Learn and practice essential geometry skills. The answer to every problem, along with helpful notes, can be found at the
218 159
English Pages 202 [340] Year 2021
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
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.
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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.
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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.
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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
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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.
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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.
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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.
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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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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.
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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.
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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
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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