LEARNING ACTIVITY

SHEETS (LAS)

MATHEMATICS 9

QUARTER 3
 DIVISION LAS DEVELOPMENT TEAM FOR

MATHEMATICS 9

Norma B. Samantela, CESO VI Schools Division Superintendent

Ma. Jeany T. Abayon, Ed.D. Assistant Schools Division Superintendent

Fatima D. Buen, CESO VI Assistant Schools Division Superintendent

DIVISION EVALUATION TEAM

CID Chief, Sancita B. Peñarubia

PSDS/Division Mathematics Coordinator, Dr. Felma A. Bonito

WRITERS:

Quarter 3 Quarter 4

Jomar A. Monforte Erlinda B. Lominario

Ma. Lyka M. Naag Victor Julius C. Perez

Cristine R. Opiana Ma. Jessa V. Cayago

Ivy A. Morco Ryan R. Raguero

Jay-ar D. Llave Ma. Donna S. Velasco

Fidelito M. Samper Mary Joy R. Pastoriza

Anabelle S. Sabaybay Angelica Q. Lovendino

Mark Anthony A. Mapula Stephanie P. Chu

Jeffrey P. Bobis

TEAM LEADER

Jennie R. Tantiado

CONTENT EDITOR

Jonathan Oscar G. Jimenez

 TABLE OF CONTENTS
GRADE 9 LAS (Learning Activity Sheets)

Week Page
Competency
No. No.

1) Determines the conditions that make a quadrilateral

1-4
a parallelograms
1
2) Uses properties to find measures of angles, sides and
5-7
other quantities involving parallelogram

Proves theorems on the different kinds of parallelogram

2 8 - 13
(rectangle, rhombus, square)

1) Proves the Midline Theorem 14 - 17

3
2) Proves theorems on trapezoids and kites 18 - 26

Solves problems involving parallelograms, trapezoids

4 27 - 30
and kites

1) Describes a proportion
5 31 - 34
2) Applies the fundamental theorems of proportionality
to solve problems involving proportion

6 Illustrates similarity of figures 35 - 37

Proves the conditions for similarity of triangles

1.1 SAS Similarity Theorem
1.2 SSS Similarity Theorem
7 38 - 45
1.3 AA Similarity Theorem
1.4 Right Triangle Similarity Theorem
1.5 Special Right Triangle Theorems
1) Applies the theorems to show that given triangles are
46 - 49
similar
8
2) Proves the Pythagorean Theorem 50 - 53

Solves problems that involve triangle similarity and right

9 54 - 58
triangles
 MATHEMATICS 9

Name: _____________________________________ Grade & Section: _____________Date:_______

LEARNING ACTIVITY SHEET

DETERMINING THE CONDITIONS WHICH GUARANTEE THAT A QUADRILATERAL

IS A PARALLELOGRAM

I. Background Information

This activity sheet shall serve as your guide to facilitate learning as it focuses on determining
the conditions that will justify that a quadrilateral is a parallelogram.

A quadrilateral is a polygon with four sides. It can be classified into different forms: trapezoid,
kite, parallelogram, rectangle, rhombus, and square.

Parallelogram is a
quadrilateral that has
two pairs of parallel
In quadrilateral ABCD:
sides. In parallelogram
ABCD, ̅̅̅̅
𝐴𝐵||𝐷𝐶̅̅̅̅ &
̅̅̅̅ ̅̅̅̅ .
𝐴𝐷||𝐵𝐶

a) the pairs of opposite sides are ̅̅̅̅

𝐴𝐵 𝑎𝑛𝑑 ̅̅̅̅
𝐷𝐶 , ̅̅̅̅
𝐵𝐶 𝑎𝑛𝑑 ̅̅̅̅
𝐴𝐷
Opposite sides of quadrilaterals are sides which do not share
As shown in the given
a common vertex. map of the family of
b) the pairs of opposite angles are ∠𝐵𝐴𝐷 𝑎𝑛𝑑 ∠𝐷𝐶𝐵, ∠𝐴𝐷𝐶 𝑎𝑛𝑑 ∠𝐶𝐵𝐴 quadrilaterals, there are
quadrilaterals that are
Opposite angles of quadrilaterals are angles which do not share
considered as special
a common side. parallelograms, namely:
c) the pairs of consecutive angles are ∠𝐴𝐵𝐶 𝑎𝑛𝑑 ∠𝐵𝐶𝐷, ∠𝐵𝐶𝐷 𝑎𝑛𝑑 ∠𝐶𝐷𝐴, rectangle, rhombus
and square.
∠𝐶𝐷𝐴 𝑎𝑛𝑑 ∠𝐷𝐴𝐵, ∠𝐷𝐴𝐵 𝑎𝑛𝑑 ∠𝐴𝐵𝐶
Consecutive angles are angles that share a common side.
d) the diagonals are ̅̅̅̅
𝐴𝐶 𝑎𝑛𝑑 ̅̅̅̅
𝐵𝐷
Diagonal of a quadrilateral is a line segment that joins any two
non-consecutive vertices.

Congruent symbol ≅

Parallel symbol ||

1
 CONDITIONS WHICH GUARANTEE THAT A QUADRILATERAL
IS A PARALLELOGRAM
A. A quadrilateral is a parallelogram if both pairs of opposite sides are congruent.
B. A quadrilateral is a parallelogram if both pairs of opposite angles are congruent.
C. A quadrilateral is a parallelogram if any two consecutive angles are supplementary.
D. A quadrilateral is parallelogram if the diagonals bisect each other.
E. A quadrilateral is a parallelogram if each diagonal divides a parallelogram into two congruent
triangles.
F. A quadrilateral is a parallelogram if one pair of opposite sides are both congruent and parallel.

II. Learning Competency: Determines the conditions that make a quadrilateral parallelogram
(M9GE-IIIa-2)
III. ACTIVITIES

ACTIVITY 1:
Directions: Study the following parallelograms. Choose from the given conditions that guarantee that
each of the quadrilaterals below is a parallelogram.
1. 2.

Answer: _______________________ Answer: _______________________

3. 4.

