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Page 1: SATPG Math Module 16 Triangles, Quadrilaterals, Circles ... · Math Module 16: Triangles, Quadrilaterals, Circles, Polygons, and 3D Shapes In this module, we will cover basic shapes

Page 1

Math Module 16: Triangles, Quadrilaterals, Circles, Polygons, and 3D Shapes

Page 2: SATPG Math Module 16 Triangles, Quadrilaterals, Circles ... · Math Module 16: Triangles, Quadrilaterals, Circles, Polygons, and 3D Shapes In this module, we will cover basic shapes

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Copyright © 2001-2014 by Steve Kirshenbaum and Celia Barranon, SAT Preparation Group, LLC. www.SATPrepGroup.com.

All rights reserved. No part of this book may be reproduced in any form or by any electronic or mechanical means including information storage and retrieval systems without express written permission from SAT Preparation Group, LLC.

This publication is designed to provide accurate and authoritative information in regard to the subject matter within this book.

DISCLAIMER: Peak performance strategies for SAT/ACT prep include physical exercise, stretching, and optimal nutrition for maximum results. It is recommend-ed that participants in this program consult with a trained medical professional prior to engaging in such activities as they relate to the services offered by SAT Preparation Group, LLC. SAT Preparation Group, LLC is not responsible for ill-ness or injury due to these activities.

*SAT is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

Your comments and corrections are welcome.

Please send them to:

SAT Preparation Group, LLC

11311 Heron Bay Blvd, Unit 2823.

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877-672-8773

www.SATPrepGroup.com

Printed and bound in the United States of America

Page 3: SATPG Math Module 16 Triangles, Quadrilaterals, Circles ... · Math Module 16: Triangles, Quadrilaterals, Circles, Polygons, and 3D Shapes In this module, we will cover basic shapes

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Math Module 16:

Triangles, Quadrilaterals, Circles, Polygons, and 3D Shapes

In this module, we will cover basic shapes covered on the SAT, ACT, GRE, and SSAT. We will delve deeply into qualities of certain geometric figures as well as

area and volume calculations. This is truly the shape of things to come...

Triangle Basics

All triangles have 3 sides The sum of the interior angles in any triangle = 180o A triangle that has all 3 angles less than 90o is called acute A triangle that has one angle that is greater than 90o is called obtuse A triangle that has 2 equal sides and 2 opposite equal angles is called isosceles A triangle that has 3 equal sides and 3 equal angles (each 60o) is equilateral A triangle that has a right angle (90o) is called a right triangle The exterior angle of a triangle is equal to the 2 remote interior angles.

Isosceles Triangle Rule

In an isosceles triangle, two equal sides are opposite two equal angles...

a

b

c a = b + c

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Characteristics of Right Triangles

In a right triangle, the legs ARE the base and height for area calculations because the legs of a right triangle are always perpendicular to each other, even in drawings that are upside down or backwards. The hypotenuse is a fancy word that describes the longest side of a right triangle, which is always opposite to the 90o angle.

Pythagorean Theorem

In a right triangle only, the square of one leg (a) plus the square of the other leg (b) is equal to the square of the hypotenuse (c). In mathematical form: a2 + b2 = c2. This is a handy way of calculating the length of one of the missing legs/hypotenuse of a right triangle. Suppose we had a right triangle with one side length of 7 and the other side length of 24. What is the length of the hypotenuse? Using the Pythagorean Theorem: 72 + 242 = x2 or x2 = 49 + 576 or x2 = 625... Taking the square root of both sides: x = 25. Now suppose that a right triangle has a hypotenuse of length 41 and one of the legs of length 9. What is the length of the other leg? Using the Pythagorean Theorem: 92 + x2 = 412 or x2 = 412 -- 92 or x2 = 1681 -- 81 or x2 = 1600. Taking the square root of both sides: x = 40.

hypotenuse

legs

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Pyth. Trip. 2x Multiple 3x Multiple 4x Multiple 5x Multiple

3, 4, 5 6, 8, 10 9, 12, 15 12, 16, 20 15, 20, 25

5, 12, 13 10, 24, 26 15, 36, 39 20, 48, 52 25, 60, 65

8, 15, 17 16, 30, 34 24, 45, 51 32, 60, 68 40, 75, 85

7, 24, 25 14, 48, 50 21, 72, 75 28, 96, 100 35, 120, 125

9, 40, 41 18, 80, 82 27, 120, 123 36, 160, 164 45, 200, 205

Pythagorean Triples There are certain whole number combinations of right triangle lengths that test writers often use that can streamline the solving of a problem. The most often utilized Pythagorean Triples are 3, 4, 5; 5, 12, 13; and 8, 15, 17. The multiples of Pythagorean Triples are also commonly used by test writers. For example, if each number in the 3, 4, 5 Pythagorean Triple is doubled to 6, 8, 10, then this would create another Pythagorean Triple, etc. In other words, if a right triangle was given and the sides of the legs were 5 and 12, then by knowing these Pythagorean Triples one could automatically know that the hypotenuse is 13 without using Pythagorean Theorem. Below is a table of common Pythagorean Triples and their multiples... Special Right Triangles There are only 2 kinds of special right triangles: 45:45:90 and 30:60:90. Special right triangles have a certain ratio of sides... 45:45:90 Special Right Triangle 30:60:90 Special Right Triangle 45o

1

1

√2 1

30o

60o

2

√3

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In a special right triangle it is favorable and convenient to put the ratios on the inside of the triangle and the actual lengths of the sides on the outside of the triangle. You’ll notice, particularly on the SAT, that in the legend at the beginning of each math section, the ratios of the special right triangles are labeled x, x, x√2 for the 45:45:90 special right triangle and s, s√3, and 2s for the 30:60:90 special right triangle and are located outside the triangle legends. My advice: lose the x’s and s’s and label the ratios (as shown on the previous page) on the inside and the actual lengths on the outside. Once a special right triangle is properly labeled, even if given just a single side, the rest of the sides can be deduced by simply multiplying or dividing by the intervening ratio. Here’s a handy mnemonic trick to remember: going toward a side with ratio 1, divide by the intervening ratio; going away from a side with ratio 1, multiply by the intervening ratio. Notice that toward ends in a d encouraging a division. A couple of example should congeal this concept...

