Geometry Handbook

Table of Contents

Page Description

Chapter 1: Basics
6 Points, Lines & Planes
7 Segments, Rays & Lines
8 Distance Between Points in 1 Dimension
8 Distances Between Collinear Points
9 Distance Between Points in 2 Dimensions
11 Partial Distances and Distance Equations
12 Distance Formula in “n” Dimensions
13 Angles
14 Types of Angles

Chapter 2: Proofs
16 Conditional Statements (Original, Converse, Inverse, Contrapositive)
17 Basic Properties of Algebra (Equality and Congruence, Addition and Multiplication)
18 Inductive vs. Deductive Reasoning
19 An Approach to Proofs

Chapter 3: Parallel and Perpendicular Lines

22 Parallel Lines and Transversals
23 Multiple Sets of Parallel Lines
24 Proving Lines are Parallel
25 Parallel and Perpendicular Lines in the Coordinate Plane
27 Proportional Segments

Chapter 4: Triangles - Basic

29 What Makes a Triangle?
31 Inequalities in Triangles
35 Types of Triangles (Scalene, Isosceles, Equilateral, Right)
37 Congruent Triangles (SAS, SSS, ASA, AAS, HL, CPCTC)
40 Centers of Triangles
42 Length of Height, Median and Angle Bisector

Chapter 5: Polygons
43 Polygons – Basic (Definitions, Names of Common Polygons)
44 Polygons – More Definitions (Definitions, Diagonals of a Polygon)
45 Interior and Exterior Angles of a Polygon
Cover art by Rebecca Williams,
Twitter handle: @jolteonkitty
 Geometry Handbook
Table of Contents

Page Description

Chapter 6: Quadrilaterals
46 Definitions of Quadrilaterals
47 Figures of Quadrilaterals
48 Amazing Property of Quadrilaterals
52 Characteristics of Parallelograms
53 Parallelogram Proofs (Sufficient Conditions)
54 Kites and Trapezoids

Chapter 7: Transformations
55 Introduction to Transformation
57 Reflection
59 Rotation
61 Translation
63 Compositions
65 Rotation About a Point Other than the Origin

Chapter 8: Similarity
68 Ratios Involving Units
69 Similar Polygons
70 Scale Factor of Similar Polygons
71 Dilations of Polygons
73 More on Dilation
74 Similar Triangles (SSS, SAS, AA)
75 Proportion Tables for Similar Triangles
78 Three Similar Triangles

Chapter 9: Right Triangles

80 Pythagorean Theorem
81 Pythagorean Triples
83 Special Triangles (45⁰-45⁰-90⁰ Triangle, 30⁰-60⁰-90⁰ Triangle)
85 Trigonometric Functions and Special Angles
86 Trigonometric Function Values in Quadrants II, III, and IV
87 Graphs of Trigonometric Functions
90 Vectors
91 Operating with Vectors
 Geometry Handbook
Table of Contents

Page Description

Chapter 10: Circles

92 Parts of a Circle
93 Angles, Arcs, and Segments
94 Circle Vocabulary
95 Facts about Circles
95 Facts about Chords
97 Facts about Tangents

Chapter 11: Perimeter and Area

98 Perimeter and Area of a Triangle
100 More on the Area of a Triangle
101 Perimeter and Area of Quadrilaterals
102 Perimeter and Area of Regular Polygons
106 Circle Lengths and Areas
108 Area of Composite Figures

Chapter 12: Surface Area and Volume

111 Polyhedra
112 A Hole in Euler’s Theorem
113 Platonic Solids
114 Prisms
116 Cylinders
118 Surface Area by Decomposition
119 Pyramids
121 Cones
123 Spheres
125 Similar Solids

127 Appendix A: Geometry Formulas

129 Appendix B: Trigonometry Formulas

131 Index
Chapter 1 Basic Geometry

Geometry
Points, Lines & Planes

Item Illustration Notation Definition

Point A location in space.

Segment A straight path that has two endpoints.

Ray A straight path that has one endpoint

and extends infinitely in one direction.

A straight path that extends infinitely in

Line l or
both directions.

m or A flat surface that extends infinitely in

Plane (points , , two dimensions.
not linear)

Collinear points are points that lie on the same line.

Coplanar points are points that lie on the same plane.

In the figure at right:

 , , , , and are points.
 l is a line
 m and n are planes.
In addition, note that:
 , , and are collinear points.
 , and are coplanar points.
 , and are coplanar points.
 Ray goes off in a southeast direction.
An intersection of geometric
 Ray goes off in a northwest direction. shapes is the set of points they
 Together, rays and make up line l. share in common.
 Line l intersects both planes m and n. l and m intersect at point E.
Note: In geometric figures such as the one above, it is
l and n intersect at point D.
important to remember that, even though planes are m and n intersect in line .
drawn with edges, they extend infinitely in the 2
dimensions shown.
Chapter 1 Basic Geometry

Geometry
Segments, Rays & Lines

Some Thoughts About …

Line Segments
 Line segments are generally named by their endpoints, so the
segment at right could be named either 𝐴𝐵 or 𝐵𝐴.
 Segment 𝐴𝐵 contains the two endpoints (A and B) and all points on line ⃖𝐴𝐵⃗ that are
between them.
 Congruent segments are segments of equal length.
 A bisector splits a segment into two congruent (equal length) segments.

Rays
 Rays are generally named by their single endpoint,
called an initial point, and another point on the ray.
 Ray 𝐴𝐵⃗ contains its initial point A and all points on line
⃖ ⃗ in the direction of the arrow.
𝐴𝐵
 Rays 𝐴𝐵⃗ and 𝐵𝐴⃗ are not the same ray.
 If point O is on line ⃖𝐴𝐵⃗ and is between points A and B,
then rays 𝑂𝐴⃗ and 𝑂𝐵⃗ are called opposite rays. They
⃖ ⃗.
have only point O in common, and together they make up line 𝐴𝐵
Lines
 Lines are generally named by either a single script letter
(e.g., l) or by two points on the line (e.g.,. 𝐴𝐵
⃖ ⃗).
 A line extends infinitely in the directions shown by its
arrows.
 Lines are parallel if they are in the same plane and they
never intersect. Lines f and g, at right, are parallel.
 Lines are perpendicular if they intersect at a 90⁰ angle. A
pair of perpendicular lines is always in the same plane. Lines
f and e, at right, are perpendicular. Lines g and e are also
perpendicular.
 Lines are skew if they are not in the same plane and they
never intersect. Lines k and l, at right, are skew.
(Remember this figure is 3-dimensional.)
Chapter 1 Basic Geometry

Geometry
Distance Between Points

Distance measures how far apart two things are. The distance between two points can be
measured in any number of dimensions, and is defined as the length of the line connecting the
two points. Distance is always a positive number.

1-Dimension (line segment)

Distance - In one dimension, the distance between two points is determined simply by
subtracting the coordinates of the points. If the endpoints are labeled, say A and B, then the
length of segment AB is shown as AB.

Example 1.1: In this segment, the length of AB, i.e., AB, is calculated as: 5 2 𝟕.
A B

Midpoint – the point equidistant from each end of a line segment. That is, the midpoint is
halfway from one end of the segment to the other. To obtain the value of the midpoint, add
the two end values and divide the result by 2.
𝟑
Example 1.2: The midpoint of segment AB in Example 1.1 is: .
𝟐

Distances Between Collinear Points

Recall that collinear points are points on the same line.

A common problem in geometry is to split a line segment into parts based on some knowledge
about the one or more of the parts.

Example 1.3: Find two possible lengths for CD if C, D, and E are collinear, and CE 15.8 cm
and DE 3.5 cm.
It is helpful to use a line diagram when dealing with midpoint problems. There are two
possible line diagrams for this problem: 1) D is between C and E, 2) E is between C and D.
In these diagrams, we show distances instead of point values:
Case 1 Case 2

𝑥 15.8 3.5 𝟏𝟐. 𝟑 𝐜𝐦 𝑥 15.8 3.5 𝟏𝟗. 𝟑 𝐜𝐦

Chapter 1 Basic Geometry

2-Dimensions

Distance – In two dimensions, the distance between two points can be calculated by
considering the line between them to be the hypotenuse of a right triangle. To determine the
length of this line:
 Calculate the difference in the 𝑥-coordinates of the points
 Calculate the difference in the 𝑦-coordinates of the points
 Use the Pythagorean Theorem.

This process is illustrated below, using the variable “d” for distance.

Example 1.4: Find the distance between (-1,1) and (2,5). Based on the
illustration to the left:
x‐coordinate difference: 2 1 3.
y‐coordinate difference: 5 1 4.

Then, the distance is calculated using the formula: d 3 4 9 16 25

We get d 25, so d √25 𝟓

If we define two points generally as (x1, y1) and (x2, y2), then the 2-dimensional distance
formula would be:
distance x x y y .

Midpoint – To obtain the value of the midpoint in two or more dimensions, add the
corresponding coordinates of the endpoints and divide each result by 2.
If you are given the value of the midpoint and asked for the coordinates of an endpoint, you
may choose to calculate a vector, which in this case is simply the difference between two
points.

Example 1.5: Find the distance between P 2, 3 and Q 3, 15 .

The formula for the distance between points is: d 𝑥 𝑥 𝑦 𝑦

Let point 1 be P 2, 3 , and let point 2 be Q 3, 15 . Then,

d 3 2 15 3 √5 12 √169 𝟏𝟑

Note that 5-12-13 is a Pythagorean Triple.

Chapter 1 Basic Geometry

Example 1.6: The midpoint of segment AD is 1, 2 . Point A has coordinates 3, 3 and point
D has coordinates 𝑥, 7 .
It is helpful to use a line diagram when dealing with midpoint problems. Label the endpoints
and midpoint, and identify the coordinates of each:

The difference between points 𝐀 and M can be expressed in two dimensions as a vector
using “〈 〉” instead of “ ”. Let’s find the difference (note: “difference” implies subtraction).
1, 2 Point 𝐌
3, 3 Point 𝐀
〈 2, 5〉 Difference vector (difference between the two points)
The difference vector can then be applied to the midpoint to get the coordinates of point 𝐃.
If I can get from A to M by moving 〈 2, 5〉, then I can get from M to D by moving 〈 2, 5〉.
1, 2 Point 𝐌
〈 2, 5〉 Difference vector
𝟏, 7 Point D. Therefore, we conclude that 𝒙 𝟏.
Note that the 𝑦-value of point 𝐃 in the solution, 7, matches the 𝑦-value of point 𝐃 in the
statement of the problem.

Example 1.7: Find the value of 𝑦 if AC 3𝑦 5, CB 4𝑦 1, AB 9𝑦 12, and C lies

between A and B.
The line diagram is crucial for this problem. It must be drawn with A and B as endpoints and C
between them.

Based on the diagram, we have: 3𝑦 5 4𝑦 1 9𝑦 12

7𝑦 4 9𝑦 12
16 2𝑦
𝟖 𝒚
Chapter 1 Basic Geometry

Partial Distances and Distance Equations

In order to find a distance part-way between two points, we need to interpolate between the
beginning and end points. We must calculate the portion of the distance covered at the desired
time, and then interpolate between the start and end points.
Let 𝑘 be the factor, representing the portion of the total distance that is of interest to us. 𝑘 is
usually given in terms of time, e.g., after 3 hours of a 10-hour journey. In general,
elapsed time
𝑘 .
total time
The formula for the interpolation, then, is:
desired point 𝑘 ∙ ending point 1 𝑘 ∙ starting point
This interpolation formula works for any number of dimensions, taking each coordinate
separately.

Example 1.8: A boat begins a journey at location 2, 5 on a grid and heads directly for point
10, 15 on the same grid. It is estimated that the trip will take 10 hours if the boat travels in a
straight line. At what point of the grid is the boat after 3 hours?
Start at: 2, 5
End at: 10, 15
3 hours → 𝑘 0.3 of the 10 hour period.
 This is the factor for the endpoint: 10, 15 .
 The staring point, 2, 5 gets a factor of 1 0.3 0.7. The factors must always add
to 1.
Ordered pair @ 𝑡 3 hours is: 2, 5 ∙ 0.7 10, 15 ∙ 0.3 𝟒. 𝟒, 𝟖. 𝟎

Note: an alternative method would be to develop separate equations for the 𝑥-variable and
𝑦-variable in terms of time, the 𝑡-variable. These are called parametric equations, and 𝑡 is
the parameter in the equations. For this problem, the parametric equations would be:
𝑡
variable start end start ∙
period length in years
𝑡
𝑥 2 10 2 ∙ 2 0.8𝑡
10
𝑡
𝑦 5 15 5 ∙ 5 𝑡
10
Note that the 10 in the denominator of these equations is the length of time, in hours,
separating the starting point and the ending point.
Solve for the required ordered pair by substituting 𝑡 3 into these equations.
Chapter 1 Basic Geometry

Geometry ADVANCED

Distance Formula in “n” Dimensions

The distance between two points can be generalized to “n” dimensions by successive use of the
Pythagorean Theorem in multiple dimensions. To move from two dimensions to three
dimensions, we start with the two-dimensional formula and apply the Pythagorean Theorem to
add the third dimension.

3 Dimensions

Consider two 3-dimensional points (x1, y1, z1) and (x2, y2, z2). Consider first the situation where
the two z-coordinates are the same. Then, the distance between the points is 2-dimensional,
i.e., d 𝑥 𝑥 𝑦 𝑦 .

We then add a third dimension using the Pythagorean Theorem:

distance d z z

distance x x y y z z
distance x x y y z z

And, finally the 3-dimensional difference formula:

distance x x y y z z

n Dimensions

Using the same methodology in “n” dimensions, we get the generalized n-dimensional
difference formula (where there are n terms beneath the radical, one for each dimension):

distance x x y y z z ⋯ w w

Or, in higher level mathematical notation:

The distance between two points A a ,a ,… ,a and 𝐵 b ,b ,… ,b is

𝑑 𝐴, 𝐵 |𝐴 𝐵| 𝑎 𝑏
Chapter 1 Basic Geometry

Geometry
Angles

Parts of an Angle
An angle consists of two rays with a common
endpoint (or, initial point).
 Each ray is a side of the angle.
 The common endpoint is called the vertex of
the angle.

Naming Angles
Angles can be named in one of two ways:
 Point-vertex-point method. In this method, the angle is named from a point on one ray,
the vertex, and a point on the other ray. This is the most unambiguous method of
naming an angle, and is useful in diagrams with multiple angles sharing the same vertex.
In the above figure, the angle shown could be named ∠BAC or ∠CAB.
 Vertex method. In cases where it is not ambiguous, an angle can be named based solely
on its vertex. In the above figure, the angle could be named ∠A.

Measure of an Angle
There are two conventions for measuring the size of an angle:
 In degrees. The symbol for degrees is ⁰. There are 360⁰ in a full circle. The angle above
measures approximately 360 8 45⁰ (one-eighth of a circle).
 In radians. There are 2𝜋 radians in a complete circle. The angle above measures

approximately radians.

Some Terms Relating to Angles

Angle interior is the area between the rays.
Angle exterior is the area not between the rays.
Adjacent angles are angles that share a ray for a side. ∠BAD and
∠DAC in the figure at right are adjacent angles.
Congruent angles are angles with the same measure.
Angle bisector is a ray that divides the angle into two congruent
angles. Ray AD⃗ bisects ∠BAC in the figure at right.
Chapter 1 Basic Geometry

Geometry
Types of Angles

C
A B D

Supplementary Angles Complementary Angles

Angles A and B are supplementary. Angles C and D are complementary.

Angles A and B form a linear pair. 𝑚∠𝐶 𝑚∠𝐷 90⁰
𝑚∠𝐴 𝑚∠𝐵 180⁰

Angles which are opposite each other when

two lines cross are vertical angles.

Angles E and G are vertical angles.

F Angles F and H are vertical angles.
E G
H 𝑚∠𝐸 𝑚∠𝐺 𝑎𝑛𝑑 𝑚∠𝐹 𝑚∠𝐻

In addition, each angle is supplementary to

the two angles adjacent to it. For example:
Vertical Angles
Angle E is supplementary to Angles F and H.

An acute angle is one that is less than 90⁰. In

the illustration above, angles E and G are
acute angles.

A right angle is one that is exactly 90⁰.

Acute Obtuse
An obtuse angle is one that is greater than
90⁰. In the illustration above, angles F and H
are obtuse angles.

A straight angle is one that is exactly 180⁰.

Right Straight
Chapter 1 Basic Geometry

Example 1.9: Two angles are complementary. The measure of one angle is 21° more than
twice the measure of the other angle. Find the measures of the angles.
Drawing the situation described in the problem is often helpful.
Let the two angles be called angle A and angle B. Let’s rewrite the problem in
terms of these two angles.
Angles A and B are complementary. 𝑚∠A 21° 2 𝑚∠B .
Let the measures of the angles be represented by the names of the angles. Then,
A B 90° 2A 2B 180° A B 90°
A 21° 2B A 2B 21° 67° B 90°
3A 201° 𝐁 𝟐𝟑°
𝐀 𝟔𝟕°
The measures of the two angles then, are, 𝟔𝟕° and 𝟐𝟑°

Example 1.10: If m∠BGC 16x 4° and m∠CGD 2x 13°,

find the value of 𝑥 so that ∠BGD is a right angle.
∠BGD is a right angle (i.e., m∠BGD 90° .

Then, 16𝑥 4° 2𝑥 13° 90°

18𝑥 9° 90°
18𝑥 81°
𝒙 𝟒. 𝟓°

Example 1.11: Find 𝑚∠1 if ∠1 is complementary to ∠2, ∠2 is supplementary to ∠3, and

𝑚∠3 126°.
Let’s turn this into equations because the English is confusing.
𝑚∠1 𝑚∠2 90° (complementary)
𝑚∠2 𝑚∠3 180° (supplementary)
𝑚∠3 126°
Working with these equations from bottom to top, we get:
𝑚∠3 126°
𝑚∠2 𝑚∠3 𝑚∠2 126° 180°, so 𝑚∠2 54°
𝑚∠1 𝑚∠2 𝑚∠1 54° 90° so 𝒎∠𝟏 𝟑𝟔°
Chapter 2 Proofs

Geometry
Conditional Statements

A conditional statement contains both a hypothesis and a conclusion in the following form:

If hypothesis, then conclusion.

Statements linked
For any conditional statement, it is possible to create three related below by red arrows
conditional statements, as shown below. In the table, p is the hypothesis must be either both
of the original statement and q is the conclusion of the original statement. true or both false.

Example
Type of Conditional Statement
Statement is:

Original Statement: If p, then q. (𝒑 → 𝒒)

 Example: If a number is divisible by 6, then it is divisible by 3. TRUE
 The original statement may be either true or false.

Converse Statement: If q, then p. (𝒒 → 𝒑)

 Example: If a number is divisible by 3, then it is divisible by 6.
FALSE
 The converse statement may be either true or false, and this does not
depend on whether the original statement is true or false.

Inverse Statement: If not p, then not q. (~𝒑 → ~𝒒)

 Example: If a number is not divisible by 6, then it is not divisible by 3.
FALSE
 The inverse statement is always true when the converse is true and
false when the converse is false.

Contrapositive Statement: If not q, then not p. (~𝒒 → ~𝒑)

 Example: If a number is not divisible by 3, then it is not divisible by 6.
TRUE
 The Contrapositive statement is always true when the original
statement is true and false when the original statement is false.

Note also that:

 When two statements must be either both true or both false, they are called equivalent
statements.
o The original statement and the contrapositive are equivalent statements.
o The converse and the inverse are equivalent statements.
 If both the original statement and the converse are true, the phrase “if and only if”
(abbreviated “iff”) may be used. For example, “A number is divisible by 3 iff the sum of
its digits is divisible by 3.”
Chapter 2 Proofs

Geometry
Basic Properties of Algebra

Properties of Equality and Congruence.

Definition for Equality Definition for Congruence

Property
For any geometric elements a, b and c.
For any real numbers a, b, and c:
(e.g., segment, angle, triangle)

Reflexive Property 𝑎 𝑎 𝑎≅𝑎

Symmetric Property 𝐼𝑓 𝑎 𝑏, 𝑡ℎ𝑒𝑛 𝑏 𝑎 𝐼𝑓 𝑎 ≅ 𝑏, 𝑡ℎ𝑒𝑛 𝑏 ≅ 𝑎

Transitive Property 𝐼𝑓 𝑎 𝑏 𝑎𝑛𝑑 𝑏 𝑐, 𝑡ℎ𝑒𝑛 𝑎 𝑐 𝐼𝑓 𝑎 ≅ 𝑏 𝑎𝑛𝑑 𝑏 ≅ 𝑐, 𝑡ℎ𝑒𝑛 𝑎 ≅ 𝑐

If 𝑎 𝑏, then either can be If 𝑎 ≅ 𝑏, then either can be

Substitution Property substituted for the other in any substituted for the other in any
equation (or inequality). congruence expression.

More Properties of Equality. For any real numbers a, b, and c:

Property Definition for Equality

Addition Property 𝐼𝑓 𝑎 𝑏, 𝑡ℎ𝑒𝑛 𝑎 𝑐 𝑏 𝑐

Subtraction Property 𝐼𝑓 𝑎 𝑏, 𝑡ℎ𝑒𝑛 𝑎 𝑐 𝑏 𝑐

Multiplication Property 𝐼𝑓 𝑎 𝑏, 𝑡ℎ𝑒𝑛 𝑎 ∙ 𝑐 𝑏∙𝑐

Division Property 𝐼𝑓 𝑎 𝑏 𝑎𝑛𝑑 𝑐 0, 𝑡ℎ𝑒𝑛 𝑎 𝑐 𝑏 𝑐

Properties of Addition and Multiplication. For any real numbers a, b, and c:

Property Definition for Addition Definition for Multiplication

Commutative Property 𝑎 𝑏 𝑏 𝑎 𝑎∙𝑏 𝑏∙𝑎

Associative Property 𝑎 𝑏 𝑐 𝑎 𝑏 𝑐 𝑎∙𝑏 ∙𝑐 𝑎∙ 𝑏∙𝑐

Distributive Property 𝑎∙ 𝑏 𝑐 𝑎∙𝑏 𝑎∙𝑐

Chapter 2 Proofs

Geometry
Inductive vs. Deductive Reasoning

Inductive Reasoning
Inductive reasoning uses observation to form a hypothesis or conjecture. The hypothesis can
then be tested to see if it is true. The test must be performed in order to confirm the
hypothesis.

Example: Observe that the sum of the numbers 1 to 4 is 4 ∙ 5/2 and that the sum of the
numbers 1 to 5 is 5 ∙ 6/2 . Hypothesis: the sum of the first n numbers is 𝑛 ∗ 𝑛 1 /2 .
Testing this hypothesis confirms that it is true.

Deductive Reasoning
Deductive reasoning argues that if something is true about a broad category of things, it is true
of an item in the category.

Example: All birds have beaks. A pigeon is a bird; therefore, it has a beak.

There are two key types of deductive reasoning of which the student should be aware:

 Law of Detachment. Given that 𝒑 → 𝒒, if p is true then q is true. In words, if one thing
implies another, then whenever the first thing is true, the second must also be true.
Example 2.1: Start with the statement: “If a living creature is human, then it has a
brain.” Then because you are human, we can conclude that you have a brain.

 Syllogism. Given that 𝒑 → 𝒒 and 𝒒 → 𝒓, we can conclude that 𝒑 → 𝒓. This is a kind of

transitive property of logic. In words, if one thing implies a second and that second
thing implies a third, then the first thing implies the third.
Example 2.2: Start with the statements: “If my pencil breaks, I will not be able to
write,” and “if I am not able to write, I will not pass my test.” Then I can conclude that
“If my pencil breaks, I will not pass my test.”
Chapter 2 Proofs

Geometry
An Approach to Proofs

Learning to develop a successful proof is one of the key skills students develop in geometry.
The process is different from anything students have encountered in previous math classes, and
may seem difficult at first. Diligence and practice in solving proofs will help students develop
reasoning skills that will serve them well for the rest of their lives.

Requirements in Performing Proofs

 Each proof starts with a set of “givens,” statements that you are supplied and from
which you must derive a “conclusion.” Your mission is to start with the givens and to
proceed logically to the conclusion, providing reasons for each step along the way.
 Each step in a proof builds on what has been developed before. Initially, you look at
what you can conclude from the” givens.” Then as you proceed through the steps in the
proof, you are able to use additional things you have concluded based on earlier steps.
 Each step in a proof must have a valid reason associated with it. So, each statement in
the proof must be furnished with an answer to the question: “Why is this step valid?”

Tips for Successful Proof Development

 At each step, think about what you know and what you can conclude from that
information. Do this initially without regard to what you are being asked to prove. Then
look at each thing you can conclude and see which ones move you closer to what you
are trying to prove.
 Go as far as you can into the proof from the beginning. If you get stuck, …
 Work backwards from the end of the proof. Ask yourself what the last step in the proof
is likely to be. For example, if you are asked to prove that two triangles are congruent,
try to see which of the several theorems about this is most likely to be useful based on
what you were given and what you have been able to prove so far.
 Continue working backwards until you see steps that can be added to the front end of
the proof. You may find yourself alternating between the front end and the back end
until you finally bridge the gap between the two sections of the proof.
 Don’t skip any steps. Some things appear obvious, but actually have a mathematical
reason for being true. For example, 𝑎 𝑎 might seem obvious, but “obvious” is not a
valid reason in a geometry proof. The reason for 𝑎 𝑎 is a property of algebra called
the “reflexive property of equality.” Use mathematical reasons for all your steps.
Chapter 2 Proofs

Proof examples (may use information presented later in this handbook)

Example 2.3: Given: 𝑚∠1 𝑚∠3 180°. Prove: ∠2 ≅ ∠3.

