MATH 15:

PLANE AND SOLID GEOMETRY

Prepared by: ALVIN C. CONSARVA, DEXTER JOHN SUPERABLE, and JAYSON
KITTS C. POQUITA

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 Math 15 – Plane and Solid Geometry
Course Description: This course covers the topic of Euclidean Geometry. The topics are
discussed using both the deductive and inductive methods to conjecture definitions,
corollaries, postulates and theorems on plane and solid geometry.

Units / Credit Equivalent: 3 units

Course Outcomes:
Upon completion of the course, the students should be able to:
1. Trace and exhibit understanding on the different axioms, theorems, postulates
and principles embodied in plane and solid geometry.
2. Use different logical methods in solving problems involving plane and solid
geometry.
3. Produce a compilation of the problem sets and examinations (Major Exam,
Quizzes, and etc.) given in the course.
4. Conduct a presentation (exhibit) of the different development and facts in
plane and solid geometry
Learning Outcomes:
a. identify and define the properties of equality and congruence;
b. determine the relationships of line, line segments, and rays;
c. define the relations involving angles and perpendicular lines;
d. define and prove an angle formed by parallel lines cut by a transversal;
e. prove statements involving angles of a triangle using theorems;
f. solving the total number of diagonals of a convex polygon using theorems;
g. find the sum of the measures of interior and exterior angles;
h. provide formal proofs of congruency of triangles and right congruent triangles;
i. provide formal proofs of problems involving theorems of parallelogram,
rhombus, square rectangle and trapezoid;
j. use theorems and postulates in solving problems concerning circles;
k. prove theorems concerning spheres and circles;
l. solve problems involving areas and volumes of basic geometric polygons ; and
m. do geometric construction with the use of any straight edge;

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Coursepack Structure
Module Intended Learning Lessons Writer
Outcomes
1  define and identify the Lesson 1: The Modern Dexter John P.
identified terms; Axiomatic System Superable
 identify and define the
Lesson 2: Review of
properties of equality and
Algebraic System
congruence;
 determine the relationships Lesson 3: Segment and
of line, line segments, and Rays
rays;
Lesson 4: Angles and
 define the relations
Perpendicular Lines
involving angles and
perpendicular lines; and Lesson 5: Parallel
 define and prove an angle Lines cut by a
formed by parallel lines Transversal
cut by a transversal.

2  classify the different kinds Lesson 1: Introduction Jayson Kitts C. Poquita

of Polygons; to Polygons
 solving the total number of
Lesson 2: Angles of a
diagonals of a convex
Triangle
polygon using theorems;
 find the unknown parts of Lesson 3: Angles of a
a given triangle using Polygon
theorem;
 prove statements involving
angles of a triangle using
theorems;
 find the sum of the
measures of interior and
exterior angles; and
 solve for the number of
sides of a polygon

3  identify the corresponding Lesson 1: Introduction Jayson Kitts C. Poquita

angles and sides of the to Triangles and
triangle to another triangle; Congruent Triangles
 name properly the two
Lesson 2: Proving
congruent triangles;
Congruent Triangles
 solve the unknown angles
or sides of the congruent Lesson 3: Triangle
triangles using the concept Congruence Theorems
of congruent triangles;

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  provide formal proofs of
congruency of triangles;
 state the postulate and
theorems right triangles
congruence; and
 provide formal proofs of
right congruent triangles.
4  solve the unknown angles Lesson 1: Properties of Jayson Kitts C. Poquita
or sides of a parallelogram, Parallelogram
rhombus, square rectangle
Lesson 2: Special
and trapezoid by applying
Parallelograms
its theorems; and
 provide formal proofs of Lesson 3: Trapezoid
problems involving
theorems of parallelogram,
rhombus, square rectangle
and trapezoid.

5  differentiate a circle and Lesson 1 – Alvin C. Consarva

the lines having a Introduction to Circles
relationship to a circle, Lesson 2 – Arcs of a
 list down the theorems circle
about circles, Lesson 3 – Inscribed
 prove theorems angles and intercepted
concerning spheres and arcs of Circles
circles, and Lesson 4 – Tangent
 use theorems and lines and inscribed
postulates in solving polygons
problems concerning
circles
6  compare and contrast the Lesson 1 – Perimeter Alvin C. Consarva
different types of solids Lesson 2 – Area of
 solve problems involving polygons
areas and volumes of Lesson 3 –
basic geometric polygons circumference and
 do geometric construction area of circles
with the use of any Lesson 4 – Definition
straight edge of Basic Solids
 prove theorems Lesson 5 – Surface
Area of Solids
Lesson 6 – Volumes

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 TABLE OF CONTENTS
Module 1: Introduction to Proving
Module Overview 1
Lesson 1: The Modern Axiomatic System 3
Lesson 2: Review of Algebra Concepts 7
Lesson 3: Segments and Rays 10
Lesson 4: Angles and Perpendicular Lines 14
Lesson 5: Parallel Lines Cut by a Transversal 20
Module Assessment 27

Module 2: Polygons
Module Overview 29
Lesson 1: Introduction to Polygons 30
Lesson 2: Angles of a Triangles 35
Lesson 3: Angles of a Polygon 41
Module Assessment 45

Module 3: Triangles and Congruent Triangles

Module Overview 47
Lesson 1: Introduction to triangles and Triangle Congruence 48
Lesson 2: Proving Congruent Triangles 52
Lesson 3: Triangle Congruence Theorems 57
Module Assessment 61

Module 4: Quadrilaterals
Module Overview 62
Lesson 1: Properties of Parallelogram 63
Lesson 2: Proving Congruent Triangles 67
Lesson 3: Triangle Congruence Theorems 72
Module Assessment 75

Module 5: Circles
Module Overview 77
Lesson 1: Introduction to Circles 78
Lesson 2: Arcs of a circle 81
Lesson 3: Inscribed angles and intercepted arcs of Circles 87
Lesson 4: Tangent lines and inscribed polygons 92
Module Assessment 103

Module 6: Perimeter, Area, and Solids

Module Overview 104
Lesson 1: Perimeter 105
Lesson 2: Area of polygons 109
Lesson 3: circumference and area of circles 119
Lesson 4: Definition of Basic Solids 125
Lesson 5: Surface Area of Solids 134
Lesson 6: Volumes 145
Module Assessment 154

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 MODULE 1
Introduction to Proving

Introduction
Nature has provided us with rich resources of shapes and sizes. Geometric shapes like
squares, rectangles, circles, etc. are found in buildings, roads and bridges, vehicles, plants,
and animals around us.
To appreciate the beauty of nature, we need basic knowledge and concepts of
geometry. In geometry, we begin with the undefined terms – points, lines, and planes, their
relationships, properties, and principle involved. These terms form the basis for defining
other geometric terms.
The development of critical thinking and reasoning is one of the main objectives in
learning geometry. To answer the “whys” will need clear understanding of postulates,
axioms, definitions, and theorems. To differentiate these, we need to understand the
mathematical reasoning called axiomatic system.
In an axiomatic system, the proof of a certain result begins from a statement that is
known to be true, followed by a series of statements each of which must follow logically
from previous ones. This type of mathematical reasoning consists of undefined terms,
definitions, axioms, and postulates, theorems, and propositions.
In an axiomatic system, the definitions make use of undefined terms, undefined terms
also make use of previously defined terms, the postulates (and axioms) are stated in terms of
the undefined terms and definitions, and each theorems that is obtained logically from the
definition, axioms, and other theorems previously proven.

At the completion of this module, you should be able to:

 define and identify the identified terms;
 identify and define the properties of equality and congruence;
 determine the relationships of line, line segments, and rays;
 define the relations involving angles and perpendicular lines; and
 define and prove an angle formed by parallel lines cut by a transversal.

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Lessons in this Module:
Lesson 1: The Modern Axiomatic System
Lesson 2: Review of Algebraic System
Lesson 3: Segment and Rays
Lesson 4: Angles and Perpendicular Lines
Lesson 5: Parallel Lines cut by a Transversal

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Lesson 1: The modern axiomatic system
Objectives: At the end of this lesson, you should be able to:

 define and identify the undefined terms; and

 identify objects involving undefined terms.
Introduction
Hello! Welcome to Lesson 1. To function effectively in the three-dimensional world,
one should have knowledge of geometric concepts of points, lines, and planes, their
relationships, and their basic properties. Knowing these would lead to a better understanding
of plane surfaces and their areas, solid and their volumes.
Our environment is rich with patterns of different shapes and sizes. These motivate
and facilitate the learning of geometry because of their appeal to our perception of beauty
and order. Recognition and analysis of their properties and measures not only develop one’s
knowledge in geometry, but also open the path for logical thinking. If you are ready, starts
the lesson now!

Activity

Tell whether each of the following suggests a point, a line, or a plane.

1. top of a box
2. a corner of a room 4. cover of a book
3. star in the sky 5. tip of a pen
6. a taut clothesline

Analysis

 How did you define a point, a line, and a plane?

Answer:

 What are the properties of point, line, and plane?

Answer:

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 Abstraction
The word geometry is derived from two Greek words: geo, meaning “earth”, and
metrein, meaning “to measure”. Thus, when taken literally, geometry involves measuring
earthly objects.
Consider the words points, lines, and planes as undefined terms. These terms are said
to be the building blocks of geometry.
Although a point has no dimension, it can be represented by a dot, and is denoted by a
capital letter, like points A, B, C, D, and E below.

B
D
A E
C

A straight path or a taut wire suggests the idea of a line. Although a line has no width and no
thickness, it can be extended infinitely in opposite directions. A line consists of infinitely
many points and between any two points on a line, there exists another point. We can name a
line by naming any two different points on the line. In the figure below, ⃡𝐴𝐵 or line AB,
identifies the line; ⃡𝐴 𝐶 𝑎𝑛𝑑 ⃡𝐵 𝐶 also identify the same line. Sometimes, lines are denoted by
m, n, or ℓ (possibly with subscripts) for simplicity.

A B C

Arrowheads are used to indicate that the line does not stop at the points where the drawing
stops.

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 A flat surface that extends infinitely
in all directions suggests the idea of a
plane. Given three points not on the same P Q
line, exactly one plane passes through
them. A plane may be denoted by a script R
Ắ Plane Ắ
letter such as Ắ in the accompanying
figure.
Now that we have described the undefined
terms, we need postulates which serve as
guiding rules or assumption from which other statements on the defined terms may be
derived.

Postulates

Two points are contained in exactly one line.

Every line contains at least two distinct points.
If two points are on a plane, then the line containing these points is also on the plane.
Every plane contains at least three nonlinear points.
(Plane Postulate) Any three points lie on at least one plane and any three nonlinear points lie in exactly one pla
If two distinct planes intersect, then their intersection is a line.

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Undefined terms summary
Undefined Dimension Representation Naming Other Information
Term
Point Dot P Has no length,
0 Read: Point P width or thickness;
occupies no space
Line Straight mark with Two capital letters Has infinite length
two arrow heads with a double but no width and
1 arrowhead above thickness; is
them or a lower straight
case letter
Plane Slanted four-sided Single capital Has infinite length
2 figure letter and width but no
thickness; is a flat
M surface

Read: Plane M

For further reference about the lesson, please watch these videos: https://www.youtube.com/watch?
v=k5etrWdIY6o https://www.youtube.com/watch?v=anRWuIq06RM

Application
Illustrate each of the following and label the diagram.
1. Point B lies in plane ℳ.
2. Lines l and m intersect at point E.
3. Plane ℱ contains line CD.
4. Plane ℰ and plane ℬ intersect at PR.

Congratulations! You have completed lesson 1 of module 1. You are now ready for Lesson 2,
which tackles about the review of algebraic Concept.

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Lesson 2: Review of Algebraic Concepts
Objectives: At the end of this lesson, you should be able to:

 identify and define the properties of equality;

 identify and define the properties of congruence; and
 solve the equation and state the reason.
Introduction
Hello! Welcome to Lesson 2. In this lesson you will learn to use the properties of
equality and congruence along with our previous learned definitions and postulate to draw
conclusions. This lesson starts by showing you how to generate reasons for solving algebraic
equations and then progress to supplying missing statements or reasons for equality and
congruence involving geometric figures like segments and angles. If you are ready, starts the
lesson now!

Activity
In this equation 4𝐴𝐵 = 16 + 2𝐴𝐵, show that 𝐴𝐵 = 8.

Analysis
 What are the steps in proving and solving the equation?
Answer:

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 Abstraction
The following are the properties of equality and congruence for real numbers. These are the
logical rules which allow you to balance, manipulate, and solve equations.

Algebraic Properties of Equality

Reflexive Property For all real numbers a,
a = a.
A number equals itself.
Symmetric property For all real numbers a and
b,
if a = b, then b = a. These three properties
Order of equality does not define an equivalence
matter. relation
Transitive property For all real numbers a, b,
and c,
if a = b and b = c, then a =
c.
Two numbers equal to the
same number are equal to
each other.
Addition property For all real numbers a, b,
and c, if a = b, then a + c =
b + c.
Subtraction Property For all real numbers a, b,
and c, if a = b, then a – c =
b – c. These properties allow you
Multiplication Property For all real numbers a, b, to balance and solve
and c, if a = b, then ac = bc. equations involving real
Division property For all real numbers a, b, numbers
and c, if a = b, and 𝑐 ≠ 0,
𝑏
then 𝑎 = .
𝑐 𝑐
Substitution Property For all real numbers a and
b, if a = b, then b can be
substituted for a in any
expression.
Distributive Property For all real numbers a, b, For more, see the section on
and c, the distributive property
a(b + c) = ab + ac

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 Properties of Congruence
Reflexive Property ̅𝐴̅̅𝐵̅ ≅ ̅𝐴̅̅𝐵̅ ∠𝐴 ≅ ∠𝐴
Symmetric Property If ̅𝐴̅̅𝐵̅ ≅ ̅𝐶̅̅𝐷̅, then ̅𝐶̅̅𝐷̅ ≅ ̅𝐴̅̅𝐵̅ .
If ∠𝐴 ≅ ∠𝐵, then ∠𝐵 ≅ ∠𝐴.
Transitive Property If ̅𝐴̅̅𝐵̅ ≅ ̅𝐶̅̅𝐷̅ 𝑎𝑛𝑑 ̅𝐶̅̅𝐷̅ ≅ 𝐸̅̅̅𝐹̅ , then ̅𝐴̅̅𝐵̅ ≅ ̅𝐸̅̅𝐹̅̅.
If ∠𝐴 ≅ ∠𝐵 𝑎𝑛𝑑 ∠𝐵 ≅ ∠𝐶 , then ∠𝐴 ≅ ∠𝐶.

Example:
In this equation 6𝐵𝐶 − 16 = 2𝐵𝐶 , prove that 𝐵𝐶 = 4.

Statement Reason
6𝐵𝐶 − 16 = 2𝐵𝐶 Given
4𝐵𝐶 = 16 Addition Property of Equality
𝐵𝐶 = 4 Division Property of Equality

For further reference about the lesson, please watch this video:
https://www.youtube.com/watch?v=OO_nIBmicKM

Application
A. Solve the following equation and state the reason for each step.
1. If 5(2̅𝐵̅̅𝐶̅ − 1) = 9̅𝐵̅̅𝐶̅ + 4, then show that ̅𝐵̅̅𝐶̅ = 9.
2. If 𝑎 = 2 𝑎𝑛𝑑 2𝑎 + 3𝑏 = 21, then show that 𝑏 = 3.
𝑏

Congratulations! You have completed lesson 2 of module 1. You are now ready for Lesson 3,
which tackles about segment and rays.

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Lesson 3: Segment and Rays
Objectives: At the end of this lesson, you should be able to:

 determine the relationships of line, line segments, and rays; and

 identify and use the segment addition postulate in solving equations.
Introduction
Hello! Welcome to Lesson 3. Relationships between and among line segments and
angles are basic concepts that must be identified and understood in the study of Geometry. If
you are ready, starts the lesson now!

Activity

Draw and label ̅𝐽̅𝐾̅ .

Analysis

 Is there another way to name ̅𝐽̅𝐾̅? Explain.

Answer:

 Explain how lines, line segments, and rays are related.

Answer:

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 Abstraction
Although we all know intuitively what a line is, it is actually difficult to give a good
mathematical definition. Roughly, we say that a line is an infinitely thin, infinitely long
collection of points extending in two opposite directions. When we draw lines in geometry,
we use an arrow at each end to show that it extends infinitely.

m
B

A line can be named either using two points on the line. For example, ⃡𝐴𝐵 or simply by
a letter usually lowercase. For example, represented by letter m.
A line segment has two endpoints. It contains these endpoints and all the points of the line
between them. You can measure the length of a segment, but not of a line.

A segment is named by its two endpoints, for example, ̅𝐴̅̅𝐵̅.

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A ray is a part of a line that has no endpoint and goes infinitely in only one direction. You
cannot measure the length of a ray.

A ray is named using its endpoints first, and then any other point on the ray. For example,
𝐵𝐴 .

Segment Addition Postulate

Consider the number line below.

A M B
-2 3 10

The points A, M, and B are called collinear points because they lie on one line. Note
also that point M is between points A and B.
Every point on a real number line corresponds to exactly one real number, and every
real number corresponds exactly one point on the real number line. In the figure above, point
A corresponds to -2. The number -2 is called the coordinate of point A.

The length of 𝐴𝐵, denoted by AB or BA, is the distance between points A and B. The
distance between two points is the absolute value of the difference of their coordinates. In
the figure above, we have

𝐴𝑀 = |−2 − 3| = |−5| = 5
𝑀𝐵 = |3 − 10| = |−7| = 7
𝐴𝐵 = |−2 − 10| = |−12| = 12.
Observe that 𝐴𝑀 + 𝑀𝐵 = 𝐴𝐵.
Postulate
Segment Addition Postulate
If point M is between points A and B, then 𝐴𝑀 + 𝑀𝐵 = 𝐴𝐵

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For further reference about the lesson, please watch this video:
https://www.youtube.com/watch?v=JcqCf762y9w

Application
Answer the following:
1. Points A, B, and C are collinear. If 𝐴𝐵 = 7 𝑐𝑚 𝑎𝑛𝑑 𝐵𝐶 = 13 𝑐𝑚, what is 𝐴𝐶?
2. On a number line, point 𝑌 is between points 𝑋 𝑎𝑛𝑑 𝑍. If 𝑋𝑍 = 18, the coordinate of
𝑌 𝑖𝑠 3 and the coordinate of 𝑍 is 7, what is the coordinate of 𝑋?

Congratulations! You have completed lesson 3 of module 1. You are now ready for Lesson 4,
which tackles about angles and perpendicular lines.

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Lesson 4: Angles and Perpendicular Lines
Objectives: At the end of this lesson, you will be able to:

 define the relations involving angles and perpendicular lines; and

 define and prove an angle formed by perpendicular lines.
Introduction

Hello! Welcome to Lesson 4. When draw a perpendicular line from a point to a line,
you can guarantee that it is perpendicular by using the construction of a line perpendicular to
a line through a point not on that line. In this lesson you will learn the importance of
perpendicular line in relation to angles. If you are ready, starts the lesson now!

Activity
Construct!
To construct a line perpendicular to the given line l and passing through the point P on l:
1. Place the needle point of the compass at P, and draw two arcs that will intersect l
at two points. Let these points be X and Y.
2. Putting the needle point of the compass at X, adjust it so that its opening is greater
than XP, and draw an arc on one half-plane created by l.
3. Using the same compass setting, place the needle point of the compass at Y, and draw
an arc that intersects the arc created in Step 2. Let this point of intersection be Q.
4. Draw ⃡𝑃 𝑄 .

Analysis

 What are the angles formed by this

construction? Answer:

 What is the relationship of ⃡𝑃 𝑄 to l at P?

Answer:

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 Abstraction
Angle Addition Postulate
A measurement relationship between the lengths of segment called Segment Addition
Postulate. It was introduced in the previous lesson. A similar relationship exists between the
measure of angles.

Postulate
Angle Addition Postulate
If point B lies in the interior of ∠𝐴𝑂𝐶, then 𝑚∠𝐴𝑂𝐵 + 𝑚∠𝐵𝑂𝐶 = ∠𝐴𝑂𝐶

Example:
A. Point U lies in the interior of ∠𝑋𝑌𝑍. If ∠𝑋𝑌𝑈 = 30 and ∠𝑈𝑌𝑍 = 40, what is ∠𝑋𝑌𝑍?
Solution:
By the Angle Addition Postulate, we have ∠𝑋𝑌𝑍 = ∠𝑋𝑌𝑈 + ∠𝑈𝑌𝑍
= 30 + 40
Therefore, ∠𝑋𝑌𝑍 = 70

B. Point B lies in the interior of ∠𝐴𝑂𝐶. Find 𝑚∠𝐴𝑂𝐵 if 𝑚∠𝐴𝑂𝐶 = 120 𝑎𝑛𝑑 𝑚∠𝐵𝑂𝐶 = 75.
Solution:
By the Angle Addition Postulate, we have 𝑚∠𝐴𝑂𝐶 = 𝑚∠𝐴𝑂𝐵 + 𝑚∠𝐵𝑂𝐶.
Thus, we have 𝑚∠𝐴𝑂𝐵 = 𝑚∠𝐴𝑂𝐶 − 𝑚∠𝐵𝑂𝐶 = 120 − 75 = 45.

Pairs of Angles
There are different ways in which pairs of angles may be classified. A pair of angles
may be related on the basis of the sum of their angle measures or of their positions.
Based on the sum of angle measures, angle pairs can be classified as either
supplementary of complementary.

Remember

Supplementary angles are two angles whose measures have a sum of 180°.
Complementary angles are two angles whose measures have a sum of 90°.

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Illustrative examples
A. ∠𝐴𝑂𝐷 𝑎𝑛𝑑 ∠𝐵𝑂𝐷 are supplementary angles.
D

120° 60°
A O B

120° + 60° = 180°

B. ∠𝑀𝑂𝑅 𝑎𝑛𝑑 ∠𝑁𝑂𝑅 are complementary angles.

M R

60° N
30°
O
60° + 30° = 90°

Remember

Adjacent angles are angles which have a common side and a common vertex but no interior points in comm

P S Q

In the figure above, ∠𝑅𝑆𝑃 𝑎𝑛𝑑 ∠𝑅𝑆𝑄 from a linear pair because they are adjacent angles
and 𝑆 𝑃 𝑎𝑛𝑑 𝑆 𝑄 are opposite rays.