Answer: _______________________ Answer: _______________________

ACTIVITY 2
Directions: Each of the following quadrilaterals being described is a parallelogram. Make an
illustration of each quadrilateral and state the reason why it is a parallelogram.
1. In quadrilateral 𝐽𝐾𝐿𝑀, ∠𝐽 ≅ ∠𝐿 and ∠𝐾 ≅ ∠𝑀.

2. In quadrilateral 𝑄𝑅𝑆𝑇, ̅̅̅̅

𝑄𝑆 and ̅̅̅̅
𝑅𝑇 intersect at point U and ̅̅̅̅
𝑄𝑈 ≅ ̅̅̅̅
𝑈𝑆 and ̅̅̅̅
𝑅𝑈 ≅ ̅̅̅̅
𝑈𝑇.

3. In quadrilateral 𝐿𝐼𝐾𝐸, 𝑚∠𝐿 + 𝑚∠𝐼 = 180, 𝑚∠𝐾 + 𝑚∠𝐸 = 180.

4. In quadrilateral 𝐽𝑂𝐾𝐸, ∆𝑂𝐾𝐸 ≅ ∆𝐸𝐽𝑂.

2
ACTIVITY 3

Directions: Answer the following questions.

1. Use the diagram of the adjustable hat rack at the right

to answer the following.
a. Draw the quadrilateral 𝐴𝐵𝐶𝐷.
b. If the hat rack were expanded outward, would 𝐴𝐵𝐶𝐷
still be a parallelogram? Explain.
2. The plan for a town park shows that the park is a quadrilateral with
straight paths along the diagonals. For what values of the variables
is the park a parallelogram?

3. What should be the measure of ∠𝐷, ∠𝐶 and ∠𝐴 in

quadrilateral 𝐴𝐵𝐶𝐷 in the diagram of a parking lot
at the right for it to become a parallelogram? Explain.

IV. Reflection

Write three words that would describe your feelings after doing the activities.

3
 4
Writer
JOMAR A. MONFORTE
Prepared by:
Answer Key:
Activity 1
1. A quadrilateral is a parallelogram if both pairs of opposite sides are congruent.
2. A quadrilateral is a parallelogram if both pairs of opposite angles are congruent/
A quadrilateral is a parallelogram if any two consecutive angles are supplementary.
3. A quadrilateral is parallelogram if the diagonals bisect each other.
4. A quadrilateral is a parallelogram if each diagonal divides a parallelogram into two
congruent triangles.
Activity 2
1. 2.
Opposite angles are congruent Diagonals bisect each other
3. 4.
Any two consecutive angles Each diagonal divides the parallelogram
are supplementary into two congruent triangles
Activity 3
1. Yes. (Evaluate student’s responses)
2. t=15, s = 10
3. ∠𝐷 = 75°, ∠𝐶 = 105°, ∠𝐴 = 105°
Deauna, Melecio C. Geometry, SIBS Publishing House Inc. 2004
Gregorio Araneta Avenue, Quezon City, 2009
Dilao, Soledad J., Bernabe, Julieta G., Orines, Fernando B., Geometry, SD Publications, Inc.,
Module 5: Quadrilaterals, First Edition, (2014), Department of Education, Vibal Group Inc.
Bryant, Merden L., Bulalayao, Leonides E., et al. Mathematics Grade 9 Learner’s Material
V. References:
 MATHEMATICS 9

Name: _____________________________________ Grade & Section: _____________Date:_______

LEARNING ACTIVITY SHEET

FINDING THE MEASURE OF ANGLES, SIDES AND OTHER QUANTITIES INVOLVING

PARALLELOGRAMS USING ITS PROPERTIES

I. Background Information

This activity sheet shall serve as your guide to facilitate learning as it focuses on using
properties to find measures of angles, sides and other quantities involving parallelogram.

PROPERTIES OF PARALLELOGRAM In the given parallelogram…

Since 𝐿𝐼 = 𝐸𝐾 and
1. In a parallelogram, any two opposite sides are
congruent. 𝐿𝐸 = 𝐼𝐾
Therefore,
̅ ≅ ̅̅̅̅
𝐿𝐼 𝐸𝐾 and
̅̅̅̅
𝐿𝐸 ≅ 𝐼𝐾̅̅̅
2. In a parallelogram, any two opposite angles Since 𝑚∠𝐾 = 𝑚∠𝑇 and
are congruent. 𝑚∠𝐴 = 𝑚∠𝐸
Therefore,
∠𝐾 ≅ ∠𝑇 and
∠𝐴 ≅ ∠𝐸
3. In a parallelogram, any two consecutive 𝑚∠𝐷 + 𝑚∠𝐼 = 180
angles are supplementary.
𝑚∠𝐼 + 𝑚∠ 𝑁 = 180
𝑚∠𝑁 + 𝑚∠𝐺 = 180
𝑚∠𝐺 + 𝑚∠𝐷 = 180
∠𝐷 & ∠ 𝐼, ∠𝐼 & ∠ 𝑁, ∠𝑁 & ∠ 𝐺, ∠𝐺 & ∠ 𝐷
are pairs of supplementary angles
4. The diagonals of a parallelogram bisect each ̅̅̅̅
𝑀𝐿 𝑎𝑛𝑑 ̅̅̅
𝐼𝐾 bisect each other at 𝑌
other.
Since 𝑀𝑌 = 𝑌𝐿 and 𝐼𝑌 = 𝑌𝐾
Therefore,
̅̅̅̅̅ ≅ 𝑌𝐿
𝑀𝑌 ̅̅̅̅ and
̅̅̅ ≅ 𝑌𝐾
𝐼𝑌 ̅̅̅̅

5. Each diagonal of a parallelogram divides the ̅̅̅̅

𝐹𝑇 𝑖𝑠 𝑎 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙
parallelogram into two congruent triangles.

∆𝐹𝐴𝑇 ≅ ∆𝑇𝐸𝐹

(Note: Always consider the corresponding parts)

5
II. Learning Competency: Uses properties to find measures of angles, sides and other quantities
involving parallelograms (M9GE-IIIb-1)

III. ACTIVITIES

ACTIVITY 1

Directions: Use the properties of parallelogram to answer the following.