Recall that congruent angles have the same measure.

Step Statement Reason

1 𝑚∠1 𝑚∠3 180° Given.

∠1 and ∠3 are supplementary. If the sum of two angles is 180°, then the
2
angles are supplementary.
3 ∠1 and ∠2 form a linear pair. Diagram.
4 ∠1 and ∠2 are supplementary. If two angles form a linear pair, then the angles
are supplementary.
5 If two angles are supplementary to the same
∠2 ≅ ∠3
angle, then they are congruent.

Example 2.4: Given: KJ ≅ MK, J is the midpoint of HK.

Prove: HJ ≅ MK.
Recall that congruent segments have the same measure.
Thought process. Based on the givens, it appears that the three segments identified in the
diagram are all congruent. That is, 𝐻𝐽 ≅ 𝐾𝐽 ≅ 𝑀𝐾 . We need to work from the congruence
we are given to the one we want to prove by considering how the segments relate to each
other one pair at a time.

Step Statement Reason

𝐾𝐽 ≅ 𝑀𝐾
1 Given
𝐽 is the midpoint of 𝐻𝐾
2 𝐾𝐽 ≅ 𝐻𝐽 A midpoint creates two congruent segments.
Transitive property of congruence (in this case,
3 𝐻𝐽 ≅ 𝑀𝐾 two segments that are each congruent to a third
segment are congruent to each other).

Note: purple text in the proof is explanatory and is not required to complete the proof.
Chapter 2 Proofs

Example 2.5: Given: ∠𝐻 ≇ ∠𝐾.

Prove: ∆𝐽𝐻𝐾 is not isosceles with base 𝐻𝐾 .
Note: the " ≇ " symbol means “is not congruent to”.
We will use proof by contradiction on this problem. In proof by
contradiction, we assume that the opposite of the conclusion is true,
then show that is impossible. This implies that the original
assumption is false, so its opposite (what we want to prove) must be
true.

Step Statement Reason

1 ∠𝐻, ∠𝐾 not congruent Given
Assume ∆𝐽𝐻𝐾 is isosceles with base Assumption intended to lead to a
2
𝐻𝐾. contradiction.
3 𝐽𝐾 𝐽𝐻 Euclid’s definition of isosceles triangle.
4 𝐽𝐾 ≅ 𝐽𝐻 Definition of congruent segments.
Angles opposite congruent sides in a
5 ∠𝐻 ≅ ∠𝐾
triangle are congruent.
6 Contradiction We are given ∠𝐻, ∠𝐾 are not congruent.
7 ∆𝐽𝐻𝐾 is not isosceles with base 𝐻𝐾. Assumption in Step 2 must be false.

Additional proofs are provided throughout this handbook.

Chapter 3 Parallel and Perpendicular Lines

Geometry
Parallel Lines and Transversals

Transversal
Alternate: refers to angles that are on
opposite sides of the transversal.

A B Consecutive: refers to angles that are

C D on the same side of the transversal.

Parallel Lines Interior: refers to angles that are

F between the parallel lines.
E
H Exterior: refers to angles that are
G
outside the parallel lines.

Corresponding Angles
Corresponding Angles are angles in the same location relative to the parallel lines and the
transversal. For example, the angles on top of the parallel lines and left of the transversal (i.e.,
top left) are corresponding angles.

Angles A and E (top left) are Corresponding Angles. So are angle pairs B and F (top right), C
and G (bottom left), and D and H (bottom right). Corresponding angles are congruent.

Alternate Interior Angles

Angles D and E are Alternate Interior Angles. Angles C and F are also alternate interior angles.
Alternate interior angles are congruent.

Alternate Exterior Angles

Angles A and H are Alternate Exterior Angles. Angles B and G are also alternate exterior
angles. Alternate exterior angles are congruent.

Consecutive Interior Angles

Angles C and E are Consecutive Interior Angles. Angles D and F are also consecutive interior
angles. Consecutive interior angles are supplementary.

Note that angles A, D, E, and H are congruent, and angles B, C, F, and G are congruent. In
addition, each of the angles in the first group are supplementary to each of the angles in the
second group.
Chapter 3 Parallel and Perpendicular Lines

Geometry
Multiple Sets of Parallel Lines

Two Transversals
Sometimes, the student is presented two sets of intersecting parallel lines, as shown above.
Note that each pair of parallel lines is a set of transversals to the other set of parallel lines.

A B I J

C D K L

E F M N
G H O P

In this case, the following groups of angles are congruent:

 Group 1: Angles A, D, E, H, I, L, M and P are all congruent.
 Group 2: Angles B, C, F, G, J, K, N, and O are all congruent.
 Each angle in the Group 1 is supplementary to each angle in Group 2.

Some Examples: In the diagram above (Two Transversals), with two pairs of parallel lines,
what types of angles are identified and what is their relationship to each other?
Example 3.1: ∠𝐷 and ∠𝐼.
These angles are alternate interior angles; they are congruent.

Example 3.2: ∠𝐶 and ∠𝐽.

These angles are alternate exterior angles; they are congruent.

Example 3.3: ∠𝐽 and ∠𝑁.

These angles are corresponding angles; they are congruent.

Example 3.4: ∠𝐹 and ∠𝑀.

These angles are consecutive interior angles; they are supplementary.

Example 3.5: ∠𝐺 and ∠𝐿.

These angles do not have a name, but they are supplementary.
Chapter 3 Parallel and Perpendicular Lines

Geometry
Proving Lines are Parallel

The properties of parallel lines cut by a transversal can be used to prove two lines are parallel.

Corresponding Angles
If two lines cut by a transversal have congruent corresponding angles,
then the lines are parallel. Note that there are 4 sets of corresponding
angles.

Alternate Interior Angles

If two lines cut by a transversal have congruent alternate interior angles
congruent, then the lines are parallel. Note that there are 2 sets of
alternate interior angles.

Alternate Exterior Angles

If two lines cut by a transversal have congruent alternate exterior
angles, then the lines are parallel. Note that there are 2 sets of
alternate exterior angles.

Consecutive Interior Angles

If two lines cut by a transversal have supplementary consecutive
interior angles, then the lines are parallel. Note that there are 2 sets of
consecutive interior angles.
Chapter 3 Parallel and Perpendicular Lines

Geometry
Parallel and Perpendicular Lines in the Coordinate Plane

Parallel Lines
Two lines are parallel if their slopes are equal.
 In 𝑦 𝑚𝑥 𝑏 form, if the values of 𝑚 are
the same.
Example 3.6: 𝑦 2𝑥 3 and
𝑦 2𝑥 1
 In Standard Form, if the coefficients of 𝑥 and
𝑦 are proportional between the equations.
Example 3.7: 3𝑥 2𝑦 5 and
6𝑥 4𝑦 7
 Also, if the lines are both vertical (i.e., their
slopes are undefined).
Example 3.8: 𝑥 3 and
𝑥 2

Perpendicular Lines
Two lines are perpendicular if the product of their
slopes is 𝟏. That is, if the slopes have different
signs and are multiplicative inverses.
 In 𝑦 𝑚𝑥 𝑏 form, the values of 𝑚
multiply to get 1..
Example 3.9: 𝑦 6𝑥 5 and
𝑦 𝑥 3

 In Standard Form, if you add the product of

the x-coefficients to the product of the y-
coefficients and get zero.
Example 3.10: 4𝑥 6𝑦 4 and
3𝑥 2𝑦 5 because 4∙3 6∙ 2 0

 Also, if one line is vertical (i.e., 𝑚 is undefined) and one line is horizontal (i.e., 𝑚 0).
Example 3.11: 𝑥 6 and
𝑦 3
Chapter 3 Parallel and Perpendicular Lines

Example 3.12: Write the equation of the perpendicular bisector of CD if C 4, 3 and

D 8, 9 .

Line containing 𝐶𝐷 :
9 3 12
𝑚 3
8 47 4
Midpoint of 4, 3 and 8, 9 is halfway between them: 6, 3

Perpendicular bisector: Slope is the “negative reciprocal” of the slope of ⃖𝐶𝐷⃗ because the
lines are perpendicular. Also, 6, 3 is a point on the perpendicular bisector.

𝑚
𝟏 𝟏 𝟏
Equation: 𝒚 𝟑 𝒙 𝟔 or 𝒚 𝒙 𝟔 𝟑 or 𝒚 𝒙 𝟓
𝟑 𝟑 𝟑
point-slope form ℎ-𝑘 form slope-intercept form

Example 3.13: Write an equation of the line that can be used to calculate the distance between
4, 3 and the line 𝑦 𝑥 9.

The distance between a point and a line is the length of the

segment connecting the point to the line at a right angle.
See the diagram to the right.
So, this question is asking for the equation of the line
perpendicular to 𝑦 𝑥 9 that contains the point
4, 3 , but is not asking us to calculate the distance.
The perpendicular line will have a slope that is the opposite
reciprocal of the original line:
1 7
𝑚
2 2
7
Then, the equation of the perpendicular line (in ℎ-𝑘 form) is:
𝟕
𝒚 𝒙 𝟒 𝟑
𝟐

Note: If we were asked to calculate the distance between Point A and the line 𝑦 𝑥
9, we would first need to find Point B at the intersection of the two lines shown, and then
measure the distance between the two points using the distance formula.
Chapter 3 Parallel and Perpendicular Lines

Geometry
Proportional Segments

Parallel Line in a Triangle

A line is parallel to one side of a triangle iff it divides the other two sides proportionately.
This if-and-only-if statement breaks down into the following two statements:
 If a line (or ray or segment) is parallel to one side of a triangle, then
it divides the other two sides proportionately.
 If a line (or ray or segment) divides two sides of a triangle
proportionately, then it is parallel to the third side.
In the diagram to the right, we see that 𝐴𝐵 ∥ 𝐸𝐷. We can conclude that:

and as well as a number of other equivalent proportion equalities.

Conversely, if we knew one of the proportions above, but were not given that the segments
were parallel, we could conclude that 𝐴𝐵 ∥ 𝐸𝐷 because of the equal proportions.

Example 3.14: Determine whether 𝐴𝐵 ∥ 𝐸𝐷 in the diagram to the right.

Let’s check the proportions. Is ?

𝐶𝐸 12 3 𝐸𝐴 6 3
𝐶𝐷 8 2 𝐷𝐵 4 2
Since the proportions of the two sides are equal, we can conclude that 𝑨𝑩 ∥ 𝑬𝑫.

Three or More Parallel Lines

Three or more parallel lines divide any transversals proportionately.
In the diagram to the right, we see that the three horizontal lines
(or rays or segments) are parallel. We can conclude that:

and .

The converse of this is not true. That is, if three or more lines
divide transversals into proportionate parts, it is not necessarily
true that the lines are parallel.
Chapter 3 Parallel and Perpendicular Lines

Example 3.15: Given that the three horizontal lines in the

diagram to the right are parallel, what is the values of 𝑥?
The three parallel horizontal lines in the diagram divide the
vertical lines into proportional segments.
25 30
10 𝑥
25𝑥 300
𝒙 𝟏𝟐

Angle Bisector
An angle bisector in a triangle divides the opposite sides into segments that are proportional
to the adjacent sides.
In the diagram to the right, we see that ∠𝐷 is bisected, crea ng
segments 𝐴𝐵 and 𝐵𝐶 opposite ∠𝐷. We can conclude that:

and .

The converse of this is also true. That is, if a line (or ray or segment) through a vertex of a
triangle splits the opposite side into segments that are proportional to the adjacent sides, then,
that line (or ray or segment) bisects the vertex angle. That is, if the above proportions are true,
then 𝐷𝐵 bisects ∠𝐷.

Example 3.16: Find the value of 𝑥 in the diagram.

An angle bisector in a triangle divides the opposite sides into segments
that are proportional to the adjacent sides. So,
18 𝑥 3
𝑥 10
𝑥 𝑥 3 18 ∙ 10
𝑥 3𝑥 180
𝑥 3𝑥 180 0
𝑥 12 𝑥 15 0 → 𝑥 12, 15
If 𝑥 15, we have negative side lengths, so we discard the solution 𝑥 15.
If 𝑥 12, the sides of ∆𝐵𝐴𝐷 would be 18, 15, 22 , which makes a valid triangle.
Conclude: 𝒙 𝟏𝟐.
Chapter 4 Triangles - Basic

Geometry
What Makes a Triangle?

Definition – A triangle is a plane figure with three sides and three angles.
 Draw three points that are not on the same line, connect them, and you have a triangle.
The three points you started with are called vertices.
 Three points determine a plane, so a triangle must have all of its parts on the same
plane.

Parts of a Triangle
 Vertices – the points where the sides intersect. In
the diagram to the right, the vertices are the red
points. Vertices are typically labeled with capital
letters.
 Legs – the sides of a triangle are also called the
triangle’s legs. In diagrams, the lengths of the legs are often represented by lower case
letters corresponding to the angles opposite them.
 Angles (interior angles) – the angles formed at each vertex are the triangle’s angles. In
the diagram above, the triangle has interior angles ∠𝐴, ∠𝐵, ∠𝐶 indicated by the green
arcs at the vertices. These angles could be named in various ways, for example:
o ∠𝐴 ∠𝐵𝐴𝐶 ∠𝐶𝐴𝐵.
o Naming the angle with a single vertex is acceptable if there is no ambiguity about
which angle is being referenced, e.g., ∠𝐴.
o If any ambiguity exists as to which angle is being referenced, the angle must be
named using three points: two of the points must be on the sides enclosing the
angle and the vertex must be in the middle, e.g., ∠𝐵𝐴𝐶 or ∠𝐶𝐴𝐵.
o Alternatively, an angle may be named with a letter or symbol next to its arc.
 Altitudes – line segments from each vertex to the opposite side of the triangle that are
perpendicular to that opposite side. In the diagram below left, an altitude is labeled h.
 Medians – line segments from each vertex to the midpoint of the opposite side of the
triangle. In the diagram below right, a median is labeled m.
Chapter 4 Triangles - Basic

Sum of Interior Angles

The sum of the interior angles of a triangle is 180°. If two
of the interior angles in a triangle have known measures,
the measure of the third can be easily calculated. For
example, in the diagram to the right, if 𝑚∠𝐴 and 𝑚∠𝐵
are known, 𝑚∠𝐶 can be calculated as:
𝑚∠𝐶 180° 𝑚∠𝐴 𝑚∠𝐵.

Third Angle Theorem: If two interior angles in one triangle are congruent to two interior angles
in another triangle, then the third interior angles in the two triangles are congruent.
This follows from the fact that the sum of the three interior angles in each triangle must be
180°.

Example 4.1: Given 𝐴𝐷 ⊥ 𝐵𝐶 , 𝐴𝐷 bisects ∠𝐵𝐴𝐶, prove ∠𝐴𝐵𝐷 ≅ ∠𝐴𝐶𝐷.

This can be proven in multiple ways. Let’s prove it with the Third Angle
Theorem.

Step Statement Reason

𝐴𝐷 ⊥ 𝐵𝐶 .
1 Given.
𝐴𝐷 bisects ∠𝐵𝐴𝐶.
∠𝐴𝐷𝐵 is a right angle. 𝐴𝐷 ⊥ 𝐵𝐶 . Perpendicular lines
2
∠𝐴𝐷𝐶 is a right angle. form right angles.
All right angles are congruent
3 ∠𝐴𝐷𝐵 ≅ ∠𝐴𝐷𝐶.
(they all measure 90°).
4 ∠𝐵𝐴𝐷 ≅ ∠𝐶𝐴𝐷. 𝐴𝐷 bisects ∠𝐵𝐴𝐶.
Third Angle Theorem (triangles
5 ∠𝐴𝐵𝐷 ≅ ∠𝐴𝐶𝐷
are ∆𝐴𝐷𝐵 and ∆𝐴𝐷𝐶).
Chapter 4 Triangles - Basic

Geometry
Inequalities in Triangles

Angles and their opposite sides in triangles are related. In fact, this is often reflected in the
labeling of angles and sides in triangle illustrations.

Angles and their opposite sides are often

labeled with the same letter. An upper case
letter is used for the angle and a lower case
letter is used for the side.

The relationship between angles and their opposite sides translates into the following triangle
inequalities:
If 𝒎∠𝑪 𝒎∠𝑩 𝒎∠𝑨, then 𝒄 𝑏 𝑎
If 𝒎∠𝑪 𝒎∠𝑩 𝒎∠𝑨, then 𝒄 𝒃 𝒂

That is, in any triangle,

 The largest side is opposite the largest angle.
 The medium side is opposite the medium angle.
 The smallest side is opposite the smallest angle.

Other Inequalities in Triangles

Triangle Inequality: The sum of the lengths of any two sides of a triangle is
greater than the length of the third side. Also, the difference of the lengths
of any two sides is smaller than the length of the third side. If 𝑎 𝑏:
𝒂 𝒃 𝒄 𝒂 𝒃 and similar for the other sides.

Exterior Angle Inequality: The measure of an external angle is greater than the measure of
either of the two non-adjacent interior angles. That is, in the figure below:
𝒎∠𝑫𝑨𝑩 𝑚∠𝑩 and 𝒎∠𝑫𝑨𝑩 𝑚∠𝐶.

Exterior Angle Equality: The measure of an external

angle is equal to the sum of the measures of the two
non-adjacent interior angles. That is, in the figure to the right:
𝒎∠𝑫𝑨𝑩 𝒎∠𝑩 𝒎∠𝑪.

Note: the Exterior Angle Equality is typically more useful than the Exterior Angle Inequality.
Chapter 4 Triangles - Basic

Sides of a Triangle
The lengths of the sides of a triangle are limited: given the lengths of any two sides, the length
of the third side must be greater than their difference and less than their sum. That is, if the
sides of a triangle have lengths 𝑎, 𝑏, and 𝑐, and you know the values of, for example, 𝑎 and 𝑏
with 𝑎 the larger of the two, then:
𝑎 𝑏 𝑐 𝑎 𝑏

Example 4.2: If a triangle has two sides with lengths 13 and 8, what are the possible lengths of
the third side?
If we let 𝑐 represent the length of the third side of a triangle, with 𝑎 13, 𝑏 8, then:
 𝑐 must be greater than the difference of 𝑎 and 𝑏: 𝑐 13 8 → 𝑐 5.
 𝑐 must be less than the sum of 𝑎 and 𝑏: 𝑐 13 8 → 𝑐 21.
If we put all of this together in a single inequality, we get:
13 8 𝑐 13 8
𝟓 𝒄 𝟐𝟏

Also, as indicated above, there are limits to the lengths of sides if the measures of the interior
angles of the triangle are known. In particular,
 The longest side of a triangle is opposite the largest
interior angle.
 The shortest side of a triangle is opposite the smallest
interior angle.
In general, if we know that 𝑚∠𝐶 𝑚∠𝐵 𝑚∠𝐴, then we know that 𝑐 𝑏 𝑎.

Example 4.3: Identify the longest segment in the diagram shown.

Let’s see what we know in each of the triangles. Note that:
 The sum of the angles in each triangle must be 180° and
 Sides across from larger angles in the same triangle are larger.

In ∆𝐴𝐵𝐶: In ∆𝐴𝐷𝐸:
In ∆𝐴𝐶𝐷:
 𝑚∠𝐵𝐴𝐶 43°  𝑚∠𝐸𝐴𝐷 38°
 𝐴𝐵 𝐵𝐶 𝐴𝐶  𝐷𝐸 𝐴𝐸 𝐴𝐷  𝐶𝐷 𝐴𝐶 𝐴𝐷

Therefore, the two candidates for longest segment are 𝐴𝐶 and 𝐴𝐷. Looking closer at the
above inequalities, we notice that in ∆𝐴𝐶𝐷, we have 𝐴𝐶 𝐴𝐷. Therefore, the longest
segment is: 𝑨𝑫.
Chapter 4 Triangles - Basic

The discussion above addresses angles within a single triangle. There is another relationship
that allows us to compare the lengths of sides in two different triangles. In particular,
If two triangles have two pairs of congruent sides, consider the angles between the
congruent sides. The triangle with the larger of these angles has the larger side
opposite that angle. This is illustrated in the next example.

Example 4.4: Find the range of values for 𝑥.

Note: never trust the relative sizes of angles and sides in a diagram. For example, the two
sides with length 9 in this diagram are drawn with different lengths!
We know two things involving 𝑥:
 The side labeled 3𝑥 4 must be positive. So, 3𝑥 4 0.
 The two angles shown (39°) and (41°) share two congruent sides
(one side with length 9 and one side of unknown length that is
shared by the two angles). Therefore, the side opposite the
smaller angle must be smaller than the side opposite the larger
angle. So, 3𝑥 4 17.
Combining these two inequalities into a single compound inequality, and solving:
Starting inequality: 0 3𝑥 4 17
Add 4: 4 3𝑥 21
𝟒
Divide by 3: 𝒙 𝟕
𝟑

Example 4.5: Given ∆ABC with A 3, 4 , B 7, 1 , C 2, 1 , and median AD, find the
coordinates of point D.
Many times, you need to draw the situation for a given problem. This is not one of those
times.
Point D is the midpoint of the side of the triangle opposite the given vertex.
In this problem, Point A is the vertex in question (it is on the median 𝐴𝐷). So, Point D is the
midpoint of the points B 7, 1 and C 2, 1 .
So, the coordinates of Point D are: 7, 1 2, 1 2 𝟒. 𝟓, 𝟎
Chapter 4 Triangles - Basic

Example 4.6: Given ∆ABC with A 2,5 , B 3,5 , C 6, 1 , and

altitude CD, find the coordinates of point D.
An altitude of a triangle is a line segment drawn from a vertex
to a point on the opposite side (extended, if necessary) that is
perpendicular to that side.
This problem is very straightforward once you graph it. To find the base point of the
altitude, we can look at the intersection of the two lines on which Point D lies.
Line containing 𝐵𝐴: 𝑦 5
Line containing 𝐶𝐷 . 𝑥 6 is perpendicular to 𝑦 5 and contains C 6, 1 .
Therefore, Point D has coordinates: 𝟔, 𝟓 .

Example 4.7: Write and solve an inequality for 𝑥.

Each side must have a positive measure, so: 𝑥 2 0
𝑥 2
Also, in the triangle on the left, we have:
7 6 𝑥 2 7 6
1 𝑥 2 13
3 𝑥 15
Next, both outside triangles have sides of length 6 and 7 with angles between them.
Since the measure of the angle in the triangle on the left 54° is less than the one in the
triangle on the right 67° , the opposite side on the left must be less than the opposite side
on the right. So, 𝑥 2 11.
𝑥 13
Putting it all together, we have: 3 𝑥, equivalent to 𝑥 3, which is more restrictive than
𝑥 2, so we use the more restrictive 3 𝑥.
We also have: 𝑥 13, which is more restrictive than 𝑥 15, so we use the more
restrictive 𝑥 13.
Finally, since 3 𝑥 and 𝑥 13, we have 𝟑 𝒙 𝟏𝟑
Chapter 4 Triangles - Basic

Geometry
Types of Triangles

Scalene Isosceles
A Scalene Triangle has 3 sides of different An Isosceles Triangle has 2 sides the same
lengths. Because the sides are of length (i.e., congruent). Because two
different lengths, the angles must also be sides are congruent, two angles must also
of different measures. be congruent.

Equilateral Right
An Equilateral Triangle has all 3 sides the A Right Triangle is one that contains a 90⁰
same length (i.e., congruent). Because all angle. It may be scalene or isosceles, but
3 sides are congruent, all 3 angles must cannot be equilateral. Right triangles
also be congruent. This requires each have sides that meet the requirements of
angle to be 60⁰. the Pythagorean Theorem.

60⁰ 60⁰

60⁰
Chapter 4 Triangles - Basic

Example 4.8: Find the values of 𝑥 and 𝑦 based on the diagram.

This problem becomes easier if we label a few more angles.
See the diagram on the right.
Angles opposite congruent sides in isosceles triangles are
congruent, which helps with our labeling.
In the triangle on the right, the sum of the interior angles must be 180°, so,
𝑏 180 37 37 106.
The adjacent angles marked 𝑎° and 𝑏° form a linear pair, so,
𝑎 180 106 74.
The center triangle has two angles of 𝑎° and one angle of
𝑦°, which must add to 180°, so,
𝒚 180 74 74 𝟑𝟐.
Finally, along the top right, angles marked 37°, 𝑎°, and 𝑥° must add to 180° in order to
form a straight angle, so,
𝒙 180 37 74 𝟔𝟗.

Example 4.9: Find the value of 𝑦 and the perimeter of the triangle.
Legs opposite congruent angles in isosceles triangles are congruent.
𝑦 5𝑦 24
𝑦 5𝑦 24 0
𝑦 8 𝑦 3 0
𝒚 𝟖, 𝟑 (2 possibilities)
If we plug each of these values into the lengths of the sides shown in the diagram, we
always get positive numbers, so there are two cases. If we had gotten a length that was
negative for either 𝑦 8 or 𝑦 3, we would have had to discard that solution.
The perimeter of the triangle is: 𝑃 𝑦 4𝑦 15 5𝑦 24 𝑦 9𝑦 39.
Case 1 (𝑦 8): 𝑃 𝑦 9𝑦 39 8 9∙8 39 𝟏𝟕𝟓. (we are not given units)
Sides of this triangle are 64, 64, 47, which gives a viable triangle.
Case 2 (𝑦 3): 𝑃 𝑦 9𝑦 39 3 9∙ 3 39 𝟐𝟏.
Sides of this triangle are 9, 9, 3, which gives a viable triangle.
Chapter 4 Triangles - Basic

Geometry
Congruent Triangles

The following theorems present conditions under which triangles are congruent.

