the Pythagorean theorem.
This formula becomes a fundamental tool in coordinate geometry for finding the lengths of line segments and is essential for solving various geometric problems, such as determining the perimeters and areas of shapes on the coordinate plane.
Class 10 Maths chapter 7 Main Points
Class 10 Maths chapter 7 focus into the section formula, which is used to find the coordinates of a point that divides a line segment into a given ratio. The chapter 7 explains the two cases of the section formula: the internal division and the external division.
This concept is essential for problems involving partitioning line segments in a specific ratio, and it has applications in finding the centroids of triangles and other geometric figures in the coordinate plane.
Chapter 7 of 10th Maths also covers the concept of the area of a triangle formed by three given points in the coordinate plane. The formula for the area is derived and presented as an absolute value to ensure the area is always a positive quantity.
This part of the chapter 7 is important as it provides a method to calculate the area of a triangle when its vertices are known, using determinants. This method is a significant departure from traditional geometry and introduces students to an algebraic approach to solving geometric problems.
Class 10 Maths Coordinate Geometry Extra Practice
The class 10 Maths coordinate geometry discusses the concept of collinearity of points. The chapter 7 explains how the area of a triangle formula can be used to determine if three points are collinear.
If the area of a triangle formed by three points is zero, it implies that the points lie on a straight line, hence are collinear. This application of the area formula is crucial for understanding more complex geometric concepts and for solving problems involving the positioning of points in a plane.
The summary of the chapter 7 Class 10 Maths
The summary of the chapter summarizes the key concepts and highlights the importance of coordinate geometry in bridging algebraic and geometric concepts. It reinforces the skills of locating points in the Cartesian plane, calculating distances and midpoints, using the section formula, finding areas of triangles, and determining collinearity.
Important Questions of the chapter 7 concludes with a variety of problems and exercises designed to test the students' understanding and application of these concepts. These exercises range from basic point plotting to more complex problems involving the calculation of areas and testing collinearity, ensuring a comprehensive grasp of coordinate geometry.
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What's hot contains a final model test examination for mathematics. It includes questions in four groups - Algebra, Geometry, Trigonometry and Mensuration, and Statistics. The Algebra section contains three multi-part questions on topics like factoring polynomials, solving systems of linear equations, and finding terms in a geometric series. The Geometry section contains two multi-part questions involving constructions related to dividing a line segment into parts and constructing triangles. The Trigonometry and Mensuration section contains one multi-part question on trigonometric ratios and properties or on mensuration formulas. The Statistics section contains one multi-part question involving concepts like frequency distribution, mean, median and mode. The document also includes 40 multiple choice questions testing is a sample test paper for mathematics for Class 10. It consists of 33 total questions divided into 4 sections - Section A has 8 one-mark questions, Section B has 9 two-mark questions, Section C has 10 three-mark questions, and Section D has 6 four-mark questions. The test covers a range of mathematics topics including trigonometry, geometry, probability, and series. Students have 3 hours to complete the paper.
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This document defines circles geometrically as the result of a cone intersecting with a plane, and algebraically as the set of points equidistant from a fixed center point. It provides the standard equation for a circle given the center (h,k) and radius r: (x-h)2 + (y-k)2 = r2. Examples are given of writing the equation of a circle and finding its center and radius given parts of the equation. The process for finding the equation of a circle given its diameter is also describedTo find the surface area and volume of composite shapes, they must first be broken down into simpler component shapes. The surface areas and volumes of each component shape are then calculated separately and summed. For surface area, overlapping portions must be accounted for to avoid overcounting. Volumes simply add together the individual volumes. The example demonstrates finding the total volume and surface area of a shape composed of a cylinder atop a prism by calculating and summing their individual volumes and surface areasThe document contains 8 problems related to polar coordinates and curves defined by polar equations. Specifically, it asks students to:
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Two rhombuses have sides with lengths of #7 #. If one rhombus has a corner with an angle of #(pi)/2 # and the other has a corner with an angle of #(3pi)/4 #, what is the difference between the areas of the rhombuses?