Suppose that the side opposite the 30o angle is 5. To determine the length of the side opposite the 90o angle, which is twice the length of the side opposite the 30o angle, simply multiply the actual length of the side opposite 30o by the intervening ratio of 2 to get 5 x 2 = 10. Similarly, to determine the length of the side opposite the 60o angle, which is √3 times the length of the side opposite the 30o angle, just multiply 5 x √3 to get 5√3. The final side lengths are shown below:

1

30o

60o

2

√3

5

1

30o

60o

2

√3

10

5

5√3

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Now let’s suppose that the length of the hypotenuse of the 45:45:90 special right triangle is 6. When going to a smaller side, divide. Divide by what, you might ask? By √2, which is the intervening ratio. The final triangle would look like this...

One more example should give you the idea...

Let’s suppose now that the side opposite the 60o angle is 9. What are the lengths of the other sides? In a 30:60:90 right triangle, it is useful to use the ratio of 1 side as a pivot. First, to go from a longer side to a shorter one, divide 9 by √3 to get 9/√3 for the side opposite the 30o angle. Then, multiply 9/√3 by 2 to determine the side opposite the 90o angle...

45o

1

1

√2 6

45o

1

1

√2 6 _6_

√2

_6_ √2

30o

60o

2

√3

9

1

30o

60o

2

√3

_9_ √3

9

1

_18_ √3

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One last paramount point about special right triangles. Test writers often get unsuspecting and/or hasty students to assume that a triangle is 30:60:90 or 45:45:90. Remember this: NEVER ASSUME ANYTHING unless pertinent information is specifically provided to be able to deduce a special right triangle. Now you try… (Please print these pages out so you can write it out by hand) Determine the missing sides: 1. 4. 2. 5.

3. 6. Answers located on page 20

45o

3

3

5

15 9

30o

8

60o

10

45o

7√3

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Legal Triangle Rules

In order for there to be a triangle in the first place, two main rules must be adhered to; I call them Legal Triangle Rule #1 and Legal Triangle Rule #2... Legal Triangle Rule #1 For this triangle to be a legal triangle, then: a + b > c a + c > b b + c > a All 3 combinations must be satisfied for legal triangle status. Suppose we have the following 3 triangle lengths: 2, 3, and 5. Even though 3 + 5 > 2 and 2 + 5 > 3, 2 + 3 is not greater than 5 and so would not reach to form a triangle in the first place; this violates Legal Triangle Rule #1. On the other hand, if the three sides of a triangle were 2, 3, and 4 where 2 + 3 > 4, 2 + 4 > 3, and 3 + 4 > 2, then this would pass the Legal Triangle Rule #1. Legal Triangle Rule #2 Legal Triangle Rule #2 applies when only two of the three sides of a particular triangle are given. Given 2 sides of a triangle, the 3rd side must be anywhere between (and NOT including) the sum and the difference of the given 2 sides. Suppose the two triangle sides given were 5 and 12. Now, it’s easy to automatically assume that the third side would be 13, but that is only true if the triangle was a right triangle. Taking the sum and the difference of 5 and 12 yields 7 < x < 17, so the third side would have to be somewhere between 7 and 17; anything from 7.0000...01 and 16.999999... would be acceptable. If the third side was be an integer, then the third side could be 8, 9, 10, 11, 12, 13, 14, 15, 0r 16.

a

b

c

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Now you try… (Please print these pages out so you can write it out by hand) 1. Determine which of the following are or are not legal triangles: a) 4, 7, 10 yes no b) 5, 6, 12 yes no c) 3, 8, 7 yes no d) 6, 10, 16 yes no e) 9, 21, 29 yes no f) 1, 2, 3 yes no 2. What are the possible lengths of the third side of a triangle given two sides lengths: a) 7, 16 ____________________________________________________________________ b) 14, 15 ___________________________________________________________________ c) 3, 4 ______________________________________________________________________ d) 11, 13 ___________________________________________________________________ 3. What is the least possible perimeter for a triangle with three integer sides if two of the sides are 6 and 19?: _______________________________________________ ____________________________________________________________________________ Answers located on page 20

Similar Triangles

Similar triangles are triangles that have the same angle measures and sides that are correspondingly proportional to a single scale factor. In other words, if a side b of triangle ABC is similar to side e of triangle DEF by 2:1, then, regardless of the length of sides a:d or c:f, both a:d and c:f must also have a scale factor of 2:1. The scale factor in a similar triangle then must be in the same proportion for all 3 corresponding sides. When dealing with similar triangles, proportions are probably the most useful tool for solving missing sides. Let’s do a couple of examples to bring the point home...

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Let’s suppose that triangles above are similar. To solve for x and y we need to set up proportions. Notice, by the way, that the scale factor for these triangles is 2:1 from left to right or 1:1/2 from right to left. There are several proportions that can be used here: 6 = x or 4 = 2 or 6 = 4 or x = 2 . 4 2 6 x x 2 6 4 Any of the above proportions would work, keeping in mind that corresponding sides are always across from each other. Cross multiplying any of these proportions yields 4x = 12 or x = 3. To find y, let’s use the proportion: y = 3 , which yields 2y = 12 or y = 6. 4 2 Embedded Similar Triangles When 2 two similar triangles are combined or smushed together into a single diagram, it is often difficult to clearly see proportions and/or scale factors. The most effective thing to do at this point is to redraw the two triangles separately and in the same orientation. The following example demonstrates this...