Side-Angle-Side (SAS) Congruence

SAS congruence requires the congruence of
two sides and the angle between those sides.
Note that there is no such thing as SSA
congruence; the congruent angle must be
between the two congruent sides.

ide-Side-Side (SSS) Congruence

SSS congruence requires the congruence of all
three sides. If all of the sides are congruent
then all of the angles must be congruent. The
converse is not true; there is no such thing as
AAA congruence.

Angle-Side-Angle (ASA) Congruence

ASA congruence requires the congruence of

two angles and the side between those angles.

Note: ASA and AAS combine to provide

congruence of two triangles whenever
any two angles and any one side of the
Angle-Angle-Side (AAS) Congruence triangles are congruent.

AAS congruence requires the congruence of

two angles and a side which is not between
those angles.

Hypotenuse Leg (HL) Congruence

HL can be used if the triangles in question
have right angles. It requires the congruence
of the hypotenuse and one of the other legs.
Chapter 4 Triangles - Basic

CPCTC
CPCTC means “corresponding parts of congruent triangles are congruent.” It is a very
powerful tool in geometry proofs and is often used shortly after a step in the proof where a pair
of triangles is proved to be congruent.

Example 4.10: Given that BE is a perpendicular bisector of CD, find ED.

In the diagram, CA ≅ DA because BE bisects CD.
So, ∆CAB ≅ ∆DAB by SAS, and ∆CAE ≅ ∆DAE by SAS.
The two hypotenuses (yep, that’s the plural form of
hypotenuse) of the triangles on the right side of the
diagram are congruent. So,
7x 10 2x 20
5x 30
x 6
CA DA, so y 4
Finally, ED EC x 2y (because ∆CAE ≅ ∆DAE, and ED and EC are corresponding
parts of those congruent triangles).
ED EC 6 2 4 𝟏𝟒

Example 4.11: Given ∆𝑃𝑄𝑅 ≅ ∆𝐽𝐾𝐿, 𝑃𝑄 9𝑥 45, 𝐽𝐾 6𝑥 15, 𝐾𝐿 2𝑥, 𝐽𝐿 5𝑥, what is
the value of 𝑥?
It’s helpful to draw a picture for this problem.
Notice that congruent segments 𝑃𝑄 and 𝐽𝐾 have
measures 9𝑥 45 and 6𝑥 15. Then:
9𝑥 45 6𝑥 15
3𝑥 60
𝒙 𝟐𝟎
We are not quite finished, even though we found a value for 𝑥. We need to check the sides
of ∆𝐽𝐾𝐿 to make sure this results in a viable triangle:
2𝑥 40, 5𝑥 100, 6𝑥 15 135
Sides of 40, 100, 135 are viable in a triangle because 40 100 135.
Note that if 𝑃𝑄 12𝑥 45, we would have calculated 𝑥 10. Then, the sides would
have been 20, 50, 75, which is not a viable triangle because 20 50 75. If this were
the case, this problem would have no solution.
Chapter 4 Triangles - Basic

Example 4.12: Given 𝐴𝐷 ⊥ 𝐵𝐶 , 𝐴𝐷 bisects ∠𝐵𝐴𝐶, prove ∠𝐵 ≅ ∠𝐶.

It looks like we want to head toward ∆𝐴𝐷𝐵 ≅ ∆𝐴𝐷𝐶, and use CPCTC.

Step Statement Reason

𝐴𝐷 ⊥ 𝐵𝐶 .
1 Given.
𝐴𝐷 bisects ∠𝐵𝐴𝐶.
∠𝐴𝐷𝐵 is a right angle. 𝐴𝐷 ⊥ 𝐵𝐶 . Perpendicular lines
2
∠𝐴𝐷𝐶 is a right angle. form right angles.
3 ∠𝐴𝐷𝐵 ≅ ∠𝐴𝐷𝐶. All right angles are congruent.
4 𝐴𝐷 ≅ 𝐴𝐷. Reflexive property of congruence.
5 ∠𝐵𝐴𝐷 ≅ ∠𝐶𝐴𝐷. 𝐴𝐷 bisects ∠𝐵𝐴𝐶.
6 ∆𝐴𝐷𝐵 ≅ ∆𝐴𝐷𝐶 ASA congruence theorem.
7 ∠𝐵 ≅ ∠𝐶 CPCTC.

Example 4.13: Given 𝐴𝐷 ∥ 𝐶𝐵 , 𝐴𝐵 ∥ 𝐶𝐷 , prove ∠𝐵 ≅ ∠𝐷

With parallel lines, we will typically look for alternate interior
angles or corresponding angles to prove things. Also, this looks
like a situation where we prove congruent triangles and can use
CPCTC.

Step Statement Reason

𝐴𝐷 ∥ 𝐶𝐵.
1 Given.
𝐴𝐵 ∥ 𝐶𝐷 .

∠𝐵𝐴𝐶 ≅ ∠𝐷𝐶𝐴. Alternate interior angles of 𝐴𝐵 ∥ 𝐶𝐷 ,

2
with 𝐴𝐶 a transversal.

3 ∠𝐵𝐶𝐴 ≅ ∠𝐷𝐴𝐶. Alternate interior angles of 𝐴𝐷 ∥ 𝐶𝐵 ,

with 𝐴𝐶 a transversal.
4 𝐴𝐶 ≅ 𝐴𝐶 . Reflexive property of congruence.
5 ∆𝐵𝐴𝐶 ≅ ∆𝐷𝐶𝐴 ASA congruence theorem.
6 ∠𝐵 ≅ ∠𝐷 CPCTC.
Chapter 4 Triangles - Basic

Geometry
Centers of Triangles

The following are all points which can be considered the center of a triangle.

Centroid (Medians)

The centroid is the intersection of the three medians of a triangle. A median is a

line segment drawn from a vertex to the midpoint of the side of the triangle that
is opposite the vertex.

 The centroid is located 2/3 of the way from a vertex to the opposite side. That is, the distance from a
vertex to the centroid is double the length from the centroid to the midpoint of the opposite line.
 The medians of a triangle create 6 inner triangles of equal area.

Orthocenter (Altitudes)

The orthocenter is the intersection of the three altitudes of a triangle. An

altitude is a line segment drawn from a vertex to a point on the opposite side
(extended, if necessary) that is perpendicular to that side.

 In an acute triangle, the orthocenter is inside the triangle.

 In a right triangle, the orthocenter is the right angle vertex.
 In an obtuse triangle, the orthocenter is outside the triangle.

Circumcenter (Perpendicular Bisectors)

The circumcenter is the intersection of the

perpendicular bisectors of the three sides of the
triangle. A perpendicular bisector is a line which Euler Line: Interestingly,
the centroid, orthocenter
both bisects the side and is perpendicular to the
and circumcenter of a
side. The circumcenter is also the center of the
triangle are collinear (i.e.,
circle circumscribed about the triangle. lie on the same line,
which is called the Euler
 In an acute triangle, the circumcenter is inside the triangle. Line).
 In a right triangle, the circumcenter is the midpoint of the hypotenuse.
 In an obtuse triangle, the circumcenter is outside the triangle.

Incenter (Angle Bisectors)

The incenter is the intersection of the angle bisectors of the three angles of
the triangle. An angle bisector cuts an angle into two congruent angles, each
of which is half the measure of the original angle. The incenter is also the
center of the circle inscribed in the triangle.
Chapter 4 Triangles - Basic

Example 4.14: Given ∆CAB, CG 3𝑥 2, GF 𝑥 3, find 𝑥 and 𝐶𝐹.

Centroid
 The centroid is the intersection of the three
medians of a triangle.
 A median is a line segment drawn from a vertex to
the midpoint of the side of the triangle that is opposite the vertex.
 The centroid is located 2/3 of the way from a vertex to the opposite side.
 The medians of a triangle create 6 inner triangles of equal area.

From the diagram, we can see that Points D, E, F are midpoints of the sides of ∆ABC. So,
AD, BE, CF are medians of ∆ABC.
Point G is the centroid of ∆ABC because it is the intersection of the three medians of the
triangle. Therefore,
CG 2 GF
3𝑥 2 2 𝑥 3
3𝑥 2 2𝑥 6
𝒙 𝟖
Then, 𝐂𝐅 CG GF 3𝑥 2 𝑥 3 4𝑥 1
4 8 1 𝟑𝟑
Chapter 4 Triangles - Basic

Geometry
Length of Altitude, Median and Angle Bisector

Altitude (Height)
The formula for the length of a height of a triangle is derived
from Heron’s formula for the area of a triangle:
𝟐 𝒔 𝒔 𝒂 𝒔 𝒃 𝒔 𝒄
𝒉
𝒄
𝟏
where, 𝒔 𝒂 𝒃 𝒄 , and
𝟐
𝒂, 𝒃, 𝒄 are the lengths of the sides of the triangle.

Median
The formula for the length of a median of a triangle is:
𝟏
𝒎 𝟐𝒂𝟐 𝟐𝒃𝟐 𝒄𝟐
𝟐
where, 𝒂, 𝒃, 𝒄 are the lengths of the sides of the triangle.

Angle Bisector
The formula for the length of an angle bisector of a triangle is:

𝒄𝟐
𝒕 𝒂𝒃 𝟏 𝟐
𝒂 𝒃

where, 𝒂, 𝒃, 𝒄 are the lengths of the sides of the triangle.

Example 4.15: Find the length of CF, if CF is a median of ∆ABC.

Point F bisects AB, so AB 2∙5 10. From the formula
above, we have:
1
𝐂𝐅 2 ∙ AC 2 ∙ CB AB
2
1 1 1
2∙4 2∙8 10 √60 ∙ 2√15 √𝟏𝟓
2 2 2
Chapter 5 Polygons

Geometry
Polygons - Basics

Basic Definitions
Polygon: a closed path of three or more line segments, where:
 no two sides with a common endpoint are collinear, and
 each segment is connected at its endpoints to exactly two other segments.
Side: a segment that is connected to other segments (which are also sides) to form a polygon.
Vertex: a point at the intersection of two sides of the polygon. (plural form: vertices)
Diagonal: a segment, from one vertex to another, which is not a side.

Vertex

Diagonal
Side

Concave: A polygon in which it is possible to draw a diagonal “outside” the

polygon. (Notice the orange diagonal drawn outside the polygon at
right.) Concave polygons actually look like they have a “cave” in them.

Convex: A polygon in which it is not possible to draw a diagonal “outside” the

polygon. (Notice that all of the orange diagonals are inside the polygon
at right.) Convex polygons appear more “rounded” and do not contain
“caves.”

Names of Some Common Polygons

Number Number Names of polygons

of Sides Name of Polygon of Sides Name of Polygon are generally formed
3 Triangle 9 Nonagon from the Greek
language; however,
4 Quadrilateral 10 Decagon
some hybrid forms of
5 Pentagon 11 Undecagon Latin and Greek (e.g.,
6 Hexagon 12 Dodecagon undecagon) have
7 Heptagon 20 Icosagon crept into common
usage.
8 Octagon n n‐gon
Chapter 5 Polygons

Geometry
Polygons – More Definitions

Definitions “Advanced” Definitions:

Equilateral: a polygon in which all of the sides are equal in length. Simple Polygon: a
Equiangular: a polygon in which all of the angles have the same polygon whose sides do
measure. not intersect at any
location other than its
Regular: a polygon which is both equilateral and equiangular. That endpoints. Simple
is, a regular polygon is one in which all of the sides have the same polygons always divide a
length and all of the angles have the same measure. plane into two regions –
one inside the polygon and
one outside the polygon.
Interior Angle: An angle formed by two sides of a polygon. The
Complex Polygon: a
angle is inside the polygon.
polygon with sides that
Exterior Angle: An angle formed by one side of a polygon and the intersect someplace other
line containing an adjacent side of the polygon. The angle is outside than their endpoints (i.e.,
the polygon. not a simple polygon).
Complex polygons do not
always have well-defined
insides and outsides.

Exterior Interior Skew Polygon: a polygon

Angle Angle for which not all of its
vertices lie on the same
plane.

How Many Diagonals Does a Convex Polygon Have?

Believe it or not, this is a common question with a simple solution. Consider a polygon with
𝒏 𝟑 sides and, therefore, 𝒏 vertices.
 Each of the n vertices of the polygon can be connected to 𝒏 𝟑 other vertices with
diagonals. That is, it can be connected to all other vertices except itself and the two to
which it is connected by sides. So, there are 𝒏 ∙ 𝒏 𝟑 lines to be drawn as diagonals.
 However, when we do this, we draw each diagonal twice because we draw it once from
each of its two endpoints. So, the number of diagonals is actually half of the number we
calculated above.
 Therefore, the number of diagonals in an n-sided polygon is:
𝑛∙ 𝑛 3
2
Chapter 5 Polygons

Geometry
Interior and Exterior Angles of a Polygon

Interior Angles
Interior Angles
The sum of the interior angles in an 𝑛-sided polygon is:
Sum of Each
Sides Interior Interior
∑ 𝑛 2 ∙ 180° Angles Angle
3 180⁰ 60⁰
If the polygon is regular, you can calculate the measure of 4 360⁰ 90⁰
each interior angle as: 5 540⁰ 108⁰
6 720⁰ 120⁰
∙ ° 7 900⁰ 129⁰
8 1,080⁰ 135⁰
9 1,260⁰ 140⁰
Notation: The Greek letter “Σ” is equivalent 10 1,440⁰ 144⁰
to the English letter “S” and is math short-hand
for a summation (i.e., addition) of things.

Exterior Angles
Exterior Angles
No matter how many sides there are in a polygon, the sum
Sum of Each
of the exterior angles is: Sides Exterior Exterior
Angles Angle
∑ 360⁰ 3 360⁰ 120⁰
4 360⁰ 90⁰
If the polygon is regular, you can calculate the measure of
5 360⁰ 72⁰
each exterior angle as:
6 360⁰ 60⁰
7 360⁰ 51⁰
⁰
8 360⁰ 45⁰
9 360⁰ 40⁰
10 360⁰ 36⁰
Chapter 6 Quadrilaterals

Geometry
Definitions of Quadrilaterals

Name Definition
Quadrilateral A polygon with 4 sides.

A quadrilateral with two consecutive pairs of congruent sides, but

Kite
with opposite sides not congruent.

Trapezoid A quadrilateral with exactly one pair of parallel sides.

Isosceles Trapezoid A trapezoid with congruent legs.

Parallelogram A quadrilateral with both pairs of opposite sides parallel.

Rectangle A parallelogram with all angles congruent (i.e., right angles).

Rhombus A parallelogram with all sides congruent.

Square A quadrilateral with all sides congruent and all angles congruent.

Quadrilateral Tree:

Quadrilateral

Kite Parallelogram Trapezoid

Rectangle Rhombus Isosceles

Trapezoid

Square
Chapter 6 Quadrilaterals

Geometry
Figures of Quadrilaterals

Kite Trapezoid Isosceles Trapezoid

 2 consecutive pairs of  1 pair of parallel sides  1 pair of parallel sides
congruent sides (called “bases”)  Congruent legs
 1 pair of congruent  Angles on the same  2 pair of congruent base
opposite angles “side” of the bases are angles
 Diagonals perpendicular supplementary  Diagonals congruent

Parallelogram Rectangle
 Both pairs of opposite sides parallel  Parallelogram with all angles
 Both pairs of opposite sides congruent congruent (i.e., right angles)
 Both pairs of opposite angles congruent  Diagonals congruent
 Consecutive angles supplementary
 Diagonals bisect each other

Rhombus Square
 Parallelogram with all sides congruent  Both a Rhombus and a Rectangle
 Diagonals perpendicular  All angles congruent (i.e., right angles)
 Each diagonal bisects a pair of  All sides congruent
opposite angles
Chapter 6 Quadrilaterals

Amazing Property of Quadrilaterals

Steps:
1. Draw any quadrilateral (green in diagram).
2. Construct squares along each side of the
quadrilateral.
3. Connect the midpoints of opposite squares
with segments.
Result: The two segments connecting the
midpoints of the squares are congruent and
perpendicular.
Diagram: In the diagram to the right, segments AC
and BD are both congruent and perpendicular. It
may not look like the segments are the same length, but they are. On a printed page, each
segment is 4.4 cm long. Amazing!

Example 6.1: Find the side length of the rhombus if its diagonals measure 14 inches and 48
inches.
Lets take one triangle from the inside of the rhombus shown
to the right. See below.
We know that the diagonals are
perpendicular, so we have a right
triangle. The two red sides of the triangle are half of the length of the
diagonals from which they come.
We have sides, then, of 𝑎 14 2 7 and 𝑏 48 2 24.
It remains for us to calculate the value of 𝑐. Let’s use the Pythagorean Theorem:
𝑐 𝑎 𝑏
𝑐 7 24
𝑐 49 576 625
𝒄 𝟐𝟓 inches (remember to use units in the answer because they are in the statement
of the problem).
Chapter 6 Quadrilaterals

Example 6.2: Find the values of 𝑥 and 𝑦 in the parallelogram below.

In a parallelogram, opposite angles are congruent and (2x + 28)o

consecutive angles are supplementary. This gives us:
𝑥 5𝑥 2𝑥 28
𝑥 3𝑥 28 0
(x2 + 5x)o (8y)o
𝑥 7 𝑥 4 0
𝑥 7 or 4
If 𝑥 7, then the angles involved are equal to 2𝑥 28 ° 2 7 28 ° 14°.
If 𝑥 4, then the angles involved are equal to 2𝑥 28 ° 2 4 28 ° 36°.
Both of these angle values are possible, so we have two cases:
If 𝑥 7, then the angles involved are 14°. Since consecutive angles are
supplementary, 14 8𝑦 180.
8𝑦 166
𝑦 20.75 and the solution is: 𝒙 𝟕, 𝒚 𝟐𝟎. 𝟕𝟓
If 𝑥 4, then the angles involved are 36°. Since consecutive angles are
supplementary, 36 8𝑦 180.
8𝑦 144
𝑦 18 and the solution is: 𝒙 𝟒, 𝒚 𝟏𝟖
Both solutions are valid because they both result in positive angle values.

Example 6.3: Find the values of 𝑥 and 𝑦 so that the figure shown is a parallelogram.

In order for the figure to be a parallelogram, opposite

sides must be congruent. So,
12
Top and bottom sides: Left and right sides:
2𝑥 4𝑦 5𝑦 𝑥 3𝑥 3𝑦 12
𝑥 𝑦 𝑥 𝑦 4
𝑦 4 𝑥 5𝑦 𝑥

Combine above results:

𝑥 4 𝑥 𝑥 𝑦
2𝑥 4 2 𝑦
𝑥 2

Solution: 𝒙 𝟐, 𝒚 𝟐
Chapter 6 Quadrilaterals

Example 6.4: Find the values of 𝑥 and 𝑦 from the rhombus below.

In a rhombus, the diagonals intersect at right angles, so:

5𝑥 5𝑦 90
𝑥 𝑦 18
𝑦 18 𝑥
In a rhombus, the sides have the same length, so:
6𝑥 23 2𝑦 3
6𝑥 20 2𝑦
3𝑥 10 𝑦
Combining the two equations:
3𝑥 10 18 𝑥 𝑦 18 𝑥
4𝑥 28 𝑦 18 7
𝒙 𝟕 𝒚 𝟏𝟏

Example 6.5: Find the values of 𝑥 and 𝑦 if 𝐴𝐷 ≅ 𝐵𝐶 and ABCD is a parallelogram.

𝐴𝐷 and 𝐵𝐶 are diagonals. Since the diagonals are congruent and
ABCD is a parallelogram, we conclude that ABCD is a rectangle.
Therefore, all four interior angles measure 90°. So,
14𝑦 20 90 12𝑥 42 90 7𝑦 55 90
14𝑦 70 12𝑥 132 7𝑦 35
𝒚 𝟓 𝒙 𝟏𝟏 𝑦 5
Note that the first and third column result in the same value for 𝑦. If this were not the
case, we would say this problem is overdetermined, and there would be no solution for 𝑦.

Example 6.6: What is the measure of 𝐻𝐽 in the parallelogram below.

First, opposite angles in a parallelogram have
equal measures, so we can find 𝑥 as follows:
3𝑥 10 46
3𝑥 36
𝑥 12
Then, opposite sides have the same length, so
𝐇𝐉 FG 𝑥 7 12 7 𝟏𝟗
Chapter 6 Quadrilaterals

Example 6.7: If a quadrilateral has congruent diagonals, is it a rectangle: never, sometimes, or

always?
For a problem like this, it is a good idea to draw the required shape, but
to put as little structure in the shape as allowed by the question.
Sometimes. Rectangles have congruent diagonals, but it is possible to
construct a quadrilateral with congruent diagonals that is not a rectangle.
See the figure to the right, which has congruent diagonals.

Example 6.8: If a quadrilateral is a rhombus, then it is a parallelogram: never, sometimes, or

always?
Always. This can be seen in the quadrilateral tree at the beginning of this chapter. A
rhombus is defined to be a parallelogram with four congruent sides.

Example 6.9: A triangle can be a kite: never, sometimes, or always?

Never. Triangles have three sides, but kites have four congruent sides.

Example 6.10: Given a trapezoid with bases of length 2𝑥 14 cm and 8𝑥 4 cm, and a
midline of length 𝑚 5𝑥 15 cm. find the length of the midline.
𝑚 is the average (mean) of 𝑏 and 𝑏 . So,
𝑏 𝑏
𝑚
2
2𝑥 14 8𝑥 4
5𝑥 15
2
10𝑥 30 2𝑥 14 8𝑥 4
Next, collect terms, all on one side of the sign.
0 2𝑥 2𝑥 40 𝑥 cannot be 4 because that would make
0 𝑥 𝑥 20 𝑚 negative ( 𝑚 5 4 15 5 ), and
0 𝑥 5 𝑥 4 negative lengths are not allowed.
𝑥 5, 4 Therefore, 𝑥 5, so
𝒎 5𝑥 15 5 5 15 𝟒𝟎 cm
Chapter 6 Quadrilaterals

Geometry
Characteristics of Parallelograms

Characteristic Square Rhombus Rectangle Parallelogram

2 pair of parallel sides    

Opposite sides are congruent    

Opposite angles are congruent    

Consecutive angles are supplementary    

Diagonals bisect each other    

All 4 angles are congruent (i.e., right angles)  

Diagonals are congruent  

All 4 sides are congruent  

Diagonals are perpendicular  

Each diagonal bisects a pair of opposite angles  

Notes: Red ‐marks are conditions sufficient to prove the quadrilateral is of the type specified.
Green ‐marks are conditions sufficient to prove the quadrilateral is of the type specified if the quadrilateral is a
parallelogram.
Chapter 6 Quadrilaterals

Geometry
Parallelogram Proofs

Proving a Quadrilateral is a Parallelogram

To prove a quadrilateral is a parallelogram, prove any of the following conditions:
1. Both pairs of opposite sides are parallel. (note: this is the definition of a parallelogram)
2. Both pairs of opposite sides are congruent.
3. Both pairs of opposite angles are congruent.
4. An interior angle is supplementary to both of its consecutive angles.
5. Its diagonals bisect each other.
6. A pair of opposite sides is both parallel and congruent.

Proving a Quadrilateral is a Rectangle

To prove a quadrilateral is a rectangle, prove any of the following conditions:
1. All 4 angles are congruent.
2. It is a parallelogram and its diagonals are congruent.

Proving a Quadrilateral is a Rhombus

To prove a quadrilateral is a rhombus, prove any of the following conditions:
1. All 4 sides are congruent.
2. It is a parallelogram and Its diagonals are perpendicular.
3. It is a parallelogram and each diagonal bisects a pair of opposite angles.

Proving a Quadrilateral is a Square

To prove a quadrilateral is a square, prove:
1. It is both a Rhombus and a Rectangle.
Chapter 6 Quadrilaterals

Geometry
Kites and Trapezoids

Facts about a Kite

To prove a quadrilateral is a kite, prove:
 It has two pair of congruent sides.
 Opposite sides are not congruent.

Also, if a quadrilateral is a kite, then:

 Its diagonals are perpendicular
 It has exactly one pair of congruent opposite angles.

Parts of a Trapezoid Base

Midsegment
Trapezoid ABCD has the following parts:
Leg
 and are bases.
 and are legs. Leg
 is the midsegment.
 and are diagonals.
 Angles A and D form a pair of base angles.
Diagonals
 Angles B and C form a pair of base angles. Base

Trapezoid Midsegment Theorem

The midsegment of a trapezoid is parallel to each of its bases and: .

Proving a Quadrilateral is an Isosceles Trapezoid

To prove a quadrilateral is an isosceles trapezoid, prove any of the following conditions:
1. It is a trapezoid and has a pair of congruent legs. (definition of isosceles trapezoid)
2. It is a trapezoid and has a pair of congruent base angles.
3. It is a trapezoid and its diagonals are congruent.
Chapter 7 Transformations

Geometry
Introduction to Transformation

A Transformation is a mapping of the pre-image of a geometric figure onto an image that

retains key characteristics of the pre-image.

Definitions
The Pre-Image is the geometric figure before it has been transformed.

The Image is the geometric figure after it has been transformed.

A mapping is an association between objects. Transformations are types of mappings. In the

figures below, we say ABCD is mapped onto A’B’C’D’, or 𝐴𝐵𝐶𝐷 ⎯ 𝐴’𝐵’𝐶’𝐷’. The order of the
vertices is critical to a properly named mapping.

An Isometry is a one-to-one mapping that preserves lengths. Transformations that are

isometries (i.e., preserve length) are called rigid transformations.

Isometric Transformations

Reflection is flipping a Rotation is turning a Translation is sliding a

figure across a line called figure around a point. figure in the plane so that
a “mirror.” The figure Rotated figures retain it changes location but
retains its size and shape, their size and shape, but retains its shape, size and
but appears “backwards” not their orientation. orientation.
after the reflection.