A. Given: BCDE is a parallelogram. B. Given: In parallelogram BEST, diagonals BS

and ET bisect each other at O.

1. ̅̅̅̅
𝐵𝐶 || ____
̅̅̅̅ ≅ ____
2. 𝐴𝐷
3. ∠𝐸𝐵𝐶 ≅ _____ 1. If ES = 10 cm, how long is ̅̅̅̅
BT? _______
4. ∆𝐵𝐶𝐷 ≅ _______ ̅̅̅? _______
2. If BE = 13 cm, how long is TS
5. If 𝑚∠𝐶𝐷𝐸 = 100, then 𝑚∠𝐷𝐸𝐵 = _____ 3. If EO = 6 cm and SO = 5 cm, what is the length
6. If 𝑚∠𝐷𝐸𝐵 = 65, then 𝑚∠𝐵𝐶𝐷 = _____ of ̅̅̅̅ ̅̅̅̅? ______, ________
ET? BS

7. If 𝐵𝐸 = 8, then 𝐶𝐷 = _____ 4. If ET + BS = 18 cm and SO = 5 cm, what is the

length of ̅̅̅̅
ET? ̅BS
̅̅̅? ______, ________
8. If 𝐸𝐴 = 10, then 𝐴𝐶 = ______

ACTIVITY 2

Directions: Quadrilateral MATH is a parallelogram. Consider each given

information and answer the questions that follow.

1. Given: MA = (𝑥 + 4) 𝑐𝑚, HT = 12 𝑐𝑚, d. What equation is to be used to solve for 𝑦?

∠𝑀𝐻𝑇 = (2𝑦)° , ∠𝑀𝐴𝑇 = 60° ______________________________________
a. What equation is to be used to solve for 𝑥? e. What is the value of 𝑦? _________________
_______________________________________ f. What is 𝑚∠𝑀𝐻𝑇? _____________________
b. What is the value of 𝑥? ___________________ g. What property of the parallelogram did you
c. How long is ̅̅̅̅̅
MA? ________________________ apply to determine the values of 𝑥 and 𝑦? Refer
to your answer in (a) and (d)
______________________________________

6
 IV. Reflection

Self Check

V. References:

Bryant, Merden L., Bulalayao, Leonides E., et al. Mathematics Grade 9 Learner’s Material
Module 5: Quadrilaterals, First Edition, (2014), Department of Education, Vibal Group Inc.
Dilao, Soledad J., Bernabe, Julieta G., Orines, Fernando B., Geometry, SD Publications, Inc.,
Gregorio Araneta Avenue, Quezon City, 2009

Deauna, Melecio C. Geometry, SIBS Publishing House Inc.

Oronce, Orlando A., Mendoza, Marilyn O., e-math, Rex Bookstore Inc.

two opposite sides/angles are congruent

Activity 2 1.a. 𝑥 + 4 = 12, b. 𝑥 = 8, c. 12 𝑐𝑚, d. 2𝑦 = 60, e. 𝑦 = 30, f. 60°, g. In a parallelogram, any
B. 1. 10 cm, 2. 13 cm, 3. 12 cm, 10 cm, 4. 8 cm, 10 cm
Activity 1 A. 1. 𝐸𝐷 ̅̅̅̅, 3. ∠𝐶𝐷𝐸, 4. ∆𝐷𝐸𝐵, 5. 𝑚∠𝐷𝐸𝐵 = 80, 6. 𝑚∠𝐵𝐶𝐷 = 65, 7. 𝐶𝐷 = 8, 8. 𝐴𝐶 = 10
̅̅̅̅, 2. 𝐵𝐴
Answer Key:

Prepared by:

JOMAR A. MONFORTE
Writer

7
 MATHEMATICS 9

Name: _____________________________________ Grade & Section: _____________Date:_______

LEARNING ACTIVITY SHEET
Proving Theorems on Rectangle, Rhombus and Square
I. Background Information:
These learning activity sheets serve as your self-learning guide. It contains three activities
which are intended to facilitate lesson comprehension on the process of proving theorems on the
different kinds of parallelogram (rectangle, rhombus, square).
The following definition of some important terms may be of help to you:

Theorem - is a statement accepted after it is proved deductively

Proof - a logical argument in which each statement is supported/justified by given
information, definitions, axioms, postulates, theorems, and previously proven statements
Reflexive Property - states that anything is equal to itself, 𝑎 = 𝑎
CPCTC - means Corresponding Parts of Congruent Triangles are Congruent
O
L
Quadrilateral - a closed plane figure
Quadrilateral consisting of four line segments or sides
E V
M A
Parallelogram Parallelogram - a quadrilateral with two
pairs of opposite sides that are parallel
H T
A I
Rectangle - a parallelogram with four right
Rectangle
angles
S O M

Rhombus - a parallelogram with all four

sides congruent
Rhombus A
G
W A
Square - a rectangle with all four Square
L sides congruent
T N
SAS (Side-Angle-Side) Congruence
Postulate - states that if the two Y R
A
sides and an included angle of one A
triangle are congruent to the A
corresponding two sides and the A
included angle of another triangle,
I MT
then the triangles are congruent A
∆𝐴𝐼𝑀 ≅ ∆𝑇𝑅𝑌
A
8 A
 (You may also
SSS (Side-Side-Side) review definition
K A of other terms,
Congruence Postulate -
algebraic
states that if the three sides properties,
of one triangle are postulates, and
congruent to the three sides L Y E S theorems that
of another triangle, then the may help you
accomplish the
triangles are congruent ∆𝐿𝑌𝐾 ≅ ∆𝑆𝐸𝐴
given activities.)

II. Learning Competency: The learner proves theorems on the different kinds of
parallelogram (rectangle, rhombus, square). (M9GE-IIIc-1)
III. Activities:

Activity 1: Complete Each Proof!

Directions: Fill in the missing statements or reasons for the following two-column proof. Choose the
letter of the correct answer found from the choices provided beside of every table. Write your answers
on a separate sheet of paper.