6

4 2

x y 3

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In the above similar triangles, if c = 8, e = 2, and f = 4, what is the sum of a + b? Step 1 with embedded similar triangles is to redraw them separately in the same orientation with labels... Because both triangles are similar 30:60:90 triangles, d = 2√3. We can now set up a proportion (or just use the 1:2:√3 ratios) to solve for a: a = 2 such that 2a√3 = 20, divide to get a = 10/√3. 10 2√3 Because of the ratios, b + f must be twice a. b + f = 20/√3. Since f = 4, b = 20/√3 -- 4. a + b = 10/√3 + 20/√3 -- 4 = 30/√3 -- 4.

a

b

c d

e f a

f = 4 e = 2

b + f = b + 4

d c + d = 10

30o

30o

1 2

√3

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Now you try… (Please print these pages out so you can write it out by hand) 1. What is one possible value for the scale factor in the similar triangles above: ____________________________________________________________________________ 2. What is the value of x in the similar triangles above?: ____________________________________________________________________________ 3. If the scale factor from the larger to the smaller similar triangles above is 2.5:1, what are the values of g and h?: ______________________________________________ ____________________________________________________________________________ Answers on page 20

The Robust World of Quadrilaterals

In the world of quadrilaterals there are many players: squares, rectangles, rhombi, trapezoids, parallelograms, and 4-sided figures in between. The sum of the angles of ANY quadrilateral = 360o. Rectangles Rectangles are 4-sided figures with 4 right angles, each 90o, and opposite sides of equal length. The diagonals of a rectangle form equal right triangles where the diagonal is also the hypotenuse.

12 x 4 3

5

g 12

h

W W L

L

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Squares Squares are special types of rectangles where all four sides are equal. The half-squares created by the diagonals of a square have interesting qualities... In a half-square, the diagonals form isosceles right triangles with side lengths in the ratio 1:1:√2 and a half-diagonal has exactly half the length of a full diagonal. By the way, ALL isosceles right triangles are half squares with side lengths in the ratio 1:1:√2. Diagonals of a square are perpendicular. Parallelograms A parallelogram has the following qualities... Both opposite sides are equal in length and parallel to each other. Each pair of opposite angles is equal Each side is a transversal that can be easily seen by extending the lines and therefore has the same properties as any parallel line with a transversal. Rhombi A rhombus is a 4-sided parallelogram that has four equal sides that does not have 90o angles between each side. The diagonals of a parallelogram are perpendicular to each other. Trapezoids Trapezoids are 4-sided figure with parallel bases. The height of a trapezoid is the perpendicular distance between the bases. An isosceles trapezoid has equal base angles and equal side lengths. A right triangle can be formed using the height of the trapezoid.

45o 1

1 √2

ao

ao bo

bo x x

y

y

b1

b2 h

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Circles

Here are some circle factoids: The diameter (d) is the maximum distance possible of a line segment through the center from one point on a circle to another point on the same circle. The radius (d) is the straight line distance from the center of a circle to any point on the circle and is exactly half the diameter. The angular measure exactly once around a circle or pivot point is 360o. The circumference is the distance exactly once around a circle. The ratio of the circumference of a circle to the diameter is called “pi” typically represented by the symbol π. π ≈ 3.14 The circumference of a circle: C = 2πr = πd The area of a circle: A = πr2 = π(d/2)2 = (πd2)/4 Triangles drawn inside circles are often isosceles triangles because two of the sides are radii, which also means that the opposite angles to the radii are equal. The central angle (<ACB) is equal to the intercepted arc AXB. The inscribed angle (<YAZ) is half the angle measure of the intercepted arc YWZ. In terms of a circle, a tangent is a line that touches a circle as it passes it on only a single point on the circle called a point of tangency; This point is automatically perpendicular from the line to the center of the circle. Keep an eye out for the word “tangent.” Thus, if the slope of one line can be determined, then the slope of the other line in automatically the negative reciprocal because perpendicular lines have negative reciprocal slopes by definition. The partial distance around a circle is called arc length (AL); it’s simply the circumference times either the fraction of the circle or degrees/360o. AL = 2πr(fraction of circle) = 2πr(degrees/360o)

C

d

r

A

B

X

Y

Z

W

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fraction of circle = degrees/360o (proportion, so cross multiply) For example, if the angle measure of arc AXB on the previous page was 70o , the length of AXB would be expressed: AL = 2πr(70o/360o). If the angle measure of <ACB on the previous page was 1/6, the arc length would be expressed: AL = 2πr(1/6). The partial area of a circle is called sector area; it’s simply the area times either the fraction of the circle or degrees/360o For example, if sector ACB from the previous page had a central angle of 70o, then Asector = πr2(70o/360o). If sector ACB from the previous page has a central angle of (1/6), then Asector = πr2(1/6).

Regular Polygons

Any multi-sided figure, including an equilateral triangle and a square, that has equal side lengths and equal individual interior angle measures (i) is known as a regular polygon. Regular Polygon Factoids The sum (∑) of the interior angles (i) = S S = 180o(n — 2), where n is the number of sides Each individual interior angle (i): i= S/n = [180o(n — 2)]/n The sum of the individual exterior angles ∑e = E = 360o i + e = 180o

e = 360o/n The perimeter of any closed 2-dimensional figure is the sum of the outside length Similar polygons also have corresponding equal scale factors and angles

e i

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Area and Volume

Shape Sides (n) S E i e

Triangle 3 180o 360o 60o 120o

Square 4 360o 360o 90o 90o

Pentagon 5 540o 360o 108o 72o

Hexagon 6 720o 360o 120o 60o

Octagon 8 1080o 360o 135o 45o

Decagon 10 1440o 360o 144o 36o

….. ….. ….. ….. ….. …..