Area of the rhombus with angle #theta=(pi)/2# and Side #a=7# is
#=a^2 sin theta#
#=7^2 sin((pi)/2)#
#=49(1)#
#=49#
Area of the rhombus with angle #theta=(3pi)/4# and Side #a=7# is
#=a^2 sin theta#
#=7^2 sin((3pi)/4)#
#=49(0.707)#
#=34.65#
So difference in Area#=49-34.65=14.35#
To find the difference between the areas of the rhombuses, calculate the area of each rhombus and then find the difference.
The area of a rhombus is given by the formula: ( \text{Area} = \frac{1}{2} \times \text{diagonal}_1 \times \text{diagonal}_2 )
Given that the side length of each rhombus is 7:
For the rhombus with a corner angle of ( \frac{\pi}{2} ):
Both diagonals are equal, and since opposite angles are equal in a rhombus, this means that the diagonals are perpendicular bisectors of each other. So, the diagonals form right angles. Hence, the length of each diagonal can be found using the Pythagorean theorem.
( \text{diagonal}_1 = 7 )
( \text{diagonal}_2 = 7 )
For the rhombus with a corner angle of ( \frac{3\pi}{4} ):
One diagonal can be found using trigonometric functions in a right triangle. Let ( x ) be the length of the diagonal opposite the ( \frac{3\pi}{4} ) angle | 677.169 | 1 |
Circles Worksheet Day #2 Answer Key
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Web put each equation in standard form and graph the circle. Web circles worksheet day ##### put each equation in standard form and graph the circle. M ∠wt and m∠rwv are vertically opposite angles.then, m∠swt = m∠rwvm∠arc st = m∠arc rvm∠arc st. Each one has model problems worked out step by step, practice problems, as well as challenge questions. Write.
Area Of Circles Worksheet With Answer Key printable pdf download
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Web 1) A Line Segment That Is Half The Diameter.
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Web Put Each Equation In Standard Form And Graph The Circle.
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The Worksheets Have Visual Simulations And Are Interactive To Improve A Student's Learning. | 677.169 | 1 |
Question 5: A and B are two non-zero vectors. (a) How can their scalar product be zero? And (b) how can their vector product be zero?
Answer
Two non-zero vectors A and B are given. We have to show how their scalar and vector product can be zero.
(a) Scalar Product
Scalar product of the given vectors can be defined as,
This means the scalar product of the vectors depends upon,
Magnitude of A
Magnitude of B
Cosine of the angle θ between them
Thus the scalar product will be zero if any of the above three quantities are zero. Since the magnitudes of A and B are given to be non-zero, therefore,
if the cosine of the angle θ is zero, then the scalar product of the given vectors can be zero. Since cos90° = 0, therefore, if the angle between A and B is 90° then the scalar product of the given vectors is zero.
Vector Product
Vector product of the given vectors can be defined as,
This means the vector product of the given vectors depends upon,
Magnitude of A
Magnitude of B
Sine of the angle θ between them
Thus the vector product will be zero if any of the above three quantities are zero. Since the magnitudes of A and B are given to be non-zero, therefore, if the sine of the angel θ is zero,
then the scalar product of the given vectors can be zero. Since sin0° = 0, therefore, if the angle between A and B is 0° then the vector product of the given vectors is zero. | 677.169 | 1 |
Who is the author of the famous theorem "a² + b² = c²" in a right-angled triangle?
Pythagoras
The Pythagorean theorem is one of the cornerstones of geometry, attributed to Pythagoras, a mathematician from ancient Greece.
This principle, which establishes a fundamental relationship between the sides of a right-angled triangle, was known in several ancient cultures well before Pythagoras, but it is he who is credited with it in the Western world.
Beyond geometry, this theorem finds applications in various fields such as physics, architecture, and navigation.
The influence of the Pythagorean theorem goes far beyond mathematics, testifying to the depth of human thought through the ages. | 677.169 | 1 |
Printable Unit Circle Chart
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30 Unit Circle Practice Worksheet Education Template
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FREE 19+ Unit Circle Charts Templates in PDF MS Word
42 Printable Unit Circle Charts & Diagrams (Sin, Cos, Tan, Cot etc)
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The Unit Circle Has A Radius Of One.
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The First Sheet Includes All The Radians, Degrees, And Tangents.