Postulate
Linear Pair Postulate
If two angles form a linear pair, then they ate supplementary.

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 A B

O
D C

In the figure above, ∠𝐴𝑂𝐵 𝑎𝑛𝑑 ∠𝐷𝑂𝐶 are vertical angles, so are ∠𝐴𝑂𝐷 𝑎𝑛𝑑 ∠𝐵𝑂𝐶. These pairs of
angles are nonadjacent.

Remember
Vertical angles are two nonadjacent angles formed by two intersecting lines

Theorem
Vertical Angle Theorem
Vertical angles are congruent

Illustrative Example A B
1 2
In the given figure, ∠𝐴𝑂𝐵 𝑎𝑛𝑑 ∠𝐶𝑂𝐷 are 4
3
right angles. O
6
Name two pairs of: 5
7 8
1. Complementary angles
C D
2. Supplementary angles
3. Vertical angles

Complementary angles: ∠3 𝑎𝑛𝑑 ∠4, ∠5 𝑎𝑛𝑑 ∠6

Supplementary angles: ∠1 𝑎𝑛𝑑 ∠2, ∠7 𝑎𝑛𝑑 ∠8
Vertical angles: ∠3 𝑎𝑛𝑑 ∠6, ∠4 𝑎𝑛𝑑 ∠5
Can you name other related angles with O as their vertex angle?

[22]
Perpendicular lines
Perpendicular lines are lines, segments or rays that intersect to form right angles. The symbol
⊥ means is perpendicular to.

S
R

P Q
In this figure, PR ⊥ QS
The right angle symbol in the figure indicates that the lines are perpendicular. In three
dimensions, you can have three lines which are mutually perpendicular.
Example:

D
E

A 35° B
O

If AB ⊥ CD, find the measure of angle EOD or m∠𝐸𝑂𝐷.

Solution:
Since AB ⊥ CD, m∠𝐵𝑂𝐷 = 90°.
By the Angle Addition Postulate, m∠𝐵𝑂𝐸 + 𝑚∠𝐸𝑂𝐷 = 𝑚∠𝐵𝑂𝐷.
Let x° be the measure of m∠𝐸𝑂𝐷.
Then,
36° + x° = 90°
Subtract 35° from each side, we get x° = 55°
Therefore, m∠𝐸𝑂𝐷 = 55°.

[23]
Practice Problem
In the accompanying figure, ⃡𝐴𝐵 ⊥ ⃡𝐹𝐶 .

B
C
A
1. Find 𝑚∠𝐴𝑂𝐹.
2. Find 𝑚∠𝐴𝑂𝐶. 123 D
O4
3. Find 𝑚∠1 + 𝑚∠2.
4. If 𝑚∠1 = 50, 𝑤ℎ𝑎𝑡 𝑖𝑠 𝑚∠2?
5. If 𝑚∠2 + 𝑚∠3 = 90, ℎ𝑜𝑤 𝑎𝑟𝑒 𝑂 𝐵 𝑎𝑛𝑑 𝑂 𝐷 𝑟𝑒𝑙𝑎𝑡𝑒𝑑? E
6. If 𝑚∠𝐴𝑂𝐷 = 130, 𝑤ℎ𝑎𝑡 𝑖𝑠 𝑚∠4? F

For further reference about the lesson, please watch these videos:
https://www.youtube.com/watch?v=92aLiyeQj0w https://www.youtube.com/watch?
v=_7aUxFzTG5w

Application
A. Form the given figure, form an equation in x and solve for the unknown measures.

2x° x°

B. In the figure, 𝑂 𝑃 ⊥ ⃡𝐴 𝐵 and 𝑚∠𝐵𝑂𝑄 = 30. What is the measure of ∠𝑃𝑂𝑄? ∠𝐴𝑂𝑄?

P
Q

A O B

Congratulations! You have completed lesson 4 of module 1. You are now ready for Lesson 5,
which tackles about Parallel lines.

[24]
Lesson 5: Parallel lines cut by a Transversal
Objectives: At the end of the lesson, you should be able to:

 define the parallel lines;

 define the parallel lines cut by a transversal; and
 define and prove an angle formed by parallel lines cut by a transversal.

Introduction
Hello! Welcome to Lesson 5. Two parallel lines cut by a transversal also form angle
pairs called corresponding angles. If two parallel lines are cut by a transversal, then the pairs
of corresponding angles have the same measure. In this lesson we will go deeper in the
discussion of parallel lines cut by a transversal. If you are ready, starts the lesson now.

Activity

Draw a two horizontal lines. Draw a slant line t cutting lines m and l. Label the angles formed indicated by 1,
With a protractor, measure the corresponding angles 1 and 5, 2 and 6, 3 and 7, 4 and
8.

Analysis

 Are the measures of each pair of corresponding angles approximately the

same? Why?
Answer:

 Measure the angles 4 and 6, 3 and 5. Are the measures approximately the
same? Why?
Answer:

[25]
 Abstraction

G H

X
A B

E F
Y

C D

The figure above is a rectangular solid. 𝐴𝐵 𝑎𝑛𝑑 𝐶𝐷 are coplanar, meaning they are in
the same plane. These segments are contained in one plane and will not meet no matter
how far they are extended. 𝐴𝐵 𝑎𝑛𝑑 𝐶𝐷 are said to be parallel. 𝐴𝐵 𝑎𝑛𝑑 𝐺𝐻 are also parallel. Can
you name other parallel segments in the figure?

𝐴𝐶 𝑎𝑛𝑑 𝐸𝐹 are not coplanar and will not meet no matter how far they are extended.
The lines containing these segments are called skew lines. Can you name other skew line
segments in the figure?
If the side of the box in the diagram is considered as part of plane X, and its opposite
side as part of plane Y, then plane X is parallel to plane Y, or X ∥ Y. Also, 𝐶⃡ 𝐷 ∥ 𝑋 𝑎𝑛𝑑 ⃡𝐸𝐹
∥
𝑌.

Remember

Two lines are parallel if and only if, they are coplanar and do not intersect. A line and a plane are para

[26]
In the figure at the right are three coplanar
lines, 𝐴⃡ 𝐵 , 𝐶⃡ 𝐷 , 𝐸⃡ 𝐹 . Notice that 𝐸⃡ 𝐹
E
intersects
⃡𝐴 𝐵 and 𝐶⃡ 𝐷 at two different points and
12
forms angles. ⃡𝐸 𝐹 is a transversal of ⃡𝐴 𝐵 and 4 3 B
A
𝐶⃡ 𝐷 .

56
C 87 D

Remember

A line that intersects two or more lines at different points is called transversal.

The angles formed by the transversal with the two other lines can be identified as follows:
Angles 1, 2, 7, 8 are exterior angles while 3, 4, 5, 6 are interior angles.
The pairs of angles, 1 and 5, 2 and 6, 3 and 7, 4 and 8 are pairs of corresponding angles.
The pairs of angles, 3 and 5, 4 and 6 are pairs of alternate interior angles, while the pairs of
angles, 1 and 7, 2 and 8 are pairs alternate exterior angles.

We will now describe the pairs of angles formed when two parallel lines are cut by a
transversal, but first, we will state our basic assumptions in the following postulates.

Postulate
Parallel Postulate
Given a line and a point on the line, there is exactly one line through the point parallel to the given line.

Given a line l and a point P not on l, the C

Parallel Postulate assumes that there is l’
exactly one line, say line l’, containing the
point P and Parallel to l. l

[27]
Postulate

[28]
 If two parallel lines are cut by a transversal, then the corresponding angles are congruent.

The accompanying figure shows that

𝑙1 𝑎𝑛𝑑 𝑙2 are parallel lines cut by a 𝑙1 𝑙2
transversal t. according to the previous
postulate, ∠1 ≅ ∠5, ∠2 ≅ ∠6, ∠4 ≅
23 67 t
∠8, 𝑎𝑛𝑑 ∠3 ≅ ∠7. 14 58

Theorem

If two parallel lines are cut by a transversal, then alternate interior angles are congruent.

Given: 𝑎 ∥ 𝑏 l
Transversal l cuts lines a and b 1
a
2
Prove: ∠2 ≅ ∠3

b 3
4
Proof:
Statement Reason
1. 𝑎 ∥ 𝑏 1. Given
2. ∠1 ≅ ∠2 2. Vertical Angle Theorem
3. ∠1 ≅ ∠3 3. If two parallel lines are cut by a
transversal, then
the corresponding angles are congruent
4. ∠2 ≅ ∠3 4. Transitive Property

Theorem
If two parallel lines are cut by transversal, then alternate exterior angles are congruent.

(Refer to the above figure.)

Given: 𝑎 ∥ 𝑏, transversal l cuts lines a and b
Prove: ∠1 ≅ ∠4

[29]
 Proof:
Statement Reason
1. : 𝑎 ∥ 𝑏 1. Given
2. ∠1 ≅ ∠2 2. Vertical Angle Theorem
3. ∠2 ≅ ∠4 3. If two parallel lines are cut by a
transversal, then the corresponding angles are
congruent.
4. ∠1 ≅ ∠4 4. Transitivity Property

Theorem
If two parallel lines are cut by transversal, then alternate interior angles on the same side of the transversal are

Given: 𝑎 ∥ 𝑏, transversal l cuts lines a and l

b
1
a
Prove: ∠2 𝑎𝑛𝑑 ∠3 are supplementary. 2

b 3
4

Proof:
Statement Reason
1. 𝑎 ∥ 𝑏 1. Given
2. ∠1 𝑎𝑛𝑑 ∠2 are supplementary. 2. Linear Pair Postulate
3. 𝑚∠1 + 𝑚∠2 = 180 3. Definition of supplementary angles
4. ∠1 ≅ ∠3 4. If two parallel lines are cut by a
transversal, then the corresponding angles are
congruent.

5. 𝑚∠1 = 𝑚∠3 5. Definition of congruent angles

6. 𝑚∠2 + 𝑚∠3 = 180 6. Substitution
7. ∠2 𝑎𝑛𝑑 ∠3 are supplementary. 7. Definition of supplementary angles

Theorem
If two parallel lines are cut by transversal, then alternate exterior angles on the same side of the transversal are s

(Refer to the above figure)

[30]
Given: 𝑎 ∥ 𝑏, transversal l cuts lines a and b
Prove: ∠1 𝑎𝑛𝑑 ∠4 are supplementary.
Proof:
Statement Reason
1. 𝑎 ∥ 𝑏 1. Given
2. ∠1 𝑎𝑛𝑑 ∠2 are supplementary. 2. Linear Pair Postulate
3. 𝑚∠1 + 𝑚∠2 = 180 3. Definition of supplementary angles
4. ∠2 ≅ ∠4 4. If two parallel lines are cut by a
transversal, then the corresponding angles are
congruent.
5. 𝑚∠2 = 𝑚∠4 5. Definition of congruent angles
6. 𝑚∠1 + 𝑚∠4 = 180 6. Substitution
7. ∠1 𝑎𝑛𝑑 ∠4 are supplementary. 7. Definition of supplementary angles

Exercises
The figure below shows 𝑚 ∥ 𝑛 𝑤𝑖𝑡ℎ 𝑙 as transversal. Name the indicated angles.
1. 4 pairs of corresponding angles
2. 2 pairs of alternate interior m
l
angles
3. 2 pairs of alternate exterior 58 7
6
angles
n
4. 2 pairs of interior angles on the
same side of the transversal 4
1
5. 2 pairs of exterior angles on the 2 3
same side of the transversal

For further reference about the lesson, please watch these videos: https://www.youtube.com/watch?
v=0eYzPVgylO8 https://www.youtube.com/watch?v=3TIgTBwmkBk
https://www.youtube.com/watch?v=dUTcNlLub7Q

[31]
 Application

Study the figure and answer the following, given that ̅𝐴̅̅𝐵̅ ∥ ̅𝐶̅̅𝐷̅ and ̅𝐴̅̅𝐷̅ ∥ ̅𝐵̅̅𝐶̅ .
1. Is ∠𝐴𝐵𝐶 ≅ ∠𝐷𝐶𝐵? Why?
2. Is ∠𝐴𝐷𝐶 ≅ ∠𝐷𝐶𝐵? Why? A
D
3. Is ∠𝐴𝐵𝐶 ≅ ∠𝐴𝐷𝐶? Why?
4. What is the relation between
∠𝐵𝐴𝐷 𝑎𝑛𝑑 ∠𝐵𝐶𝐷?

C
B

Congratulations! You have completed all 5 Lessons in module 1. You are now ready for
Module 2, which tackles about the polygons.

[32]
 Module Assessment

A. Name the property illustrated below.

1. If 1 𝑚∠𝐷 = 45, then 𝑚∠𝐷 = 90.
2
2. If 𝑏 = 12 − 𝑎 𝑎𝑛𝑑 2𝑎 + 3𝑏 = 10, then 2𝑎 + 3(12 − 𝑎) = 10.
3. If 𝑚∠1 = 40 𝑎𝑛𝑑 𝑚∠2 = 𝑚∠1 = 50, then 𝑚∠2 = 90.
4. If 𝑚∠1 + 𝑚∠2 ≅ 𝑚∠4 + 𝑚∠2, then 𝑚∠1 ≅ 𝑚∠4.
5. If 1 ̅𝐴̅̅𝐵̅ = 6 𝑎𝑛𝑑 ̅𝐶̅̅𝐸̅ = 10, then ̅𝐴̅̅𝐵̅ + ̅𝐶̅̅𝐸̅ = 28.
3
B. Using the number line, find:

6. AB 9. AD
7. BC 10. BE
8. DE 11. CF
C. Using the same figure above, determine which of the following relationship is
true. 12. 𝐶𝐷 < 𝐴𝐶
13. 𝐴𝐵 > 𝐶𝐷
14. 𝐵𝐶 > 𝐶𝐷
15. 𝐴𝐵 = 𝐵𝐶
16. 𝐴𝐵 + 𝐵𝐶 = 𝐴𝐶
17. 𝐵𝐶 + 𝐶𝐷 = 𝐴𝐷

[33]
D. The figure below shows 𝑚 ∥ 𝑛 𝑤𝑖𝑡ℎ 𝑙 as transversal. Name the indicated angles.

18. 4 pairs of corresponding

angles m
19. 2 pairs of alternate interior
l
angles
20. 2 pairs of alternate exterior 58 7
6
angles
n
21. 2 pairs of interior angles on
the same side of the 4
1
transversal 2 3
22. 2 pairs of exterior angles on
the same side of the
transversal

References:
Dilao Ed.D S.E., Bernabe, J., Geometry III, SD Publications, Revised Edition.
https://www.radford.edu/~wacase/Geometry%20Notes%201.pdf
https://catalogimages.wiley.com/images/db/pdf/0471799408.excerpt.pdf

Page | 28
Module 2: Polygons
Module Overview: Hello! Welcome to Module 2. In this module, you will be able to explore the geometric fig

Module Objectives/Outcomes:
 classify the different kinds of Polygons;
 solving the total number of diagonals of a convex polygon using theorems;
 find the unknown parts of a given triangle using theorem;
 prove statements involving angles of a triangle using theorems;
 find the sum of the measures of interior and exterior angles; and
 solve for the number of sides of a
polygon Lesson/s in the Module:
Lesson 1: Introduction to Polygons
Lesson 2: Angles of a Triangle
Lesson 3: Angles of a Polygon

Page | 29
Module 2: Polygons
Lesson 1: Introduction to Polygons
Learning Outcomes
At the end of the lesson, you will be able to:
define polygons;
classify the different kinds of Polygons; and
solving the total number of diagonals of a convex polygon using theorems.

Duration: 1 hr.

Introduction:
Hello! Welcome to lesson 1. As we walk, we see polygons around us, either on the roads,
houses, buildings or even in nature. In this lesson, we will wonder about the world of
polygons and solving its total number of diagonals using theorems and properties. Let’s do
this.

ACTIVITY

Let’s Categorize it!

In this activity, classify the following polygons if regular, irregular, convex, concave, or not a
polygon.
1.

4.

2.

3. 5.

Page | 30
 ANALYSIS

How did you find out that it is a convex, concave or not a polygon from the given activity?

In item no. 5, if you draw a diagonal in the polygon, how many would it create?

ABSTRACTION

Polygons
The term polygon was derived from two Greek words, polus, meaning many and gonos,
meaning angled. Combining those terms, polygon literally means many angles.

 AThe segments
polygon formed
is a close by the
plane points
figure are called
whose theline
sides are sides of a polygon.
segments that Its assigned symbol
is n.
intersect only at the endpoints.
 The points are referred to as the vertices of the polygon.

Naming of a polygon
We can name a polygon by naming all its vertices in consecutive order. For example,
refer in the figure below.
A B

D C

Page | 31
 We name this polygon as polygon ABCD. Another acceptable name is DCBA. We
cannot name the polygon as DBAC.

Regular polygons – all sides and angles are congruent—both equilateral and equiangular.
Irregular polygons – is any polygon that is not a regular polygon.

Convex Polygons – its angles measures are between 0° to 180°.

(triangle, quadrilateral, pentagon, hexagon, heptagon, octagon, nonagon, decagon…)
Concave Polygons – if any internal angle is 180°. A line segment joining two points of a
concave polygon can contain points in the exterior of the polygon.

Table 2.1 This table show some special names for polygons with fixed numbers of sides.
Number of Sides Name of a Polygon
3 Triangle
4 Quadrilateral
5 Pentagon
6 Hexagon
7 Heptagon
8 Octagon
9 Nonagon
10 Decagon
n n-gon

Diagonals of a Polygon
A diagonal of a polygon is a line segment that joints two non-consecutive vertices.

Page | 32
 This table describes the polygons number of side and their corresponding total numbers of
diagonals.

 Table 2.2
Z
 C H G

T Y



F W X
 A B E
Triangle Quadrilateral Pentagon
Sides = 3 Sides = 4 Sides = 5
Diagonals = 0 Diagonals = 2 Diagonals = 5

Theorem 2.1.1
The total number of diagonals D in a polygon in a polygon of n sides given by
𝑛 (𝑛−3)
the formula D =
2

Example 1: Find the number of diagonals of a quadrilateral and a dodecagon.

Quadrilatera Dodecagon
l
12 (12−3)
𝑛 (𝑛−3) D=
D= 2
2
12 (12−3)
4 (4−3) D=
D= 2
2
12 (9)
4 (1) D=
D= 2
2
108
4 D=
D= 2
2
108
D=
D=2 2

D = 54

Page | 33
Example 2: Given the number of diagonals formed, determine what polygon is involved.
1. 44 2. 35
𝑛 (𝑛−3) 𝑛 (𝑛−3)
Solution: Diagonals = Diagonals =
2 2
𝑛 (𝑛−3) 𝑛 (𝑛−3)
44 = 35 =
2 2

88 = n(n-3) 70 = n(n-3)

88 = n2 – 3n 70 = n2 – 3n

0 = n2 – 3n – 88 0 = n2 – 3n – 70

0 = (n + 8)(n - 11) 0 = (n + 7)(n - 10)

n= - 8, n= 11 n= - 7, n= 10

we choose n= 11 ⇒ the we choose n= 10 ⇒ the

polygon is undecagon polygon is decagon
Video Recommendations

- https://www.youtube.com/watch?v=xtIUOcK0yH4

APPLICATION

A. Determine whether the following are polygons or not. If they are polygons, determine
whether they are convex or concave.
1. 2. 3. 4. 5.

B. Determine the number of diagonals given the polygon.

1. 16 – gon 2. Nonagon 3. 13 - gon
C. Determine the number of sides of the polygon given the number of
diagonals. 1. 0 2. 104 3. 5

Congratulations! You have completed lesson 1 of module 2. You are now ready for Lesson 2,
which tackles about angles of a triangle.

Page | 34
Lesson 2: Angles of a Triangle
Learning Outcomes
At the end of the lesson, you will be able to:
define and identify different types of triangles;
find the unknown parts of a given triangle using theorem; and
prove statements involving angles of a triangle using theorems.

Duration: 1 hr.

Introduction:
Hello! Welcome to lesson 2. Triangles are among the most important objects studied in
mathematics. That’s why in this lesson, we will talk about different types of triangles and
their theorems. Let’s do this.

ACTIVITY

Cut it!
Follow the following steps and answer questions in the analysis.
1. Prepare a bond paper, scissors, and protractor.
2. Form the bond paper into a triangle. Make sure that every side of a triangle is straight.
3. Cut the angles of angles from the corner of the triangle.
4. Lastly, put the angles together at the same vertex as shown below

Page | 35
 ANALYSIS

From the given activity, what is the sum of the measures of the three angles?

ABSTRACTION

A
A triangle (symbol △) F
is the unionof three-line
Each pointsegments
A, B, andthat
C is the vertex of the
triangle.
are determined Eby three noncollinear points.
D  Point D is in the interior of the triangle.
 Point E is on the triangle.
B C  Point F is on the exterior of the triangle.
△ABC

Example 1: In △RST, RS = 4, ST = 4, and m∠S = 90°. Describe completely the type of

triangle represented.

Solution: △RST is an isosceles triangle.

Theorem 2.2.1
The sum of the measures of the angles of a triangle is 180°.

Page | 36
Example 2: In △ABC, m∠A = 54° and m∠B = 74°. Find m∠C.
Solution: m∠A + m∠B + m∠C = 180°

m∠C = 180° - m∠A -

m∠B m∠C = 180° - 54° -
74°

m∠C = 52°

Example 3: Solve for x.

(2x)°

(4x-8)° (5x-10)°
Solution: 2x + (4x-8) + (5x-10) = 180°
11x – 18 = 180
11x = 180 + 18
11x = 198
x = 18

Theorem 2.2.2
The measures of each angle in an equiangular triangle is 60°.

Theorem 2.2.3
The acute angles of a right triangle are complementary.

Theorem 2.2.4
If two angles of a triangle are congruent to two triangles of another
triangle, then the remaining pair of angles are also congruent.