Any Polygon n 180o(n — 2) 360o [180o(n — 2)]/n or 180o -- e

180o — i

Shape Image Area Volume Surface Area

Rectangle L x W

Square side2

Triangle (1/2)bh

Equilateral Triangle (side2√3)/4

Trapezoid (1/2)(b1 + b2)h

Parallelogram bh

Rhombus (1/2)d1d2

Circle πr2

Polygon (1/2)ap (a = apothem and p = perimeter)

Cube side3 6 x side2

Rectangular Solid L x W x H 2LW + 2LH + 2WH

Right Circ. Cylinder πr2h 2πr2 + 2πrh

Sphere (4/3)πr3 4πr2

h b

side

W L

side

h b2 b1

a

h

d1 d2

r

b

side

H W

L r h

r

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Shaded Regions

Shaded regions are mainstays particularly on the SAT, where literally every published test has at least one. The formula for shaded regions is as follows... Ashaded = Awhole -- Aunshaded One caveat is that the Awhole must be at the boundary to avoid any unnecessary calculations. Let’s do an example for clarity... In the figure above, rectangle ABCD has midpoints W, X, Y, and Z. If AC = 24 and CD = 10, what is the area of the shaded region? The area of the whole, Awhole = L x W = 24 x 10 = 240. The area of the unshaded two right triangles, Aunshaded = 2(1/2)(12)(5) = 60 Ashaded = Awhole -- Aunshaded = 240 -- 60 = 180 Occasionally, a problem might ask for the area of an unshaded region. In that case reverse the formula: Aunshaded = Awhole -- Ashaded

How Many Fit?

Whenever you see the phrase “how many” in a math problem, you can typically deduce that it is a “how many fit?” type of question. How many fit problems either have to do with area or volume...

how many fit = Abigger or Vbigger_, where A = area and V = volume Asmaller Vsmaller

A

B

C

D

W A

X

Y

Z

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Let’s do a couple of examples to bring the point home... If an equilateral triangle has it’s sided quadrupled, how many times greater in ar-ea is the larger triangle? Remember that the area of a right triangle is: (side2√3)/4. If we assume that the smaller triangle has a side length of 2 and the larger triangle has a side length of 8... (82√3)/4 = 64 = 16 times the area (22√3)/4 4 How many times the volume is a cube with length 10 than a cube with length 5? The volume of a cube is side3. So, ... 103 = 1000 = 8 times the volume 53 125

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Answers

Special Rt. Triangles & Pyth. Triples

1.

2.

3.

4.

5.

6.

Legal Triangles

1. a) yes b) no c) yes d) no e) yes f) no 2. a) 9 < x < 23 b) 1 < x < 29 c) 1 < x < 7 d) 2 < x < 24 3. Third side: 13 < x < 25, so least third side = 14 Perimeter = 6 + 14 + 19 = 39

Similar Triangles

1. x = 9

2. g = 7.5 h = 4.8

45o

3√2

3

3

3

5

√34

15 9

12

30o

8

16

8√3

60o

10

(7√3)/√2

20/√3

45o

7√3

(7√3)/√2

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Triangles, Quadrilaterals, Circles, and Polygon SAT Questions

1: If all angles in the figure above are right angles, what is the area of the figure?

(A) 33

(B) 44

(C) 68

(D) 77

(E) 88

2: The lengths of two sides of a triangle are 13 and 3. What is one possible length of the third side?

3: If the area of each face of a cube is 121 square centimeters, what is the volume of the cube in cubic centimeters?

4: In the figure above? RT and SU intersect at point V and RS || UT. If a = 65 and b = 35, what is the value of c?

(A) 90

(B) 80

(C) 65

(D) 35

(E) Unable to be determined from the information given.

3

4 8

11

ao bo

co

R

S T

U V

Note: Figure not drawn to scale.

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Triangles, Quadrilaterals, Circles, and Polygon SAT Questions

5: Right circular cylinders G and H have the same volume. The radius of cylinder H is twice the radius of cylinder G. What is the ratio of the height of cylinder H to the height of cylinder G?

(A) 4 to 1

(B) 2 to 1

(C) 1 to 1

(D) 1 to 2

(E) 1 to 4

6: In the circle with center C and diameter ST above, the two semicircles have diameters CS and CT. If the circumference of the circle is 64π, what is the length of the curved path from S to T through C?

(A) 4π

(B) 8π

(C) 16π

(D) 32π

(E) 64π

7: In the figure above, a small square is in-side a larger square. In terms of x, what is the area of the shaded region?

(A) 36 -- x2

(B) x2 -- 36

(C) 36 -- 2x

(D) 12 -- 2x

(E) 2x -- 12

8: The figure above is a right triangle. What is the value of 81 + x2?

(A) 98

(B) 99

(C) 121

(D) 144

(E) 196

C

S

T

6

x

9 -- x

9 + x

14

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Triangles, Quadrilaterals, Circles, and Polygon SAT Questions

9: The circle graph above shows the re-sults when 10,000 people were asked “how many movie theater movies do you attend per month?” The answer they gave is represented by x. How many people replied 5 or more?

(A) 2,500

(B) 5,500

(C) 6,000

(D) 9,000

(E) 10,000

10: How many rectangular blocks measuring 4 inches by 3 inches by 2 inches can be packed into a cube-shaped box whose interior measures 1 foot on each edge? (1 foot = 12 inches)

(A) 72

(B) 64

(C) 48

(D) 24

(E) 6

11: In the figure above, triangle ABC is similar to triangle DEF. What is the length of side FE?