Theorem 1: If a parallelogram has a right angle, then it has four right angles, and the
parallelogram is a rectangle.

Given: 𝐴𝐼𝑀𝑆 is a parallelogram with ∠𝐴 as a right angle. A I

Prove: ∠𝐼, ∠𝑀, and ∠𝑆 are right angles.

Proof: S M
Statements Reasons
1. 𝐴𝐼𝑀𝑆 is a parallelogram 1. Given Choices:
with ∠𝐴 as a right angle
2. 𝑚∠𝐴 = 90 2.
3. 3. In a parallelogram, opposite angles a. ∠𝐴 ≅ ∠𝑀 and
are congruent. ∠𝐼 ≅ ∠𝑆
4. 𝑚∠𝐴 = 𝑚∠𝑀 4. Definition of congruent angles b. 90 = 90
𝑚∠𝐼 = 𝑚∠𝑆 c. Any two
5. 𝑚∠𝑀 = 90 5. consecutive
6. 𝑚∠𝐴 + 𝑚∠𝐼 = 180 6. angles of a
7. 90 + 𝑚∠𝐼 = 180 7. Substitution (Statement Numbers 2 parallelogram
and 4) are
8. 8. Reflexive Property supplementary.
9. 𝑚∠𝐼 = 90 9. Subtraction Property (Statement d. Definition of
Numbers 7 and 8) right angle
10. 10. Substitution (Statement Numbers 4 e. Substitution
and 9) (Statement
11. ∠𝐼, ∠𝑀, and ∠𝑆 are right 11. If the measure of an angle is 90, Numbers 2 and 4)
angles. then it is a right angle. f. 𝑚∠𝑆 = 90
12. 𝐴𝐼𝑀𝑆 is a rectangle 12. Definition of rectangle

9
 Theorem 2: The diagonals of a rectangle are congruent.

Given: ̅̅̅̅̅ and 𝐼𝑆

𝐴𝐼𝑀𝑆 is a rectangle with diagonals 𝐴𝑀 ̅̅̅̅ A I
̅̅̅̅̅ ≅ 𝐼𝑆
Prove: 𝐴𝑀 ̅̅̅̅

S M

Before writing a two-column proof for Theorem 2, you may isolate two overlapping triangles
(∆𝐴𝑆𝑀 and ∆𝐼𝑀𝑆) from rectangle 𝐴𝐼𝑀𝑆. These two triangles will serve as guide in writing the proof.
Figure Isolation:
A I
A A
A A
A A
S M S M
Proof:
Statements Reasons
Choices:
1. 1. Given
2. ̅̅̅̅ ̅̅̅̅
𝐴𝑆 ≅ 𝐼𝑀 2. In a parallelogram,
opposite sides are a. ∆𝐴𝑆𝑀 ≅ ∆𝐼𝑀𝑆
congruent. b. ∠𝐴𝑆𝑀 ≅ ∠𝐼𝑀𝑆
c. 𝐴𝐼𝑀𝑆 is a
3. ∠𝐴𝑆𝑀 and ∠𝐼𝑀𝑆 are right 3.
angles rectangle with
4. 4. All right angles are diagonals ̅̅̅̅̅
𝐴𝑀
̅̅̅̅
and 𝐼𝑆
congruent.
d. CPCTC
5. ̅̅̅̅
𝑆𝑀 ≅ ̅̅̅̅
𝑆𝑀 5. Reflexive Property
e. Theorem 1
6. 6. SAS Congruence Postulate
̅̅̅̅̅ ≅ ̅̅̅̅
7. 𝐴𝑀 𝐼𝑆 7.

Theorem 3: The diagonals of a rhombus are perpendicular.

O
Figure Isolation:
Given: Rhombus 𝐺𝑂𝐴𝐿 O O
S A
Prove: ̅̅̅̅
𝐺𝐴 ⊥ ̅̅̅̅
𝐿𝑂 G

G S S A
L

Before writing a two-column proof for Theorem 3, you may isolate two overlapping triangles
(∆𝐺𝑆𝑂 and ∆𝐴𝑆𝑂) from rhombus 𝐺𝑂𝐴𝐿. These two triangles will serve as your guide in writing the
proof.

10
Proof:
Statements Reasons
1. 1. Given Choices:
̅̅̅̅ ≅ 𝐴𝑂
2. 𝐺𝑂 ̅̅̅̅ 2. Definition of rhombus
3. 𝐺𝐴 and ̅̅̅̅
̅̅̅̅ 𝐿𝑂 bisect each other. 3. The diagonals of a
parallelogram bisect
each other.
4. S is the midpoint of ̅̅̅̅
𝐺𝐴 4. ̅̅̅̅ ⊥ 𝐿𝑂
a. 𝐺𝐴 ̅̅̅̅
5. 5. Definition of midpoint ̅̅̅̅
b. 𝐺𝑆 ≅ 𝐴𝑆̅̅̅̅
6. ̅̅̅̅
𝑆𝑂 ≅ ̅̅̅̅
𝑆𝑂 6. c. ̅̅̅̅
𝐿𝑂 bisects ̅̅̅̅
𝐺𝐴
7. 7. SSS Congruence at S.
Postulate
d. ∆𝐺𝑆𝑂 ≅ ∆𝐴𝑆𝑂
8. ∠𝐺𝑆𝑂 ≅ ∠𝐴𝑆𝑂 8. CPCTC
e. Reflexive
9. ∠𝐺𝑆O and ∠𝐴𝑆𝑂 are right 9. ∠𝐺𝑆O and ∠𝐴𝑆𝑂 form
Property
angles a linear pair and are
congruent. f. Rhombus 𝐺𝑂𝐴𝐿
10. 10. Perpendicular lines
meet to form right
angles.

Theorem 4: Each diagonal of a rhombus bisects opposite angles.

