Plane Geometry Practice Workbook with Answers: Circles, Chords, Secants, and Tangents (Master Essential Geometry Skills) 1941691897, 9781941691892

Table of contents :
Contents
Introduction
1 Radius and Circumference
2 Area of a Circle
3 Central Angles and Arc Length
4 Inscribed Angles
5 Thales's Theorem
6 Chords and Circular Segments
7 Secants
8 Tangents
9 Inscribed and Circumscribed Shapes
10 Beyond the Plane
Answers to Chapter 1
Answers to Chapter 2
Answers to Chapter 3
Answers to Chapter 4
Answers to Chapter 5
Answers to Chapter 6
Answers to Chapter 7
Answers to Chapter 8
Answers to Chapter 9
Answers to Chapter 10
Glossary
Notation and Symbols
Greek Alphabet

Citation preview

Plane Geometry Practice Workbook with Answers Volume 2: Circles, Chords, Secants, and Tangents

Chris McMullen, Ph.D.

Copyright © 2021 Chris McMullen, Ph.D.

www.improveyourmathfluency.com

www.monkeyphysicsblog.wordpress.com www.chrismcmullen.com

All rights are reserved. However, educators or parents who purchase one copy of this workbook (or who

borrow one physical copy from a library) may make and distribute photocopies of selected pages for

instructional (non-commercial) purposes for their own students or children only.

Zishka Publishing

Paperback ISBN: 978-1-941691-89-2

Mathematics > Geometry

Contents Introduction

iv

2 Area of a Circle

17

1 Radius and Circumference

3 Central Angles and Arc Length

4 Inscribed Angles

5 Thales’s Theorem

6 Chords and Circular Segments

7 Secants

6

24

33

48

60

84

8 Tangents

95

10 Beyond the Plane

146

9 Inscribed and Circumscribed Shapes Answer Key Glossary

Notation and Symbols Greek Alphabet

114 188

300

323

326

Introduction The goal of this workbook is to help students master essential geometry skills through explanations, examples, and practice.

• The second volume of this series focuses on circles, while the first volume (sold separately) focused on triangles, quadrilaterals, and other polygons.

• Each chapter begins with an introduction to the pertinent concepts and includes examples

illustrating how the concepts can be applied.

• Several exercises in each chapter offer ample practice.

• The first chapter begins with the relationship between the radius, the diameter, and the

circumference, and also discusses the significance of π.

• Chapter 2 introduces the basic formula for the area of a circle. Chapters 3 and 6 extend this idea to sectors, circular segments, and other regions.

• Chapters 3-4 cover central angles, inscribed angles, and arc length.

• Chapter 5 introduces Thales’s theorem.

• Chapters 6-8 discuss chords, segments, secants,

and tangents, including some important theorems.

• Chapter 9 explores shapes that are inscribed in circles or circumscribed about circles.

• The final chapter surveys shapes beyond the plane, including the cube, prism, pyramid, sphere,

cylinder, and cone.

• Answer key. Practice makes permanent, but not

necessarily perfect. Check the answers at the back of the book and strive to learn from any mistakes. This will help to ensure that practice makes

perfect.

Radius and Circumference This chapter presents basic properties of circles,

including the center, radius, diameter, circumference, and the significance of 𝜋𝜋. These concepts set the

foundation from which a deeper understanding of circles may be developed.

6

Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 1 Concepts A circle is a curve for which every point that lies on the curve is equidistant from a special point called the

center of the circle. For example, point O lies at the

center of the circle below, and every point that makes up the circle is equidistant from point O. The distance

from the center of the circle to any point on the circle is called the radius. The diameter is a line segment that

passes through the center of the circle and which joins two points that lie on the circle.

7

Chapter 1 – Radius and Circumference

The diameter (D) of a circle is twice as long as the

radius (R) of the circle: D = 2R. Put another way, the D

radius is one-half as long as the diameter: R = 2. If a

problem gives the radius of the circle, the diameter can be found by doubling the radius. If a problem gives the

diameter of the circle, the radius can be found by cutting the diameter in half.

The circumference of a circle equals the total distance around the circle. The circumference is basically the

same thing as perimeter, except that the term perimeter is used for polygons whereas the term circumference is used for circles.

8

Plane Geometry Practice Workbook with Answers, Volume 2

For any circle, regardless of how small or large the

circle is, if its circumference and diameter are each

measured, the circumference divided by the diameter always equals the same value. A larger circle has a

larger diameter and a larger circumference, while a smaller circle has a smaller diameter and a smaller

circumference. The diameter and circumference always come in the same proportion. The ratio of the

circumference of a circle to its diameter is called π, the

lowercase Greek letter pi. Numerically, π equals

3.141592653589793… The digits continue on forever

without a repeating pattern to them. Since π is slightly

larger than the number 3, the circumference of any

circle is slightly larger than three times its diameter.

9

Chapter 1 – Radius and Circumference

The significance of the number π is that the ratio of the circumference of any circle to its diameter numerically equals π. Since π is infinitely long when expressed in decimal form, it is often rounded to 3.14 when

performing numerical calculations (unless additional precision is needed). For some calculations, it is common to leave π in the answer. For example, exact value, whereas

3(3.14) 4

3π 4

is an

≈ 2.36 is the same quantity

approximated to three significant figures. (Technically,

3(3.14) 4

= 2.355, but since π was rounded to 3.14 for this

calculation, it does not make sense for the answer to have more significant digits than 3.14.)

10

Plane Geometry Practice Workbook with Answers, Volume 2

The circumference (C) of a circle is related to the

diameter (D) of the circle according to the following

formula: C = πD. Since diameter is twice the radius, the circumference of a circle is related to the radius (R) of

the circle according to the following formula: C = 2πR.

The symbol ⊙ (which is a small circle with a dot in the

center) is sometimes used to represent a circle. When a

letter follows the symbol for a circle, the letter indicates the point that lies at the center of the circle. For

example, ⊙A represents a circle centered about the

point A, whereas ⊙B represents a circle centered about the point B. When one diagram contains multiple

circles, this notation helps to distinguish one circle from another.

Note that some texts and instructors use lowercase symbols for diameter and radius (d = 2r), whereas

other texts and instructors use uppercase symbols (D = 2R). Both conventions are in common usage. 11

Chapter 1 – Radius and Circumference

Chapter 1 Examples Example 1. The radius of a circle is 5. What are the diameter and circumference?

The diameter is twice the radius: D = 2R = 2(5) = 10. The circumference is π times the diameter: C = πD =

10π is the exact value of the circumference, while

10(3.14) ≈ 31.4 is the circumference rounded to three

significant figures.

Example 2. The circumference of a circle is 8. What is

the radius?

The circumference is related to the radius by C = 2πR. Divide both sides by 2π. 8 4 C = = =R 2π 2π π 4

The exact answer is π. Using a calculator, this is 4

approximately equal to 3.14 ≈ 1.27 to three significant

figures.

12

Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 1 Problems 1. The radius of a circle is 20.

(A) What is the diameter of the circle?

(B) Give an exact value for the circumference in terms of the constant π.

(C) Use a calculator to approximate the circumference to three significant figures.

2. The diameter of a circle is 8.

(A) What is the radius of the circle?

(B) Give an exact value for the circumference in terms of the constant π.

(C) Use a calculator to approximate the circumference to three significant figures.

13

Chapter 1 – Radius and Circumference

3. The circumference of a circle is 1.

(A) Give an exact value for the diameter in terms of the constant π.

(B) Use a calculator to approximate the diameter to

three significant figures.

(C) Give an exact value for the radius in terms of the constant π.

(D) Use a calculator to approximate the radius to three

significant figures.

4. The radius of ⊙A equals the circumference of ⊙B. (A) Express the ratio of ⊙A’s circumference to ⊙B’s

circumference as an exact value in terms of the constant π.

(B) Express the ratio of ⊙A’s circumference to ⊙B’s radius as an exact value in terms of the constant π.

14

Plane Geometry Practice Workbook with Answers, Volume 2

5. Connie and Danny each travel from point A to point B in the diagram below. Connie travels 36 yards along the straight line from A to B, whereas Danny travels along the semicircular arc from A to B. How much farther

does Danny travel compared to Connie? Express the

answer both as an exact value in terms of the constant π and also as an approximate value rounded to three significant figures.

6. Marty and Nancy begin running from the same

position on a circular track. Marty jogs with a constant speed of 2.4 meters per second in one direction while Nancy jogs with a constant speed of 3.6 meters per

second in the opposite direction. Marty and Nancy first meet up exactly 4 minutes after they started. What is

the diameter of the track? Express the answer both as

an exact value in terms of the constant π and also as an

approximate value rounded to three significant figures. 15

Chapter 1 – Radius and Circumference

7. Find a variety of circular objects with different sizes such as a quarter, hamburger bun, basketball, bicycle wheel, and hula hoop. Measure the diameter straight

across the center using a ruler, meterstick, or similar

straight measuring device. Then wrap string or thread

around the circular object to measure its circumference directly. (Do not use a calculator or the known value of

π to determine diameter or circumference.) Measure all

of distances in the same units (such as centimeters). Make a coordinate graph with diameter on the

horizontal axis and circumference on the vertical axis.

(If using Excel, choose the option for an 𝑥𝑥𝑥𝑥 scatter plot.)

Predict what the slope of the graph should be and compare with the actual slope.

16

Area of a Circle Pay close attention to the differences between the

formulas for the circumference and area of a circle.

Thinking about the units can help to avoid getting these similar formulas mixed up. Circumference is a distance, so the length is not squared in the formula for

circumference: C = 2πR. In contrast, area is measured in square units (like square feet or square meters). Since

area is measured in square units, the radius is squared in the formula for area: A = πR2 .

17

Chapter 2 – Area of a Circle

Chapter 2 Concepts The area (A) of a circle is related to its radius (R) by the formula A = πR2 . Remember that the radius is squared in the formula for area (but is NOT squared in the

formula for circumference, C = 2πR). Recall that the

diameter of a circle is equal to twice the radius of the circle: D = 2R.

To find the area of a sector, see Chapter 3. For other

regions involving circles, such as circular segments, see Chapter 6.

18

Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 2 Examples Example 1. The radius of a circle is 5. What is the area of

the circle?

Use the formula for the area of a circle: A = πR2 = π(5)2 = 25π is the exact value, while 25(3.14) ≈ 78.5 is rounded to three significant figures.

Example 2. The ratio of the area of a circle to its

circumference is 6. What is the radius of the circle? Divide the formula for area by the formula for

circumference. Since the formula for area involves

radius, also use the formula for circumference that involves radius.

A πR2 = C 2πR

Recall from algebra that

𝑥𝑥 2 𝑥𝑥

= 𝑥𝑥. Note that 0π cancels.

A R = C 2 The problem states that A:C = 6. R 6= 2 Multiply both sides of the equation by 2. 12 = R 19

Chapter 2 – Area of a Circle

Example 3. Determine the area of the semicircle shown

above.

Since the shape is a semicircle, the bottom side is the

diameter: D = 20. The radius is one-half the diameter:

R=

D 2

=

20 2

circle: A =

= 10. A semicircle has one-half the area of a πR2 2

=

π(10)2 2

=

100π 2

= 50π is the exact value,

while 50(3.14) ≈ 157 is rounded to three significant figures.

20

Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 2 Problems 1. The diameter of a circle is 12. Find the exact area of

the circle in terms of the constant π and also

approximate the answer to three significant figures. 2. The area of a circle is 16. Find the exact

circumference of the circle in terms of the constant π and also approximate the answer to three significant figures.

3. The diameter of ⊙A is twice the diameter ⊙B. Find the ratio of their areas.

4. Find the exact area of the quarter circle shown below in terms of the constant π and also approximate the answer to three significant figures.

21

Chapter 2 – Area of a Circle

5. Point O lies at the center of the large circle below,

such that the diameter of the small circle is the radius of the large circle. Find the ratio of the area of the shaded region to the area of the small circle.

6. The circumference of a circle, the perimeter of a

square, and the perimeter of an equilateral triangle are

the same. Find the ratio of the area of the circle to the

area of the square, and also find the ratio of the area of the circle to the area of the triangle. Find the exact

answers and also approximate each answer to three significant figures.

22

Plane Geometry Practice Workbook with Answers, Volume 2

7. In the diagram below, points D, E, and F lie at the

vertices of a right triangle and also lie at the centers of ��� is the radius the three circles. ���� DF is the radius of ⊙D, �EF

of ⊙E, and ���� DE happens to be equal to the radius of ⊙F.

If the area of ⊙D is 16 and the area of ⊙F is 36, what is the area of ⊙E?

23

Central Angles and Arc Length A circular arc length represents a fraction of the

circumference of a circle. The central angle similarly

represents a fraction of 360° (or 2π radians). These two

fractions lead to an important formula: arc length equals radius times the central angle in radians.

24

Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 3 Concepts An arc length is the distance along a portion of the circle. An arc length represents a fraction of the

circumference. A sector is a region enclosed by an arc

length and the two radii that connect to its endpoints. A sector looks like a slice of pie that is cut from the center

(right figure below). A central angle is an angle that is

formed by two radii. For example, in the diagram below,

point O lies at the center of the circle, s represents the

arc length between points A and B, θ is the central angle

corresponding to the arc length s, and the shaded region bounded by s and the two radii is a sector.

There are actually two arc lengths, two central angles,

and two sectors in the diagram above. The shortest arc length from A to B is called the minor arc, while the

longest arc length from A to B is called the major arc. In the diagram on the following page, s is the minor arc 25

Chapter 3 – Central Angles and Arc Length

whereas t is the major arc. Note that the minor arc and

major arc add up to the circumference: s + t = C. Also, θ is the central angle corresponding to the minor arc,

whereas φ is the central angle corresponding to the

major arc. Note that these two central angles add up to

360°, which is equivalent to 2π radians: θ + φ = 360° =

2π rad (Volume 1, Chapter 1).

An important property of a circle is that the arc length is proportional to the central angle. A smaller central

angle has a smaller corresponding arc length, while a larger central angle has a larger corresponding arc

length. They come in a direct proportion. The arc length

(s) is a fraction of the circumference (C), and the central angle θ is a fraction of 2π radians (which equates to

360°). This leads to the following proportion: θ s = C 2π 26

Plane Geometry Practice Workbook with Answers, Volume 2

Since the angle was put in as 2π, the equation above is only valid if θ is expressed in radians. Since the

circumference is related to the radius by C = 2πR, the

above formula can be rewritten as follows. θ s = 2πR 2π Multiply both sides by 2πR. s = Rθ

The equation above is known as the arc length formula. The arc length is equal to the radius times the angle, provided that the angle is expressed in radians (not degrees). Recall (from Volume 1) that π radians

corresponds to 180°. To convert an angle from degrees π

to radians, multiply by 180°. For example, 45° becomes π

π

45° 180° = 4 rad. To convert an angle from radians to degrees, multiply by

π 180° 6

π

= 30°.

180° π

π

. For example, 6 rad becomes

27

Chapter 3 – Central Angles and Arc Length

Chapter 3 Examples Example 1. The diagram below shows a sector with a radius of 8. Determine s. π

Multiply by 180° to convert the central angle from

degrees to radians.

π π = rad 180° 4 Now that the central angle is in radians, the arc length 45°

formula may be used.

π s = Rθ = 8 � � = 2π 4 The exact value of the arc length is 2π, which is approximately equal to 2(3.14) = 6.28 to three significant figures.

28

Plane Geometry Practice Workbook with Answers, Volume 2

Example 2. Determine the area of the sector shown

above.

Since 360° ÷ 45° = 8, the sector shown above has 1/8

of the area of a full circle.

πR2 π82 64π = = = 8π A= 8 8 8

The exact area of the sector is 8π, which is

approximately equal to 8(3.14) ≈ 25.1 to three

significant figures.

Notation: In this book, a lowercase letter of the English

alphabet (like s or q) labeling an arc refers to a distance equal to the arc length. This book uses one of the

following three common methods to indicate an angle:

an angle symbol followed by three uppercase English letters (like ∠ABC), an angle symbol followed by a

number (like ∠1), or a lowercase Greek letter (like θ). A lowercase English letter (like s or q) in this book can only be a distance.

29

Chapter 3 – Central Angles and Arc Length

Chapter 3 Problems Note: The diagrams are not drawn to scale.

1. The diagram below shows a sector with a radius of 12.

(A) Determine the exact value of s in terms of the constant π.

(B) Determine the exact area of the sector in terms of the constant π.

2. The diagram below shows a sector with a radius of 18.

(A) Determine the exact value of s in terms of the constant π.

(B) Determine the exact area of the sector in terms of the constant π.

30

Plane Geometry Practice Workbook with Answers, Volume 2

3. The diagram below shows a sector with a radius of 20 and an arc length of 8π. Find the value of θ in degrees.

4. The diagram below shows a sector with a central

angle of 120° and an arc length of 21π. Determine R.

5. The diagram below shows a sector with a radius of

90. The area of the sector is 5805π. Find the value of θ in degrees and the exact value of s in terms of the constant π.

31

Chapter 3 – Central Angles and Arc Length

6. In the diagram below, two diameters divide the circle into four sectors. The diameter of the circle is 48.

Determine α, β, and γ. Also determine the exact values of p, q, r, s, and the circumference in terms of the

constant π.

7. (A) Derive a formula for the area of a sector in terms of the radius, the central angle (expressed in radians),

and the constant π. The arc length may not appear in the final answer.

(B) Derive a formula for the area of a sector in terms of

the radius and the arc length. The central angle may not appear in the final answer.

8. Derive a formula for the arc length of a circular arc in terms of the radius, the central angle, and the constant π, where the central angle is expressed in degrees rather than radians.

32

Inscribed Angles Inscribed angles are fundamental to the geometry of

circles. Inscribed angles relate to a variety of important properties of circles, which will be explored in this and the following chapters.

33

Chapter 4 – Inscribed Angles

Chapter 4 Concepts The vertex of an inscribed angle lies on the

circumference of a circle and the two sides that form the angle subtend an arc. (This means that two sides cut off an arc opposite to the vertex.) For example, in the

diagram below, angle α is an inscribed angle. Point C,

the vertex of angle α, lies on the circumference. Sides ���� AC and ���� BC of angle α subtend arc length s between points A

and B.

One useful property of inscribed angles is that if two

different inscribed angles subtend the same arc length in the same circle, then the inscribed angles are

congruent. For example, in the diagram on the following page, α and β are both inscribed angles since their

vertices lie on the circumference of the circle, and α and β subtend the same arc length (s). Therefore, α ≅ β. 34

Plane Geometry Practice Workbook with Answers, Volume 2

Another useful property of inscribed angles is the

inscribed angle theorem. According to the inscribed

angle theorem, an inscribed angle has one-half of the angular measure of a central angle that subtends the

same arc. For example, in the diagram below, point O ���� are lies at the center of the circle such that ���� AO and BO radii. Angle α is an inscribed angle, whereas θ is a

central angle. Since α and θ intercept the same arc

length (s), according to the inscribed angle theorem, θ

α= . 2

35

Chapter 4 – Inscribed Angles

Note: Except when otherwise noted, an arc will refer to the minor arc and a central angle will refer to the

central angle corresponding to the minor arc. When it would be desirable to refer to a major arc, this will be

made clear (for example, by calling it a “major” arc). Arc

length s shown on the previous page is the minor arc from A to B. In contrast, the major arc from A to B

includes A to C plus C to B. The major arc and minor arc

added together form the circumference. The “arc length from A to B” will automatically mean the minor arc s in the diagram above. The arc length from A to C plus the arc length from C to B could be called “the major arc from A to B” to distinguish it with the minor arc length s.

36

Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 4 Examples Example 1. Determine ∠BDC in the diagram below.

Since ∠BAC and ∠BDC are inscribed angles that subtend

the same arc (between points B and C), these angles are

congruent: ∠BAC ≅ ∠BDC. Therefore, ∠BDC = 42°.

37

Chapter 4 – Inscribed Angles

Example 2. In the diagram below, point O lies at the

center of the circle. Find ∠ACB.

Since point O lies at the center of the circle, ∠AOB is a

central angle. Since central angle ∠AOB and inscribed angle ∠ACB subtend the same arc (between points A and B), the inscribed angle theorem may be applied. ∠AOB 140° = = 70° ∠ACB = 2 2

38

Plane Geometry Practice Workbook with Answers, Volume 2

Example 3. In the diagram below, ���� AC is a diameter and O lies at the center. Find ∠CBO.

Since point O lies at the center of the circle, ∠AOB is a

central angle. Since central angle ∠AOB and inscribed angle ∠ACB subtend the same arc (between points A

and B), the inscribed angle theorem may be applied. ∠AOB 60° = = 30° ∠ACB = 2 2 ���� ≅ ���� Since point O lies at the center of the circle, BO CO since both are radii. Therefore, ∆BCO is an isosceles

triangle. Since ∆BCO is an isosceles triangle, ∠BCO ≅

∠CBO. The final answer is ∠CBO = 30°.

Tip: It is often useful to show that a triangle is isosceles because two sides are radii of the same circle.

39

Chapter 4 – Inscribed Angles

Example 4. Prove the inscribed angle theorem for the

case shown below, where the center of the circle (which is point O) lies inside ∆ABC. (Other cases will be explored in problems at the end of the chapter.)

Since point O lies at the center of the circle, ���� AO ≅ ���� BO ≅ ���� CO since these are all radii. Therefore, ∆ACO and ∆BCO are both isosceles triangles. Since ∆ACO is an isosceles

triangle, σ ≅ τ. Since ∆BCO is an isosceles triangle, α ≅

β. The three interior angles of any triangle add up to

180°. For ∆ACO, σ + τ + ρ = 180°. For ∆BCO, α + β +

γ = 180°. Since σ ≅ τ and α ≅ β, these formulas become 2σ + ρ = 180° and 2β + γ = 180°. Add these equations

together: 2σ + ρ + 2β + γ = 360°. Since ρ, γ, and ∠AOB form a full circle, these angles add up to 360°: ρ + γ +

∠AOB = 360°. Combine these equations. 2σ + ρ + 2β + γ = ρ + γ + ∠AOB 2σ + 2β = ∠AOB 40

Plane Geometry Practice Workbook with Answers, Volume 2

Since σ + β = ∠ACB, this proves the inscribed angle

theorem for the case shown above. ∠AOB ∠ACB = 2

41

Chapter 4 – Inscribed Angles

Chapter 4 Problems Note: The diagrams are not drawn to scale.

1. In the diagram below, point E does NOT lie at the center and BD is NOT a diameter.

(A) Which pairs of angles are congruent in the diagram above? For each pair of angles, state the reason that they are congruent.

(B) If ∠CBD = 39° and ∠ADB = 65°, find ∠ACB, ∠CAD, ∠AED, and ∠BEC.

42

Plane Geometry Practice Workbook with Answers, Volume 2

2. In the diagram below, point O lies at the center of the circle and ∠BOA = 76°. Find ∠ACB and ∠ADB.

3. In the diagram below, point O lies at the center of the circle and ∠EFH = 68°. Find ∠EGH and ∠EOH.

43

Chapter 4 – Inscribed Angles

4. In the diagram below, point O lies at the center of the circle and ∠BOD = 150°. Find ∠BCD and ∠BAD.

5. In the diagram below, point O lies at the center of the circle, ∠QPR = 60°, and OQ = 2. Find ∠QOR, ∠OQR, ∠ORQ, OR, and QR.

44

Plane Geometry Practice Workbook with Answers, Volume 2

6. In the diagram below, point O lies at the center of the circle and ∠AOD = 36°. Find ∠ACD and ∠BDC.

7. In the diagram below, ⊙O and ⊙Q intersect at points

P and R, point O lies on the circumference of ⊙Q, and ∠POR = 74°. Find ∠OPQ, ∠OQP, and ∠PQR.

45

Chapter 4 – Inscribed Angles

8. In the diagram below, point O lies at the center of the θ

circle. Show that β = 180° − 2 and that α + β = 180°.

9. Prove the inscribed angle theorem for the case shown below, where the center of the circle (which is point O) lies on diameter ���� AC.

46

Plane Geometry Practice Workbook with Answers, Volume 2

10. In the diagram below, point O lies at the center of

the circle. Show that α + δ = 90°. Do NOT assume that any of the lines are parallel.

11. Prove the inscribed angle theorem for the case

shown above, where the center of the circle (which is point O) lies outside of ∆ABC. 47

Thales’s Theorem Thales’s theorem concerns a right triangle that is

inscribed in a circle. When Thales’s theorem applies, all of the properties of right triangles may be useful,

including the Pythagorean theorem, the LH congruence test, the LH similarity test, and special right triangles (45°-45°-90° and 30°-60°-90°). It may be helpful to

review Chapter 5 from Volume 1 before proceeding.

48

Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 5 Concepts According to Thales’s theorem (sometimes called

Thales theorem), if all three interior angles of a triangle are inscribed angles (which means that the triangle is

inscribed in the circle) and if one side of the triangle is a diameter of the circle, the angle opposite to the

diameter is a right angle. Thales’s theorem has a number of consequences:

• If one side of a triangle that is inscribed in a circle is a diameter, the triangle is a right triangle. The

diameter is the hypotenuse of the triangle. The two acute angles are complementary angles.

• If a right triangle is inscribed in a circle, the hypotenuse will be a diameter of the circle.

• If an inscribed angle subtends an arc equal to a semicircle, the inscribed angle must be a right angle.

• If Thales’s theorem applies to a triangle inscribed

in a circle, the Pythagorean theorem applies to the triangle. The sum of the squares of the legs of the triangle is equal to the square of the diameter of the circle.

49

Chapter 5 – Thales’s Theorem

In the diagram above, point O lies at the center of the circle and also lies on ���� AC such that ���� AC is a diameter of

the circle. The angle opposite to the diameter (∠ABC) is a right angle according to Thales’s theorem. Angles α

and β are complements: α + β = 90°. ∆ABC is a right AB triangle. Side ���� AC is the hypotenuse of ∆ABC; sides ����

���� are legs. According to the Pythagorean theorem, and BC AB 2 + BC 2 = AC2 .

The diagram above shows angle ∠ABC inscribed in a

semicircle. ∠ABC is a right angle according to Thales’s theorem.

50

Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 5 Examples Example 1. In the diagram below, ���� AC is a diameter. Find α.

Since ���� AC is a diameter, ∠ABC is a right angle. Since

∆ABC is a right triangle, the two acute angles of ∆ABC are complementary. α + 63° = 90° α = 90° − 63° = 27°

Example 2. In the diagram below, ���� AC is a diameter. Find BC.

51

Chapter 5 – Thales’s Theorem

Since ���� AC is a diameter, ∠ABC is a right angle. Since

∆ABC is a right triangle, the lengths of the sides are related by the Pythagorean theorem. AB 2 + BC 2 = AC 2 122 + BC 2 = 132

144 + BC 2 = 169

BC 2 = 169 − 144 = 25 BC = √25 = 5

52

Plane Geometry Practice Workbook with Answers, Volume 2

Example 3. In the semicircle below, AB = 4. Find θ, BC,

and AC.

Since ���� AC is the base of a semicircle, ∠ABC is a right

angle. Since ∆ABC is a right triangle with a 30° angle, ∆ABC is a 30°-60°-90° triangle and θ = 60°. Recall

(from Volume 1, Chapter 5) that the sides of a 30°-60°-

90° triangle come in the ratio 1:√3:2, with the short side

opposite to the 30° angle and the long side opposite to the right angle. Since AB = 4 is opposite to the 30°

angle, BC = AB√3 = 4√3 and AC = 2AB = 2(4) = 8.

Observe that AB:BC:AC = 4:4√3:8, which reduces to 1:√3:2 (if you divide each side by 4).

Example 4. Discuss how Thales’s theorem relates to the

circumcenter of a triangle.

53

Chapter 5 – Thales’s Theorem

Recall (from Volume 1, Chapter 7) that:

• The three perpendicular bisectors of a triangle intersect at the circumcenter.

• The circumcenter is equidistant from the three vertices of the triangle.

• For a right triangle, the circumcenter lies at the midpoint of the hypotenuse.

In the diagram on the previous page, ∆ABC is a right triangle; ∠ABC is a right angle and ���� AC is the hypotenuse. Point O is the circumcenter of the triangle, which lies at the midpoint of the hypotenuse. Point O is equidistant from vertices A, B, and C. Since ���� OA ≅ ���� OB ≅ ���� OC, a circle can be circumscribed about the triangle as shown

above, which means that a circle can be drawn that

passes through all three vertices. Point O lies at the center of the circle and ���� AC is a diameter of the circle.

Whereas this example is a discussion of Thales theorem, Problems 10-11 explore proofs of Thales theorem. For more about circumscribed (and inscribed) circles, see Chapter 9.

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Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 5 Problems Note: The diagrams are not drawn to scale. 1. In the diagram below, ���� AC is a diameter. Find θ and φ.

2. In the diagram below, points D, O, and F are collinear

and point O lies at the center of the circle. Find α, β, γ, δ, and ε.

55

Chapter 5 – Thales’s Theorem

3. In the diagram below, ���� QR is a diameter. Find QR.

��� are diameters, 4. In the diagram below, ���� AD and �CF BC = 9, BF = 13, and DE = 5. Find AE.

5. In the semicircle below, JL = 6. Find θ, JK, and KL.

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Plane Geometry Practice Workbook with Answers, Volume 2

6. In the diagram below, ���� PR is a diameter and PQ = 12. Find χ, QR, and PR.

7. Find the combined area of the two shaded regions in the semicircle below.

8. In the diagram below, ���� DF and ���� EG are diameters. Prove ��� ∥ ���� that �EF DG and that the two triangles are congruent.

57

Chapter 5 – Thales’s Theorem

9. In the diagram below, ���� PR and ���� QS are diameters. Prove

that PQRS is a rectangle.

10. Use the diagram below to prove Thales’s theorem.

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Plane Geometry Practice Workbook with Answers, Volume 2

11. Use the diagrams below, where ���� BO ⊥ ���� AC, to prove Thales’s theorem. First, prove directly that ∆ABC is a

45°-45°-90° triangle. (Do NOT use Thales’s theorem to

prove this. Do NOT use a method similar to Problem 10 to prove this. Do NOT use the idea of the circumcenter ���� ⊥ ���� AC and basic to prove this. Simply prove it using BO

properties of circles.) Now use the result that ∆ABC is a

45°-45°-90° triangle to prove that ∠ADC is a right angle.

59

Chords and Circular Segments This chapter discusses chords, including theorems that

relate to intersecting chords, and also explores the area of a circular segment which is formed by a chord.

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Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 6 Concepts A chord (pronounced with a silent “h” so that it sounds like “cord”) is a line segment that connects two points that lie on the circumference of a circle. A circular

segment is a region enclosed by an arc and a chord. For example, in the diagram on the left below, ���� AB is a chord

and the shaded region between ���� AB and arc length s is a

circular segment.