12: In the figure above, C is the center of the circle and segment XZ is tangent to the circle at point Y. If the measure of < XYC is wo, how many possible values are there for w?

(A) More than four

(B) Four

(C ) Three

(D) Two

(E) One

x < 5 30% 5 ≤ x ≤ 10

25%

x > 10 35%

x = 0 10%

10 18

3 7 A

B

C

D

E

F

Z

X

Y

C

Page 24: SATPG Math Module 16 Triangles, Quadrilaterals, Circles ... · Math Module 16: Triangles, Quadrilaterals, Circles, Polygons, and 3D Shapes In this module, we will cover basic shapes

Page 24

Triangles, Quadrilaterals, Circles, and Polygon SAT Questions

13: A right circular cylinder has a base circumference of 8π. If the volume of the cylinder is 256π, what is the height?

(A) 4

(B) 16

(C) 24

(D) 32

(E) 64

14: In the figure above, w > 90o and x = y + 3. If w, x, and y are integers, what is the greatest possible value of x?

(A) 30

(B) 45

(C) 46

(D) 63

(E) 89

15: In triangle RST above, what is the length of side ST?

(A) 6

(B) 8

(C) √40 (approximately 6.32)

(D) √61 (approximately 7.81)

(E) 25/√34 (approximately 4.29)

16: The circle above with radius r is inscribed in a square with side x. What is the value of r/x?

(A) 1/4

(B) 1/π

(C ) 1/2

(D) 2/π

(E) 1/(√2)

wo

xo

yo

√34

9

5

R

S

T U

r x

Page 25: SATPG Math Module 16 Triangles, Quadrilaterals, Circles ... · Math Module 16: Triangles, Quadrilaterals, Circles, Polygons, and 3D Shapes In this module, we will cover basic shapes

Page 25

Triangles, Quadrilaterals, Circles, and Polygon SAT Questions

17: In the figure above, points W, Z and T lie on the same line. What is the value of a?

18: Which of the following could be the lengths of the sides of a triangle?

(A) 3, 4, and 1

(B) 6, 2, and 9

(C) 4, 10, and 6

(D) 4, 6, and 3

(E) 7, 8, and 1

19: In the figure above, the radius of the circle with center P is twice the radius of the circle with center O. What is the radius of the circle with center P?

(A) 3

(B) 4

(C ) 5

(D) 8

(E) 10

20: The circle above has center C. Which of the following measures would be sufficient by itself to determine the radius of the circle?

I. The length of arc DEF II. The perimeter of triangle CDF III. The length of chord DF

(A) None

(B) I only

(C ) II only

(D) III only

(E) I, II, and III

ao W

X Y

Z 55o

110o 130o

T

2 x 7

O P

C

D

E

F

Page 26: SATPG Math Module 16 Triangles, Quadrilaterals, Circles ... · Math Module 16: Triangles, Quadrilaterals, Circles, Polygons, and 3D Shapes In this module, we will cover basic shapes

Page 26

Triangles, Quadrilaterals, Circles, and Polygon SAT Questions

21: In the figure above, if ACEF is a quadrilateral and BDG is a triangle, what is the sum of the degree measures of the angles that are marked?

(A) 480

(B) 540

(C ) 600

(D) 660

(E) 720

22: Two spheres, one with radius 11 and one with radius 3 are tangent to each other. If A is any point on one of the spheres and B is any point on the other sphere, what is the greatest possible length of AB?

(A) 28

(B) 24

(C) 20

(D) 16

(E) 14

23: If the area of the triangle above is 120, what is the approximate length of side YX?

(A) 38

(B) 36

(C ) 33

(D) 32

(E) 31

24: What is the area of a right triangle whose perimeter is 72 and whose sides are 2x, 2x + 6, and 2x + 12?

(A) 54

(B) 81

(C ) 216

(D) 270

(E) 324

C

A D

F

Z

B

E

G

X

Y

30

Page 27: SATPG Math Module 16 Triangles, Quadrilaterals, Circles ... · Math Module 16: Triangles, Quadrilaterals, Circles, Polygons, and 3D Shapes In this module, we will cover basic shapes

Page 27

Triangles, Quadrilaterals, Circles, and Polygon SAT Questions

25: In triangle XYZ above, what is the value of a + b + c + d?

26: In the figure above, C is the center of both circles. If the circumference of the larg-er circle is 48 and the radius of the smaller circle is half the radius of the larger circle, what is the length of arc AXB ? (A) 2

(B) 4

(C) 6

(D) 8

(E) 10

27: In triangle RTV above, RV = RT, U is the midpoint of RV, and S is the midpoint of RT. If RU = y and US = 8, what is the length of VT?

(A) 12

(B) 16

(C ) 4y

(D) 8y

(E) 8y2

28: In rectangle PQRS above, the area of the shaded region is (πxy)/4. If the area of the shaded region is 14π, what is the best whole number approximation of the unshaded region of rectangle PQRS?

(A) 8

(B) 10

(C ) 12

(D) 14

(E) 16

ao

110o Z

P

Y

X

R

U

bo

co

do

C 60o

A

B

X

T

V

Note: Figure not drawn to scale. T

Q R

S

x

y

Page 28: SATPG Math Module 16 Triangles, Quadrilaterals, Circles ... · Math Module 16: Triangles, Quadrilaterals, Circles, Polygons, and 3D Shapes In this module, we will cover basic shapes

Page 28

Triangles, Quadrilaterals, Circles, and Polygon SAT Questions

29: In the figure above, the circle has center C and radius r. Lines XY and XZ are tangent to the circle. If P is the midpoint of line segment XZ and the measure of <CPZ equals the measure of < PCZ, what is the length of segment CX in terms of r?