Example 4: Solve for x. (3 𝑥-6)°

4
(3𝑥 + 14 )°
Solution: Since the two triangles are right angles and another given pair of angles are
congruent. Then that makes the remaining angles are also congruent.
4
x + 14 = 3x – 6
3
4
x - 3x = – 6 – 14
Page | 37
3

Page | 38
 5
− x = - 20
3
3
x = - 20 (− )
5

x = 12

Example 5: Given: ̅𝑂̅̅𝑁̅ bisects ∠DNA N

̅𝑂̅̅𝑁̅ ⊥̅𝐴̅̅𝐷̅

Prove: ∠D ≅ ∠A
D A
Proof O
Statement Reasons

1. ̅𝑂̅̅𝑁̅ bisects ∠DNA Given

2. ∠DNO and ∠ANO are congruent Definition of angle bisector
3. ̅𝑂̅̅𝑁̅ ⊥̅𝐴̅̅𝐷̅ Given

4. ∠DON and ∠AON are right angles Definition of ⊥

5. ∠DON ≅ ∠AON All right angles are congruent

6. ∠D ≅ ∠A If two angles of a triangle are congruent to
two angles of another triangle, then the
remaining angle pair is congruent.

 Angle formed by a side and extension of the

adjacent side is an exterior angle.
 ∠ACD is an exterior angle of △ABC.
 A triangle has a total of 6 exterior angles;
two at each vertex.
 ∠A and ∠B are the two nonadjacent interior
(a) angles or remote interior angles for
exterior
Theorem 2.2.5
The measure of an exterior angle is equal to the sum of the measures of
the two nonadjacent interior angles.

Page | 39
 Q
Example 6: Solve for x. R
12
m∠1 = x2 + 2x

m∠S = x2 - 2x
S T
m∠T = 3x + 10

Solution: By theorem 2.6.

m∠1 = m∠S + m∠T

x2 + 2x = (x2 - 2x) + (3x + 10)

2x = x + 10
x = 10

Video Recommendations

- https://www.khanacademy.org/math/geometry/hs-geo-congruence/hs-geo-
congruence-theorems/v/proof-sum-of-measures-of-angles-in-a-triangle-are-180 (proof
of the sum of the angles of a triangle is equal to 180°)
- https://www.khanacademy.org/math/geometry/hs-geo-congruence/hs-geo-
congruence-theorems/v/triangle-angle-example-1 (Triangle exterior angle example)
- https://youtu.be/zWECG5Kig7U

APPLICATION

A. Solve for x.
1. 2. 2x° 55°
x°

55° 2x°
(x+25)°

Page | 40
B. Complete the following.
1. Proof of the Exterior Angle of a Triangle Theorem (EATT)
Given: ∠4 is an exterior angle to ∠3.
2
∠1 and ∠2 are the remote interior angles of
∠4.
Prove: m∠4 = m∠1 + m∠2 1 3 4

Proof:

Statements Reasons
1. ∠4 is an exterior angle to ∠3 Given
2. ∠4 and ∠3 form a linear pair
3. Linear Pair Postulate
4. m∠4 + m∠3 = 180°
5. m∠1 + m∠2 + m∠3 = 180°
6. Transitive property
7. m∠4 = m∠1 + m∠2

Congratulations! You have completed lesson 2 of module 2. You are now ready for Lesson 3,
which tackles about angles of a polygon.

Page | 41
Lesson 3: Angles of a Polygon
Learning Outcomes
At the end of the lesson, you will be able to:
find the sum of the measures of interior and exterior angles; and
solve for the number of sides of a polygon.

Duration: 1 hr.

Introduction:
Hello! Welcome to lesson 3. Since we already know the sum of the measures of the angles for
a triangle. Let’s talk about the other angles of a polygon.

ACTIVITY

Let’s Categorize it!

In this activity, you will draw a regular polygon. Then divide the polygon into triangles. Shade
the triangles with different colors to each other. A sample of it is shown below.

ANALYSIS

Page | 42
From the given activity, how many triangles did your polygon have? If the sum of each
triangle’s angles is 180° based on lesson 2 theorem, what is the sum of interior angles of the
polygon?

ABSTRACTION

Summarized observations from dividing polygons into triangles.

All polygons can be divided into triangles. For each triangle is
180°.Polygon Sides Number of triangles Sum of Angles
the polygons can be
divided
Triangle 3 1 180°
Quadrilateral 4 2 2(180°) = 360°
Pentagon 5 3 3(180°) = 540°
Hexagon 6 4 4(180°) = 720°
Heptagon 7 5 5(180°) = 900°
⋮ ⋮ ⋮ ⋮
n-gon n-gon n-2 (n-2)(180°)

Sum of the Interior Angles of a Polygon

Theorem 2.3.1
The sum of the interior angles of any polygon having n sides is 180°(n-2).

Reminder:
The sum of the interior angles of a triangle is180°.

Page | 42
 Proof.
268
T Y
Consider TWXYZ pentagon with auxiliary 1 9

segments (diagonals from vertex to another)

3 7
as shown. 4 5
WX
m∠7 + m∠9 + m∠8 = 180°
m∠4 + m∠5 + m∠6 = 180°
m∠1 + m∠3 + m∠2 = 180°
m∠T + m∠W + m∠X + m∠Y + m∠Z + = 540°

For pentagon TWXYZ, which n = 5, the sum of the measures of interior angles is
180° (n - 2), which is equal to 540°.
When drawing diagonals from one vertex of a polygon n sides, we always form (n -
2) triangles.

Example 1: Give the sum of the interior angles of a polygon with the following number of
sides.

a. 15 b. 12
(applying SUM = 180° (n-2) SUM = 180° (n-2)
theorem = 180° (15-2) = 180° (12-2)
2.3.1) = 2,340° = 1,800°

Example 2: Given the sum of the measures of the interior angles of a polygon, give the
number of sides.
a. 1,080° b. 1,980°
SUM= 180° (n-2) SUM= 180° (n-2)
1,080= 180° (n-2) 1,080= 180° (n-2)
1,080 1,980
180 = n-2 180 = n-2
6 = n-2 11 = n-2
6+2 =n 11+2 = n
8 =n 13 =n

Theorem 2.3.2
𝟏𝟖𝟎°(𝒏−𝟐)
The measure I of one angle of a regular polygon is given by 𝑰 = .
𝒏
Page |
Example 3: Find the measure of each interior angle of a floor tile in the shape of an
equiangular hexagon.
Solution:
For a hexagon, n = 6.

(6−2) . 180°
I= 6

4 . 180°
I= 6

720°
I= 6

I = 120°

Theorem 2.3.3
The sum of measures of the exterior angles of a polygon, one at each vertex, is
360°.

Theorem 2.3.4
The measure E of each exterior angle of a regular polygon or equiangular
polygon of n sides is E = 𝟑𝟔𝟎°
𝒏

Example 3: Find the number of sides of a regular polygon if each interior angle measures
144°. Use Theorem 2.3.4
Solution:
If each interior angle measures is 144°, then each exterior angle measures 36° (they
are supplementary because exterior sides of these adjacent angles form a straight
line)
Now each of the n exterior angles has the measure
360°
𝑛

In this case, E = 360° (using Theorem 2.3.4)

𝑛

360°
36° = 𝑛

36°n = 360°

360°
n= 36°
Page | 44
 n = 10
The polygon (decagon) has 10 sides

Video Recommendations

- https://www.khanacademy.org/math/geometry-home/geometry-shapes/angles-with-
polygons/v/sum-of-interior-angles-of-a-polygon
- https://www.khanacademy.org/math/geometry-home/geometry-shapes/angles-with-
polygons/v/sum-of-the-exterior-angles-of-convex-polygon
- https://www.youtube.com/watch?v=ccbEQgRAx_k

APPLICATION

Answer the following questions.

1. Given: ̅𝐴̅̅𝐵̅ || ̅𝐷̅̅𝐶̅ , ̅𝐴̅̅𝐷̅ || ̅𝐵̅̅𝐶̅
̅𝐴̅̅𝐸̅ || ̅𝐹̅̅𝐶̅ With angle measures as 36°

indicated. Find x, y, and z.

z
77° x y

2. Find the sum of the measures of the interior angles of a polygons n side if:
a) n = 7 b) n = 14
3. Find the number of sides that a polygon has if the sum of the measures of its interior
angles is:
a) 2340° b) n = 1260°
4. Find the number of sides in a regular polygon whose exterior angles each
measure: a) 24° b) n = 18°

Congratulations! You have completed lesson 3 of module 2. You are now ready for Module 3,
which tackles about triangles and congruent triangles.

MODULE
ASSESSMENT

A. Determine the number of diagonals given the polygon.

Page | 45
1. 20 – gon 2. Heptagon 3. 25 - gon
B. Determine the number of sides of the polygon given the number of diagonals.
1. 27 2. 90 3. 104
C. Solve for x, y and z.
1. 65° 2.
124°

x°
30° x° y° z°

D. Complete the following: U V

Given: ̅𝑈̅̅𝑁̅ || ̅𝑉̅̅𝑅̅ L

N R
Prove: ∠U ≅ ∠R
Proof:
Statements Reasons

1. ̅𝑈̅̅𝑁̅ || ̅𝑉̅̅𝑅̅ Given

2. ∠N and ∠V are alternate interior angles
3. If two lines are ||, then alternate interior
angles are ≅
4. Vertical angle theorem
5. If two angles of a ∆ are ≅ to two angles of
another ∆, then the remaining angle pair is ≅

E. Find the sum of the measures of the interior angles of a polygons n side if:
1. n = 17 2. n = 24
F. Find the number of sides that a polygon has if the sum of the measures of its interior angles
is:

Page | 46
1. n = 5040° 2. n = 1260°
G. Find the number of sides in a regular polygon whose exterior angles each
measure: 1. n = 72° 2. n = 90°

References
Alexander, D.C. and Koeberlein, G. M. (2011), Geometry for College Students, Belmont CA,
USA.CENCAGE Learning.
Rich, B. and Thomas, C. (2009). Schaums outlines: Geometry. USA. McGraw-Hill
Companies, Inc.

Module 3: Triangles and Congruent Triangles

Module Overview: Hello! Welcome to Module 3. In this module, you will be able to deepen your understan

Module Objectives/Outcomes:
 identify the corresponding angles and sides of the triangle to another triangle;
 name properly the two congruent triangles;
 solve the unknown angles or sides of the congruent triangles using the concept of
congruent triangles;
 state the postulate and theorems on congruency of triangles;
 provide formal proofs of congruency of triangles;
 state the postulate and theorems right triangles congruence; and
 provide formal proofs of right congruent triangles.

Lesson/s in the Module:

Lesson 1: Introduction to Triangles and Congruent Triangles
Lesson 2: Proving Congruent Triangles
Lesson 3: Triangle Congruence Theorems

Page | 47
Module 3: Triangles and Congruent Triangles
Lesson 1: Introduction to Triangles and Congruent Triangles
Learning Outcomes
At the end of the lesson, you will be able to:
discuss congruence between two geometric figures;
identify the corresponding angles and sides of the triangle to another triangle;
name properly the two congruent triangles; and
solve the unknown angles or sides of the congruent triangles using the concept of congruent triangles.

Duration: 1 hr.

Introduction:
Hello! Welcome to lesson 1. Did you know the difference between equality and congruence?
This lesson will teach you about the difference between the two. But this discussion will
focus on congruent triangles and its theorems.

ACTIVITY

Page | 48
Let’s Compare it!

In this activity, you will analyze and compare the given figures below. And answer the
following questions in the Analysis.
Figures 1

Area of the ∆= 4 Area of the square = 4

Figures 2 A D
53°53°
55
3 3

53° 53°
C 4 B F 4 E

ANALYSIS

From the given activity, Does the illustrations in figure 1 is equal? congruent? In what way?

Do the illustrations in figure 2 is equal? congruent? Why?

ABSTRACTION

Triangle Congruence

Two triangles are congruent if the six parts of the first triangle are congruent
to the six corresponding parts of the second triangle.

Page | 49
In other words, triangles are congruent if they have the same shape and size.
(Congruence symbol is ≅)

 Corresponding angles are pairs of angles whose vertices are paired in a given
correspondence between two triangles.
 Corresponding sides are pairs of sides whose endpoints are the vertices that are paired in
a given correspondence between two triangles.
We say that ∆ABC ≅ ∆DEF

* The angles and sides of

the triangles that have the
same markings have the
same measurement.

It is important to write the vertices in proper order such that the vertices that correspond to
each other are in the same position in naming the triangle.

∆ABC≅∆DEF
From the given figure, we can also pair the vertices and sides of the triangles. The following
correspondence may be expressed as:
A↔D (vertex A corresponds to vertex D)
̅A̅̅B̅ ↔̅D̅̅E̅ (side AB corresponds to side DE)
B↔E (vertex B corresponds to vertex E)
̅B̅̅C̅ ↔̅E̅̅F̅ (side BC corresponds to side EF)
C↔F (vertex C corresponds to vertex F)
̅A̅̅C̅ ↔̅D̅̅F̅ (side AC corresponds to side DF)
Since the corresponding points are the vertices of the triangles, we can also express the
correspondence of the triangles as ∆ABC↔∆DEF.

Example 1:
The claim, ∆RST ≅ ∆WXY name all pairs of corresponding sides and
angles.
Solution: R↔W, S↔X, and T↔Y
̅R̅̅S̅ ↔̅W̅̅̅X̅, S̅̅T̅ ↔̅X̅̅Y̅, and ̅R̅̅T̅ ↔̅W̅̅̅Y̅

For two congruent triangles, the correspondence vertices are given by A↔D,
B↔E, and C↔F. Complete the following statement:
a. ∆BCE ≅ ? b. ∆DEF ≅ ?
Solution: a. ∆BCE ≅ ∆EFD b. ∆DEF ≅ ∆ABC

Page | 50
(important terms)
 Included angle for the two sides of a triangle is an angle whose rays contain
two indicated sides.
 Included side for the two angles is the side whose endpoints are the vertices of the
two angle indicated angles. B

A C
∠A is the included angle of sides ̅𝐴̅̅𝐵̅ ̅𝐴̅̅𝐶̅ . ̅𝑨̅̅𝑪̅ is the included side ∠A and ∠C.
and

Some properties of congruent triangles that are useful in the next discussions:

Theorem 3.1.1
Congruence of triangles is reflexive, symmetric and transitive.

1. ∆ABC ≅ ∆ABC (Reflexive property of Congruence)

2. If ∆ABC ≅ ∆DEF, then ∆DEF ≅ ∆ABC. (Symmetric Property of Congruence)
3. If ∆ABC ≅ ∆DEF and ∆DEF ≅ ∆HIG, then ∆ABC ≅ ∆HIG. (Transitive Property
of Congruence)

Video Recommendations

- https://www.khanacademy.org/math/geometry/hs-geo-congruence/xff63fac4:hs-geo-
congruent-triangles/v/congruent-triangles-and-sss

APPLICATION

A. Given that ∆ABC ≅ ∆XYZ. ∆ABC is marked as shown. How should ∆XYZ?
B X

A C Y Z

Page | 51
 1. What side should have two 4. Which side is congruent to ̅A̅̅B̅ ?
marks? 5. If BC = 3.5 units, is XY also
2. What angle should have a equal to 3.5?
single marking? 6. If m∠Z = 88°, is m∠C = 88°?
3. Which angle is congruent ∠C?
B. Solve for x and y. Given that the two indicated triangles are congruent, give
the congruence statement.

(5y-2) °47°
1. 26 2.
23°(3x-4) °

3y

Congratulations! You have completed lesson 1 of module 3. You are now ready for Lesson 2,
which tackles proving congruent triangles.
Lesson 2: Proving Congruent Triangles
Learning Outcomes
At the end of the lesson, you will be able to:
state the postulate and theorems on congruency of triangles; and
provide formal proofs of congruency of triangles.

Duration: 1 hr.

Introduction:
Hello! Welcome to lesson 2. Since you know already the concept of congruent triangles from
the previous discussion, in this lesson, we will talk about proving congruent triangles. Let’s
do some proving.

ACTIVITY

Guess what?
In this activity you will guess or identify if the given triangles are congruent by SAS(Side-
Angle-Side), ASA (Angle-Side-Angle), SSS (Side-Side-Side) and SAA(Side-Angle-Angle).
Just write SAS, ASA, SSS, or SAA.
1. 3.

Page | 52
2. 4.

ANALYSIS

From the given activity, how did you answer? did you see any pattern?

ABSTRACTION

Postulate 3.2.1
SAS Postulate(Side-Angle-Side): If two sides and the included angle of one triangle are congruent to the cor

Thus b = b1, c = c1, and ∠A ≅ ∠A1 in figure 3-1, then ∆ABC ≅ ∆A1B1C1
B B1

c ≅ c1

A C A1 C1
b b1
Fig .3-1

Postulate 3.2.2
ASA Postulate(Angle-Side-Angle): If two angles and the included side of one triangle are congruent to the c

Thus ∠A ≅ ∠A1, ∠C ≅ ∠C1, and b = b1 in figure 3-2, then ∆ABC ≅ ∆A1B1C1

B B1

A C A1 C1
1
b b
Fig .3-2

Page | 53
 Postulate 3.2.3
SSS Postulate(Side-Side-Side): If three sides of one triangle are congruent to the corresponding sides of an

Thus a = a1, b = b1, and c = c1 in figure 3-3, then ∆ABC ≅ ∆A1B1C1

B B1
a a1
c ≅ c1

A C A1 C1
b b1
Fig .3-3

Example 1: Determine the reason for congruency of the triangle.

In each part of fig. 3-4, ∆I can be proved congruent of ∆II. Make a diagram showing the
equal parts of both triangles and state the congruency postulate that is involved.

Solution:
a. AC is a common side of both triangles in figure (a). ∆I ≅∆II by ASA.
b. ∠1 and ∠2 are vertical angles in figure (b). ∆I ≅∆II by SAS.
c. BD is a common side of both triangles in figure (c). ∆I ≅∆II by SSS.

Example 2: P
For SAS Postulate
Given :
̅𝑃̅̅𝑁̅ ⊥ ̅𝑀̅̅̅𝑄̅

Page | 54
 ̅𝑀̅̅𝑁̅̅ ≅ ̅𝑁̅̅𝑄̅
(see figure in the right)
Prove: ∆PNM ≅ ∆PNQ 12 Q
M N

Proof
Statement Reason
1. ̅𝑃̅̅𝑁̅ ⊥ ̅𝑀̅̅̅𝑄̅ Given

2. ∠1 ≅ ∠2 If two lines are ⊥, they meet to form ≅

adjacent angles

3. ̅𝑀̅̅𝑁̅̅ ≅ ̅𝑁̅̅𝑄̅ Given

4. ̅𝑃̅̅𝑁̅ ≅ ̅𝑃̅̅𝑁̅ Reflexive property

5. ∆PNM ≅ ∆PNQ SAS Postulate

Example 3: For ASA Postulate

Given :
̅𝐴̅̅𝐶̅ ≅ ̅𝐷̅̅𝐶̅

∠1 ≅ ∠2
(see figure in the right)
Prove: ∆ACE ≅ ∆DCB
Proof
Statement Reason
1. ̅𝐴̅̅𝐵̅ and ̅𝐶̅̅𝐷̅ bisect each other at M Given

2. ∠1 ≅ ∠2 Given
3. ∠C ≅ ∠C Reflexive property
4. ∆AMC ≅ ∆BMD SSS Postulate

Example 4: D
For SSS Postulate A

Given :
̅𝐴̅̅𝐵̅ and ̅𝐶̅̅𝐷̅ bisect each other at M.
̅𝐴̅̅𝐶̅ ≅ ̅𝐷̅̅𝐵̅ M

(see figure in the right)

Page | 55
C
B

Page | 56
 Prove: ∆AMC ≅ ∆BMD
Proof
Statement Reason
1. ̅𝐴̅̅𝐵̅ and ̅𝐶̅̅𝐷̅ bisect each other at M Given

2. ̅𝐴̅̅𝑀̅̅ ≅ If a segment is bisected, the segments

̅𝑀̅̅̅𝐵̅ formed are ≅
̅𝐶̅̅𝑀̅̅ ≅
̅𝑀̅̅̅𝐷̅
3. ̅𝐴̅̅𝐶̅ ≅ ̅𝐷̅̅𝐵̅ Given
4. ∆AMC ≅ ∆BMD SSS Postulate

Video Recommendations

- https://www.khanacademy.org/math/geometry/hs-geo-congruence/xff63fac4:hs-geo-
congruent-triangles/v/other-triangle-congruence-postulates
- https://youtu.be/3V7S803lVQg
- https://www.youtube.com/watch?v=jWHOF6cFbpw

APPLICATION

Proving B C
3 4
1. Supply the reasons for the following

proof. Given : ̅𝐴̅̅𝐵̅ || ̅𝐶̅̅𝐷̅, ̅𝐴̅̅𝐵̅ ≅ ̅𝐶̅̅𝐷̅

1
Prove: ∆ABC ≅ ∆CDA A 2 D
Proof:
Statement Reason

1. ̅𝐴̅̅𝐵̅ ≅ ̅𝐶̅̅𝐷̅ 1.

2. ̅𝐴̅̅𝐵̅ || ̅𝐶̅̅𝐷̅ 2.
3. ∠1 and ∠4 are alternate interior ∠s 3.
4. ∠1 ≅ ∠4 4.

5. ̅𝐴̅̅𝐶̅ ≅ ̅𝐴̅̅𝐶̅ 5.
6. ∆ABC ≅ ∆CDA 6.

J N
2. Complete the following proof.

Given : ̅𝐽̅𝐸̅ and ̅𝑂̅̅𝑁̅ bisect each other at A. A

Page | 57
Prove: ∆JAO ≅ ∆EAN
Proof: O E

Statement Reasons
1. ̅𝐽̅𝐸̅ and ̅𝑂̅̅𝑁̅ bisect each other at A
2. Definition of segment bisector
3. Vertical angle Theorems

4. ̅𝑂̅̅𝐴̅ ≅ ̅𝐴̅̅𝑁̅
5. SAS Postulate
3. Complete the following proof.