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Theorem 2:
If A, B, C are collinear, then real numbers x, y, z not all zero such that (bidirectional)
x+y+z=0 and xA + yB + zC = 0
Dot (Inner) Product:
A . B is equal to |A| |B| cos(angle between). Which is essentially the product of the projection of one vector onto the other.
Exterior Product (wedge product):
u ^ u = 0
u ^ v + v ^ u = 0
Length: |u| * |v| * sin (angle)
Geometrically: it's the area of the quadrilateral created by the two vectors.
Algebraically, determinant with z column zero: u1*v2 - u2*v1.
Law of Cosines:
Given an angle, theta, between b and a, c^2 = a^2 + b^2 - 2abcos(theta).
REVISIT TRIPLE PRODUCT OF 3 3D VECTORS.
Also written wedge (^). Is equal to twice the area of the triangle formed by the two vectors.
A ^ B ^ C = 1/2 the area of the triangle ABC in the Z=1 plane.
if u,v,w in R^2, then u ^ v ^ w = 0.
A,B,C collinear (3 points in the plane) <==> A ^ B + B^C + C^A = 0.
Need to get the whole collinearity and determinant down.
Barycentric Coordinates:
A, B, C three non-collinear points, any vector P may be expressed as P = xA + yB + zC, where x + y + z = 1.
Altitudes of triangles are concurrent proof: Draw the triangle, draw two of the altitudes and the point that they intersect at, and then you'll say "let's show that the line that goes through this intersection from the remaining vertex is actually an altitude". So you just need to show that some dot product is zero. Should be algebra.
Centroid proof: So like, for the centroid of a triangle being the average of the three sides, you connect the edges to the midpoint of their opposite side, and you notice that the intersection seems to divide these lines in a ration of 2:1, so you write the midpoint as two of the sides and then you're tasked with incorporating the last side, so you do so by considering this point on the line with a ratio of 2:1, and then you see that the algebra simplifies to sum of all the sides over three.
Medians of a triangle are concurrent proof: Really, you just observe the median for one of the sides, and you consider the point along the anchor that is 2:1 farther from the vertex it drops from. Then you'll see that because it bisects the opposite side, you can write it as sum of halves of the other two sides. It's algebra from there. This is the triangle's centroid.
Black hole problem: two lines and a point, all intersecting at a point, describe that point of intersection by using Disargues. Short version: create two triangles with these lines, show that they're perspective with respect to a line and therefore they are perspective with respect to a point. Essentially what you do is create one full triangle with your lines and then another side of a triangle. You show that these two have an axis of perspective, MAY WANT TO REVISIT.
REVISIT THEOREMS FROM CLASS I MISSED.
feuerbach's circle theorem: Just take a brief look.
Menelaus Proof: Use Theorem 2. | 677.169 | 1 |
Vector Algebra is particularly useful in situations that involve force and velocity. They are also useful in calculating angles and distances, building network pipes, measuring distances between aircraft, and so on. In civil engineering, vector algebra is widely used.
As a result, students must thoroughly practise this chapter to strengthen its concepts.
Vector Algebra covers a wide range of topics that may be asked on the board exam. The following are the main topics covered in Chapter 10 "Vector Algebra":
Types of Vectors
Addition of Vectors
Multiplication of a vector by a scalar
Components of a vector
Vector joining two points
Section formula
Product of two vectors
Get Access to CBSE Class 12 Maths Important Questions for the Academic Year 2023-24
You can also find CBSE Class 12 Maths Important Questions Chapter-by-Chapter Important Questions here:
Vector Algebra is part of the unit Vectors and Three-Dimensional Geometry, which accounts for 14 of the total marks in the exam. Important Questions for "Vector Algebra" include questions with various marking schemes that may appear in board exams.
There are 57 important questions covering the entire chapter of "Vector Algebra" here. This chapter's practice questions provide you with perfect and thorough practice. These are also beneficial for revision.
In this chapter, Extramarks provides important questions worth 1 and 4 marks. Typically, only such questions are asked in "Vector Algebra" board exams.
This chapter contains four exercises, as well as a miscellaneous exercise, to help students clearly understand the concepts of Vectors and Vector Algebra.