A diameter is a special chord. A diameter is a chord that passes through the center of the circle. A chord that

does not pass through the center of the circle is shorter

than the diameter. For example, in the diagram above on the right, point O lies at the center of the circle, ���� BC is a diameter, ���� AD is a chord that does not pass through O,

and AD < BC.

61

Chapter 6 – Chords and Circular Segments

If two chords in the same circle are congruent, the ���� ≅ ���� chords subtend congruent arcs. For example, if AD BC in the diagram below, then arc lengths s and t are equal.

Tip: When working with chords or circular segments, it is often helpful to draw line segments connecting the

endpoints of the chord to the center of the circle. Since the endpoints of the chord lie on the circumference of

the circle, each line segment that connects an endpoint

of a chord to the center of a circle is a radius. A triangle bounded by a chord and two radii is an isosceles

triangle. For example, in the diagram on the next page, points A and B are the endpoints of chord ���� AB, point O

���� are radii, lies at the center of the circle, ���� AO and BO ∠AOB is a central angle, and ∆ABO is isosceles. 62

Plane Geometry Practice Workbook with Answers, Volume 2

Note the distinction between a circular segment and a sector (Chapter 3). A circular segment refers to the

region between a chord and an arc (left figure on the

next page), whereas a sector is a pie slice cut from the

center of the circle (right figure on the next page). The area of the circular segment shown below on the left

can be found by subtracting the area of ∆ABO from the area of the sector shown on the right. Recall from Chapter 3, Problem 7 that the area of a sector is

θR2 2

,

provided that θ (which is ∠AOB in the diagram below) is expressed in radians. Recall that π rad = 180°. Also

recall that the area of a triangle is

Chapter 6).

63

bh 2

(Volume 1,

Chapter 6 – Chords and Circular Segments

When two chords of the same circle intersect, the angle between the chords is equal to the average value of the two central angles that subtend the same arcs. For example, chords ���� AC and ���� BD intersect at point E in the

diagram below, α ≅ γ (vertical angles), point O lies at the center of the circle, and central angles θ and φ

subtend the same arcs (s and t) as α and γ. Angles α, γ,

θ, and φ are related by α = γ =

64

θ+φ 2

.

Plane Geometry Practice Workbook with Answers, Volume 2

According to the intersecting chords theorem, the

distances in the left diagram above are related by the

following formula, which involves multiplying distances together: (AE)(CE) = (BE)(DE) = R2 − EO2 . As shown below, EO is the distance from the point where the

chords intersect (E) to the center of the circle (O). The radius of the circle is R.

Notation: In this book, a lowercase letter of the English

alphabet (like s or q) labeling an arc refers to a distance equal to the arc length. This book uses one of the

following three common method to indicate an angle: an angle symbol followed by three uppercase English letters (like ∠ABC), an angle symbol followed by a

number (like ∠1), or a lowercase Greek letter (like θ). A lowercase English letter (like s or q) in this book can only be a distance.

65

Chapter 6 – Chords and Circular Segments

When reading a book that labels an angle on an arc, this is usually the central angle for the arc (even if the central angle is not drawn).

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Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 6 Examples Example 1. In the diagram below, AB = 5√2 and the

diameter of the circle is 10. Find the central angle that subtends the (minor) arc from A to B.

First, draw the center of the circle (point O) and draw ����. Since D = 10, the radius is line segments ���� AO and BO R=

D 2

=

10 2

= 5. Note that AO = BO = R = 5. The sides of

∆ABO come in the ratio AO:BO:AB = 5:5:5√2 = 1:1:√2. (The ratio was reduced by dividing each side by 5.)

Since the sides of a 45°-45°-90° triangle come in the

ratio 1:1:√2 (see Volume 1, Chapter 5), this shows that

∆ABO is a 45°-45°-90° triangle. The hypotenuse (AB) is

opposite to the right angle. Therefore, ∠AOB = 90°. 67

Chapter 6 – Chords and Circular Segments

Example 2. Find the area of the shaded circular segment

in the diagram below. As in Example 1, AB = 5√2 and the diameter of the circle is 10.

First find the area of the sector (Chapter 3). Recall from Example 1 that ∠AOB = 90°.

Since 360° ÷ 90° = 4, the sector has 1/4 of the area of a

full circle:

πR2 4

=

π52 4

=

25π 4

. Since ∆ABO is a right triangle,

the two legs (AO and BO) serve as the base and height. 1

1

The area of ∆ABO equals 2 (AO)(BO) = 2 (5)(5) =

25 2

.

Subtract the area of ∆ABO from the area of the sector to

find the area of the circular segment: A =

25 4 4

4

−

(π − 2) is the exact area, which was found by

factoring out 25

25π

(π − 2) =

25

4 25 4

25 2

. Check the answer by distributing:

π−

25 4

(2) =

25 4

π−

25 2

. The area of the

circular segment is approximately equal to 68

=

Plane Geometry Practice Workbook with Answers, Volume 2

25 4

(3.14 − 2) ≈ 7.13 when rounded to three significant

figures.

Example 3. The same circle is shown twice below. On

the right, point O lies at the center of the circle. Determine α and γ.

First, α ≅ γ because α and γ are vertical angles. Second, ���� ���� are intersecting chords, which means that α AC and BD

and γ are each equal to the average of the central angles that subtend the same arcs as α and γ. ∠AOB + ∠COD 30° + 60° 90° = = = 45° α=γ= 2 2 2

69

Chapter 6 – Chords and Circular Segments

Example 4. In the diagram below, AE = 2, BE = 4, and

CE = 10. Find DE.

Apply the intersecting chords theorem: (AE)(CE) = (BE)(DE) (2)(10) = (4)(DE) 20 = 5 = DE 4

Example 5. In the diagram below, AD = 9, CD = 32, and the radius of the circle equals 22. Find DO.

Apply the intersecting chords theorem: (AD)(CD) = R2 − DO2 (9)(32) = 222 − DO2 70

Plane Geometry Practice Workbook with Answers, Volume 2

288 = 484 − DO2

DO2 + 288 = 484

DO2 = 484 − 288 = 196 DO = √196 = 14

Check: (BD)(DE) = (22 − 14)(22 + 14) = (8)(36) =

288.

Example 6. In the diagram below, point O lies at the center of the circle and ���� AD ≅ ���� BC. Prove that s ≅ t.

���� ���� ≅ ���� ���� ≅ AO ≅ ���� DO ≅ BO CO because each is a radius and AD ���� BC is given. ∆ADO ≅ ∆BCO according to SSS (recall Volume 1, Chapter 3). ∠AOD ≅ ∠BOC via the CPCTC.

Since arc length equals radius times the central angle (Chapter 3), it follows that s ≅ t (because ∠AOD and

∠BOC are the central angles that subtend arc lengths s

and t). This proves that if two chords are congruent, the corresponding minor arcs are also congruent. 71

Chapter 6 – Chords and Circular Segments

Chapter 6 Problems Note: The diagrams are not drawn to scale.

1. In the diagram below, point O lies at the center of the circle, AO = 12, and ∠AOB = 60°. Determine the length of chord AB and the area of the shaded circular segment.

2. The diagram below shows a quarter circle. Determine the length of chord CD and the area of the shaded circular segment.

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Plane Geometry Practice Workbook with Answers, Volume 2

3. In the diagram below, EF = √3 and the radius of the

circle is 1. Determine the central angle that subtends the minor arc from E to F and the area of the shaded circular segment.

4. In the diagram below, DF is a diameter of the circle. Determine the length of chord DE and the combined area of the two shaded circular segments.

73

Chapter 6 – Chords and Circular Segments

5. In the diagram below, point O lies at the center of the circle and the major arc length from K to L is 10π.

Determine the length of chord KL and the area of the shaded region.

6. In the diagram below, point O lies at the center of the

circle, AO = 2, and ∠AOB = 30°. Determine the length of chord AB and the area of the shaded circular segment.

Tip: Draw a vertical line segment down from A to make

two right triangles. Draw these triangles in a larger diagram off to the side.

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Plane Geometry Practice Workbook with Answers, Volume 2

7. The diagram below shows square EFGH and two ���� as diameters. Find the ���� and GH semicircles that have EF combined area of the shaded regions.

8. The diagram below shows square ABCD and two

quarter circles that have points A and C as centers. Find the area of the shaded region.

9. In the diagram below, ⊙W and ⊙X intersect at points Y and Z, ⊙W and ⊙X have the same radius (which is 6),

and the radius of ⊙W and ⊙X is equal to YZ. Find the area of the shaded region.

75

Chapter 6 – Chords and Circular Segments

10. In the diagram below, ⊙P, ⊙Q, and ⊙R each have a

radius of 3, ⊙P and ⊙Q touch only at point T, ⊙Q and

⊙R touch only at point U, and ⊙P and ⊙R touch only at

point S. Find the area of the shaded region.

11. The diagram below shows ∆ABC and three arcs of

circles that have points D, E, and F as centers. Points D, E, and F are midpoints of their respective sides and AB = AC = BC = 2. Find the area of the shaded region.

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Plane Geometry Practice Workbook with Answers, Volume 2

12. The crescent-shaped region formed between two

circular arcs such as the shaded region below is called a

lune. In the diagram below, LK and MN are the

diameters of the two semicircles. Find the area of the shaded lune.

13. In the diagram below, point O lies at the center of ���� is a diameter, and AB = CD = 7. Determine the circle, AD

α, β, γ, δ, ε, φ, η, and κ.

77

Chapter 6 – Chords and Circular Segments

14. The same circle is shown twice below. On the right,

point O lies at the center of the circle. Determine α, β, γ,

and δ.

15. The same circle is shown twice below. On the right, point O lies at the center of the circle. Determine α.

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Plane Geometry Practice Workbook with Answers, Volume 2

16. In the diagram below, ∠AFB = 74°, ∠COD = 108°, and point O lies at the center of the circle. Determine ∠AOF.

17. In the diagram below, KN = 8, LN = 9, MN = 27, and the diameter of the circle is 36. Find JN and the distance from N to the center of the circle.

79

Chapter 6 – Chords and Circular Segments

18. In the diagram below, VX = 6, OV = 7, VZ = 12, and point O lies at the center of the circle. Find VY, OY, and WY.

19. In the diagram below, point O lies at the center of the circle, AE = 1, DO = √2, and ���� AC ⊥ ���� BD. Find ∠AOE, ∠COD, AO, EO, and BE without using the formulas

below. Use these answers and the information given to verify that (BE)(DE) = (AE)(CE) = R2 − EO2 and that

∠BEA =

∠AOE+∠COD 2

.

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Plane Geometry Practice Workbook with Answers, Volume 2

����� ≅ ���� ����� ∥ XY ����. 20. In the diagram below, VW VZ. Prove that WZ

Note: Point V does NOT lie at the center of the circle and NONE of the chords shown is a diameter.

21. Prove that ∆ABE~∆CDE and ∆ADE~∆BCE in the

diagram below. Notes: Point E may NOT lie at the center

of the circle. Do NOT assume that any of the chords are parallel. Do NOT use the intersecting chords theorem.

Hint: Review Chapter 4.

81

Chapter 6 – Chords and Circular Segments

22. Prove that (AE)(CE) = (BE)(DE) for the diagram

below. Notes: Point E may NOT lie at the center of the circle. Do NOT assume that any of the chords are parallel.

23. In the diagram below, point O lies at the center of the circle. Prove that (AE)(CE) = (BE)(DE) = R2 − EO2 .

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Plane Geometry Practice Workbook with Answers, Volume 2

24. In the diagram below, point O lies at the center of the circle. Prove that ∠AEB = ∠CED =

83

∠AOB+∠COD 2

.

Secants A secant is similar to a chord in that it intersects a circle at two points, but a secant is different from a chord in

that a secant extends outside of the circle. Whereas two chords can only intersect inside of a circle (or on the

circumference of the circle), two secants can intersect outside of the circle (in addition to inside or on the circle).

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Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 7 Concepts A secant is a line (or a line segment) that intersects a circle at two points and which extends outside of the

circle. Any secant is an extension of a chord outside of ����, ���� the circle. For example, AD AC, and ���� BD below are

secants because each line segment intersects the circle at two points (B and C) and extends outside of the ���� also intersects the circle at two circle. Although BC

���� is a chord because ���� points (B and C), BC BC does not ����, ���� ���� extend outside of the circle. The secants AD AC, and BD are extensions of chord ���� BC.

If two secants intersect inside of a circle, the theorems from Chapter 6 for intersecting chords apply. If two secants intersect outside of a circle, the theorems described in the following paragraphs apply. 85

Chapter 7 – Secants

When two secants intersect outside of the circle, the

angle between the secants is one-half of the difference

of the two central angles that subtend the same arcs. ���� and ���� DE intersect at point E in For example, secants CE the diagram below, point O lies at the center of the

circle, and central angles θ and φ subtend the same arcs

(s and t) as α. Angles α, θ, and φ are related by α =

φ−θ 2

.

According to the intersecting secants theorem, the

distances in the diagram above are related by the following formula: (AE)(DE) = (BE)(CE) = EO2 − R2 , where EO is the distance from the point where the

secants intersect (E) to the center of the circle (O) and R

is the radius of the circle. It is important to note that all four line segments in the intersecting secants theorem

involve the same point (E), which is the point where the

secants intersect. It is a common mistake for students to

use AE and AD when they should instead use AE and DE, 86

Plane Geometry Practice Workbook with Answers, Volume 2

and similarly to use BE and BC when they should

instead use BE and CE. Study the diagram and formula closely to see what is wrong with using AE and AD (or

BE and BC). If all four line segments do not contain the same letter, a big mistake is being made.

The intersecting chords theorem and the intersecting

secants theorem are two forms of a theorem called the power of a point. Given a circle and given any point X,

for any secant that passes through point X, the product

of the two distances from X to the two points where the secant intersects the circle equals the same value. For

the case where X lies inside of the circle, the power of a point theorem becomes the intersecting chords

theorem. For the case where X lies outside of the circle, the power of a point theorem becomes the intersecting secants theorem. In each case, the product of the distances from X to the points where the secant intersects the circle equals a fixed value. 87

Chapter 7 – Secants

Chapter 7 Examples Example 1. In the diagram below, point O lies at the

center of the circle. Determine α.

���� are intersecting secants, which means that α ���� DE and CE is equal to the difference of the central angles that

subtend the same arcs as α. ∠COD − ∠AOB 144° − 44° 100° = = = 50° α= 2 2 2

Example 2. In the diagram below, AE = 3, AD = 9, BE = 4, the radius of the circle is 6, and point O lies at the center of the circle. Find BC and EO.

88

Plane Geometry Practice Workbook with Answers, Volume 2

Apply the intersecting secants theorem. Note that DE =

AE + AD = 3 + 9 = 12. (AE)(DE) = (BE)(CE) (3)(12) = 4CE 36 = 9 = CE 4

Note that BE + BC = CE such that BC = CE − BE = 9 − 4 = 5.

(AE)(DE) = EO2 − R2 (3)(12) = EO2 − 62 36 + 36 = EO2 72 = EO2

√72 = �(36)(2) = √36√2 = 6√2 = EO

89

Chapter 7 – Secants

Chapter 7 Problems Note: The diagrams are not drawn to scale.

1. In the diagram below, point O lies at the center of the circle. Determine α.

2. In the diagram below, point O lies at the center of the circle. Determine ∠LOM.

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Plane Geometry Practice Workbook with Answers, Volume 2

3. In the diagram below, point O lies at the center of the ����� is a diameter. Determine ∠XOY. circle and WZ

4. In the diagram below, AE = 10, BE = 9, BC = 15, and the diameter of the circle is 19. Find AD, EO, and EF.

91

Chapter 7 – Secants

5. In the diagram below, TU = 8, PS = 14, RU = 15, and

the radius of the circle is 11. Find QR and SU.

6. In the diagram below, VW = 4, VX = 6, and XY = 3. Find the radius of the circle.

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Plane Geometry Practice Workbook with Answers, Volume 2

7. Prove that (AE)(DE) = (BE)(CE) for the diagram below.

8. In the diagram below, point O lies at the center of the circle. Prove that (AE)(DE) = (BE)(CE) = EO2 − R2 .

93

Chapter 7 – Secants

9. In the diagram below, point O lies at the center of the circle. Prove that α =

φ−θ 2

.

94

Tangents Whereas a secant intersects the circumference of a

circle at two points, a tangent is a line that intersects the circumference of a circle only at a single point. Some important theorems involve the intersection of a

tangent and a chord, the intersection of a tangent and a secant, or the intersection of two tangent lines.

95

Chapter 8 – Tangents

Chapter 8 Concepts A tangent is a line that touches a circle only at a single point. Recall from Volume 1, Chapter 1 that a line is

infinite whereas a line segment is finite. In order to tell whether or not a line segment is tangent to a circle,

imagine extending the line segment in each direction to

make an infinite line; the infinite line will only intersect the circumference of the circle at a single point if the

AC, ���� AB, and line is tangent to the circle. For example, ⃖���⃗ AC, ���� ���� BC are tangent to the circle below because they only ⃖����⃗, �EF ���, etc. are NOT tangent to intersect the circle at B. DG the circle because they intersect the circle at E and F. ��� HI is NOT tangent to the circle; if extended, it would intersect the circle at two points.

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Plane Geometry Practice Workbook with Answers, Volume 2

Any line that is tangent to a circle is perpendicular to the radius that passes through the point where the tangent intersects the circle. For example, in the

diagram below, tangent line ⃖���⃗ AC is perpendicular to radius ���� BO which intersects the tangent at point B.

According to the tangent-chord theorem, if a tangent

and chord intersect, the angle between the tangent and chord is one-half of the central angle formed by the

chord. For example, in the diagram on the next page,

BD at point B, ∠CBD is the tangent ⃖���⃗ AC intersects chord ����

BD, and ∠BOD is angle between tangent ⃖���⃗ AC and chord ����

the central angle formed by the chord. According to the tangent-chord theorem, ∠CBD =

∠BOD 2

. If a tangent and

chord intersect, the point of intersection is an endpoint of the chord. For example, in the diagram on the next page, the point of intersection is B. 97

Chapter 8 – Tangents

If a tangent and secant intersect, the angle between the tangent and secant is one-half of the difference of the

two central angles that subtend the same arcs. For

⃖���⃗ intersect at point D example, tangent ⃖����⃗ BD and secant CD

in the diagram on the following page, point O lies at the center of the circle, and central angles θ and φ subtend the same arcs (s and t) as α. Angles α, θ, and φ are

related by α =

φ−θ 2

. According to the tangent-secant

theorem, the distances are related by the following

formula: (AD)(CD) = BD2 = DO2 − R2 , where DO is the distance from the point where the tangent and secant

intersect (D) to the center of the circle (O) and R is the radius of the circle.

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Plane Geometry Practice Workbook with Answers, Volume 2

If two tangents (of the same circle) intersect, the two

distances from the point where the tangents intersect to the points where the tangents touch the circle are

⃖���⃗ intersect at congruent. For example, tangents ⃖���⃗ AC and BC

point C in the diagram on the following page, and point O lies at the center of the circle. Line segments ���� AC and

���� are congruent: ���� BC AC ≅ ���� BC. The angle between the two intersecting tangents is a supplement to the central

angle that subtends the same (minor) arc. For example,

in the diagram on the following page, α + θ = 180°. The distances are related by the following formula: AC2 =

BC 2 = CO2 − R2 , where CO is the distance from the

point where the tangents intersect (C) to the center of

the circle (O) and R is the radius of the circle. If α is NOT a right angle, then quadrilateral ABCO is a kite. Recall

from Volume 1, Chapter 9 that a kite has two pairs of

congruent edges, but does not have any parallel edges. A 99

Chapter 8 – Tangents

kite is NOT a parallelogram. For the special case where

α is a right angle, in that special case quadrilateral AOBC is a parallelogram (more precisely, in that special case AOBC is a square).

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Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 8 Examples Example 1. In the diagram below, point O lies at the

center of the circle and line ⃖���⃗ AC is tangent to the circle at point B. Determine ∠CBD.

Apply the tangent-chord theorem. ∠BOD 60° ∠CBD = = = 30° 2 2

Example 2. In the diagram below, point O lies at the

⃖����⃗ is tangent to the circle at center of the circle and line BD

point B. Determine α.

101

Chapter 8 – Tangents

Use the formula that relates angles for a tangent that intersects a secant. ∠BOC − ∠AOB 158° − 86° 72° = = = 36° α= 2 2 2

Example 3. In the diagram below, AD = 4, AC = 5, the

radius of the circle is 3, and point O lies at the center of the circle. Find BD and DO.

Apply the tangent-secant theorem. Note that CD =

AD + AC = 4 + 5 = 9. (AD)(CD) = BD2

(4)(9) = 36 = BD2 √36 = 6 = BD

BD2 = DO2 − R2 62 = DO2 − 32 36 = DO2 − 9 45 = DO2 102

Plane Geometry Practice Workbook with Answers, Volume 2

√45 = �(9)(5) = √9√5 = 3√5 = DO

Example 4. In the diagram below, point O lies at the

⃖���⃗ are tangent to center of the circle and lines ⃖���⃗ AC and BC the circle at points A and B. Determine α.

Use the formula that relates angles for intersecting tangents.

α + θ = 180°

α = 180° − θ = 180° − 130° = 50°

103

Chapter 8 – Tangents

Chapter 8 Problems Note: The diagrams are not drawn to scale.

1. In the diagram below, point O lies at the center of the circle and line ⃖���⃗ AC is tangent to the circle at point B. Determine ∠CBD, ∠CBO, ∠BDO, and ∠DBO.

2. In the diagram below, ∠EFH = 82°, point O lies at the

center of the circle, and line ⃖���⃗ EG is tangent to the circle at

point F. Determine ∠HFO, ∠FOH, ∠FHO, and ∠GFO.

104

Plane Geometry Practice Workbook with Answers, Volume 2

3. In the diagram below, point O lies at the center of the ⃖�����⃗ is tangent to the circle at point X. circle and line WY Determine ∠WXZ, ∠OXZ, ∠OZX, ∠OXW, OX, and XZ.

4. In the diagram below, ∠AOB = 43°, ∠BOC = 77°,

point O lies at the center of the circle, and line ⃖����⃗ BD is

tangent to the circle at point B. Determine α.

105

Chapter 8 – Tangents

5. In the diagram below, point O lies at the center of the circle and line ⃖�����⃗ KM is tangent to the circle at point M.

Determine ∠LOM, ∠LON, ∠NLO, and ∠LNO.

6. In the diagram below, point O lies at the center of the circle and line ⃖����⃗ RU is tangent to the circle at point U. Determine ∠SOU, ∠SOT, ∠OSU, ∠OST, ∠RUS, and ∠RSU.

106

Plane Geometry Practice Workbook with Answers, Volume 2

7. In the diagram below, AB = 12, AC = 8, the radius of

the circle is 6, and point O lies at the center of the circle. Find CD and AO.

8. In the diagram below, JK = 1 + √3, LM = √3, the

diameter of the circle is 2, and point O lies at the center of the circle. Find JM, JO, KM, ∠KJM, ∠KOL, ∠LOM, and ∠MLO.

107

Chapter 8 – Tangents

9. In the diagram below, point O lies at the center of the ⃖���⃗ are tangent to the circle at circle and lines ⃖���⃗ AC and BC points A and B. Determine α, ∠CAO, and ∠CBO.

10. In the diagram below, point O lies at the center of ⃖����⃗ and ⃖���⃗ the circle and lines DE EF are tangent to the circle at points D and F. Determine ∠DOF, ∠DFO, ∠FDO, and ∠DFE.

108

Plane Geometry Practice Workbook with Answers, Volume 2

11. In the diagram below, point O lies at the center of

⃖����⃗ and TV ⃖���⃗ are tangent to the circle the circle and lines TU at points U and V. Determine TV, OU, and OT.

12. In the diagram below, ∠QPR = 60°, point O lies at ⃖���⃗ and PR ⃖���⃗ are tangent the center of the circle, and lines PQ to the circle at points Q and R. Determine ∠QOR, ∠QRO, ∠RQO, ∠PQR, ∠PRQ, OP, PQ, PR, QR, minor arc length s, and major arc length t.

109

Chapter 8 – Tangents

⃖����⃗ and BO ⃖����⃗ are 13. In the diagram below, ∠AOB = 60°, AO tangent to ⊙Q at points A and B, ⊙O and ⊙Q touch

only at point C, and the radius of ⊙Q is 1. Find the area of the shaded region.

⃖����⃗ is tangent to ⊙O and ⊙Q 14. In the diagram below, MP at points N and P, ⊙O has a diameter of 6, ⊙Q has a

diameter of 10, ⊙O and ⊙Q touch only at point R, and

points M, O, R, and Q are collinear. Find MN, NP, MP, MO,

OQ, and MQ.

110

Plane Geometry Practice Workbook with Answers, Volume 2

15. In the diagram below, point O lies at the center of the circle and line ⃖���⃗ AC is tangent to the circle at point B. BO. Prove that ⃖���⃗ AC ⊥ ����

16. In the diagram below, point O lies at the center of ⃖���⃗ are tangent to the circle at the circle, lines ⃖���⃗ AC and BC points A and B, and ∠ACB is acute. Show that quadrilateral AOBC is a kite.

111

Chapter 8 – Tangents

17. In the diagram below, point O lies at the center of the circle and line ⃖���⃗ AC is tangent to the circle at point B. Show that ∠CBD =

∠BOD 2

.

18. In the diagram below, point O lies at the center of

⃖����⃗ is tangent to the circle at point B. the circle and line BD Show that α =

φ−θ 2

.

112

Plane Geometry Practice Workbook with Answers, Volume 2

19. In the diagram below, point O lies at the center of

⃖����⃗ is tangent to the circle at point B. the circle and line BD Show that (AD)(CD) = BD2 = DO2 − R2 .

20. In the diagram below, point O lies at the center of the circle and lines ⃖���⃗ AC and ⃖���⃗ BC are tangent to the circle at points A and B. Show that α = α + 180° = φ.

φ−θ 2

, α + θ = 180°, and

21. In the diagram above, point O lies at the center of BC are tangent to the circle at the circle and lines ⃖���⃗ AC and ⃖���⃗ points A and B. Show that AC 2 = BC 2 = CO2 − R2 . 113

Inscribed and Circumscribed Shapes This chapter discusses polygons that are inscribed in

circles or that are circumscribed about circles, including triangles, cyclic quadrilaterals, and regular polygons.

114

Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 9 Concepts A triangle is inscribed in a circle if all three vertices lie on the circumference of the circle. A triangle is

circumscribed about a circle if all three edges are

tangent to the circle. For example, the left diagram

below shows a triangle that is inscribed in a circle and the right diagram below shows a triangle that is circumscribed about a circle.

Stating that a circle is inscribed in a triangle is

equivalent to stating that the triangle is circumscribed about the circle. For example, the right diagram above may either be described as a triangle that is

circumscribed about a circle or as a circle that is inscribed in a triangle. Stating that a circle is

circumscribed about a triangle is equivalent to stating

that the triangle is inscribed in the circle. For example, the left diagram above may either be described as a 115

Chapter 9 – Inscribed and Circumscribed Shapes

triangle that is inscribed in a circle or as a circle that is circumscribed about a triangle. The two perspectives are equivalent.

If a triangle is inscribed in a circle (which is equivalent to a circle that is circumscribed about a triangle), the circumcenter of the triangle lies at the center of the

circle. Recall from Volume 1, Chapter 7 that the three perpendicular bisectors of a triangle intersect at the

circumcenter. A perpendicular bisector is a line segment that is perpendicular to the side of a triangle and which

passes through the midpoint of the side. Also recall that the circumcenter is equidistant from the vertices of the

triangle. The distance from the circumcenter to a vertex

is called the circumradius. In the diagram on the next page, D, E, and F are the midpoints of sides ���� AB, ���� BC, and

���� AC. Perpendicular bisectors ���� DO, ���� EO, and ���� FO intersect at

the circumcenter (O). Since point O is the circumcenter, AO = BO = CO = R is the radius of ⊙O and is also the

circumradius of ∆ABC. Point O lies at the center of the ����, ���� EO, and ���� FO) circle. The perpendicular bisectors (DO ����, ���� and circumradii (AO BO, and ���� CO) divide ∆ABC into

three pairs of congruent triangles. 116

Plane Geometry Practice Workbook with Answers, Volume 2

If a right triangle is inscribed in a circle, the hypotenuse of the triangle is a diameter of the circle (according to Thales’s theorem) and the circumcenter lies on the

midpoint of the hypotenuse (left diagram below). If an acute triangle is inscribed in a circle, the circumcenter

lies inside of the triangle (middle diagram below). If an obtuse triangle is inscribed in a circle, the circumcenter lies outside of the triangle (right diagram below).

If a triangle is circumscribed about a circle (which is

equivalent to a circle that is inscribed in a triangle), the incenter of the triangle lies at the center of the circle. Recall from Volume 1, Chapter 7 that the three angle 117

Chapter 9 – Inscribed and Circumscribed Shapes

bisectors of a triangle intersect at the incenter. An angle bisector is a line segment that bisects an interior angle. Also recall that the incenter is equidistant from the

sides of the triangle. The distance from the incenter to a side is called the inradius. In the diagram on the next page, angle bisectors ��� AI, ���� B I, and �C���I intersect at the

incenter (I). Since point I is the incenter, DI = EI = FI = r is the radius of ⊙I and is also the inradius of ∆ABC.

Point I lies at the center of the circle. The angle bisectors ���, ���� ���, �E���I, and F ���I) divide ∆ABC B I, and �C���I) and inradii (DI (AI

into three pairs of congruent triangles. It should be

obvious that (unlike the circumcenter) the incenter

always lie inside of the triangle, regardless if it is acute, right, or obtuse. Recall from Volume 1, Chapter 7 that

the area of a triangle equals one-half of the perimeter Pr

times the inradius: A = 2 . Volume 2 of this book will use lowercase r for inradius and uppercase R for

circumradius (or the radius of a circle in general), but

note that many books and instructors do NOT follow any standard convention regarding this notation. When

working with a triangle that is circumscribed about a

circle (which is equivalent to a circle that is inscribed in a triangle), it may be helpful to apply the triangle 118

Plane Geometry Practice Workbook with Answers, Volume 2

bisector theorem. Recall from Volume 1, Chapter 7 that an angle bisector divides the opposite side of the

triangle into segments in proportion to the lengths of

the other two sides. For example, in ∆ABC below, since BE

CE

∠BAE ≅ ∠CAE, it follows that AB = AC, which may be BE

AB

equivalently expressed as CE = AC. Similar relations apply to the other angle bisectors of ∆ABC below.