Given : ̅𝐴̅̅𝐼 ≅ ̅𝐺̅̅𝐿̅ , ̅𝐵̅̅𝐺̅ ≅ ̅𝐼̅𝐸̅ , ̅𝐵̅̅𝐴̅ ≅ ̅𝐸̅̅𝐿̅ E

Prove: ∆ABG ≅ ∆LEI

I
A L
G

Proof: B
Statement Reason

1. ̅𝐵̅̅𝐺̅ ≅ ̅𝐼̅𝐸̅ Given

2. ̅𝐵̅̅𝐴̅ ≅ ̅𝐸̅̅𝐿̅ Given

3. ̅𝐴̅̅𝐼 ≅ ̅𝐺̅̅𝐿̅ Given

4. ̅𝐴̅̅𝐼 = ̅𝐺̅̅𝐿̅
5.
6. AI + IG = IG + AI
7. AG = LI
8. Definition of ≅ segements
9.
Congratulations! You made it. You have completed lesson 2 of module 3. You are now ready
for Lesson 3, which tackles triangle congruence theorems.
Lesson 3: Triangle Congruence Theorems
Learning Outcomes
At the end of the lesson, you will be able to:
state the postulate and theorems right triangles congruence; and
provide formal proofs of right congruent triangles.

Page | 58
Duration: 1 hr.

Introduction:
Hello! Welcome to lesson 3. Being a student, it is not a surprise that we will meet the
words right triangles in some math subjects even though you are not a mathematics major
student. Right triangle is essential not only in school but also for carpenters’ engineers and
can find it everywhere. Now let’s our discussion to deepen your understanding about the right
triangle congruence. Let’s do this.

ACTIVITY

Draw me!
In this activity, draw two right triangles and name the its parts. Then answer the following
questions in analysis.

ANALYSIS

From the given activity, what if the legs of a right triangle are congruent to the corresponding
legs of another right triangle, does it make the triangles congruent by SAS Postulate? If yes,
why?

ABSTRACTION

We are not only limited to postulates SAS, ASA, and SSS.

Theorem 3.3.1
SSS Postulate(Side-Side-Side): If two angles and the side opposite one of the
angles in one triangle are congruent to the corresponding parts of another triangle,
Page |
then the triangles are congruent.
Right Triangle Congruence
Acute
angle

leg

Right Acute
angle leg angle

Theorem 3.3.1
LL Congruence Theorem: If the legs of the right triangle are congruent to the
corresponding legs of another right triangle, then the triangles are congruent.
Thus b = b1, c = c1, and ∠A, ∠A1 are right angles as shown in the figure below, then ∆ABC ≅
∆A1B1C1. B 1
B
c
c1
≅
A
b C A1 C1
b1
Theorem 3.3.1
LA Congruence Theorem: If a leg and an acute angle of a right triangle are
congruent to the corresponding leg and acute angle of another right triangle, then
Thus b = bthe
1 triangles are congruent.
, ∠C ≅ ∠C 1, and ∠A ≅ ∠A1 in figure 3-1, then ∆ABC ≅ ∆A1B1C1
B B1

A b C A1 b1 C1

Theorem 3.3.1
HA Congruence Theorem: If the hypotenuse and one acute angle of a right
triangle is congruent to the corresponding hypotenuse and acute angle of another
right triangle, then the triangles are congruent.

Page | 59
Thus a = a1, ∠C ≅ ∠C 1, and ∠A, ∠A1 are right angles as shown in the figure below, then
∆ABC ≅ ∆A1B1C1. B B1

a a1

≅
A C A1 C1

Example 1: For SAA Theorem

M
Given: ̅𝑌̅̅𝑇̅ || ̅𝑀̅̅̅𝐸̅ , ̅𝑌̅̅𝐸̅ || ̅𝑀̅̅̅𝐴̅, and ̅𝑌̅̅𝑇̅ ≅ ̅𝑀̅̅̅𝐸̅
A

Y
E
T
Prove: ∆YET ≅ ∆MAE
Proof
Statement Reason
1. ̅𝑌̅̅𝑇̅ ≅ ̅𝑀̅̅̅𝐸̅ Given
2. ̅𝑌̅̅𝑇̅ || ̅𝑀̅̅̅𝐸̅ Given
3. ∠YTE≅ ∠MEA If lines cut by a transversal, corresponding
∠s are congruent.
4. ̅𝑌̅̅𝐸̅ || ̅𝑀̅̅̅𝐴̅ Given
5. ∠YTE≅ ∠MEA If lines cut by a transversal, corresponding
∠s are congruent.
6. ∆PNM ≅ ∆PNQ SAA Theorem

Example 2: Given the marked parts to be congruent, are the given pair of triangles
congruent?
1. - Yes. The common leg is congruent to itself, two
corresponding angles are congruent, and both triangles
are right. The congruence is justified by the LA
theorem.

- Yes. We have two right triangles, their hypotenuses are

2.
congruent, and a pair of congruent corresponding angles.
The congruency is justified by the HA Theorem of
SAA Theorem.

Page | 60
3. - No. We don’t have an AAA Congruence Theorem.

Page | 61
Example 3: Given: ̅𝑄̅̅𝑇̅ ⊥ ̅𝑈̅̅𝑅̅, 𝑉̅̅̅𝑆̅ ⊥ ̅𝑈̅̅𝑅̅, ̅𝑈̅̅𝑄̅ || 𝑅̅̅̅𝑉̅ , ̅𝑄̅̅𝑈̅ ≅ ̅𝑉̅̅𝑅̅
Prove: ∆QUT ≅ ∆VRS
Q
R
S

T
U

Proof
Statements Reason

1. ̅𝑄̅̅𝑇̅ ⊥ ̅𝑈̅̅𝑅̅ Given

2. ∠QTU is a right ∠ Definition of ⊥

3. ∆QTU is a right ∆ Definition of right ∆

4. 𝑉̅̅̅𝑆̅ ⊥ ̅𝑈̅̅𝑅̅ Given

5. ∠VSR is a right ∠ Definition of ⊥

6. ∆VSR is a right ∆ Definition of right ∆

7. ̅𝑄̅̅𝑈̅ ≅ ̅𝑉̅̅𝑅̅ Given

8. ̅𝑈̅̅𝑄̅ || ̅𝑅̅̅𝑉̅ Given

9. ∠QUT ≅ ∠VRS If || lines are cut by a transversal, the
alternate interior angles are ≅
9. ∆QUT ≅ ∆VRS HA Congruence Theorem

Video Recommendations

- https://study.com/academy/lesson/congruency-of-right-triangles.html
- https://study.com/academy/lesson/the-ha-theorem.html

Page | 62
 APPLICATION

B
Proving
1. Supply the reasons of the following proof.
D C
Given: ̅𝐴̅̅𝐵̅ ⊥̅𝐶̅̅𝐷̅, ∠A ≅ ∠B
Prove: ∆ACD ≅ ∆BCD
A
Proof:
Statements Reasons
1. CD ≅ CD Reflexive property

2. ̅𝐴̅̅𝐵̅ ⊥̅𝐶̅̅𝐷̅

3. ∠CDA and ∠CDB are right ∠s

4. ∠CDA and ∠CDB are right ∆s
5. ∠A ≅ ∠B
6. ∆ACD ≅ ∆BCD
Congratulations! You made it. You have completed lesson 3 of module 3. You are now ready
for Module 4, which tackles about Quadrilaterals.

MODULE
ASSESSMENT

Proving: Provide a Two Column Proof. L

1. Proving Congruent Triangles

Given: 𝐿̅𝐼 and ̅𝐴̅̅𝐸̅ are the perpendicular bisectors of each other. F
A E
Prove: ∆AFL ≅ ∆EFI

I
2. Triangle Congruence Theorem
Draw and provide proof.

Page | 63
Given: ̅𝐵̅̅𝐶̅ bisects ̅𝐴̅̅𝐷̅ at E, ∠B
≅ ∠C Prove: ∆ABE ≅ ∆DCE
References
Alexander, D.C. and Koeberlein, G. M. (2011), Geometry for College Students, Belmont CA,
USA.CENCAGE Learning.
Rich, B. and Thomas, C. (2009). Schaums outlines: Geometry. USA. McGraw-Hill
Companies, Inc.

Module 4: Quadrilaterals
Module Overview: Hello! Welcome to Module 4. In this module, you will be able to solve and provide proofs

Module Objectives/Outcomes:
 solve the unknown angles or sides of a parallelogram, rhombus, square rectangle
and trapezoid by applying its theorems; and
 provide formal proofs of problems involving theorems of parallelogram, rhombus,
square rectangle and trapezoid.
Lesson/s in the Module:
Lesson 1: Properties of Parallelogram
Lesson 2: Special Parallelograms
Lesson 3: Trapezoid

Page | 64
Module 4: Quadrilaterals
Lesson 1: Properties of Parallelogram
Learning Outcomes
At the end of the lesson, you will be able to:
solve the unknown angles or sides of a parallelogram by applying its theorems; and
provide formal proofs involving theorems of parallelogram.

Duration: 1 hr.

Introduction:
Hello! Welcome to lesson 1. In today’s lesson we are going to use the properties of
parallelograms to uncover the missing sides and angles. Let’s do this.

ACTIVITY

Give it to me!
In this activity, you will provide the properties of a parallelogram you can think of. Give at
least 3.
1.
2.
3.

ANALYSIS

Page | 65
From the given activity, did you answer that the consecutive angles of a parallelogram are
supplementary? And if we draw diagonals does it bisect each other? Support your answers

Quadrilateral
ABSTRACTION Classification
Quadrilaterals are classified according to the number of pairs of its parallel sides.
 If quadrilaterals do not have any pair of parallel sides, it is called a
trapezium.
 If a quadrilateral has only one pair of parallel lines, it is called a trapezoid.
 If a quadrilateral has two pairs of parallel lines, it is called a parallelogram.
Special Parallelograms
 If all the interior angles of a parallelogram are right angles, it is called a
rectangle.
 If all the sides of a parallelogram are congruent to each other, it is called a
rhombus.
 If a parallelogram is both a rectangle and a rhombus, it is called a square.

Quadrilaterals

Parallelograms Trapezoid Trapezium

Rectangle Rhombus

Square

Parallelogram Properties

Theorem 4.1.1
Consecutive angles of a parallelogram are supplementary.

Theorem 4.1.2
Opposite angles of a parallelogram are congruent.

Page | 66
 Theorem 4.1.3
In a parallelogram, either diagonal separates the parallelogram into two congruent
triangles.

Theorem 4.1.4
Opposite sides of a parallelogram are congruent.

Theorem 4.1.5
The diagonals of a parallelogram bisect each other.

Example 1:
Assuming ABCD is a parallelogram, find x and y in each part of the figures
below.

Solutions:
a. By theorem 4.1.4, BC = AD = 3x and CD = AB = 2x

b. By P
theorem
= 2 (l +4.1.5,
w) x + 2y = 15, and x =BC
3y = AD, By theorem
40= 2 (3x + 4.1.4 3x = 2y – 2
Substitute
2x) 3y for x in the first 3(4) = 2yx–=23y
equation:
40= 2 (5x)x + 2y = 15 12 = 2yx –=2
3y + 2y
40= 10x = 15 12+2= 2y3(3)
5y = 15 14 = 2yx = 9
y=3 7 =y

c.
By theorem 4.1.2, 3x – 20 = x + 40 By theorem
3x – 20 = x + 40 4.1.1 y + (x +
3x – x = 40 + 20 40) = 180
2x = 60 y + (30 + 40) = 180
x = 30 y + 70 = 180 Page | 66
y = 180 – 70
Example 2: Given: ▱ ABCD, ̅𝐴̅̅𝐹̅ ≅ ̅𝐸̅̅𝐶̅ B E
C

Prove: ∠BFA ≅ ∠DEC

F
A D
Proof:
Statements Reasons
1. ▱ ABCD is a parallelogram Given
2. ̅𝐴̅̅𝐵̅ ≅ ̅𝐶̅̅𝐷̅ Opposite sides of a parallelogram are
congruent
3. ̅𝐴̅̅𝐵̅ || ̅𝐶̅̅𝐷̅ Definition of a parallelogram
4. ∠BAF ≅ ∠DCE If || lines are cut by a transversal, the
alternate interior angles are ≅
5. ̅𝐴̅̅𝐹̅ ≅ ̅𝐸̅̅𝐶̅ Given
6. ∆BEF ≅ ∆DCE SAS Postulate
7. ∠BFA ≅ ∠DEC Corresponding Parts of a Congruent
Triangles are Congruent (CPCTC)

Video Recommendations

- https://www.khanacademy.org/math/geometry/hs-geo-congruence/hs-geo-
quadrilaterals-theorems/v/proof-opposite-sides-of-parallelogram-congruent
- https://www.youtube.com/watch?v=t42sLuns4Qs
- https://www.youtube.com/watch?v=MQJnPfNb2Og

APPLICATION

A. Solve for x and y, given the parallelograms.

1. G H 2. B C
(9y-6)°
5x+2 6x-3

I (7y+12)°
J A D

A B

Page | 67
B. Proving
Given: ̅𝐴̅̅𝐵̅ ≅ ̅𝐷̅̅𝐶̅, ̅𝐴̅̅𝐷̅ ≅ ̅𝐵̅̅𝐶̅
Prove: ABCD is a parallelogram
D C

Congratulations! You have completed lesson 1 of module 4. You are now ready for Lesson 2,
which tackles special parallelograms.

Lesson 2: Special Parallelograms

Learning Outcomes
At the end of the lesson, you will be able to:
solve the unknown angles or sides of a parallelogram by applying its theorems; and
provide formal proofs involving special parallelograms.

Duration: 1 hr.

Introduction:
Hello! Welcome to lesson 2. Since you know the different kinds of quadrilaterals and the
properties of a parallelogram, in this lesson we will focus on special parallelograms and its
properties. Let’s do this.

ACTIVITY

Venn Diagram
In this activity, you will differentiate the three geometric figures base on their properties
using Venn diagram.

rectangle rhombus

Page | 68
 square

ANALYSIS

From the given activity about, if the square is both a rectangle and a rhombus, does the
rectangle can be a square? Does the rhombus can be a square?

ABSTRACTION

Rectangles, Rhombuses, and Squares belong to the set of parallelograms.

1. A rectangle is an equiangular parallelogram.

2. A rhombus is an equilateral parallelogram.
3. A square is an equilateral and equiangular parallelogram. Thus, square is both a
rectangle and a rhombus.
Parallelograms

Rectangles Squares Rhombuses

Theorem 4.2.1
The diagonals of a rectangle are congruent.

Theorem 4.2.2
The diagonals of a rhombus are perpendicular to each other.

Page | 69
 Theorem 4.2.3
A diagonal of a rhombus bisects a pair of opposite angles.

Theorem 4.2.4
The diagonals of a square are congruent, perpendicular to each other, and they bisect
the four angles of the square.

Property Rectangle Rhombus Square

1. All the properties of a parallelogram. yes yes yes
2. Equiangular yes no yes
3. All interior angles are right angles yes no yes
4. Equilateral no yes yes
5. Congruent diagonals yes no yes
6. Diagonals bisect each vertex angle no yes yes
7. Perpendicular no yes yes
8. Diagonals form 2 pairs of congruent yes yes yes
triangles

Example 1: Assuming ABCD is a rhombus, find x and y in each part of the figures
below.

a.
̅𝐴̅̅𝐵̅ ≅ ̅𝐴̅̅𝐷̅ because Since ∆ABD is equiangular
rhombus is equilateral, it is equilateral, so y = 20.
then
3x – 7 = 20
3x = 20 + 7
3x = 27

Page | 70
b.
Since ̅𝐵̅̅𝐶̅ ≅ Since ̅𝐶̅̅𝐷̅ ≅
̅𝐴̅̅𝐵̅, 5y + 6 ̅𝐵̅̅𝐶̅ , x = y +
= y + 20 20
5y – y = 20 – x = 3.5 + 20
6 x = 23.5

c.
Since ̅𝐴̅̅𝐶̅ bisects 2x + 15 Since ∠A and ∠B
∠A,
4x – 5 = 2x + 15 2(10) + 15 = 35 are supplementary
4x – 2x = 15 + 5 ∠A + ∠B = 180
2x = 20 ∠A = 2 (35) 70 + y = 180

I
Example 2: Given: ▱ALCM is a rectangle, ̅𝐴̅̅𝐼 ≅ ̅𝐼̅𝑀̅̅
F E
Prove: ̅𝐿̅̅𝐹̅ ≅ ̅𝐸̅̅𝐶̅ L C

Proof:
A M

Statements Reasons
1. ▱ALCM is a rectangle Given
2. ̅𝐿̅̅𝐴̅ ≅ ̅𝐶̅̅𝑀̅̅ Opposite sides of a rectangle are ≅
3. ∠L, ∠C, ∠LAM, ∠CMA, are right ∠s All angles of a rectangle are right angles
4. ∠L ≅ ∠C, ∠LAM ≅ ∠CMA All right angles are congruent
5. m∠LAM = m∠CMA Definition of ≅ angles
6. ̅𝐴̅̅𝐼 ≅ ̅𝐼̅𝑀̅̅ Given
7. m∠FAM = m∠EMA Isosceles Triangle Theorem and Definition
of congruent angles
8. m∠LAF + m∠FAM = m∠LAM Angle Addition Postulate
m∠CME + m∠EMA = m∠CMA
9. m∠LAF + m∠FAM = m∠CME+ m∠EMA Transitivity
10. m∠LAF = m∠CME Cancelation property
11. ∆LAF ≅ ∆CME LA Congruence theorem

Page | 71
 12. ̅𝐿̅̅𝐹̅ ≅ ̅𝐸̅̅𝐶̅ CPCTC

Video Recommendations

- https://www.youtube.com/watch?v=3i2yp-lI_V4
- https://www.youtube.com/watch?v=EtoP5G98sok

APPLICATION

Proving: Supply the reasons of the following proof.

1. Given: Rectangle ABCD

E is midpoint of ̅𝐵̅̅𝐶̅ .

Prove:
̅𝐴̅̅𝐸̅ ≅ ̅𝐸̅̅𝐷̅

Plan: Prove ∆AEB ≅ ∆DEC

Statements Reasons
1. ABCD is a rectangle
2. E is a midpoint of ̅𝐵̅̅𝐶̅
3. ̅𝐵̅̅𝐸̅ ≅ ̅𝐸̅̅𝐶̅
4. ∠B ≅ ∠C
5. ̅𝐴̅̅𝐵̅ ≅ ̅𝐶̅̅𝐷̅
6. ∆AEB ≅ ∆DEC

7. ̅𝐴̅̅𝐸̅ ≅ ̅𝐸̅̅𝐷̅

2. Given: Rhombus ABCD

̅𝐴̅̅𝐶̅ is a diagonal
Page | 72
 Prove: ̅𝐴̅̅𝐶̅ bisect ∠A and ∠C
Plan: Prove (1) ∠1 and ∠2 are congruent to ∠3
(2) ∠3 and ∠4 are congruent to ∠1
Statements Reasons
1. ABCD is a rhombus
2. ̅𝐴̅̅𝐵̅ ≅ ̅𝐵̅̅𝐶̅
3. ∠1≅∠3
4. ̅𝐵̅̅𝐶̅ || ̅𝐴̅̅𝐷̅, ̅𝐴̅̅𝐵̅ ||
̅𝐶̅̅𝐷̅
5. ∠2≅∠3, ∠1≅∠4
6. ∠1≅∠2, ∠3≅∠4
7. ̅𝐴̅̅𝐶̅ bisect ∠A and ∠C

Congratulations! You made it. You have completed lesson 2 of module 4. You are now ready
for Lesson 3, which tackles about trapezoid.

Lesson 3: Trapezoid
Learning Outcomes
At the end of the lesson, you will be able to:
solve the unknown angles or sides of a trapezoid by applying its theorems; and
provide formal proofs involving trapezoids.

Duration: 1 hr.

Introduction:
Hello! Welcome to lesson 3. In this lesson we will discuss another quadrilateral which is
trapezoid. Let’s do this.

ACTIVITY

Let’s compare!
In this activity, you will compare and contrast Trapezium and Trapezoid

Page | 73
 Trapezoid Trapezium

ANALYSIS

From the given activity, what could be the properties of a trapezoid?

ABSTRACTION

The bases of the trapezoid are its parallel sides; the legs are its non-parallel sides.
Trapezoid is a quadrilateral with exactly two parallel
sides.

When identifying a trapezoid, it is important that we must indicate which sides are parallel.

In trapezoid ABCD, ̅𝐵̅̅𝐶̅ || ̅𝐴̅̅𝐷̅. The legs are ̅𝐴̅̅𝐵̅ and ̅𝐶̅̅𝐷̅, and the bases are ̅𝐵̅̅𝐶̅

and ̅𝐴̅̅𝐷̅.

Theorem 4.3.1
The base angles of an isosceles trapezoid are congruent.

Theorem 4.3.2
The diagonals of an isosceles trapezoid are congruent.

Page | 74
 Theorem 4.3.3
If the base angles of a trapezoid are congruent, then the trapezoid is isosceles.

Theorem 4.3.4
If the diagonals of a trapezoid are congruent, then the trapezoid is isosceles.

Theorem 4.3.5
The median of a trapezoid is parallel to the bases and its measure is half the sum of
the measures of the bases.

Example 1: Given the trapezoid SRTY, with ̅𝑆̅̅𝑌̅ || ̅𝑅̅̅𝑇̅. ̅𝑃̅̅𝑄̅ is a median.
R T
1. If RT = 10 and SY = 18, then PQ = .