OO
O Figure Isolation: 1 2
1 2
Given: Rhombus 𝐺𝑂𝐴𝐿
G A
Prove: ∠1 ≅ ∠2 S A
G
∠3 ≅ ∠4 3 4
3 4
LL
L

Before writing a two-column proof for Theorem 4, you may isolate two overlapping triangles
(∆𝐺𝐿𝑂 and ∆𝐴𝐿𝑂) from rhombus 𝐺𝑂𝐴𝐿. These two triangles will serve as your guide in writing the
proof.

Proof:
Statements Reasons
Choices:
1. 1. Given
a. ̅̅̅̅
𝐿𝑂 ≅ ̅̅̅̅
𝐿𝑂
2. ̅̅̅̅
𝐺𝐿 ≅ ̅̅̅̅ ̅̅̅̅ ≅ 𝐴𝑂
𝐴𝐿; 𝐺𝑂 ̅̅̅̅ 2. Definition of rhombus b. CPCTC
3. 3. Reflexive Property c. Rhombus
𝐺𝑂𝐴𝐿
4. ∆𝐺𝐿𝑂 ≅ ∆𝐴𝐿𝑂 4. d. SSS congruence
Postulate
5. ∠1 ≅ ∠2; ∠3 ≅ ∠4 5.

11
 Activity 2: Show Me Your Proof!
Directions: Write a two-column proof. (You may isolate two overlapping triangles from
rhombus 𝐺𝑂𝐴𝐿. These two triangles will serve as your guide in writing the proof.)

Theorem 4: Each diagonal of a rhombus bisects opposite angles.

O
Given: Rhombus 𝐺𝑂𝐴𝐿
Prove: ∠6 ≅ ∠7 S
G 7 9 A
∠8 ≅ ∠9 6 8

Rubrics for scoring (Activity 2)

0 point 1 point 2 points 3 points
The learner draws
The learner draws the The learner draws the
Learner the diagram correctly
diagram correctly but diagram correctly and
shows no and has filled it
Amount has not filled it with has filled it partially
attempt completely with
of Work to do
appropriate number of with appropriate
appropriate number
statements and number of statements
activity 2. of statements and
reasons. and reasons.
reasons.
The learner’s proof The learner’s proof
The learner’s proof
uses sound reasoning uses sound reasoning
Learner uses sound reasoning
based on algebraic based on algebraic
shows no based on algebraic
properties, definitions, properties,
attempt properties, definitions,
postulates, and definitions,
Flow to write a postulates, and
theorems. But there postulates, and
two- theorems. But there
are more than three theorems. All reasons
column are one to three invalid
invalid reasons written written in the two-
proof. reasons written in the
in the two-column column proof are
two-column proof.
proof. valid.
There are more than There are one to five
All statements and
Learner five statements and/or statements and/or
reasons written by
shows no reasons written by the reasons written by the
the learner in the
attempt learner in the two- learner in the two-
two-column proof
Clarity to write a column proof that are column proof that are
are written using
two- written without using written without using
accepted geometric
column accepted geometric or accepted geometric or
or algebraic terms
proof. algebraic terms and algebraic terms and
and notation.
notation. notation.

Congratulations! You were able to prove theorems on rectangle and rhombus. These theorems on
rectangle and rhombus are true to all squares. You can apply these theorems in answering the next
activity.

12
 Activity 3: Yes, You Can!
F

A B I J
X Y
E G

C L K
D
H
Directions: Refer to the given figures above (rectangle ABCD, rhombus EFGH, square IJKL) and
answer the following:

1. What is the 𝑚∠𝐵𝐶𝐷? _____ 6. What is the 𝑚∠𝐼𝐽𝐾? _____

̅̅̅̅? _____
2. If 𝐴𝐶 = 4.62 𝑐𝑚, how long is 𝐵𝐷 7. If 𝐽𝐿 = 1.8 𝑐𝑚, how long is ̅̅̅
𝐼𝐾 ? _____
3. What is the 𝑚∠𝐸𝑋𝐹? _____ 8. What is the 𝑚∠𝐼𝑌𝐽? _____
4. If 𝐸𝑋 = 2.38 𝑐𝑚, how long is ̅̅̅̅
𝐸𝐺 ? _____ 9. What is the 𝑚∠𝐽𝐿𝐾? _____
5. If 𝑚∠𝐸𝐻𝐹 = 57, what is 𝑚∠𝐺𝐻𝐹? ____ 10. If 𝐽𝐿 = 1.8 𝑐𝑚 and ̅̅̅
𝐼𝐾 = 𝑥 + 1, what is
the value of 𝑥? _____

IV. Reflection References

1
Draw a heart if you were happy while
Abuzo, Emmanuel P., et al. (2013).
doing the activities. Mathematics Learner’s Module 8.
Department of Education, pp. 333-334
2
Draw a star if you have learned
something from the activities.
Bryant, Merden L., et al. (2014).
Draw a question mark if you have Mathematics Learner’s Material 9.
Department of Education, pp. 320-325, p.
3
question/s about the activities that
you need to ask your teacher. (You 345
may write your question/s, too.)
https://lrmds.deped.gov.ph/detail/4769
(You may draw or write your reflection/s on your
https://www.rcampus.com/rubricshowc.c
paper.) 😊 fm?sp=yes&code=G8X62C&
cm, 8.) 90, 9.)
6.) 90, 7.) 1.8
4.76 cm, 5.) 57,
cm, 3.) 90, 4.)
1.) 90, 2.) 4.62
Activity 3:

vary)
(Answers may
Activity 2:

5.) b
1.) c, 3.) a, 4.) d,
Theorem 4
6.) e, 7.) d, 10.) a
1.) f, 4.) c, 5.) b,
Theorem 3
6.) a, 7.) d
1.) c, 3.) e, 4.) b,

6.) c, 8.) b, 10.) f

2.) d, 3.) a, 5.) e,
Theorem 1

Activity 1:

Answer Key
Theorem 2

Prepared by:
MA. LYKA M. NAAG
Writer

13
 MATHEMATICS 9
Name: ___________________________________Grade&Section:_____________Date: __________
LEARNING ACTIVITY SHEET
PROVING THE MIDLINE THEOREM
I. Background Information for Learners
These activity sheets serve as your self-learning guide and will facilitate the lesson on proving
the midline theorem.
Proving the midline theorem is very useful in helping us understand some of the concepts in
geometry especially in connection with trapezoid and similarity. For us to prove the midline theorem,
let’s first know what the midline theorem is.