A vector's scalar components are its direction ratios, which represent its projections along the respective axes.
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If a^ is a unit vector and (x)(x+a^)=15, then the value of |x| is ____.afv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2Daebbfv3ySLgzGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=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@7F17@
Opt.
Since a^ is a unit vector, so |a^|=1.(xa^)(x+a^)=15x.x+x.<m
Ans.
Since a^ is a unit vector, so |a^|=1.(xa)(x+a^)=15x .x+x.a^a^.x ˆa^.a^=15 |x|21=15[µx .a^=a^.x and a^.a^=1] |x|2=16 |x|=4 If a^ is a unit vector and (x 'a^)(x +a^)=15, then the value of |x| is 4¯.
FAQs (Frequently Asked Questions)
1. Explain the different types of vectors.
The different types of vectors include the following:
Zero Vector: A Zero Vector (or null vector) is a vector whose initial and terminal points coincide and are denoted as 0. Because it has no magnitude, a zero vector cannot be assigned a definite direction. Alternatively, it could be regarded as having no direction. The zero vector is represented by the vectors AA and BB
Unit Vector: A Unit Vector is a vector with the magnitude of one (i.e., one unit). a denotes the unit vector in the direction of a given vector.
Coinitial Vectors: Coinitial vectors are two or more vectors that have the same initial point.
Collinear Vectors: Two or more vectors are said to be collinear if their magnitudes and directions are parallel to the same line.
Equal Vectors: Two vectors are said to be equal if they have the same magnitude and direction regardless of their initial point positions and are written as.
Negative of a Vector: A vector with the same magnitude as a given vector (say,) but the opposite direction is referred to as the negative of the given vector. Vector, for example, is the inverse of vector and is denoted as = -.
2. What are the observations of the scalar product of two vectors?
The scalar product of two vectors observations are-
A real number is the dot product of two vectors.
If a and b are non-zero vectors, their dot product is zero only if they are perpendicular to each other.
If the angle between a and b vectors is zero, the dot product is equal to the magnitude of the vectors' individual products.
If the angle between a and b vectors is 180 degrees, the dot product is equal to the magnitude of the vectors' individual products but in the opposite direction.
3. What are the observations of the cross product of two vectors?
The observations of the cross product of two vectors are as follows-
A vector quantity is always the cross product of two vectors.
If a and b vectors are not equal to zero vectors, their cross product equals the product of the magnitudes of the individual vectors, the sin angle between the two vectors, and the normal vector.
If the angle formed by two vectors, a and b, is 90 degrees, sin 90 equals 1. As a result, the vector product in this case equals the product of the magnitudes of the individual vectors and the normal vector. | 677.169 | 1 |
polygon
A polygon is a 2-dimensional geometric shape that is made up of straight lines connected end to end
A polygon is a 2-dimensional geometric shape that is made up of straight lines connected end to end. It is a closed figure with any number of sides greater than 2. The sides of a polygon do not cross each other.
Polygons can vary in shape and the number of sides they have. Some common polygons include triangles (3 sides), quadrilaterals (4 sides), pentagons (5 sides), hexagons (6 sides), heptagons (7 sides), octagons (8 sides), nonagons (9 sides), and decagons (10 sides). However, there are many other polygons that can exist based on the number of sides.
In addition to the number of sides, polygons are also classified based on their angles. There are two types of polygons: regular and irregular.
1. Regular Polygon: In a regular polygon, all the sides have the same length, and all the angles have the same measure. For example, a regular hexagon has six sides of equal length and six interior angles of equal measure.
2. Irregular Polygon: In an irregular polygon, the sides may have different lengths, and the angles may have different measures. The irregular polygon does not follow a specific pattern or rule for its sides and angles.
To find various properties of polygons, you can use different formulas:
1. Perimeter: The perimeter of a polygon is the total length of its sides. To find the perimeter, you add up the lengths of all the sides. For irregular polygons, you would measure the length of each side and add them together.
2. Area: The area of a polygon is the amount of space enclosed by its sides. The formula to find the area of various polygons depends on the type of polygon. For example, to find the area of a triangle, you can use the formula A = 1/2 * base * height. For regular polygons, there are different formulas available.