Given any triangle, there exists exactly one circle that

the triangle can be inscribed in such that every vertex of

the triangle lies on the circumference of the circle. The

center of the circle that the triangle is inscribed in lies at the circumcenter of the triangle. There also exists

exactly one circle that the triangle can be circumscribed about such that every side of the triangle is tangent to the circle. The center of the circle that the triangle is

circumscribed about lies at the incenter of the triangle. 119

Chapter 9 – Inscribed and Circumscribed Shapes

An incircle refers to a circle that is inscribed inside of a polygon (which means that the polygon is

circumscribed about the circle), while a circumcircle refers to a circle that can be circumscribed about a

polygon (which means that the polygon is inscribed in

the circle). For example, the smaller circle below (⊙I) is the incircle of ∆ABC (since ⊙I is inscribed in ∆ABC)

and the larger circle (⊙O) is the circumcircle of ∆ABC

(since ⊙O is circumscribed about ∆ABC).

Given a quadrilateral, it is NOT always possible to find a circle that the quadrilateral can be inscribed in with all four vertices lying on the circumference of the circle, and it is NOT always possible to find a circle that the

quadrilateral can be circumscribed about with all four sides tangent to the circle. Quadrilaterals that can be

fully inscribed in a circle or that can be fully circumscribed about a circle are special. 120

Plane Geometry Practice Workbook with Answers, Volume 2

To understand why some quadrilaterals are unable to be

inscribed in a circle, consider the left diagram below.

Points A, B, and C are three vertices of a quadrilateral. Since the three vertices can form ∆ABC, there exists

exactly one circle for which points A, B, and C all lie on

its circumference. In order for the quadrilateral to have all four vertices lying on the same circle, the fourth

vertex must lie on the circumcircle of ∆ABC, like the

middle diagram below. If the fourth vertex does not lie

on the circumcircle of ∆ABC, like the right diagram, then

the quadrilateral can NOT be fully inscribed in a circle. A similar argument applies to circumscribing a quadrilateral about a circle.

If it is possible to draw a circle around a quadrilateral

with all four vertices lying on the circumference of the

circle, the quadrilateral is called a cyclic quadrilateral. A

cyclic quadrilateral can be fully inscribed in a circle. If it

is possible to draw a circle inside of a quadrilateral with all four sides tangent to the circle, the quadrilateral is 121

Chapter 9 – Inscribed and Circumscribed Shapes

called a tangential quadrilateral. If a quadrilateral is both a cyclic quadrilateral and a tangential quadrilateral, it is called a bicentric quadrilateral. For example, in the

diagrams below, ABCD is a cyclic quadrilateral (vertices A, B, C, and D lie on the circumcircle), EFGH is a ����, ���� FG, ���� GH, and ���� EH are tangential quadrilateral (EF

tangents to the incircle), and JKLM is a bicentric quadrilateral (it is both cyclic and tangential).

Following are some properties of cyclic quadrilaterals (which are quadrilaterals that can be inscribed in

circles with all four vertices lying on the circumcircle). • A parallelogram is a cyclic quadrilateral if it has four 90° angles. All rectangles and squares are

cyclic. If a rhombus or any other parallelogram

does not have 90° interior angles, it is NOT cyclic.

• A trapezoid is a cyclic quadrilateral if is isosceles (meaning that the two legs are congruent). If a trapezoid is not isosceles, it is NOT cyclic. 122

Plane Geometry Practice Workbook with Answers, Volume 2

• A kite is cyclic if two opposite interior angles are

both right angles. (Recall from Volume 1, Chapter 9 that a kite has two pairs of congruent edges, but

does not have any parallel edges.) If a kite does not have two right angles, it is NOT cyclic.

• Any pair of opposite interior angles of a convex cyclic quadrilateral add up to 180°. They are

supplements. For example, in the diagram on the

following page, ∠ABC + ∠ADC = 180° and ∠BAD + ∠BCD = 180°.

• The diagonals of a convex cyclic quadrilateral

satisfy the intersecting chords theorem (Chapter

6). For example, in the diagram on the next page,

(AE)(CE) = (BE)(DE) = R2 − EO2 , where point O is the circumcenter.

• According to Ptolemy’s theorem, the product of the

diagonals of a convex cyclic quadrilateral equals the sum of the products of the opposite sides. For

example, (AC)(BD) = (AD)(BC) + (AB)(CD) for the diagram on the next page.

• Inscribed angles that subtend the same arc length are congruent (Chapter 4). For example, in the diagram on the next page, ∠ACD ≅ ∠ABD. 123

Chapter 9 – Inscribed and Circumscribed Shapes

• For a cyclic quadrilateral, all four perpendicular

bisectors are concurrent at the circumcenter. They intersect at a single point, which lies at the center of the circle.

Following are some properties of tangential

quadrilaterals (which are quadrilaterals that can be

circumscribed about circles with all four sides tangent

to the incircle).

• All kites and all squares are tangential

quadrilaterals. (Recall from Volume 1, Chapter 9 that a kite has two pairs of congruent edges, but

does not have any parallel edges.) If a rectangle or other parallelogram is not equilateral, it is NOT a tangential quadrilateral.

• Some trapezoids are tangential quadrilaterals, but others are not. (There is NO correlation between a

trapezoid being tangential and being isosceles. 124

Plane Geometry Practice Workbook with Answers, Volume 2

Some isosceles trapezoids are tangential, but others are not. Some tangential trapezoids are isosceles, but others are not.)

• According to the Pitot theorem, the sum of the lengths of the opposite sides of a tangential

quadrilateral equals one-half of the perimeter. For

example, in the diagram below, P

AB + CD = BC + AD = 2.

• The incircles of the triangles formed by a diagonal of a tangential quadrilateral touch only at a single point. For example, see points E and F below.

• For a tangential quadrilateral, all four angle

bisectors are concurrent at the incenter. They

intersect at a single point, which lies at the center of the circle.

A polygon is a cyclic polygon if it is possible to draw a circle around it with all of its vertices lying on the

circumference of the circle; such a circle is called a 125

Chapter 9 – Inscribed and Circumscribed Shapes

circumcircle. A polygon is a tangential polygon if it is

possible to draw a circle inside of it with all of its sides tangent to the circle; such a circle is called an incircle. For example, the left diagram below shows a cyclic

octagon inscribed in a circumcircle (or equivalently, a

circumcircle that is circumscribed about an octagon), and the right diagram below shows a tangential octagon

circumscribed about an incircle (or equivalently, an incircle that is inscribed in an octagon).

Every regular polygon has both a circumcircle and an incircle. (Recall from Volume 1, Chapter 10 that a

regular polygon is both equilateral and equiangular; all

of its sides are congruent and all of its interior angles are congruent.) Some irregular polygons do not have

circumcircles and some irregular polygons do not have incircles.

126

Plane Geometry Practice Workbook with Answers, Volume 2

It may be helpful to review Chapter 7 (regarding the incenter and circumcenter of a triangle), Chapter 9

(regarding quadrilaterals), and Chapter 10 (regarding polygons with more than four sides) from Volume 1.

Tip: When working with tangents, if the line segment

that connects the center of the circle to the point where the tangent touches the circle is not already drawn, it often helps to draw this line segment.

127

Chapter 9 – Inscribed and Circumscribed Shapes

Chapter 9 Examples Example 1. If ∆ABC below is inscribed in a circle, find the

circumradius and the distance from the circumcenter to the midpoint of each side.

The diagram to the right above shows ∆ABC inscribed in a circle. Point O lies at the center of the circle. Points D AC. Since ∆ABC is a and F are the midpoints of ���� AB and ����

���� is a diameter of the circle and point O right triangle, BC ���� according to Thales’s theorem is a midpoint of BC

(Chapter 5). Since point O lies at the center of the circle, ���� BO ≅ ���� CO (since each is a radius). This agrees with AO ≅ ���� Volume 1, Chapter 7, Problem 27 which states that the circumcenter of a right triangle lies at the midpoint of the hypotenuse. Apply the Pythagorean theorem to determine BC.

AB 2 + AC 2 = BC 2

82 + 62 = 64 + 36 = 100 = BC2 128

Plane Geometry Practice Workbook with Answers, Volume 2

√100 = 10 = BC

The diameter is BC = 10 such that the circumradius is R=

BC 2

=

10 2

= 5.

The three perpendicular bisectors are concurrent at the circumcenter (point O). For a right triangle, the

circumcenter lies at the midpoint of the hypotenuse, so ���� is zero. Since the distance from O to the midpoint of BC

8 point D is the midpoint of ���� AB, BD = AD = 2 = 4. Since 6 point F is the midpoint of ���� AC, AF = CF = 2 = 3. The

distance from D to O is DO = AF = 3 and the distance from F to O is FO = AD = 4.

Example 2. If ∆ABC below is circumscribed about a circle, find the inradius and the distance from the incenter to vertex A. The diagram to the right above shows ∆ABC circumscribed about a circle. Point I lies at the center of BC, and ���� AC are tangent to the circle the circle. Sides ���� AB, ����

at points D, E, and F. Since point I lies at the center of 129

Chapter 9 – Inscribed and Circumscribed Shapes

��� ≅ �E���I ≅ ��� the circle, DI F I (since each is a radius). Note

that BE = CE = √3 such that BC = BE + CE = 2√3. Apply the Pythagorean theorem to find AE. BE 2 + AE 2 = AB 2 2

�√3� + AE 2 = 22 3 + AE 2 = 4

AE 2 = 4 − 3 = 1 AE = √1 = 1

Beware that EI is NOT equal to one-half of AE. Although

����I is an angle bisector, such that ∠ABI = ∠EBI = ∠ ABE, B 2 ���� AE. (It may help to review Volume B I does NOT bisect ����

1, Chapter 7.) It should be clear in the diagram above that EI = DI = FI = r is the radius of the circle, and that AI is larger than the radius. Since AI + EI = AE and since AI > EI, it follows that EI is less than half of AE.

One way to solve for EI is to first find the perimeter: P = 2 + 2 + 2√3 = 4 + 2√3. The base is 2√3 and the height is 1. The area is A =

bh 2

=

�2√3�(1) 2

= √3. The area also

equals one-half of the perimeter times the inradius: A = Pr 2

. Plug P = 4 + 2√3 and A = √3 into this formula: √3 =

�4+2√3�r 2

. Multiply by 2 on both sides: 2√3 = �4 + 2√3�r. 130

Plane Geometry Practice Workbook with Answers, Volume 2

2√3

Divide by 4 + 2√3 on both sides: 4+2

√3

= r. Divide the √3

numerator and denominator each by 2: 2+

√3

= r. It is

customary to rationalize the denominator. To do this, multiply the numerator and denominator each by 2 − √3 (which is the conjugate of 2 + √3).

Note that √3√3 = 3. Recall the “foil” method from

algebra: (a + b)(c − d) = ac − ad + bc − bd. 2 − √3 2√3 − 3 2√3 − 3 √3 r= � �= = 4−3 2 + √3 2 − √3 4 − 2√3 + 2√3 − 3 2√3 − 3 = = 2√3 − 3 1

Since EI + AI = 1 and r = EI, the distance from the

incenter to vertex A can be found by: AI = 1 − EI = 1 − �2√3 − 3� = 1 − 2√3 − (−3) = 1 − 2√3 + 3 = 4 − 2√3

An alternative way to find that r = 2√3 − 3 is to apply

the triangle bisector theorem (Volume 1, Chapter 7) to ∠ABI of ∆ABE:

EI

AI

EI

AI

1−EI

. Cross

= AB. Plug in numbers to get BE

= 2. 3

√

Since EI + AI = AE, it follows that AI = AE − EI =

1 − EI. The previous equation becomes: 131

EI

= 3

√

2

Chapter 9 – Inscribed and Circumscribed Shapes

multiply (or multiply both sides by 2 and by √3): 2EI =

√3(1 − EI). Distribute: 2EI = √3 − EI√3. Add EI√3 to

both sides of the equation: 2EI + EI√3 = √3. Factor:

EI�2 + √3� = √3. Divide both sides of the equation by √3

2 + √3 to get EI = 2+ 3. As shown previously, this is √

equivalent to EI = 2√3 − 3 when the denominator is rationalized.

132

Plane Geometry Practice Workbook with Answers, Volume 2

Example 3. If square ABCD below is inscribed in a circle, find the circumradius.

The left diagram above indicates that AB = BC = CD = AD

= 2 for the square. The diagram to the right above shows

square ABCD inscribed in a circle. Point O lies at the center of the circle. Since ∆ABC is a right triangle with

two congruent sides, ∆ABC is a 45°-45°-90°. Since the sides of a 45°-45°-90° triangle come in the ratio 1:1:√2

(recall Volume 1, Chapter 5) and the legs are AB = BC =

2, the hypotenuse is AC = 2√2. The circumradius is AO = BO = CO = DO = R =

AC 2

=

2√2 2

133

= √2.

Chapter 9 – Inscribed and Circumscribed Shapes

Example 4. The regular hexagon below has edge length 2

and is circumscribed about a circle. Point I lies at the center of the circle. Find the inradius.

For the regular hexagon, ∠AIF =

360° 6

= 60°. Divide ∆AFI

into two 30°-60°-90° triangles. Since the sides of a 30°-

60°-90° triangle come in the ratio 1:√3:2 (recall Volume

1, Chapter 5) with the short side opposite to the 30°

angle, AI = 2 and GI = √3. Since GI is the inradius, r = GI

= √3.

134

Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 9 Problems Note: The diagrams are not drawn to scale.

1. The same right triangle, ∆ABC, is drawn twice below.

In the left diagram, ∆ABC is circumscribed about ⊙I. In the right diagram, ∆ABC is inscribed in ⊙O. Find the

inradius and the circumradius. Note: AB = 24 and BC = 10.

135

Chapter 9 – Inscribed and Circumscribed Shapes

2. The same equilateral triangle, ∆ABC, which has an edge length of 2, is drawn twice below. In the left

diagram, ∆ABC is circumscribed about ⊙I. In the right

diagram, ∆ABC is inscribed in ⊙O. Find the inradius and

the circumradius.

3. The same triangle, ∆ABC, is drawn twice below. In the left diagram, ∆ABC is circumscribed about ⊙I. In the right diagram, ∆ABC is inscribed in ⊙O. Find the

inradius and the circumradius. Note: AB = AC = 5 and BC = 8.

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Plane Geometry Practice Workbook with Answers, Volume 2

4. The same square, ABCD, which has an edge length of

√2, is drawn twice below. In the left diagram, ABCD is circumscribed about ⊙I. In the right diagram, ABCD is inscribed in ⊙O. Find the inradius and the circumradius.

5. The same trapezoid, ABCD, is drawn twice below. In

the left diagram, ABCD is circumscribed about ⊙I. In the right diagram, ABCD is inscribed in ⊙O. Find the inradius and the circumradius.

137

Chapter 9 – Inscribed and Circumscribed Shapes

6. The same kite, ABCD, is drawn twice below. In the left diagram, ABCD is circumscribed about ⊙I. In the right diagram, ABCD is inscribed in ⊙O. Find the inradius

and the circumradius. Notes: ∠BAD = ∠BCD = 90°, but ���� and BC ���� are NOT ∠ADC > 90° and ∠ABC < 90°. Sides AD parallel, and sides ���� AB and ���� CD are NOT parallel.

7. The same regular octagon, which has an edge length

of 2, is drawn twice below. In the left diagram, the

octagon is circumscribed about ⊙I. In the right diagram, the octagon is inscribed in ⊙O. Find the inradius and the circumradius.

138

Plane Geometry Practice Workbook with Answers, Volume 2

8. The same regular hexagon, which has an edge length of √3, is drawn twice below. In the left diagram, the hexagon is circumscribed about ⊙I. In the right

diagram, the hexagon is inscribed in ⊙O. Find the area of the shaded region in each diagram.

139

Chapter 9 – Inscribed and Circumscribed Shapes

9. The same triangle, ∆ABC, is drawn twice below. In the left diagram, the sides of ∆ABC are labeled a, b, and c. In

the right diagram, ∆ABC is circumscribed about ⊙I. The

circle is tangent to a, b, and c at points D, E, and F.

Symbols d, e, and f are defined as d = AD, e = BE, and f = CF.

(A) Show that a = d + e, b = e + f, and c = d + f. P

(B) Show that 2 = d + e + f, where P is the perimeter of

∆ABC.

P

P

P

(C) Show that d = 2 − b, e = 2 − c, and f = 2 − a. (D) Show that A = �

Pdef 2

, where A is the area of ∆ABC

and Pdef means to multiply the four distances (P, d, e,

and f) together.

140

Plane Geometry Practice Workbook with Answers, Volume 2

10. The same right triangle, ∆ABC, is drawn twice

below. In the left diagram, ∆ABC is circumscribed about ⊙I. In the right diagram, ∆ABC is inscribed in ⊙O.

(A) Show that AB + BC − AC equals the “indiameter”

(the diameter of the incircle).

(B) Show that AC equals the “circumdiameter” (the diameter of the circumcircle).

11. Show that ∠ABC + ∠ADC = 180° and

∠BAD + ∠BCD = 180° for the cyclic quadrilateral

below. Note: Do NOT assume that any sides are parallel.

141

Chapter 9 – Inscribed and Circumscribed Shapes

12. In the diagram below, ∠ABD ≅ ∠CBX. Show that (AC)(BD) = (AD)(BC) + (AB)(CD) for cyclic

quadrilateral ABCD. Note: Do NOT assume that any sides are parallel.

13. All four of the perpendicular bisectors of a cyclic

quadrilateral are concurrent at the circumcenter.

Discuss key ideas that would be involved in proving this statement.

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Plane Geometry Practice Workbook with Answers, Volume 2

14. Give the reason for each of the following statements.

(A) A parallelogram is only a cyclic quadrilateral if it is a rectangle or a square, and a rhombus is only a cyclic quadrilateral if it is a square.

(B) A trapezoid is only a cyclic quadrilateral if it is isosceles.

(C) A kite is only a cyclic quadrilateral if one pair of its opposite interior angles are both right angles. P

15. Show that AB + CD = BC + AD = 2 for the tangential quadrilateral below, where P is the perimeter. Note: Do NOT assume that any sides are parallel.

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Chapter 9 – Inscribed and Circumscribed Shapes

16. All four of the angle bisectors of a tangential

quadrilateral are concurrent at the incenter. Discuss key ideas that would be involved in proving this statement.

17. Give the reason for each of the following statements. (A) A rectangle is only a tangential quadrilateral if it is a square.

(B) A parallelogram is only a tangential quadrilateral if it is a rhombus.

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Plane Geometry Practice Workbook with Answers, Volume 2

18. (A) Can a trapezoid with a right angle be a

tangential quadrilateral? Show/explain. If so, is every trapezoid that has a right angle a tangential quadrilateral? Explain.

(B) Can a trapezoid with a right angle be a cyclic

quadrilateral? Show/explain. If so, is every trapezoid that has a right angle a cyclic quadrilateral? Explain. (C) Can an isosceles trapezoid be a tangential

quadrilateral? Show/explain. If so, is every isosceles

trapezoid a tangential quadrilateral?

(D) Can an isosceles trapezoid be a cyclic quadrilateral? Show/explain. If so, is every isosceles trapezoid a cyclic quadrilateral?

19. Prove that every kite is a tangential quadrilateral.

Note: The diagram below shows a general kite that has NO right angles and NO parallel sides.

145

Beyond the Plane This chapter briefly explores a variety of common

three-dimensional solids and their surfaces, including

cubes, prisms, pyramids, spheres, cylinders, and cones.

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Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 10 Concepts A cube is a three-dimensional solid that is bounded by

six square faces that meet at right angles, like the figure below. A cube has 8 vertices, 12 edges, and 6 square

faces.

A rectangular prism (also called cuboid) is a three-

dimensional solid that is bounded by six rectangular

faces that meet at right angles, like the figure below. A rectangular prism has 8 vertices, 12 edges, and 6

rectangular faces. The 12 edges generally come in 3

different lengths (there are 4 edges of each length). The 6 rectangular faces generally come in 3 different sizes

(there are 2 faces of each size).

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Chapter 10 – Beyond the Plane

The volume of a three-dimensional solid provides a

measure of the amount of space contained within the

surface of a closed 3D object. One way to visualize the

volume of a rectangular prism is to divide it up into unit cubes, as shown below.

The volume of a rectangular prism equals the product of its length, width, and height: V = LWH. This can be seen in the diagram above. The rectangular prism above has

a (horizontal) length of L = 6, a width (going back) of W

= 5, and a (vertical) height of H = 3. The volume of the

rectangular prism above is V = LWH = (6)(5)(3) = 90 (in cubic units). Since a cube has square faces, the

volume of a cube is V = L3 , where L3 means to multiply L times L times L.

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Plane Geometry Practice Workbook with Answers, Volume 2

The surface area of a three-dimensional solid provides a measure of the total area of the surface of the solid.

Since the area of a square is L2 and a cube is bounded by 6 square faces, the surface area of a cube is S = 6L2 . A

rectangular prism has 2 faces with area of LW (top and

bottom), 2 faces with area of WH (right and left), and 2

faces with area of LH (front and back). The surface area of a rectangular prism is:

S = 2LW + 2WH + 2LH

A rectangular prism has two different kinds of

diagonals. A body diagonal joins two opposite corners, whereas a face diagonal lies within one face, as shown

on the next page. The length of a diagonal can be found

by applying the Pythagorean theorem. For example, the face diagonal shown on the next page has length

√L2 + W 2 , while the body diagonal is

d = √L2 + W 2 + H 2 (it makes a right triangle with the face diagonal).

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Chapter 10 – Beyond the Plane

The right figure below combines the face diagonal from the left figure with the body diagonal from the middle figure. The face diagonal has a length of √L2 + W 2

according to the Pythagorean theorem. Since the height

H is perpendicular to the bottom face and the face

diagonal in the left figure lies on the bottom face, the

Pythagorean theorem can be applied to find the length of the body diagonal.

2

d = ���L2 + W 2 � + H 2 = �L2 + W 2 + H 2

A prism is a three-dimensional solid bounded by two

congruent parallel polygons on its two ends and which has parallelograms along its body, like the examples

below. For a right prism, the parallelograms along the body are rectangles. An oblique prism is slanted. The

diagram on the following page shows a triangular prism (left figure), a hexagonal prism (middle figure), and an

oblique prism (right figure).

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The volume of a prism equals the area of the base

(either of the two congruent parallel polygons on its

two ends) times the height (the perpendicular distance

between the two bases): V = Bh. For an oblique prism,

the height is NOT equal to the length of the prism, but is instead measured perpendicularly to the bases, as

shown in the right figure below. If a prism is oblique, remember that h is perpendicular to the bases.

The diagram on the next page illustrates why h is

measured perpendicular to the base for an oblique prism. Imagine slicing off one end as shown and

reattaching it onto the other side of the prism. This

transforms the oblique prism into a rectangular prism

of equal volume that is not slanted. For the equivalent

rectangular prism that is not slanted, it should be clear

why h is measured perpendicular to the top and bottom faces. Even for the oblique prism shown at the left, h is 151

Chapter 10 – Beyond the Plane

measured perpendicular to the top and bottom bases because the volumes of the two prisms below are the same (the only difference is that a wedge was moved from one side to the other). In this book, the term

rectangular prism will mean a prism that is not slanted

unless stated otherwise. The word “oblique” will be used to indicate that a prism is slanted. The left prism below

would be called an oblique (or slanted) prism, whereas

the reformed prism on the right below would be called a rectangular prism. (A rectangular prism that is not

slanted could be called a right rectangular prism, but the convention is for the term rectangular prism to mean a right rectangular prism by default.)

A special property of a rectangular prism (regardless of whether it is right or oblique) is that any pair of

opposite faces may serve as the two bases because in

each case the opposite faces are congruent and the faces along the body are either rectangles or parallelograms. For example, the same oblique rectangular prism is 152

Plane Geometry Practice Workbook with Answers, Volume 2

shown twice below. The volume can be found by multiplying the area of the top face and the

perpendicular distance h1 (left figure) or by multiplying the area of the left face and the perpendicular distance h2 (right figure). Yet another option is to multiply the area of the front face and the depth (measured

perpendicular to the front and back faces). Either method gives the same volume.

A pyramid is a three-dimensional solid that has a base in the shape of a polygon and a point called the apex that does not lie in the same plane as the polygon,

where edges connect the apex to every vertex of the

polygon, like the figure on the next page. The faces of a pyramid are triangles. A regular pyramid is a pyramid

where the base is a regular polygon, which means that the polygon at the base is both equilateral and

equiangular. A right pyramid means that the apex is

directly above the centroid of the polygon. For a right regular pyramid, the lateral edges (the edges that are 153

Chapter 10 – Beyond the Plane

not part of the base) are congruent. Conventionally, a

pyramid is assumed to be a right pyramid unless stated otherwise. The diagram below shows a pyramid with a square base (left figure) and a pyramid with a

hexagonal base (right figure). An oblique pyramid is slanted.

The volume of a pyramid equals one-third of the area of the base times the height: V =

Bh 3

, where B is the base

area. This formula applies even to oblique or irregular pyramids, provided that the height is measured

perpendicular to the base, as shown above on the left

(where h is the perpendicular distance from the apex to the base).

A polyhedron is a three-dimensional solid that is

bounded by faces that are polygons. Polyhedra include the cube, prism, pyramid, and many other threedimensional shapes, such as the tetrahedron,

octahedron, dodecahedron, and icosahedron. The nature of a polyhedron depends on which type of 154

Plane Geometry Practice Workbook with Answers, Volume 2

polygon is used as the face in addition to the number of

faces. For example, both polyhedra shown below have 6

sides, but differ in the polygons used for the faces. The

left figure is a cube, which has 6 square faces. The right figure is a triangular bipyramid, which has 6 triangular faces (it can be formed by attaching two triangular

pyramids, called tetrahedra, together at one face; since these are triangular pyramids, NOT square pyramids, there are 6 outside faces).

A regular polyhedron is a highly symmetric polyhedron with congruent faces arranged in a similar manner at

each vertex. The Schläfli symbol is an ordered pair in

the form {n, m} that helps to mathematically label the

different kinds of regular polyhedra. The first number

(n) represents the number of edges of the polygons that are used as the faces, while the second number (m)

represents the number of faces that meet at each vertex.

For example, the Schläfli symbol for the cube is {4, 3}

because a cube has square faces (and a square has n = 4

edges) and m = 3 faces meet at every vertex (inspect the 155

Chapter 10 – Beyond the Plane

diagram on the previous page). There are five regular

convex polyhedra, which are referred to as the Platonic solids: the tetrahedron {3, 3}, cube {4, 3}, octahedron

{3, 4}, dodecahedron {5, 3}, and icosahedron {3, 5}. The

cube and octahedron are a pair of dual polyhedra in the sense that their Schläfli numbers are swapped. The

dodecahedron and icosahedron are another pair of dual polyhedra. When two polyhedra are dual to one

another, the roles of the vertices and faces are swapped between the polyhedra. For example, a cube has square faces (with 4 edges) and 3 faces meet at each vertex,

whereas its dual (the octahedron) has triangular faces (with 3 edges) and 4 faces meet at each vertex.

• a tetrahedron {3, 3} has 4 vertices, 6 edges, and 4

triangular faces. In the left figure below, 2 faces are in front, 1 face is in back, and 1 face is at the bottom.

• a cube {4, 3} has 8 vertices, 12 edges, and 6 square faces.

• an octahedron {3, 4} has 6 vertices, 12 edges, and 8 triangular faces. This can be formed by attaching two square pyramids at their square bases. (The 156

Plane Geometry Practice Workbook with Answers, Volume 2

square base does NOT count as one of the faces; only its edges touch the outside.)

• a dodecahedron {5, 3} has 20 vertices, 30 edges, and 12 pentagonal faces. In the fourth figure below, 6 faces are in front and 6 faces are in back.

• an icosahedron {3, 5} has 12 vertices, 30 edges, and 20 triangular faces. In the right figure below, one vertex is in the front center, one vertex is in the

back center, 10 faces are in front, and 10 faces are in back.

For a convex polyhedron, the number of vertices (V),

the number of edges (E), and the number of faces (F) are related by Euler’s formula, which states that the

number of vertices plus the number of faces is two more than the number of edges.

V+F=E+2

For example, a cube has V = 8 vertices, E = 12 edges,

and F = 6 square faces. For a cube, Euler’s number gives 8 + 6 = 12 + 2 = 14.

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Chapter 10 – Beyond the Plane

Not all three-dimensional objects have flat sides. Some,

like spheres, are curved.

A sphere is a three-dimensional curved surface where every point on the surface is equidistant from the

center. A sphere with radius R has an inside volume of

V=

4πR3 3

and a surface area of S = 4πR2 . Since volume is

measured in cubic units whereas area is measured in

square units, it is easy to remember that volume has R3

whereas surface area has R2 . (In mathematics, the terms

sphere and ball have different meanings. A sphere refers to the points on the surface, whereas ball is a solid that

includes a sphere and all of the points inside of it. The

volume inside of a sphere equals the volume of a ball of the same radius.)

A cylinder is similar to a prism, except that each end of a cylinder is a curve (like a circle or ellipse) instead of a polygon. A circular cylinder is a cylinder with two

circles on its ends. For a right cylinder, the axis of the cylinder is perpendicular to each base, like the left

figure below. An oblique cylinder is slanted, like the 158

Plane Geometry Practice Workbook with Answers, Volume 2

right figure below. A right circular cylinder has circles on its ends and an axis that is perpendicular to the circles, like the left figure below. Conventionally, a

cylinder is assumed to be right (not oblique) unless

stated otherwise. For example, the left figure below is called a circular cylinder, whereas the right figure is called an oblique cylinder.

Since a cylinder is similar to a prism (except that the

ends of a cylinder are curves instead of polygons), the

volume of a cylinder can be found using the formula for

the volume of a prism: V = Bh, where B is the area of the base and h is measured the same way as it is for a prism

(perpendicular to the bases). For a circular cylinder, the base area is the area of a circle. The volume of a circular

cylinder is V = πR2 h (regardless of whether the cylinder is right or oblique, provided that h is measured

perpendicularly to the bases). The surface area of a

cylinder can be found by unrolling the surface of its body, as shown on the following page. When a right

circular cylinder is unrolled, its body is transformed into a rectangle with a length equal to the circumference of 159

Chapter 10 – Beyond the Plane

the circle and the same height as the cylinder. The total surface area of a right circular cylinder is S = 2πRh + 2πR2 = 2πR(h + R). The first term, 2πRh,

represents the surface area of the body (called the

lateral surface area); it equals the area of the rectangle formed by unrolling the cylinder (since the

circumference of a circle is 2πR). The second term,

2πR2 , represents the area of the two circular bases. The

total surface area of a cylinder includes the surface area

of the body plus the area of the two bases.

A cone is similar to a pyramid, except that the base of a

cone is a closed curve (like a circle or ellipse) instead of a polygon. A circular cone is a cone with a circle as its base. For a right cone, the axis of the cone is

perpendicular to the base, like the left figure on the next page. An oblique cone is slanted, like the right figure on the next page. A right circular cone has a circle as its

base and an axis that is perpendicular to the circular 160

Plane Geometry Practice Workbook with Answers, Volume 2

base, like the left figure below. Conventionally, the term cone implies a right cone unless stated otherwise.