MIDLINE THEOREM: In ∆ABC, D and E

The segment that joins are midpoints of ̅̅̅̅
𝐴𝐵 and ̅̅̅̅
𝐴𝐶
the midpoints of two sides of a respectively. These midpoints
triangle is parallel to the third side when joined makes segment
and half as long. ̅̅̅̅
𝐷𝐸
which is the midline of ∆ABC.

The following definitions of some important terms may be of help to you:

Definition of Midpoint - A point on a line Definition of Midpoint AM =
segment that divides it into two equal parts. MB
The halfway point of a line segment.
SAS (Side-Angle-Side) Congruence Postulate
- states that if the two sides and an included
angle of one triangle are congruent to the
corresponding two sides and the included
angle of another triangle, then the triangles
are congruent
The AIP Theorem – If
two lines form congruent alternate interior
angles with a transversal, then the lines are AIP
parallel. THEOREM
CPCTC - Corresponding Parts of Congruent
Triangles are Congruent
Vertical Angles - two nonadjacent angles CPCTC
formed by two intersecting lines.
Transitive property of congruence -
̅̅̅̅ ≅ 𝐶𝐷
𝐴𝐵 ̅̅̅̅ 𝑎𝑛𝑑 𝐶𝐷
̅̅̅̅ ≅ 𝐸𝐹
̅̅̅̅ , 𝑡ℎ𝑒𝑛 𝐴𝐵
̅̅̅̅
̅̅̅̅
≅ 𝐸𝐹
Segment addition postulate - states that if
we are given two points on a line
segment, A and C, a third point B lies on the
line segment AC if and only if the distances
between the points meet the requirements
of the equation AB + BC = AC.
(You may also review definition of other terms,
algebraic properties, postulates, and theorems
that may help you accomplish the given activities.)
14
 II. Learning Competency: Proves the midline theorem (for triangles). (M9GE-IIId-1)

ACTIVITY 1: It’s Paperellelogram!

Let three family members or friends join when you do the activity. Prepare the materials needed.
Follow the procedure and answer the questions given.
Materials: 4 pieces of short bond paper, pencil, ruler, adhesive tape, protractor, and scissors.

Procedure:
1. Each member shall draw and cut a different kind of triangle out of a bond
paper. (equilateral triangle, right triangle, obtuse triangle, and acute triangle
that is not equiangular.)
2. Choose a third side of a triangle. Mark each midpoint of the other two sides
then connect the midpoints to form a segment.
3. Measure the segment drawn and the third side you chose.
Q:Compare the lenghts of the segments drawn and the third side you
chose.What did you observe?
4. Cut the triangle along the segment drawn.
Q:What two figures are formed after cutting the triangle along the segment
drawn?
5. Use an adhesive tape to reconnect the triangle with the other figure in such
a way that their common vertex was a midpoint and that congruent segments
formed by a midpoint coincide.
Q:After reconnecting the cutouts, what figure new figure is formed? Why?
Q:What can you say about your findings in relation to those of your family
members/friends?
Q:Do you think that the findings apply to all kinds of triangles? Why?

ACTIVITY 2: Show Me!

Directions: To prove the midline theorem, analyze the given situation below and provide the reasons
to the given statements that follow. You can choose the correct answers from the box of choices at
the right.
MIDLINE THEOREM:
The segment that joins the midpoints of two sides of a triangle is parallel to the third
side and half as long.

̅̅̅̅̅,
Given: ∆HNS, O is the midpoint of 𝐻𝑁
̅̅̅̅
E is the midpoint of 𝑁𝑆
1
Prove: ̅̅̅̅
𝑂𝐸 ∥ ̅̅̅̅
𝐻𝑆 , 𝑂𝐸 = 2 𝐻𝑆

15
Proof:

Statements Reasons CHOICES

̅̅̅̅̅,
1. ∆HNS, O is the midpoint of 𝐻𝑁 1.Given
̅̅̅̅
E is the midpoint of 𝑁𝑆 a. Definition of
2. When 𝑂𝐸 ̅̅̅̅̅is extended. There is a 2.In a line there is a exactly one point between parallelogram.
point T such that OE = ET two other points that is considered as the
midpoint of the line. And these two points are b. SAS
equidistant to the midpoint. Congruence
̅̅̅̅ ≅ 𝐸𝑆
3. 𝐸𝑁 ̅̅̅̅ 3. Postulate
4. ∠2 ≅ ∠3 4.Vertical Angles are Congruent
5. ∆𝐎𝐍𝐄 ≅ ∆𝐓𝐒𝐄 5. c. Parallelogra
6. ∠1 ≅ ∠4 6.CPCTC m Property.
̅̅̅̅̅ ∥ ̅̅
7. 𝐻𝑁 ̅̅
𝑆𝑇 7.
d. Definition
8. ̅̅̅̅
𝑂𝐻 ≅ ̅̅̅̅
𝑂𝑁 8.Definition of Midpoint
of Midpoint
9. 𝑂𝑁 ≅ ̅̅
̅̅̅̅ ̅̅
𝑇𝑆 9. CPCTC(Statement 5)
10. ̅̅̅̅
𝑂𝐻 ≅ ̅̅
̅̅
𝑆𝑇 10. e. Substitution
11. Quadrilateral HOTS is a 11. [statement
parallelogram. (SN) 2]
12. ̅̅̅̅
𝑂𝐸 ∥ ̅̅̅̅
𝐻𝑆 ̅̅̅̅ is on the side of ̅̅̅̅
12.𝑂𝐸 𝑂𝑇 of HOTS
13. OE + ET = OT 13.Segment Addition Postulate (SAP) f. Addition
14. OE + OE = OT 14.
g. By AIP
15. 2OE = OT 15.
Theorem
̅̅̅̅ ∥ 𝑂𝑇
16. 𝐻𝑆 ̅̅̅̅ 16.
17. 2OE = HS 17. By Substitution h. Substitution
1
18. 𝑂𝐸 = 2 𝐻𝑆(The segment joining 18. (SN 14 and 15)
the midpoints of two sides of a
triangle is half as long as the third i. Transitive
side.) Property

Congratulations! You’ve just completed the proof of the midline theorem. This theorem can be
applied to solve problems. Try the activity that follows.
ACTIVITY 3: Go for It!
̅̅̅̅̅ and 𝐶𝐺
Directions: In ∆MCG, A and I are the midpoints of 𝑀𝐺 ̅̅̅̅ ,respectively. Consider each given

information and answer the questions that follow.