Table of Characteristics of Isometric Transformations

Transformation Reflection Rotation Translation

Isometry (Retains Lengths)? Yes Yes Yes
Retains Angles? Yes Yes Yes
Retains Orientation to Axes? No No Yes
Chapter 7 Transformations

Geometry
Introduction to Transformation (cont’d)

Transformation of a Point
A point is the easiest object to transform. Simply reflect, rotate or translate it following the
rules for the transformation selected. By transforming key points first, any transformation
becomes much easier.

Transformation of a Geometric Figure

To transform any geometric figure, it is only necessary to transform the items that define the
figure, and then re-form it. For example:
 To transform a line segment, transform its two endpoints, and then connect the
resulting images with a line segment.
 To transform a ray, transform the initial point and any other point on the ray, and then
construct a ray using the resulting images.
 To transform a line, transform any two points on the line, and then fit a line through the
resulting images.
 To transform a polygon, transform each of its vertices, and then connect the resulting
images with line segments.
 To transform a circle, transform its center and, if necessary, its radius. From the
resulting images, construct the image circle.
 To transform other conic sections (parabolas, ellipses and hyperbolas), transform the
foci, vertices and/or directrix. From the resulting images, construct the image conic
section.

Example 7.1: Reflect Quadrilateral ABCD over the mirror shown.

To reflect a point over a mirror:
 Connect the point to the mirror with a segment that is perpendicular to the mirror.
 Draw the segment again, in the same direction, beyond the mirror.
 Place the image point at the end of the second segment.
See the diagrams below.
Chapter 7 Transformations

Geometry
Reflection

Definitions
Reflection is flipping a figure across a mirror.
The Line of Reflection is the mirror through which the
reflection takes place.
Note that:
 The line segment connecting corresponding points in
the image and pre-image is bisected by the mirror.
 The line segment connecting corresponding points in
the image and pre-image is perpendicular to the mirror.

Reflection through an Axis or the Line 𝒚 𝒙

Reflection of the point (a, b) through the x‐ or y‐axis or the line 𝑦 𝑥 gives the following
results:

Pre-Image Mirror Image

Point Line Point
(a, b) x-axis (a, -b)
(a, b) y-axis (-a, b)
(a, b) the line: 𝑦 𝑥 (b, a)

If you forget the above table, start with a point such as 3, 2 on a set of coordinate axes.
Reflect the point through the selected line and see which set of “a, b” coordinates works.

Line of Symmetry
A Line of Symmetry is any line through which a figure can be mapped onto itself. The thin black
lines in the following figures show their axes of symmetry:
Chapter 7 Transformations

Example 7.2: Which of the following quadrilaterals has line symmetry? Square, rectangle,
isosceles trapezoid, rhombus?
A figure has line symmetry if it is possible to draw a line so that the image looks the same
when reflected over the line.
In drawing the figures to help answer this problem, it is important to draw them in their
most general form. For example, when considering a rhombus, we would not want to draw
a square (even though a square is a type of rhombus) to analyze because a rhombus is not
required to have the right angles contained in a square. Doing so could lead us to the
wrong conclusions.
In the figures below, lines of symmetry are drawn as dashed green segments.

Answer: all of the quadrilaterals mentioned have line symmetry.

Example 7.3: Reflect ∆𝐴𝐵𝐶 over the 𝑥-axis and over the 𝑦-axis. What are the 𝑥 and 𝑦
coordinates after reflection?
Starting coordinates (black in the diagram):
2, 1 , 3, 4 , 5, 2
After reflection over the 𝑥-axis (orange in the diagram):
𝑥-values are unchanged. 𝑦-values change sign.
2, 1 , 3, 4 , 5, 2
After reflection over the 𝑦-axis (magenta in the diagram):
𝑥-values change sign. 𝑦-values are unchanged.
2, 1 , 3, 4 , 5, 2

Notice the symmetry in the diagram. Symmetry is often noticed because it looks “pretty.”
Chapter 7 Transformations

Geometry
Rotation

Definitions
Rotation is turning a figure by an angle about a fixed point.
The Center of Rotation is the point about which the figure is
rotated. Point P, at right, is the center of rotation.
The Angle of Rotation determines the extent of the rotation.
The angle is formed by the rays that connect the center of
rotation to the pre-image and the image of the rotation. Angle
P, at right, is the angle of rotation. Though shown only for
Point A, the angle is the same for any of the figure’s 4 vertices.

Note: In performing rotations, it is important to indicate the direction of the rotation –

clockwise or counterclockwise.

Rotation about the Origin

Rotation of the point (a, b) about the origin (0, 0) gives the following results:

Pre-Image Clockwise Counterclockwise Image

Point Rotation Rotation Point
(a, b) 90⁰ 270⁰ (b, -a)
(a, b) 180⁰ 180⁰ (-a, -b)
(a, b) 270⁰ 90⁰ (-b, a)
(a, b) 360⁰ 360⁰ (a, b)

If you forget the above table, start with the point 3, 2 on a set of coordinate axes. Rotate the
point by the selected angle and see which set of “a, b” coordinates works.

Rotational Symmetry
A figure in a plane has Rotational Symmetry if it can be mapped onto itself by a rotation of
180⁰ or less. Any regular polygon has rotational symmetry, as does a circle. Here are some
examples of figures with rotational symmetry:
Chapter 7 Transformations

Example 7.4: Which of the following quadrilaterals has rotational symmetry? Square,
rectangle, isosceles trapezoid, rhombus?
A figure has rotational symmetry if it is possible to rotate the image and get a result that
looks the same. The order of a rotational symmetry is the number of positions the shape
can take (within a 360˚ rotation) and look the same.
In the figures below, lines of symmetry are drawn as dashed green segments.

Answer: A rectangle has rotational symmetry of order 2 (0˚ and 180˚ rotations).
A square has rotational symmetry of order 4 (0˚, 90˚, 180˚ and 270˚ rotations).
An isosceles trapezoid does not have rotational symmetry.
A rhombus has rotational symmetry of order 2 (0˚ and 180˚ rotations).

Example 7.5: Rotate ∆𝐴𝐵𝐶 counterclockwise by 90° about the origin and, separately,
clockwise by 90° about the origin. What are the 𝑥 and 𝑦 coordinates after rotation?
Rotating “about” a point means that the point is the center of rotation.
Starting coordinates (black in the diagram):
2, 1 , 3, 4 , 5, 2
A rotation of 90° counterclockwise about the origin produces
a mapping of 𝑎, 𝑏 → 𝑏, 𝑎 . That is, the 𝑥 and 𝑦
coordinates switch and the new 𝑥-value changes its sign.
After rotation about the origin, (orange in the diagram):
1, 2 , 4, 3 , 2, 5
A rotation of 90° clockwise about the origin produces a mapping of 𝑎, 𝑏 → 𝑏, 𝑎 . That
is, the 𝑥 and 𝑦 coordinates switch and the new 𝑦-value changes its sign.
After reflection over the 𝑦-axis (magenta in the diagram):
1, 2 , 4, 3 , 2, 5

Notice that the two rotations produce coordinates that are a 180° rotation from each
other. That is, rotating 1, 2 , 4, 3 , 2, 5 by 180° gives 1, 2 , 4, 3 , 2, 5 , and
rotating 1, 2 , 4, 3 , 2, 5 by 180° gives 1, 2 , 4, 3 , 2, 5 . That’s because a
point rotated 90° counterclockwise the same point rotated 90° clockwise are 180° apart.
Chapter 7 Transformations

Geometry
Translation

Definitions
Translation is sliding a figure in the plane. Each
point in the figure is moved the same distance in
the same direction. The result is an image that
looks the same as the pre-image in every way,
except it has been moved to a different location
in the plane.
Each of the four orange line segments in the
figure at right has the same length and direction.

When Two Reflections One Translation

If two mirrors are parallel, then reflection through
one of them, followed by a reflection through the
second is a translation.
In the figure at right, the black lines show the paths
of the two reflections; this is also the path of the
resulting translation. Note the following:
 The distance of the resulting translation
(e.g., from A to A’’) is double the distance
between the mirrors.
 The black lines of movement are perpendicular to both mirrors.

Defining Translations in the Coordinate Plane (Using Vectors)

A translation moves each point by the same distance in the same direction. In the coordinate
plane, this is equivalent to moving each point the same amount in the x‐direction and the same
amount in the y‐direction. This combination of x‐ and y‐direction movement is described by a
mathematical concept called a vector.

In the above figure, translation from A to 𝑨′′ moves 10 in the x‐direction and the -3 in the y‐
direction. In vector notation, this is: 𝑨𝑨′′⃑ 〈10, 3〉. Notice the “half-ray” symbol over the
two points and the funny-looking brackets around the movement values.

So, the translation resulting from the two reflections in the above figure moves each point of
the pre-image by the vector 𝑨𝑨′′⃑. Every translation can be defined by the vector representing
its movement in the coordinate plane.
Chapter 7 Transformations

Translation Coordinate Form

Translations are often shown as coordinates with an enclosed mapping, so a translations of
𝑥, 𝑦 → 𝑥 2, 𝑦 5 means decrease the 𝑥-values of translated points by 2 and increase the
𝑦-values of translated points by 5.

Example 7.6: Translate the triangle shown in the diagram according to the mapping:
𝑥, 𝑦 → 𝑥 6, 𝑦 2 .
Starting coordinates (black in the diagram):

2, 1 , 3, 4 , 5, 2
After translation (orange in the diagram):
𝑥-values increase by 6. 𝑦-values decrease by 2.
4, 3 , 3, 6 , 1, 4
When you look at the result of a translation in a graph, it
often looks like we just slid the figure from one place to
another (which we did). The shape retains its shape and orientation.

Example 7.7: If a point 3, 6 is translated so that its image is 1, 12 , what is the translation
coordinate form of the translation?
This question boils down to asking how far 𝑥 moved and how far 𝑦 moved, from preimage
3, 6 to image 1, 12 . The easiest way to answer this is to subtract the two points to
obtain the movement vector, then convert that to the desired form.
Image: 1, 12
Preimage: 3, 6
Movement vector: 〈 4, 6〉
To obtain the translation coordinate form, add the movement vector 〈 4, 6〉 to the
general point 𝑥, 𝑦 .
𝒙, 𝒚 → 𝒙 𝟒, 𝒚 𝟔
Chapter 7 Transformations

Geometry
Compositions

When multiple transformations are combined, the result is called a Composition of the
Transformations. Two examples of this are:
 Combining two reflections through parallel mirrors to generate a translation (see the
previous page).
 Combining a translation and a reflection to generate what is called a glide reflection.
The glide part of the name refers to translation, which is a kind of gliding of a figure on
the plane.

Note: In a glide reflection, if the line of reflection is parallel to the direction of the
translation, it does not matter whether the reflection or the translation is performed first.

Figure 1: Translation followed by Reflection. Figure 2: Reflection followed by Translation.

Composition Theorem
The composition of multiple isometries is as Isometry. Put more simply, if transformations that
preserve length are combined, the composition will preserve length. It is also true that if
transformations that preserve angle measure are combined, the composition will preserve
angle measure.

Order of Composition
Order matters in most compositions that involve more than one class of transformation. If you
apply multiple transformations of the same kind (e.g., reflection, rotation, or translation), order
generally does not matter; however, applying transformations in more than one class may
produce different final images if the order is switched.
Chapter 7 Transformations

Example 7.8: Translate the triangle shown in the diagram according to the mapping:
𝑥, 𝑦 → 𝑥 6, 𝑦 2 , followed by a counterclockwise rotation of 90°.
Starting coordinates (black in the diagram):
2, 1 , 3, 4 , 5, 2
After translation (orange in the diagram):
𝑥-values increase by 6. 𝑦-values decrease by 2.
4, 3 , 3, 6 , 1, 4
A rotation of 90° counterclockwise about the origin
produces a mapping of 𝑎, 𝑏 → 𝑏, 𝑎 . That is, the 𝑥
and 𝑦 coordinates switch and the new 𝑥 value changes its sign.
After a subsequent rotation about the origin, (magenta in the diagram):
3, 4 , 6, 3 , 4, 1

Example 7.9: Reverse the order of the transformations in the previous example. That is,
Rotate the triangle shown in the diagram counterclockwise by 90°, followed by translation
according to the mapping: 𝑥, 𝑦 → 𝑥 6, 𝑦 2 .
Starting coordinates (black in the diagram):
2, 1 , 3, 4 , 5, 2
A rotation of 90° counterclockwise
about the origin produces a mapping of
𝑎, 𝑏 → 𝑏, 𝑎 . That is, the 𝑥 and 𝑦
coordinates switch and the new 𝑥 value
changes its sign.
After rotation about the origin, (orange
in the diagram):
1, 2 , 4, 3 , 2, 5
After a subsequent translation (magenta in the diagram):
𝑥-values increase by 6. 𝑦-values decrease by 2.
7, 4 , 10, 5 , 8, 7

Notice that the examples above involved performing the same transformations on the same
starting triangle, but in a different order. The results are very different, illustrating that order
matters in compositions that involve more than one class of transformation.
Chapter 7 Transformations

Geometry
Rotation About a Point Other than the Origin

Rotating an object about a point involves rotating each point of the object by the same angle
about that point. For a polygon, this is accomplished by rotating each vertex and then
connecting them with segments, so you mainly have to worry about the vertices, which are
points. An example of the process of rotating a point about another point is described below.
It is a good example of what can be accomplished with a composition of transformations.

Let’s define the following points:

 The point about which the rotation will take place, i.e., the center of rotation: 𝑥 , 𝑦 .
 The initial point (before rotation), i.e., the preimage: 𝑥 , 𝑦 .
 The final point (after rotation), i.e. the image: 𝑥 , 𝑦 .

The problem is to determine 𝑥 , 𝑦 if we are given 𝑥 , 𝑦 and 𝑥 , 𝑦 . It involves 3 steps:

1. Convert the problem to one of rotating a point about the origin (a much easier
problem).
2. Perform the rotation.
3. Reverse the translation in Step 1.

We consider each step separately, algebraically and geometrically, in the following example:

Example 7.10: Rotate a point by 90⁰ about another point.

Step 1: Convert the problem to one of rotating a point about the

origin:
First, we translate our point 2, 1 and the center of rotation
2, 3 so that the center of rotation moves to 0, 0 . That
involves subtracting (2, 3) from both the point and the center.
General Situation Example
Points in this step Points in this step
 Rotation Center: 𝑥 , 𝑦  Rotation Center: 2, 3
 Initial point: 𝑥 , 𝑦  Initial point: 2, 1
 Image of translation  Image of translation
Translate our point by subtracting 𝑥 , 𝑦 Translate our point by subtracting 2, 3
from 𝑥 , 𝑦 . The resulting image is: from 2, 1 . The resulting image is:
𝑥 𝑥 ,𝑦 𝑦 4, 2
The next steps depend on whether we are making a clockwise or counter clockwise
rotation.
Chapter 7 Transformations

Example 7.10a: Clockwise Rotation:

Step 2: Perform the rotation about the origin:

Rotating 90⁰ clockwise about the origin 0, 0 is simply a
process of switching the 𝑥- and 𝑦-values of a point and
negating the new 𝑦-term. That is, 𝑥, 𝑦 becomes 𝑦, 𝑥
after clockwise rotation by 90⁰.
General Situation Example
Pre-rotated point Pre-rotated point
(from Step 1): (from Step 1):
𝑥 𝑥 ,𝑦 𝑦 4, 2
Point after rotation: Point after rotation:
𝑦 𝑦 , 𝑥 𝑥 2, 4

Step 3: Reverse the translation performed in Step 1.

To do this, simply translate the image of rotation by the coordinates of the center of
rotation (adding back what was subtracted in Step 1).
General Situation Example
Point after rotation (from Step 2): Point after rotation (from Step 2):
𝑦 𝑦 , 𝑥 𝑥 2, 4
Add back the point of rotation 𝑥 , 𝑦 Add back the center of rotation 2, 3 :
𝑦 𝑦 𝑥 , 𝑥 𝑥 𝑦 0, 7
which gives us the final image: 𝒙𝟐 , 𝒙𝟐

Finally, here are the formulas for 𝒙𝟐 and 𝒚𝟐 :

Notice that the formulas for clockwise

Clockwise Rotation
rotation (this page) and counter-clockwise
𝒙𝟐 𝒚𝟏 𝒚𝟎 𝒙𝟎 rotation (next page) by 90⁰ are the same
except the terms in magenta are negated
𝒚𝟐 𝒙𝟏 𝒙𝟎 𝒚𝟎
between the formulas.

Interesting note: If you are asked to find the point about which a rotation occurred, you can
substitute the values for the starting point 𝑥 , 𝑦 and the ending point 𝑥 , 𝑦 in the above
equations and solve for 𝑥 and 𝑦 .
Chapter 7 Transformations

Example 7.10b: Counterclockwise Rotation:

Step 2: Perform the rotation about the origin:

Rotating 90⁰ counterclockwise about the origin
0, 0 is simply a process of switching the 𝑥- and
𝑦-values of a point and negating the new 𝑥-term.
That is, 𝑥, 𝑦 becomes 𝑦, 𝑥 after
counterclockwise rotation by 90⁰.
General Situation Example
Pre-rotated point (from Step 1): Pre-rotated point (from Step 1):
𝑥 𝑥 ,𝑦 𝑦 4, 2
Point after rotation: Point after rotation:
𝑦 𝑦 ,𝑥 𝑥 2, 4

Step 3: Reverse the translation performed in Step 1.

To do this, simply translate the image of rotation by the coordinates of the center of
rotation (adding back what was subtracted in Step 1).
General Situation Example
Point after rotation (from Step 2): Point after rotation (from Step 2):
𝑦 𝑦 ,𝑥 𝑥 2, 4
Add back the point of rotation 𝑥 , 𝑦 Add back the point of rotation (2, 3):
𝑦 𝑦 𝑥 ,𝑥 𝑥 𝑦 4, 1
which gives us the final image: 𝒙𝟐 , 𝒙𝟐

Finally, here are the formulas for 𝒙𝟐 and 𝒚𝟐 :

Notice that the formulas for clockwise

Counterclockwise Rotation
rotation (this page) and counter-clockwise
𝒙𝟐 𝒚𝟏 𝒚𝟎 𝒙𝟎 rotation (next page) by 90⁰ are the same
𝒚𝟐 𝒙𝟏 𝒙𝟎 𝒚𝟎 except the terms in magenta are negated
between the formulas.

Interesting note: The point half-way between the clockwise and counterclockwise rotations of
90⁰ is the center of rotation, 𝑥 , 𝑦 . In the example, halfway between 0, 7 and 4, 1 is
2, 3 .
Chapter 8 Similarity

Geometry
Ratios Involving Units

Ratios Involving Units

Example 8.1:
When simplifying ratios containing the same units:
3 inches 𝟏
 Simplify the fraction.
12 inches 𝟒
 Notice that the units disappear. They behave
Note: the unit “inches” cancels out,
just like factors; if the units exist in the
so the answer is , not 𝑖𝑛𝑐ℎ.
numerator and denominator, the cancel and are
not in the answer.
When simplifying ratios containing different units:
 Adjust the ratio so that the numerator and denominator have the same units.
 Simplify the fraction.
 Notice that the units disappear.

Example 8.2:
3 inches 3 inches 3 inches 𝟏
2 feet 2 feet ∙ 12 inches⁄foot 24 inches 𝟖

Dealing with Units

Notice in the above example that units can be treated the same as factors; they can be used in
fractions and they cancel when they divide. This fact can be used to figure out whether
multiplication or division is needed in a problem. Consider the following:
Example 8.3: How long did it take for a car traveling at 48 miles per hour to go 32 miles?

Consider the units of each item: 32 miles 48

 If you multiply, you get: 32 miles ∙ 48 1,536 . This is clearly wrong!

 If you divide, you get: 32 miles 48 miles ∙ hour. Now, this

looks reasonable. Notice how the "miles" unit cancel out in the final answer.
We could have solved this problem by remembering that 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑟𝑎𝑡𝑒 ∙ 𝑡𝑖𝑚𝑒, or 𝑑 𝑟𝑡.
However, paying close attention to the units also generates the correct answer. In addition, the
”units” technique always works, no matter what the problem!
Chapter 8 Similarity

Geometry
Similar Polygons

In similar polygons,
 Corresponding angles are congruent, and
 Corresponding sides are proportional.

Both of these conditions are necessary for two

polygons to be similar. Conversely, when two
polygons are similar, all of the corresponding
angles are congruent and all of the sides are proportional.

Naming Similar Polygons

Similar polygons should be named such that corresponding angles are in the same location in
the name, and the order of the points in the name should “follow the polygon around.”
Example 8.4: The polygons above could be shown similar with the following names:
𝐴𝐵𝐶𝐷𝐸𝐹𝐺𝐻𝐼 ~ 𝑆𝑇𝑈𝑉𝑊𝑋𝑌𝑍
It would also be acceptable to show the similarity as:
𝐷𝐸𝐹𝐺𝐻𝐼𝐴𝐵𝐶 ~ 𝑉𝑊𝑋𝑌𝑍𝑆𝑇𝑈
Any names that preserve the order of the points and keeps corresponding angles in
corresponding locations in the names would be acceptable.

Proportions
One common problem relating to similar polygons is to present three side lengths, where two
of the sides correspond, and to ask for the length of the side corresponding to the third length.

Example 8.5: In the above similar polygons, if 𝐵𝐶 20, 𝐸𝐹 12, 𝑎𝑛𝑑 𝑊𝑋 6, 𝑤ℎ𝑎𝑡 𝑖𝑠 𝑇𝑈?
This problem is solvable with proportions. To do so properly, it is important to relate
corresponding items in the proportion:

𝐵𝐶 𝐸𝐹 20 12
⎯ ⎯ 𝑇𝑈 10
𝑇𝑈 𝑊𝑋 𝑇𝑈 6

Notice that the left polygon is represented on the top of both proportions and that the left-
most segments of the two polygons are in the left fraction.
Chapter 8 Similarity

Geometry
Scale Factors of Similar Polygons

From the similar polygons below, the following is known about the lengths of the sides:

𝐴𝐵 𝐵𝐶 𝐶𝐷 𝐷𝐸 𝐸𝐹 𝐹𝐺 𝐺𝐻 𝐻𝐴
𝑘
𝑆𝑇 𝑇𝑈 𝑈𝑉 𝑉𝑊 𝑊𝑋 𝑋𝑌 𝑌𝑍 𝑍𝐴

That is, the ratios of corresponding sides in the

two polygons are the same and they equal
some constant 𝒌, called the scale factor of the
two polygons. The value of 𝑘, then, is all you
need to know to relate corresponding sides in
the two polygons.

Finding the Missing Length

Any time the student is asked to find the missing length in similar polygons:
 Look for two corresponding sides for which the values are known.
 Calculate the value of 𝑘.
 Use the value of 𝑘 to solve for the missing length.

𝑘 is a measure of the relative size of the two polygons. Using this knowledge, it is possible to
put into words an easily understandable relationship between the polygons.
 Let Polygon 1 be the one whose sides are in the numerators of the fractions.
 Let Polygon 2 be the one whose sides are in the denominators of the fractions.
 Then, it can be said that Polygon 1 is 𝒌 times the size of the Polygon 2.

Example 8.6: In the above similar polygons, if 𝐵𝐶 20, 𝐸𝐹 12, 𝑎𝑛𝑑 𝑊𝑋 6, 𝑤ℎ𝑎𝑡 𝑖𝑠 𝑇𝑈?
Seeing that 𝐸𝐹 and 𝑊𝑋 relate, calculate:

𝐸𝐹 12
2 𝑘
𝑊𝑋 6

Then solve for 𝑇𝑈 based on the value of 𝑘:

𝐵𝐶 20
𝑘 → 2 → 𝑇𝑈 10
𝑇𝑈 𝑇𝑈

Also, since 𝑘 2, the length of every side in the blue polygon is double the length of its
corresponding side in the orange polygon.
Chapter 8 Similarity

Geometry
Dilation of Polygons

A dilation is a special case of transformation involving similar polygons. It can be thought of as

a transformation that creates a polygon of the same shape but a different size from the original.
Key elements of a dilation are:
 Scale Factor – The scale factor of similar polygons is the constant 𝑘 which represents the
relative sizes of the polygons.
 Center – The center is the point from which the dilation takes place.

Note that 𝑘 0 and 𝑘 1 in order to generate a second polygon. Then,

 If 𝑘 1, the dilation is called an “enlargement.”
 If 𝑘 1, the dilation is called a “reduction.”

Dilations with Center (0, 0)

In coordinate geometry, dilations are often performed with the center being the origin 0, 0 .
In that case, to obtain the dilation of a polygon:
 Multiply the coordinates of each vertex by the scale factor 𝑘, and
 Connect the vertices of the dilation with line segments (i.e., connect the dots).

Examples:
In the following examples:
 The green polygon is the original.
 The blue polygon is the dilation.
 The dashed orange lines show the movement away from
(enlargement) or toward (reduction) the center, which is
the origin in all 3 examples.

Notice that, in each example:

distance from center distance from center

to a vertex of the 𝑘∙ to a vertex of the
dilated polygon original polygon

This fact can be used to construct dilations when coordinate axes

are not available. Alternatively, the student could draw a set of
coordinate axes as an aid to performing the dilation.
Chapter 8 Similarity

Example 8.7: Given that , what is the scale factor of ∆𝐴𝐵𝐸 to ∆𝐷𝐵𝐶?

There is only one set of corresponding sides to work with in this

diagram, so there is only one ratio we can calculate directly from

the diagram. Fortunately, we are given , so

that’s all we need.

2.