(A) r + √2

(B) 2r

(C ) r√5

(D) r√3

(E) r√2

30: The pyramid above has a square base and four congruent triangular faces. The height of the pyramid is 10 centimeters and each edge of the square base is 8 centimeters long. What is the length of WY? (A) 2√29

(B) 2√33

(C) 2√39

(D) 2√43

(E) 2√57

31: In the figure above, a polygon with equal sides and equal angles is partially hidden by a piece of cardboard. If w + z = 90, how many sides does the polygon have?

(A) Five

(B) Six

(C ) Seven

(D) Eight

(E) Nine

32: A sphere with radius r fits snugly inside a cube that touches the center of each face. What is the volume of the cube in terms of r?

(A) 8r3

(B) 4r3

(C ) 2r3

(D) (4/3)r3

(E) r3

Z

Y

X

wo C

P

C

W

Y

Cardboard

zo

Page 29: SATPG Math Module 16 Triangles, Quadrilaterals, Circles ... · Math Module 16: Triangles, Quadrilaterals, Circles, Polygons, and 3D Shapes In this module, we will cover basic shapes

Page 29

Solutions to Triangles, Quadrilaterals, Circles, and Polygon SAT Questions

1: If all angles in the figure above are right angles, what is the area of the figure?

(A) 33

(B) 44

(C) 68

(D) 77

(E) 88

Solution to 1 (level 1)

The trick on these types of questions is to break up the drawing into ones in which the areas can be determined more easily, in this case two separate rectangles that can bee added together to determine the combined area... 3 x 4 + 8 x 7 = 12 + 56 = 68 Answer (C)

2The lengths of two sides of a triangle are 13 and 3. What is one possible length of the third side?

Solution to 2 (level 2)

The third side can be anywhere between the sum and difference of the two given sides by Legal Triangle Rule #2. Answer: 10 < x < 16 So, 10.1, 10.2, 10.3,... 15.7, 15.8, 15.9 would be correct. (Note that 10 and 16 are NOT acceptable answers because they are outside of the range.)

3: If the area of each face of a cube is 121 square centimeters, what is the volume of the cube in cubic centimeters?

Solution to 3 (level 2)

Since every cube has equal side lengths, taking the square root of 121 yields a side length of 11. Vcube = side3 = 113 = 1331 Answer = 1331

4: In the figure above? RT and SU intersect at point V and RS || UT. If a = 65 and b = 35, what is the value of c?

(A) 90

(B) 80

(C) 65

(D) 35

(E) Unable to be determined from the information given.

Solution to 4 (level 3)

The alternate interior angle to a is also 65o. Since the 2 angles in triangle TUV is 100o, angle c must be 80o. Answer (B)

3

4 8

11

ao bo

co

R

S T

U V

Note: Figure not drawn to scale.

35o 65o

Page 30: SATPG Math Module 16 Triangles, Quadrilaterals, Circles ... · Math Module 16: Triangles, Quadrilaterals, Circles, Polygons, and 3D Shapes In this module, we will cover basic shapes

Page 30

Solutions to Triangles, Quadrilaterals, Circles, and Polygon SAT Questions

5: Right circular cylinders G and H have the same volume. The radius of cylinder H is twice the radius of cylinder G. What is the ratio of the height of cylinder H to the height of cylinder G?

(A) 4 to 1

(B) 2 to 1

(C) 1 to 1

(D) 1 to 2

(E) 1 to 4

Solution to 5 (level 3)

VG = πr2hG VH = π(2r)2hH = 4πr2hH VG = VH , which means... πr2hG = 4πr2hH , leaving: hG = 4hH hH /hG = 1/4 Answer (E)

6: In the circle with center C and diameter ST above, the two semicircles have diameters CS and CT. If the circumference of the circle is 64π, what is the length of the curved path from S to T through C?

(A) 4π

(B) 8π

(C) 16π

(D) 32π

(E) 64π

Solution to 6 (level 3)

Given the circumference of the larger circle is 64π, we can find the radius: 64π = 2πr, r = 32. This means the radius of the combined 2 semicircles is 16. 2π16 = 32π Answer (D)

7: In the figure above, a small square is in-side a larger square. In terms of x, what is the area of the shaded region?

(A) 36 -- x2

(B) x2 -- 36

(C) 36 -- 2x

(D) 12 -- 2x

(E) 2x -- 12

Solution to 7 (level 2)

Ashaded = Awhole -- Aunshaded Ashaded = 62 -- x2 = 36 -- x2. Answer (A)

8: The figure above is a right triangle. What is the value of 81 + x2?

(A) 98

(B) 99

(C) 121

(D) 144

(E) 196

Solution to 8 (level 3)

Set up Pythagorean Theorem... (9 -- x)2 + (9 + x)2 = 142 81 -- 18x + x2 + 81 + 18x + x2 = 196 162 + 2x2 = 196 Factor out a 2 and divide... 81 + x2 = 98 Answer (A)

r 2r

hG hH

S

T

C

6

x

9 -- x

9 + x

14

Page 31: SATPG Math Module 16 Triangles, Quadrilaterals, Circles ... · Math Module 16: Triangles, Quadrilaterals, Circles, Polygons, and 3D Shapes In this module, we will cover basic shapes

Page 31

Solutions to Triangles, Quadrilaterals, Circles, and Polygon SAT Questions

9: The circle graph above shows the re-sults when 10,000 people were asked “how many movie theater movies do you attend per month?” The answer they gave is represented by x. How many people replied 5 or more?