1. Given: AI=5.5
Question: What is the measure of ̅̅̅̅̅? _______
𝑀𝐶
2.Given GC= 8
Question: What is the measure of ̅̅̅ ? _______
𝐶𝐼
3. Given GI= 4 and MA = 5
Question: What is the measure of MG? GC? MG + GC ? _____
4.Given AI= 17
Question: What is the measure of AI + MC? _______

16
 RUBRIC FOR ACTIVITY 1: It’s Paperellelogram!
4 points 3 point 2 points 1 point
A complete response to Good solid response
Explanation is Misses key
Understanding questions with detailed with clear
unclear. points.
explanation. explanation.
Shows complete Response shows
Shows substantial Response shows
understanding of the a complete lack
Demonstrated understanding of the some
questions, of
knowledge problem, ideas and understanding of
mathematical ideas and understanding
processes. the task.
processes. of the task.
Does not meet
Goes beyond the Meets the Hardly meets the
the
Requirements requirements of the requirements of the requirements of
requirements of
task. task. the task.
the task.

References
Reflection
Bryant, M.L.,Bulalayao, L. E., Callanta, M.M.,Cruz,
J.D.,De Vera, R.F., Garcia, G.T., Javier, S.E.,Lazaro, R.A.,
Mesterio, B.J. &Saladino, R.H.A,(2014),
What is your “hugot line” to
Mathematics Learner’s Material 9, FEF Printing Corp.
express your feelings while
https://ncalculators.com/geometry/mid-points-
answering the activities. calculator.htm
https://virtualnerd.com/geometry/congruent-
triangles/proof-sss-sas/definition-sas-triangle-
congruence-postulate
https://slideplayer.com/slide/14709210/
https://www.learner.org/series/learning-math-
geometry/dissections-and-proof/part-c-the-midline-
theorem-55-minutes/
MG+GC=18

1. MG=10
2.4
1.11
Activity 3:

18.h
16.c
15.f
14.e
11.a
10.i
7.g
5.b
3.d
Activity 2:

vary)
(Answers may
Activity 1:
Answer Key
GC=8

Prepared by:
CRISTINE R. OPIANA
Writer

17
 MATHEMATICS 9
Name: ___________________________________Grade&Section:_____________Date: __________
LEARNING ACTIVITY SHEET
PROVING THEOREMS ON TRAPEZOIDS
I. Background Information for Learners
These learning activity sheets serve as your self-learning guide. It contains activities which are
intended to facilitate lesson comprehension on the process of proving theorems on trapezoids.
Trapezoid is a quadrilateral with exactly one pair of parallel sides. The non-parallel sides are
called the legs and the angles are called base angles. If the legs of a trapezoid are congruent, then the
trapezoid is an isosceles trapezoid. Let’s take a look at the figures below.

Fig.1 Trapezoid with its parts Fig.2 Isosceles Trapezoid with congruent legs
Similar to triangle midline theorem is a theorem for trapezoids called the
midsegment(median) theorem of trapezoid. A median or midsegment of a trapezoid is the line
segment connecting the midpoints of the two non-parallel sides (legs) of a trapezoid.

MIDSEGMENT(MEDIAN) THEOREM ON
TRAPEZOID:
A line connecting the midpoints of the
two legs of a trapezoid is parallel to the
bases, and its length is equal to half the
sum of lengths of the bases. ̅̅̅̅̅
𝑀𝑁 is the midsegment or median of the
trapezoid ABCD where

The other theorems on trapezoid are related to isosceles trapezoid. The first theorem involves the
base angles of an isosceles trapezoid.

Theorem 1: The base angles of an isosceles trapezoid are

congruent.
In isosceles trapezoid ABCD, there are two pairs of base angles.
∠𝐀 ≅ ∠𝐁 and ∠D ≅ ∠C
The second theorem involves the pair of opposite angles of an isosceles trapezoid.

Theorem 2: Opposite angles of an isosceles trapezoid are

supplementa supplementary.
ry
In isosceles trapezoid ABCD, there are two pairs of opposite angles.
∠𝐀 + ∠𝐂 = 𝟏𝟖𝟎° ∠𝐃 + ∠𝐁 = 𝟏𝟖𝟎°

18
 ̅̅̅̅ ≅ 𝐵𝐷
In isosceles trapezoid ABCD, there are two diagonals which are congruent: 𝐴𝐶 ̅̅̅̅

Theorem 3: The diagonals of an isosceles trapezoid are

congruent.

The following definitions of some important terms may be of help to you:

Midline Theorem - The segment that joins the midpoints

of two sides of a triangle is parallel to the third side and
half as long.
Definition of Parallelogram - a quadrilateral with
two pairs of opposite sides that are parallel

Substitution- When two things are equal, we can replace one with the other, and we know
that the equation will still be true.
Distributive Property : a(b+c) = ab+ac
Parallel Postulate - It states that through any given point not on a line there passes exactly
one line parallel to that line in the same plane.
Symmetric Property of Congruence - If 𝐴𝐵 ̅̅̅̅ ≅ 𝐶𝐷
̅̅̅̅ , 𝑡ℎ𝑒𝑛 𝐶𝐷
̅̅̅̅ ≅ 𝐴𝐵
̅̅̅̅ or if ∠ABC≅∠DEF, then
∠DEF ≅∠ABC.
Same Side Interior Angles Theorem - If two parallel lines are cut by a transversal, then
the same side interior angles are supplementary.
Definition of Isosceles Triangle - A triangle with two equal sides.
Isosceles Triangle Theorem - If two sides of a triangle are congruent, then angles opposite
those sides are congruent.