Therefore, the scale factor of ∆𝐴𝐵𝐸 to ∆𝐷𝐵𝐶 is 𝟐.

Example 8.8: Given a triangle with vertices at 1, 4 , 5, 2 , 2, 3 , what are the coordinates
of the vertices of the triangle after dilation 𝐷: 𝑥, 𝑦 → 2𝑥, 3𝑦 ?
The coordinates of the preimage are 1, 4 , 5, 2 , 2, 3 .
The dilation doubles all 𝑥- values and triples all 𝑦-values. So, the coordinates of the image
are:
𝟐, 𝟏𝟐 , 𝟏𝟎, 𝟔 , 𝟒, 𝟗

Example 8.9: Given two similar cubes have a scale factor of 4: 3, what is the ratio of their
volumes?
Volumes exist in three dimensions, so the ratio of their volumes would be the third power
(i.e., the cube) of the scale factor. In fact, that’s why the third power of a number is
referred to as the “cube” of the number.
4 𝟔𝟒
𝐑𝐚𝐭𝐢𝐨 .
3 𝟐𝟕
Chapter 8 Similarity

ADVANCED
Geometry
More on Dilation

Dilations of Non-Polygons
Any geometric figure can be dilated. In the dilation of the
green circle at right, notice that:
 The dilation factor is 2.
 The original circle has center 7, 3 and radius 5.
 The dilated circle has center 14, 6 and radius 10.
So, the center and radius are both increased by a factor of 𝑘 2. This is true of any figure in a
dilation with the center at the origin. All of the key elements that define the figure are
increased by the scale factor 𝑘.

Dilations with Center 𝒂, 𝒃

In the figures below, the green quadrilaterals are dilated to the blue ones with a scale factor of
𝑘 2. Notice the following:
In the figure to the left, the dilation has center
0, 0 , whereas in the figure to the right, the
dilation has center 4, 3 . The size of the
resulting figure is the same in both cases
(because 𝑘 2 in both figures), but the
location is different.

Graphically, the series of transformations that is equivalent to a dilation from a point 𝑎, 𝑏

other than the origin is shown below. Compare the final result to the figure above (right).
 Step 1: Translate the original figure by 𝑎, 𝑏 to reset the center at the origin.
 Step 2: Perform the dilation.
 Step 3: Translate the dilated figure by 𝑎, 𝑏 . These steps are illustrated below.

Step 1 Step 2 Step 3

Chapter 8 Similarity

Geometry
Similar Triangles

The following theorems present conditions under which triangles are similar.

Side-Angle-Side (SAS) Similarity

SAS similarity requires the proportionality of
two sides and the congruence of the angle
between those sides. Note that there is no such
thing as SSA similarity; the congruent angle must
be between the two proportional sides.

Side-Side-Side (SSS) Similarity

SSS similarity requires the proportionality of all
three sides. If all of the sides are proportional,
then all of the angles must be congruent.

Angle-Angle (AA) Similarity

AA similarity requires the congruence of two
angles and the side between those angles.

Similar Triangle Parts

In similar triangles,
 Corresponding sides are proportional.
 Corresponding angles are congruent.

Establishing the proper names for similar triangles is crucial to line up corresponding vertices.
In the picture above, we can say:
∆𝐴𝐵𝐶~∆𝐷𝐸𝐹 or ∆𝐵𝐶𝐴~∆𝐸𝐹𝐷 or ∆𝐶𝐴𝐵~∆𝐹𝐷𝐸 or
∆𝐴𝐶𝐵~∆𝐷𝐹𝐸 or ∆𝐵𝐴𝐶~∆𝐸𝐷𝐹 or ∆𝐶𝐵𝐴~∆𝐹𝐸𝐷

All of these are correct because they match corresponding parts in the naming. Each of these
similarities implies the following relationships between parts of the two triangles:
∠𝐴 ≅ ∠𝐷 and ∠𝐵 ≅ ∠𝐸 and ∠𝐶 ≅ ∠𝐹
Chapter 8 Similarity

Geometry
Proportion Tables for Similar Triangles

Setting Up a Table of Proportions

It is often useful to set up a table to identify the proper proportions

in a similarity. Consider the figure to the right. The table might look
something like this:

Triangle Left Side Right Side Bottom Side

Top ∆ AB BC CA

Bottom ∆ DE EF FD

The purpose of a table like this is to organize the information you have about the similar
triangles so that you can readily develop the proportions you need.

Developing the Proportions

To develop proportions from the table:

 Extract the columns needed from the table:

AB BC
DE EF Also from the above
table,
 Eliminate the table lines.
𝐴𝐵 𝐶𝐴
 Replace the horizontal lines with “division lines.”
𝐷𝐸 𝐹𝐷
 Put an equal sign between the two resulting fractions: 𝐵𝐶 𝐶𝐴
𝐸𝐹 𝐹𝐷

Solving for the unknown length of a side:

You can extract any two columns you like from the table. Usually, you will have information on
lengths of three of the sides and will be asked to calculate a fourth.

Look in the table for the columns that contain the 4 sides in question, and then set up your
proportion. Substitute known values into the proportion, and solve for the remaining variable.
Chapter 8 Similarity

Example 8.10: Are the triangles in the diagram similar? If so, write the similarity statement
and state the theorem used to determine the similarity.
We only have the sides to work with, so we must check
proportions. The easiest way to do this is by increasing
the sizes of the sides of the triangles as you move from
left to right in the proportions. So, we want to know if:
smallest value middle value largest value
smallest value middle value largest value
Side lengths for one triangle go in the numerators of the fractions and side lengths for the
other triangle go in the denominators of the fractions. So, we want to know if:
8 9 10
?
24 27 30
Simplifying the fractions, we get: . Then, by the SSS Similarity Theorem,
∆𝑭𝑫𝑬~∆𝑨𝑪𝑩.

Example 8.11: If ∆𝐸𝐷𝐹~∆𝐵𝐶𝐴, what is the value of 𝑥?

Let’s be careful with letter order in setting up our

proportion for this problem. In identifying
proportions, refer to the names of the triangles that
the lengths are coming from.

first letter 𝐸 , last letter 𝐹 first letter 𝐵 , last letter 𝐴

first letter 𝐸 , second letter 𝐷 first letter 𝐵 , second letter 𝐶
𝐸𝐹 𝐵𝐴
𝐸𝐷 𝐵𝐶
𝑥 5 13
5 𝑥 3
𝑥 5 𝑥 3 13 ∙ 5
𝑥 2𝑥 15 65
𝑥 2𝑥 80 0
𝑥 10 𝑥 8 0
𝑥 10, 8.
Notice that 𝑥 8 would give negative lengths in the diagram, so we discard that
solution. So, 𝒙 𝟏𝟎.
Chapter 8 Similarity

⃖ ⃗ ∥ ⃖𝐶𝐷⃗ and 𝐴𝐷 intersects 𝐶𝐸 at point 𝐵. Find the

Example 8.12: In the figure to the right, 𝐴𝐸
length of 𝐶𝐸 .

First, we need to find the similarity in the diagram, then find the
appropriate proportion.
 ∠𝐴𝐵𝐸 ≅ ∠𝐷𝐵𝐶 because they are vertical angles.
 ∠𝐴 ≅ ∠𝐷 because they are alternate interior angles of
parallel lines ⃖𝐴𝐸⃗ and ⃖𝐶𝐷⃗ with transversal 𝐴𝐷.
 ∆𝐴𝐵𝐸~∆𝐷𝐵𝐶, then, by the AA Similarity Theorem.
The proportion we want must follow the lettering in the similarity.

, with the large triangle in the numerator of the fractions and the small
triangle in the denominator of the fractions in the proportion.
10 8
5 𝐶𝐵
10 ∙ 𝐶𝐵 40
𝐶𝐵 4
𝑪𝑬 𝐸𝐵 𝐶𝐵 8 4 𝟏𝟐

Example 8.13: Given: ∠𝐶𝐸𝐴 ≅ ∠𝐶𝐵𝐷.

Prove: 𝐴𝑍 ∙ 𝑋𝑌 𝐴𝐵 ∙ 𝑍𝑌

Step Statement Reason

1 ∠𝐶𝐸𝐴 ≅ ∠𝐶𝐵𝐷 Given.
2 ∠𝐶 ≅ ∠𝐶 Reflexive property of congruence.
AA Similarity Theorem.
3 ∆𝐶𝐸𝐴 ≅ ∆𝐶𝐵𝐷
Angles in Steps 1 and 2.
𝐴𝐶 𝐷𝐶 Corresponding sides in similar
4
𝐴𝐸 𝐷𝐵 triangles are proportional.
Multiplicative property of
5 𝑨𝑪 ∙ 𝑫𝑩 𝑫𝑪 ∙ 𝑨𝑬
equality (applied twice).
Chapter 8 Similarity

Geometry
Three Similar Triangles

A common problem in geometry is to find the missing value in proportions based on a set of
three similar triangles, two of which are inside the third. The diagram often looks like this:

Pythagorean Relationships

Inside triangle on the left: 𝒅𝟐 𝒉𝟐 𝒂𝟐

Inside triangle on the right: 𝒉𝟐 𝒆𝟐 𝒃𝟐
Outside (large) triangle: 𝒂𝟐 𝒃𝟐 𝒄𝟐

Similar Triangle Relationships

Because all three triangles are similar, we have the relationships in the table below. These
relationships are not obvious from the picture, but are very useful in solving problems based on
the above diagram. Using similarities between the triangles, 2 at a time, we get:

From the inside triangle on From the inside triangle on

From the two inside triangles the left and the outside the right and the outside
triangle triangle

𝒉 𝒆 𝒂 𝒄 𝒃 𝒄
𝒅 𝒉 𝒅 𝒂 𝒆 𝒃
or or or

𝒉𝟐 𝒅∙𝒆 𝒂𝟐 𝒅∙𝒄 𝒃𝟐 𝒆∙𝒄

The left side squared The right side squared

The height squared
= the product of: = the product of:
= the product of:
the part of the base below it the part of the base below it
the two parts of the base
and the entire base and the entire base
Chapter 8 Similarity

Example 8.14: Solve for the value of 𝑥 in the diagram.

From the chart on the previous page:
The height squared the product of the two parts of the
base.
15 5𝑥
225
𝒙 𝟒𝟓
5

Example 8.15: Solve for the value of 𝑥 in the diagram.

From the chart on the previous page:
The left side squared the product of the part of the base
below it and the entire base.
𝑥 4 𝑥 𝑥 10
𝑥 8𝑥 16 𝑥 10𝑥
16 2𝑥
16
𝒙 𝟖
2
Chapter 9 Right Triangles

Geometry
Pythagorean Theorem

In a right triangle, the Pythagorean Theorem says:

𝒂𝟐 𝒃𝟐 𝒄𝟐

where,
 a and b are the lengths of the legs of a right
triangle, and
 c is the length of the hypotenuse.

Right, Acute, or Obtuse Triangle?

In addition to allowing the solution of right triangles, the Pythagorean Formula can be used to
determine whether a triangle is a right triangle, an acute triangle, or an obtuse triangle.

To determine whether a triangle is obtuse, right, or acute:

 Arrange the lengths of the sides from low to high; call them a, b, and c, in increasing order
 Calculate: 𝑎 , 𝑏 , and 𝑐 .
 Compare: 𝑎 𝑏 vs. 𝑐
 Use the illustrations below to determine which type of triangle you have.

Obtuse Triangle Right Triangle Acute Triangle

𝒂𝟐 𝒃𝟐 𝒄𝟐 𝒂𝟐 𝒃𝟐 𝒄𝟐 𝒂𝟐 𝒃𝟐 𝒄𝟐

Example 9.1: Example 9.2: Example 9.3:

Triangle with sides: 7, 9, 12 Triangle with sides: 6, 8, 10 Triangle with sides: 5, 8, 9
7 9 𝑣𝑠. 12 6 8 𝑣𝑠. 10 5 8 𝑣𝑠. 9
49 81 144 36 64 100 25 64 81
→ 𝑶𝒃𝒕𝒖𝒔𝒆 𝑻𝒓𝒊𝒂𝒏𝒈𝒍𝒆 → 𝑹𝒊𝒈𝒉𝒕 𝑻𝒓𝒊𝒂𝒏𝒈𝒍𝒆 → 𝑨𝒄𝒖𝒕𝒆 𝑻𝒓𝒊𝒂𝒏𝒈𝒍𝒆
Chapter 9 Right Triangles

Geometry
Pythagorean Triples

Pythagorean Theorem: 𝒂𝟐 𝒃𝟐 𝒄𝟐

Pythagorean triples are sets of 3 positive integers that meet the requirements of the
Pythagorean Theorem. Because these sets of integers provide “pretty” solutions to geometry
problems, they are a favorite of geometry books and teachers. Knowing what triples exist can
help the student quickly identify solutions to problems that might otherwise take considerable
time to solve.

3-4-5 Triangle Family 7-24-25 Triangle Family

Sample Sample
Triples Triples
3-4-5 7-24-25
6-8-10 14-48-50
9-12-15 21-72-75
12-16-20 ...
30-40-50 70-240-250

𝟑𝟐 𝟒𝟐 𝟓𝟐 𝟕𝟐 𝟐𝟒𝟐 𝟐𝟓𝟐
9 16 25 49 576 625

5-12-13 Triangle Family 8-15-17 Triangle Family

Sample Sample
Triples Triples
5-12-13 8-15-17
10-24-26 16-30-34
15-36-39 24-45-51
... ...
50-120-130 80-150-170

𝟓𝟐 𝟏𝟐𝟐 𝟏𝟑𝟐 𝟖𝟐 𝟏𝟓𝟐 𝟏𝟕𝟐

25 144 169 64 225 289
Chapter 9 Right Triangles

Example 9.4: Find the value of 𝑥.

𝑥 15 36
𝑥 225 1296
𝑥 1521
𝒙 𝟑𝟗

Example 9.5: M is the midpoint of 𝑃𝑄 in rectangle 𝑃𝑄𝑅𝑆. What is

the perimeter of ∆𝑀𝑆𝑇.
The measures in black in the diagram are given, so we add the
ones in magenta. Then,

𝑆𝑇 5 12 13
𝑇𝑀 6 8 10
𝑀𝑆 6 13 √205
𝑷 ∆𝑴𝑺𝑻 𝑆𝑇 𝑇𝑀 𝑀𝑆

13 10 √205

𝟐𝟑 √𝟐𝟎𝟓

Example 9.6: A treasure is buried 16 paces north and 30 paces west of a

landmark. How many paces is the treasure from the landmark via a direct
route?
𝑥 16 30 1156
𝒙 √1156 𝟑𝟒 paces
Chapter 9 Right Triangles

Geometry
Special Triangles

The relationship among the lengths of the sides of a triangle is dependent on the measures of
the angles in the triangle. For a right triangle (i.e., one that contains a 90⁰ angle), two special
cases are of particular interest. These are shown below:

45⁰-45⁰-90⁰ Triangle
In a 45⁰-45⁰-90⁰ triangle, the congruence of two
angles guarantees the congruence of the two
√𝟐
legs of the triangle. The proportions of the three
1
sides are: 𝟏 ∶ 𝟏 ∶ √𝟐. That is, the two legs have
the same length and the hypotenuse is √𝟐 times
1 as long as either leg.

30⁰-60⁰-90⁰ Triangle

In a 30⁰-60⁰-90⁰ triangle, the proportions of the

three sides are: 𝟏 ∶ √𝟑 ∶ 𝟐. That is, the long leg
2
√𝟑 is √𝟑 times as long as the short leg, and the
hypotenuse is 𝟐 times as long as the short leg.

In a right triangle, we need to know the lengths of two sides to determine the length of the
third. The power of the relationships in the special triangles lies in the fact that we need only
know the length of one side of the triangle to determine the lengths of the other two sides.

Example Side Lengths

45⁰-45⁰-90⁰ Triangle 30⁰-60⁰-90⁰ Triangle

𝟏 ∶ 𝟏 ∶ √𝟐 𝟐 ∶ 𝟐 ∶ 𝟐√𝟐 𝟏 ∶ √𝟑 ∶ 𝟐 𝟐 ∶ 𝟐√𝟑 ∶ 𝟒

√𝟐 ∶ √𝟐 ∶ 𝟐 √𝟑 ∶ √𝟑 ∶ √𝟔 √𝟐 ∶ √𝟔 ∶ 𝟐√𝟐 √𝟑 ∶ 𝟑 ∶ 𝟐√𝟑
𝟑√𝟐 ∶ 𝟑√𝟐 ∶ 𝟔 𝟐𝟓 ∶ 𝟐𝟓 ∶ 𝟐𝟓√𝟐 𝟑√𝟐 ∶ 𝟑√𝟔 ∶ 𝟔√𝟐 𝟐𝟓 ∶ 𝟐𝟓√𝟑 ∶ 𝟓𝟎
Chapter 9 Right Triangles

Example 9.7: Find the values of 𝑥 and 𝑦.

𝒙 12√2 √2 𝟏𝟐

𝒚 𝑥 𝟏𝟐

Example 9.8: Find the values of 𝑥 and 𝑦.

4 4 √3 𝟒√𝟑
𝒙 4 √3 ∙
√3 √3 √3 𝟑
4√3 𝟖√𝟑
𝒚 2∙
3 𝟑

Example 9.9: Find the area of the isosceles trapezoid shown. All measures are in meters (m).
𝑚 is the midsegment of the trapezoid.
In the figure:
 BF and CE are drawn perpendicular to
both BC and FE.
 ∆ABF ≅ ∆DCE, both are right triangles.
 BCEF is a rectangle.
We want the total area of the trapezoid. The formula for this is:
𝑏 𝑏
𝐴𝑟𝑒𝑎 ∙ℎ 𝑚∙ℎ
2
7 19
𝑚 13
2
ℎ is determined using the proportions of a 30° 60° 90° (1: √3: 2) triangle: ∆ABF.
a is the length of the long side of ∆ABF.
19 7
a 6
2
a 6
ℎ 2√3
√3 √3
Finally, 𝑨𝒓𝒆𝒂 13 ∙ 2√3 𝟐𝟔√𝟑 m2
Chapter 9 Right Triangles

Geometry
Trig Functions and Special Angles

Trigonometric Functions

SOH-CAH-TOA

sin sin 𝐴 sin 𝐵

cos cos 𝐴 cos 𝐵

tan tan 𝐴 tan 𝐵

Special Angles

Trig Functions of Special Angles

Radians Degrees 𝐬𝐢𝐧 𝜽 𝐜𝐨𝐬 𝜽 𝐭𝐚𝐧 𝜽

0 0⁰ √0 √4 √0
0 1 0
2 2 √4

𝜋 30⁰ √1 1 √3 √1 √3
6
2 2 2 √3 3

𝜋 45⁰ √2 √2 √
4 1
2 2 √

𝜋 60⁰ √3 √1 1 √3
3 √3
2 2 2 √1

𝜋 90⁰ √4 √0 undefined
2 1 0
2 2
Chapter 9 Right Triangles

Geometry
Trigonometric Function Values in Quadrants II, III, and IV

In quadrants other than Quadrant I, trigonometric values for angles are calculated in the
following manner:
 Draw the angle θ on the Cartesian Plane.
 Calculate the measure of the angle from the x-
axis to θ.
 Find the value of the trigonometric function of
the angle in the previous step.
 Assign a “ ” or “ “ sign to the trigonometric
value based on the function used and the
quadrant θ is in.

Examples:

Example 9.10: 𝜽 in Quadrant II – Calculate: 180⁰ 𝑚∠𝜃

For 𝜃 120⁰, base your work on 180° 120° 60°
√ √𝟑
sin 60° , so: 𝐬𝐢𝐧 𝟏𝟐𝟎°
𝟐

Example 9.11: 𝜽 in Quadrant III – Calculate: 𝑚∠𝜃 180⁰

For 𝜃 210⁰, base your work on 210° 180° 30°
√ √𝟑
cos 30° , so: 𝐜𝐨𝐬 𝟐𝟏𝟎°
𝟐

Example 9.12: 𝜽 in Quadrant IV – Calculate: 360⁰ 𝑚∠𝜃

For 𝜃 315⁰, base your work on 360° 315° 45°
tan 45° 1, so: 𝐭𝐚𝐧 𝟑𝟏𝟓° 𝟏
Chapter 9 Right Triangles

Geometry
Graphs of Trigonometric Functions

The sine and cosecant functions are inverses. So:

1 1
sin 𝜃 and csc 𝜃
csc 𝜃 sin 𝜃

The cosine and secant functions are inverses. So:

1 1
cos 𝜃 and sec 𝜃
sec 𝜃 cos 𝜃

The tangent and cotangent functions are inverses. So:

1 1
tan 𝜃 and cot 𝜃
cot 𝜃 tan 𝜃
Chapter 9 Right Triangles

Example 9.13: Find the values of 𝑥 and 𝑦. Round values to 2 decimal places.

𝑥 5
tan 44° cos 44°
5 𝑦
5
𝒙 5 ∙ tan 44° 𝟒. 𝟖𝟑 𝒚 𝟔. 𝟗𝟓
cos 44°

Example 9.14: Find the values of 𝑥 and 𝑦. Round values to 2 decimal places.

16
sin 25° 𝑦 25° 90°
𝑥
16
𝒙 𝟑𝟕. 𝟖𝟔 𝒚 90° 25° 𝟔𝟓°
sin 25°

Example 9.15: cos 𝑥 0.5. What is sec 𝑥? csc 𝑦 4. What is sin 𝑦?

cos 𝑥 0.5 csc 𝑦 4

1 1 1 1
𝐬𝐞𝐜 𝒙 𝟐 sin 𝑦 𝟎. 𝟐𝟓
cos 𝑥 0.5 csc 𝑦 4

Example 9.16: sin 𝜃 , tan 𝜃 0. Find the values of sec 𝜃 and cot 𝜃.

Notice that sin 𝜃 0 , tan 𝜃 0. Therefore, 𝜃 is in 𝑄3, so we draw the angle in that
quadrant.

In 𝑄3, 𝑦 is negative; 𝑟 is always positive. Since sin 𝜃 , we let

𝑦 2, 𝑟 3.
Using the Pythagorean Theorem, we calculate the length of the
horizontal leg of the triangle: 3 2 √5. Since the angle is in
𝑄3, 𝑥 is negative, so we must have 𝑥 √5.
√
Then, sec 𝜃
√

√ √
And, cot 𝜃
Chapter 9 Right Triangles

Example 9.17: cot 𝜃 , cos 𝜃 0. Find the value of csc 𝜃 and cos 𝜃.

Notice that cot 𝜃 0 , cos 𝜃 0. Therefore, 𝜃 is in 𝑄2, so we draw the angle in that
quadrant.

In 𝑄2, 𝑥 is negative, and 𝑦 is positive. Since cot 𝜃 , we let 𝑥 9, 𝑦 4.

Using the Pythagorean Theorem, we can calculate the length of the

hypotenuse of the triangle: 𝑟 9 4 √97.
√
Then, csc 𝜃

√
And, cos 𝜃
√
Chapter 9 Right Triangles

Geometry
Vectors

Definitions
 A vector is a geometric object that has both
magnitude (length) and direction.
 The Tail of the vector is the end opposite the arrow.
It represents where the vector is moving from.
 The Head of the vector is the end with the arrow. It 𝐯 𝑨𝑩⃑
represents where the vector is moving to.
 The Zero Vector is denoted 0. It has zero length and
all the properties of zero.
 Two vectors are equal is they have both the same magnitude and the same direction.
 Two vectors are parallel if they have the same or opposite directions. That is, if the angles
of the vectors are the same or 180⁰ different.
 Two vectors are perpendicular if the difference of the angles of the vectors is 90⁰ or 270⁰.

Magnitude of a Vector
The distance formula gives the magnitude of a vector. If the head and tail of vector v are the
points 𝐴 𝑥 , 𝑦 and 𝐵 𝑥 , 𝑦 , then the magnitude of v is:

|𝐯| 𝑨𝑩⃑ 𝒙𝟐 𝒙𝟏 𝟐 𝒚𝟐 𝒚𝟏 𝟐

Note that 𝑨𝑩⃑ 𝑩𝑨⃑ . The directions of the two vectors are opposite, but their magnitudes
are the same.

Direction of a Vector
The direction of a vector is determined by the angle it makes
with a horizontal line. In the figure at right, the direction is the
angle 𝛉. The value of 𝛉 can be calculated based on the lengths
of the sides of the triangle the vector forms.

𝟑 𝟏
𝟑
𝐭𝐚𝐧 𝜽 or 𝜽 𝐭𝐚𝐧
𝟒 𝟒

where the function tan‐1 is the inverse tangent function. The second equation in the line above
reads “𝜃 is the angle whose tangent is .”
Chapter 9 Right Triangles

Geometry
Operations with Vectors

It is possible to operate with vectors in some of the same ways we operate with numbers. In
particular:

Adding Vectors
Vectors can be added in rectangular form by separately adding their x‐ and y‐components. In
general,
𝐮 〈𝑢 , 𝑢 〉
𝐯 〈𝑣 , 𝑣 〉
𝐮 𝐯 〈𝑢 , 𝑢 〉 〈𝑣 , 𝑣 〉 〈𝑢 𝑣 ,𝑢 𝑣 〉

Example 9.18: In the figure at right,

𝐮 〈4, 3〉
𝐯 〈2, 6〉
𝐰 𝐮 𝐯 〈4, 3〉 〈2, 6〉 〈6, 3〉

Vector Algebra
𝐮 𝐯 𝐯 𝐮 𝐮 𝐮 𝟎 a∙ 𝐮 𝐯 a∙𝐮 a∙𝐯
𝐮 𝐯 𝐰 𝐮 𝐰 𝐯 𝟎∙𝐮 𝟎 a b ∙𝐮 a∙𝐮 b∙𝐮
𝐮 𝟎 𝐮 1∙𝐮 𝐮 ab ∙ 𝐮 a∙ b∙𝐮 b∙ a∙𝐮

Scalar Multiplication
Scalar multiplication changes the magnitude of a vector, but not the direction. In general,
𝐮 〈𝑢 , 𝑢 〉
𝑘∙𝐮 〈𝑘 ∙ 𝑢 , 𝑘 ∙ 𝑢 〉

Example 9.19: In the figure at right,

𝐮 〈4, 3〉
2∙𝐮 2 ∙ 〈4, 3〉 〈8, 6〉
Chapter 10 Circles

Geometry
Parts of Circles

Center – the middle of the circle. All points on the circle

are the same distance from the center.
Radius – a line segment with one endpoint at the center
and the other endpoint on the circle. The term “radius” is
also used to refer to the distance from the center to the
points on the circle.
Diameter – a line segment with endpoints on the circle
that passes through the center.