2. If m∠RPQ = 87°, then m∠RSY = . P Q

3. If RT = 2x + 5, SY = 3x + 6, and PQ = 38, Find x.
S Y
Solution:
𝑅𝑇+𝑆𝑌 10+18
1. PQ = 2 = 2 = 14
2. ∠RPQ is congruent to ∠RSY, because they are corresponding angles of parallel lines.
m∠RPQ = m∠RSY = 87°.
3. 𝑅𝑇+𝑌𝑆
PQ =
2
2𝑥+5+3𝑥+6
38 =
2
76 = 5x + 11
-5x= - 65
x = 13

Example 2: Given: ▱CHNG is an isosceles trapezoid, H N

with ̅𝐻̅̅𝑁̅̅ || ̅𝐶̅̅𝐺̅, ̅𝐶̅𝐻̅̅ ≅ ̅𝑁̅̅𝐺̅ E

Prove: ∆CEG is an isosceles triangle. C G
Statements Reasons
1. ̅𝐶̅̅𝐻̅ ≅ ̅𝑁̅̅𝐺̅ Given
2. ∠HCG≅∠NGC Base angles of an isosceles trapezoid are
congruent.

3. ̅𝐶̅̅𝐺̅ ≅ ̅𝐶̅̅𝐺̅ Reflexive Property

Page | 75
 4. ∆CHG≅∆GNC SAS Postulate
5. ∠HGC≅∠NCG CPCTC
6. ̅𝐸̅̅𝐶̅ ≅ ̅𝐸̅̅𝐺̅ Converse of the Isosceles Triangle Theorem
(CITT)
7. ∆CEG is an isosceles triangle Definition of Isosceles Triangle.

Video Recommendations

- https://www.youtube.com/watch?v=EvTHRcNBAc8
- https://www.youtube.com/watch?v=rZ4ukEqbaYE

APPLICATION

A. Given the trapezoid ▱QERY with ̅𝐸̅̅𝑅̅ || ̅𝑄̅̅𝑌̅, ̅𝑊̅̅̅𝑇̅ is the median.
1. Find WT, if ER = 7, QY = 17.
2. Find WT, if ER = x + 7, WT = 2x – 1, QY = 18.
3. Find m∠QER, if m∠EWT = (8x + 16)°, m∠EQY = (4x + 20)°.
B. Complete the following proof.
Given: Isosceles trapezoid ABCD

(̅𝐵̅̅𝐶̅ || ̅𝐴̅̅𝐷̅, ̅𝐶̅̅𝐷̅)

̅𝐴̅̅𝐵̅ ≅
Prove: ∠A ≅ ∠D
Plan: Draw ⊥s to base from B and C.
Prove ∆I ≅ ∆II
Statements Reasons

1. Draw ̅𝐵̅̅𝐸̅ ⊥̅𝐴̅̅𝐷̅ and ̅𝐶̅̅𝐹̅ A ⊥ may be drawn to a line outside point.
⊥̅𝐴̅̅𝐷̅
2. ̅𝐵̅̅𝐶̅ || ̅𝐴̅̅𝐷̅, ̅𝐴̅̅𝐵̅ ≅ ̅𝐶̅̅𝐷̅
3. Parallel lines are everywhere equidistant.
Definition of congruent segments.

Page | 76
 4. Perpendiculars form right angles. All right
angles are congruent.
5. ∆I ≅ ∆II
6. ∠A ≅ ∠D

Congratulations! You made it. You have completed lesson 2 of module 4. You are now ready
for Module.

MODULE
ASSESSMENT

A. Find the value of x and y that ensure each quadrilateral is a parallelogram.

1. 2.

B. Find x and y.
1. 2.

ABCD is a trapezoid. ABCD is an isosceles trapezoid.

C. Proving
Given: ̅𝑊̅̅̅𝑋̅ || ̅𝑍̅̅𝑌̅ and ∠s Z and Y are supplementary
W X

Prove: WXYZ is a parallelogram

Z Y
Proofs
Statements Reasons

1. ̅𝑊̅̅̅𝑋̅ || ̅𝑍̅̅𝑌̅
2. Given
3. If two lines cut by a transversal line so that
interior angles on the same side of the

Page | 77
 transversal are supplementary, these lines
are paraWllel X

4. If both pairs of opposite sides of a

quadrilateral are ||, the quadrilateral is a
parallelogram

Given: Rectangle WXYZ with diagonals ̅𝑊̅̅̅𝑌̅ and ̅𝑋̅̅𝑍̅

V
Prove: ∠1 ≅ ∠2
Z 1 2 Y

Proofs
Statements Reasons
1. Given
2. The diagonals of a rectangle are ≅

3. ̅𝑊̅̅̅𝑍̅ ≅ ̅𝑋̅̅𝑌̅ The opposite sides of a rectangle are ≅

4. ̅𝑍̅̅𝑌̅ ≅ ̅𝑍̅̅𝑌̅
5. ∆XZY ≅ ∆WYZ
6.

Given: ▱PLMN is an isosceles trapezoid with ̅𝐿̅̅𝑃̅ || ̅𝑀̅̅𝑁̅̅

∠MLQ ≅ ∠MQL
M N
Prove: ▱PLMN is a parallelogram

References L Q P

Alexander, D.C. and Koeberlein, G. M. (2011), Geometry for College Students, Belmont CA,
USA.CENCAGE Learning.
Rich, B. and Thomas, C. (2009). Schaums outlines: Geometry. USA. McGraw-Hill
Companies, Inc.
Good Day! Welcome to our 5th module, the circle. Since you have reached this far, I
assumed that you have mastered the concepts of the previous modules. In this module we are
going to explore the parts of the circles and its relationship with other lines. Also, this module
can help us in proving different theorems involving circles and its postulates.

Page | 78
 In order to achieve the competency of this module, this module is divided into four
lessons.
Lesson 1 – Introduction to Circles
Lesson 2 – Arcs of a circle
Lesson 3 – Inscribed angles and intercepted arcs of Circles
Lesson 4 – Tangent lines and inscribed polygons
With this, upon completion of this module, the learner should be to:

 differentiate a circle and the lines having a relationship to a circle,

 list down the theorems about circles,
 prove theorems concerning spheres and circles, and
 use theorems and postulates in solving problems concerning circles

Module 5: circles
Lesson 1. Introduction to circles
Objectives:

At the end of the lesson, you will be able to:

Page | 79
 define circle and its parts,
 identify different parts of a circle, and
 prove theorems involving parts of the circle.
Duration: 1 hr.
Introduction:
Hello! Welcome to the first lesson of module 5. As we roam around, circles appear
everywhere from the things we use up to the food we ate. In this lesson we will deal with
circles, its related terms and its definitions.

ACTIVITY

Materials: illustration board, yarn, pencil, push pin, ruler.

Procedures:
 Mark a point on the center of the illustration board.
 Place the push pin on the point
 Tie the yarn to the push pin
 Tie the yarn to the pencil at 5 to 6 inches
 Hold the pencil and turn the pencil
 Label the point in the center as “O” and label the any side of the shape as
A, B, C, & D

ANALYSIS

From the given activity, what kind of shape you created? Is there any difference from other
polygons? List down any observations from the shape you created.

In the activity, draw a segment passes through point O from one side of the shape to another
side. Draw a segment from point O to points A, B, C, & D. Measure its lengths. List down
your observations.

Page | 80
 ABSTRACTION

A circle is the set of all points in a plane that are at a fixed distance from a given point
known as the center of the circle.

A circle is named by its center point. In the activity, point O is the center of the circle.
The symbol for circle is , so the circle in activity is O. Points A, B, C, and D are
points of (or on) the circle. Points O (the center) and any other points inside the circle are the
interior of circle O and points outside the circle are the exterior.

In the activity, segment OA is a radius of the circle. A radius is a segment that joins
the center of the circle to a point on the circle. Segments OA, OB, OC, and OD and are radii
(plural of radius) of a circle O. By definition, OA = OB = OC = OD, thus,
All radii of a circle are congruent.

A line segment that joins two points of a circle (such as in Figure 5.1) is a chord of the e.
A diameter of a circle is a chord that contains the center of the circle; in circl Figure, TW
diameter of circle Q.

DEFINITION
Congruent circles are two or more circles that have congruent Figure 5.1
radii.

In Figure 5.2, circles P and Q are congruent because their radii have equal lengths.
We can slide circle P to the right to coincide with circle Q.

In Figure 6.6, inter sect at O, the center of the circle. Name:

Example 1:
In Figure 5.3, segm ent MP and NQ intersect at point O,
center of the circle. Name the fol lowin
Figure 5.2
a) All four radii (shown)
b) Both diameters (shown)
c) All four chords (shown)

Solution Figure 5.3

Page | 81
a) segments OM, ON, OP, and OQ
b) segments MP and QN
c) segments MP, QN, QP, and NP

Theorem 5.1.1
A radius that is perpendicular to a chord bisects the chord.

Figure 5.4

APPLICATION

1. Draw a circle O, OA is a radius, AB is a diameter, and AC is a

chord.
a) How does OA compare to AB?
b) How does AC compare to AB?
c) How does AC compare to OA?

2. In the figure 5.5. list down the following:

a) radii of a circle
Figure 5.5
b) diameter of a circle
c) interior of a circle
d) chords of a circle
e) points on a circle

Congratulations, you have just completed the lesson 1 of the module 5. You are now ready for
lesson. Which tackles another lines related to the circle.

Lesson 2. Arcs of a circle

Objectives:
At the end of the lesson, you will be able to:
define arcs of a circle,
illustrate the characteristics of an arc, and
apply theorems in solving the unknown parts of a circle.


Page | 82
Duration: 1 hr.
Introduction:
Hello! Welcome to the second lesson of module 5. As we go deepen to the other
related concepts of a circle, we tend to wonder how this concept is being used in real life.
Engineers used the concepts in building infrastructures and architects uses the concepts for
their designs. In this lesson we will dig deeper to the concepts of ARCs of a circle.

ACTIVITY

Materials: illustration board, yarn, pencil, push pin, ruler, protractor.

Procedures:
 Mark a point on the center of the illustration board.
 Place the push pin on the point
 Tie the yarn to the push pin
 Tie the yarn to the pencil at 5 to 6 inches
 Hold the pencil and turn the pencil in a complete revolution
 Label the point in the center as “O”
 Draw diameter AB
 Draw radius OC

ANALYSIS

From the given activity, what the angle measure of AOC? What is the angle measure of
AOB? How about angle COB? Compare the angle measure of two angles. List down any
observations.

In the activity, consider the side of a circle connecting BAC? Is there any relationship to the
radaii of a circle? List down your observations below.

ABSTRACTION

Page | 83
DEFINITION
An arc of a circle is any portion of the circumference of a circle. To
recall, the circumference of a circle is the perimeter or distance around
a circle. In of Figure 5.5, the part of the circle shown from point A to
point B is arc AB symbolized by AB. If is AC a diameter, then (three
letters are used for clarity) is a semicircle. In Figure 5.5, a minor arc
like AB is part of a semicircle; a major arc such as ABCD (also
denoted by ACD or ABD) is more than a semicircle but less than the
entire circle. Therefore, we can say that the circumference of a circle is
the full arc of the circle itself. In the activity, BAC is an arc of a circle Figure 5.6
O.

DEFINITION
Concentric circles are coplanar circles that have a common center.

The concentric circles in Figure 5.7 have the common center O.

DEFINITION
Figure 5.7
A central angle of a circle is an angle whose vertex is the center of
the circle and whose sides are radii of the circle.

In Figure 5.8, angle NOP is a central angle of circle O. The intercepted

arc of angle NOP is NP. The intercepted arc of an angle is determined
by the two points of intersection of the angle with the circle and all
points of the arc in the interior of the angle.

Figure 5.8
Example 1:
In Figure 5.8, segment MP and NQ intersect at point O, the
center of the circle. Name the following:
a) One semicircle
b) One major arc
c) Intercepted arc of angle MON
d) One central angle
e) One minor arc
f) Central angle that intercepts NP Figure 5.8

Solution
a) MQP (other answers QPN, PNM, NMQ)
b) arc MPN or MQN (other answer arcs QPM or QNM, PNQ or PMQ, NMP or NQP can also
be named using the four letters)
c) arc MN
d) angle QOP (other answers angles PON, NOM, MOQ)
e) arc NP (other answers arcs MN, MQ, & QP)
f) angle NOP (also called angle 2)

Page | 84
the following statements is the consequence of the Segment-Addition Postulate

In a circle, the length of a diameter is twice that of a radius.

Example
Segment QN2:is a diameter of circle O in the figure 5.7 and PN = ON = 12. Find the length of
chord QP

Suggested videos:
https://www.khanacademy.org/math/geometry-home/cc-geometry-circles/arc-measures/v/intro-arc-
measure?modal=1

In Central
a circle,Angle
the degree measure of a central angle is equal to the degree
Postulate.
measure of its intercepted arc.

In the figure 5.9, if the measure of arc AB = 90 degrees, then the

measure of angle AOB = 90 degrees. The reflex angle that intercepts
arc BCA and that is composed of three right angles measures 270 Figure 5.9
degrees.
If the measure of arc AB is 90 degrees, arc BCD measures 180 degrees and arc DA measures
90 degrees. It follows that arcs AB + BCD + AD is 360 degrees. Consequently, we have the
following generalization.

The sum of the measures of the consecutive arcs that form a circle is 360°.

In the figure 5.9 (a), if the measure of angles XYZ = 76°, then arc XZ = 76° by the Central
Angle Postulate. If two arcs have equal degree measures [Figures 5.9(b) and (c)] but are parts
of two circles with unequal radii, then these arcs will not coincide.

Page | 85
This observation leads to the following definition

Figure 5.9

DEFINITION
In a circle or congruent circles, congruent arcs are arcs with equal
measures.

NOTE:
To clarify the definition of congruent arcs, consider the concentric
circles (having the same center) in Figure 5.10. Here the degree measure of
ang AOB of the smaller circle is the same as the degree measure of angle Figure 5.10
COD of the larger circle. Even though the measure of arc AB is equal to the measure of arc
CD, AB is not congruent to CD, because the arcs would not coincide.

Suggested videos:
https://www.khanacademy.org/math/geometry-home/cc-geometry-circles/central-angles-and-arc-length-in-
degrees/v/finding-central-angle-measure-given-arc-length?modal=1

In Example 3:
circle of figure 5.11, OE bisects angle AOD. Using the measures
indicated, find:

Figure 5.11
SOLUTIONS
a) 105°
b) 70°
c) 105°
d) 150°
e) 75° - (360° – (105° + 70° + 35°))
f) 285° - (360° – 75°)
g) The arcs are congruent because both measure 75° and both are found in the same circle

Page | 86
 h) 210° (from 105° + 70° + 35°)

Arc-Angle Postulate.

The drawing in figure 5.12 (a), further supports the claim of Arc-Angle postulate.
Given points A, B, & C on circle O in figure 5.12 (a). suppose the radaii OA, OB, &
OC are drawn. Because the measure of angle AOB + BOC = angle AOC. By Angle Addition
Postulate, it follows that measure of arc AB + BC = arc ABC. We shoul write the three letter
(ABC) because arc AC can be the major arc AC.

Figure 5.12

Examplethe
Determine 4: measure of the angle formed by the hands of a
clock at 3:12 P.M. (see figure 5.13)

Figure 5.13
SOLUTIONS
The minute hand moves through 12 minutes, which is 12 or 1 Of an hour. Thus, the minute
60 5
hand points in a direction whose angle measure from the vertical is 1 (360°) or 72°. At exactly
5
3 P.M., the hour hand would form an angle of 90° with the vertical. However, gears inside the
clock also turn the hour hand through 1 of the 30° arc from the 3 toward the 4; that is, the hour
5
hand moves another 1 (30°) or 6° to form an angle of 96° with the vertical. The angle between
5
the hands must measure 96°-72° or 24°.

Page | 87
 APPLICATION

1. Refer to figure 5.14

Find the following: (show your solution) Figure 5.14

a. AB
b. BC

2. Refer to figure 5.15

Find the following: (show your solution)

a. RV
b. TV Figure 5.15

Congratulations, you have just completed the lesson 2 of the module 5. You are now ready for
lesson 3 which tackles another relationship of arcs and angles of a circle.
Lesson 3. Inscribed angles and Intercepted Arcs of a circles
Objectives:
At the end of the lesson, you will be able to:
illustrate the characteristics of an inscribed and intercepted arc,
use theorems in solving problems, and
prove theorems involving arcs of a circle

Duration: 1 hr.
Introduction:
Hello! Welcome to the third lesson of module 5. As we as we discuss the arcs of a
circle, we discover that the measure of an arc can be used to measure the corresponding
central angle. It can be used to measure other types of angles related to the circle, including
the inscribed angle.

Page | 88
 ACTIVITY

In figure 5.16, Angle B is inscribed angle whose sides are

chords BA & BC

1. Use a protractor to find the measure of the following: Figure 5.16

a. Angle AOC
b. Angle B
2. What is the measure of arc AC?

ANALYSIS

From the given activity, how is the measure of the inscribed angle B related to the measure of
intercepted arc AC?

ABSTRACTION

DEFINITION
An inscribed angle of a circle is an angle whose vertex is a point on the circle and whose
sides are chords of the circle. The word inscribed is often linked to the word inside.

Theorem 5.3.1
The measure of an inscribed angle of a circle is one-half the measure of its intercepted arc.

The proof of Theorem 5.3.1 must be divided into three cases:

CASE 1. One side of the inscribed angle is a diameter. See Figure 5.18.

Page | 89
CASE 2. The diameter to the vertex of the inscribed angle lies in the interior of
the angle. See Figure 5.17(a).
CASE 3. The diameter to the vertex of the inscribed angle lies in the exterior of
the angle. See Figure 5.17(b).

Figure 5.18 Figure 5.17

Theorem 5.3.2
In a circle (or in congruent circles), congruent minor arcs have congruent central angles.

If arc AB is Congruent to arc CD in

congruent circles O & P, then
angle 1 is congruent to angle 2. By
theorem 5.3.2

Figure 5.19

Theorem 5.3.3
In a circle (or in congruent circles), congruent central angles have congruent arcs.

Page | 90
 Theorem 5.3.4
In a circle (or in congruent circles), congruent chords have congruent minor (major) arcs.

Theorem 5.3.5
In a circle (or in congruent circles), congruent arcs have congruent chords.

On the basis of an earlier definition, we define the distance from the center of a circle
to a chord to be the length of the perpendicular segment joining the center to that
chord. Congruent triangles are used to prove the next two theorems.

Theorem 5.3.6
Chords that are at the same distance from the center of a circle are congruent.
DEFINITION
Concentric circles are coplanar circles that have a common

Figure 5.20

Proofs of the remaining theorems are left as module assessment.

Theorem 5.3.7
Congruent chords are located at the same distance from the center of a circle.

Page | 91
Create a drawing to illustrate Theorem 5.3.7.

Theorem
An 5.3.8 in a semicircle is a right angle.
angle inscribed

Theorem 5.3.8 is illustrated in Figure 5.21, where ∠S is inscribed in the semicircle RST. Note
that ∠S also intercepts semicircle RVT.

Theorem 5.3.9
If two inscribed angles intercept the same arc, then these angles are congruent.
In Figure 5.8, angle NOP is a central angle of circle O. The

Theorem 5.3.9 is illustrated in Figure 5.22. Note that ∠1 & ∠2 both intercept XY. Because
m∠1=1mXY and m∠2 = 1 mXY, ∠∠1≈∠∠2.
2 2

Figure 5.22
Figure 5.21
Suggested videos:

https://www.khanacademy.org/math/geometry-home/cc-geometry-circles/inscribed-angles/v/inscribed-and-
central-angles?modal=1
https://www.khanacademy.org/math/geometry-home/cc-geometry-circles/inscribed-angles/v/inscribed-angles-exercise-
example?modal=1

APPLICATION

1.

Page | 92
 2.

Congratulations, you have just completed lesson 3 of the module 5. You are now ready for the
next lesson which tackles about the tangent lines and its theorems and postulates.

Lesson 4. Tangents lines and Inscribed polygons

Objectives:

At the end of the lesson, you will be able to:

discuss the characteristics of an inscribed polygon,
list and use theorems in solving problems, and
prove theorems involving tangent of a circle

Page | 93
Duration: 1 hr.
Introduction:
Hello! Welcome to lesson 4 of module 5. As we learn the relationships of chords to
the arc and its theorems, have we ever thought of another relationship of circle to the lines
passes through the points of a circle at one point? At two points? Is it possible that a straight
line intersects 3 points in the circle? In this lesson, we are going to discuss different lines
passing through the circle.

ACTIVITY

Draw a circle named O. Identify 4 points on the circle called A, B, C, & D. Join the points to
form a quadrilateral inside the circle. Using a protractor, measure each angle (∠A, ∠B, ∠C, &
∠D). Draw a line l passes through point A only.

3. Find the sum of m∠A + m∠C.

4. Find the sum of m∠B + m∠D.

ANALYSIS

1. How are angles A & C related?

2. How are angles B & D related?
3. What have you observed about the line l?
4. Draw a radius OA and find the angle measure formed by the radius and line l.

ABSTRACTION

We need to assume that the lines and circles are coplanar

Page | 94
DEFINITION

A tangent is a line that intersects a circle at exactly one point;

the intersection point is the point of contact or point of
tangency.

In the activity, line l is tangent to circle O, point A is the point of

tangency.

The term tangent also applies to a segment or ray that is part of a

tangent line to a circle. In each case, the tangent touches only one
point.

DEFINITION

A secant is a line (or segment or ray) that intersects a circle at

exactly two points.

In Figure 5.23 (a), line s is a secant to circle O also; line t is a

tangent to circle O, and point C is its point of contact.
In Figure 5.23(b), segment AB is a tangent to circle O and point
T is its point of tangency; ray CD is secant with points of
intersection at E and F. Figure 5.23

DEFINITION

A polygon is inscribed in a circle if its vertices are points on the circle and its sides are
chords of the circle. Equivalently, the circle is said to be circumscribed about the
polygon. The polygon inscribed in a circle is further described as a cyclic polygon.

In Figure 5.23, triangle ABC is inscribed in

circle O and quadrilateral RSTV is inscribed
In circle Q. Conversely, circle O is
circumscribed about triangle ABC and circle
Q is circumscribed about quadrilateral RSTV.
Note that segments AB, BC, & CD, and are
chords of circle O and that segments RS, ST,
TV, and RV are chords of circle Q.
Quadrilateral RSTV and triangle ABC are
cyclic polygons. Figure 5.23

Theorem 5.4.1
If a quadrilateral is inscribed in a circle, the opposite angles are supplementary.
Alternative Form: The opposite angles of a cyclic quadrilateral are supplementary.