3. Interior Angles: The sum of the interior angles of a polygon can be found using the formula (n-2) * 180 degrees, where n represents the number of sides. For example, a quadrilateral (4 sides) has interior angles summing up to (4-2) * 180 = 360 degrees.
4. Exterior Angles: The exterior angle of a polygon is the angle formed between one side and the extension of an adjacent side. The sum of the exterior angles of any polygon is always 360 | 677.169 | 1 |
Question 2.
Give three examples of 3-dimensional shapes around you which are the combinations of 2 or more 3-dimensional shapes.
Solution:
3-dimensional shapes which are the combination of
2 or more 3-dimensional shapes.
(i) A funnel: Combination of cone and cylinder.
(ii) A toy: Combination of a cone and hemisphere.
(iii) An ice-cream cone: Combination of a cone and hemisphere.
(iv) A circus tent: Combination of a cylinder and a cone.
Question 3.
Give two examples of solids which are not polyhedrons.
Solution:
Sides which are not polyhedron:
(i) Cylinder
(ii) Sphere
(iii) Cone
Question 4.
Why a pentagonal pyramid having all its edges congruent cannot be a regular polyhedron?
Solution:
A pentagonal pyramid having all its edges congruent cannot be a regular
polyhedron because all the vertices of it are not formed by the same number of faces.
Question 6.
Verify Euler's formula for the following figures:
Solution:
Question 7.
For each of the given solid, the three views are given. Identify for each solid the corresponding top, front and side views:
Solution:
(a) A television
(i) Front view
(ii) Side view
(iii) Top view | 677.169 | 1 |
Polygon
A polygon is a closed figure that is composed of straight sides, known as line segments
A polygon is a closed figure that is composed of straight sides, known as line segments. Each line segment intersects with exactly two other segments, with no crossings or self-intersections. The word polygon is derived from the Greek words "poly" meaning "many" and "gonia" meaning "angle".
Polygons can be classified based on the number of sides they have. Here are some common types of polygons:
1. Triangle: A triangle is a polygon with three sides. It can be further classified into different types based on its angles and side lengths, such as equilateral, isosceles, or scalene triangles.
2. Quadrilateral: A quadrilateral is a polygon with four sides. Some examples of quadrilaterals are squares, rectangles, parallelograms, and trapezoids.
3. Pentagon: A pentagon is a polygon with five sides.
4. Hexagon: A hexagon is a polygon with six sides.
5. Heptagon: A heptagon is a polygon with seven sides.
6. Octagon: An octagon is a polygon with eight sides.
7. Nonagon: A nonagon is a polygon with nine sides.
8. Decagon: A decagon is a polygon with ten sides.
The sum of the interior angles of an n-sided polygon can be found using the formula: (n – 2) * 180 degrees. For example, a triangle has three sides, so the sum of its interior angles would be (3 – 2) * 180 = 180 degrees. Similarly, a quadrilateral has four sides, so the sum of its interior angles would be (4 – 2) * 180 = 360 degrees.
To calculate the exterior angle of a regular polygon, you can use the formula: 360 degrees / n, where n represents the number of sides. For instance, for a hexagon, the exterior angle would be 360 / 6 = 60 degrees.
Polygons have various properties and characteristics. They can be classified based on their symmetry, regularity, and the lengths of their sides and angles. Additionally, polygons can be used in real-life applications such as measuring land area, creating architectural designs, and solving geometric problems in mathematics | 677.169 | 1 |
Some applications of trigonometry
Trigonometry studies triangles and relationships between sides and angles. This document discusses using trigonometric ratios to calculate heights and distances, including the angles of elevation and depression. It provides examples of using trigonometry to find the height of a tower from the angle of elevation measured 30 meters away (30 meters high), and the height of a pole from the angle made by a rope tied to its top (10 meters high). It also explains calculating the length of a kite string from the angle of elevation.
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Trigonometry deals with relationships between sides and angles of triangles, especially right triangles. It has many applications in fields like architecture, astronomy, engineering, and more. The document provides background on trigonometry, defines trigonometric functions and ratios, discusses right triangles, and gives several examples of how trigonometry is used in areas like navigation, construction, and digital imaging.