(A) 2,500

(B) 5,500

(C) 6,000

(D) 9,000

(E) 10,000

Solution to 9 (level 3)

The percent of 5 or more movies is both the 25% of 5 ≤ x ≤ 10 and the 35% of x > 10, which = 60%. 60/100 = x/10,000, x = 6,000. Answer (C)

10: How many rectangular blocks measuring 4 inches by 3 inches by 2 inches can be packed into a cube-shaped box whose interior measures 1 foot on each edge? (1 foot = 12 inches)

(A) 72

(B) 64

(C) 48

(D) 24

(E) 6

Solution to 10 (level 4)

This is a “how many fit” question. How many fit = Vbigger = Vsmaller 123 = 1728 = 72 (4)(3)(2) 24 Answer (A)

11: In the figure above, triangle ABC is similar to triangle DEF. What is the length of side FE?

Solution to 11 (level 3)

Similar triangles tend to be proportions. 10/18 = 7/x Cross multiply... 10x = 126 x = 12.6 Answer = 12.6

12: In the figure above, C is the center of the circle and segment XZ is tangent to the circle at point Y. If the measure of < XYC is wo, how many possible values are there for w?

(A) More than four

(B) Four

(C ) Three

(D) Two

(E) One

Solution to 12 (level 3)

Since a tangent to a circle is always perpendicular to the center, angle XYC can only be 90o. Answer (E)

x < 5 30% 5 ≤ x ≤ 10

25%

x > 10 35%

x = 0 10%

10 18

3 7 A

B

C

D

E

F

X

Y

C Z

Page 32: SATPG Math Module 16 Triangles, Quadrilaterals, Circles ... · Math Module 16: Triangles, Quadrilaterals, Circles, Polygons, and 3D Shapes In this module, we will cover basic shapes

Page 32

Solutions to Triangles, Quadrilaterals, Circles, and Polygon SAT Questions

13: A right circular cylinder has a base circumference of 8π. If the volume of the cylinder is 256π, what is the height?

(A) 4

(B) 16

(C) 24

(D) 32

(E) 64

Solution to 13 (level 3)

First find the radius using circumference formula: 8π = 2πr, r = 4 Now solve for h... 256π = π42h 256π = 16πh h = 16 Answer (B)

14: In the figure above, w > 90o and x = y + 3. If w, x, and y are integers, what is the greatest possible value of x?

(A) 30

(B) 45

(C) 46

(D) 63

(E) 89

Solution to 14 (level 3)

For x to be the greatest possible, w must be the least possible, which is 91o. So, 91 + x + y = 180 x + y = 89 and x = y + 3 or y = x -- 3 2x -- 3 = 89: x = 46. Answer (C)

15: In triangle RST above, what is the length of side ST?

(A) 6

(B) 8

(C) √40 (approximately 6.32)

(D) √61 (approximately 7.81)

(E) 25/√34 (approximately 4.29)

Solution to 15 (level 3)

Use Pyth. Theorem to determine RU: 52 + (UR)2 = (√34)2, RU = 3 If RU = 3, then UT = 6... 52 + 62 = (ST)2, ST = √61 Answer (D)

16: The circle above with radius r is inscribed in a square with side x. What is the value of r/x?

(A) 1/4

(B) 1/π

(C ) 1/2

(D) 2/π

(E) 1/(√2)

Solution to 16 (level 3)

Since the side of the square is twice the length of the radius, r/x = 1/2 Answer (C)

wo

xo

yo

√34

9

5

R T U

S

r x

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Page 33

Solutions to Triangles, Quadrilaterals, Circles, and Polygon SAT Questions

17: In the figure above, points W, Z and T lie on the same line. What is the value of a?

Solution to 17 (level 3)

The sum of the angles of any quadrilateral = 360o. 55 + 110 + 130 + x = 360 x = 360 -- 295 = 65 If x = 65, then a = 180 -- 65 = 115 Answer = 115

18: Which of the following could be the lengths of the sides of a triangle?

(A) 3, 4, and 1

(B) 6, 2, and 9

(C) 4, 10, and 6

(D) 4, 6, and 3

(E) 7, 8, and 1

Solution to 18 (level 3)

In a legal triangle, each 2 sides must sum to greater than the value of the 3rd side, all combinations: 4 + 6 > 3 4 + 3 > 6 3 + 6 > 4 Answer (D)

19: In the figure above, the radius of the circle with center P is twice the radius of the circle with center O. What is the radius of the circle with center P?

(A) 3

(B) 4

(C ) 5

(D) 8

(E) 10

Solution to 19 (level 3)

2(x + 2) = x + 7 2x + 4 = x + 7 Combining like terms... x = 3, so 3 + 7 = 10 Answer (E)

20: The circle above has center C. Which of the following measures would be sufficient by itself to determine the radius of the circle?

I. The length of arc DEF II. The perimeter of triangle CDF III. The length of chord DF

(A) None

(B) I only

(C ) II only

(D) III only

(E) I, II, and III

Solution to 20 (level 3)

I. Knowing the length of arc DEF allows for the radius to be known. II. Since the triangle is an isosceles right triangle, the radius can thus be found. III. Same as reason II. Answer (E)

ao W

Z 55o

110o 130o

T

X Y

x 2 x 7

O P

C

D

E

F

Page 34: SATPG Math Module 16 Triangles, Quadrilaterals, Circles ... · Math Module 16: Triangles, Quadrilaterals, Circles, Polygons, and 3D Shapes In this module, we will cover basic shapes

Page 34

Solutions to Triangles, Quadrilaterals, Circles, and Polygon SAT Questions

21: In the figure above, if ACEF is a quad-rilateral and BDG is a triangle, what is the sum of the degree measures of the angles that are marked?