Arc – a path along a circle.

Minor Arc – a path along the circle that is less than 180⁰.
Major Arc – a path along the circle that is greater than
180⁰.
Semicircle – a path along a circle that equals 180⁰.
Sector – a region inside a circle that is bounded by two
radii and an arc.

Secant Line – a line that intersects the circle in

exactly two points.
Tangent Line– a line that intersects the circle
in exactly one point.
Chord – a line segment with endpoints on the
circle that does not pass through the center.
Chapter 10 Circles

Geometry
Angles, Arcs, and Segments

Central Angle Inscribed Angle

𝟏
𝒎∠𝑨 𝒎 𝑹𝑺 𝒎∠𝑨 𝒎 𝑹𝑺
𝟐

Vertex inside the circle Vertex outside the circle

𝟏 𝟏
𝒎∠𝑨 𝒎 𝑹𝑺 𝒎 𝑴𝑵 𝒎∠𝑨 𝒎 𝑹𝑺 𝒎 𝑴𝑵
𝟐 𝟐
𝑹𝑨 ∙ 𝑨𝑵 𝑺𝑨 ∙ 𝑨𝑴 𝑨𝑴 ∙ 𝑨𝑹 𝑨𝑵 ∙ 𝑨𝑺

Tangent on one side Tangents on two sides

𝟏 𝟏
𝒎∠𝑨 𝒎 𝑹𝑺 𝒎 𝑹𝑵 𝒎∠𝑨 𝒎 𝑹𝑻𝑺 𝒎 𝑹𝑳𝑺
𝟐 𝟐
𝑨𝑹𝟐 𝑨𝑵 ∙ 𝑨𝑺 𝑨𝑹 𝑨𝑺
Chapter 10 Circles

Circle Vocabulary:
 Subtended angle: an angle whose two rays pass through the
endpoints of a geometric object (e.g., an arc on a circle).
 An arc subtends an angle. An angle is subtended by an arc.
 In the diagram to the right, AC subtends both ∠𝐴BC and ∠AOC.
Both ∠𝐴BC and ∠AOC are subtended by AC.
 Naming a circle: Circles are typically named by their centers. In the diagram above, we
would refer to the circle as Circle O. Typically, the point at the center of a circle is
named O or a letter close to O in the English alphabet.
 Interior point: a point whose distance from the center of the circle is less than the
radius of the circle. That is, the point is inside the circle.
 Exterior point: a point whose distance from the center of the circle is more than the
radius of the circle. That is, the point is outside the circle.
 Central angle: An angle with its vertex at the center of a circle. In the diagram above,
∠AOC is a central angle.
 Inscribed angle: An angle with its vertex on a circle and its rays passing through the
circle. In the diagram above, ∠ABC is an inscribed angle.
 Tangent-chord angle: An angle with its vertex on a circle, one ray tangent to the circle,
and one ray passing through the circle. In the diagram above, line l is tangent to Circle
O at Point B. ∠ABD and ∠CBD are tangent-chord angles.
 Circumscribed polygon: A polygon outside a circle, with all of the sides of the polygon
tangent to the circle. Circumscribed polygons are typically regular (i.e., they have equal
angle measures and equal side lengths).
 Inscribed polygon: A polygon inside a circle, with all of its vertices on the circle.

Example 10.1: Given: AB ≅ BC and 𝑚∠A 70°, find 𝑚 ABC.

The top diagram to the right is given with this problem. In order to solve
the problem, we add a few things to the top diagram to get the bottom
diagram.
In the bottom diagram, items in black are given, and items in blue are
derived as follows:
𝑚∠C 𝑚∠A 70° because they are opposite congruent sides in a
triangle.
𝑚 BC 𝑚 AB 2 ∙ 70° 140° because the arcs subtend angles of 70°.
𝒎 𝐀𝐁𝐂 𝑚 AB 𝑚 BC 140° 140° 𝟐𝟖𝟎°
Chapter 10 Circles

Example 10.2: Solve for 𝑥 in the circle provided.

H is a vertex inside the circle, so we have the relationship:
1
140° 𝑚 JM 𝑚 𝐾𝐿
2
1
140 6𝑥 5 10𝑥 3
2
140 8𝑥 4
136 8𝑥
𝟏𝟕 𝒙

Facts about Circles

 The circumference of a circle is: 𝐶 𝜋𝑑 2𝜋𝑟, where 𝑑 is the diameter of the circle
and 𝑟 is the radius of the circle.
 The area of a circle is 𝐶 𝜋𝑟 , where 𝑟 is the radius of the circle.
 A diameter spits a circle into two arcs, each of measure 180°.
 All radii in a circle are congruent. Likewise, all radii in congruent circles are congruent.
 If a quadrilateral is inscribed in a circle, then opposite angles of the quadrilateral are
supplementary.

Example 10.3: Solve for 𝑥 and 𝑦.

Opposite angles in a quadrilateral inscribed in a circle add to 180°.
𝑥 85 180 → 𝒙 𝟏𝟎𝟓
𝑚∠𝐴 𝑥 15 ° 105 10 ° 120°
𝑦 120 180 → 𝒚 𝟔𝟎

Facts about Chords

 The distance of a chord from the center of a circle is measured from the center of the
circle to the midpoint of the chord. The radius extending through the midpoint of the
chord is perpendicular to the chord.
 The perpendicular bisector of a chord passes through the center of the circle.
 Chords that are the same distance from the center of the same circle or congruent
circles are congruent.
Chapter 10 Circles

Example 10.4: A square with area 100 cm is inscribed in a circle. Find the exact value of the
area of the circle.
If a square has an area of 100, it must have a side length of:
𝑠 √100 10.
We create ∆OAB in the diagram to find the radius of Circle O.
∆OAB is a 45°-45°-90° triangle with sides of length 5, so the
hypotenuse, OB 5√2.

The radius of the circle is the length of OB. OB 5√2.

Finally, the area requested is:
𝟐
𝑪 𝜋𝑟 𝜋 ∙ 5√2 𝟓𝟎𝝅 𝐜𝐦𝟐 .

Example 10.5: Given three tangent circles with distances between their radii of 9, 17, 22, find
the radii of the circles.
Let’s call the radii of the three circles 𝑎, 𝑏, 𝑐. Then,
𝑎 𝑏 9, 𝑏 𝑐 22, 𝑎 𝑐 17
Solve.
𝑎 𝑐 17 𝑏 𝑐 8
𝑎 𝑏 9 𝑏 𝑐 22
𝑐 𝑏 8 2𝑐 30
𝒄 𝟏𝟓
With 𝑐 15, we get 𝒃 𝟕, 𝒂 𝟐 from the starting equations.

Example 10.6: Find the length of a chord that is 15 cm from the center of a circle with a
diameter of 34 cm.
The figure to the left diagrams this problem. All radii of the circle
are 17 cm in length. The distance from the center to the chord (AC)
is 15 cm, and AC is perpendicular to the segment drawn from the
center to the chord, OB.
𝐀𝐂 2 ∙ AB

2 ∙ 17 15
2∙8 𝟏𝟔
Chapter 10 Circles

Example 10.7: Given 𝑚∠𝑃 48° and 𝑚 𝐴𝐶 80°, what is 𝑚 𝐴𝐵?

1
48° 𝑚 BC 𝑚 AC °
2
1
48° 𝑚 BC 80 °
2
96° 𝑚 BC 80 °

𝑚 BC 176°
𝑚 AB 360° 𝑚 AC 𝑚 BC
𝒎 𝐀𝐁 360° 80° 176° 𝟏𝟎𝟒°

Facts about Tangents

 Tangents to a circle from an external point are congruent.
 A tangent to a circle is perpendicular to the radius of the circle that intersects the
tangent at the point of tangency.
 If two lines that are tangent to a circle intersect at an external point, then the line
containing the point of intersection and the center of the circle bisects the angle formed
by the two tangents.
 All of these facts about tangents are illustrated in the example below.

Example 10.8: PB and PA are tangent to Circle O, PA 40 and PO 41. Find PB and the
radius of the circle.

The above left diagram is given with this problem. In order to solve the problem, we add a
few things to get the above right diagram.
Tangents to a circle from an external point are congruent, so 𝑷𝑩 𝑃𝐴 𝟒𝟎.
There are right angles at the points of tangency. Pythagoras will help us get the radius.

𝒓 41 40 𝟗
Chapter 11 Perimeter and Area

Geometry
Perimeter and Area of a Triangle

Perimeter of a Triangle
The perimeter of a triangle is simply the sum of the measures of the three sides of the triangle.

𝑷 𝒂 𝒃 𝒄

Area of a Triangle
There are two formulas for the area of a triangle, depending on what information about the
triangle is available.
Formula 1: The formula most familiar to the student can be used when the base and height of
the triangle are either known or can be determined.
𝟏
𝑨 𝒃𝒉
𝟐
where, 𝑏 is the length of the base of the triangle.
ℎ is the height of the triangle.

Note: The base can be any side of the triangle. The height is the measure of the altitude of
whichever side is selected as the base. So, you can use:

or or

Formula 2: Heron’s formula for the area of a triangle can be used when
the lengths of all of the sides are known. Sometimes this formula,
though less appealing, can be very useful.

𝑨 𝒔 𝒔 𝒂 𝒔 𝒃 𝒔 𝒄
𝟏 𝟏
where, 𝒔 𝑷 𝒂 𝒃 𝒄 . Note: 𝑠 is sometimes called the semi-perimeter of the triangle.
𝟐 𝟐

𝒂, 𝒃, 𝒄 are the lengths of the sides of the triangle.

Chapter 11 Perimeter and Area

Example 11.1: C, B, D are midpoints. BD 12 cm, DF 11 cm, CD 10.4 cm. Find the
perimeter of ∆𝐴𝐸𝐹.
The four small triangles formed by connecting midpoints C, B, D are all
congruent. The perimeter of the ∆𝐴𝐸𝐹 will be double the perimeter
of any of the four interior triangles.
We are given the three lengths shown in magenta in the diagram.
Let’s use the perimeter of ∆𝐷𝐵𝐹 as our basis to calculate the perimeter of ∆𝐴𝐸𝐹.
𝑃 ∆𝐷𝐵𝐹 𝐵𝐷 𝐵𝐹 𝐷𝐹
Of the three distances in the formula, we are missing 𝐵𝐹, but fortunately we know that
𝐵𝐹 𝐶𝐷 10.4. Then,
𝑃 ∆𝐷𝐵𝐹 𝐵𝐷 𝐵𝐹 𝐷𝐹 12 10.4 11 33.4.
𝑷 ∆𝑨𝑬𝑭 2 ∙ 𝑃 ∆𝐷𝐵𝐹 2 ∙ 33.4 𝟔𝟔. 𝟖 cm.

Example 11.2: Given a triangle with vertices at 1, 4 , 5, 2 , 2, 3 , what is the perimeter of

the triangle after dilation? Round to 2 decimals.
The coordinates of the preimage are 1, 4 , 5, 2 , 2, 3 .
The dilation doubles all 𝑥- values and triples all 𝑦-values. So, the coordinates of the image
are: 2, 12 , 10, 6 , 4, 9
Distances:
From 2, 12 to 10, 6 , the distance is: 10 2 6 12 √388 19.70.

From 10, 6 to 4, 9 , the distance is: 4 10 9 6 √421 20.52.

From 4, 9 to 2, 12 , the distance is: 2 4 12 9 √45 6.71.

Perimeter 19.70 20.52 6.71 𝟒𝟔. 𝟗𝟑

Example 11.3: If a triangle has lengths of 8, 9, and 15 m, what is its area? Round to 2 decimals.
Using Heron’s formula,
8 9 15
𝑠 16
2
𝑨 16 16 8 16 9 16 15 √16 ∙ 8 ∙ 7 ∙ 1 √896 𝟐𝟗. 𝟗𝟑 𝐦𝟐
Chapter 11 Perimeter and Area

ADVANCED
Geometry
More on the Area of a Triangle

Trigonometric Formulas
The following formulas for the area of a triangle come from trigonometry. Which one is used
depends on the information available:

Two angles and a side:

𝟏 𝒂𝟐 ∙ 𝐬𝐢𝐧 𝑩 ∙ 𝐬𝐢𝐧 𝑪 𝟏 𝒃𝟐 ∙ 𝐬𝐢𝐧 𝑨 ∙ 𝐬𝐢𝐧 𝑪 𝟏 𝒄𝟐 ∙ 𝐬𝐢𝐧 𝑨 ∙ 𝐬𝐢𝐧 𝑩

𝑨 ∙ ∙ ∙
𝟐 𝐬𝐢𝐧 𝑨 𝟐 𝐬𝐢𝐧 𝑩 𝟐 𝐬𝐢𝐧 𝑪

Two sides and an angle:

𝟏 𝟏 𝟏
𝑨 𝒂𝒃 𝐬𝐢𝐧 𝑪 𝒂𝒄 𝐬𝐢𝐧 𝑩 𝒃𝒄 𝐬𝐢𝐧 𝑨
𝟐 𝟐 𝟐

Coordinate Geometry

If the three vertices of a triangle are displayed in a coordinate plane, the formula below, using a
determinant, will give the area of a triangle.
Let the three points in the coordinate plane be: 𝒙𝟏 , 𝒚𝟏 , 𝒙𝟐 , 𝒚𝟐 , 𝒙𝟑 , 𝒚𝟑 . Then, the area of
the triangle is one half of the absolute value of the determinant below:

𝒙𝟏 𝒚𝟏 𝟏
𝟏
𝑨 ∙ 𝒙𝟐 𝒚𝟐 𝟏
𝟐
𝒙𝟑 𝒚𝟑 𝟏

Example: For the triangle in the figure at right, the area is:

𝟏 𝟐 𝟒 𝟏
𝑨 ∙ 𝟑 𝟐 𝟏
𝟐
𝟑 𝟏 𝟏
𝟏 𝟐 𝟏 𝟑 𝟏 𝟑 𝟐 𝟏 𝟐𝟕
∙ 𝟐 𝟒 ∙ 𝟐𝟕
𝟐 𝟏 𝟏 𝟑 𝟏 𝟑 𝟏 𝟐 𝟐
Chapter 11 Perimeter and Area

Geometry
Perimeter and Area of Quadrilaterals

Name Illustration Perimeter Area

1
Kite 2 2
2

1
Trapezoid
2

Parallelogram 2 2

Rectangle 2 2

1
Rhombus 4
2

1
Square 4
2
Chapter 11 Perimeter and Area

Geometry
Perimeter and Area of Regular Polygons

Definitions – Regular Polygons

 The center of a polygon is the center of its circumscribed
circle. Point O is the center of the hexagon at right.
 The radius of the polygon is the radius of its
circumscribed circle. 𝑶𝑨 and 𝑶𝑩 are both radii of the
hexagon at right.
 The apothem of a polygon is the distance from the center
to the midpoint of any of its sides. a is the apothem of
the hexagon at right.
 The central angle of a polygon is an angle whose vertex is the center of the circle and whose
sides pass through consecutive vertices of the polygon. In the figure above, ∠𝑨𝑶𝑩 is a
central angle of the hexagon.

Area of a Regular Polygon

𝟏
𝑨 𝒂𝑷 where, 𝑎 is the apothem of the polygon
𝟐
𝑃 is the perimeter of the polygon

Perimeter and Area of Similar Figures

Let k be the scale factor relating two similar geometric figures F1 and F2 such that 𝐅𝟐 𝐤 ∙ 𝐅𝟏 .

Then,

𝐏𝐞𝐫𝐢𝐦𝐞𝐭𝐞𝐫 𝐨𝐟 𝐅𝟐
𝐤
𝐏𝐞𝐫𝐢𝐦𝐞𝐭𝐞𝐫 𝐨𝐟 𝐅𝟏

and

𝐀𝐫𝐞𝐚 𝐨𝐟 𝐅𝟐
𝐤𝟐
𝐀𝐫𝐞𝐚 𝐨𝐟 𝐅𝟏
Chapter 11 Perimeter and Area

Example 11.4: The scale factor of two similar polygons is 5: 2. The perimeter of the larger
polygon is 40 ft and its area is 100 ft . What are the perimeter and area of the smaller
polygon?
Scale factors and perimeter are both linear measures.
For perimeter, we have the proportion: For area, we have the proportion:
5 40 5 100
2 𝑃 2 𝐴
40 ∙ 2 25 100
𝑷 𝟏𝟔 𝐟𝐭
5 4 𝐴
100 ∙ 4
𝑨 𝟏𝟔 𝐟𝐭 𝟐
25

Example 11.5: Two similar figures have areas of 80 and 180. Find the ratio of their perimeters:
Area ratios are the squares of the corresponding linear ratios. Perimeters are linear
measures. Therefore, we have the proportion:

Area ratio: 𝑘

The small figure’s area is in the numerator of the above fraction and the large figure’s area
is in the denominator of the above fraction. Then,

𝟐
Perimeter ratio: 𝒌 √𝑘
𝟑

Example 11.6: What is the length of the apothem of a regular

hexagon with side length 12 cm?
The apothem splits the bottom side of the hexagon in half, i.e.,
into two segments of length 6.
Each interior angle in a regular hexagon is 120°, so half of that is
60°. This gives us a 30°-60°-90° triangle, with one side of the
triangle being the apothem. We can calculate, then:

𝒂 6 ∙ √3 𝟔√𝟑 𝐜𝐦
Chapter 11 Perimeter and Area

Example 11.7: What is the area of a regular hexagon with side length 12 cm?
The perimeter of this regular hexagon is:
𝑃 6 sides ∙ 12 cm per side 72 cm
The length of the apothem is 6√3 from the previous example.
The area of the regular hexagon in the diagram is:
1 1
𝑨 𝑎𝑃 6√3 ∙ 72 𝟐𝟏𝟔√𝟑 𝐮𝐧𝐢𝐭𝐬𝟐
2 2

Example 11.8: What is the area of the kite in the diagram? All measurements are in inches.
We need the lengths of the diagonals of the kite.
The vertical diagonal has length 𝑑 8 8 16.
To find the horizontal diagonal, we need the help of Pythagoras.
𝑥 8 17 → 𝑥 15
𝑑 15 6 21
Finally, we have:
1 1
𝐴 𝑑 𝑑 16 21 𝟏𝟔𝟖 𝐢𝐧𝟐
2 2

Example 11.9: Derive a formula for the area of an equilateral triangle with side length 𝑠.
Let the height of the equilateral triangle be ℎ. We need to find 𝑏.
We draw an altitude from the top of the triangle to the base, creating
a pair of congruent interior triangles. This results in 30°-60°-90°
triangles, each with base . The length of the height, then, is √3.
𝑠
The length of the whole base is: 2 ∙ 𝑠.
2
Finally,
1 1 𝑠 √𝟑 𝟐
𝑨 𝑏ℎ 𝑠∙ √3 𝒔
2 2 2 𝟒
Chapter 11 Perimeter and Area

Example 11.10: Successive squares are formed by joining the midpoints of each side. If the
outermost square has a side length of 20 m, what is the area of the shaded square?
Notice that we are able to create a 45°-45°-90° triangle in the upper right corner of the
diagram. Working in from the outer square to the next inner square, we see that the side
lengths of the squares shrink by a factor of √2.
Since the side lengths shrink by a factor of √2, the areas of
successive squares must shrink by a factor of √2 2.
The outer square has an area of: 𝐴 20 400 units .
The shaded square is three squares in from the outer square, so
its area must be:
1
𝑨 400 ∙ 𝟓𝟎 𝐦𝟐
2

Example 11.11: If ∆𝐴𝐵𝐶~∆𝐷𝐸𝐹, 𝐴𝐶 22 𝑎𝑛𝑑 𝐷𝐹 55, what is the ratio of the area of ∆𝐴𝐵𝐶
to the area of ∆𝐷𝐸𝐹.
The ratio of the areas is the square of the ratio of the linear measures.
∆𝐴𝐵𝐶 area 22 2 𝟒
𝒓
∆𝐷𝐸𝐹 area 55 5 𝟗

Example 11.12: If the ratios of the areas of two similar polygons is , what is the ratio of
their perimeters?
The ratio of the areas is the square of the ratio of the linear measures. So, the ratio of
linear measures (e.g., perimeter) is the square root of the ratio of the areas.

121 𝟏𝟏
𝒓
196 𝟏𝟒
Chapter 11 Perimeter and Area

Geometry
Circle Lengths and Areas

Circumference and Area

𝑪 𝟐𝝅 ∙ 𝒓 is the circumference (i.e., the perimeter) of the circle.

𝑨 𝝅𝒓𝟐 is the area of the circle.

where: 𝑟 is the radius of the circle.

Length of an Arc on a Circle

A common problem in the geometry of circles is to measure the length of an arc on a circle.
Definition: An arc is a segment along the circumference of a circle.

𝒎𝐀𝐁
𝒂𝒓𝒄 𝒍𝒆𝒏𝒈𝒕𝒉 ∙𝑪
𝟑𝟔𝟎

where: 𝑚∠AB is the measure (in degrees) of the arc. Note that
this is also the measure of the central angle ∠𝐴𝑂𝐵.
𝐶 is the circumference of the circle.

Area of a Sector of a Circle

Another common problem in the geometry of circles is to measure the area of a sector a circle.
Definition: A sector is a region in a circle that is bounded by two radii and an arc of the circle.

𝒎𝐀𝐁
𝒔𝒆𝒄𝒕𝒐𝒓 𝒂𝒓𝒆𝒂 ∙𝑨
𝟑𝟔𝟎

where: 𝑚∠AB is the measure (in degrees) of the arc. Note that
this is also the measure of the central angle ∠𝐴𝑂𝐵.
𝐴 is the area of the circle.
Chapter 11 Perimeter and Area

Example 11.13: What is the area of the shaded region if 𝑚∠AOC 95° and 𝑚 AB 53𝜋 m?

The length of the arc is of the circumference of the

circle.
53
𝐶 53𝜋 72𝜋 2𝜋𝑟 → 𝑟 36
72
53 53 53
𝑨𝒓𝒆𝒈𝒊𝒐𝒏 ∙𝐴 ∙ 𝜋𝑟 ∙ 𝜋 ∙ 36 𝟗𝟓𝟒𝝅 𝐦𝟐
72 72 72

Example 11.14: What is the length of major arc DPJ if 𝑚∠DOJ 135° and the diameter of the
circle is 16 meters.
The circumference of the circle is: 𝐶 𝜋𝑑 16𝜋 m.
DPJ has the same measure as the central angle subtended by it.
So, 𝑚 DPJ 360° 135° 225°. Then,
225
𝐥𝐞𝐧𝐠𝐭𝐡 𝐨𝐟 𝐃𝐏𝐉 ∙ 16𝜋 𝟏𝟎 m.
360

Example 11.15: Find the length of minor arc DJ if 𝑚∠DOJ 135° and the area of the circle is
25𝜋 cm .
𝐴 𝜋𝑟 25𝜋 → 𝑟 5
𝐶 2𝜋𝑟 2∙𝜋∙5 10𝜋 m.
DJ has the same measure as the central angle subtended by it.

So, 𝑚 DJ 135°, or of the circumference of the circle.

The length of major arc 𝐷𝐽, then is:

3 𝟏𝟓
𝐥𝐞𝐧𝐠𝐭𝐡 𝐨𝐟 𝐃𝐉 ∙ 10𝜋 cm.
8 𝟒
Chapter 11 Perimeter and Area

Geometry
Area of Composite Figures

To calculate the area of a figure that is a composite of shapes, consider each shape separately.

Example 11.16:
Calculate the area of the blue region in the figure to the right.
To solve this:
 Recognize that the figure is the composite of a
rectangle and two triangles.
 Disassemble the composite figure into its components.
 Calculate the area of the components.
 Subtract to get the area of the composite figure.

𝟏
𝑨𝒓𝒆𝒂 𝒐𝒇 𝑹𝒆𝒈𝒊𝒐𝒏 𝟒∙𝟔 𝟐 ∙𝟒∙𝟑 𝟐𝟒 𝟏𝟐 𝟏𝟐
𝟐

Example 11.17:
Calculate the area of the blue region in the figure to the right.
To solve this:
 Recognize that the figure is the composite of a square and a
circle.
 Disassemble the composite figure into its components.
 Calculate the area of the components.
 Subtract to get the area of the composite figure.

𝑨𝒓𝒆𝒂 𝒐𝒇 𝑹𝒆𝒈𝒊𝒐𝒏 𝟖𝟐 𝟒 𝝅 ∙ 𝟐𝟐 𝟔𝟒 𝟏𝟔𝝅 ~ 𝟏𝟑. 𝟕𝟑

Chapter 11 Perimeter and Area

Example 11.18: Two congruent semicircles and a full circle are arranged inside a large
semicircle as shown in the diagram. The radius of the smaller semicircles is 𝑥. The radius of the
full circle is 3. Find the total area of the aqua-colored shaded regions.