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Similar to Some applications of trigonometry
Trigonometry studies triangles and relationships between sides and angles. It uses trigonometric ratios to calculate heights and distances. There are two types of angles used - angle of elevation, which is the angle formed between the line of sight and horizontal when looking up, and angle of depression, which is the angle formed when looking down. Some example problems calculate heights and distances using trigonometric ratios and the given angles of elevation or depression attached to the top of the poleHello everyone...There are many teachers in the schools which gives students to make a powerpoint presentation on Maths topic...most of students get confused that what to make...so now no need to worry about...you can download it!
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The document defines and provides examples of calculating angles of elevation and depression. It explains that angle of elevation is formed when looking at an object higher than the observer, while angle of depression is formed when looking at an object lower than the observer. Examples are provided to demonstrate calculating unknown lengths using trigonometric functions like tangent, sine and cosine based on the angles of elevation or depression and known lengths.
This document provides solutions to 16 questions about applications of trigonometry involving angles of elevation and depression. The questions calculate heights and distances using trigonometric ratios in right triangles formed by towers, poles, buildings, and other objects viewed from different points. The solutions demonstrate setting up and solving right triangle trig problems systematically to find the requested unknown values in each scenario.
Here are the steps to solve this problem using trigonometry:
1. Draw a sketch of the situation showing the hot air balloon, observer, and relevant angles and distances.
2. Label the given information: The observer is 20 m from the base of the balloon and the angle of elevation is 35°.
3. Identify the trigonometric ratio to use based on the given and missing information. Since we are given an angle of elevation and the distance from the observer to the base of the balloon, we will use tangent.
4. Set up and solve the trigonometric equation:
Tan 35° = Opposite/Adjacent
Tan 35° = Height/20
Height = 20 * Tan
This document provides an overview of trigonometry. It defines trigonometry as dealing with relationships between sides and angles of triangles, particularly right triangles. The origins of trigonometry can be traced back 4000 years to ancient civilizations. Key concepts discussed include right triangles, the Pythagorean theorem, trigonometric ratios like sine, cosine and tangent, and applications of trigonometry in fields like construction, astronomy, and engineering. Examples are provided for using trigonometric functions to solve problems involving heights and distances.
Trigonometry is a branch of Mathematics that deals with the distances or heights of objects which can be found using some mathematical techniques. The word 'trigonometry' is derived from the Greek words 'tri' (meaning three) , 'gon' (meaning sides) and 'metron' (meaning measure). Historically, it was developed for astronomy and geography, but scientists have been using it for centuries for other purposes, too. Besides other fields of mathematics, trigonometry is used in physics, engineering, and chemistry. Within mathematics, trigonometry is used primarily in calculus (which is perhaps its greatest application), linear algebra, and statistics. Since these fields are used throughout the natural and social sciences, trigonometry is a very useful subject to know
1) This document provides lesson objectives and examples for solving right triangles using trigonometric functions, the Pythagorean theorem, and angle relationships. It defines trigonometric ratios, angle of elevation/depression, bearing, and course.
2) Examples are provided to solve right triangles, find missing angles and sides, and solve real-world problems involving width of a stream, height of a flagpole, camera angle of depression, and height of a tower.
3) Additional examples solve problems involving slant distance to a sunken ship, plane bearings and courses between locations, and references are provided for further reading.
The document discusses trigonometry and its uses in navigation, measuring heights and distances, and astronomical studies. It provides examples of trigonometric ratios like sine, cosine, and tangent for common angles. It then explains concepts like line of sight, angle of elevation and depression. It gives two word problems as examples - one calculating the height of a building given the observer's distance and angle of elevation, and another calculating the height a bird is flying given the angle of depression and distance to an object on the ground.
The document contains 19 math word problems involving angles of elevation, depression, and shadows. It provides the questions, diagrams illustrating the problems, and step-by-step solutions. The problems cover topics like determining the height of towers, pillars, and hills using trigonometric ratios when given angles and distances.
The document is about angles of elevation and depression. It provides examples to illustrate and differentiate between angles of elevation and depression. It gives practice problems identifying whether angles are of elevation or depression and solving word problems involving these angle types. Examples include finding distances given angles of elevation or depression from a plane or tower, or finding angles given distances and heights in situations involving the sun, kites, flagpoles, etc. Diagrams are provided with the problems to illustrate the geometry involved.