Since a cone is similar to a pyramid (except that the

base of a cone is a closed curve instead of a polygon),

the volume of a cone can be found using the formula for the volume of a pyramid: V =

Bh 3

, where B is the area of

the base and h is measured the same way as it is for a pyramid (perpendicular to the base). For a circular

cone, the base area is the area of a circle. The volume of a circular cone is V = right circular cone is

πR2 h 3

. The total surface area of a

S = πR2 + πR√R2 + h2 = πR�R + √R2 + h2 �

The first term, πR2 , represents the area of the base. The

second term, πR√R2 + h2 , represents the surface area of the body (called the lateral surface area). The total

surface area of a cone includes the surface area of the body plus the area of the base.

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Chapter 10 – Beyond the Plane

A cross section is the region of intersection of a plane

and a three-dimensional object. For example, in the left figure below, a horizontal plane intersects a sphere to make a circular cross section. The middle and right

figures below show how the cross section of a pyramid can be a square or a triangle (though these are not the

only possibilities), depending on the orientation of the plane.

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Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 10 Examples Example 1. Find the volume and surface area of the

rectangular prism below.

Recall that it will be implied that a rectangular prism is a right rectangular prism unless stated otherwise. Find the appropriate formulas for the volume and surface

area of a (right) rectangular prism. For this problem, it does not matter which sides are called L, W, and H, as long as one is horizontal, one is vertical, and one is depth. The solution below uses L = 10, W = 4, and H = 5.

V = LWH = (10)(4)(5) = 200 S = 2LW + 2WH + 2LH

S = 2(10)(4) + 2(4)(5) + 2(10)(5) S = 80 + 40 + 100 = 220 163

Chapter 10 – Beyond the Plane

Example 2. Find the lengths of all of the face diagonals

for the rectangular prism above. Also find the length of the body diagonal.

In the diagram above, BC = FG = AD = EH = 10, CG = BF = DH = AE = 4, and GH = CD = EF = AB = 5. There are

three different types of face diagonals: those like AC,

those like CF, and those like CH. Use the Pythagorean theorem to find the face diagonals.

AC = √AB 2 + BC 2 = √52 + 102 = √25 + 100 = √125 = �(25)(5) = √25√5 = 5√5

CF = √BC 2 + BF 2 = √102 + 42 = √100 + 16 = √116 = �(4)(29) = √4√29 = 2√29

CH = √CG 2 + GH 2 = √42 + 52 = √16 + 25 = √41 One way to find the body diagonal is to use the formula below. (An alternative way is to use the Pythagorean

theorem with CD and CF. It will give the same answer.) DF = √BF 2 + BC 2 + CD2 = √42 + 102 + 52 DF = √16 + 100 + 25 = √141 164

Plane Geometry Practice Workbook with Answers, Volume 2

Example 3. Find the volume and surface area of the

(right) square pyramid below.

First draw and label relevant shapes. Given the figure at the right, square BCDE, ∆ACG, and ∆ACD are redrawn

and labeled off to the side. These shapes will be helpful for solving this problem. Note that AB = AC = AD = AE

= 10, point G lies at the center of the square, and lateral faces ∆ABC, ∆ACD, ∆ADE, and ∆ABE are congruent

because the pyramid is right and the base polygon is

regular . Point A is the apex, square BCDE is the base,

and h = AG = 8 is the height. Since AG, CG, and AC form a right triangle, the Pythagorean theorem can be applied to find CG.

CG = �AC2 − AG 2 = �102 − 82 = √100 − 64 = √36 = 6

Since CG is one-half of the diagonal of square BCDE, CE = BD = 2CG = 2(6) = 12. Since the diagonals of a

square are angle bisectors and are also perpendicular to each other (Volume 1, Chapter 9), ∆CDG is a 45°-45°-90°

triangle such that CD = CG√2 = 6√2. The area of the 165

Chapter 10 – Beyond the Plane

2

2

square base is B = CD2 = �6√2� = 62 �√2� = 36(2) = 72. The volume of the pyramid is:

V=

Bh 3

=

(72)(8) 3

= 24(8) = 192

To find the surface area, first find the area of ∆ACD.

Since ∆ACD is isosceles, it may be divided into two

congruent right triangles (right figure above). Note that CF =

CD 2

find AF.

=

6√2 2

= 3√2. Use the Pythagorean theorem to 2

AF = �AC 2 − CF 2 = �102 − �3√2� 2

= �100 − 32 �√2�

AF = �100 − (9)(2) = √100 − 18 = √82

The area of ∆ACD is 6

(CD)(AF)

1

= 2 �6√2��√82� =

2

�2� �2(82) = 3√164 = 3�(4)(41) = 3√4√41 =

3(2)√41 = 6√41. The total surface area equals the area

of square BCDE plus 4 times the area of ∆ACD. S = 72 + 4�6√41� = 72 + 24√41 = 24�3 + √41�

For an alternative formula involving “slant height” (ℓ =

AF = √82), see Problem 31.

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Plane Geometry Practice Workbook with Answers, Volume 2

Example 4. Find the volume and surface area of the

regular tetrahedron below.

First draw and label relevant shapes. Given the figure at the left, ∆ACD, ∆BCD, and ∆ADG are redrawn and

labeled off to the side. These shapes will be helpful for solving this problem. A tetrahedron has 4 equilateral

triangles as its faces. Therefore, AB = AC = AD = BC =

BD = CD = 12. Point A is the apex, ∆BCD is the base, h = AG is the height, and point G is the centroid of ∆BCD

since the apex lies directly above the centroid of the base of a (right) regular pyramid (meaning that the apex and

centroid both lie on the altitude). The second figure

above divides ∆ACD into two right triangles. Line segment ���� AE is a median (Volume 1, Chapter 7) since it bisects ���� CD. Note that CE = DE =

CD 2

=

12 2

= 6. Use the

Pythagorean theorem to find the length of the median.

AE = �AC 2 − CE 2 = �122 − 62 = √144 − 36 = √108 = �(36)(3) = √36√3 = 6√3 167

Chapter 10 – Beyond the Plane

1

The area of ∆BCD (or any other face) is 2 (12)�6√3� =

36√3. The total surface area of the tetrahedron is 4 times this (since it is bounded by 4 congruent faces). S = 4�36√3� = 144√3

Recall from Volume 1, Chapter 7 that the centroid of a

triangle lies two-thirds of the length of a median from a 2

2

vertex: BG = 3 (AE) = 3 �6√3� = 4√3. (Note that twothirds of the length of a median from a vertex is

equivalent to one-third of the length of a median from a midpoint: 2/3 + 1/3 = 1.) Note that BG = CG = DG =

4√3. To find the height of the tetrahedron (AG), apply the Pythagorean theorem to ∆ADG. 2

AG = �AD2 − DG 2 = �122 − �4√3� 2

= �144 − 42 �√3� = �144 − (16)(3)

AG = √144 − 48 = √96 = √16√6 = 4√6

Use the base area (36√3) and h = AG to find the volume of the tetrahedron: V=

Bh 3

=

�36√3��4√6� 3

=

144√18 3

= 48�(9)(2) = 48√9√2 =

48(3)√2 = 144√2 168

Plane Geometry Practice Workbook with Answers, Volume 2

Note that the height of the tetrahedron (AG) is different from the height used to find the base area (since the

tetrahedron’s height is not the same as the height of one of its faces).

Example 5. In the diagram below, point O lies at the

center of the sphere and point P lies on the surface. Find the volume inside and the surface area of the sphere.

The radius of the sphere is R = OP = 2. Find the formulas for the volume and surface area of a sphere. 4πR3 4π(2)3 4π8 32π = = = V= 3 3 3 3 S = 4πR2 = 4π(2)2 = 4π4 = 16π

Example 6. In the diagram below, the dotted line

represents the axis of the (right) circular cylinder. Find

the volume inside and total surface area of the cylinder.

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Chapter 10 – Beyond the Plane

The radius of the cylinder is R = OP = 3 and the height

is h = 10. Find the formulas for the volume and surface area of a right circular cylinder. V = πR2 h= π(3)2 (10) = π9(10) = 90π

S = 2πR(h + R) = 2π3(10 + 3) = 6π(13) = 78π

Example 7. In the diagram below, the dotted line

represents the axis of the (right) circular cone. Find the volume inside and total surface area of the cone.

The radius of the cone is R = OP = 5 and the height is h

= 12. Find the formulas for the volume and surface area of a right circular cone. V=

πR2 h 3

=

π(5)2 12 3

= π(25)(4) = 100π

S = πR�R + √R2 + h2 � = π5�5 + √52 + 122 � =

5π�5 + √25 + 144� S = 5π�5 + √169� = 5π(5 + 13) = 5π(18) = 90π

For an alternative formula involving “slant height,” see Problem 32.

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Plane Geometry Practice Workbook with Answers, Volume 2

Example 8. In the left figure below, a vertical plane that

is parallel to the axis of a prism intersects the prism. In the right figure below, a vertical plane that is

perpendicular to the axis of the prism intersects the

prism. Draw the cross section of the prism for each case.

Note that the gray rectangles that represent the planes

are longer, wider, and taller than the prism. For the left figure, the cross section is a rectangle. For the right

figure, the cross section is a triangle. Imagine that the prism is made of cheese and that the cheese is being

sliced according to the orientation of each plane. When

the cheese is sliced into two pieces, the cross section will be the matching face of each piece of cheese.

171

Chapter 10 – Beyond the Plane

Chapter 10 Problems Note: The diagrams are not drawn to scale.

1. Find the volume, surface area, face diagonal, and body diagonal of the cube below.

2. The diameter of the sphere below is 12. Find the

volume inside and the surface area of the sphere.

3. Find the volume, surface area, face diagonals, and body diagonal of the rectangular prism below.

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Plane Geometry Practice Workbook with Answers, Volume 2

4. The (right) triangular prism below has front and rear bases that are right triangles with a 30° interior angle. Find the volume and surface area of the prism.

5. The diagram below shows a hemisphere (which is

one-half of a sphere). Point O lies at the center of the flat

circular base and point P lies on the curved surface. Find the volume inside and total surface area of the hemisphere below.

6. A submarine travels horizontally 16 miles to the east, then travels horizontally 15 miles to the north, and then descends 12 miles straight down. How far is the submarine from its starting position? 173

Chapter 10 – Beyond the Plane

7. Find the volume and surface area of the (right) square pyramid below.

8. The rectangular prism below was made by stacking cubes. The edge length of each cube is 0.5. Find the

volume and surface area of the rectangular prism.

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Plane Geometry Practice Workbook with Answers, Volume 2

9. The pyramid below was made by stacking cubes. The edge length of each cube is 2. Find the volume of the pyramid.

10. In the diagram below, the dotted line represents the axis of the (right) circular cylinder. Find the volume inside and total surface area of the cylinder.

11. In the diagram below, ∠ABC = 60° and the dotted

line represents the axis of the (right) circular cone. Find the volume inside and total surface area of the cone.

175

Chapter 10 – Beyond the Plane

12. The (right) triangular prism below has front and rear bases that are equilateral triangles. Find the volume and surface area of the prism.

13. In the regular tetrahedron below, point G is the

centroid of the base and DG = 6. Find the volume and surface area of the tetrahedron.

14. Find the volume and surface area of the regular

octahedron below.

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Plane Geometry Practice Workbook with Answers, Volume 2

15. Each face of the regular dodecahedron below has an area of 5. Find the surface area of the dodecahedron.

16. Find the surface area of the icosahedron below.

17. The metal pipe below is hollow at both ends and

throughout its interior. The inside surface of the pipe and the outside surface of the pipe are coaxial right circular

cylinders. The inner diameter is 6, the outer diameter is 8, and the length of the pipe is 15. Find the volume of the metal. (The question is looking for the amount of metal in the pipe itself, NOT the amount of air in the

hollow region.) Also find the total surface area of the

pipe (including the inside surface, outside surface, and the ends).

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Chapter 10 – Beyond the Plane

18. The sphere on the left below has twice the inside

volume as the sphere on the right. Find the ratio of their radii. Also find the ratio of their surface areas.

19. The prism below has parallelograms on the front

and back, rectangles on the left and right, and rectangles on the top and bottom. The 60° angle labeled is one of the interior angles of the parallelograms. Find the

volume and surface area of the prism.

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Plane Geometry Practice Workbook with Answers, Volume 2

20. In the diagram below, a horizontal thin black ring

was placed on top of a sphere. Points A and B are opposite points on the ring, such that ���� AB is a diameter of the ring. Point O lies at the center of the sphere. The

radius of the sphere is 24. What is the radius of the ring? Note: Only the front half of the ring is visible in the diagram.

21. The axis of the oblique cylinder below passes

through point O, the ends are circles, AO = 8, AB = 20,

and ∠BAO = 60°. Find the volume inside of the cylinder.

179

Chapter 10 – Beyond the Plane

22. The (right) hexagonal prism below has front and

rear bases that are regular hexagons. Find the volume and surface area of the prism.

23. The (right) hexagonal pyramid below has a regular

hexagon for its base. The apex is 26 units from each of

the other vertices. Find the volume and surface area of the pyramid.

24. A sphere has a volume of 288π cubic units. What is

the diameter of the sphere? What is the surface area of the sphere?

25. A cube has a surface area of 73.5 square units. What

is the volume of the cube?

180

Plane Geometry Practice Workbook with Answers, Volume 2

26. The diagram below shows a rectangular prism that

has been unfolded. The dotted lines indicate the fold

marks. Determine the dimensions, volume, and surface area of the rectangular prism (before it was unfolded). Note: 14 is the total unfolded height.

27. The rectangle below can be rolled into the shape of a right circular cylinder so that its length will be 6 or so that its length will be 15 (depending on whether it is rolled right or down). For each case, determine the

radius, volume, and surface area of the resulting right circular cylinder (if there is NO overlap and circular ends are added).

181

Chapter 10 – Beyond the Plane

28. The diagram below shows a truncated cone

(meaning that the top of the cone was cut off and

removed). The top and bottom bases are parallel circles, which are 4 units apart vertically. The top circle has a diameter of 8, and the bottom circle has a diameter of

12. Determine the volume inside of the truncated cone.

29. A sphere, right circular cylinder, and right circular

cone each have the same radius. The cylinder and cone each have a height equal to their diameter. Show that the

ratio of their volumes in the order cone:sphere:cylinder equals 1:2:3 and that the sum of the volumes of the cone

and sphere equals the volume of the cylinder. Note: This idea dates back to Archimedes (circa 225 BCE).

182

Plane Geometry Practice Workbook with Answers, Volume 2

30. How many vertices, edges, and faces does a

triangular prism have? Verify that the answers satisfy Euler’s formula. For the polyhedron that is dual to the

triangular prism, how many vertices, edges, and faces does it have? Draw the dual polyhedron.

31. The slant height of a pyramid with a base that is a

regular polygon extends from the apex to the midpoint

of an edge of the base. For example, in the diagram below, point F lies at the midpoint of ���� CD such that AF is the slant height. Show that the lateral surface area

(which is the area of the triangular, non-base faces only) of a right pyramid with a base that is a regular polygon (not necessarily a square like the pyramid below) equals

Pℓ 2

, where ℓ is the slant height and P is the

perimeter of the base. (It is NOT an oblique pyramid.

The slant height refers to the slant of AF, not the axis of

the pyramid.)

183

Chapter 10 – Beyond the Plane

32. The slant height of a right circular cone extends

from the apex to a point on the circumference of the

base. For example, in the diagram below, ℓ is the slant height. Show that the lateral surface area (which

excludes the area of the base) of a right circular cone equals πRℓ, where ℓ is the slant height and R is the

radius. Note: The cone is NOT oblique. The slant height

refers to the slant of the distance, not the axis of the cone.

33. The diagram below illustrates a right circular cone

that has had its body unrolled into the shape of a sector (Chapter 3). The circular base is not shown in the

diagram below. The radius of the cone (before it was

unrolled) was 4. Find θ in degrees. Also find the volume and total surface area of the cone (before unrolling). Include the base.

184

Plane Geometry Practice Workbook with Answers, Volume 2

34. The cube below on the left was cut along the plane

that contains ∆CFH to create the (irregular) tetrahedron below on the right. Show that the volume of the

tetrahedron is one-sixth of the volume of the cube.

35. Prove that ∠EBG = 60° in the diagram below.

185

Chapter 10 – Beyond the Plane

36. The right circular cone below on the left was cut

along a horizontal plane to make the truncated cone

below on the right. The two bases of the truncated cone are parallel. The truncated cone has one-half as much

volume as the cone on the left. Find the ratio of radii of the bottom and top bases of the truncated cone.

37. The diagram below shows one octant (or one eighth) ����, ���� of a sphere: AO = BO = CO = R, ���� AO ⊥ ���� BO, ���� AO ⊥ CO BO ⊥

����, and the three circular arc lengths are congruent. CO Find the volume inside and the total surface area in

terms of R and the constant π.

186

Plane Geometry Practice Workbook with Answers, Volume 2

38. Draw diagrams to show how the cross section of a

cube could be a triangle, quadrilateral, pentagon,

hexagon, line segment, or a point.

39. Draw diagrams to show how the cross section of a

right circular cylinder could be a circle, ellipse, or a rectangle.

40. Draw diagrams to show how the cross section of the

(surface of the) right circular double cone below could

be a circle, ellipse, parabola, hyperbola, a single straight line, a pair of straight lines, or a point.

187

Answer Key Chapter 1 Answers 1. (A) D = 40. Check: D = 2R = 2(20) = 40.

(B) C = 40π. Check: C = πD = π(40) = 40π. (C) C ≈ 126

to three significant figures.

2. (A) R = 4. Check: D = 2R = 2(4) = 8.

(B) C = 8π. Check: C = πD = π(8) = 8π. (C) C ≈ 25.1 to three significant figures. 1

1

3. (A) D = π. Check: C = πD = π �π� = 1. (B) D ≈ 0.318 to three significant figures. 1

1

(C) R = 2π. Check: C = 2πR = 2π �2π� = 1. (D) R ≈ 0.159 to three significant figures.

Note: Be careful to enter 1÷2÷π or to enter 1÷(2π) on the calculator. (Many students make the mistake of

entering 1÷2π, which is incorrect. This gives an answer

greater than 1. Why? Most calculators interpret 1÷2π to

mean one-half times pi rather than one divided by two pi. That is, most calculators interpret 1÷2π to mean

1÷2×π.)

188

Plane Geometry Practice Workbook with Answers, Volume 2

4. (A) 2π. Note: R A = CB = 2πR B , such that C

(B) 4π2 . Note: RA = B

2πR A RB

=

2π(2πR B ) RB

4π2 because R A = CB = 2πR B .

RA RB

= 2π.

= (2π)2 = 22 π2 =

5. 18(π − 2) yards is the exact value and 20.5 yards is the approximate value rounded to three significant

figures. Notes: The semicircular arc has a distance equal to one-half of the circumference:

πD 2

=

π(36) 2

= 18π ≈

18(3.14) ≈ 56.5. Subtract the diameter to find the

difference between the two distances: 18π − 36 = 18(π − 2) ≈ 56.5 − 36 = 20.5. The form 18(π − 2) was obtained by factoring out the 18.

189

Answer Key

6. D =

1440 π

≈ 458 meters. Notes: Every second, Marty

and Nancy travel a combined 2.4 + 3.6 = 6 meters.

Convert 4 minutes to 4 × 60 = 240 seconds so that the units of time are consistent. Since Marty and Nancy

travel a combined distance of 6 meters each second, in 240 seconds they travel a total of 6 × 240 = 1440

meters, which equals the circumference of the track. 𝐶𝐶

D=π =

1440 π

. Side note: The approximate answer is 458

if you use several digits for π and then round after using

the calculator, but 459 if you enter π as 3.14 on the

calculator. It is considered better practice to use extra digits throughout the calculation and only round the final answer.

Notes: In the diagram above, Marty and Nancy each begin at “start” and meet up at “finish.” Since they each travel 190

Plane Geometry Practice Workbook with Answers, Volume 2

for a total time of 240 seconds, during this time Marty travels 2.4 × 240 = 576 meters and Nancy travels

3.6 × 240 = 864 meters. As a check, 576 + 864 = 1440

meters equals the circumference of the circle.

7. The slope is expected to equal π, which is

approximately equal to 3.14. The actual slope may be slightly different due to limited precision of the

measurements. Notes: The equation for a straight line is

𝑦𝑦 = m𝑥𝑥 + b. Since circumference is plotted on the 𝑦𝑦-axis

and diameter is plotted on the 𝑥𝑥-axis, for this plot the

equation becomes C = mD + b. Compare this to the

equation C = πD to see that m = π and b = 0. That is, the

rise over the run should equal the circumference divided by the diameter, which is π.

191

Answer Key

Chapter 2 Answers 1. A = 36π ≈ 113. Notes: R =

π(6)2 = 36π.

D 2

2. C = 8√π ≈ 14.2. Check: R = because 4

2

π

= 4

√

12 2

= 6 and A = πR2 =

, C = 2πR = π

8π

√π

= 8√π

√

2 = π (because π = π), and A = πR = π √ √ √ π 16

π � π� = π � π � = 16. Note: Use a calculator to find the √

approximate answer to three significant figures. A

3. AA = 4. Notes: Plug DA = 2R A and DB = 2R B into DA = B

2DB to get 2R A = 2(2R B ). This simplifies to 2R A = 4R B and further to R A = 2R B . (This should make sense: If

⊙A has twice the diameter as ⊙B, it follows that ⊙A also has twice the radius as ⊙B.) Plug R A = 2R B into

AA = πR2A to get AA = π(2R B )2 = 4πR2B , which equals 4AB since AB = πR2B . It is possible to reason out the

answer with very little work. The main idea is that since the radius is squared, the 2 from R A = 2R B gets

squared: 22 = 4.

192

Plane Geometry Practice Workbook with Answers, Volume 2

4. A = 225π ≈ 707. Note: A = 225π.

πR2 4

=

π(30)2 4

=

900π 4

=

5. 3:1. Notes: Let the radius of the small circle be R. The area of the small circle is then Asmall = πR2 . The radius

of the large circle is 2R. The area of the large circle is Alarge = π(2R)2 = 4πR2 . Recall from algebra that (ab)n = an bn . To find the area of the shaded region,

subtract the area of the small circle from the area of the large circle: Ashaded = Alarge − Asmall = 4πR2 − πR2 = 3πR2 . The ratio of the area of the shaded circle to the area of the small circle is

Ashaded Asmall

193

=

3πR2 πR2

= 3.

Answer Key

4

6. Acir :Asqr = 4:π ≈ 4:3.14 (which is equivalent to π :1 ≈ π

1.27:1 and to 1: 4 ≈ 1:0.785) and Acir :Atri = 9:π√3 ≈

9:5.44 (which is equivalent to and to 1:

π√3 9

9

3√3

:1 = 3

π√

π

:1 ≈ 1.65:1

≈ 1:605). Notes: Let d be the edge length of

the equilateral triangle, L be the edge length of the

square, and R be the radius of the circle. The perimeter of the triangle is Ptri = 3d. The perimeter of the square is Psqr = 4L. The circumference of the circle is Ccir =

2πR. According to the problem, Ptri = Psqr = Ccir such 3d

4L

that 3d = 4L = 2πR. Thus, R = 2π = 2π =

2L π

. The area of

the equilateral triangle is (recall Volume 1, Chapter 6) Atri =

d2 √3 4

. The area of the square is Asqr = L2 . The area 3d

of the circle is Acir = πR2 . Since R = = 2π

2L π

, the area of 2L 2

the circle can also be expressed as Acir = π � π � = 4L2

π � π2 � =

4L2 π

3d 2

9d2

or as Acir = π �2π� = π �4π2 � =

desired ratios are:

194

9d2 4π

. The

Plane Geometry Practice Workbook with Answers, Volume 2

4 9 Acir 9d2 /4π 9d2 d2 √3 9d2 = = ÷ = × = Atri d2 √3/4 4π 4 4π d2 √3 π√3 =

9 √3

=

9√3 3√3 = 3π π

π√3 √3 Acir 4L2 /π 4 = = Asqr L2 π

For a fixed boundary length (namely, the perimeter or circumference), the circle is the shape with the

maximum area. (If instead the area were fixed, the circle would be the shape with the minimum boundary length. Similarly, in three dimensions, the sphere is the shape for a given volume that minimizes the surface area, which is why water droplets tend to be spherical.)

7. AE = 20. Notes: AD = π(DF)2 , AE = π(EF)2 , and AF =

π(DE)2 . The Pythagorean theorem gives DF 2 + EF 2 = DE 2 . Multiply both sides by π to get π(DF)2 + π(EF)2 = π(DE)2 , which shows that AD + AE = AF . Thus, 16 + AE = 36 and AE = 36 − 16 = 20.

195

Answer Key

Chapter 3 Answers π

π

1. (A) s = 4π. (B) A = 24π. Notes: 60° = 3 rad, 12 �3� = 4π, and

360° 60°

= 6 so that A =

πR2 6

.

2. (A) s = 21π. (B) A = 189π. Notes: 210° = 7π

360°

18 � � = 21π, and = 6 210° 3. θ = 72°. Notes: 72° = 4. R = 21π.

63 2

12

2π 5

7

so that A =

90 �

30

12

2π

= 31.5. Notes: 120° =

360°

7πR2

6

.

rad,

rad and 20 � 5 � = 8π. 2π 3

rad and

5. θ = 258° and s = 129π. Notes: 258° = 43π

7π

60

2

43π 30

� = 129π, and 258° = 43 so that A =

196

63 2π

�3�=

rad,

43πR2 60

.

Plane Geometry Practice Workbook with Answers, Volume 2

6. α = 105°, β = 75°, γ = 105°, p = 10π, q = 14π, r =

10π, s = 14π, and C = 48π. Notes: D = 48 and R = 24. α

and 75° are supplements. α and γ, and β and 75° are two pairs of vertical angles. Checks: C = 2πR = 2π(24) = 48π, 75° + α + β + γ = 75° + 105° + 75° + 105° =

360°, and p + q + r + s = 10π + 14π + 10π + 14π =

48π = C.

7. (A) A =

θR2 2

. (B) A =

sR

θ

. Notes: 2π is the ratio of the 2

area of the sector to the area of the full circle. Multiply θ

2 by πR to get the area of the sector. Since s = Rθ, it 2π s

s

follows that θ = . Plug θ = into A = R R

θR2 2

to get A =

sR 2

.

Checks: For a full circle, θ = 2π rad (which corresponds to 360°) and A =

2πR2 2

= πR2 . For a semicircle, θ = π rad

which (corresponds to 180°) and A = π

π R2

1B, θ = 3 rad and A = 3

rad and A = A=

43π R2 30 2

=

7π R2 6 2

43πR2 60

=

.

7πR2 12

2

=

πR2 6

πR2 2

. For Problem

. For Problem 2B, θ =

. For Problem 5, θ = 197

43π 30

7π 6

rad and

Answer Key

πRθ

8. s = 180° if θ is in degrees instead of radians. Notes: π

Multiply by 180° to convert from degrees to radians. This

formula incorporates the conversion factor. Checks: For a full circle, θ = 360° and s = semicircle, θ = 180° and s =

πR360°

180° πR180°

198

180°

= 2πR = C. For a C

= πR = 2.

Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 4 Answers 1. (A) ∠ADB ≅ ∠ACB (both subtend the arc from A to

B), ∠CAD ≅ ∠CBD (both subtend the arc from C to D), and ∠AED ≅ ∠BEC as well as ∠AEB ≅ ∠CED (vertical angles).

(B) ∠ACB = 65°, ∠CAD = 39°, ∠AED = 76°, and

∠BEC = 76°. Note: ∠CBD + ∠ACB + ∠BEC = 180°

because the three interior angles of any triangle add up to 180°. Check: 39° + 65° + 76° = 180°.

2. ∠ACB = ∠ADB = 38°. Notes: Apply the inscribed

angle theorem. Central angle ∠BOA and inscribed angles ∠ACB and ∠ADB all subtend the arc from A to B.

3. ∠EGH = 68° and ∠EOH = 136°. Notes: Apply the inscribed angle theorem. Central angle ∠EOH and

inscribed angles ∠EFH and ∠EGH all subtend the arc

from E to H.

199

Answer Key

4. ∠BCD = 75° and ∠BAD = 105°. Notes: Central angle ∠BOD and inscribed angle ∠BCD both subtend the

minor arc from B to D (which is the arc from B to A plus the arc from A to D). The reflex angle shown below is 210° because the reflex angle and ∠BOD add up to

210° + 150° = 360°. Inscribed angle ∠BAD and the

reflex angle both subtend the major arc from B to D

(which is the arc from B to C plus the arc from C to D). Thus, ∠BAD =

210° 2

Problem 8: 180° −

. This agrees with the formula from 150° 2

= 180° − 75°.

200

Plane Geometry Practice Workbook with Answers, Volume 2

5. ∠QOR = 120°, ∠OQR = 30°, ∠ORQ = 30°, OR = 2, and QR = 2√3. Notes: Central angle ∠QOR and inscribed

angle ∠QPR both subtend the arc from Q to R. Since OR

and OQ are each radii, OR = OQ = 2. Therefore, ∆QOR is isosceles, such that ∠OQR ≅ ∠ORQ. The three interior

angles of ∆QOR add up to 120° + 30° + 30° = 180°. The

ratio of the sides of a 30°-30°-120° triangle is 1:1:√3

(Volume 1, Chapter 5, Problem 12).

6. ∠ACD = 18° and ∠BDC = 18°. Notes: Central angle

∠AOD and inscribed angle ∠ACD both subtend the arc

from A to D. ∠BOC = 36° because ∠BOC and ∠AOD are

vertical angles. Central angle ∠BOC and inscribed angle ∠BDC both subtend the arc from B to C.

201

Answer Key

7. ∠OPQ = 37°, ∠OQP = 106°, and ∠PQR = 148°. Notes: Since point O lies at the center of ⊙O, ���� OP ≅ ���� OR because

each equals the radius of ⊙O. Since point Q lies at the center of ⊙Q, ���� OQ ≅ ���� PQ ≅ ���� QR because each equals the

radius of ⊙Q. ∆OPQ ≅ ∆OQR according to SSS (Volume 1, Chapter 3). Therefore, ∠POQ = ∠QOR =

∠POR 2

=

74° 2

=

37°. Since ���� OQ ≅ ���� PQ, ∆OPQ is isosceles: ∠POQ = ∠OPQ =

37°. The three angles of ∆OPQ add up to 180°:

∠POQ + ∠OPQ + ∠OQP = 37° + 37° + ∠OQP = 180°. It follows that ∠OQP = 106°. Similarly, ∠OQR = 106°. ∠OQP, ∠OQR, and ∠PQR form a full circle:

∠OQP + ∠OQR + ∠PQR = 106° + 106° + ∠PQR = 360°. It follows that ∠PQR = 148°.