First, let’s find 𝑥:

∆𝐴𝐷𝐶 is a right triangle, so using the
Pythagorean Theorem:
𝐴𝐷 𝐶𝐷 𝐴𝐶
𝑥 𝐶𝐷 𝑥 3
𝑥 𝐶𝐷 𝑥 6𝑥 9
𝐶𝐷 6𝑥 9
We also know that 2𝑥 is the radius of the large (outer) semicircle.
Then, on line segment 𝐷𝐸 (also a radius of the large semicircle):
2𝑥 𝐶𝐷 3
𝐶𝐷 2𝑥 3
𝐶𝐷 2𝑥 3 4𝑥 12𝑥 9
Then, set the two expressions for 𝐶𝐷 equal to each other:
4𝑥 12𝑥 9 6𝑥 9
4𝑥 18𝑥 0
2𝑥 2𝑥 9 0
9
𝑥 0,
2
The answer 𝑥 0 makes no sense, so we must have: 𝑥

The shaded area of the diagram is developed as follows:

1 9 9 81
𝐴 𝜋 𝜋
2 2 2 2
1 9 81
𝐴 𝜋 𝜋
2 2 8
𝐴 𝜋 3 9𝜋
𝐴 𝐴 2∙𝐴 𝐴
81 81 𝟒𝟓
𝑨𝐬𝐡𝐚𝐝𝐞𝐝 𝜋 2∙ 𝜋 9𝜋 𝝅
2 8 𝟒
Chapter 11 Perimeter and Area

Example 11.19: What is the area of the region shaded in the diagram? All measurements are
in feet.
Shaded area sector area – triangle area.

Sector area ∙ 𝜋 ∙ 12 24𝜋.

The orange triangle is equilateral with sides of length 12 ft. This

allows us to complete the its measurements as shown below. Then,
∙ √
Triangle area bh 36√3

Shaded area 𝟐𝟒𝝅 – 𝟑𝟔√𝟑 ft2

Example 11.20: What is the area of the annulus shaded in the diagram?

An annulus is the area between two circles, so its area is the

difference of the areas of the two circles:
8
𝐴 𝜋𝑟 𝜋∙ 8 2 100𝜋 2

𝐴 𝜋𝑟 𝜋∙8 64𝜋
𝑨𝐚𝐧𝐧𝐮𝐥𝐮𝐬 𝐴 𝐴 100𝜋 64𝜋 𝟑𝟔𝝅 𝐮𝐧𝐢𝐭𝐬 𝟐
Chapter 12 Surface Area and Volume

Geometry
Polyhedra

Definitions
Faces
 A Polyhedron is a 3-dimensional solid bounded by a series
of polygons.
 Faces are the polygons that bound the polyhedron.
 An Edge is the line segment at the intersection of two faces.
 A Vertex is a point at the intersection of two edges. Edges

 A Regular polyhedron is one in which all of the faces are the

Vertices
same regular polygon.
 A Convex Polyhedron is one in which all diagonals are contained within the interior of the
polyhedron. A Concave polyhedron is one that is not convex.
 A Cross Section is the intersection of a plane with the polyhedron.

Euler’s Theorem
Let: 𝐹 the number of faces of a polyhedron. Example 12.1: Euler’s Theorem
𝑉 the number of vertices of a polyhedron. The cube above has …
𝐸 the number of edges of a polyhedron.  6 faces
 8 vertices
Then, for any polyhedron that does not intersect itself,
 12 edges

𝟔 𝟖 𝟏𝟐 𝟐 
𝑭 𝑽 𝑬 𝟐

Calculating the Number of Edges

The number of edges of a polyhedron is one-half the number of sides in the polygons it
comprises. Each side that is counted in this way is shared by two polygons; simply adding all
the sides of the polygons, therefore, double counts the number of edges on the polyhedron.
Example 12.2: Consider a soccer ball. It is polyhedron made up of 20
hexagons and 12 pentagons. Then the number of edges is:
1
𝑬 ∙ 20 ∙ 6 12 ∙ 5 𝟗𝟎
2
Note: use of the factor reflects each edge being counted twice in the sides of the polygons.
Chapter 12 Surface Area and Volume

Geometry
A Hole in Euler’s Theorem

Topology is a branch of mathematics that studies the properties of objects that are preserved
through manipulation that does not include tearing. An object may be stretched, twisted and
otherwise deformed, but not torn. In this branch of mathematics, a donut is equivalent to a
coffee cup because both have one hole; you can deform either the cup or the donut and create
the other, like you are playing with clay.

All of the usual polyhedra have no holes in them, so Euler’s Equation holds. What happens if
we allow the polyhedra to have holes in them? That is, what if we consider topological shapes
different from the ones we normally consider?

Euler’s Characteristic
When Euler’s Equation is rewritten as 𝑭 𝑬 𝑽 𝟐, the left-hand side of the equation is
called the Euler Characteristic.

The Euler Characteristic of a shape is: 𝑭 𝑬 𝑽

Generalized Euler’s Theorem

Let: 𝐹 the number of faces of a polyhedron.
𝑉 the number of vertices of a polyhedron.
𝐸 the number of edges of a polyhedron.
𝑔 the number of holes in the polyhedron. 𝑔 is
called the genus of the shape.

Then, for any polyhedron that does not intersect itself,

Example 12.3:

𝑭 𝑬 𝑽 𝟐 𝟐𝒈 The cube with a tunnel in it has …

𝐹 16, 𝐸 32, 𝑉 16
so, 𝑭 𝑬 𝑽 𝟎
Note that the value of Euler’s Characteristic can be
negative if the shape has more than one hole in it (i.e., if Then, 0 2 2𝑔
𝑔 2)! 𝒈 𝟏 hole 
Chapter 12 Surface Area and Volume

Geometry
Platonic Solids

A Platonic Solid is a convex regular polyhedron with faces composed of congruent convex
regular polygons. There five of them:

Key Properties of Platonic Solids

It is interesting to look at the key properties of these regular polyhedra.

Name Faces Vertices Edges Type of Face

Tetrahedron 4 4 6 Triangle
Cube 6 8 12 Square
Octahedron 8 6 12 Triangle
Dodecahedron 12 20 30 Pentagon
Icosahedron 20 12 30 Triangle

Notice the following patterns in the table:

 All of the numbers of faces are even. Only the cube has a number of faces that is not a
multiple of 4.
 All of the numbers of vertices are even. Only the octahedron has a number of faces that
is not a multiple of 4.
 The number of faces and vertices seem to alternate (e.g., cube 6-8 vs. octahedron 8-6).
 All of the numbers of edges are multiples of 6.
 There are only three possibilities for the numbers of edges – 6, 12 and 30.
 The faces are one of: regular triangles, squares or regular pentagons.
Chapter 12 Surface Area and Volume

Geometry
Prisms

Definitions
 A Prism is a polyhedron with two congruent polygonal faces
that lie in parallel planes.
 The Bases are the parallel polygonal faces.
 The Lateral Faces are the faces that are not bases.
 The Lateral Edges are the edges between the lateral faces.
 The Slant Height is the length of a lateral edge. Note that
all lateral edges are the same length.
 The Height is the perpendicular length between the bases.
 A Right Prism is one in which the angles between the bases and the
lateral edges are right angles. Note that in a right prism, the height and
the slant height are the same.
 An Oblique Prism is one that is not a right prism.
Right Hexagonal
 The Surface Area of a prism is the sum of the areas of all its faces. Prism

 The Lateral Area of a prism is the sum of the areas of its lateral faces.

Surface Area and Volume of a Right Prism

Surface Area: 𝑺𝑨 𝑷𝒉 𝟐𝑩 where, 𝑃 𝑡ℎ𝑒 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒
ℎ 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑖𝑠𝑚
Lateral SA: 𝑺𝑨 𝑷𝒉
𝐵 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒
Volume: 𝑽 𝑩𝒉

Cavalieri’s Principle
If two solids have the same height and the same cross-sectional area at every level, then they
have the same volume. This principle allows us to derive a formula for the volume of an
oblique prism from the formula for the volume of a right prism.

Surface Area and Volume of an Oblique Prism

Surface Area: 𝑺𝑨 𝑳𝑺𝑨 𝟐𝑩 where, 𝐿𝑆𝐴 𝑡ℎ𝑒 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎
ℎ 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑖𝑠𝑚
Volume: 𝑽 𝑩𝒉
𝐵 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒
The lateral surface area of an oblique prism is the sum of the
areas of the faces, which must be calculated individually.
Chapter 12 Surface Area and Volume

Example 12.4: Find the volume of the triangular prism.

This is a right prism, with a triangle for a base. First, find 𝐵, the area of
the triangular base.
1
𝐵 12 16 96
2
The height is the length perpendicular to the base. So, ℎ 10.
Finally, 𝑽 𝐵ℎ 96 ∙ 10 𝟗𝟔𝟎

Example 12.5: Find the lateral surface area and the total surface area of the
triangular prism.
The formula for the surface area of a prism is:
𝑆𝐴 Ph 2B, where P is the perimeter of the base, h is the height of
the prism, and B is the area of one base. Ph is also called the lateral
surface area of the prism.
The height is the length of a segment perpendicular to the base. So, ℎ 10.
The base is a triangle, so we need to calculate the length of its hypotenuse in order to
calculate the perimeter, P. Pythagoras will help us with this; the hypotenuse has length:

𝑐 12 16 20
We can now calculate: P 12 16 20 48. Therefore,
𝑳𝑺𝑨 𝑃∙ℎ 48 ∙ 10 𝟒𝟖𝟎
The area of one triangular base of the prism is: 𝐵 96 from the prior example.
The total surface area of the triangular prism, then, is:
𝑺𝑨 480 2 ∙ 96 𝟔𝟕𝟐.
Chapter 12 Surface Area and Volume

Geometry
Cylinders

Definitions
 A Cylinder is a figure with two congruent circular bases in parallel planes.
 The Axis of a cylinder is the line connecting the centers of the circular bases.
 A cylinder has only one Lateral Surface. When deconstructed, the lateral surface of a
cylinder is a rectangle with length equal to the circumference of the base.
 There are no Lateral Edges in a cylinder.
 The Slant Height is the length of the lateral side between the bases. Note
that all lateral distances are the same length. The slant height has
applicability only if the cylinder is oblique.
 The Height is the perpendicular length between the bases.

 A Right Cylinder is one in which the angles between the bases and the lateral side are right
angles. Note that in a right cylinder, the height and the slant height are the same.
 An Oblique Cylinder is one that is not a right cylinder.

 The Surface Area of a cylinder is the sum of the areas of its bases and its lateral surface.
 The Lateral Area of a cylinder is the areas of its lateral surface.

Surface Area and Volume of a Right Cylinder

Surface Area: 𝑺𝑨 𝑪𝒉 𝟐𝑩 where, 𝐶 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒
𝟐𝝅𝒓𝒉 𝟐𝝅𝒓𝟐 ℎ 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟
𝐵 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒
Lateral SA: 𝑺𝑨 𝑪𝒉 𝟐𝝅𝒓𝒉 𝑟 𝑡ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒

Volume: 𝑽 𝑩𝒉 𝝅𝒓𝟐 𝒉

Surface Area and Volume of an Oblique Cylinder

Surface Area: 𝑺𝑨 𝑷𝒍 𝟐𝑩 where, 𝑃 𝑡ℎ𝑒 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑎 right section*
𝑜𝑓 𝑡ℎ𝑒 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟
Volume: 𝑽 𝑩𝒉 𝝅𝒓𝟐 𝒉 𝑙 𝑡ℎ𝑒 𝑠𝑙𝑎𝑛𝑡 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟
ℎ 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟
* A right section of an oblique cylinder is 𝐵 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒
a cross section perpendicular to the axis 𝑟 𝑡ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒
of the cylinder.
Chapter 12 Surface Area and Volume

Example 12.6: Find the volume of a right cylinder that has a diameter of 6 cm
and a height of 10 cm.
For a cylinder, 𝑉 𝜋𝑟 ℎ. In this case, 𝑟 6 2 3, ℎ 10.

𝑉 𝜋𝑟 ℎ 𝜋 ∙ 3 ∙ 10 90𝜋 cm

Example 12.7: Find the lateral surface area and the total surface area of a right
cylinder that has a diameter of 6 cm and a height of 10 cm
The formula for the surface area of a cylinder is:
𝑆𝐴 2πrh 2πr , where r is the radius of the base, h is the height of the cylinder,
and πr is the area of one base. 2πrh is also called the lateral surface area of the right
cylinder.
The radius is half the diameter: 𝑟 6 2 3
The height is the length of the side perpendicular to the base. So, ℎ 10.
Therefore,
𝑳𝑺𝑨 2πrh 2π ∙ 3 ∙ 10 𝟔𝟎𝝅
The area of one circular base of the cylinder is: πr π 3 9𝜋.
The total surface area of the right cylinder, then, is:
𝑺𝑨 60𝜋 2 ∙ 9𝜋 𝟕𝟖𝝅.
Chapter 12 Surface Area and Volume

Geometry
Surface Area by Decomposition

Sometimes the student is asked to calculate the surface are of a prism that does not quite fit
into one of the categories for which an easy formula exists. In this case, the answer may be to
decompose the prism into its component shapes, and then calculate the areas of the
components. Note: this process also works with cylinders and pyramids.

Decomposition of a Prism
To calculate the surface area of a prism, decompose it and look at each of the prism’s faces
individually.
Example 12.8: Calculate the surface area of the triangular prism:
To do this, first notice that we need the value of the hypotenuse of the
base. Use the Pythagorean Theorem or Pythagorean Triples to
determine the missing value is 10. Then, decompose the figure into its
various faces:

The surface area, then, is calculated as:

𝑆𝐴 2 𝐵𝑎𝑠𝑒𝑠 𝐹𝑟𝑜𝑛𝑡 𝐵𝑎𝑐𝑘 𝑆𝑖𝑑𝑒
1
𝑺𝑨 2∙ ∙6∙8 10 ∙ 7 8∙7 6∙7 𝟐𝟏𝟔
2

Decomposition of a Right Cylinder

Example 12.9: Calculate the surface area of the cylinder:
The cylinder is decomposed into two circles (the bases) and a
rectangle (the lateral face).
The surface area, then, is calculated as:
𝑆𝐴 2 𝑡𝑜𝑝𝑠 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑓𝑎𝑐𝑒
𝑆𝐴 2∙ 𝜋∙3 6𝜋 ∙ 5
𝑺𝑨 𝟒𝟖𝝅 𝟏𝟓𝟎. 𝟖𝟎
Chapter 12 Surface Area and Volume

Geometry
Pyramids

Pyramids
 A Pyramid is a polyhedron in which the base is a polygon and
the lateral sides are triangles with a common vertex.
 The Base is a polygon of any size or shape.
 The Lateral Faces are the faces that are not the base.
 The Lateral Edges are the edges between the lateral faces.
 The Apex of the pyramid is the intersection of the lateral
edges. It is the point at the top of the pyramid.
 The Slant Height of a regular pyramid is the altitude of one of
the lateral faces.
 The Height is the perpendicular length between the base and the apex.

 A Regular Pyramid is one in which the lateral faces are congruent triangles. The height of a
regular pyramid intersects the base at its center.
 An Oblique Pyramid is one that is not a right pyramid. That is, the
apex is not aligned directly above the center of the base.

 The Surface Area of a pyramid is the sum of the areas of all its
faces.
 The Lateral Area of a pyramid is the sum of the areas of its lateral
faces.

Surface Area and Volume of a Regular Pyramid

𝟏 where, 𝑃 𝑡ℎ𝑒 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒
Surface Area: 𝑺𝑨 𝑷𝒔 𝑩
𝟐 𝑠 𝑡ℎ𝑒 𝑠𝑙𝑎𝑛𝑡 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑦𝑟𝑎𝑚𝑖𝑑
𝟏
Lateral SA: 𝑺𝑨 𝑷𝒔 ℎ 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑦𝑟𝑎𝑚𝑖𝑑
𝟐
𝟏 𝐵 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒
Volume: 𝑽 𝑩𝒉
𝟑

Surface Area and Volume of an Oblique Pyramid

where, 𝐿𝑆𝐴 𝑡ℎ𝑒 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎
Surface Area: 𝑺𝑨 𝑳𝑺𝑨 𝑩
ℎ 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑦𝑟𝑎𝑚𝑖𝑑
𝟏
Volume: 𝑽 𝑩𝒉 𝐵 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒
𝟑

The lateral surface area of an oblique pyramid is the sum of

the areas of the faces, which must be calculated individually.
Chapter 12 Surface Area and Volume

Example 12.10: Calculate the volume of the square pyramid shown if the
perimeter of the base is 64 and the height is 15.

For a square pyramid, we introduce a factor of into the volume

calculation, relative to a prism. The same factor is used in the
calculation of a cone, relative to a cylinder. The origins of the factor
come from Calculus and the fact that we are working in 3 dimensions.

If the perimeter of the base is 64, then the length of one base edge is: 64 4 16.
Our base is a square with area: 𝐵 16 256. ℎ 15.
1 1
𝑨 𝐵ℎ 256 15 𝟏𝟐𝟖𝟎
3 3

Example 12.11: Calculate the slant height of the face of the square pyramid in
the previous example.
If we look inside the pyramid, we can see a triangle that has a height of
length ℎ 15, a leg that is half the length of a base edge of the pyramid
(16 2 8) and a hypotenuse of the slant height (s). Use the Pythagorean
Theorem, then, to determine: 𝐬 √15 8 𝟏𝟕

Example 12.12: Calculate the lateral surface area and the total surface area of the square
pyramid in the previous example.
The formula for the surface area of a square pyramid is:

𝑆𝐴 Ps B, where P is the perimeter of the base, s is the slant

height of the pyramid, and B is the area of the base. 𝑃𝑠 is also
called the lateral surface area of the pyramid.
From the previous two examples, we know that P 64 and s 17. Therefore,
1 1
𝑳𝑺𝑨 ∙𝑃∙𝑠 ∙ 64 ∙ 17 𝟓𝟒𝟒
2 2
The base length is 16, so the area of the base is: 𝐵 16 256.
The total surface area of the square pyramid, then, is:
𝑺𝑨 544 256 𝟖𝟎𝟎.
Chapter 12 Surface Area and Volume

Geometry
Cones

Definitions
 A Circular Cone is a 3-dimensional geometric figure with a circular base which tapers
smoothly to a vertex (or apex). The apex and base are in different planes. Note: there is
also an elliptical cone that has an ellipse as a base, but that will not be considered here.
 The Base is a circle.
 The Lateral Surface is area of the figure between the base and the apex.
 There are no Lateral Edges in a cone.
 The Apex of the cone is the point at the top of the cone.
 The Slant Height of a cone is the length along the lateral surface from the apex to the base.
 The Height is the perpendicular length between the base and the apex.

 A Right Cone is one in which the height of the cone intersects the base at
its center.
 An Oblique Cone is one that is not a right cone. That is, the apex is not
aligned directly above the center of the base.

 The Surface Area of a cone is the sum of the area of its lateral surface
and its base.
 The Lateral Area of a cone is the area of its lateral surface.

Surface Area and Volume of a Right Cone

Surface Area: 𝑺𝑨 𝝅𝒓𝒔 𝝅𝒓𝟐 where, 𝑟 𝑡ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒
𝑠 𝑡ℎ𝑒 𝑠𝑙𝑎𝑛𝑡 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑛𝑒
Lateral SA: 𝑺𝑨 𝝅𝒓𝒔
ℎ 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑛𝑒
𝟏 𝟏
Volume: 𝑽 𝑩𝒉 𝝅𝒓𝟐 𝒉 𝐵 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒
𝟑 𝟑

Surface Area and Volume of an Oblique Cone

Surface Area: 𝑺𝑨 𝑳𝑺𝑨 𝝅𝒓𝟐 where, 𝐿𝑆𝐴 𝑡ℎ𝑒 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎
𝟏 𝟏 𝑟 𝑡ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒
Volume: 𝑽 𝑩𝒉 𝝅𝒓𝟐 𝒉 ℎ 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑛𝑒
𝟑 𝟑

There is no easy formula for the lateral surface area of an

oblique cone.
Chapter 12 Surface Area and Volume

Example 12.13: Calculate the exact volume of the right cone shown.

For a cone, 𝑉 𝜋𝑟 ℎ. In this case, 𝑟 18 2 9, ℎ 12.

1 1
𝑉 𝜋𝑟 ℎ 𝜋 ∙ 9 ∙ 12 324𝜋 cm
3 3

Example 12.14: Find the lateral surface area and the total surface
area of a right cone shown.
The formula for the surface area of a cone is:
𝑆𝐴 πr𝑙 πr , where r is the radius of the base, 𝑙 is the slant height of the cone, and
πr is the area of the base. πr𝑙 is also called the lateral surface area of the right cone.
The radius is half the diameter: 𝑟 18 2 9
The height is given in the diagram. ℎ 12.
A cross-sectional view of a cone is a triangle. We want to examine the
right triangle in the cross-section to determine the slant height, 𝑙.
Pythagoras will help us with this; the hypotenuse has length:

𝑐 9 12 15
Therefore,
𝑳𝑺𝑨 πr𝑙 π ∙ 9 ∙ 15 𝟏𝟑𝟓𝝅
The area of the circular base of the cone is: πr π 9 81𝜋.
The total surface area of the right cone, then, is:
𝑺𝑨 135𝜋 81𝜋 𝟐𝟏𝟔𝝅.
Chapter 12 Surface Area and Volume

Geometry
Spheres

Definitions
 A Sphere is a 3-dimensional geometric figure in which all
points are a fixed distance from a point. A good example of
a sphere is a ball.
 Center – the middle of the sphere. All points on the sphere
are the same distance from the center.
 Radius – a line segment with one endpoint at the center and
the other endpoint on the sphere. The term “radius” is also
used to refer to the distance from the center to the points
on the sphere.
 Diameter – a line segment with endpoints on the sphere
that passes through the center.
 Great Circle – the intersection of a plane and a sphere
that passes through the center.
 Hemisphere – half of a sphere. A great circle separates a
plane into two hemispheres.
 Secant Line – a line that intersects the sphere in exactly
two points.
 Tangent Line– a line that intersects the sphere in exactly
one point.
 Chord – a line segment with endpoints on the sphere that does not pass through the center.

Surface Area and Volume of a Sphere

Surface Area: 𝑺𝑨 𝟒𝝅𝒓𝟐
𝟒
Volume: 𝑽 𝝅𝒓𝟑
𝟑

where, 𝑟 𝑡ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝ℎ𝑒𝑟𝑒

Chapter 12 Surface Area and Volume

Example 12.15: Find the volume of a sphere with radius 9.

The volume of a sphere is: 𝑉 𝜋𝑟 . In this case, 𝑟 9.

4
𝑽 𝜋 9 𝟗𝟕𝟐𝝅
3

Example 12.16: Find the surface area of a sphere with radius 9.

The surface area of a sphere is: 𝑆𝐴 4𝜋𝑟 . In this case, 𝑟 9.
𝑺𝑨 4𝜋 9 𝟑𝟐𝟒𝝅

Interestingly, in Calculus, you will learn that the formula for the surface area of a sphere is the
derivative of the formula for the volume of a sphere. That is:
4 𝑑𝑉
𝑉 𝜋𝑟 4𝜋𝑟 𝑆𝐴
3 𝑑𝑟
This also occurs with the formulas for the area and circumference of a circle.
𝑑𝐴
𝐴 𝜋𝑟 2𝜋𝑟 𝐶
𝑑𝑟

Example 12.17: The Earth has a volume is approximately 1.08 trillion km3. Assuming that the
Earth is a sphere, estimate its radius to the nearest kilometer and to the nearest mile.
4
The volume of a sphere is: 𝑉 𝜋𝑟 . In this case, 𝑉 1,080,000,000,000.
3
Get your calculator ready.
4
1,080,000,000,000 𝜋𝑟
3
257,831,007,809 𝑟
𝒓 257,831,007,809 6,364.7065 𝟔, 𝟑𝟔𝟓 𝐤𝐦
km
𝒓 6,364.7065 km ∙ 0.62137119 3,954.8453 𝟑, 𝟗𝟓𝟓 𝐦𝐢𝐥𝐞𝐬
mile

Example 12.18: Approximate the circumference of the Earth in kilometers and miles.
Using the radius estimates from the prior example:
Kilometers: 𝑪 2𝜋𝑟 2𝜋 ∙ 6,364.7065 km 𝟑𝟗, 𝟗𝟗𝟏 𝐤𝐦 or about 40,000 km.
Miles: 𝑪 2𝜋𝑟 2𝜋 ∙ 3,954.8453 miles 𝟐𝟒, 𝟖𝟒𝟗 𝐦𝐢𝐥𝐞𝐬 𝐤𝐦 or about 25,000 miles.

Given the accuracy of our starting values, two significant digits in our answers is about the
best we can hope for. 40,000 km and 25,000 miles are real estimates of the
circumference of the Earth to two significant digits.
Chapter 12 Surface Area and Volume

Geometry
Similar Solids

Similar Solids have equal ratios of corresponding linear measurements (e.g., edges, radii). So,
all of their key dimensions are proportional.