This document provides information about angles of elevation and depression. It defines key terms like line of sight, angle of elevation, and angle of depression. It presents examples of how to classify angles as elevation or depression and solve problems involving right triangles using trigonometric ratios. The document also discusses applications of elevation and depression angles in fields like engineering and gives sample evaluation questions.
Trigonometry is a branch of mathematics that deals with relationships between the sides and angles of triangles, especially right triangles. It has many applications in fields like astronomy, navigation, engineering, and more. Some key uses of trigonometry include measuring inaccessible heights and distances by using trigonometric functions and properties of triangles formed by angles of elevation or depression. For example, trigonometry can be used to calculate the height of a building or tree by measuring the angle of elevation from a known distance away. It also has applications in measuring distances in astronomy, designing curved architectural structures, and calculating road grades. The document provides examples of various real-world applications of trigonometric concepts Angle of Elevation
The angle which the line of sight
makes with a horizontal line drawn
away from their eyes is called the
angle of Elevation of aero plane from
them.
Angel of Elevation
8. • Angle of Elevation: In the picture below, an
observer is standing at the top of a building is
looking straight ahead (horizontal line). The
observer must raise his eyes to see the airplane
(slanting line). This is known as the angle of
elevation.
9. • Angle of Depression: The angle below horizontal that an
observer must look to see an object that is lower than
the observer. Note: The angle of depression is congruent
to the angle of elevation (this assumes the object is close
enough to the observer so that the horizontals for the
observer and the object are effectively parallel).
13. The angle of elevation of the top of a tower from a
point on the ground, which is 30 m away from the
foot of the tower is 30°. Find the height of the tower.
.
Let AB be the tower and the angle of elevation from point C
(on ground) is
30°.
In ΔABC,
Therefore, the height of the tower is
14. A circus artist is climbing a 20 m long rope, which is tightly
stretched and tied from the top of a vertical pole to the
ground. Find the height of the pole, if the angle made by the
rope with the ground level is 30 °.
Sol:- It can be observed from the figure that AB is the pole.
In ΔABC,
Therefore, the height of the pole is 10 m.
15. .
Let K be the kite and the string is tied to point P on
the ground.
In ΔKLP,
Hence, the length of the string is
21. Questions based on trigonometry :• The angle of elevation of the top of a pole measures 48° from a point on
the ground 18 ft. away from its base. Find the height of the flagpole.
• Solution
Step 1: Let's first visualize the situation
Step 2: Let 'x' be the height of the flagpole.
STEP 3: From triangle ABC,tan48=x/18
Step 4: x = 18 × tan 48° = 18 × 1.11061… = 19.99102…» 20
Step 5: So, the flagpole is about 20 ft. high.
22. C
A
50
D
45
A hoarding is fitted above a building. The height of the
building is 12 m. When I look at the lights fitted on top
of the hoarding, the angle of elevation is 500 and when I
look at the top of the building from the same place, the
angle is 450. If the height of the flat on each floor is
equal to the height of the hoarding, the max floors on
the building are? (Tan 500=1.1917)
B
ANSWER : Let AB denote the height of the building,
Let AC denote the height of the hoarding on top of the building
Thus, Tan500 = (12 + AC) ÷ 12
1.1917 = 1 + (AC ÷ 12)
1.1917 – 1 = AC ÷ 12
12 ÷ AC = 1 ÷ 0.1917 ~ 5 | 677.169 | 1 |
About This Lesson
Attached are some of the assignments we use to for our Geometry students to practice applying ratios by calculating the balance point for balanced systems. Also attached; an activity homework sheet for when we take turns to try to balance on a board on different sized round cylinders and a triangular prism, and a copy of the instruction sheet we hand out. For the activity we use a board about 18 inches long, a round cylindrical tube about 3.5 inches outside diameter, a round cylindrical wood dowel about one inch diameter and a short piece of 1-1/2" angle iron | 677.169 | 1 |
I we know $\hat C$ and length of the bisector of $\hat C$ and side c then how can we construct our triangle?
my attempt:
if $\hat C = \alpha$, now I draw this.
each point that I choose on two arcs (point C) and connect it to A and B, make triangle ABC with angle $\hat C = \alpha$ and AB = c. But I need some of points that make bisector angle $\hat C$ with my own length of the bisector of $\hat C$. How can I find that?