Alternative solution: Show that point Q is the

circumcenter of ∆OPR, where the three perpendicular

bisectors of the triangle intersect (Volume 1, Chapter 7). Since ∆OPR is isosceles, ���� OQ is both a perpendicular

bisector and an angle bisector (see Volume 1, Chapter 7, Example 6). (However, this is NOT true of ���� PQ and ���� QR.) ���� is an angle bisector, ∠POQ = ∠QOR = Since OQ 74° 2

= 37°.

202

∠POR 2

=

Plane Geometry Practice Workbook with Answers, Volume 2

8. Main ideas: Central angle θ and inscribed angle α

both subtend the minor arc from B to D (which is the

arc from B to A plus the arc from A to D). Therefore, α = θ

2

. Central angle φ and inscribed angle β both subtend

the major arc from B to D (which is the arc from B to C φ

plus the arc from C to D). Therefore, β = 2 . Since θ and

φ form a full circle, θ + φ = 360°. Subtract θ from both sides of the equation: φ = 360° − θ. Plug this into β =

to get β =

360°−θ 2

=

360°

θ

2

θ

θ

− 2 = 180° − 2. For the second

proof, rewrite α = 2 and β =

φ 2

203

as 2α = θ and 2β = φ.

φ 2

Answer Key

Plug these into θ + φ = 360° to get 2α + 2β = 360°. Divide by 2 on both sides: α + β = 180°.

���� ≅ CO ���� because each is a radius. 9. Main ideas: BO

Therefore, ∆BCO is isosceles. Since ∆BCO is isosceles,

α ≅ γ. The three interior angles of ∆BCO add up to 180°:

α + γ + φ = 180°. Since α ≅ γ, this becomes 2α + φ =

180°. Since θ and φ are supplementary angles, θ + φ = 180°. Combine the last two equations: 2α + φ = θ + φ.

Subtract φ from both sides: 2α = θ. Divide by 2 on both θ

sides: α = 2.

10. Main ideas: ���� AO ≅ ���� BO ≅ ���� CO because each is a radius. Therefore, ∆ABO, ∆BCO, and ∆ACO are isosceles: δ =

φ + χ, γ = α + σ, and σ ≅ φ. The three interior angles of

∆ABC add up to 180°. α + γ + δ + χ = 180°

Plug γ = α + σ and δ − φ = χ into the above equation. α + α + σ + δ + δ − φ = 180° 2α + σ + 2δ − φ = 180°

Plug σ ≅ φ into the above equation. 2α + φ + 2δ − φ = 180° 2α + 2δ = 180° 204

Plane Geometry Practice Workbook with Answers, Volume 2

Divide both sides of the equation by 2. α + δ = 90°

11. Main ideas: The three interior angles of ∆ABO add up to 180°.

θ + δ + χ + φ = 180°

Recall δ = φ + χ from the solution to Problem 10. θ + δ + δ = 180° θ + 2δ = 180°

Recall α + δ = 90° from Problem 10. Subtract α from both sides: δ = 90° − α. θ + 2(90° − α) = 180° θ + 180° − 2α = 180° θ − 2α = 0

θ = 2α θ =α 2 Central angle θ and inscribed angle α each subtend the arc from A to B.

205

Answer Key

Chapter 5 Answers 1. θ = 31° and φ = 48°. Check: 31° + 59° = 90° and 48° + 42° = 90°.

2. α = 38°, β = 76°, γ = 38°, δ = 52°, and ε = 52°.

Notes: 104° and β are supplements. The inscribed angle

theorem applies to α and β. It also applies to 104° and ε. ���� ���� since each is a radius. ∆FEO and ∆DEO EO ≅ FO DO ≅ ���� are isosceles: α ≅ γ and δ ≅ ε. The sum of the three interior angles of each triangle (including ∆DEF) is

180°. γ + δ = 90°.

3. QR = 2.5. Check: 1.52 + 22 = 2.25 + 4 = 6.25 = 2.52 . 4. AE = 15. Notes: BC 2 + BF 2 = CF 2 = AD2 = ��� ≅ ���� DE 2 + AE 2 since �CF AD.

Check: 92 + 132 = 81 + 169 = 250 agrees with 52 + 152 = 25 + 225 = 250.

206

Plane Geometry Practice Workbook with Answers, Volume 2

5. θ = 30°, JK = 3√3, and KL = 3. Notes: KL:JK:JL =

3:3√3:6 reduces to 1:√3:2. The hypotenuse (JL) is twice the side opposite to the 30° angle (which is KL), and the side opposite to the 60° angle (which is JK) is √3 larger

than the side opposite to the 30° angle (which is KL). (Recall Volume 1, Chapter 5.)

6. χ = 45°, QR = 12, and PR = 12√2. PQ:QR:PR =

12:12:12√2 reduces to 1:1:√2. The legs (PQ and QR) are

congruent, and the hypotenuse (PR) is √2 times larger than either leg. (Recall Volume 1, Chapter 5.)

207

Answer Key

7. A =

289π 8

− 60 exactly (in terms of the constant π),

which is approximately equal to 53.5 to three significant figures. Notes: The area of the semicircle is 1

πR2 2

. The area

of the right triangle is 2 (AB)(BC). Use the Pythagorean

theorem to find that AB = 8. To find the combined area of the shaded regions, subtract the area of the triangle from the area of the semicircle. R =

17 2

.

Check: 82 + 152 = 64 + 225 = 289 = 172 .

208

Plane Geometry Practice Workbook with Answers, Volume 2

8. Main ideas: ���� DF and ���� EG intersect at the center of the ���� ≅ ���� circle (point O) above. ���� DO ≅ GO EO ≅ ���� FO since each is a radius. ∠DOG ≅ ∠EOF (vertical angles). ∆DOG ≅ ∆EOF via SAS. ∠GDO ≅ ∠EFO and ∠DGO ≅ ∠FEO

according to the CPCTC. These two congruent pairs of ����. (Recall ���� ∥ DG alternate interior angles show that EF

Volume 1, Chapters 1-3.) Side note: ∠GDO ≅ ∠EFO ≅

∠DGO ≅ ∠FEO since ∆DOG and ∆EOF are isosceles.

9. Main ideas: PQRS has two pairs of parallel edges according to Problem 8. PQRS is a parallelogram. ���� PR ≅

���� ���� QS since each is a diameter of the same circle. ���� PR and QS

are the diagonals of parallelogram PQRS. Recall from

Volume 1, Chapter 9, Problem 17 that if the diagonals of a parallelogram are congruent, the parallelogram is a

rectangle. An alternative proof is to use Thales’s

theorem to show that the four interior angles are all right angles.

209

Answer Key

10. Main ideas: α + β + γ = 180° and θ + φ + ε = 180° (the three interior angles of a triangle add up to 180°). α ≅ γ and φ ≅ ε (∆ABO and ∆BCO are isosceles; ���� AO ≅

���� BO ≅ ���� CO since each is a radius). Therefore, β + 2γ =

180° and θ + 2φ = 180°. Rewrite these as β = 180° − 2γ and θ = 180° − 2φ. β + θ = 180°

(supplementary angles). Plug the previous equations into this equation. 180° − 2γ + 180° − 2φ = 180° 360° = 180° + 2γ + 2φ 180° = 2γ + 2φ 90° = γ + φ

11. Main ideas: ���� AO ≅ ���� BO ≅ ���� CO (each is a radius). ���� ⊥ ���� BO AC (given). ∠BOC = 90° and ∠AOB = 90° (since ���� BO ⊥ ���� AC). ∆ABO and ∆BCO are each 45°-45°-90°

triangles (in each case, one angle is a right angle and the two legs are congruent). Therefore, ∠BCO = 45°,

∠CBO = 45°, ∠ABO = 45°, and ∠BAO = 45°. Also,

∠CBO + ∠ABO = 45° + 45° = 90°. This shows directly that ∆ABC is a 45°-45°-90° triangle.

Part 2: Since ∠ABC = ∠CBO + ∠ABO = 90°, any other

angle inscribed in the same circle that subtends the arc from A to C is congruent with ∠ABC. (This is one of the 210

Plane Geometry Practice Workbook with Answers, Volume 2

theorems from Chapter 4.) Therefore, ∠ABC ≅ ∠ADC, and any other angle inscribed in the same circle that

subtends the arc from A to C is a right angle, proving Thales’s theorem.

211

Answer Key

Chapter 6 Answers 1. AB = 12 and A = 24π − 36√3 = 12�2π − 3√3� ≈ 13. Notes: Since 360° ÷ 6 = 60°, the area of the sector is

πR2 6

=

π122 6

=

144π 6

= 24π. Since AO = BO = R and

∠AOB = 60°, it follows that ∆ABO is an equilateral

triangle. The edge length of ∆ABO is the radius of the

circle: L = R = 12. Recall from Volume 1, Chapter 6 that the area of an equilateral triangle with edge length L is

L2 √3 4

=

122 √3 4

=

144√3 4

= 36√3. The area of the sector

minus the area of the triangle equals the area of the circular segment:

A = 24π − 36√3 = 12�2π − 3√3�

In the last step, 12 was factored out. (Check this by distributing the 12.)

2. CD = 6√2 and A = 9π − 18 = 9(π − 2) ≈ 10.3. Notes: The area of the quarter circle is

πR2 4

=

π62 4

=

36π 4

= 9π.

Since CO = DO = R and ∠COD = 90°, it follows that

∆CDO is a 45°-45°-90° triangle. The area of ∆CDO is 1

1

1

2 (6)2 = bh = R = 2 2 2

36 2

= 18. The area of the sector 212

Plane Geometry Practice Workbook with Answers, Volume 2

minus the area of the triangle equals the area of the circular segment:

A = 9π − 18 = 9(π − 2)

In the last step, 9 was factored out. (Check this by distributing the 9.)

3. ∠EOF = 120° (see the diagram on the next page) and π

A= − 3

√3 4

≈ 0.614. Notes: Since EO:FO:EF = 1:1:√3,

∆EOF (where O lies at the center) is a 30°-30°-120°

triangle (recall Volume 1, Chapter 5, Problem 12). In the right diagram on the next page, ∆DEO and ∆DEF are

30°-60°-90° triangles. Since the sides of a 30°-60°-90°

triangle come in the ratio 1:√3:2, DE =

√3 2

(since DE is

1

opposite to 30°, DE must be one-half of EF) and DO = 2 (since DO is opposite to 30°, it must be one-half EO).

The base of ∆EFO is FO = 1 and the height of ∆EFO is DE =

√3 . 2

√3

1

The area of ∆EFO is 2 (1) � 2 � =

= 120°, the area of the sector is

πR2 3

=

√3 . 4

π12 3

Since 360° ÷ 3 π

= 3. The area

of the sector minus the area of the triangle equals the area of the circular segment:

π √3 A= − 4 3 213

Answer Key

4. DE = 2√5 and A =

9π 2

− 4√5 ≈ 5.19. Notes: ∆DEF is a

right triangle according to Thales’s theorem (Chapter 5). Use the Pythagorean theorem to check that DE =

2√5:

2

2

DE 2 + EF 2 = �2√5� + 42 = 22 �√5� + 42 = 4(5) + 16 = 20 + 16 = 36 = 62 = DF 2 1

1

1

The area of ∆DEF is 2 bh = 2 (DE)(EF) = 2 �2√5�(4) =

4√5. The radius of the circle is R = of the semicircle is

πR2 2

=

π32 2

=

9π 2

DF 2

6

= 2 = 3. The area

. The area of the sector

minus the area of the triangle equals the combined area

of the two circular segments: 9π A= − 4√5 2

214

Plane Geometry Practice Workbook with Answers, Volume 2

5. KL = 6 and A = 30π ≈ 94.2. Notes: 300° = π

300° × 180° =

30π 18

=

5π 3

rad (since 30 and 18 are each

evenly divisible by 6) is the central angle for the major

arc. The given value 10π is the corresponding major arc length: s = 10π. (The problem clearly states that 10π is

an arc length, NOT an angle.) Use the arc length formula

(Chapter 3).

s = Rθ 5π 10π = R 3 30π = 5πR 6=R

The minor central angle ∠KOL = 360° − 300° = 60°.

Since KO = LO = R = 6 and ∠KOL = 60°, ∆KLO is an

300°

equilateral triangle. Therefore, KL = R = 6. Since 360° = 5 6

, the shaded region is 5/6 of a full circle. The area of the

shaded region (which is a sector, NOT a circular segment) is

5πR2 6

=

5π62 6

=

5π36 6

= 5π6 = 30π.

6. AB = �8 − 4√3 = 2�2 − √3 = √6 − √2 ≈ 1.04 (all three answers are equivalent, but the form √6 − √2 is

preferred because it does not have nested square roots) 215

Answer Key

π

and A = 3 − 1 =

π−3 3

≈ 0.0472. Notes: ∆ABO is isosceles:

AO = BO = R = 2. (∆ABO is NOT a right triangle.) Draw vertical line segment ���� AC, as shown on the next page, to create right triangles ∆ACO and ∆ABC. Since ∆ACO is a

30°-60°-90° triangle, AC:CO:AO = 1:√3:2. Since CO = √3

and BO = 2, it follows that BC = 2 − √3. Use the Pythagorean theorem to find AB. BC 2 + AC 2 = AB 2 2

�2 − √3� + 12 = AB 2

Apply the “foil” method: (t − u)2 = t 2 − 2tu + u2 . 4 − 4√3 + 3 + 1 = AB 2

8 − 4√3 = 4�2 − √3� = AB 2

�8 − 4√3 = 2�2 − √3 = AB

Following is a clever way to avoid the nested square roots given above.

8 − 4√3 = 8 − 2(2)√3 = 8 − 2√4√3 = 6 − 2√12 + 2 2

= �√6 − √2�

30°

1

AB = �8 − 4√3 = √6 − √2

Since 360° = 12, the area of the sector is 1/12 the area of a full circle:

216

Plane Geometry Practice Workbook with Answers, Volume 2

πR2 π22 4π π = = = 12 12 12 3

Since ∆ABO has a base of BO = 2 and a height of AC = 1,

the area of ∆ABO is 1 1 1 bh = (BO)(AC) = (2)(1) = 1 2 2 2 and the area of the circular segment is π−3 π A= −1= 3 3

7. A = 4 − π ≈ 0.858. Notes: The area of the square is

22 = 4. The combined area of the two semicircles equals 2

the area of one circle of radius R = 2 = 1. The combined area of the two semicircles is π𝑅𝑅2 = π12 = π. To find

the combined area of the shaded regions, subtract the

area of the two semicircles from the area of the square.

8. A = 2π − 4 = 2(π − 2) ≈ 2.28. Method 1: Cut the

square in half along a diagonal as shown below. Find the 217

Answer Key

area of each shaded circular segment shown below (recall the solution to Problem 2). The area of the

quarter circle is 1

πR2 4 1

=

π22 4

1

=

4π 4

= π. The area of the right 4

triangle is bh = R2 = (2)2 = = 2. The area of the 2 2 2 2

sector minus the area of the triangle equals the area of

the circular segment: π − 2. The two circular segments

shown below have equal area. The total area is A = 2π − 4 = 2(π − 2).

Method 2: The area of the two quarter circles shown

below added together equals the area of the square plus the area of the desired shaded region.

Therefore, the area of the two quarter circles minus the

area of the square equals the area of the desired shaded region.

218

Plane Geometry Practice Workbook with Answers, Volume 2

πR2 πR2 + − R2 = A 4 4 4π 4π π22 π22 + − 22 = + − 4 = 2π − 4 A= 4 4 4 4 = 2(π − 2)

9. A = 12π − 18√3 = 6�2π − 3√3� ≈ 6.52. Notes: WY = WZ = YZ = XY = XZ = R = 6. ∆WYZ ≅ ∆XYZ and both are equilateral triangles. Find the area of the circular

segment shown below (like the solution to Problem 1) and then double this. The area of the sector is

π62 6

=

36π 6

πR2 6

=

= 6π. Recall from Volume 1, Chapter 6 that the

area of an equilateral triangle with edge length L is

L2 √3 4

=

62 √3 4

=

36√3 4

= 9√3. The area of the sector minus

the area of the triangle equals the area of the circular

segment: 6π − 9√3 = 3�2π − 3√3�. Double this to find the total area: A = 12π − 18√3 = 6�2π − 3√3�.

219

Answer Key

10. A = 9√3 −

9π 2

π

= 9 �√3 − 2� ≈ 1.45. Notes: ∆PQR is

an equilateral triangle with edge length PQ = PR = QR

= 2R = 2(3) = 6. Recall from Volume 1, Chapter 6 that

the area of an equilateral triangle with edge length L is

L2 √3 4

=

62 √3 4

=

36√3 4

= 9√3. Each of the three sectors (for

vertices P, Q, and R) has a 60° angle (like Problems 1 and 9). Each sector has area

πR2 6

=

π32 6

=

9π 6

=

3π 2

. (Note

that the radius of each sector is R = 3, whereas the edge length of equilateral triangle ∆PQR is 2R = 6.) The 3π

combined area of the three sectors is 3 � � = 2

9π 2

. The

area of ∆PQR minus the area of the three sectors equals

the area of the shaded region. 9π π A = 9√3 − = 9 �√3 − � 2 2

π

11. A = 2 −

√3 2

=

π−√3 2

≈ 0.705. Notes: Draw ∆DEF as

shown on the next page. ∆DEF is an equilateral triangle

with edge length DE = DF = EF = 220

AB 2

=

AC 2

=

BC 2

2

=2=1

Plane Geometry Practice Workbook with Answers, Volume 2

(recall Volume 1, Chapter 7, “Tips for Working with Medians”). The area of ∆DEF is

L2 √3 4

=

12 √3 4

=

√3 4

(recall

Volume 1, Chapter 6). The area of the sector from vertex E that includes both ∆DEF and the shaded circular

segment in the diagram on the right below is π 6

πR2 6

=

π12 6

=

(like the solution to Problem 1). The area of the sector

minus the area of ∆DEF equals the area of the shaded

circular segment shown in the diagram on the right π

below: 6 −

√3 . 4

To find the shaded region given in the

problem, add the area of ∆DEF to three times the area of

the circular segment shown below on the right. π √3 √3 √3 3π 3√3 3π 2√3 + 3� − � = + − = − 4 4 4 4 4 6 6 6 π √3 π − √3 = − = 2 2 2

221

Answer Key

π

12. A = √3 − 6 ≈ 1.21. Notes: The area of the small

semicircle shaded in the diagram on the left on the next page minus the area of the circular segment shaded in

the middle diagram equals the area of the lune shaded in the diagram on the right. The radius of the small 2

semicircle is R small = 2 = 1. The area of the small

semicircle is

πR2small 2

=

π12 2

π

= 2. Each sector in the middle

diagram on the following page has a 60° central angle, a 4

radius of R large = 2 = 2, and an area of

4π 6

=

2π 3

πR2large 6

=

π22 6

=

. ∆MNO is an equilateral triangle. The area of

∆MNO is

L2 √3 4

=

22 √3 4

=

4√3 4

= √3 (Volume 1, Chapter 6).

Subtract the area of ∆MNO from the area of the sector to

find the area of the circular segment shaded in the middle diagram on the following page:

2π 3

− √3. To find

the area of the lune, subtract the area of the circular

segment from the area of the small semicircle shaded in the left diagram. π 2π π 2π 3π 4π − � − √3� = − + √3 = √3 + − 2 3 2 3 6 6 π = √3 − 6 222

Plane Geometry Practice Workbook with Answers, Volume 2

The minus sign was distributed. For example, 𝑎𝑎 − (𝑏𝑏 − 𝑐𝑐 ) = 𝑎𝑎 − 𝑏𝑏 − (−𝑐𝑐 ) = 𝑎𝑎 − 𝑏𝑏 + 𝑐𝑐.

This problem involved a lune based on an equilateral

triangle (∆MNO). Don’t confuse this lune with another common and important lune, which instead features a

45°-45°-90° triangle. The lune based on the 45°-45°-90°

triangle is important because the π’s cancel out for that lune. Compare the two lunes shown on the next page.

The lune on the left was featured in this problem, and

involves an equilateral triangle. The lune on the right involves a 45°-45°-90° triangle. For the lune on the

right, the π’s cancel out. Try it! 223

Answer Key

13. α = 62°, β = 62°, γ = 56°, δ = 68°, ε = 56°, φ = 56°,

η = 62°, and κ = 62°. Notes: Central angles that subtend congruent arcs are congruent. Since congruent chords subtend congruent arcs, central angles that subtend

congruent chords are congruent. Therefore, since AB =

CD = 7, φ = 56°. AO = BO = CO = DO = R. Although the value of R is not given, it is clear that ∆ABO, ∆BCO, and

∆CDO are isosceles triangles, such that α = β, γ = ε, and

η = κ. Checks: 56° + δ + φ = 56° + 68° + 56° = 180°

(straight angle), α + β + 56° = 62° + 62° + 56° = 180°, δ + γ + ε = 68° + 56° + 56° = 180°, φ + η + κ =

56° + 62° + 62° = 180°. Parallel notes: It turns out that ���� BC ∥ ���� AD (which is why γ = 56° and ε = φ are two pairs of congruent alternate interior angles), but ���� BO is NOT parallel to ���� CD (η does NOT equal δ) and ���� AB is NOT ���� (δ does NOT equal β). parallel to CO 224

Plane Geometry Practice Workbook with Answers, Volume 2

14. α = γ = 61° and β = δ = 119°. Notes: α = γ = 48°+74° 2

=

122° 2

= 61° (intersecting chords) and β = δ

(vertical angles). Check: α + β + γ + δ =

61° + 119° + 61° + 119° = 360°.

15. α = 31°. Check: ∠PTQ = ∠STR = 48° 2

α+17° 2

=

31°+17°

= 24°. Note: To solve for α, first write 24° =

and then multiply both sides of the equation by 2. 16. ∠AOF = 40°. Check: ∠AFB = ∠CFD = 40°+108° 2

=

148° 2

=

2 α+17° 2

∠AOB+∠COD 2

=

= 74°. Notes: ∠AOB ≅ ∠AOF. ���� AC and ���� BD

���� happens are the intersecting chords. In this problem, BD

to be a diameter.

17. JN = 24 and NO = 6√3. Checks: (JN)(LN) = (24)(9) = 216, (KN)(MN) = (8)(27) = 216, and 2

2

2

2

2

R − NO = 18 − �6√3� = 324 − (6)2 �√3� = 324 − 36(3) = 324 − 108 = 216. 225

Answer Key

18. VY = 4, OY = 11, and WY = 22. Checks: (VX)(VZ) = (6)(12) = 72 and R2 − OV 2 = 112 − 72 = 121 − 49 = 72. Notes: Apply the intersecting chords theorem to ����� WY ����. In this problem, ����� WY happens to be a diameter. and XZ VY + OV = OY = R.

19. ∠AOE = 45°, ∠COD = 135°, AO = √2, EO = 1, and

BE = √2 − 1. Notes: AO = BO = CO = DO = R = √2.

∆AEO and ∆CEO are 45°-45°-90° triangles. AE:EO:AO = CE:EO:CO = 1:1:√2. ∠AOE = ∠COE = 45°.

∠COE + ∠COD = 180°. BE + EO = BO. DO + EO = DE = 1 + √2. Checks: (BE)(DE) = �√2 − 1��1 + √2� =

√2 + 2 − 1 − √2 = 2 − 1 = 1, (AE)(CE) = (1)(1) = 1, 2

R2 − EO2 = �√2� − 12 = 2 − 1 = 1, and

45°+135° 2

=

180° 2

= 90° = ∠BEA. 226

∠AOE+∠COD 2

=

Plane Geometry Practice Workbook with Answers, Volume 2

����� and XZ ���� 20. Main ideas: (VW)(VY) = (VZ)(VX) since WY VZ, this becomes are intersecting chords. Since ����� VW ≅ ����

(VW)(VY) = (VW)(VX), which reduces to VY = VX.

∆WVZ~∆VXY according to SAS similarity (VW/VZ = 1,

VY/VX = 1, and the angle between the sides is

congruent; recall Volume 1, Chapter 4). ∠WZV ≅ ∠VXY

are congruent alternate interior angles (because ∆WVZ is similar to ∆VXY and each triangle is isosceles).

Note: Recall from Volume 1 that ~ means similarity, whereas ≅ means congruence.

21. Main ideas: ∠AEB ≅ ∠CED (vertical angles) and

∠ABD ≅ ∠ACD (these inscribed angles subtend the same arc length from A to D; recall Chapter 4).

∆ABE~∆CDE according to AA (recall Volume 1, Chapter

4). It can similarly be shown that ∆ADE~∆BCE.

22. Main ideas: ∆ABE~∆CDE (see the solution to

Problem 21). AE:BE = DE:CE because the

corresponding sides of similar triangles come in the

same proportion (Volume 1, Chapter 4). AE DE = BE CE Cross multiply. (Multiply both sides by BE and by CE.) (AE)(CE) = (BE)(DE) 227

Answer Key

23. Main ideas: (AE)(CE) = (BE)(DE) was proven in

Problem 22. Apply the intersecting chords theorem FG: (proven in Problems 21-22) to chords ���� AC and ����

(AE)(CE) = (EF)(EG). Since EF + EO = FO = GO = R, it

follows that EF = R − EO and EG = R + EO. Therefore, (AE)(CE) = (EF)(EG) may be rewritten as: (AE)(CE) = (R − EO)(R + EO) Apply the “foil” method: (a + b)(c + d) =

ac + ad + bc + bd. (AE)(CE) = R2 + R(EO) − (EO)R − EO2 = R2 − EO2

����, as shown on the 24. Main ideas: Draw line segment AD

next page. ∠AED + ∠DAE + ∠ADE = 180° (interior

angles of ∆ADE). ∠AED + ∠CED = 180° (supplementary angles). ∠CED = ∠DAE + ∠ADE (set the first two

equations equal and subtract ∠AED from both sides.) ∠DAE = ∠DAC =

∠COD

∠COD

∠AOB

2

and ∠ADE = ∠ADB =

∠AOB 2

inscribed angle is one-half of a central angle that

(an

subtends the same arc; recall Chapter 4). Plug ∠DAE = 2

and ∠ADE =

get ∠CED =

∠COD 2

+

into ∠CED = ∠DAE + ∠ADE to

2 ∠AOB 2

=

∠COD+∠AOB 2

228

.

Plane Geometry Practice Workbook with Answers, Volume 2

229

Answer Key

Chapter 7 Answers 1. α = 39°. Note: α =

∠COD−∠AOB 2

2. ∠LOM = 10°. Check: α =

=

107°−29°

∠JOK−∠LOM 2

2

=

=

78° 2

58°−10° 2

= 39°.

=

24°. Notes: To solve for ∠LOM, first write 24° =

58°−∠LOM 2

48° 2

=

and then multiply both sides of the equation

by 2. If it does not “look” like ∠LOM = 10°, it is because the diagrams are not drawn to scale.

3. ∠XOY = 92°. Check: α =

∠YOZ−∠WOX 2

=

72°−16° 2

=

28°. Notes: To solve for ∠WOX, first write 28° = 72°−∠WOX 2

56° 2

=

and then multiply both sides of the equation by

2. Three angles form a straight angle:

∠WOX + ∠XOY + ∠YOZ = 16° + 92° + 72° = 180°.

230

Plane Geometry Practice Workbook with Answers, Volume 2

4. AD = 11.6, EO = 17.5, and EF = 8. Notes: DE =

AE + AD = 10 + 11.6 = 21.6, CE = BE + BC = 9 + 15 = 24, and EO = EF + FO = 8 + 9.5 = 17.5. Checks: (AE)(DE) = (10)(21.6) = 216, (BE)(CE) =

(9)(24) = 216, and EO2 − R2 = 17.52 − 9.52 =

306.25 − 90.25 = 216. The intersecting secants

theorem can even be applied to the extension of ���� EO

(add FO to EO so that this secant reaches the other side of the circle): (EF)(EF + 2FO) = (8)(8 + 19) = 8(27) =

216.

5. QR = 1 and SU = 10. Notes: OU = OT + TU = 11 + 8

= 19, PU = PS + SU = 14 + 10 = 24, and QU = QR + RU = 1 + 15 = 16. To solve for SU, first write SU(SU + 14)

= 240. Distribute: SU 2 + 14SU = 240. Subtract 240 from both sides: SU 2 + 14SU − 240 = 0. This is a

quadratic equation. Either use the quadratic formula or factor this quadratic as (SU + 24)(SU − 10) = 0.

Checks: (SU)(PU) = (10)(24) = 240, (RU)(QU) = (15)(16) = 240, and OU 2 − R2 = 192 − 112 =

361 − 121 = 240. The intersecting secants theorem can even be applied to the extension of ���� OU (add OT to OU so that this secant reaches the other side of the circle): (TU)(TU + 2OT) = (8)(8 + 22) = 8(30) = 240. 231

Answer Key

6. OW = OZ = R =

19 4

= 4.75. Notes: VY = VX + XY =

6 + 3 = 9 and OV = VW + OW = 4 + 4.75 = 8.75. Checks: (VX)(VY) = (6)(9) = 54 and OV 2 − R2 = 8.752 − 4.752 = 76.5625 − 22.5625 = 54. The

intersecting secants theorem can even be applied to ���� VZ: (VW)(VZ) = (4)(4 + 9.5) = 4(13.5) = 54. 7. Main ideas: ∠ACB ≅ ∠ADB (these inscribed angles

subtend the same arc length from A to B; recall Chapter

4). ∆ACE and ∆BDE share ∠CED. ∆ACE~∆BDE according

to AA (recall Volume 1, Chapter 4). AE:CE = BE:DE

because the corresponding sides of similar triangles

come in the same proportion (Volume 1, Chapter 4). AE BE = CE DE Cross multiply. (Multiply both sides by CE and by DE.) (AE)(DE) = (BE)(CE)

232

Plane Geometry Practice Workbook with Answers, Volume 2

8. Main ideas: (AE)(DE) = (BE)(CE) was proven in

Problem 7. An analogous relation can be written for ���� and EG ����: (BE)(CE) = (EF)(EG). intersecting secants CE Since FO = GO = R, EF + FO = EF + R = EO, and

EO + GO = EO + R = EG, it follows that EF = EO − R and EG = EO + R. Therefore, (BE)(CE) = (EF)(EG) may be rewritten as: (BE)(CE) = (EO − R)(EO + R)

Apply the “foil” method: (a + b)(c + d)

= ac + ad + bc + bd. (BE)(CE) = EO2 + (EO)R − R(EO) − R2 = EO2 − R2 9. Main ideas: α + ∠CAE + ∠ACE = 180° (interior

angles of ∆ACE). ∠CAE + ∠CAD = 180° (supplementary angles). α + ∠ACE = ∠CAD (set the first two equations

equal and subtract ∠CAE from both sides.) Subtract

θ

∠ACE from both sides: α = ∠CAD − ∠ACE. ∠ACE = 2 and ∠CAD =

φ 2

(an inscribed angle is one-half of a

central angle that subtends the same arc; recall Chapter θ

4). Plug ∠ACE = 2 and ∠CAD = ∠ACE to get α =

φ−θ 2

.