Edges, Surface Area and Volume of Similar Figures

Let k be the scale factor relating two similar geometric solids F1 and F2 such that 𝐅𝟐 𝐤 ∙ 𝐅𝟏 .
Then, for corresponding parts of F1 and F2,

𝐄𝐝𝐠𝐞 𝐨𝐟 𝐅𝟐
𝐤
𝐄𝐝𝐠𝐞 𝐨𝐟 𝐅𝟏

and

𝐒𝐮𝐫𝐟𝐚𝐜𝐞 𝐀𝐫𝐞𝐚 𝐨𝐟 𝐅𝟐
𝐤𝟐
𝐒𝐮𝐫𝐟𝐚𝐜𝐞 𝐀𝐫𝐞𝐚 𝐨𝐟 𝐅𝟏

And

𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐅𝟐
𝐤𝟑
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐅𝟏

These formulas hold true for any corresponding portion of the

figures. So, for example:

k k
Chapter 12 Surface Area and Volume

Example 12.19: Two similar octahedrons have edges of lengths 4 and 12. Find the ratio of their
volumes.
Volume ratio Linear ratio
4 1 𝟏
𝐕𝐨𝐥𝐮𝐦𝐞 𝐫𝐚𝐭𝐢𝐨
12 3 𝟐𝟕

Example 12.20: Two similar icosahedrons have volumes of 250 and 686. Find the ratio of their
surface areas.
Call the linear ratio between similar objects 𝑘. Then:
Linear measure : area : volume have relative ratios of 𝑘 ∶ 𝑘 ∶ 𝑘 . To get from a volume
ratio to a surface area ratio, we need to take the cube root of the volume ratio (to get from
volume to linear) and square the result (to get from linear to area). Alternatively, we could
take the 2/3 power of the volume relativities to get the same answer.

250 125 5 𝟐𝟓
𝐀𝐫𝐞𝐚 𝐫𝐚𝐭𝐢𝐨
686 343 7 𝟒𝟗

Alternative Method:
250 𝟐𝟓
𝐀𝐫𝐞𝐚 𝐫𝐚𝐭𝐢𝐨
686 𝟒𝟗
Appendix A Geometry Formulas

Geometry
Summary of Perimeter and Area Formulas – 2D Shapes
Shape Figure Perimeter Area
𝟏
𝑷 𝟐𝒃 𝟐𝒄 𝑨 𝒅 𝒅
Kite 𝟐 𝟏 𝟐
𝑏, 𝑐 𝑠𝑖𝑑𝑒𝑠
𝑑 ,𝑑 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙𝑠

𝟏
𝑷 𝒃𝟏 𝒃𝟐 𝒄 𝒅 𝐀 𝐛 𝐛𝟐 𝐡
Trapezoid 𝟐 𝟏
𝑏 ,𝑏 𝑏𝑎𝑠𝑒𝑠
𝑐, 𝑑 𝑠𝑖𝑑𝑒𝑠 b ,b bases
h height

𝐀 𝐛𝐡
𝑷 𝟐𝒃 𝟐𝒄
Parallelogram 𝑏 𝑏𝑎𝑠𝑒
𝑏, 𝑐 𝑠𝑖𝑑𝑒𝑠
ℎ ℎ𝑒𝑖𝑔ℎ𝑡

𝑷 𝟐𝒃 𝟐𝒄 𝐀 𝐛𝐡
Rectangle 𝑏 𝑏𝑎𝑠𝑒
𝑏, 𝑐 𝑠𝑖𝑑𝑒𝑠
ℎ ℎ𝑒𝑖𝑔ℎ𝑡

𝟏
𝑷 𝟒𝒔 𝑨 𝒃𝒉 𝒅 𝒅
Rhombus 𝟐 𝟏 𝟐
𝑠 𝑠𝑖𝑑𝑒 𝑑 ,𝑑 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙𝑠

𝟏
𝑷 𝟒𝒔 𝑨 𝒔𝟐 𝒅 𝒅
Square 𝟐 𝟏 𝟐
𝑠 𝑠𝑖𝑑𝑒 𝑑 ,𝑑 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙𝑠

𝟏
𝑷 𝒏𝒔 𝑨 𝒂∙𝑷
Regular Polygon 𝟐
𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑖𝑑𝑒𝑠
𝑠 𝑠𝑖𝑑𝑒 𝑎 𝑎𝑝𝑜𝑡ℎ𝑒𝑚
𝑃 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟

𝑪 𝟐𝝅𝒓 𝝅𝒅
𝑨 𝝅𝒓𝟐
Circle 𝑟 𝑟𝑎𝑑𝑖𝑢𝑠
𝑟 𝑟𝑎𝑑𝑖𝑢𝑠
𝑑 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟

𝑷 𝟐𝝅
𝟏
𝒓𝟏 𝟐 𝒓𝟐 𝟐 𝑨 𝝅𝒓𝟏 𝒓𝟐
𝟐
Ellipse 𝑟 𝑚𝑎𝑗𝑜𝑟 𝑎𝑥𝑖𝑠 𝑟𝑎𝑑𝑖𝑢𝑠 𝑟 𝑚𝑎𝑗𝑜𝑟 𝑎𝑥𝑖𝑠 𝑟𝑎𝑑𝑖𝑢𝑠
𝑟 𝑚𝑖𝑛𝑜𝑟 𝑎𝑥𝑖𝑠 𝑟𝑎𝑑𝑖𝑢𝑠 𝑟 𝑚𝑖𝑛𝑜𝑟 𝑎𝑥𝑖𝑠 𝑟𝑎𝑑𝑖𝑢𝑠
Appendix A Geometry Formulas

Geometry
Summary of Surface Area and Volume Formulas – 3D Shapes

Shape Figure Surface Area Volume

𝟒 𝟑
Sphere 𝑺𝑨 𝟒𝝅𝒓𝟐 𝑽 𝝅𝒓
𝟑
𝑟 𝑟𝑎𝑑𝑖𝑢𝑠
𝑟 𝑟𝑎𝑑𝑖𝑢𝑠

𝑺𝑨 𝟐𝝅𝒓𝒉 𝟐𝝅𝒓𝟐 𝑽 𝝅𝒓𝟐 𝒉

Right
ℎ ℎ𝑒𝑖𝑔ℎ𝑡 ℎ ℎ𝑒𝑖𝑔ℎ𝑡
Cylinder
𝑟 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑏𝑎𝑠𝑒 𝑟 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑏𝑎𝑠𝑒

𝟏 𝟐
𝑺𝑨 𝝅𝒓𝒍 𝝅𝒓𝟐 𝑽 𝝅𝒓 𝒉
Cone 𝟑
𝑙 𝑠𝑙𝑎𝑛𝑡 ℎ𝑒𝑖𝑔ℎ𝑡
ℎ ℎ𝑒𝑖𝑔ℎ𝑡
𝑟 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑏𝑎𝑠𝑒
𝑟 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑏𝑎𝑠𝑒

𝟏 𝟐
Square 𝑺𝑨 𝟐𝒔𝒍 𝒔𝟐 𝑽 𝒔 𝒉
𝟑
Pyramid 𝑠 𝑏𝑎𝑠𝑒 𝑠𝑖𝑑𝑒 𝑙𝑒𝑛𝑔𝑡ℎ
𝑠 𝑏𝑎𝑠𝑒 𝑠𝑖𝑑𝑒 𝑙𝑒𝑛𝑔𝑡ℎ
𝑙 𝑠𝑙𝑎𝑛𝑡 ℎ𝑒𝑖𝑔ℎ𝑡
ℎ ℎ𝑒𝑖𝑔ℎ𝑡

𝑺𝑨 𝟐 ∙ 𝒍𝒘 𝒍𝒉 𝒘𝒉 𝑽 𝒍𝒘𝒉
Rectangular
Prism 𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑙 𝑙𝑒𝑛𝑔𝑡ℎ
𝑤 𝑤𝑖𝑑𝑡ℎ 𝑤 𝑤𝑖𝑑𝑡ℎ
ℎ ℎ𝑒𝑖𝑔ℎ𝑡 ℎ ℎ𝑒𝑖𝑔ℎ𝑡

𝑺𝑨 𝟔𝒔𝟐 𝑽 𝒔𝟑
Cube
𝑠 𝑠𝑖𝑑𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑎𝑙𝑙 𝑠𝑖𝑑𝑒𝑠 𝑠 𝑠𝑖𝑑𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑎𝑙𝑙 𝑠𝑖𝑑𝑒𝑠

𝑺𝑨 𝑷𝒉 𝟐𝑩 𝑽 𝑩𝒉
General
𝑃 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝐵𝑎𝑠𝑒 𝐵 𝑎𝑟𝑒𝑎 𝑜𝑓 𝐵𝑎𝑠𝑒
Right Prism ℎ ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑟 𝑙𝑒𝑛𝑔𝑡ℎ ℎ ℎ𝑒𝑖𝑔ℎ𝑡
𝐵 𝑎𝑟𝑒𝑎 𝑜𝑓 𝐵𝑎𝑠𝑒
Appendix B Trigonometry Formulas
Trigonometry Reference

Function Relationships Opposite Angle Formulas Cofunction Formulas (in Quadrant I)

1 1 sin 𝜃 sin 𝜃 𝜋 𝜋
sin θ csc θ sin 𝜃 cos 𝜃 cos 𝜃 sin 𝜃
csc θ sin θ 2 2
cos 𝜃 cos 𝜃 𝜋 𝜋
1 1 tan 𝜃 cot 𝜃 cot 𝜃 tan 𝜃
cos θ sec θ 2 2
sec θ cos θ tan 𝜃 tan 𝜃
𝜋 𝜋
1 1 cot 𝜃 cot 𝜃 sec 𝜃 csc 𝜃 csc 𝜃 sec 𝜃
tan θ cot θ 2 2
cot θ tan θ
sec 𝜃 sec 𝜃
sin θ cos θ Angle Addition Formulas
tan θ cot θ
cos θ sin θ csc 𝜃 csc 𝜃
sin 𝐴 𝐵 sin 𝐴 cos 𝐵 cos 𝐴 sin 𝐵
Pythagorean Identities Half Angle Formulas sin 𝐴 𝐵 sin 𝐴 cos 𝐵 cos 𝐴 sin 𝐵

sin 𝜃 cos 𝜃 1 cos 𝐴 𝐵 cos 𝐴 cos 𝐵 sin 𝐴 sin 𝐵

𝜃 1 cos 𝜃
sin cos 𝐴 𝐵 cos 𝐴 cos 𝐵 sin 𝐴 sin 𝐵
tan 𝜃 1 sec 𝜃 2 2
tan 𝐴 tan 𝐵
cot 𝜃 1 csc 𝜃 tan 𝐴 𝐵
𝜃 1 cos 𝜃 1 – tan 𝐴 tan 𝐵
cos tan 𝐴 tan 𝐵
2 2
Double Angle Formulas tan 𝐴 𝐵
1 tan 𝐴 tan 𝐵
sin 2𝜃 2 sin 𝜃 cos 𝜃 𝜃 1 cos 𝜃
tan
2 1 cos 𝜃 Product-to-Sum Formulas
cos 2𝜃 cos 𝜃 sin 𝜃
1 2 sin 𝜃 1
1 cos 𝜃 sin 𝐴 ∙ sin 𝐵 cos 𝐴 𝐵 cos 𝐴 𝐵
2 cos 𝜃 1 2
sin 𝜃
1
2 tan 𝜃 sin 𝜃 cos 𝐴 ∙ cos 𝐵 cos 𝐴 𝐵 cos 𝐴 𝐵
tan 2𝜃 2
1 tan 𝜃 1 cos 𝜃
1
sin 𝐴 ∙ cos 𝐵 sin 𝐴 𝐵 sin 𝐴 𝐵
2
Triple Angle Formulas Power Reducing Formulas 1
cos 𝐴 ∙ sin 𝐵 sin 𝐴 𝐵 sin 𝐴 𝐵
sin 3𝜃 3 sin 𝜃 4 sin 𝜃 1 cos 2𝜃 2
sin 𝜃
2
cos 3𝜃 4 cos 𝜃 3 cos 𝜃 1 cos 2𝜃 Sum-to-Product Formulas
cos 𝜃
3 tan 𝜃 tan 𝜃 2 𝐴 𝐵 𝐴 𝐵
tan 3𝜃 1 cos 2𝜃 sin 𝐴 sin 𝐵 2 ∙ sin ∙ cos
1 3 tan 𝜃 tan 𝜃 2 2
1 cos 2𝜃 𝐴 𝐵 𝐴 𝐵
sin 𝐴 sin 𝐵 2 ∙ sin ∙ cos
2 2
Arc Length Law of Sines 𝐴 𝐵 𝐴 𝐵
𝑆 𝑟𝜃 cos 𝐴 cos 𝐵 2 ∙ cos ∙ cos
2 2
𝐴 𝐵 𝐴 𝐵
Law of Cosines Law of Tangents cos 𝐴 cos 𝐵 2 ∙ sin ∙ sin
2 2
𝑎 𝑏 𝑐 2𝑏𝑐 cos 𝐴 1
𝑎 𝑏 tan 𝐴 𝐵
𝑏 𝑎 𝑐 2𝑎𝑐 cos 𝐵 2
𝑎 𝑏 1 Mollweide’s Formulas
𝑐 𝑎 𝑏 2𝑎𝑏 cos 𝐶 tan 𝐴 𝐵 mathguy.us
2 1
𝑎 𝑏 cos 𝐴 𝐵
2
Euler’s Formula Polar Multiplication and Division 𝑐 1
sin 𝐶
2
𝑒 cos 𝜃 𝑖 sin 𝜃 cis 𝜃 Let: 𝑎 𝑟 cis 𝜃 𝑏 𝑟 cis 𝜑 1
𝑎 𝑏 sin 𝐴 𝐵
𝑎 𝑟 2
DeMoivre’s Formula 𝑎∙𝑏 𝑟 𝑟 cis 𝜃 𝜑 cis 𝜃 𝜑 𝑐 1
𝑏 𝑟 cos 𝐶
2
𝑟 cis 𝜃 𝑟 cis 𝑛𝜃
Appendix B Trigonometry Formulas
Trigonometry Reference

Period 2𝜋 Period 2𝜋 Period 𝜋 𝒚 𝑨 ∙ 𝒇 𝑩𝒙 𝑪 𝑫

Amplitude: |𝑨|
"𝒇"
Period:
𝑩
𝑪
Phase Shift:
𝑩
Vertical Shift: 𝑫

Period 2𝜋 Period 2𝜋 Period 𝜋 Harmonic Motion

𝑑 𝑎 cos 𝜔𝑡 or
𝑑 𝑎 sin 𝜔𝑡

𝜔 2𝜋𝑓, 𝜔 0

Trig Functions of Special Angles (Unit Circle)

𝜽 Rad 𝜽° 𝐬𝐢𝐧 𝜽 𝐜𝐨𝐬 𝜽 𝐭𝐚𝐧 𝜽

0 0⁰ 0 1 0
𝜋 1/2
6 30⁰ √3/2 √3/3
𝜋 1
4 45⁰ √2/2 √2/2
𝜋 1/2
3 60⁰ √3/2 √3
𝜋 90⁰ 1 0 undefined
2

Rectangular/Polar Conversion Triangle Area Vector Properties

Rectangular Polar 1
𝐴 𝑏ℎ 0 𝐮 𝐮 0 𝐮
𝑥, 𝑦 𝑟, 𝜃 2
𝐮 𝐮 𝐮 𝐮 0
𝑟 𝑥𝑦 𝐴 𝑠 𝑠 𝑎 𝑠 𝑏 𝑠 𝑐
𝑥 𝑟 cos 𝜃
𝑦 1 1 𝐮 𝐯 𝐯 𝐮
𝑦 𝑟 sin 𝜃 𝜃 tan 𝑠 𝑃 𝑎 𝑏 𝑐
𝑥 2 2 𝐮 𝐯 𝐰 𝐮 𝐯 𝐰
𝑟 cos 𝜃 𝑖 sin 𝜃 1 𝑎 sin 𝐵 sin 𝐶
𝑎 𝑏𝑖 𝐴 𝑚 𝑛𝐮 𝑚𝑛 𝐮
or 𝑟 𝑐𝑖𝑠𝜃 2 sin 𝐴
𝑚 𝐮 𝐯 𝑚𝐮 𝑚𝐯
𝑎 𝑟 cos 𝜃 𝑟 𝑎 𝑏 1
𝐴 𝑎𝑏 sin 𝐶
𝑏 2 𝑚 𝑛 𝐮 𝑚𝐮 𝑛𝐮
𝑏 𝑟 sin 𝜃 𝜃 tan
𝑎 𝑥1 𝑦1 1
1 1 𝐯 𝐯
𝑎𝐢 𝑏𝐣 ‖𝐯‖ ∠𝜃 𝐴 𝑥2 𝑦2 1
2 𝑥3 𝑦3 1 ‖𝑚𝐯‖ |𝑚| ‖𝐯‖
𝑎 ‖𝐯‖ cos 𝜃 ‖𝐯‖ 𝑎 𝑏 𝐯
𝑏 1 Unit Vector:
𝑏 ‖𝐯‖ sin 𝜃 𝐴 ‖𝐮‖ ‖𝐯‖ sin θ ‖𝐯‖
𝜃 tan 2
𝑎

Vector Dot Product Vector Cross Product Angle between Vectors

u u 𝐮∘𝐯 ‖𝐮 𝐯‖
𝐮∘𝐯 𝑢 ∙𝑣 𝑢 ∙𝑣 𝐮x𝐯 u v u v cos 𝜃 sin 𝜃
v v ‖𝐮‖ ‖𝐯‖ ‖𝐮‖ ‖𝐯‖
𝐮∘ 𝐯 𝐰 𝐮∘𝐯 𝐮∘𝐰 𝐮x 𝐯 𝐰 𝐮x𝐯 𝐮x𝐰 ⊥ iff 𝐮 ∘ 𝐯 0 ∥ iff 𝐮 x 𝐯 0
 Geometry Handbook
Index

Page Subject

22 Alternate Exterior Angles

22 Alternate Interior Angles
29, 42 Altitude of a Triangle
28, 42 Angle Bisector Length in a Triangle
Angles
13 Angles - Basic
14 Angles - Types
110 Annulus
102 Apothem
92 Arcs
106 Arc Length
Area
108 Area - Composite Figures
102 Area - Polygons
101 Area - Quadrilaterals
106 Area - Region of a Circle
98, 100 Area - Triangle
127 Area Formulas - Summary for 2D Shapes
116 Axis of a Cylinder
114 Cavalieri's Principle
92 Center of a Circle
102 Center of a Regular Polygon
Centers of Triangles
40 Centroid
40 Circumcenter
40 Incenter
40 Orthocenter
94, 102 Central Angle
40 Centroid
92 Chord
95 Chord Facts
Circles
106 Circles - Arc Lengths
92 Circles - Definitions of Parts
95 Circles - Facts
106 Circles - Region Areas
93 Circles - Related Angles
93 Circles - Related Segments
40 Circles and Triangles
 Geometry Handbook
Index

Page Subject

40 Circumcenter
94 Circumscribed Polygon
108 Composite Figures
111 Concave
16 Conditional Statements (Original, Converse, Inverse, Contrapositive)
Cones
121 Cones - Definitions
121 Cones - Surface Area and Volume
37 Congruent Triangles
16 Contrapositive of a Statement
16 Converse of a Statement
43, 111 Convex
22 Corresponding Angles
87 Cosecant Function
85 - 87 Cosine Function
87 Cotangent Function
38 CPCTC
111 Cross Section
113 Cube (Hexahedron)
Cylinders
116 Cylinders - Definitions
116 Cylinders - Surface Area and Volume
118 Decomposition
18 Deductive Reasoning
43 Diagonal
92 Diameter of a Circle
71 - 73 Dilation
Distance
8 Collinear Points
11 Distance Equations
8 Distance Formula in 1 Dimension
9 Distance Formula in 2 Dimensions
12 Distance Formula in “n” Dimensions
11 Partial Distances
113 Dodecahedron
111 Edge
44 Equiangular
44 Equilateral
35 Equilateral Triangle
 Geometry Handbook
Index

Page Subject

111, 112 Euler’s Theorem

44 - 45 Exterior Angle
94 Exterior Point of a Circle
111 Face
123 Great Circle
42 Height Length in a Triangle
123 Hemisphere
98 Heron's Formula - Area of a Triangle
37 Hypotenuse Leg Theorem (triangle congruence)
113 Icosahedron
40 Incenter
18 Inductive Reasoning
94 Inscribed Angle
94 Inscribed Polygon
44 - 45 Interior Angle
94 Interior Point of a Circle
16 Inverse of a Statement
35 Isosceles Triangle
55 Isometric Transformations
47, 54 Kites
114 Lateral Edge
114 Lateral Face
29 Legs of a Triangle
6, 7 Line
Logic
16 Contrapositive of a Statement
16 Converse of a Statement
16 Inverse of a Statement
92 Major Arc
55 Mapping
29, 42 Median - Length in a Triangle
54 Midsegment of a Trapezoid
92 Minor Arc
114 Oblique
113 Octahedron
40 Orthocenter
Parallel Lines
22, 23 Parallel Lines and Transversals
25 Parallel Lines in the Coordinate Plane
 Geometry Handbook
Index

Page Subject

Parallelogram
52 Parallelograms - Characteristics
53 Parallelograms - Proofs (Sufficient Conditions)
Perimeter
106 Perimeter - Arc Length of a Circle
102 Perimeter - Polygons
101 Perimeter - Quadrilaterals
98 Perimeter - Triangle
#REF! Perimeter Formulas - Summary for 2D Shapes
25 Perpendicular Lines in the Coordinate Plane
6 Plane
113 Platonic Solids
6 Points
Polygons
43, 44 Polygons - Definitions
71, 73 Polygons - Dilation
71, 73 Polygons - Dilations of Polygons
45 Polygons - Exterior Angles
45 Polygons - Interior Angles
43 Polygons - Names
44 Polygons - Number of Diagonals in a Polygon
102 Polygons - Perimeter and Area
70 Polygons - Scale Factor of Similar Polygons
69 Polygons - Similarity
Polyhedra
111 Polyhedra - Definitions
111, 112 Polyhedra - Euler's Theorem
111 Polyhedra - Number of Edges
55 Preimage
Prisms
114 Prisms - Definitions
114 Prisms - Surface Area and Volume
Proofs
24 Proofs - Parallel Lines
53 Proofs - Parallelograms
19 Proofs - Requirements
19 Proofs - Tips for Success
Properties
17 Properties of Addition and Multiplication
 Geometry Handbook
Index

Page Subject

17 Properties of Algebra
17 Properties of Equality and Congruence
27 Proportional Segments
28 Angle Bisector
27 Parallel Line in a Triangle
27 Three or More Parallel Lines
Pyramids
119 Pyramids - Definitions
119 Pyramids - Surface Area and Volume
80 Pythagorean Theorem
81 Pythagorean Triples
Quadrilaterals
47 Quadrilaterals - Characteristics
46 Quadrilaterals - Definitions
47 Quadrilaterals - Figures
101 Quadrilaterals - Perimeter and Area
92 Radius of a Circle
102 Radius of a Regular Polygon
68 Ratios - Dealing with Units
6, 7 Ray
18 Reasoning - Inductive vs. Deductive
47 Rectangle
55, 57 Reflection
47 Rhombus
35 Right Triangle
55, 59 Rotation
35 Scalene Triangle
87 Secant Function
92 Secant Line
92 Sector
106 Sector Area
6, 7 Segment
Segment, Proportional
28 Angle Bisector
27 Parallel Line in a Triangle
27 Three or More Parallel Lines
92 Semicircle
43 Side
Similarity
 Geometry Handbook
Index

Page Subject

69 - 73 Similar Polygons
74 - 78 Similar Triangles
125 Similarity - Solids
85 - 87 Sine Function
114 Slant Height
125 Solids - Similarity
Sphere
123 Spheres - Definitions
123 Spheres - Surface Area and Volume
47 Square
94 Subtend (Arc, Angle)
Surface Area
121 Surface Area - Cones
116 Surface Area - Cylinders
114 Surface Area - Prisms
119 Surface Area - Pyramids
123 Surface Area - Spheres
118 Surface Area - Using Decomposition
128 Surface Area Formulas - Summary for 3D Shapes
97 Tangent Facts
92 Tangent Line
85 - 87 Tangent Function
113 Tetrahedron
30 Third Angle Theorem
Transformation
55 Image
55 Preimage
66 Transformation - Composition
55 Transformation - Definitions
55 Transformation - Isometric
57 Transformation - Reflection
59 Transformation - Rotation
61 Transformation - Rotation by 90⁰ about a Point (x0, y0)
64 Transformation - Translation
55, 64 Translation
65 Translation Coordinate Form
54 Trapezoid
Triangles
40 Centers of Triangles
 Geometry Handbook
Index

Page Subject

29 Legs of a Triangle
30 Sum of Interior Angles
37 Triangle Congruence (SAS, SSS, ASA, AAS, HL, CPCTC)
31 Triangle Inequalities
74 Triangle Similarity (SSS, SAS, AA)
35 Triangles - General
98, 100 Triangles - Perimeter and Area
75 Triangles - Proportion Tables for Similar Triangles
83 Triangles - Special (45⁰-45⁰-90⁰ Triangle, 30⁰-60⁰-90⁰ Triangle)
78 Triangles - Three Similar Triangles
29 Vertices
29 What Makes a Triangle?
Trigonometric Functions
87 Cosecant Function
85 - 87 Cosine Function
87 Cotangent Function
87 Secant Function
85 - 87 Sine Function
85 - 87 Tangent Function
85 Trigonometric Functions - Definition
87 Trigonometric Functions - Graphs
85 Trigonometric Functions - Special Angles
86 Trigonometric Functions - Values in Quadrants II, III, and IV
129 Trigonometry Formulas - Summary
Vectors
90 Vectors - Definitions
90 Vectors - Direction
90 Vectors - Magnitude
91 Vectors - Operations
29, 43 Vertex
Volume
121 Volume - Cones
116 Volume - Cylinders
114 Volume - Prisms
119 Volume - Pyramids
123 Volume - Spheres
128 Volume Formulas - Summary for 3D Shapes