(A) 480

(B) 540

(C ) 600

(D) 660

(E) 720

Solution to 21 (level 3)

The sum of the angles of any quadrilateral = 360o. The sum of the angles of a triangle = 180o. 180 + 360 = 540 Answer (B)

22: Two spheres, one with radius 11 and one with radius 3 are tangent to each oth-er. If A is any point on one of the spheres and B is any point on the other sphere, what is the greatest possible length of AB?

(A) 28

(B) 24

(C) 20

(D) 16

(E) 14

Solution to 22 (level 3)

Answer (A)

23: If the area of the triangle above is 120, what is the approximate length of side YX?

(A) 38

(B) 36

(C ) 33

(D) 32

(E) 31

Solution to 23 (level 3)

A = (1/2) bh 120 = (1/2)b(30) b = 8 Using Pythagorean Theorem: 82 + 302 = (YX)2 YX = 31.05 Answer (E)

24: What is the area of a right triangle whose perimeter is 72 and whose sides are 2x, 2x + 6, and 2x + 12?

(A) 54

(B) 81

(C ) 216

(D) 270

(E) 324

Solution to 24 (level 3)

P = 2x + 2x + 6 + 2x + 12 72 = 6x + 18 6x = 54 x = 9 base = 2(9) + 6 = 24 height = 2(9) = 18 A = (1/2)(24)(18) = 216 Answer (C)

C

A D

F

B

E

G

22 6 A B

X

Y

30

2x

2x + 6

2x + 12

Page 35: SATPG Math Module 16 Triangles, Quadrilaterals, Circles ... · Math Module 16: Triangles, Quadrilaterals, Circles, Polygons, and 3D Shapes In this module, we will cover basic shapes

Page 35

Solutions to Triangles, Quadrilaterals, Circles, and Polygon SAT Questions

25: In triangle XYZ above, what is the value of a + b + c + d?

Solution to 25 (level 3)

Since each triangle = 180o, a + b = 180 -- 110 = 70 and c + d = 180 -- 110 = 70 70 + 70 = 140 Answer = 140

26: In the figure above, C is the center of both circles. If the circumference of the larger circle is 48 and the radius of the smaller circle is half the radius of the larger circle, what is the length of arc AXB ? (A) 2

(B) 4

(C) 6

(D) 8

(E) 10

Solution to 26 (level 4)

C = 2πr 48 = 2πr or r = 48/(2π) = 24/π Since rsmaller = (1/2)rlarger... rsmaller = (1/2)(24/π) = 12/π ALAXB = 2πrsmaller(deg/360) ALAXB = 2π(12/π)(60/360) ALAXB = 24/6 = 4 Answer (B)

27: In triangle RTV above, RV = RT, U is the midpoint of RV, and S is the midpoint of RT. If RU = y and US = 8, what is the length of VT?

(A) 12

(B) 16

(C ) 4y

(D) 8y

(E) 8y2

Solution to 27 (level 4)

Redraw triangles separately... Scale factor = 2:1, VT = 16 Answer (B)

28: In rectangle PQRS above, the area of the shaded region is (πxy)/4. If the area of the shaded region is 14π, what is the best whole number approximation of the unshaded region of rectangle PQRS?

(A) 8

(B) 10

(C ) 12

(D) 14

(E) 16

Solution to 28 (level 4)

Set the two areas together... πxy = 14π and cross multiply... 4 1 xy = 56, which is the Arectangle Aunshaded = Awhole -- Ashaded Aunshaded = 56 -- 14π ≈ 12 Answer (C)

ao

110o Z

Y

X

bo

co

do

C

A

B

X

60o

S

R

U

T

V

Note: Figure not drawn to scale.

8

y

R

T

V

R

S

U y

8

P

Q R

S

x

y

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Page 36

Solutions to Triangles, Quadrilaterals, Circles, and Polygon SAT Questions

29: In the figure above, the circle has center C and radius r. Lines XY and XZ are tangent to the circle. If P is the midpoint of line segment XZ and the measure of <CPZ equals the measure of < PCZ, what is the length of segment CX in terms of r?

(A) r + √2

(B) 2r

(C ) r√5

(D) r√3

(E) r√2

Solution to 29 (level 4)

Since <CPZ = < PCZ = 45o and P is the midpoint of XZ, PZ = r and XP = r... Pyth. Theorem: r2 + (2r)2 = (XC)2, so r2 + 4r2 = (XC)2 (XC)2 = 5r2 or XC = r√5 Answer (B)

Solution to 30 (level 5)

Draw an isosceles triangle from the center of the base to the corner forming a right triangle. 102 + (4√2)2 = WY2 WY = √(100 + 32) = 2√33 Answer (B)

31: In the figure above, a polygon with equal sides and equal angles is partially hidden by a piece of cardboard. If w + z = 90, how many sides does the polygon have?

(A) Five

(B) Six

(C ) Seven

(D) Eight

(E) Nine

Solution to 31 (level 5)

The hidden regular polygon forms a quadrilateral, which = 360o. 2i + 90 = 360, i = 135. i + e = 180, so e = 45, e = 360/n, 45 = 360/n, n = 8 Answer (D)

32: A sphere with radius r fits snugly inside a cube that touches the center of each face. What is the volume of the cube in terms of r?

(A) 8r3

(B) 4r3

(C ) 2r3

(D) (4/3)r3

(E) r3

Solution to 32 (level 5)

If the radius of the sphere is r, then the side length of the cube is double or 2r. Vcube = side3 Vcube = (2r)3 = 8r3 Answer (A)

Z

Y

X

C

P

r

r r

W

Y

30: The pyramid above has a square base and four congruent triangular faces. The height of the pyramid is 10 centimeters and each edge of the square base is 8 centimeters long. What is the length of WY? (A) 2√29

(B) 2√33

(C) 2√39

(D) 2√43

(E) 2√57

4

4√2

h = 10

wo

Cardboard

zo io io