233

φ 2

into α = ∠CAD −

Answer Key

Chapter 8 Answers 1. ∠CBD = 18.5°, ∠CBO = 90°, ∠BDO = ∠DBO = 71.5°. Notes: ∠CBD =

∠BOD 2

. Since ⃖���⃗ AC⊥ ���� BO, ∠CBO = 90°.

∠BDO ≅ ∠DBO because ∆BDO is isosceles (since BO = DO = R). Checks: ∠BDO + ∠DBO + ∠BOD =

71.5° + 71.5° + 37° = 180° and ∠DBO + ∠CBD = 71.5° + 18.5° = 90°.

2. ∠HFO = 8°, ∠FOH = 164°, ∠FHO = 8°, and ∠GFO = 90°. Notes: ∠EFH =

∠FOH 2

. Since ⃖���⃗ EG ⊥ ���� FO, ∠GFO = 90°.

∠HFO ≅ ∠FHO because ∆FHO is isosceles (since FO =

HO = R). Checks: ∠HFO + ∠FHO + ∠FOH =

8° + 8° + 164° = 180° and ∠EFH + ∠HFO = 82° + 8° = 90°.

3. ∠WXZ = 45°, ∠OXZ = 45°, ∠OZX = 45°, ∠OXW = 90°, OX = 5, and XZ = 5√2. Notes: ∠WXZ =

∠XOZ 2

. Since

⃖�����⃗ ⊥ ���� WY OX, ∠OXW = 90°. ∠OXZ ≅ ∠OZX because ∆OXZ is isosceles (since OX = OZ = R = 5). Since ∆OXZ is a

45°-45°-90° triangle, OX:OZ:XZ = 5:5:5√2 = 1:1:√2

(Volume 1, Chapter 5). Checks: ∠OXZ + ∠OZX + ∠XOZ = 45° + 45° + 90° = 180° and ∠WXZ + ∠OXZ = 234

Plane Geometry Practice Workbook with Answers, Volume 2

45° + 45° = 90°.

4. α = 17°. Notes: φ = 77°, θ = 43°, and α =

φ−θ 2

5. ∠LOM = 176°, ∠LON = 80°, ∠NLO = 50°, and ∠LNO = 50°. Notes: α = 36° and α =

∠LOM−∠MON 2

.

.

∠NLO ≅ ∠LNO because ∆LNO is isosceles (since LO = NO = R). Checks: ∠NLO + ∠LNO + ∠LON =

50° + 50° + 80° = 180° and ∠LOM + ∠MON + ∠LON =

176° + 104° + 80° = 360°.

6. ∠SOU = 98°, ∠SOT = 68°, ∠OSU = 41°, ∠OST = 56°, ∠RUS = 49°, and ∠RSU = 83°. Notes: α = 48° and α =

194°−∠SOU 2

. ∠OSU ≅ ∠OUS because ∆OSU is isosceles

(since OS = OU = R) and ∠OST ≅ ∠OTS because ∆OST is isosceles (since OS = OT = R). Since ⃖����⃗ RU ⊥ ���� OU, ∠OUR = 90°. Checks: ∠OSU + ∠OUS + ∠SOU = 41° + 41° + 98° = 180°, ∠OST + ∠OTS + ∠SOT = 56° + 56° + 68° =

180°, ∠SRU + ∠RUS + ∠RSU = 48° + 49° + 83° = 180°,

∠RUS + ∠OUS = 49° + 41° = 90°, and

∠SOU + ∠SOT + 194° = 98° + 68° + 194° = 360°. 235

Answer Key

7. CD = 10 and AO = √180 = 6√5. Notes: AD = AC + CD = 8 + 10 = 18, (AC)(AD) = AB 2 = AO2 − R2 , and

√180 = �(36)(5) = √36√5 = 6√5.

8. JM = 2, JO = �5 + 2√3, KM = √2, ∠KJM = 30°,

∠KOL = 150°, ∠LOM = 120°, and ∠MLO = 30°. Notes:

JL = JM + ML = 2 + √3, (JM)(JL) = JK 2 = JO2 − R2 , R =

D 2

2

= 2 = 1, α = 30°, and α =

∠KOL−∠KOM 2

. ∠LMO ≅ ∠MLO

because ∆LMO is isosceles (since LO = MO = R). Since ∆LMO is a 30°-30°-120° triangle, LO:MO:LM = 1:1:√3

(Volume 1, Chapter 5, Problem 12). One way to solve for 2

JM is to first write JM(JM + √3) = �1 + √3� = 4 + 2√3

(see the checks that follow). Distribute: JM 2 + JM√3 =

4 + 2√3. Subtract 4 and 2√3 from both sides:

JM 2 + JM√3 − 4 − 2√3 = 0. This is quadratic equation. Use the quadratic formula. Alternatively, there is an

easier to way to find that JM = 2. Draw line segment ����� MN, as shown on the next page. Quadrilateral KNMO is a square. MN = KO = R = 1, such that JN = √3. Use the 2

Pythagorean theorem. Checks: JK 2 = �1 + √3� = 236

Plane Geometry Practice Workbook with Answers, Volume 2

2

2

1 + 2(1)√3 + �√3� = 1 + 2√3 + 3 = 4 + 2√3 agrees 2

with JO − R = ��5 + 2√3� − 12 = 5 + 2√3 − 1 = 2

2

4 + 2√3. ∠MLO + ∠LMO + ∠LOM = 30° + 30° + 120° = 180° and ∠KOL + ∠LOM + ∠KOM = 150° + 120° + 90° = 360°.

9. α = 36°, ∠CAO = 90°, and ∠CBO = 90°.

⃖���⃗ ⊥ ���� Notes: α + 144° = 180°, ⃖���⃗ AC ⊥ ���� AO, and BC BO.

10. ∠DOF = 122°, ∠DFO = 29°, ∠FDO = 29°, and

DO, ∠DFE = 61°. Notes: ∠DOF + ∠DEF = 180°, ⃖����⃗ DE ⊥ ���� ����. Checks: ∠DFO + ∠FDO + ∠DOF = ⃖���⃗ ⊥ FO and EF 29° + 29° + 122° = 180°, ∠DFE + ∠FDE + ∠DEF =

61° + 61° + 58° = 180°, ∠EDF + ∠FDO = 61° + 29° = 90°, and ∠DFE + ∠DFO = 61° + 29° = 90°. 237

Answer Key

11. TV = 12, OU = 5, and OT = 13. Note: TU 2 = TV 2 = OT 2 − R2 .

12. ∠QOR = 120°, ∠QRO = 30°, ∠RQO = 30°, ∠PQR = 60°, ∠PRQ = 60°, OP = 12, PQ = 6√3, PR = 6√3, QR =

6√3, s = 4π, and t = 8π. Notes: ∠QPR + ∠QOR = 180°, ⃖���⃗ ⊥ ���� OQ, and ⃖���⃗ PR ⊥ ���� OR. ∠OQR ≅ ∠ORQ because ∆OQR is PQ isosceles (since OQ = OR = R). It follows that ∠PQR ≅

∠PRQ (since ∠PQR and ∠PRQ are complements to ∠OQR and ∠ORQ). Since ∆OQR is a 30°-30°-120°

triangle, OQ:OR:QR = 6:6:6√3 = 1:1:√3 (Volume 1,

Chapter 5, Problem 12). Since ∆POR is a 30°-60°-90°

triangle, OR:PR:OP = 6:6√3:12 = 1:√3:2 (Volume 1,

Chapter 5). Recall the arc length formula from Chapter 3. Convert 120° to

2π 3

rad to get s = 4π. Note that s + t =

4π + 8π = C = 2πR = 12π.

238

Plane Geometry Practice Workbook with Answers, Volume 2

13. A =

15π 2

≈ 23.6. Notes: First determine the radius of

the large circle. Draw ���� CO, as shown below. Since ∆BOQ

is a 30°-60°-90° triangle, BQ:BO:OQ = 1:√3:2 (Volume

1, Chapter 5, Problem 12), such that OQ = 2. Then CO =

OQ + CQ = 2 + 1 = 3 is the radius of the large circle. A=

5πR2 6

=

5π32 6

=

45π 6

=

15π 2

.

14. MN = 3√15, NP = 2√15, MP = 5√15, MO = 12, OQ =

8, and MQ = 20. Notes: ∆MNO~∆MPQ according to AA

(Volume 1, Chapter 4) since ∠NMO is shared and ∠MNO = ∠MPQ = 90°. Therefore, NO:MN:MO = 6

PQ:MP:MQ. Since NO = 2 = 3 and PQ =

10 2

= 5, NO:MN

= PQ:MP becomes 3:MO = 5:MQ. Write this in the form 3

5

= MQ. Since MQ = MO + OR + RQ = MO + 3 + 5 = MO

MO + 8, the ratio becomes

3

= MO

5

. Cross multiply

MO + 8

to get 3(MO + 8) = 5MO. Distribute: 3MO + 24 = 5MO. Subtract 3MO from both sides: 24 = 2MO. Divide by 2 239

Answer Key

3

5

on both sides: 12 = MO. Plug this into MO = MQ to get 3

5

= MQ. Cross multiply: 3MQ = 60. Divide by 3 on both 12

sides: MQ = 20. To find MN and MP, use the

Pythagorean theorem. Note that MN + NP = MP and √135 = √9√15 = 3√15.

2

Checks: MN 2 + NO2 = �3√15� + 32 = 32 (15) + 9 = 9(15) + 9 = 135 + 9 = 144 = 122 = MO2 and 2

MP 2 + PQ2 = �5√15� + 52 = 52 (15) + 25 =

25(15) + 25 = 375 + 25= 400 = 202 = MQ2 .

15. Main ideas: Of all the points on ⃖���⃗ AC, point B is the

closest to point O. (Since B lies on the circumference of the circle, BO equals the radius. The distance from any

other point on ⃖���⃗ AC to point O is clearly greater than the

radius because those points lie outside of the circle.) The shortest distance from point O to ⃖���⃗ AC is along a line segment that is perpendicular to ⃖���⃗ AC (recall Volume 1,

Chapter 7, Example 9). This method uses a proof from Volume 1 (Chapter 7, Example 9) to prove that

⃖���⃗ AC ⊥ ���� BO. Notes: There are other ways to prove this. However, beware that there are also a variety of

common ways to attempt this proof that are incorrect. 240

Plane Geometry Practice Workbook with Answers, Volume 2

Some incorrect attempts involve circular logic (that is, effectively assuming the end result to prove it). Other incorrect attempts omit an important step or

oversimplify the solution. It may also seem tempting to try to take advantage of symmetry, but it is difficult to formulate a proof based on this idea that is thorough and logically sound.

16. Main ideas: AO = BO = R. AO2 + AC 2 = CO2 and BO2 + BC 2 = CO2 according to the Pythagorean

theorem. Combine these to get AO2 + AC 2 = BO2 + BC 2 . Since AO = BO, it follows that AC = BC. Recall the definition of a kite from Chapter 1, Volume 9.

17. Main ideas: ∠BDO + ∠DBO + ∠BOD = 180°.

∠BDO ≅ ∠DBO since ∆BDO is isosceles (since BO = DO = R). Replace ∠BDO with ∠DBO in the previous

equation: 2∠DBO + ∠BOD = 180°. Divide by 2 on both sides: ∠DBO +

∠BOD 2

= 90°. Since ⃖���⃗ AC ⊥ ���� BO, ∠CBO = 90°,

such that ∠CBD + ∠DBO = 90°. Combine the two previous equations together: ∠DBO +

∠BOD 2

=

∠CBD + ∠DBO. Subtract ∠DBO from both sides: ∠CBD.

241

∠BOD 2

=

Answer Key

18. Main ideas: Draw line segment ���� BC and angle β, as

shown below. α + ∠CBD + ∠BCD = 180° (interior

angles of ∆BCD). ∠CBD + β = 180° (supplementary angles). α + ∠BCD = β (set the first two equations

equal and subtract ∠CBD from both sides). Subtract θ

∠BCD from both sides: α = β − ∠BCD. ∠BCD = 2 and β=

φ 2

(an inscribed angle is one-half of a central angle

that subtends the same arc; recall Chapter 4). Plug θ

∠BCD = 2 and β =

φ

into α = β − ∠BCD to get α = 2

φ−θ 2

.

19. Main ideas: Consider the diagram below, where ⃖����⃗ DE

is a secant (instead of a tangent). Apply the intersecting secants theorem from Chapter 7: (AD)(CD) =

(BD)(DE). Now imagine redrawing the secant such that

points B and E are closer together. As points B and E get very close together, the distances BD and DE get closer. In the limit that B and E actually merge together, BD = 242

Plane Geometry Practice Workbook with Answers, Volume 2

DE such that (BD)(DE) = BD2 . Note that thinking of

mathematical ideas in terms of limits is a very powerful tool. In fact, this form of thinking serves as the basis for

calculus. Many important formulas in higher-level math,

physics, engineering, and other fields are ultimately

derived by applying limits or by applying an operation that is defined in terms of a limit (namely, by either

applying the calculus operation known as a derivative

or by working with differential elements). It is also possible to show that (AD)(CD) = BD2 by working with

geometric principles and triangles in the given diagram. The second part of the proof, to show that BD2 =

DO2 − R2 , can be performed efficiently by applying the Pythagorean theorem to ∆BOD in the diagram given in the problem (not the diagram below).

243

Answer Key

20. Main ideas: α + 90° + θ + 90° = 360° because the interior angles of quadrilateral AOBC add up to 360°

(recall Volume 1, Chapter 9). Subtract 180° from both

sides to get α + θ = 180°. Central angles θ and φ form a full circle: θ + φ = 360°. Subtract φ from both sides:

θ = 360° − φ. Substitute this into the previous

equation: α + 360° − φ = 180°. Add φ to both sides:

α + 360° = 180° + φ. Subtract 180° from both sides:

α + 180° = φ. Add the equations α + θ = 180° and

α + 180° = φ together: 2α + θ + 180° = 180° + φ.

Subtract 180° from both sides: 2α + θ = φ. Subtract θ

from both sides: 2α = φ − θ. Divide by 2 on both sides:

α=

φ−θ 2

.

21. Main ideas: Draw line segment ���� CO to make two right triangles. The Pythagorean theorem gives R2 + BC 2 =

CO2 and R2 + AC2 = CO2 . It follows that BC = AC and that AC 2 = BC 2 = CO2 − R2 .

244

Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 9 Answers 1. r = 4 and R = 13. Inradius: First use the Pythagorean

theorem to see that AC = 26: BC 2 + AB 2 = 102 + 242 =

100 + 576 = 676 = 262 = AC 2 . The perimeter is P =

AB + BC + AC = 24 + 10 + 26 = 60. The base is BC = 10 and the height is AB = 24. The area is A = (10)(24) 2

=

to 120 =

240

2 60r 2

(BC)(AB)

Pr

2

=

= 120. The area is also A = 2 , which leads

. Multiply by 2 on both sides: 240 = 60r.

Divide by 60 on both sides 4 = r.

Circumradius: According to Thales’s theorem (Chapter

5), since ∠ABC is a right angle, AC is the diameter of the

circumcircle: AC = 2R, such that 26 = 2R and 13 = R.

Note: In the left diagram, ∆ABC is circumscribed about ⊙I, yet r is the inradius (not the circumradius). Why? The name of the radius makes sense from the

perspective that ⊙I is inscribed in ∆ABC (which is 245

Answer Key

equivalent to stating that ∆ABC is circumscribed about

⊙I). Similarly, in the right diagram, ∆ABC is inscribed in ⊙O, yet R is the circumradius (not the inradius). In this case, the radius is named from the perspective that ⊙O is circumscribed about ∆ABC (which is equivalent to

stating that ∆ABC is inscribed in ⊙O). In each case, the

name of the radius comes from the point of view of the

circle. If the circle is inscribed in the triangle, it is called the inradius (and the circle is called the incircle). If the

circle is circumscribed about the triangle, it is called the circumradius (and the circle is called the circumcircle). 2. r =

1

√

= 3

√3 3

and R =

2

√

= 3

2√3 . Inradius: 3

����I and Draw C

��� F I. Recall from Volume 1, Chapter 7 that the incenter is

located where the angle bisectors intersect. Therefore, ����I is an angle bisector. This shows that ∆ABC is a C 30°-60°-90° triangle. Recall from Volume 1, Chapter 5

that the sides of a 30°-60°-90° triangle come in the ratio

1:√3:2, meaning that the side opposite to the 60° angle is √3 times larger than the side opposite to the 30°

angle. This means that CF is √3 times larger than IF. Since CF = 1, this means that IF = 246

1

√3

. Multiply the

Plane Geometry Practice Workbook with Answers, Volume 2

numerator and denominator each by √3 to rationalize

the denominator: IF = =

1

= 3

√

√3 . 3

1

√

= 3

1 √3 √3 √3

√3 . 3

=

Note that r = IF

����. Recall from Volume 1, Circumradius: Draw ���� CO and FO Chapter 7 that the circumcenter is located where the perpendicular bisectors intersect. Therefore, ���� FO is a

perpendicular bisector. (For an equilateral triangle, an

angle bisector and perpendicular bisector are the same. However, for scalene triangles, angle bisectors and

perpendicular bisectors are different.) This shows that ∆ABC is a 30°-60°-90° triangle. Recall from Volume 1,

Chapter 5 that the sides of a 30°-60°-90° triangle come in the ratio 1:√3:2, meaning that the hypotenuse is

twice as large as the side opposite to the 30° angle. This

means that CO is 2 times larger than FO. Since OF = IF = 1

√

= 3

√3 , 3

this means that R = CO =

247

2

√

= 3

2√3 . 3

Answer Key

4

3. r = 3 and R =

25 6

. Inradius: Draw point D as the

���� (see the diagram that midpoint of ���� BC and draw AD

follows.) The perimeter is P = 5 + 5 + 8 = 18. The base is b = 8. Apply the Pythagorean theorem to find the

height: BD2 + AD2 = AB 2 . Plug BD = 4 and AB = 5 into

this equation: 42 + AD2 = 52 . Simplify: 16 + AD2 = 25. Subtract 16 from both sides: AD2 = 25 − 16 = 9.

Square root both sides: AD = √9 = 3 = h is the height.

The area is A = Pr

bh 2

=

(8)(3) 2

=

24 2

= 12. Area also equals A

= 2 . Plug A = 12 and P = 18 into this formula: 12 =

18r 2

Multiply by 2 on both sides: 24 = 18r. Divide by 18 on 24

4

.

both sides: 18 = 3 = r. An alternative solution is to use

the triangle bisector theorem (Volume 1, Chapter 7): DI 4

AI

= 5 . Note that AI + DI = AD = 3, such that AI =

3 − DI. Plug this into the previous equation:

DI

9DI = 12. Divide by 9 on both sides: DI =

= 3.

4

=

3−DI 5

.

Cross multiply: 5DI = 3(4) − 4DI. Add 4DI to both sides: 12 9

4

Circumradius: Draw points P and R as the midpoints of ���� AC (see the diagram that follows). Apply the AB and ���� Pythagorean theorem: BD2 + DO2 = BO2 . The 248

Plane Geometry Practice Workbook with Answers, Volume 2

circumradius is the distance from point O to each

vertex: AO = BO = CO = R. Note that AD + DO = AO = R. Plug AD = 3 (see the inradius notes) into this

equation: 3 + DO = R. Subtract 3 from both sides: DO =

R − 3. Plug this, BD = 4, and BO = R into BD2 + DO2 =

BO2 .

42 + (R − 3)2 = R2 Apply the “foil” method: (a − b)2 = a2 − 2ab + b2 . 16 + R2 − 6R + 9 = R2

Subtract R2 from both sides: 16 − 6R + 9 = 0. Combine like terms and add 6R to both sides: 25 = 6R. The circumradius is

4. r =

√2 2

25 6

= R.

and R = 1. Inradius: The diameter of the

incircle is AD = √2. Divide the diameter by 2 to find the

inradius: r =

since

1 √2 √2 √2

=

AD 2

√2 . 2

=

√2 . 2

The answer is equivalent to 249

1

√2

Answer Key

Circumradius: The diagonal of the square equals the diameter of the circumcircle. Use the Pythagorean

theorem to find the diagonal: AB 2 + BC 2 = AC 2 . Plug AB 2

2

= BC = √2 into this equation: �√2� + �√2� = 2 + 2 =

4 = AC 2 . Square root both sides: √4 = 2 = AC. Divide

the diameter of the circumcircle by 2: R = 28

7

5. r = √3 and R = � 3 = 2�3 =

2√21 . 3

AC 2

2

= 2 = 1.

Inradius: Draw

vertical line segments ���� AE and ���� DF to divide the trapezoid into two triangles and a rectangle (see the diagram on

the next page). Since the top side is AD = 2, the base is

BC = 6, and 6 − 2 = 4, it follows that BE = EF = CF = 2. Use the Pythagorean theorem to find the height:

AE 2 + BE 2 = AB 2 . Plug BE = 2 and AB = 4 into this equation: AE 2 + 22 = 42 . Simplify: AE 2 + 4 = 16.

Subtract 4 from both sides: AE 2 = 16 − 4 = 12. Square root both sides: AE = √12 = �(4)(3) = √4√3 = 2√3.

The inradius is r = IM =

AE 2

=

2√3 2

= √3.

Circumradius: Apply the Pythagorean to ∆BMO in the

diagram on the following page: BM 2 + MO2 = BO2 . The

circumradius is AO = BO = CO = DO = R. Plug BM = 3 250

Plane Geometry Practice Workbook with Answers, Volume 2

and BO = R into the equation: 32 + MO2 = 9 + MO2 =

R2 . Apply the Pythagorean to ∆DJO in the diagram

below: DJ 2 + JO2 = DO2 . Plug DJ = 1 and DO = R into

the equation: 1 + JO2 = R2 . Note that JO + MO = JM.

Recall from the inradius notes that JM = AE = 2√3, such

that JO + MO = 2√3. Subtract MO from both sides: JO =

2√3 − MO. Plug this into 1 + JO2 = R2 . 2

1 + �2√3 − MO� = R2

Apply the “foil” method: (a − b)2 = a2 − 2ab + b2 . 2

1 + (2)2 �√3� − 2MO�2√3� + MO2 = R2 1 + 4(3) − 4MO√3 + MO2 = R2 1 + 12 − 4MO√3 + MO2 = R2 13 − 4MO√3 + MO2 = R2

Subtract the equation 9 + MO2 = R2 (from earlier) from

the equation above.

13 − 4MO√3 + MO2 − 9 − MO2 = R2 − R2 4 − 4MO√3 = 0 4MO√3 = 4

MO =

MO√3 = 1 1

√3

=

1 √3

√3 √3

251

=

√3 3

Answer Key

Plug MO =

�

1

√

2 2 into 9 + MO = R (from earlier). 3 2

1

1 27 1 28 9+� � =9+ = + = = R2 3 3 3 3 √3

7 28 7 2√7 2√7 √3 2√21 � � = √4 =2 = = = =R 3 3 3 3 √3 √3 √3

The answer R =

2√21

28

equivalent to � 3 .

6. r = 12 and R =

3

35 2

has a rational denominator and is

= 17.5. Inradius: Draw �E���I and ��� F I,

���� and ���� which are perpendicular to AD AB (but which are NOT perpendicular bisectors; finding the inradius ���, not perpendicular involves angle bisectors, like AI

bisectors, whereas finding the circumradius involves ����), as shown in the perpendicular bisectors, like KO

diagram that follows. Note that AEIF is a square: AE =

FI = AF = EI = r. Show that ∆DEI~∆BFI~∆ABD (Volume 252

Plane Geometry Practice Workbook with Answers, Volume 2

1, Chapter 4). It follows that AD:AB:BD = DE:EI:DI. Express AD:AB = DE:EI as follows: AD DE = AB EI

Note that AE + DE = AD. Since AE = FI = r and AD = 21,

this may be written as r + DE = 21. Subtract r from both

sides: DE = 21 − r. Plug AD = 21, AB = 28, EI = FI = r, and DE = 21 − r into the previous ratio. 21 21 − r = r 28

Note that 21/28 reduces to 3/4. (Divide 21 and 28 each by 7.)

3 21 − r = r 4

Cross multiply: 3r = 84 − 4r. Add 4r to both sides: 7r =

84. The inradius is r = 12.

Circumradius: Use the Pythagorean theorem to find BD: AD2 + AB 2 = BD2 . Plug AD = 21 and AB = 28 into this equation: 212 + 282 = 441 + 784 = 1225 = BD2 .

Square root both sides: √1225 = 35 = BD. The

circumradius (point O) is equidistant from the vertices:

R = AO = BO = CO = DO. Since DO = BO and DO + BO = BD, it follows that R = BO =

BD 2

253

=

35 2

= 17.5.

Answer Key

7. r = 1 + √2 ≈ 2.41 and R = �4 + 2√2 =

�2�2 + √2� ≈ 2.61.

Octagon: Before finding either the inradius or the circumradius, it is useful to find the height of the

octagon. See the middle diagram on the following page,

where the top and right edges of the octagon have been extended. The height of the octagon is PQ. Note that

∆BCP and ∆DFQ are each 45°-45°-90° triangles with a

hypotenuse of 2. Since the hypotenuse of a 45°-45°-90°

triangle is √2 times larger than either leg, PC = DQ = √2 such that the hypotenuse is BC = DF = √2√2 = 2. The

height of the octagon is PQ = CP + CD + DQ = √2 + 2 + √2 = 2 + 2√2 = 2�1 + √2�.

Inradius: The inradius is one-half of the height of the octagon: r = TI =

PQ 2

=

2�1+√2� 2

254

= 1 + √2.

Plane Geometry Practice Workbook with Answers, Volume 2

Circumradius: In the diagram on the right below, OU =

1 + √2 is one-half of the height of the octagon and BU = AU = 1. Use the Pythagorean theorem. BU 2 + OU 2 = BO2 2

12 + �1 + √2� = BO2

Apply the “foil” method: (a + b)2 = a2 + 2ab + b2 . 1 + 1 + 2√2 + 2 = BO2

4 + 2√2 = 2�2 + √2� = BO2

�4 + 2√2 = �2�2 + √2� = BO

8. A = 3π −

9 √3 2

9 √3 2

−

9π 4

9

π

= 2 �√3 − 2� ≈ 0.726 and A =

= 3 �π −

3√3 � 2

≈ 1.63.

Left diagram: Draw and label point K as the midpoint of �EF ��� (see the diagram on the next page). ∆EIK is a 30°60°-90° triangle. The hypotenuse is EI = √3 (because

∆EFI is an equilateral triangle; recall Volume 1, Chapter

10, Example 4). Since EK is opposite to the 30° angle, EK 255

Answer Key

is one-half of the hypotenuse: EK =

EI

=

2

√3 . 2

Use the

Pythagorean theorem to find the inradius: IK 2 + EK 2 = 2

EI . Plug in EI = √3 and EK = 3

√3 : 2 3

√3 � 2

2

IK + �

2

2

= �√3� .

Simplify: IK 2 + 4 = 3. Subtract 4 from both sides: IK 2 = 3

3−4= √9 √4

12

3

4

3

9

9

− 4 = 4. Square root both sides: IK = �4 =

= 2. Since ∆EIK has a base of EK = 1 √3

3

3

√3 2

and a height of

3√3 . 8

The hexagon

36√3

=

IK = , the area of ∆EIK is � � � � = 2 2 2 2

can be divided into 12 triangles congruent with ∆EIK.

The area of the hexagon is 12 � agrees with the formula

3L2 √3 2

3√3 � 8

=

=

2

8

3�√3� √3 2

=

9√3 2

. This

3(3)√3 2

=

9√3 2

from Volume 1, Chapter 10, Example 4. The inradius is r 3

2

3 2

= IK = 2. The area of the incircle is πr = π �2� = 9

π �4� =

9π 4

. Subtract the area of the incircle from the

area of the hexagon to find the area of the shaded region:

π 9√3 9π 9 − = �√3 − � ≈ 0.726 A= 2 4 2 2 256

Plane Geometry Practice Workbook with Answers, Volume 2

Right diagram: The circumradius is R = OE = OF = √3

(see the diagram below). The area of the circumcircle is 2

πR2 = π�√3� = π(3) = 3π. This is the same hexagon,

so it has the same area found previously:

9√3 2

the area of the hexagon from the area of the

. Subtract

circumcircle to find the area of the shaded region: 9√3 ≈ 1.63 A = 3π − 2

9. (A) a = d + e, b = e + f, and c = d + f. Notes: AD = AF = d, BD = BE = e, and CE = CF = f. Recall this property of intersecting tangents from Chapter 8. a = AB = AD + BD = d + e, b = BC = BE + CE = e + f, and c = AC = AF + CF = d + f. P

(B) 2 = d + e + f. Notes: P = a + b + c =

(d + e) + (e + f) + (d + f) = 2d + 2e + 2f. Divide by 2 P

on both sides to get 2 = d + e + f. 257

Answer Key

P

P

P

(C) d = 2 − b, e = 2 − c, and f = 2 − a. Notes: Since P

P

b = e + f, 2 = d + e + f = d + b, such that d = 2 − b. P

P

Since c = d + f, 2 = d + e + f = c + e, such that e = 2 − c. P

P

Since a = d + e, 2 = d + e + f = a + f, such that f = 2 − a.

(D) A = �

Pdef

Chapter 6). P

2

. Notes: Recall Heron’s formula (Volume 1,

P P P P A = � � − a� � − b� � − c� 2 2 2 2 P

P

Since d = 2 − b, e = 2 − c, and f = 2 − a, this becomes A=�

Pdef 2

.

10. 2r = AB + BC − AC and 2R = AC. Incircle: Label the

left diagram the same way as Problem 9. Since ∠ABC is a right triangle, BDIE is a square and r = BD = EI = BE =

DI. Using the left figure on the next page, AB + BC − AC

= d + e + e + f − d − f = 2e = 2r.

Circumcircle: Apply Thales’s theorem (Chapter 5) to see

that AC is the circumdiameter. 258

Plane Geometry Practice Workbook with Answers, Volume 2

11. Main ideas: This was actually proven in Chapter 4. See the solution to Problem 8 of Chapter 4: α + β =

180°. A similar solution also works for the other pair of opposite interior angles.

12. Main ideas: Part 1: ∠ABD ≅ ∠CBX (given) and

∠ADB ≅ ∠ACB (each inscribed angle subtends the arc

from A to B; recall Chapter 4). ∆ABD~∆BCX according to AA (recall Volume 1, Chapter 4). The sides of similar

triangles come in the same ratio: AD:BD:AB = CX:BC:BX.

Express AD:BD = CX:BC as fractions. AD CX = BD BC Cross multiply: (AD)(BC) = (BD)(CX).