Page | 95
Proof of Theorem 5.4.1. In the proof, we show that ∠R and ∠T are supplementary. In a similar
proof, we could also have shown that ∠S and ∠V are supplementary as well.

Given: RSTV is inscribed in circle Q. (see figure 5.24) Prove: ∠R and ∠T are supplementar

Proof:

From lesson 3, an inscribed angle is equal in measure to th

2 2
mSRV , it follows that Figure 5.24

The proof of theorem 5.4.1

DEFINITION
A polygon is circumscribed about a circle if all sides of the
polygon are line Segments tangent to the circle; also, the circle is
said to be inscribed in the polygon.

In Figure 5.25(a), triangle ABC is circumscribed about circle D. In

Figure 5.25(b), square MNPQ is circumscribed about circle T.
Furthermore, circle D is inscribed in triangle ABC, and circle T is
inscribed in square MNPQ.
Figure 5.25

Page | 96
Note that segments AB, BC, & CA are tangents to circle D and that
segments MN, NP, PQ, & QM are tangents to circle T.
We know that a central angle has a measure equal to the measure of
its intercepted arc and that an inscribed angle has a measure equal to
one- half the measure of its intercepted arc.
Now we consider another type of angle in the circle.
Figure 5.25

TheTheorem 5.4.2
measure of an angle formed by two chords that intersect within a circle is one-half the sum of the measur
angle.

In the figure 5.26(a), ∠1 intercepts arc DB and ∠AEC intercepts arc

AC. According to theorem

To prove theorem 5.4.2, we need to draw an auxiliary line CB. [see figure 5.26(b)]

Figure 5.26

Page | 97
Example 1.
In the figure 5.26(a), measure of arc AC is 85° and arc DB is 62°. Find ∠1.

Solution: by theorem 5.4.2,

Figure 5.27 shows the different locations of points. It is determined

by the distance from the center. Points are inside the circle if the
distance is lesser than the radius; points that are found on the circle if
the distance between the point and the center is equal to the radius.
Outside points of the circle if the distance between center and the
point is larger than the radius.
Figure 5.27

TheTheorem
radius (or5.4.3
any other line through the center of a circle) drawn to a tangent at the point of tangency is perp

Figure 5.27

Page | 98
Example 2.
A shuttle going to the moon has reached a position that is 5 mi
above its surface. If the radius of the moon is 1080 mi, how far to
the horizon can the NASA crew members see?

Solutions

According to Theorem 5.4.3, the tangent determining the line of

sight and the radius of the moon form a right angle. In the right
triangle determined, let t represent the desired distance.

Using the Pythagorean Theorem

10802 + 𝑡2 = 10852
1, 166, 4002 + 𝑡 2 = 1,177,2252
2
t2 = 1,177,2252 -1,166,400
t2 = 10,825 t = √10,825 t ≈ 104 mi
corollary 5.4.4 Figure
The measure of an angle formed by a tangent and a chord drawn to the point of tangency is one-half the meas

By theorem 5.4.3

Figure 5.28

Example 3.

In figure 5.29, circle O with diameter DB, tangent line AC, and arc
measure DE = 84°.

Find:

a. m∠1
b. m∠2
Figure 5.29
c. m∠ABD
d. m∠ABE
solutions
:

Page | 99
 Theorem 5.4.5
The measure of an angle formed when two secants intersect at a point outside the circle is one-half the diffe

Figure 5.30

Remember:
In an application of Theorem 5.4.5, one subtracts the smaller c measure
ar
Example 4.
In the figure 5.31, m∠AOB = 136°. If m∠DOC =

46° Find: m∠E

Figure 5.31

Theorems 5.4.5–5.4.7 show that any angle formed by two lines that intersect outside
a circle has a measure equal to one-half of the difference of the measures of the two
intercepted arcs. The next two theorems are not proved, but the auxiliary lines shown
in Figures 5.32 and 5.33(a) will help complete the proofs.

Theorem 5.4.6
If an angle is formed by a secant and a tangent that intersect in the exterior of a circle, then the measure of the
intercepted arcs.

Page | 100
 According to theorem 5.4.6

in Figure 5.32. Again, we must subtract the measure of the smaller

arc from the measure of the larger arc.
A quick study of the figures that illustrated in the previous Figure 5.32
Theorems
shows that the smaller arc is “nearer” the vertex of the angle and that the larger arc is “farther
from” the vertex.

If Theorem 5.4.7
an angle is formed by two intersecting tangents, then the measure of the angle is one-half the difference of th

In the figure 5.32(a), ∠ABC intercepts the two arcs determined

by points A & C. AC is a minor arc and ADC is a major arc.
According to theorem 5.4.6, we subtract the measure of the minor
arc from the measure of the major arc as shown below.

Example 5. Refer to the figure 5.32(b)

Given:

Find:

Figure 5.32

Solution.

Page | 101
Example 6

In the figure 5.33, m∠1 = 46°, find the measure of arc AB and arc ACB

solution: Figure 5.33

Table 5.4.1

Page | 102
 Location of the Vertex of the Angle Rule for Measuring the Angle
Center of the circle The measure of the intercepted arc
One-half the sum of the measures of the
In the interior of the circle
intercepted arcs
One-half the measure of the intercepted
On the circle
arc
One-half the difference of the measures
In the exterior of the circle
of the two intercepted arcs

If Theorem 5.4.7
two parallel lines intersect a circle, the intercepted arcs between these lines are congruent.

In figure 5.34, line AB is parallel to line CD, it follows that

arc AC is congruent to arc BD. Equivalently, measure of arc
AC is equal to measure of arc BD.

Figure 5.34

Page | 103
 APPLICATION

3. Explain the following. Use postulate, theorems, or corollaries to validate your

argument.
a. Is it possible for:
i. a rectangle inscribed in a circle to have a diameter for a side? Explain.
ii. a rectangle circumscribed about a circle to be a square? Explain.
4. Complete the two-column proof.

5.

Page | 104
MODULE ASSESSMENT

I. Use theorems and postulates to solve the following:

a. State whether the statements are always true (A), sometimes true (S), or
never true (N).
i. In a circle, congruent chords are equidistant from the center.
ii. If a triangle is inscribed in a circle and one of its sides is a diameter,
then the triangle is an isosceles triangle.
iii. Two concentric circles have at least one point in common.
iv. Tangents to a circle at the endpoints of a diameter are parallel

b. Solve the following:

i. Two circles intersect and they have a common chord 12 cm long. The
measure of the angles formed by the common chord and a radius of
each circle to the points of intersection of the circles is 45°. Find the
length of the radius of each circle.
ii. Two circles are concentric. A chord of the larger circle is also tangent
to the smaller circle. The lengths of the radii are 20 and 16,
respectively. Find the length of the chord.

II. Construct proof of the following theorems:

a. Theorem 5.3.7
b. Theorem 5.3.8

Congratulations, you have just completed module 5. You are now ready for the next module
which tackles about the different kinds of solids and its characteristics.

References:

- Alexander, D.C. and Koeberlein, G. M. (2011), Geometry for College Students, Belmont
CA, USA.CENCAGE Learning.
- Rich, B. and Thomas, C. (2009). Schaums outlines: Geometry. USA. McGraw-Hill Companies, Inc.

Supplemental videos

https://www.khanacademy.org/math/geometry-home/cc-geometry-circles
https://www.khanacademy.org/math/geometry-home/cc-geometry-circles/arc-measures/v/intro-arc-
measure?modal=1
https://www.khanacademy.org/math/geometry-home/cc-geometry-circles/central-angles-and-arc-length-
in- degrees/v/finding-central-angle-measure-given-arc-length?modal=1
https://www.khanacademy.org/math/geometry-home/cc-geometry-circles/inscribed-angles/v/inscribed-
angles- exercise-example?modal=1
https://www.khanacademy.org/math/geometry-home/cc-geometry-circles/inscribed-angles/v/inscribed-and-
central-angles?modal=1

Page | 105
Good Day! Welcome to our last module, the Perimeter, Area, Surface Area and Volume of
Solids figures. In this module we are going to explore the world of three-dimensional space.
Also, this module can help us in understand and apply the concepts in real life
In order to achieve the competency of this module, this module is divided into six
lessons.
Lesson 1 – Perimeter
Lesson 2 – Area of polygons
Lesson 3 – circumference and area of
circles Lesson 4 – Definition of Basic
Solids Lesson 5 – Surface Area of Solids
Lesson 6 – Volumes
With this, upon completion of this module, the learner should be to:
 compare and contrast the different types of solids
 solve problems involving areas and volumes of basic geometric polygons
 do geometric construction with the use of any straight edge
 prove theorems

Page | 106
Module 6: Perimeter, Area, surface Area and Volume of Solids
Lesson 1. Perimeter of a polygons
Objectives:

At the end of the lesson, you will be able to:

define perimeter, and
solve problems involving perimeter.

Duration: 1 hr.
Introduction:
Hello! Welcome to lesson 1 of module 6. In the previous lessons, we learned different
kinds of polygon and its properties. Now let’s continue learning different concepts in plane
geometry, in this lesson we will explore the measurements around the polygon.

ACTIVITY

Materials: meter stick, clean sheet of paper, pen.

Procedures:
 Sketch the floor plan of your room
 Use meter stick to measure the side of your room.
 List down all gathered data.
 Find the sum of all data

ANALYSIS

From the given activity, what kind of polygon your room look like? What is the total measure
of the side of your room? List down your observations below.

Page | 107
 ABSTRACTION

DEFINITION

The perimeter of a polygon is the sum of the lengths of all sides of the polygon.

In the activity, the sum of the length of your room is the perimeter of your room. In finding
the perimeter of any polygon, just simply add the length of all sides. To have a clear view, the
table below is the summary of formula in finding the perimeter of a polygon.

Example 1.

Find the perimeter of triangle ABC as shown in figure 6.1 if:

a. AB = 5 in., AC = 6 in., and BC = 7 in.
b. Altitude AD = 8 cm, BC = 6 cm, and AB ≅ AC

Solutions
a. Since the triangle is scalene, we use the
formula Figure 6.1
P=a+b+c
P = 5 in. + 6 in. + 7 in.
P = 18 in.

Note:
Always indicate the unit being used.

Page | 108
 b.

Because x + x = 2x, we have √73 + √73 = 2√73

Example 2.

While remodeling, Mr. De Guzman have decided to replace

the old woodwork with Colonial-style oak woodwork.

a) Using the floor plan provided in Figure 6.2, find the

amount of baseboard (in linear feet) needed for the
room. Do not make any allowances for doors!
b) Find the cost of the baseboard if the price is per 50.00
linear foot.
Figure 6.2

Solutions

a. Dimensions not shown measure 20 - 12 or 8 ft and 18 - 12 or 6 ft. The perimeter,

or “distance around,” the room is
12+ 6 + 8 + 12 + 20 + 18 = 76 linear ft

b. The cost is 76 X 50 = 3,800.

Suggested videos:
https://youtu.be/LoaBd-sPzkU

https://www.khanacademy.org/math/geometry-home/geometry-area-perimeter/geometry-
perimeter/v/perimeter-and-area-basics

Page | 109
 APPLICATION

a. Find the perimeter of each figure.

Congratulations, you have just completed the lesson 1 of the module 6. You are now ready for
the next lesson which tackles the area of polygons.

Lesson 2. Area of a polygons

Page | 110
Objectives:
At the end of the lesson, you will be able to:
define area,
derive the formula of area, and
solve problems involving area of polygons.

Duration: 1 hr.
Introduction:
Hello! Welcome to lesson 2 of module 6. In the previous lesson, we discussed the
measurements around the polygons. Now let’s dig into the inside of the polygons.

ACTIVITY

Materials: ruler, clean sheet of paper, pen.

Procedures:
 Use the to draw a square with a side of 4 inches
 Mark every inch of all sides of the square.
 In every inch, draw a segment connecting the opposite side.
 Draw another rectangle with a dimension of 4 cm & 3 cm
 Repeat the steps 2 & 3

ANALYSIS

In the first polygon (square), count the number of squares created. How is it related to the
length of sides? Consider the second polygon, do the same observation. List down your
observations below.

ABSTRACTION
Page | 111
lines are one-dimensional, we consider only length when
measuring a line segment. A line segment is measured in
linear units such as inches, centimeters, or yards. When a line
segment measures 5 centimeters, we write AB = 5 cm or
simply AB = 5 (if the units are apparent or are not stated). The
instrument of measure is the ruler. A plane is an infinite two-
dimensional surface. A closed or bounded portion of the plane
is called a region. When a region such as R in plane M [see
Figure 6.3(a)] is measured, we call this measure the “area of
the plane region.” The unit used to measure area is called a
square unit because it is a square with each side of length 1
[see Figure 6.3(b)]. The measure of the area of region R is the
number of non-overlapping square units that can be placed
adjacent to each other in the region. Figure 6.3

Square units (not linear units) are used to measure area. Using an exponent, we write square
inches as in2. The unit represented by Figure 6.3(b) is 1 square inch or 1 in2.

Figure 6.4
We can measure the area of the region within a triangle [see Figure 6.4(b)]. However, we
cannot actually measure the area of the triangle itself (three line segments do not have area).
Nonetheless, the area of the region within a triangle is commonly referred to as the area of
the triangle.

Area Postulate

Corresponding to every bounded region is a unique positive

number A, known as the area of that region.

Figure 6.5

One way to estimate the area of a region is to place it in a grid, as shown in Figure 6.5.
Counting only the number of whole squares inside the region gives an approximation that is
less than the actual area. On the other hand, counting squares that are inside or partially inside
provides an approximation that is greater than the actual area.

Page | 112
To develop another property of area, we consider triangle ABC and DEF (which are
congruent) in Figure 6.6. One triangle can be placed over the other so that they coincide. How
are the areas of the two triangles related? The answer is found in the following postulate.

Figure 6.6

If two closed plane figures are congruent, then their areas are equal.

Area Addition Postulate

Let R and S be two enclosed regions that do not overlap.

Then A = AR + AS

Example 1 Figure 6.7

It is convenient to provide a subscript for A (area) that names the figure whose area is
indicated. The principle used in Example 2 is conveniently and compactly stated in the form
𝐴𝐴𝐵𝐶𝐷𝐸 = 𝐴𝐴𝐵𝐶𝐷 + 𝐴𝐴𝐷𝐸

Page | 113
Area of Rectangle

Rectangle MNPQ in the figure 6.8 has the dimensions of 3

cm and 4 cm. the number of squares, 1 cm on a side, in the
rectangle is 12. Rather than counting the number of squares
in the figure, just multiply the dimension to calculate the
area. the unit of area is 𝑐𝑚2.

Multiplication of dimensions is handled like algebraic Figure 6.8

multiplication. Compare

3x x 4x = 12𝑥2 and 3 cm x 4 cm = 12 𝑐𝑚2

If the units used to measure the dimensions of a region are not the same, then they must be
converted into like units in order to calculate area

The area A of a rectangle whose base has length b and whose altitude has length h is
given by A = bh. It is also common to describe the dimensions of a rectangle as its length l and
its width w. The area of the rectangle is then written A = lw.

Example 2

Find the area of rectangle ABCD in Figure 8.10 if AB =

12 cm and AD = 7 cm.

Solution

Because it makes no difference which dimension is

chosen as base b and which as altitude h, we arbitrarily Figure 6.9
choose AB = b = 12 cm and AD = h _ 7 cm. Then

A = bh
A = 12 cm x 7 cm
A = 84 𝑐𝑚2

If units are not provided for the dimensions of a region, we assume that they are alike. In such
a case, we simply give the area as a number of square units.

Theorem 6.2.1
The area A of a square whose sides are each of length s is given by
A = 𝑠2.

Page | 114
 Area of Parallelogram

A rectangle’s altitude is one of its sides, but that is not true of a parallelogram. An altitude of
a parallelogram is a perpendicular segment from one side to the opposite side, known as the
base. A side may have to be extended in order to show this altitude-base relationship in a
drawing. In Figure 6.10(a), if RS is designated as the base, then any of the segments ZR, VX,
or YS is an altitude corresponding to that base (or, for that matter, to base VT).

Figure 6.10

Another look at parallelogram RSTV [in Figure 6.10(b)] shows that (ST or VR) could just as
well have been chosen as the base. Possible choices for the corresponding altitude in this case
include VH and GS. In the theorem that follows, it is necessary to select a base and an altitude
drawn to that base.

Theorem 6.2.2
The area A of a parallelogram with a base of length b and with corresponding altitude of length h is given by A

Given.

Prove

Proof: as shown in figure 6.11(b)

Figure 6.11 P e | 115

Example 3.

Given that all dimensions in Figure 6.12 are in inches, find

the area parallelogram MNPQ of by using base
a. MN b. PN

Solutions

a. MN_= QP =b = 8, and the corresponding altitude Figure 6.12

is of length QT = h = 5. Then
A = 8 in. x 5 in.
A = 40 𝑖𝑛2
2
b. PN = b = 6, so the corresponding altitude length is MR = h = 6 . then
3
2
A=6x6
3
A=6x 20
3
A = 40 𝑖𝑛2

In Example 3, the area of was not changed when a different base and its
corresponding altitude were used to calculate its area

Example 4

Given: In Figure 6.13, parallelogram with PN = 8 and

QP = 10
Altitude QR to base MN has length QR = 6

Find: SN, the length of the altitude between QM and PN

Figure 6.13
solution.

Choosing MN = b = 10 and QR = h = 6, we see that

A = bh = 10 x 6 = 60

Now we choose PN = b = 8 and SN = h, so A = 8h. Because the area of the

parallelogram is unique, it follows that
8h = 60
h = 60/8 = 7.5
that is, SN = 7.5

Suggested videos:
https://youtu.be/hm17lVaor0Q
https://www.khanacademy.org/math/geometry-home/geometry-area-perimeter/geometry-
parallelogram-area/v/intuition-for-area-of-a-parallelogram

Page | 116
Area of a triangle

The formula used to calculate the area of a triangle follows easily from the formula for
the area of a parallelogram. In the formula, any side of the triangle can be chosen as
its base; however, we must use the length of the corresponding altitude for that base.

Theorem 6.2.3
The area A of a triangle whose base has length b and whose corresponding altitude has length h is given by A

Given: in the figure 6.14(a), triangle ABC with CD perpendicular to AB

AB = b & CD = h

Prove: A = ½ bh

Proof:

In figure 6.14(b)

Figure 6.14

Page | 117
Example 4

In the figure 6.15,

find the area of triangle ABC if AB = 10 cm and CD = 7 cm.

Solution

With AB as base, b = 10 cm.

The corresponding altitude for base AB is CD, so
h = 7 cm. Now Figure 6.15
A =1/2 bh
becomes

A =1/2 x 10 cm x 7 cm
A = 35 cm2

Corollary 6.2.3
The area of a right triangle with legs of lengths a and b is given by
A = 1/2ab.

In the proof of Corollary 6.2.3, the length of either leg can be chosen as the base; in turn, the
length of the altitude to that base is the length of the remaining leg. This follows from the fact
“The legs of a right triangle are perpendicular.”

Given: In figure 6.16

Figure 6.16
Find: Area of triangle MNP

Solution:

Page | 118
If the lengths of the sides of a triangle are known, the formula generally used to calculate the
area is Heron’s Formula (named in honor of Heron of Alexandria, circa A.D. 75). One of
the numbers found in this formula is the semiperimeter of a triangle, which is defined as one-
half the perimeter. For the triangle that has sides of lengths a, b, and c, the semiperimeter is 𝐴
=
√𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐). We apply Heron’s Formula in Example below.

Heron’s Formula
If the three sides of a triangle have lengths a, b, and c, then the area
A of the triangle is given by
𝐴 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐), where the semiperimeter of the triangle is
1
𝑠 = 2 ( 𝑎 + 𝑏 + 𝑐)

Example 5

Find the area of a triangle which has sides of lengths 4, 13, and 15.

solution
If we designate the sides as a = 4, b = 13, and c = 15, then semiperimeter of the triangle is
given by .
1 1
(4 + 13 + 15) (32) = 16
𝑠= 2
Therefore, 2=

𝐴 = √16(16 − 4)(16 − 13)(16 − 15)

𝐴 = √16(12)(3)(1) = √576 = 24 𝑢𝑛𝑖𝑡𝑠2

Suggested Videos:
https://www.khanacademy.org/math/geometry-home/geometry-area-perimeter/geometry-
area-triangle/v/intuition-for-area-of-a-triangle

https://www.khanacademy.org/math/geometry-home/geometry-volume-surface-area/heron-
formula-tutorial/v/heron-s-formula
https://www.khanacademy.org/math/geometry-home/geometry-volume-surface-area/heron-
formula-tutorial/v/part-1-of-proof-of-heron-s-formula
https://www.khanacademy.org/math/geometry-home/geometry-volume-surface-area/heron-
formula-tutorial/v/part-2-of-the-proof-of-heron-s-formula

Page | 119
 APPLICATION

b. Find the area of the following figure.

1. A rectangle’s length is 6 cm, and its width is 9 cm.
2. A right triangle has one leg measuring 20 in. and a hypotenuse measuring
29 in.

3.

4.

Congratulations, you have just completed the lesson 2 of the module 6. You are now ready for
the next lesson which tackles the perimeter and area of circles.

Page | 120
Lesson 3. Circumference and Area of a Circle
Objectives:
At the end of the lesson, you will be able to:
derive the formula of area,
prove theorems involving circumference and area of circle, and
solve problems involving circumference and area of circle.

Duration: 1 hr.
Introduction:
Hello! Welcome to lesson 3 of module 6. In the previous lesson, we discussed the
measurements around the polygons and its area. Now let’s explore the length around the
circle and its area.

ACTIVITY

Materials: string, ruler, pencil, paper.

Procedures:
 Draw 3 circles with different radius
 Trace the circle using the string around the circle.
 Measure the length.
 Repeat the steps to other circles

ANALYSIS

In the activity, compare the length of the circle, how is it related to its diameter. List down
your observations below.