Update 1:
Hi. Thanks RicardoCruz for answer. I get some help from my teacher. He said:
we assume that we draw our $\triangle ABH$. now we construct bisector of $\angle H$ that intersect with arc in G. G is in middle of arc AGB. so for construct our real triangle we can find middle of this arc then we make circle with center of G and radius $(n + x)$ to find place of point H. so only thing that we must find length of $x$. He said that you can use similarity of triangle.
I use this way in picture so we have a quadratic equation and we can find $x$. can you have any other better way for find $x$?
$\begingroup$Your question is not clear. If you know the measure of angle $\hat C$ then of course you know the measure of the bisector of that angle. Please be more clear on the information given you at the beginning.$\endgroup$
$\begingroup$@RoryDaulton: I think the OP meant that the length of the bisector of $\hat C$ is known (together with $\hat C$ and the opposite side $c$). English is probably not the OP's native language.$\endgroup$
EDIT: It turns out that this post has an error, namely that one of my algebraic manipulations is incorrect. (I knew this was too good to be true....) I haven't been able to modify the solution to make it still work because now the degrees of the relative terms don't yield to the same techniques, but I still leave the solution up hopefully to serve as inspiration for other possible solutions.
For ease of typesetting, let $\triangle ABC$ be the triangle in question, and let $d$ be the length of the angle bisector from $A$. Note that by the angle bisector theorem it's easy to prove that this bisector divides $BC$ into two segments of length $\frac{ac}{a+b}$ and $\frac{bc}{a+b}$.
Now recall Stewart's Theorem, which states that whenever $D$ is a point on $BC$, if $BC=a$, $AC=b$, $AB=c$, $AD=d$, $BD=m$, and $CD=n$, the relationship $$b^2m+c^2n=d^2a+amn$$ holds. (The easiest way to remember this is to use the mnemonic "A man and his dad put a bomb in the sink" :) ) Plugging this into the result in question gives $$a^2\left(\dfrac{bc}{a+b}\right)+b^2\left(\dfrac{ac}{a+b}\right)=d^2c+c\left(\dfrac{ac}{a+b}\right)\left(\dfrac{bc}{a+b}\right).$$ Note that the left hand side simplifies to $abc$, so dividing through by $c$ yields $$ab=d^2+\dfrac{abc^2}{(a+b)^2}.$$ Now some algebraic manipulation gives $$d^2=\dfrac{ab(1-c^2)}{(a+b)^2}\implies \dfrac{1-c^2}{d^2}=\dfrac{(a+b)^2}{ab}=\dfrac ab+\dfrac ba + 2.$$ This is actually a quadratic in $a/b$, so we can solve for the ratio $a/b$. This is actually great news, because the segment $AD$ divides $BC$ into two segments whose lengths are in the ratio $a/b$!
With this in mind, I propose the following construction:
Construct the angle $C$. Extend the sides of $\angle C$ far out with a straightedge.
Construct the angle bisector of $\angle C$. Extend it to a point $D$ so that the length of $CD$ is $d$ (which is already given to you).
Pick an arbitrary point $P$ on one of the sides of $\angle C$. Now construct a point $Q$ on this same side so that $CP/CQ=a/b$. (This is the part which I'm a bit unsure about, but considering that $a/b$ can be expressed entirely in terms of the lengths of $c$ and $d$ combined with nothing more complicated than square roots and squares, I'm pretty sure there exists a routine but tedious algorithm to do this.)
Construct a circle with center $C$ and radius $CQ$. This intersects the other side of the angle $C$ at some point $R$.
Draw a line through $D$ parallel to $PR$. Let it intersect the sides of $\angle C$ at $A$ and $B$.
$\begingroup$You're correct. This solution doesn't work out as intended. :( I don't know how to fix it, but I think some of the ideas might be helpful in later developing a correct solution.$\endgroup$ | 677.169 | 1 |
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