Part 2: ∠BAC ≅ ∠BDC (each inscribed angle subtends the arc from B to C) and ∠ABX ≅ ∠CBD (start with

∠ABX = ∠ABD + ∠DBX and ∠CBD = ∠CBX + ∠DBX,

then plug in ∠ABD ≅ ∠CBX to see that ∠ABX ≅ ∠CBD). 259

Answer Key

∆ABX~∆BCD according to AA. The sides of similar

triangles come in the same ratio: AB:AX:BX = BD:CD:BC. Express AB:AX = BD:CD as fractions. AB BD = AX CD

Cross multiply: (AB)(CD) = (AX)(BD).

Part 3: AX + CX = AC, such that AX = AC − CX. Plug this

into the equation from Part 2: (AB)(CD) = (AC − CX)(BD). Distribute: (AB)(CD) =

(AC)(BD) − (CX)(BD). Recall from Part 1 that (AD)(BC) = (BD)(CX). Plug this into the previous equation:

(AB)(CD) = (AC)(BD) − (AD)(BC). Add (AD)(BC) to both sides: (AB)(CD) + (AD)(BC) = (AC)(BD). Note:

The proof remains valid even if BX is removed from the diagram.

13. Notes: This problem asks for a discussion of ideas, which is NOT the same as a full, structured proof.

Although proofs are a common and important aspect of geometry, it is also a useful skill to be able to discuss

ideas. There are multiple ways to answer this question

satisfactorily. Following is a sample of a few ideas that might be discussed.

260

Plane Geometry Practice Workbook with Answers, Volume 2

• O is the circumcenter. P, Q, R, and S below are the

midpoints of the sides. • ���� OP, ���� OQ, ���� OR, and ���� OS are the perpendicular bisectors. These are concurrent at O.

• AO = BO = CO = DO = R is the circumradius.

• For any triangle, the three perpendicular bisectors

are concurrent (Volume 1, Chapter 7). • In ∆ABC (note that side ���� AC is not drawn), perpendicular bisectors ���� OP and ���� OQ intersect at O. ���� and ���� • In ∆BCD, perpendicular bisectors OQ OR intersect at O.

���� • In ∆ACD, perpendicular bisectors ���� OR and OS intersect at O.

���� • In ∆ABD, perpendicular bisectors ���� OP and OS intersect at O.

• These four triangles have the same circumradius: AO = BO = CO = DO = R.

• These four triangles have ���� OP, ���� OQ, ���� OR, and ���� OS in common.

261

Answer Key

14. (A) Opposite interior angles of a parallelogram are congruent (Volume 1, Chapter 9). Opposite interior

angles of a cyclic quadrilateral add up to 180° (Problem 11). In order to meet both criteria, a cyclic

parallelogram must have four 90° interior angles

(which is a rectangle or square). A cyclic rhombus must therefore be a square.

(B) Consider the cyclic trapezoid below. A trapezoid has ���� ∥ BC ����. According to the one pair of parallel sides: AD

parallel postulate (Volume 1, Chapter 1), the same-side

interior angles are supplementary: ψ + χ = 180° and

φ + θ = 180°. For a cyclic quadrilateral, opposite

interior angles of a cyclic quadrilateral add up to 180° (Problem 11): ψ + θ = 180° and φ + χ = 180°.

Compare ψ + χ = 180° with φ + χ = 180° to see that

ψ = φ. Compare ψ + χ = 180° with ψ + θ = 180° to see 262

Plane Geometry Practice Workbook with Answers, Volume 2

that χ = θ. The conditions ψ = φ and χ = θ are characteristic of an isosceles trapezoid.

(C) One pair of opposite interior angles are congruent for a kite (Volume 1, Chapter 9). Opposite interior

angles of a cyclic quadrilateral add up to 180° (Problem 11). In order to meet both criteria, one pair of opposite interior angles must be congruent right angles for a cyclic kite.

15. Main ideas: In the diagram that follows, AD = d, AP

= AS = e, BP = BQ = f, CQ = CR = g, and DR = DS = h. Recall this property of intersecting tangents from

Chapter 8. Compare AB + CD = e + f + g + h with

BC + AD = f + g + e + h to see that AB + CD = BC + AD. Since P = AB + CD + BC + AD and AB + CD = BC + AD,

P

it follows that AB + CD and BC + AD are each equal to 2. 263

Answer Key

16. Notes: This problem asks for a discussion of ideas, which is NOT the same as a full, structured proof.

Although proofs are a common and important aspect of geometry, it is also a useful skill to be able to discuss

ideas. There are multiple ways to answer this question satisfactorily. Following is a sample of a few ideas that might be discussed.

• I is the incenter. P, Q, R, and S below are the points

where the sides are tangent to the circle. (They are NOT midpoints of the sides, in contrast to Problem

13.) • ��� AI, ���� B I, �C���I, and ���� D I are the angle bisectors. These are concurrent at I.

• IP = IQ = IR = IS = r is the inradius.

• Recall from Chapter 8, regarding intersecting ���� ≅ �AS ���, ���� ����, ���� BP ≅ BQ CQ ≅ ���� CR, and tangents, that AP 264

Plane Geometry Practice Workbook with Answers, Volume 2

���� DR ≅ ���� DS. (Also recall Volume 1, Chapter 7, Problem

25.)

���� ≅ �AS ��� and IP = IS = r. • In quadrilateral APIS, AP • In quadrilateral BQIP, ���� BP ≅ ���� BQ and IQ = IP = r. This is the same point I and inradius since IP is common.

• In quadrilateral CRIQ, ���� CQ ≅ ���� CR and IR = IQ = r. This is the same point I and inradius since IQ is common.

���� ≅ ���� • In quadrilateral DSIR, DR DS and IS = IR = r. This is the same point I and inradius since IR is common.

Note: Beware that, in the general case depicted below, A, I, and C are NOT collinear, and B, I, and D are NOT

collinear. However, A, P, and B are collinear, B, Q, and C are collinear, C, R, and D are collinear, and D, S, and A

are collinear. (If it seems like any of the four sides may be bent, it is an optical illusion.) 265

Answer Key

17. (A)/(B) Main ideas: Both are a direct consequence

of the result of Problem 15. Each pair of opposite sides must add up to one-half of the perimeter.

18. (A) Yes. A tangential trapezoid can have two right

angles, as shown below. In order for a trapezoid to be a tangential trapezoid, AB + CD must equal BC + AD (Problem 15).

(B) No. A cyclic trapezoid can not have a right angle. (A

cyclic trapezoid is isosceles. If an isosceles trapezoid has

one right angle, it would technically be a rectangle

instead of a trapezoid.) See Problem 14B. This follows from Problem 11.

266

Plane Geometry Practice Workbook with Answers, Volume 2

(C) Yes. An isosceles trapezoid can be a tangential

quadrilateral. An example can be found in Problem 5. In order for a trapezoid to be a tangential trapezoid, AB + CD must equal BC + AD (Problem 15).

(D) Yes. An isosceles trapezoid can be a cyclic

quadrilateral. An example can be found in Problem 5.

Every isosceles trapezoid is a cyclic trapezoid. See the solution to Problem 14B.

19. Main ideas: Draw angle bisectors ��� AI, ���� B I, �C���I, and ���� D I, ����, �I��X�, �I��� Y, and ��� I Z are NOT as shown below. Note that IW

perpendicular bisectors (but they are perpendicular to the sides of the kite; they just do not bisect the sides).

Note that ∠WAX, ∠WIX, ∠YCZ, and ∠YIZ are generally

NOT right angles (but ∠AWI, ∠AXI, ∠CYI, and ∠CZI are right angles). ∆AIW ≅ ∆AIX according to AAS (∠AWI

and ∠AXI are each right angles, ∠WAI ≅ ∠XAI because ��� AI is shared). According AI is an angle bisector, and side ���

���� ≅ �I��X�. to the CPCTC (recall Volume 1, Chapter 3), IW ����I and �C���I can be used to Similarly, angle bisectors B

show that pairs of triangles are congruent, and the Y and I���� Y ≅ ��� I Z. CPCTC can be used to show that I���X� ≅ �I��� This allows a circle to be drawn with an incenter at

point I and an inradius of r = IW = IX = IY = IZ which is 267

Answer Key

tangent to the sides at points W, X, Y, and Z (where line Y, and ��� I Z are perpendicular to the segments ���� IW, �I��X�, �I���

sides).

268

Plane Geometry Practice Workbook with Answers, Volume 2

Chapter 10 Answers 1. V = 64, S = 96, f = 4√2, and d = 4√3. Notes: V = L3 , S = 6L2 , f = √L2 + L2 = √2L2 = L√2, and d =

√L2 + L2 + L2 = √3L2 = L√3. Use the Pythagorean

theorem to find the diagonals. Volume is measured in cubic units (like meters cubed or feet cubed) and

surface area is measured in square units (like square meters or square feet). Most of the problems in this book do not state specific units (they are just

mathematical units), but it is a good habit to think about whether the units would be cubed or squared for

volume or surface area (this helps to ensure that the

formulas do not get mixed up, and including units with the final answer would be worth points in a science course).

2. V = 288π and S = 144π. Notes: V = D

S = 4πR2 . R = 2 =

12 2

4πR3 3

and

= 6. The volume inside of a sphere

equals the volume of a ball that has the same radius. 269

Answer Key

3. V = 144, S = 192, a = 3√17, b = 5, c = 4√10, and d = 13. Notes: V = LWH, S = 2LW + 2WH + 2LH,

a = √L2 + W 2 , b = √W 2 + H 2 , c = √L2 + H 2 , and

d = √L2 + W 2 + H 2 . One possibility is L = 12, W = 3,

and H = 4 (but any combination would suffice, such as

L = 4, W = 12, and H = 3).

a = �122 + 32 = √144 + 9 = √153 = �(9)(17) = √9√17 = 3√17

b = �32 + 42 = √9 + 16 = √25 = 5

c = �122 + 42 = √144 + 16 = √160 = �(16)(10) = √16√10 = 4√10

d = �122 + 32 + 42 = √144 + 9 + 16 = √169 = 13

4. V = 3√3 and S = 18 + 7√3. Notes: V = Bh. The

triangle is a 30°-60°-90° triangle. The sides of the 30°-

60°-90° triangle are b = 1, a = √3, and c = 2 (Volume 1, 1

Chapter 5). The area of the triangle is B = 2 (1)�√3� =

√3 . 2

The height of the prism (NOT the height of the 270

Plane Geometry Practice Workbook with Answers, Volume 2

triangle) is h = 6 (the perpendicular distance between the triangular faces). To find the surface area, find the area of all 5 faces and add them together: • front face: A = • back face: A =

√3 2

√3 2

• left face: A = 6a = 6√3

• right face: A = 6c = 6(2) = 12

• bottom face: A = 6b = 6(1) = 6 S=

5. V =

√3 2

1024π 3

+

√3 2

+ 6√3 + 12 + 6 = √3 + 6√3 + 18 = 18 + 7√3

1 4πR3

and S = 192π. Notes: V = 2

3

=

4πR3 6

because a hemisphere has one-half the volume of a sphere and S =

4πR2 2

+ πR2 = 2πR2 + πR2 = 3πR2

because the top of the hemisphere has one-half the

surface area of a sphere and the base of the hemisphere is a circle (so the area of the circle, πR2 , must be 271

Answer Key

included in order to find the total surface area of the hemisphere).

6. d = 25 mi. Notes: d = √162 + 152 + 122 =

√256 + 225 + 144 = √625 = 25. Make a rectangular prism with L = 16, W = 15, and H = 12. Find the body diagonal. 4

7. V = 3 and S = 8. Notes: The base area is the area of 2

2

the square: B = CD = �√2� = 2. The volume is V = Bh 3

=

(2)(2) 3

4

= 3. ∆CDG is a 45°-45°-90° triangle. CG = DG

= BG = EG = 1 such that CG:DG:CD = 1:1:√2. Use the Pythagorean theorem to find AC.

AC = √AG 2 + CG 2 = √22 + 12 = √4 + 1 = √5

∆ACD is isosceles. Divide ∆ACD into two right triangles

(right diagram on the next page). Point F is the midpoint of CD: CF = DF = theorem to find AF. AF =

√AC2 �

−

10 2

CF 2 1

CD 2

=

√2 . 2

Use the Pythagorean 2

2

2 √2 = ��√5� − � 2 � = �5 − 4 = 9

− 2 = �2 =

3

√

= 2

272

3 √2 √2 √2

=

3√2 2

Plane Geometry Practice Workbook with Answers, Volume 2

1

1

The area of ∆ACD is 2 (CD)(AF) = 2 �√2� � 3√2√2 (2)(2)

=

3(2) 4

6

3

3√2 � 2

=

= 4 = 2. The total surface area of the

pyramid equals the area of the base plus 4 times the area of ∆ACD.

3

S = 2 + 4 �2� = 2 +

8. V =

45 2

12 2

=2+6=8

= 22.5 and S = 54. Notes: First find the

number of cubes (which is NOT the final answer). The

rectangular prism is 10 cubes wide by 6 cubes deep by

3 cubes high: (10)(6)(3) = 180. The length of each cube 1

3

1 3

1

is L = 0.5 = 2. The volume of one cube is L = �2� = 8.

Multiply the number of cubes by the volume of each cube to find the volume of the rectangular prism: 1

(180) � � = 8

180 8

=

45 2

= 22.5. To find the surface area of

the rectangular prism, first determine how many

squares are showing on its surface (which is NOT the final answer): 2(10)(6) + 2(10)(3) + 2(6)(3) = 273

Answer Key

120 + 60 + 36 = 216. The length of each square is 1

1 2

2

1

L = 0.5 = 2. The area of one square is L = �2� = 4.

Multiply the number of squares on the surface by the area of each square to find the surface area of the 1

rectangular prism: (216) �4� =

216 4

= 54.

9. V = 1120. Notes: First find the number of cubes in the pyramid: 12 + 22 + 32 + 42 + 52 + 62 + 72 =

1 + 4 + 9 + 16 + 25 + 36 + 49 = 140. The length of

each cube is L = 2. The volume of one cube is L3 = 23 = 8. Multiply the number of cubes by the volume of each cube to find the volume of the pyramid: (140)(8) =

1120.

10. V = 36π and S = 44π. Notes: V = πR2 h and S = 2πRh + 2πR2 = 2πR(h + R). R = 2 and h = 9. 11. V =

8π√3 3

and S = 12π. Notes: ∠CBO =

∠ABC 2

=

30°. ∆BCO is a 30°-60°-90° triangle: CO:BO:BC =

60° 2

=

2:2√3:4 (Volume 1, Chapter 5). R = CO = 2 and h = BO = 2√3.

274

Plane Geometry Practice Workbook with Answers, Volume 2

V=

πR2 h 3

=

π(2)2 �2√3� 3

=

π(4)�2√3� 3

=

8π√3 3

S = πR2 + πR√R2 + h2 = πR�R + √R2 + h2 � 2

S = π2 �2 + �22 + �2√3� � 2

S = 2π �2 + �4 + (2)2 �√3� �

S = 2π �2 + �4 + (4)(3)� = 2π�2 + √4 + 12�

S = 2π�2 + √16� = 2π(2 + 4) = 2π(6) = 12π

12. V = 24√3 and S = 72 + 8√3. Notes: V = Bh. Split

∆ABC into two 30°-60°-90° triangles, as shown on the

next page. The sides of each 30°-60°-90° triangle are BM = MC = 2, AM = 2√3, and AB = AC = 4 (Volume 1, 1

Chapter 5). The area of ∆ABC is B = (4)�2√3� = 4√3. 2

The height of the prism (NOT the height of the triangle) is h = 6 (the perpendicular distance between the

triangular faces). To find the surface area, find the area of all 5 faces and add them together: • triangular faces: A = 4√3

• rectangular faces: A = 4(6) = 24 275

Answer Key

S = 2�4√3� + 3(24) = 72 + 8√3 = 8�9 + √3�

13. V = 54√6 and S = 108√3. Notes: BG = CG = DG = 6 is given. ∆ABC ≅ ∆ACD ≅ ∆ABD ≅ ∆BCD for a regular 2

tetrahedron. Since ∆ACD ≅ ∆BCD, BG = 3AE in the

figures on the next page (because the centroid is located 2/3 of the length of a median, such as AE, from a 3

vertex). Multiply by 3/2 on both sides: 2BG = AE. Plug 3

in BG = 6 to get AE = 2 (6) =

18 2

= 9. ∆ACE is a 30°-60°-

90° triangle. The side opposite to 60° (AE) is √3 times

larger than the side opposite to 30° (CE). Therefore, CE =

AE

√

= 3

9

√

= 3

9 √3 √3 √3

=

9√3 3

= 3√3 = CE and CD = CE + DE

= 3√3 + 3√3 = 6√3. Since the tetrahedron is regular, AB = AC = AD = BC = BD = CD = 6√3. Use the

Pythagorean theorem to find AG. 276

Plane Geometry Practice Workbook with Answers, Volume 2

2

AG = √AD2 − DG 2 = ��6√3� − 62 = 2

�(6)2 �√3� − 36 = �(36)(3) − 36

AG = √108 − 36 = √72 = �(36)(2) = √36√2 = 6√2 1

1

∆ACE has area 2 (CD)(AE) = 2 �6√3�(9) = 27√3. Since

∆ACE ≅ ∆BCD, the base area is B = 27√3. The height is h = AG = 6√2. Note that the height of the tetrahedron (AG) is different from the height used to find the base

area (since the tetrahedron’s height is not the same as the height of one of its faces). The volume of the tetrahedron is:

V=

Bh 3

=

�27√3��6√2� 3

= 54√6

The total surface area of the tetrahedron is 4 times the base area (since the 4 faces of the tetrahedron are congruent).

S = 4�27√3� = 108√3

277

Answer Key

14. V =

8√2 3

and S = 8√3. Notes: Divide the octahedron

into two congruent square pyramids, as shown below. (Point E is in front of the page, whereas point F is

behind the page.) The 8 triangular faces are equilateral triangles. Divide ∆ADE into two 30°-60°-90° triangles: DM = 1, AM = √3, and AD = 2. The area of ∆ADE is 1 2

(2)�√3� = √3. The surface area of the octahedron is S

= 8√3. ∆DGE is a 45°-45°-90° triangle: DG = GE = √2

and DE = 2 (since √2√2 = 2). Use the Pythagorean

theorem to find AG.

2

AG = √AE 2 − GE 2 = �22 − �√2� = √4 − 2 = √2 = h The pyramid has base area (of the square) B = L2 = DE 2 = 22 = 4 and volume

Bh 3

=

(4)�√2� 3

=

4√2 . 3

The

volume of the octahedron is twice this value: V = 2�

4√2 � 3

=

8 √2 3

.

278

Plane Geometry Practice Workbook with Answers, Volume 2

15. S = 60. Note: The dodecahedron has 12 sides: S = 12A = 12(5) = 60.

16. S = 20√3. Notes: Every edge of the regular

icosahedron has length 2. Every face is an equilateral triangle, like ∆ABC below. Divide ∆ABC into two 30°-

60°-90° triangles: BM = 1, AM = √3, and AB = 2. The 1

1

area of ∆ABC is 2 (BC)(AM) = 2 (2)�√3� = √3. Since an

icosahedron has 20 faces, the surface area of the icosahedron is S = 20√3.

17. V = 105π and S = 224π. Notes: To find the volume

of the metal, subtract the volume of the inner cylinder from the volume of the outer cylinder. The radii are 3

and 4. The volume of each cylinder is V = πR2 h, where

the corresponding radius is used. For both, h = 15. V = π(4)2 (15) − π(3)2 (15) = π(16)(15) − π(9)(15) = 240π − 135π = 105π 279

Answer Key

To find the total surface area of the pipe, first find the surface area of each part:

• outside surface: 2πRh = 2π(4)(15) = 2π(60) = 120π

• inside surface: 2πRh = 2π(3)(15) = 2π(45) = 90π

• left end: π(4)2 − π(3)2 = 16π − 9π = 7π (subtract the inner circle from the outer circle)

• right end: π(4)2 − π(3)2 = 16π − 9π = 7π

The total surface area of the pipe is: S = 120π + 90π + 7π + 7π = 224π b

S

3

18. a = 21/3 = √2 ≈ 1.26 and S left = 22/3 ≈ 1.59. Notes: right

Let b = the radius of the left sphere and a = the radius of the right sphere. The volumes are Vleft =

Vright =

4πa3 3

4πb3 3

and

. The left volume is twice the right volume:

Vleft = 2Vright . Combine these equations together: 2� 3

4πa3 3

4πb3 3

=

�. Multiply both sides by 3 and divide by 4π to get 3

3

b 3

b = 2a . Divide by a on both sides: � � = 2. Cube a b

3

root both sides: = 21/3 = √2. The surface areas are a 280

Plane Geometry Practice Workbook with Answers, Volume 2

S

4πb2

Sleft = 4πb2 and Sright = 4πa2 . Divide: S left = 4πa2 = b 2

right

2

�a� = �21/3 � = 22/3 .

19. V = 300√3 and S = 230 + 120√3. Notes: ∆ABC is a

30°-60°-90° triangle. BC:AC:AB = 4:4√3:8. The base

area is B = (5)(15) = 75. The height is h = AC = 4√3. The volume of the prism is V = Bh = (75)(4√3) =

300√3. To find the surface area of the prism, first find the area of each side: • left/right rectangles: (5)(8) = 40 each.

• top/bottom rectangles: (15)(5) = 75 each.

• front/back parallelograms: (base)(height) = (15)(4√3) = 60√3.

The total surface area of the prism is:

S = 2(40) + 2(75) + 2(60√3) = 80 + 150 + 120√3 = 230 + 120√3

281

Answer Key

20. R = 12. Notes: The diameter of the ring (AB) and the radii of the sphere (AO and BO) form an equilateral

triangle, as shown below. Since AO = BO = 24, it follows

that AB = 24. Since AB = 24, the radius of the ring is R = AM = BM =

24 2

= 12.

21. V = 640π√3. Notes: R = AO = 8. Draw right triangle

∆ABC, as shown below. Since ∠BAO = 60°, ∆ABC is a

30°-60°-90° triangle. The hypotenuse (AB) is twice the

side opposite to the 30° angle, such that AC =

AB 2

=

20 2

=

10. The side opposite to the 60° angle is √3 times larger than the side opposite to the 30° angle: BC = AC√3 =

10√3. The height of the cylinder is the line segment that is perpendicular to the ends and which joins the ends: h

= BC = 10√3. Use V = πR2 h with R = 8.

282

Plane Geometry Practice Workbook with Answers, Volume 2

22. V = 54√3 and S = 108 + 12√3. Notes: Recall from Volume 1, Chapter 10, Example 4 that the area of a regular hexagon with edge length L is B = L = 2 to get B =

3(2)2 √3 2

=

3(4)√3 2

=

12√3 2

3L2 √3 2

. Plug in

= 6√3. Since the

front and back bases are hexagons, the base area is 6√3.

The volume of the prism is V = Bh = (6√3)(9) = 54√3.

To find the surface area of the prism, add the area of the 2 hexagons to the area of the 6 rectangles: S =

2(6√3) + 6(9)(2) = 12√3 + 108.

23. V = 1200√3 and S = 30√651 + 150√3. Notes: A

regular hexagon can be divided into 6 equilateral

triangles (Volume 1, Chapter 10, Example 4). Therefore, BO = 10 in the diagram on the next page. Note that

∠AOB = 90° because h = AO is perpendicular to the base. Use the Pythagorean theorem to find AO:

h = AO = √AB 2 − BO2 = √262 − 102 = √676 − 100 =

√576 = 24 Recall from Volume 1, Chapter 10, Example 4 that the area of a regular hexagon with edge length L is B = 3L2 √3 2

. Plug in L = 10 to get B = 283

3(10)2 √3 2

=

3(100)√3 2

=

Answer Key

150√3. The volume of the pyramid is V = 150√3(24) 3

Bh

= 1200√3. Note that BM = CM =

Use the Pythagorean theorem to find AM:

3

=

BC 2

10

=

2

= 5.

AM = √AB 2 − BM 2 = √262 − 52 = √676 − 25 = √651 1

1

The area of ∆ABC is 2 (BC)(AM) = 2 (10)�√651� =

5√651. To find the surface area of the pyramid, add the

area of the hexagon to the area of the 6 triangles: S = 150√3 + 6(5√651) = 150√3 + 30√651.

24. D = 12 and S = 144π. Notes: V = 288π =

4πR3 3

.

Multiply by 3 and divide by 4π on both sides: 216 = R3 . 3

Cube root both sides: √216 = 6 = R. The diameter is D = 2R = 2(6) = 12. The surface area is S = 4πR2 . 25. V =

343 8

= 42.875. Notes: S = 6L2 = 73.5 =

Divide by 6 on both sides: L2 =

147 12

3

7 3

49 4

284

.

. Square root 73

both sides: L = = 3.5. V = L = � � = 3 = 2 2 2

42.875.

2

. Divide 147 and 12

each by 3 to reduce the fraction: L2 = 7

147

343 8

=

Plane Geometry Practice Workbook with Answers, Volume 2

26. L = 12 (horizontal), W = 4 (depth), H = 6 (vertical), V = 288, and S = 288 (but S has square units whereas V has units cubed). Notes: See the diagram below.

L = AD = EH = BC = FG = 12 is given in the diagram.

AD + AE + EH + DH = 32 is given. Plug in AD = EH = 12 to get 12 + AE + 12 + DH = 32. Simplify:

24 + AE + DH = 32. Subtract 24 from both sides: 285

Answer Key

AE + DH = 8. Note that AE = DH, such that 2AE = 8.

Divide by 2 on both sides: AE = 4 = DH = BF = CG = W. CG + CD + DH = 14 is given. Plug in CG = DH = 4 to get 4 + CD + 4 = 14. Subtract 8 from both sides: CD = 6 = AB = EF = GH = H. The volume is V = LWH =

(12)(4)(6) = 288. The surface area is: S = 2LW + 2WH + 2LH = 2(12)(4) + 2(4)(6) + 2(12)(6) = 96 + 48 + 144 = 288. 3

15

27. R = π or R = 2π, V =

S = 90 +

225 2π

135 π

or V =

675

, and S = 90 + 2π

18 π

. Notes: See the diagram on the following

page. If the rectangle is rolled down to form the

or

horizontal cylinder at the bottom, h = 15 and 2πR = 6, 6

3

3 2

2

9

for which R = 2π = π, V = πR h = π �π� 15 = π π2 15 =

135 π

, and S = 2πRh + 2πR2 = 2πR(h + R) 3

3

3

= 2π π �15 + π� = 6 �15 + π� = 90 +

18 π

. If the rectangle

is rolled right to form the vertical cylinder at the right, h = 6 and 2πR = 15, for which R = 15 2

225

π �2π� 6 = π 4π2 6 = 2π

15

15

1350 4π

=

15

675 2π

15

2 , V = πR h= 2π

, and S = 2πR(h + R) =

�6 + 2π� = 15 �6 + 2π� = 90 + 2π 286

225 2π

.

Plane Geometry Practice Workbook with Answers, Volume 2

28. V =

304π 3

. Notes: The diagram below shows how the

truncated cone can be formed by subtracting a short

cone from a tall cone. CD = 4, BO = 6, and CO = DE = 4. Note that CDOE is a square. Show that ∆ACD~∆DEB

such that AC:CD = DE:BE (Volume 1, Chapter 4).

OE + BE = BO such that BE = 6 − 4 = 2. AC:CD = DE:BE

becomes

AC 4

4

= 2. Simplify:

AC 4

= 2. Multiply by 4 on both

sides: AC = 2(4) = 8. The tall cone has a radius of BO =

6 and a height of AO = AC + CO = 8 + 4 = 12. The short cone has a radius of CD = 4 and a height of AC = 8. To find the volume of the truncated cone, subtract the volumes of the cones. Recall that V =

πR2 h 3

for a right

circular cone. π(6)2 (12) π(4)2 (8) π(36)(12) π(16)(8) V= − = − 3 3 3 3 128π 432π 128π 304π V = 144π − = − = 3 3 3 3 287

Answer Key

Vcone

29. V

cylinder

1 Vsphere

= 3, V

cylinder

2

= 3, and Vcone + Vsphere = Vcylinder .

Notes: Each shape has the same radius (R). The cone and cylinder have a height of h = D = 2R. • cone: V =

πR2 h 3

=

πR2 (2R) 3

=

2πR3 3

• cylinder: V = πR2 h = πR2 (2R) = 2πR3 • sphere: V =

4πR3 3

3 2πR3 2πR 1 1 Vcone = ÷ 2πR3 = × = Vcylinder 3 3 2πR3 3 3 Vsphere 4πR3 4πR 1 2 = ÷ 2πR3 = × = Vcylinder 3 3 2πR3 3

Vcone + Vsphere

2πR3 4πR3 6πR3 = + = = 2πR3 3 3 3

= Vcylinder

30. V = 6 (3 for each triangle), E = 9 (3 for the front triangle, 3 for the back triangle, and 3 heights),

F = 5 (2 triangles + 3 rectangles), V + F = 6 + 5 = 11 agrees with E + 2 = 9 + 2 = 11, the dual polyhedron 288

Plane Geometry Practice Workbook with Answers, Volume 2

has V and F reversed, and the dual polyhedron is the

triangular bipyramid shown below on the right (which can be formed by joining two tetrahedra at one

triangle). Note: The word “dual” has slightly different usages in the context of polyhedra. The most strict usage applies to Platonic solids and carries strong

symmetry requirements. A common loose usage simply refers to swapping V and F (as was done here).

31. Main ideas: ���� CD ⊥ ���� AF. The area of ∆ACD is 1

1

(CD)(AF) = Lℓ, where L = CD is the length of each 2 2 side of the base (which is a regular polygon with an unknown number of sides) and ℓ = AF is the slant

height (which is NOT the same as the height of the

pyramid, which is AG). Since the regular polygon has N sides, there will be N triangular faces congruent with ∆ACD. The lateral surface area thus is A =

NLℓ 2

. The

perimeter of the regular polygon is P = NL. Plug this into the previous equation: A = 289

Pℓ 2

. This is the lateral

Answer Key

surface area (it does NOT include the area of the base). Note: In print, L and l are uppercase and lowercase

versions of the same letter. Note that l could easily be

confused with the number 1 or the uppercase letter I.

With this in mind, it is common to use a script or cursive lowercase ℓ instead of a printed l (to help avoid confusion).

32. Main ideas: The total surface area of a right circular cone is S = πR2 + πR√R2 + h2 , where the second term,

πR√R2 + h2 , is the lateral surface area (which excludes the area of the base). Use the Pythagorean theorem to show that ℓ = √R2 + h2 .