Page | 121
 ABSTRACTION

In geometry, any two figures that have the same shape are
described as similar. For this reason, we say that all circles are
similar to each other. Just as a proportionality exists among the
sides of similar triangles, experimentation shows that there is a
proportionality among the circumferences (distances around)
and diameters (distances across) of circles. Representing the
circumferences of the circles in Figure 6.16 by C1, C2, and C3
and their respective lengths of diameters by d1, d2, and d3, we
claim that
𝐶1 𝐶2 𝐶3
= = =𝑘
𝑑1 𝑑2 𝑑3

where k is the constant of proportionality.

The ratio of the circumference of a circle to the length of its

diameter is a unique positive constant.

The constant of proportionality k described in the

opening paragraph is represented by the Greek letter Figure 6.16
𝜋 (pi).

DEFINITION
𝜋 (pi) is the ratio between the circumference C and the diameter length d of any circle;
𝐶
thus, 𝜋 = in any circle.
𝑑

the lengths of the diameter and radius of the circle are represented by d and r respectively.

Theorem 6.3.1
The circumference of a circle is given by the formula
C = 𝜋𝑑 or C = 2𝜋𝑟.

Given: In figure 6.17, Circle O with length of diameter d and length of

radius r.
Prove: C = 2𝜋𝑟

Proof: Figure 6.17

VALUE
OF 𝜋 (pi)

Page | 122
 In calculating the circumference of a circle, we generally leave the symbol 𝜋 in the answer in
order to state an exact result. However, the value of 𝜋 is irrational and cannot
be represented exactly by a common fraction or by a terminating decimal. When an
approximation is needed for 𝜋, we use a calculator.
22
Approximations of 𝜋 that have been
commonly used throughout history include 𝜋 ≈ , and
7 ,𝜋 ≈
𝜋 ≈ 3.1416. Although these approximate values have been used for
centuries, your calculator provides greater accuracy. A calculator will
show that 𝜋 ≈ 3.141592654.

Example 1
22
Circle O in figure 6.18, OA = 7 cm; using 𝜋 ≈
7 , Figure 6.18
a. find the approximate circumference C of circle O.
b. find the approximate length of the minor arc AB.

Solution

Example 2

The exact circumference of a circle is 17 𝜋 in.

a) Find the length of the radius.

b) Find the length of the diameter.

Solution

a) C = 2 𝜋 r b) Because d = 2r, d = 2(8.5), or d = 17 in.

17 𝜋 = 2 𝜋 r

17𝜋 2𝜋𝑟
2𝜋 = 2𝜋
17
𝑟= = 8.5 𝑖𝑛
2

Page | 123
Example 3

A thin circular rubber gasket is used as a seal to prevent oil from leaking
from a tank (see Figure 6.19). If the gasket has a radius of 2.37 in., use the
value of 𝜋 provided by your calculator to find the circumference of the
gasket to the nearest hundredth of an inch.
Figure 6.19

Solution

Using the calculator with C = 2 𝜋 r, we have C = 2 * 𝜋 * 2.37 in

C ≈ 14.8911491
C ≈ 14.89 in.

Area of Circle

Now consider the problem of finding the area of a circle. To do so, let a
regular polygon of n sides be inscribed in the circle. As we allow n to
grow larger (often written as 𝑛 → ∞ and read “n approaches infinity”),
two observations can be made:

1. The length of an apothem of the regular polygon approaches the

length of a radius of the circle as its limit (𝑎 → 𝑟). Figure 6.20
2. The perimeter of the regular polygon approaches the
circumference of the circle as its limit (𝑃 → 𝐶).

In Figure 6.20, the area of an inscribed regular polygon with n sides approaches the area of
the circle as its limit as n increases. Using observations 1 and 2, we make the following
claim. Because the formula for the area of a regular polygon is
1
𝐴 = 𝑎𝑃
2
the area of the circumscribed circle is given by the limit
1
𝐴 = 𝑟𝐶
Because C = 2 𝜋 r, this formula becomes 2

1
𝐴 = 𝑟(2𝜋𝑟)
2
Or
𝐴 = 𝜋𝑟2

Theorem 6.3.2
The area A of a circle whose radius has length r is given by
𝐴 = 𝜋𝑟2.
Page | 124
Example 4

Find the approximate area of a circle whose radius has a length of 10 in. (use 𝜋 = 3.14)

Solution
𝐴 = 𝜋𝑟2.
𝐴 = (3.14)(102 )
𝐴 = 3.14(100)
𝐴 = 314 𝑖𝑛2.

.
Example 5

The approximate area of a circle is 38.5 cm2. Find the length of the radius of the
22
circle. (use 𝜋 ≈ )
7

solution
By substituting known values, the formula 𝐴 = 𝜋𝑟2 becomes
22 2
38.5 = 𝑟
7
22 2
38.5 = 𝑟
7
Multiplying each side of the equation by 7 , we have
22

Or

49
𝑟2 = 4

Taking the positive square root for the approximate length of radius,

49 √49 7
𝑟=√ = = = 3.5 𝑐𝑚
4 √4 2

A plane figure bounded by concentric circles is known as a ring or Figure 6.21

annulus (see Figure 6.21). The piece of hardware known as a washer has the shape of an
annulus.

Page | 125
Suggested videos:
https://www.khanacademy.org/math/cc-seventh-grade-math/cc-7th-geometry/cc-7th-area-
circumference/v/area-of-a-circle
https://www.khanacademy.org/math/cc-seventh-grade-math/cc-7th-geometry/cc-7th-area-
circumference/v/relating-circumference-and-area?modal=1
https://www.khanacademy.org/math/cc-seventh-grade-math/cc-7th-geometry/cc-7th-area-
circumference/v/finding-radius-and-diameter-from-circumference?modal=1

APPLICATION

Solve the following:

1. Find the circumference and area of a circle whose radius has length 8 cm.
2. Find the approximate circumference and area of a circle whose radius has length 10 1
2
22
in. Use 𝜋 ≈ .
7
3. Find the approximate lengths of a radius and a diameter of a
circle whose circumference is:
22
a. 88 in. (Use 𝜋 ≈ )
7
b. 157 m (Use 𝜋 ≈ 3.14)
4. In figure 6.22, concentric circles with radii of lengths R and
r, where R > r, explain why 𝐴𝑟𝑖𝑛𝑔 = 𝜋(R + r)(R - r).
Figure 6.22

Congratulations, you have just completed the lesson 3 of the module 6. You are now ready for
the next lesson which tackles the basic solids.

Page | 126
Lesson 4. Definitions of Basic Solids
Objectives:
At the end of the lesson, you will be able to:
define different solids, and
illustrate the characteristics of each solids.

Duration: 1 hr.
Introduction:
Hello! Welcome to lesson 4 of module 6. In the previous lesson, we discussed the
measurements of a polygon and circle inside and around them. Now let’s explore the three-
dimensional space.

ACTIVITY

Materials: ruler, pencil, paper.

Procedures:
 Draw two congruent circles and congruent polygon
 In a circle, connect the sides by parallel segment
 In a polygon, connect the corresponding vertices by parallel line

ANALYSIS

In the activity, figure did you created? What is the role of parallel line in producing a unique
figure? List down your observations below.

Page | 127
 ABSTRACTION

Suppose that two congruent polygons lie in parallel planes in such a way that their
corresponding sides are parallel. If the corresponding vertices of these polygons [such as A
and A’ in [Figure 6.23(a)] are joined by line segments, then the “solid” that results is a prism.
The congruent figures that lie in the parallel planes are the bases of the prism. The parallel
planes need not be shown in the drawings of prisms. Suggested by an empty box, the prism is
like a shell that encloses a portion of space by the parts of planes that form the prism; thus, a
prism does not contain interior points. In practice, it is sometimes convenient to call a prism
such as a brick a solid; of course, this interpretation of prism would contain its interior points.

Figure 6.23

In Figure 6.23(a), segments AB, AC, BC, A’B’, A’C’, and B’C’ are base edges, AA’, BB’ and
CC’, and are lateral edges of the prism. Because the lateral edges of this prism are
perpendicular to its base edges, the lateral faces (like quadrilateral ACC’A’) are rectangles.
Points A, B, C, A’, B’, and C’ are the vertices of the prism. In Figure 6.23(b), the lateral edges
of the prism are not perpendicular to its base edges; in this situation, the lateral edges are
often described as oblique (slanted). For the oblique prism, the lateral faces are
parallelograms. Considering the prisms in Figure 6.23, this leads to the following definitions.

DEFINITION

A right prism is a prism in which the lateral edges are perpendicular to the base edges at
their points of intersection. An oblique prism is a prism in which the parallel lateral edges
are oblique to the base edges at their points of intersection.

Part of the description used to classify a prism depends on its base. For instance, the prism in
Figure 6.23(a) is a right triangular prism; in this case, the word right describes
the prism, whereas the word triangular refers to the triangular base. Similarly, the prism
in Figure 6.23(b) is an oblique square prism, if we assume that the bases are squares. Both
prisms in Figure 6.23 have an altitude (a perpendicular segment joining italics on height the
planes that contain the bases) of length h, also known as the height of the prism.

Example 1

Page | 128
Name each type of prism in Figure 6.24

Figure 6.24

Solutions
a. The lateral edges are perpendicular to the base edges of the hexagonal base. The prism is
a right hexagonal prism.
b. The lateral edges are oblique to the base edges of the pentagonal base. The prism is an
oblique pentagonal prism.
c. The lateral edges are perpendicular to the base edges of the triangular base. Because the
base is equilateral, the prism is a right equilateral triangular prism.

Figure 6.25
Figure 6.26

The solids shown in Figure 6.25 are pyramids. In Figure 6.25(a), point A is noncoplanar with
square BCDE. In Figure 6.25(b), F is noncoplanar with triangle GHJ. In these pyramids, the
noncoplanar point has been joined (by drawing line segments) to each vertex of the square
and to each vertex of the triangle, respectively. Every pyramid has exactly one base. Square
BCDE is the base of the first pyramid, and triangle GHJ is the base of the second pyramid.
Point A is known as the vertex of the square pyramid; likewise, point F is the vertex of the
triangular pyramid.

The pyramid in Figure 6.26 is a pentagonal pyramid. It has vertex K, pentagon LMNPQ for
its base, and lateral edges KL, KM, KN KP, and KQ. Although K is called the vertex of the
pyramid, there are actually six vertices: K, L, M, N, P, and Q. The sides of the base LM, MN,
NP, PQ, and QL are base edges. All lateral faces of a pyramid are triangles; triangle KLM is
one of the five lateral faces of the pentagonal pyramid. Including base LMNPQ, this pyramid

Page | 129
has a total of six faces. The altitude of the pyramid of length h, is the line segment from the
vertex K perpendicular to the plane of the base.

DEFINITION

A regular pyramid is a pyramid whose base is a regular polygon and whose lateral edges are
all congruent

Suppose that the pyramid in Figure 6.26 is a regular pentagonal pyramid. Then the lateral
faces are necessarily congruent to each other; by SSS, ∆𝐾𝐿𝑀 ≅ ∆𝐾𝑀𝑁 ≅ ∆𝐾𝑁𝑃 ≅ 𝐾𝑃𝑄 ≅
∆𝐾𝑄𝐿. Each lateral face is an isosceles triangle. In a regular pyramid,
the altitude joins the vertex of the pyramid to the center of the regular polygon that is the base
of the pyramid. The length of the altitude is height h.

DEFINITION

The slant height of a regular pyramid is the altitude from the vertex of the pyramid to the
base of any of the congruent lateral faces of the regular pyramid.

Among pyramids, only a regular pyramid has a slant height.

In our formulas and explanations, we use l to represent the length of the slant height of a
regular pyramid. See figure 6.27(c)

Figure 6.27

Example 2

For a regular square pyramid with height 4 in. and base edges of length 6 in. each, find the
length of the slant height l.

Solution

Page | 130
In Figure 6.27, it can be shown that the apothem to any side has length 3 in. (one-half the
length of the side of the square base). Also, the slant height is the hypotenuse of a right
triangle with legs equal to the lengths of the altitude and the apothem. By the Pythagorean
Theorem, we have
𝑙2 = 𝑎2 + ℎ2
𝑙 2 = 32 + 42
𝑙 2 = 9 + 16
𝑙2 = 25
𝑙 = 5 𝑖𝑛𝑐ℎ𝑒𝑠

Theorem 6.4.1
In a regular pyramid, the length a of the apothem of the base, the altitude h, and the slant height l satisfy the P
ℎ2 in every regular pyramid.

CYLINDER

Consider the solids in Figure 6.28, in which congruent circles lie in parallel planes. For the
circles on the left, suppose that centers O and O’ are joined to form OO’; similarly, suppose
that segments QQ’ joins the centers of the circles on the right. Let segments such as XX’ join
two points of the circles on the left, so that XX’ is parallel to OO’. If all such segments (like
XX’, YY’ and ZZ’) are parallel to each other, then a cylinder is generated. Because OO’ is not
perpendicular to planes P and P’, the solid on the left is an oblique circular cylinder. With
QQ’ perpendicular to planes P and P’, the solid on the right is a right circular cylinder. For
both cylinders, the distance h between the planes P and P’ is the length of the altitude of the
cylinder; h is also called the height of the cylinder. The congruent circles are known as the
bases of each cylinder.

Figure 6.28

Page | 131
A right circular cylinder is shown in Figure 6.29; however,
the parallel planes (such as P and P’ in Figure 6.28) are not
pictured. The line segment joining the centers of the two
circular bases is known as the axis of the cylinder. For a
right circular cylinder, it is necessary that the axis be
perpendicular to the planes of the circular bases; in such a
case, the length of the altitude h is the length of the axis.

Figure 6.29

POLYHEDRONS

When two planes intersect, the angle formed by two half-

planes with a common edge (the line of intersection) is a
dihedral angle. The angle shown in Figure 6.30 is such an
angle. In Figure 9.30, the measure of the dihedral angle is the
same as that of the angle determined by two rays that

1. have a vertex (the common endpoint) on the edge.

2. lie in the planes so that they are perpendicular to
the edge.
Figure 6.30

A polyhedron (plural polyhedrons or polyhedra) is a solid bounded by plane regions.

Polygons form the faces of the solid, and the segments common to these polygons are the
edges of the polyhedron. Endpoints of the edges are the vertices of the polyhedron. When a
polyhedron is convex, each face determines a plane for which all remaining faces lie on the
same side of that plane. Figure 6.31(a) illustrates a
convex polyhedron, and Figure
6.31(b) illustrates a concave
polyhedron; as shown in Figure
6.31(b), a line segment
containing the two uppermost
vertices lies in the exterior of
the concave polyhedron

Figure 6.31
The prisms and pyramids
discussed earlier were special types of polyhedrons. For instance, a pentagonal pyramid can
be described as a hexahedron because it has six faces. Because some of their surfaces do not
lie in planes, the cylinders and cones are not polyhedrons.

Leonhard Euler (Swiss, 1707–1763) found that the number of vertices, edges, and faces of
any polyhedron are related by Euler’s equation. This equation is given in the following
theorem, which is stated without proof.

Page | 132
 Euler’s Equation
The number of vertices V, the number of edges E, and the number of faces F of a polyhedron are related by th
V+F=E+2

Example 3

Verify Euler’s equation for the (a) tetrahedron and (b) square pyramid
shown in
Figure 6.32.

Solution

a. The tetrahedron has four vertices (V = 4), six edges (E = 6), and
four faces (F = 4). So Euler’s equation becomes 4 + 4 = 6 + 2,
which is true.
b. The pyramid has five vertices (“vertex” + vertices from the base),
eight edges (4 base edges + 4 lateral edges), and five faces (4
triangular faces 1 square base). Now V + F = E + 2 becomes 5 + 5 =
8 + 2, which is also true. Figure 6.32

DEFINITION

A regular polyhedron is a convex polyhedron whose faces are congruent regular polygons
arranged in such a way that adjacent faces form congruent dihedral angles.

There are exactly five regular polyhedrons, named as follows:

1. Regular tetrahedron, which has 4 faces (congruent equilateral triangles)

2. Regular hexahedron (or cube), which has 6 faces (congruent squares)
3. Regular octahedron, which has 8 faces (congruent equilateral triangles)
4. Regular dodecahedron, which has 12 faces (congruent regular pentagons)
5. Regular icosahedron, which has 20 faces (congruent equilateral

triangles) CONES

Figure 6.33

Page | 133
In Figure 6.34, consider point P, which lies outside the plane
containing circle O. A surface known as a cone results when line
segments are drawn from P to points on the circle. However, if P is
joined to all possible points on the circle as well as to points in the
interior of the circle, a solid is formed. If segment PO is not
perpendicular to the plane of circle O in Figure 6.34, the cone is an
oblique circular cone.

In Figures 6.34 and 6.35, point P is the vertex of the cone, and
circle O is the base. The segment PO, which joins the vertex to the
center of the circular base, is the axis of the cone. If the axis is
perpendicular to the plane containing the base, as in Figure 5.35, the Figure 6.34
cone is a right circular cone. In any cone, the perpendicular segment
from the vertex to the plane of the base is the altitude of the cone. In
a right circular cone, the length h of the altitude equals the length of
the axis. For a right circular cone, and only for this type of cone, any
line segment that joins the vertex to a point on the circle is a slant
height of the cone; we will denote the length of the slant height by l
as shown in Figure 6.35.

Figure 6.35

SPHERE

Another type of solid with which you are familiar is the sphere. Although the surface of a
basketball correctly depicts the sphere, we often use the term sphere to refer to a solid like a
baseball as well. The sphere has point symmetry about its center. It is also defined as a locus
of points.

In space, the sphere is characterized in three ways:

1. A sphere is the locus of points at a fixed distance r from a given point O. Point O is
known as the center of the sphere, even though it is not a part of the spherical surface.

2. A sphere is the surface determined when a circle (or semicircle) is rotated about
any of its diameters.
3. A sphere is the surface that represents the theoretical limit of an “inscribed” regular
polyhedron whose number of faces increases without limit.

Page | 134
In Figure 6.32, a sphere was generated as the locus of points in
space at a distance r from point O. The line segment OP is a
radius of sphere O, and QP is a diameter of the sphere. The
intersection of a sphere and a plane that contains its center is a
great circle of the sphere. For the earth, the equator is a great
circle that separates the earth into two hemispheres.

Figure 6.32

APPLICATION

Solve the following:

1. A regular polyhedron has 12 edges and 8 vertices.

a. Use Euler’s equation to find the number of faces.
b. Use the result from part (a) to name the regular polyhedron.

2. Consider the solid shown.

a. Does it appear to be a prism?
b. Is it right or oblique?
c. What type of base(s) does the solid have?
d. Name the type of solid. for item # 2
e. What type of figure is each lateral face?

3. In the solid shown, base ABCD is a square.

a. Is the solid a prism or a pyramid?
b. Name the vertex of the pyramid. for item # 3
c. Name the lateral edges.
d. Name the lateral faces.
e. Is the solid a regular square pyramid?

Congratulations, you have just completed the lesson 4 of module 6. You are now ready for the
next lesson which tackles the surface area of solids.

Page | 135
Lesson 5. Surface Area of Basic Solids
Objectives:
At the end of the lesson, you will be able to:
describe the characteristics of different surface area of a solid,
derive the formula in solving surface area of a solid, and
solve problems involving surface area.

Duration: 1 hr.
Introduction:
Hello! Welcome to lesson 5 of module 6. In the previous lesson, we discussed and
define different solid. Now let’s explore the facial area of different solids.

ACTIVITY

Materials: illustration board, packaging tape, ruler, pencil, paper.

Procedures:
 In the illustration board, draw 6 pieces of squares with equal dimension
 Cut out the squares from the illustration board
 Find the area of its square
 Using a packaging tape, connect the sides of the squares, this leads to
creation of a cube.
 Add up the area of a square found in step 3.

ANALYSIS

In the activity, what is the dimension of your square? What is the area of each square? What
is the total area of all squares? How is it related to square’s dimensions? List down your
observations below.

Page | 136
 ABSTRACTION

DEFINITION

The lateral area L of a prism is the sum of the areas of all

lateral faces.

In the right triangular prism of Figure 6.33, a, b, and c are

the lengths of the sides of either base. These dimensions are
used along with the length of the altitude (denoted by h) to
calculate the lateral area, the sum of the areas of rectangles
ACC’A’, ABB’A’, and BCC’B’. The lateral area L of the
right triangular prism can be found as follows:

𝐿 = 𝑎ℎ + 𝑏ℎ + 𝑐ℎ Figure 6.33
𝐿 = ℎ (𝑎 + 𝑏 + 𝑐 )
𝐿 = ℎ𝑃

where P is the perimeter of a base of the prism. This formula, 𝐿 = ℎ𝑃, is valid for finding
the lateral area of any right prism. Although lateral faces of an oblique prism are
parallelograms, the formula L = hP is also used to find its lateral area.

TheTheorem
lateral area
6.5.1L of any prism whose altitude has measure h and whose base has perimeter P is given by L =

Example 1.

The bases of the right prism shown in Figure 6.34 are equilateral
pentagons with sides of length 3 in. each. If the altitude measures 4 in.,
find the lateral area of the prism.

Solution

Each lateral face is a rectangle with dimensions 3 in. by 4 in. The area
of each rectangular face is 3 𝑖𝑛 𝑥 4 𝑖𝑛 = 12 𝑖𝑛2. Because there are five Figure 6.34
congruent lateral faces, the lateral area of the pentagonal prism is
5 𝑥 12 𝑖𝑛2 = 60 𝑖𝑛2.

If we apply the L = hP in Example 1, this leads to

𝐿 = 4 𝑖𝑛 𝑥 15 𝑖𝑛
𝐿 = 60 𝑖𝑛2

Page | 137
DEFINITION

For any prism, the total area T is the sum of the lateral area and the areas of the bases.

The total area of the prism is also known as its surface area.

Both bases and lateral faces are known as faces of a prism.