33. θ = 120°, V =

128π√2 , 3

and S = 64π. Notes: When the

cone is unfolded (see below), the lateral surface

becomes a sector. The radius of the sector is the slant height of the cone (ℓ = 12) and the arc length of the

sector is the circumference of the cone (2πR). Note that

every point on the circumference of the base of the cone and of the arc length of the sector is equidistant from

the apex of the cone. Recall from Chapter 3 that the arc length of a sector equals the radius of the sector times 290

Plane Geometry Practice Workbook with Answers, Volume 2

the central angle in radians: 2πR = ℓθ (where the

circumference of the base of the cone was put in place of the arc length and the slant height of the cone was

put in place of the radius of the sector). Plug in R = 4

and ℓ = 12 to get 2π4 = 12θ, which simplifies to 8π = 8π

12θ. Divide by 12 on both sides: 12 =

Multiply by 2π 3

=

2π 180° 3

π

180° π

2π 3

= θ in radians.

to convert from radians to degrees: θ =

= 120°. Use the Pythagorean theorem to

find the height of the cone (right diagram below). h = �ℓ2 − R2 = �122 − 42 = √144 − 16 = √128 = �(64)(2) = √64√2 = 8√2

πR2 h π(4)2 �8√2� π(16)�8√2� 128π√2 V= = = = 3 3 3 3

S = πR2 + πR√R2 + h2 = πR2 + πRℓ = πR(R + ℓ) = π4(4 + 12) = π4(16) = 64π

34. Main ideas: Let L be the edge length of the cube: CG = FG = GH = L. ∆CGH, ∆FGH, and ∆CFG are all 45°-45°-

90° right triangles (but ∆CFH is NOT a right triangle; see 291

Answer Key

Problem 35). CF = CH = FH = L√2 because each is the hypotenuse of a 45°-45°-90° right triangle (Volume 1,

Chapter 5). The right figure is an oblique pyramid (because its apex, point H, does not lie above the

centroid of the base) with a triangular base (∆CFG). The 1

1

base area (the area of ∆CFG) is 2 (CG)(FG) = 2 (L)(L) = L2 2

. The height is measured perpendicular to the base: h

= GH = L. The volume of the figure on the right is V = Bh 3

1 L2

L3

= 3 � 2 � (L) = 6 . Note: For students who try to

visualize first cutting the cube in half along rectangle

CDEF, the resulting figure has 3 times the volume as the

figure on the right (NOT 2 times the volume): This gives

11

1

11

1

= 6 (NOT 2 2 = 4). One way to realize this is to note 32 that CDEFH has one rectangular side (CDEF), so it is

clearly NOT congruent with figure CFGH on the right which only has triangular sides.

35. Main ideas: BE = BG = GE because each is the

diagonal of a congruent square face, which shows that ∆BEG is an equilateral triangle. Notes: Although ���� ��� ⊥ ���� BF ⊥ �EF FG and faces AEFB, EFGH, and BCGF are

perpendicular to one another, it is possible for a line 292

Plane Geometry Practice Workbook with Answers, Volume 2

segment lying in one face to NOT be perpendicular to a line segment lying in a perpendicular face. This point

can be grasped easily as follows. Draw a line segment from point E to a point just right of F (on ���� FG), so this line segment lies on face EFGH. Draw another line

segment from point E to a point just in front of F (on ���� BF), so this line segment lies on face AEFB. These two

line segments lie on two perpendicular faces, yet they are obviously NOT perpendicular line segments, as

shown below. This note is only for students who did not think to approach the problem by trying to show that

∆BEG is equilateral, and who were focused on the faces

being perpendicular to one another (which is a mistake

commonly encountered in the solution of this problem).

The easy way to solve this problem is to show that

∆BEG is equilateral, such that it has three 60° interior angles.

293

Answer Key

BO

3

36. CD = 21/3 = √2 ≈ 1.26. Notes: The original cone

(left figure on the following page) minus the short cone

(middle figure) equals the truncated cone (right figure). Since the truncated cone has half the volume of the

original cone, the small cone below must also have half the volume of the original cone: Vsmall =

volume of a right circular cone is

previous equation:

πR2 h 3

Voriginal 2

. The

. Plug this into the 2

π(R small )2 hsmall 1 π�R original � horiginal = 3 3 2 ∆ACD~∆AOB on the next page, such that AC:CD =

AO:BO (Volume 1, Chapter 4). Since AC = hsmall , AO =

horiginal , CD = R small , and BO = R original , AC:CD = AO:BO

can be written as:

hsmall horiginal = R small R original

Multiply both sides of the equation by R small . horiginal R small hsmall = R original Plug this equation into

π(R small )2 hsmall 3

294

2

1 π�R original � horiginal

=2

3

.

Plane Geometry Practice Workbook with Answers, Volume 2

2

π(R small )2 horiginal R small 1 π�R original � horiginal = 3R original 3 2

Apply the algebra rule 𝑥𝑥 2 𝑥𝑥 = 𝑥𝑥 3 .

2

π(R small )3 horiginal 1 π�R original � horiginal = 3R original 3 2

Multiply both sides by 3R original and divide both sides by πhoriginal .

3

�R original � (R small )3 = 2 Multiply both sides by 2 and divide both sides by (R small )3 . 3

�R original � 2= (R small )3

Cube root both sides of the equation. R original BO 3 = √2 = R small CD

295

Answer Key

37. V =

πR3 6

and S =

5πR2 4

. Notes: This is one-eighth of a

4πR3

sphere. The volume inside of a complete sphere is

3

.

Divide by 8 to find the volume inside of the given figure:

4πR3 3(8)

=

4πR3 24

=

πR3 6

. The area of the curved surface is one-

eighth of the surface area of a complete sphere: πR2 2

4πR2 8

. The three flat surfaces each have an area of one-

fourth of a circle: 3 �

πR2 4

�=

3πR2 4

=

. To find the total

surface area, add the area of the curved surface to the area of the flat surfaces: 5πR2 4

.

πR2 2

+

3πR2 4

=

2πR2 4

+

3πR2 4

=

38. See the diagrams on the following page. Notes:

• top left: the plane touches the cube only at a single point (vertex H).

• top middle: the plane intersects the cube along a ����). single edge (GH • top right: the plane intersects the cube at ∆BDE.

• bottom left: the plane intersects the cube only at a single face (CDHG).

• bottom middle: the intersection is a pentagon. 296

Plane Geometry Practice Workbook with Answers, Volume 2

• bottom right: the intersection is a hexagon (with one edge on every face).

39. See the diagrams below. Notes: For a rectangle, the

plane is parallel to the axis of the cylinder (left). For a circle, the plane is perpendicular to the axis of the

cylinder (middle). For an ellipse, the plane is tilted (right).

297

Answer Key

40. See the diagrams that follow. Notes: These are the classic conic sections.

• top left: the plane is perpendicular to the axis of the cone and passes through (only) the apex of the cones.

• top right: the plane is perpendicular to the axis of the cone, intersecting the cone to make a circle.

• middle left: the plane contains the axis of the cone,

intersecting the cone to make a pair of intersecting lines.

• middle right: the plane is tilted compared to the case that made a circle.

• bottom left: the plane intersects (only) a line along the surface of the cones and passing through the apex.

• bottom right: the plane is parallel to the plane that made a single line, so as to intersect the cone to make a parabola.

• next page: the plane is parallel to the axis of the

cone, intersecting the cone to make two branches of a hyperbola.

298

Plane Geometry Practice Workbook with Answers, Volume 2

299

Glossary 3D: three-dimensional. A 3D object extends along all

three independent directions of space. A 3D object has length (horizontal), height (vertical), and depth (front/back).

AA: angle-angle. If two interior angles are congruent for

two triangles, the triangles are similar.

AAS: angle-angle-side. If two interior angles and a side

that is not between those angles are congruent for two triangles, the triangles are congruent.

acute angle: an angle that is less than 90°.

acute triangle: a triangle with three acute interior

angles.

angle: the shape made when two line segments are joined at a common endpoint.

angle bisector: a line segment that cuts an angle into two smaller congruent angles.

apex: the farthest point from the base of a pyramid or cone.

300

Plane Geometry Practice Workbook with Answers, Volume 2

arc length: the distance along a circular arc.

ASA: angle-side-angle. If two interior angles and a side

that is between those angles are congruent for two triangles, the triangles are congruent.

axis: a straight line through the center a 3D object, for which the object is symmetrically arranged.

ball: a 3D solid that includes a sphere and all of the

points inside of it.

base: the side of a triangle or pyramid opposite to the

apex, either of the parallel sides of a trapezoid, either of

the parallel sides of a prism or cylinder, or the flat side of a cone.

bicentric: a polygon that is both a cyclic polygon and a tangential polygon.

bisect: cut a shape into two congruent parts.

bisector: a line that bisects a shape. See angle bisector and perpendicular bisector.

body diagonal: a line segment that joins two opposite corners of a rectangular prism. 301

Glossary

central angle: an angle where the two sides are radii of

the same circle.

centroid: the point where the three medians of a triangle intersect.

chord: a line segment that connects two points that lie

on the circumference of a circle, like the line segment in the left figure below.

circle: a curve for which every point on the curve is

equidistant from its center.

circular segment: a region enclosed by an arc and a chord, like the shaded region above.

circumcenter: the point where the perpendicular bisectors of a polygon intersect. 302

Plane Geometry Practice Workbook with Answers, Volume 2

circumcircle: a circle that can be circumscribed about a polygon such that every vertex of the polygon lies on the circumference of the circle.

circumdiameter: twice the circumradius.

circumference: the total distance around a complete circle.

circumradius: the distance from the circumcenter to any

vertex of a cyclic polygon.

circumscribe: to draw one shape around another either

so that every edge of the outer shape is tangent to the

inner shape (like a polygon that is circumscribed about a circle) or so that every vertex of the inner shape lies on the outer shape (like a circle that is circumscribed

about a polygon). In the left figure below, the hexagon is circumscribed about the circle. In the right figure, the circle is circumscribed about the hexagon.

coaxial: having the same axis of symmetry. collinear: lying on the same line. 303

Glossary

complementary angles: two angles that together form a 90° angle.

concurrent: three or more lines that intersect at a single

point.

cone: a 3D object that is similar to a pyramid, except

that the base of a cone is a closed curve (like a circle or ellipse) instead of a polygon, like the figure below.

congruent: having the same shape and size (even if one

shape is rotated relative to the other, and even if one shape is the mirror image of the other).

conic section: a shape that can be obtained as a cross

section of a cone. Conic sections include a circle, ellipse, parabola, hyperbola, line, intersecting lines, and point. conjugate: when two irrational expressions are

multiplied together to form a rational product, the expressions are considered to be conjugates. For

example, 2 − √3 is the conjugate of 2 + √3 because

�2 + √3��2 − √3� = 4 − 2√3 + 2√3 − 3 = 4 − 3 = 1.

convex: a polygon where every interior angle is less than 180° (or a similar polyhedron). 304

Plane Geometry Practice Workbook with Answers, Volume 2

CPCTC: corresponding parts of congruent triangles are congruent.

cross section: the region of intersection of a plane and a three-dimensional object.

cube: a polyhedron with 8 vertices, 12 edges, and 6

square faces that meet at right angles, like the figure below.

cuboid: see rectangular prism.

cyclic polygon: a polygon for which it is possible to

draw a circle around the polygon with all of its vertices lying on the circumference of the circle.

cylinder: a 3D object that is similar to a prism, except that each end of a cylinder is a curve (like a circle or ellipse) instead of a polygon, like the figure below.

degree: 1/360th of a circle, as measured from the center. diagonal: a line segment that joins two non-adjacent

vertices in a polygon or polyhedron. 305

Glossary

diameter: a line segment that passes through the center

of a circle and which joins two points that lie on the

circle. A diameter is a special chord that passes through the center.

dimension: a measure of extent.

dodecahedron: a polyhedron with 20 vertices, 30 edges,

and 12 faces, where each face is a pentagon and where three pentagons meet at each vertex.

dual polyhedra: two polyhedra where the numbers of

vertices and faces are swapped (such that their Schläfli symbols are reversed).

edge: a line segment that connects two adjacent vertices of a polygon or polyhedron.

306

Plane Geometry Practice Workbook with Answers, Volume 2

ellipse: a symmetric closed curve that could be formed

by stretching or squeezing a circle. Whereas every point on a circle is equidistant from the center, the sum of the

distances from two fixed points (called foci) is the same

for every point on an ellipse.

equiangular: a polygon where all of the interior angles

are congruent.

equilateral: a polygon where all of the sides are congruent.

Euler’s formula: for a polyhedron, the number of

vertices plus the number of faces is two more than the number of edges: V + F = E + 2.

face: a flat side of a polyhedron.

face diagonal: a diagonal that lies within one face (in contrast to a body diagonal).

height: a line segment that is perpendicular to the base (of a triangle, pyramid, cylinder, or cone, for example) and connects to the apex or to the opposite base.

hemisphere: one-half of a sphere.

Heron’s formula: an equation to find the area of a

triangle in terms of the lengths of the three sides, where P = a + b + c is the perimeter. 307

Glossary

hexagon: a polygon with six sides.

horizontal line: a line that runs across to the left and

right.

hyperbola: a symmetric curve with two branches where the difference between the distances from two fixed

points (called foci) is the same for every point on the curve.

hypotenuse: the longest side of a right triangle, which is opposite to the 90° angle.

icosahedron: a polyhedron with 12 vertices, 30 edges, and 20 faces, where each face is a triangle and where five triangles meet at each vertex.

incenter: the point where the angle bisectors of a polygon intersect.

incircle: a circle that can be inscribed in a polygon such that every edge of the polygon is tangent to the circle.

indiameter: twice the inradius.

inradius: the shortest distance from the incenter to any side of a tangential polygon.

308

Plane Geometry Practice Workbook with Answers, Volume 2

inscribe: to draw one shape inside of another either so

that every edge of the outer shape is tangent to the

inner shape (like a circle that is inscribed in a polygon) or so that every vertex of the inner shape lies on the

outer shape (like a polygon that is inscribed in a circle). In the left figure below, the circle is inscribed in the

hexagon. In the right figure, the hexagon in inscribed in the circle.

inscribed angle: an angle with a vertex that lies on the

circumference of a circle and with sides that are chords, like α in the figure below.

309

Glossary

inscribed angle theorem: an inscribed angle has one-

half of the angular measure of a central angle that subtends the same arc.

interior angle: an angle that is formed by two sides that meet at a vertex of a polygon, and which lies inside of the polygon.

intersect: when two shapes cross paths. isosceles: having two congruent sides.

kite: a quadrilateral with no parallel sides that has two pairs of congruent sides.

lateral edge: an edge that is not part of the base of a polyhedron.

lateral face: a face that is not a base of a polyhedron.

lateral surface area: the surface area of a polyhedron,

cylinder, or cone excluding the area of the base.

leg: either of the shorter sides of a right triangle, which touch the 90° angle, or the non-parallel sides of a

trapezoid.

line: a straight path that is infinite in each direction.

line segment: a straight, finite path that connects two endpoints.

310

Plane Geometry Practice Workbook with Answers, Volume 2

lune: the crescent-shaped region lying between two

intersecting circular arcs, such as the shaded region below.

major arc: the longest arc length between two points on

the circumference of a circle. The corresponding central angle is greater than (or equal to) 180°.

median: a line segment that joins one vertex to the

midpoint of the opposite side of a triangle, or a line segment that joins the midpoints of the legs of a trapezoid.

midpoint: a point that bisects a line segment.

minor arc: the shortest arc length between two points

on the circumference of a circle. The corresponding central angle is less than (or equal to) 180°.

oblique: slanted. A prism, pyramid, cylinder, or cone is

oblique if the axis is not perpendicular to the base. obtuse angle: an angle that is greater than 90°. 311

Glossary

obtuse triangle: a triangle with one obtuse angle and

two acute angles.

octagon: a polygon with eight sides.

octahedron: a polyhedron with 6 vertices, 12 edges, and 8 triangular faces. It can be formed by joining two square pyramids together at their square bases.

octant: any of the eight regions of 3D space that are

formed when three mutually perpendicular planes intersect.

parabola: a symmetric curve where every point on the

curve is equidistant from a point (called the focus) and a straight line (called the directrix). The equation 𝑦𝑦 = 𝑎𝑎𝑥𝑥 2 represents a simple parabola through the origin

that is symmetric about the 𝑦𝑦-axis.

parallel: lines that extend in the same direction and are the same distance apart at any position such that they will never intersect.

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Plane Geometry Practice Workbook with Answers, Volume 2

parallel postulate: if the same-side interior angles do ⃖����⃗ and CD ⃖���⃗ not add up to exactly 180°, then lines AB intersect (and therefore are not parallel), but if the

same-side interior angles (either ∠3 + ∠5 or ∠4 + ∠6) ⃖����⃗ and CD ⃖���⃗ must be do add up to exactly 180°, then lines AB ⃖����⃗ and CD ⃖���⃗ do not intersect). parallel (in this case, AB

parallelogram: a quadrilateral with two pairs of opposite sides that are parallel and congruent.

pentagon: a polygon with five sides.

perimeter: the total distance around the edges of a

polygon. The sum of the side lengths.

perpendicular: shapes that meet at a 90° angle. perpendicular bisector: a line that passes

perpendicularly through a line segment and cuts the line segment in half.

pi: the ratio of the circumference of any circle to its

diameter, which approximately equals 3.14159 (with the digits continuing forever without repeating). 313

Glossary

Pitot theorem: the sum of the lengths of the opposite

sides of a tangential quadrilateral equals one-half of the perimeter.

plane: a flat surface that extends infinitely in two independent directions.

Platonic solids: the five regular convex polyhedra, which include the tetrahedron, cube, octahedron, dodecahedron, and icosahedron.

polygon: a closed plane figure that is bounded by straight sides.

polyhedron: a three-dimensional solid that is bounded

by faces that are polygons.

power of a point theorem: given a circle and given any

point X, for any secant that passes through point X, the product of the two distances from X to the two points

where the secant intersects the circle equals the same value. Special cases of the power of a point theorem include intersecting chords, intersecting secants,

intersecting tangents, a chord intersecting a tangent, and a secant intersecting a tangent. 314

Plane Geometry Practice Workbook with Answers, Volume 2

prism: a three-dimensional solid bounded by two

congruent parallel polygons on its two ends and which has parallelograms along its body, like the examples below.

Ptolemy’s theorem: the product of the diagonals of a convex cyclic quadrilateral equals the sum of the products of the opposite sides.

pyramid: a three-dimensional solid that has a base in

the shape of a polygon and a point called the apex that does not lie in the same plane as the polygon, where

edges connect the apex to every vertex of the polygon, like the examples below.

315

Glossary

Pythagorean theorem: the sum of the squares of the

lengths of the two legs of a right triangle equals the

square of the hypotenuse. For the right triangle below, a2 + b2 = c 2 .

quadrilateral: a polygon with four sides.

radian: a unit of angular measure equivalent to

180° π

,

such that the angular measure of one full circle equals 2π radians (equivalent to 360°).

radius: the distance from the center of a circle to any

point on the circle.

rectangle: an equiangular parallelogram. Every interior angle is 90°.

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Plane Geometry Practice Workbook with Answers, Volume 2

rectangular prism: a three-dimensional solid that is

bounded by six rectangular faces that meet at right

angles, like the figure below. A rectangular prism is also called a cuboid.

regular: a polygon that is both equilateral and

equiangular, or a highly symmetric polyhedron with

congruent faces arranged in a similar manner at each vertex.

rhombus: an equilateral parallelogram.

right angle: an angle measuring exactly 90°.

right cone: a cone where the axis is perpendicular to the base.

right cylinder: a cylinder where the axis is perpendicular to the ends.

right prism: a prism where the lateral parallelograms are rectangles.

right pyramid: a pyramid where the apex lies directly above the centroid of the base. 317

Glossary

right triangle: a triangle with one right angle and two acute angles.

SAS: side-angle-side. If two sides and the interior angle that is formed by those sides are congruent for two triangles, the triangles are congruent.

scalene: a triangle that does not have any two sides with the same length.

Schläfli symbol: an ordered pair in the form {n, m} that

helps to mathematically label the different kinds of

regular polyhedra. The first number (n) represents the number of edges of the polygons that are used as the faces, while the second number (m) represents the number of faces that meet at each vertex.

secant: a line (or a line segment) that intersects a circle at two points and which extends outside of the circle.

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Plane Geometry Practice Workbook with Answers, Volume 2

sector: a region enclosed by an arc and two radii, like

the shaded region in the left figure below.

segment, circular: a region enclosed by an arc and a

chord, like the shaded region in the right figure above. semicircle: one-half of a circle.

side: a line segment that joins two adjacent vertices of a polygon, or a face of a polyhedron. The sides of a

polygon or polyhedron form its boundary.

similar: having the same shape, but not the same size.

slant height: the distance from the apex of a pyramid to a midpoint of an edge of the base or the distance from

the apex of a cone to a point on the circumference of the base.

slope: the rise over the run of a line plotted on a

coordinate graph.

319

Glossary

sphere: a three-dimensional curved surface where every point on the surface is equidistant from the center.

square: a regular quadrilateral. It is both equilateral and

equiangular. Like a rectangle, every interior angle is 90°.

SSS: side-side-side. If every side is congruent for two triangles, the triangles are congruent.

subtend: to be opposite in position to (said of an angle that subtends an arc).

supplementary angles: two angles that together form a 180° angle.

surface area: the area corresponding to the surface of an object.

tangent: a line that touches a circle only at a single point.

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Plane Geometry Practice Workbook with Answers, Volume 2

tangential polygon: a polygon for which it is possible to

draw a circle inside of the polygon with all of its sides tangent to the circle.

tetrahedron: a polyhedron bounded by four triangular faces.

Thales’s theorem: if all three interior angles of a triangle

are inscribed angles (which means that the triangle is

inscribed in the circle) and if one side of the triangle is a diameter of the circle, the angle opposite to the diameter is a right angle.

trapezoid: a quadrilateral with one pair of parallel sides. triangle: a polygon with three sides.

triangle bisector theorem: an angle bisector divides the opposite side of the triangle into segments in

proportion to the lengths of the other two sides. For BD

CD

example, in the diagram below, AB = AC (which is BD

AB

equivalent to CD = AC).

321

Glossary

truncated cone: the remains of a cone after the portion

that includes the apex has been sliced off, like the figure below.

unit cube: a cube that is 1 unit long, 1 unit wide, and 1

unit high.

vertex: the point where the sides of an angle intersect.

The plural form is vertices.

vertical angles: angles that appear on opposite sides of a

vertex where two lines intersect.

vertical line: a line that runs up and down without any slant.

vertices: the plural form of vertex.

volume: a measure of the amount of space contained

within the surface of a closed 3D object.

322

Notation and Symbols ⊙O AB

���� AB

⃖����⃗ AB �����⃗ AB ⃖����� AB

ABC

∆ABC

∠ABC ∠1 α

a single point labeled A

a circle centered about point O

the straight-line distance between points A and B

a finite line segment connecting point A to point B an infinite line passing through points A and B a semi-infinite ray starting at A and passing through B and beyond a semi-infinite ray starting at B and passing through A and beyond

an infinite plane containing points A, B, and C

a (finite) triangle with vertices at points A, B, and C an angle with point B at the vertex, with sides ���� AB and ���� BC a numbered angle

an angle indicated by using a lowercase Greek letter 323

Notation and Symbols

ABCD °

rad π

⊥ ∥

∦

= ≅ ~

a:b

a quadrilateral with vertices at points A, B, C, and D degrees radians

the constant pi, which is approximately equal to 3.14159 perpendicular to parallel to

not parallel to is equal to

is congruent with is similar to

the ratio of a to b

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Plane Geometry Practice Workbook with Answers, Volume 1

A

area (or a point labeled A)

S

surface area (or a point labeled S)

B

the area of the base of a 3D object (or a point labeled B)

V s

volume (or a point labeled V) arc length

b

the base of a triangle

C

circumference (or the centroid of a triangle or a point labeled C)

I

the incenter of a polygon

h

P

O

height

perimeter (or a point labeled P)

the center of a circle or the circumcenter of a polygon

r

inradius

D

diameter (or a point labeled D)

R

the radius of a circle (or the circumradius or a point labeled R)

325

Greek Alphabet α β γ

δ ε

ϵ

ζ η

θ ι

κ λ

uppercase alpha lowercase beta Β uppercase beta lowercase uppercase Γ gamma gamma uppercase lowercase delta Δ delta lowercase uppercase Ε epsilon epsilon a common variation of lowercase epsilon lowercase zeta Ζ uppercase zeta lowercase eta Η uppercase eta uppercase lowercase theta Θ theta lowercase iota Ι uppercase iota lowercase uppercase Κ kappa kappa lowercase uppercase Λ lambda lambda

lowercase alpha

326

Α

Plane Geometry Practice Workbook with Answers, Volume 1

μ ν ξ ο

π ρ σ τ

υ

φ ϕ χ ψ

ω

lowercase mu Μ uppercase mu lowercase nu Ν uppercase nu lowercase xi Ξ uppercase xi lowercase uppercase Ο omicron omicron lowercase pi Π uppercase pi lowercase rho Ρ uppercase rho lowercase uppercase Σ sigma sigma lowercase tau Τ uppercase tau lowercase uppercase Υ upsilon upsilon lowercase phi Φ uppercase phi a common variation of lowercase phi lowercase chi Χ uppercase chi lowercase psi Ψ uppercase psi lowercase uppercase Ω omega omega

327

Greek Alphabet

Notes:

• ε and ϵ are two common variations of lowercase epsilon.

• φ and ϕ are two common variations of lowercase phi.

• xi (ξ and Ξ) is pronounced zi or si (with a long i sound, like “lie”) in English. In Greek, it is

pronounced ksee (with a long e sound, like “bee”).

However, there are math and science teachers who speak English who pronounce xi as ksee.

• Ξ is uppercase xi, whereas ≡ means “is defined as.” The middle line is shorter in Ξ, whereas all three bars have equal length in ≡.

• phi (φ and Φ) and chi (χ and Χ) may be

pronounced in English with a long e (like “bee”) or a long i (like “lie”). Although the long i may make

more sense from the perspective that many ancient Greek and Latin words in the English language

ending with “i” tend to end with a long i (like “lie”) sound, such as “cacti” (the plural form of cactus),

“octopi,” and “alibi,” the long e (like “bee”) sound

preserves the original pronunciation. Both

pronunciations are in common use today. In 328

Plane Geometry Practice Workbook with Answers, Volume 1

contrast, pi (π and Π) is only pronounced with a

long i (like “lie”).

• psi (ψ and Ψ) is usually pronounced like the word

“sigh” in English. This letter is common in quantum mechanics.

• π (lowercase pi) is usually reserved for the

constant that approximately equals 3.14159 (but which continues forever without repeating). It

equals the ratio of the circumference of any circle to its diameter.

• θ (lowercase theta) and φ (lowercase phi) are

commonly used for angles in many applications of geometry (such as physics and engineering).

• δ (lowercase delta) and ε (lowercase epsilon) are commonly used in calculus.

• ϵ (a variation of lowercase epsilon) is common in set theory notation.

• Δ (uppercase delta) often means “change in.” Example: ΔU = the change in U.

• Σ (uppercase sigma) is commonly used as a summation symbol.

• σ (lowercase sigma) is frequently used in statistics. 329

Greek Alphabet

• Ω (uppercase omega) is the SI unit of resistance (the “Ohm”).

• Some Greek letters (such as Α, ο, and Ν) are best

avoided since they resemble letters in the English alphabet.

330

Did You Miss the First Volume? The Learning Began in Volume 1 Triangles, Quadrilaterals, and Other Polygons

WAS THIS BOOK HELPFUL? Much effort and thought was put into this book, such as: • Providing more than just the answers in the answer key. • Careful selection of examples and problems for their instructional value. • Numerous detailed illustrations to help visualize the principles. • Coverage of a variety of essential geometry topics. • A concise review of relevant concepts at the beginning of each chapter. If you appreciate the effort that went into making this book possible, there is a simple way that you could show it: Please take a moment to post an honest review.

For example, you can review this book at Amazon.com or Goodreads.com.

Even a short review can be helpful and will be much appreciated. If you are not sure what to write, following are a few ideas, though it is best to describe what is important to you. • How much did you learn from reading and using this workbook? • Was the information in the answer key helpful? • Were you able to understand the ideas? • Was it helpful to follow the examples while solving the problems? • Would you recommend this book to others? If so, why?

Do you believe that you found a mistake? Please email the author, Chris McMullen, at [email protected] to ask about it. One of two things will happen: • You might discover that it wasn’t a mistake after all and learn why. • You might be right, in which case the author will be grateful and future readers will benefit from the correction. Everyone is human.

ABOUT THE AUTHOR Dr. Chris McMullen has over 20 years of experience

teaching university physics in California, Oklahoma,

Pennsylvania, and Louisiana. Dr. McMullen is also an

author of math and science workbooks. Whether in the classroom or as a writer, Dr. McMullen loves sharing knowledge and the art of motivating and engaging students.

The author earned his Ph.D. in phenomenological

high-energy physics (particle physics) from Oklahoma State University in 2002. Originally from California, Chris McMullen earned his Master’s degree from

California State University, Northridge, where his thesis was in the field of electron spin resonance.

As a physics teacher, Dr. McMullen observed that

many students lack fluency in fundamental math skills.

In an effort to help students of all ages and levels master basic math skills, he published a series of math

workbooks on arithmetic, fractions, long division, word problems, algebra, geometry, trigonometry, logarithms, and calculus entitled Improve Your Math Fluency. Dr.

McMullen has also published a variety of science books, including astronomy, chemistry, and physics workbooks.

Author, Chris McMullen, Ph.D.

MATH This series of math workbooks is geared toward practicing essential math skills: • Prealgebra and algebra

• Geometry and trigonometry

• Logarithms and exponentials • Calculus

• Fractions, decimals, and percentages • Long division • Arithmetic

• Word problems

• Roman numerals

• The four-color theorem and basic graph theory www.improveyourmathfluency.com

PUZZLES The author of this book, Chris McMullen, enjoys solving puzzles. His favorite puzzle is Kakuro (kind of like a

cross between crossword puzzles and Sudoku). He once taught a three-week summer course on puzzles. If you

enjoy mathematical pattern puzzles, you might appreciate:

300+ Mathematical Pattern Puzzles

Number Pattern Recognition & Reasoning

• Pattern recognition and visual discrimination

• Analytical skills

• Logic and reasoning • Analogies

• Mathematics

THE FOURTH DIMENSION Are you curious about a possible fourth dimension of space?

• Explore the world of hypercubes and hyperspheres.

• Imagine living in a two-dimensional world.

• Try to understand the fourth dimension by analogy.

• Several illustrations help to try to visualize a fourth dimension of space.

• Investigate hypercube patterns.

• What would it be like to be a 4D being living in a 4D world?

• Learn about the physics of a possible fourdimensional universe.

SCIENCE Dr. McMullen has published a variety of science books, including:

• Basic astronomy concepts • Basic chemistry concepts

• Balancing chemical reactions

• Calculus-based physics textbooks

• Calculus-based physics workbooks

• Calculus-based physics examples • Trig-based physics workbooks

• Trig-based physics examples

• Creative physics problems • Modern physics

www.monkeyphysicsblog.wordpress.com