Thus, the total area T of the prism is the sum of the areas of
all its faces.
Recalling Heron’s Formula, we know that the base area B of
the right triangular prism in Figure 6.35 can be found by the
formula

𝐵 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) Figure 6.35

in which s is the semiperimeter of the triangular

base

Example 2

Find the surface area of the right triangular prism with an altitude of length 8 in. if the
sides of the triangular bases have lengths of 13 in., 14 in., and 15 in.

solution

The lateral area is found by adding the areas of the three rectangular lateral faces. That is,

𝐿 = 8(13 + 14 + 15)
𝐿 = 8 𝑖𝑛 (42 𝑖𝑛)
𝐿 = 336 𝑖𝑛2
1
We use Heron’s Formula to find the area of each base. Where 𝑠 = (13 + 14 + 15) or s =
2
21, 𝐵 = √21(21 − 13)(21 − 14)(21 − 15), 𝐵 = √21(8)(7)(6),
𝐵 = √7056, 𝐵 = 84
Calculating the total area (or surface area) of the triangular prism, we have
𝑇 = 336 + 2(84), 𝑜𝑟 𝑇 = 504 𝑖𝑛2

The total area

Theorem T of any prism with lateral area L and base area B is given by T = L + 2B.
6.5.2

Page | 138
GIVEN: The pentagonal prism of
Figure 6.36

PROVE:
T = L + 2B

PROOF:
When the prism is “taken apart” and
Figure 6.36
laid flat, as shown in Figure 6.36(b),
we see that the total area depends
upon the lateral area (shaded darker)
and the areas of the two bases; that is,
T = L + 2B

DEFINITION

A regular prism is a right prism whose bases are regular polygons

Henceforth, the prism in Figure 6.24(c) in lesson 4 will be called a regular triangular prism.
In the next example, the base of the prism is a regular hexagon. Because the prism is a right
prism, the lateral faces are congruent rectangles.

Example 3

Find the lateral area L and

the surface area T of the
regular hexagonal prism in
Figure 6.37(a).

Figure 6.37
Solution

In Figure 6.37(a), there are six congruent lateral faces, each rectangular and with dimensions
of 4 in. by 10 in. Then

𝐿 = 6(4𝑥10), 𝐿 = 240 𝑖𝑛2

For the regular hexagonal base [see Figure 6.37(b)], the apothem measures 𝑎 = 2√3 𝑖𝑛.
and the perimeter is 24 in. Then the area B of each base is given by the formula for the area of
a regular polygon.

Page | 139
 1 1
𝐵 = 𝑎𝑃, (2√3)(24), 𝐵 = 24√3, 𝐵 ≈ 41.57 𝑖𝑛
𝐵= 2
2 total Area is
Now the

𝑇 = 𝐿 + 2𝐵, 𝑇 = 240 + 2(24√3), 𝑇 = 48√3, 𝑇 ≈ 323. 14 𝑖𝑛2

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SURFACE AREA OF A PYRAMID

To lay the groundwork for the next theorem, we justify the

result by “taking apart” one of the regular pyramids and
laying it out flat. Although we use a regular hexagonal
pyramid for this purpose, the argument is similar if the base
is any regular polygon. When the lateral faces of the regular
pyramid are folded down into the plane, as shown in Figure
6.38, the shaded lateral area is the sum of the areas of the
triangular
lateral faces. Using 𝐴 = 1 𝑏ℎ , we find that the area of each
2
triangular face is 1 ∗ 𝑠 ∗ 𝑙 (each side of the base of the pyramid
2 Figure 6.38
has length s, and the slant height has length). The combined
areas of the triangles give the lateral area. Because there are n
triangles

1
𝐿 =𝑛 ∗ ∗ 𝑠∗ 𝑙
2
1
𝐿 = 𝑙(𝑛 ∗ 𝑠)
21
𝐿 = 𝑙𝑃
2
Where P is the perimeter of the base

Theorem 6.5.3The lateral area L of a regular pyramid with slant height of length and perimeter P of the base is given by
1
𝐿 = 2 𝑙𝑃

Page | 140
Example 4

Find the lateral area of a regular

pentagonal pyramid if the sides of the
base measure 8 cm and the lateral
edges measure 10 cm each [see
Figure 6.39(a)].

Figure 6.39
Solution

For the triangular lateral face [see Figure 9.18(b)], the slant height bisects the base edge as
indicated. Applying the Pythagorean Theorem, we have

42 + 𝑙2 = 102, 𝑠𝑜 16 + 𝑙2 = 100, 𝑙2 = 84, 𝑙 = √84,

𝑙 = √84 = √4 ∗ 21 = √4 ∗ √21 = 2√21

1 1 1
Now, 𝐿 = 𝑙𝑃 becomes 𝐿 = (2√21)(5 ∗ 8) = 2√21(40),
2 2 2
𝐿 = 40√21 ≈ 183.30 𝑐𝑚2

It may be easier to find the lateral area of a regular pyramid without using the formula of
Theorem 6.5.3; simply find the area of one lateral face and multiply by the number of faces.

Theorem
The 6.5.4
total area (surface area) T of a pyramid with lateral area L and base area B is given by T = L + B.

The formula for the total area T of the pyramid can be written T = 1 𝑙𝑃 + B.
2

Example 5

Find the total area of a regular square pyramid

that has base edges of length 4 ft and
lateral edges of length 6 ft. [See Figure
6.40(a).]

Solution
Figure 6.40
To determine the lateral area, we need the
length of the slant height. [See Figure 6.40(b)]

Page | 141
𝑙 2 + 22 = 62 , 𝑜𝑟 𝑙 2 + 4 = 36, → 𝑙 2 = 32, 𝑙 = √32, 𝑙 = √16 ∗ 2 → 𝑙 = 4√2

The lateral area is

𝐿 = 𝑃𝑙 . Therefore,
1
2

1
𝐿= 4√2(16), 𝑜𝑟 32√2 𝑓𝑡2
2
Because the area of the square base is or 𝐵 = 42 , 𝑜𝑟 16 𝑓𝑡2 the total area is
𝑇 = 16 + 32√2 ≈ 61.25 𝑓𝑡2

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SURFACE AREA OF CONE

Recall now that the lateral area for a regular pyramid is given by 𝐿 =
1
𝑙𝑃. For a right circular cone, consider an inscribed regular pyramid
2
as in Figure 6.41. As the number of sides of the inscribed polygon’s
base grows larger, the perimeter of the inscribed polygon approaches
the circumference of the circle as a limit. In addition, the slant height
of the congruent triangular faces approaches that of the slant height of
the cone. Thus, the lateral area of the right circular cone can be
1
compared to 𝐿 = 𝑙𝑃 ; for the cone, we have
2

1 Figure 6.41
𝐿= 𝐶𝑙
2
in which C is the circumference of the base. The fact that 𝐶 = 2𝜋𝑟 leads to
1
𝐿 = 𝑙(2𝜋𝑟)
2
So
𝐿 = 𝜋𝑟𝑙

Theorem 6.5.5
The lateral area L of a right circular cone with slant height of
length and circumference C of the base is given by 𝐿 = 1 𝑙𝐶.
2
Alternative Form: Where r is the length of the radius of the base, L = 𝜋rl.

Page | 142
Example 6

For the right circular cone in which r = 3 cm and h = 6 cm (see Figure

6.42), find the

a. exact and approximate lateral area L.

b. exact and approximate total area T.

Figure 6.42
Solution
a. We need the length of the slant height h for each problem part, so we apply the
Pythagorean Theorem:

𝑙 2 = 𝑟 2 + ℎ2
𝑙 2 = 32 + 62
𝑙2 = 9 + 36
𝑙2 = 45
𝑙 = √45 → 𝑙 = √9 ∗ 5 → 𝑙 = √9 ∗ √5 → 𝑙 = 3√5

Using L = 𝜋rl., we have

L = 𝜋 ∗ 3 ∗.3√5
L = 9𝜋√5 𝑐𝑚2 → 𝐿 ≈ 63.22 𝑐𝑚2

b. We also have T = B + L

𝑇 = 𝜋𝑟2 + 𝜋𝑟𝑙
𝑇 = 𝜋32 + 𝜋 ∗ 3 ∗ 3√5
𝑇 = 9𝜋 + 9𝜋√5 → 𝑇 ≈ 91.50 𝑐𝑚2

InTheorem 6.5.6 cone, the lengths of the radius r (of the base), the altitude h, and the slant height satisfy the
a right circular
that is, 𝑙2 = 𝑟2 + ℎ2in every right circular cone.

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Page | 143
SURFACE AREA OF CYLINDER

Think of the aluminum can pictured as a right circular

cylinder. The cylinder’s circular bases are the lid and
bottom of the can, and the lateral surface is the “label” of
the can. If the label were sliced downward by a
perpendicular line between the planes, removed, and
rolled out flat, it would be rectangular in shape. As
shown in figure 6.43, that rectangle would have a length Figure 6.43
equal to the circumference of the circular base and a
width equal to the height of the cylinder. Thus,
the lateral area is given by
A = bh, which becomes L = Ch, or L = 2𝜋rh.

The formula for the lateral area of a right circular cylinder

should be compared to the formula L=hP, the lateral area of a right prism whose base has
perimeter P.

Theorem
The 6.5.7L of a right circular cylinder with altitude of length
lateral area
h and circumference C of the base is given by L = hC.
Alternative Form: The lateral area of the right circular cylinder can be expressed in the form L = 2𝜋rh, where

Theorem 6.5.8
The total area T of a right circular cylinder with base area B and lateral area L is given by T = L + 2B.
Alternative Form: Where r is the length of the radius of the base and h is the length of the altitude of the cylin

Example 6

For the right circular cylinder shown in Figure 6.44,

find the

a. exact lateral area L.

b. exact surface area T.

Figure 6.44

Page | 144
Solution:

a. L = 2𝜋rh b. T = L + 2B
L = 2(𝜋)(5)(12) T = 2𝜋rh + 2𝜋𝑟 2.
L = 120 𝜋 𝑖𝑛2 T = 2*𝜋*5*12 + 2*𝜋 ∗ 52
T = 120 𝜋 + 50 𝜋
T = 170 𝜋

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SURFACE AREA OF SPHERE

The next theorem claims that the surface area of a sphere equals four times the area of a great
circle of that sphere. This theorem, which is proved in calculus, treats the sphere as a surface
of revolution.

Theorem 6.5.9
The surface area S of a sphere whose radius has length r is given by
S = 4𝜋 𝑟2.

Example 7

Find the surface area of a sphere whose radius is r = 7 in. Use your calculator to approximate
the result.

Solution
S = 4𝜋 72 → 𝑆 = 196𝜋 𝑖𝑛2

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Page | 145
 APPLICATION

Solve the following:

1. In the accompanying regular pentagonal prism,

suppose that each base edge measures 6 in.
and that the apothem of the base measures 4.1
in. The altitude of the prism measures 10 in.
a. Find the lateral area of the prism.
b. Find the total area of the prism.
for item # 1

2. Suppose that the base of the hexagonal pyramid

in Exercise 6 has an area of 41.6 cm2 and that
each lateral face has an area of 20 cm2. Find for item # 2
the total (surface) area of the pyramid.

3. Suppose that and in the right circular cylinder.

Find the exact and approximate

a. lateral area.
b. total area.

Congratulations, you have just completed the lesson 5 of module. You are now ready for the
next lesson which tackles the volume of solids.

Page | 146
Lesson 5. Volume of a Solid
Objectives:

At the end of the lesson, you will be able to:

derive the formula of volume of different solids,
prove theorems involving volumes of a solid, and
use theorems in solve problems involving volume of a solid.
Duration: 1 hr.

Introduction:
Hello! Welcome to the last lesson of module 6. In the previous lessons, we discussed
lateral and surface area of different solid. Now let’s explore the capacity of different solids.

ACTIVITY

Figure 6.45

In figure 6.45, compare figure a to figure b.

ANALYSIS

In the activity, is figure a related to figure b? why? How may cubes are there in figure a? is it
equal to the number of cubes in figure b? why?

ABSTRACTION

Page | 147
To introduce the notion of volume, we realize that a prism encloses a portion of space.
Without a formal definition, we say that volume is a number that measures the amount
of enclosed space. To begin, we need a unit for measuring volume. Just
as the meter can be used to measure length and the square yard can be
used to measure area, a cubic unit is used to measure the amount of
space enclosed within a bounded region of space.

DEFINITION
A cube is a right square prism whose edges are congruent (see figure
6.46)
Figure 6.46
The volume enclosed by the cube shown in Figure 6.46 is 1
cubic inch or 1 𝑖𝑛2. The volume of a solid is the number of
cubic units within the solid. Thus, we assume that the volume
of any solid is a positive number of cubic units.

Figure 6.47

Volume Postulate
Corresponding to every solid is a unique positive number V
known as the volume of that solid.
The simplest figure for which we can determine volume is the right rectangular prism.
Such a solid might be described as a parallelpiped or as a “box.” Because boxes
are used as containers for storage and shipping (such as a boxcar), it is important to calculate
volume as a measure of capacity. Aright rectangular prism is shown in Figure 6.47; its
dimensions are length l, width w, and height (or altitude) h.

In the activity, the volume of a right rectangular prism in figure 6.46(a) with the length of 4
in., width 3 in., and height 2 in. is easily shown to be 24 𝑖𝑛 3. The volume is the product of the
three dimensions of the given solid. We see not only that 4*3*2 = 24 but also that the units of
volume are in*in*in = 𝑖𝑛 3.

Volume of Prism
The volume of a right rectangular prism is given by
V = lwh
Where l measures the length, w the width, and h the altitude of the prism.

In order to apply the formula, the units used for dimensions l, w, and h must be alike.

Page | 148
Example 1

Find the volume of a box whose dimensions are 1 ft, 8 in., and
10 in. (see figure 6.48)

Solution: Figure 6.48

Although it makes no difference which dimension is chosen for or w or h, it is most important

that the units of measure be the same. Thus, 1 ft is replaced by 12 in. in the formula for
volume.

V = lwh
V = 12 in * 8 in * 10 in
V = 960 𝑖𝑛3

Note that the formula for the volume of the right rectangular prism V = lwh could be replaced
by the formula V = Bh, where B is the area of the base of the prism; that is B = lw. This
volume relationship is true for right prisms in general.

In the next example, we use the fact that 1 𝑦𝑑3 = 27 𝑓𝑡 3. In the cube
shown in Figure 6.49, each dimension measures 1 yd, or 3 ft. The cube’s
volume is given by 1 yd * 1 yd * 1 yd = 1 𝑦𝑑3 or 3 ft * 3 ft 8 3 ft = 27
𝑓𝑡 3.

Example 2
Figure 6.49
Ms. Ivana is having a concrete driveway poured at her house. The section
to be poured is rectangular, measuring 12 ft by 40 ft, and is 4 in. deep. How many cubic yards
of concrete are needed?

Solution:
Using V = lwh, we must be consistent with units. Thus, l = 12 ft, w = 40 ft, and h = 1 ft (from
3
4 in.). Now

V = 12 ft * 40 ft * 1 ft
3
V = 160 𝑓𝑡3
25
To change 160 𝑓𝑡3 to cubic yards, we divide by 27 to obtain 5 𝑦𝑑3
27

Thus, Ms. Ivana will be charged for 6 𝑦𝑑3 of concrete, the result of rounding upward

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Page | 149
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Page | 150
VOLUME OF A PYRAMID

There are kits that contain a hollow pyramid and a hollow prism that have congruent bases
and the same altitude. Using a kit, fill the pyramid with water and then empty the water into
the prism.
a) How many times did you have to empty the pyramid in order to fill the prism?
b) As a fraction, the volume of the pyramid is what part of the volume of the prism?

Theorem 6.6.1
The volume V of a pyramid having a base area B and an altitude of length h is given by
V = 1/3 Bh

Example 3

Find the volume of the regular square pyramid with height h = 4 in and base
edges of length s = 6 in

solution

the area of the square base is B = (6 𝑖𝑛)2 𝑜𝑟 36 𝑖𝑛2. Because h = 4 in, the formula
1
𝑉 = (36 𝑖𝑛2)(4 𝑖𝑛) → 𝑉 = 48 𝑖𝑛3 3

In a Theorem 6.6.2 the lengths of altitude h, radius r of the base, and lateral edge e satisfy the Pythagorean T
regular pyramid,
that is, 𝑒2 = ℎ2 + 𝑟2

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Page | 151
VOLUME OF A CYLINDER

In considering the volume of a right circular cylinder, recall

that the volume of a prism is given by V = Bh, where B is the
area of the base. In Figure 6.50, we inscribe a prism in the
cylinder as shown. Suppose that the prism is regular and that
the number of sides in the inscribed polygon’s base becomes
larger and larger; thus, the base approaches a circle in this
limiting process. The area of the polygonal base also
approaches the area of the circle, and the volume of the prism
approaches that of the right circular cylinder

Figure 6.50
Theorem 6.6.3
The volume V of a right circular cylinder with base area B and altitude of length h is given by V = Bh.
Alternative Form: Where r is the length of the radius of the base, the volume for the right circular cylinder c

Example 4
If d = 4 cm and h = 3.5 cm, find the approximate volume of the right
circular cylinder as shown in Figure 6.51. Give your answer in two
decimal places.

Solution

since d = 4, so r = 2. Thus, V = Bh or 𝜋𝑟2ℎ becomes

𝑉 = 𝜋𝑟2ℎ Figure 6.51

𝑉 = 𝜋22 (3.5)

𝑉 = 𝜋(4)(3.5)
𝑉 = 14 𝜋 ≈ 43.98 𝑐𝑚3

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Page | 152
VOLUME OF CONE

Recall that the volume of a pyramid is given by the formula V = 1/3Bh.

Consider a regular pyramid inscribed in a right circular cone. If its
number of sides increases indefinitely, the volume of the pyramid
approaches that of the right circular cone (see figure 6.52).
Then the volume of the right circular cone is V = 1/3Bh. Because the
area of the base of the cone is B = 𝜋𝑟 2 , as an alternative formula for the
volume of cone is
Figure 6.52
1
𝑉= 𝜋𝑟 ℎ 3
2

Theorem 6.6.4
The volume V of a right circular cone with base area B and altitude of length h is given by V = 1/3Bh
Alternative Form: Where r is the length of the radius of the base, the formula for the volume of the cone is u
1

𝑉 =𝜋𝑟2ℎ
3

NOTE:
The formulas that contain the slant height are used only with the regular pyramid and the
right circular cone (see table in the next page)

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Page | 153
VOLUME OF A SPHERE

In the third description of the sphere enables us to find its

volume. To accomplish this, we treat the sphere as the
theoretical limit of an inscribed regular polyhedron whose
number of faces n increases without limit. The polyhedron
can be separated into n pyramids; the center of the sphere is
the vertex of each pyramid. As n increases, the altitude of
each pyramid approaches the radius of the sphere in length.
Next, we find the sum of the volumes of these pyramids,
the limit of which is the volume of the sphere.
In Figure 6.52, one of the pyramids described in the
preceding paragraph is shown. We designate the height of
each and every pyramid by h. Where the areas of the bases
Figure 6.52
of the pyramids are written 𝐵1, 𝐵2, 𝐵3 , and so on, the sum of
the volumes of the n pyramids forming the polyhedron is

1 1 1 1
𝐵1ℎ + 𝐵2ℎ + 𝐵3ℎ + ⋯ + 𝐵𝑛ℎ
3 3 3 3
Next, we write the volume in polyhydron form

1
ℎ(𝐵1 + 𝐵2 + 𝐵3 + ⋯ + 𝐵𝑛 )
3
As n increases, ℎ → 𝑟 and 𝐵1 + 𝐵2 + 𝐵3 + ⋯ + 𝐵𝑛 → 𝑆, the surface area of a sphere.
Because the surface area of a sphere is S = 4𝜋𝑟 2, the sum approaches the following limit as
the volume of the sphere
1 4 3
1 ℎ (𝐵 + 𝐵 + 𝐵 ) 1 𝑟(4𝜋𝑟 2)
= 𝜋𝑟
+⋯+𝐵
3 1
3
2 𝑛 → 𝑟𝑆 𝑜𝑟
3 3 3

Theorem 6.6.5
The volume V of a sphere with a radius of length r is given by
4
𝑉 =𝜋𝑟3
3

Example 5

Find the exact volume of a sphere whose length of radius is 1.5 in.

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Solutions

This calculation can be done more easily if we replace 1.5 by 3

2

4
𝑉= 𝜋𝑟3
3
4 3 3
3 ∗ ∗
2 2
𝑉= ∗𝜋∗
3 3
2 ∗
2
4
9
𝑉= ∗𝜋∗
3
4

3∗ 9
3 ∗𝜋 𝜋 𝑖𝑛3
𝑉= 2
2 →

Suggested Videos:
https://youtu.be/7vojilWXkMI
https://youtu.be/xuPl_8o_j7k

APPLICATION

Solve the following:

3
1. A hollow steel door is 32 in. wide by 80 in. tall by 1 in. thick. How many cubic
8
inches of foam insulation are needed to fill the door?
2. In the pentagonal pyramid, suppose that each base edge measures 9.2 cm and that the
apothem of the base measures 6.3 cm. The altitude of the pyramid measures 14.6 cm.
a. Find the base area of the pyramid.
b. Find the volume of the pyramid.
3. If the exact volume of a right circular cylinder is 200𝜋 𝑐𝑚3 and its altitude measures 8
cm, what is the measure of the radius of the circular base?
4. A sphere has a volume equal to 99 𝑖𝑛3. Determine the length of the radius of the sphere
7
22
(let 𝜋 ≈ )
7

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Congratulations, you have just completed the lesson 6 of module 6. You are now ready for the
next module assessment in the next page.

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Module Assessment

1. Solve the following

a. The base of a right prism is a triangle whose sides measure 7 cm, 8 cm, and 12
cm. The altitude of the prism measures 11 cm. Calculate the lateral area of the
right prism.
b. Find the volume of cement used in the block shown below

c. A spherical storage tank filled with water has a radius of 10 ft. Use the
calculator’s stored value of _ to find to nearest tenth of unit the approximate
a. surface area of the sphere.
b. volume of the sphere.

d. Recall Euler’s Formula. For a certain polyhedron, there are eight faces and six
vertices. How many edges does it have?

2. Solve and Explain if the radius of one sphere is three times as long as the radius of
another sphere, how do the surface areas of the spheres compare? How do the
volumes compare?

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