(You may also review definition of other terms, algebraic properties, postulates, and theorems
that may help you accomplish the given activities.)

II. Learning Competency: Proves the theorems on trapezoids and kites. (M9GE-IIId-2)

ACTIVITY 1: Show Me!

MIDSEGMENT(MEDIAN) THEOREM ON TRAPEZOID:

A line connecting the midpoints of the two legs of a trapezoid is parallel to the
bases, and its length is equal to half the sum of lengths of the bases.

Directions: Prove the midsegment(median) theorem by analyzing the given situation below and
providing the reasons to the given statements that follow. You can choose the correct answers from
the box of choices at the right.
Given: Trapezoid MINS with median TR
1
Prove: ̅̅̅̅
𝑇𝑅 ∥ ̅𝐼𝑁
̅̅̅, ̅̅̅̅
𝑇𝑅 ∥ ̅̅̅̅
𝑀𝑆, TR = 2 (MS + IN)

19
Proof:
Statements Reasons
1 1
1. TR = 2 MS + 2 𝐼𝑁 1. Given CHOICES
̅ , with P as its midpoint.
2. Draw 𝐼𝑆 2.Two points determine exactly one line. a. Midline
1
̅̅̅̅ ∥ 𝑀𝑆
3.TP = 2 MS and 𝑇𝑃 ̅̅̅̅ 3.Midline theorem on triangle IMS Theorem on
triangle INS
1
4.PR = IN and ̅̅̅̅
𝑃𝑅 ∥ ̅𝐼𝑁
̅̅̅ 4. b. Definition of
2
̅̅̅̅ ∥ 𝐼𝑁
5. 𝑀𝑆 ̅̅̅̅ 5. Trapezoid
c. If 2 lines are ||
6. ̅̅̅̅
𝑇𝑃 ∥ ̅𝐼𝑁
̅̅̅ 6.
to the same line,
̅̅̅̅ and 𝑃𝑅
7.𝑇𝑃 ̅̅̅̅ are both parallel to 𝐼𝑁
̅̅̅̅. ̅̅̅̅ and 𝑃𝑅
7.𝑇𝑃 ̅̅̅̅ are either parallel or the same they are || to
Thus, T,P and R are collinear. line (definition of parallel). Since they contain each other.
a common point P, then 𝑇𝑃 ̅̅̅̅ and 𝑃𝑅
̅̅̅̅ are d. Segment
contained in the same line.
Addition Postulate
8.TR = TP + PR 8.
e. Distributive
1 1
9.TR = MS + IN 9.Substitution property of
2 2
1 equality
10.TR = 2 (MS + IN) 10.

Let’s Do Some Exercise V.1!

Directions: Answer the following questions. Refer to the figure at the right.

1. If BA=24cm and DC=26cm, find the length of ̅̅̅̅

𝐾𝐹 ? _____
2. If BA=10cm and DC=12cm, find the length of 𝐾𝐹 ̅̅̅̅ ? _____
3. Find the length of the middle base of the trapezoid whose
parallel bases measures 14 cm and 46 cm. ______
4. Suppose that DC=26cm and KF=25cm.What is the
measure of BA ? _______

ACTIVITY 2: Show Me!

Theorem 1: The base angles of an isosceles trapezoid are congruent.

Directions: Analyze the given situation and provide the reasons to the given statements below. Choose
the correct answers from the box of choices at the right.
Given: Isosceles Trapezoid AMOR , MO ∥ AR
Prove: ∠A =∠R, ∠AMO =∠O
Proof:
Statements Reasons
1.Isosceles Trapezoid 1.Given CHOICES
AMOR, MO ∥ AR
2.AM ≅ OR; MO || AR 2. a.Symmetric Property
3.From M, draw ME || 3.Parallel Postulate b.Substitution
OR where E lies on AR. c.Base angles of an isosceles trapezoid are ≅
4. MORE is a 4.Definition of d.Definition of an isosceles triangle
parallelogram. parallelogram e.Parallelogram Property
f. Definition of Isosceles Trapezoid
g.Supplements of congruent angles are ≅
20
 ̅̅̅̅ ≅ OR
5.ME ̅̅̅̅ 5.
̅̅̅̅
6.OR ≅ ME̅̅̅̅ 6.Symmetric Property
̅̅̅̅̅ ≅ ̅̅̅̅
7.AM ME 7.Transitive Property (Statement 2 and 6)
8.∆AME is an isosceles triangle. 8.
9.∠1 ≅ ∠A 9.
10.∠1 ≅ ∠R 10.Corresponding angles are congruent
11.∠R ≅ ∠A 11.
12.∠A ≅ ∠R 12.
13. ∠A and ∠AMO are supplementary angles. 13.Interior angles, MO || AR
∠O and ∠R are supplementary angles.
14.∠AMO ≅ ∠O 14.

Let’s Do Some Exercise V.2!

Directions: Let’s apply the theorem on isosceles trapezoid by answering the following questions.

1. In isosceles trapezoid TRAP, if the degree measure

of ∠P = 60 degrees. What is the measure of ∠A?
2. Refer to isosceles trapezoid TRAP. The measure
of ∠T is 115 degrees, what is the measure of ∠R?
3. Refer to isosceles trapezoid TRAP.
If the measure of ∠P is 35 degrees, then the measure
of ∠A is _________.
4. If ∠P measures 30 degrees, which of the other angles
also measures 30 degrees?

ACTIVITY 3: Show Me!

Theorem 2: Opposite angles of an isosceles trapezoid are supplementary.

Given: Isosceles Trapezoid ARTS

Prove: ∠ART and ∠S are supplementary CHOICES
a. Definition of Isosceles
∠A and ∠T are supplementary trapezoid
Proof: b.parallelogram property
c.isosceles triangle theorem
d.base angles of an isosceles
Statements Reasons triangle are congruent
1.Isosceles Trapezoid ARTS 1.Given e.substitution
̅̅̅̅ ≅ TS
2.AR ̅̅̅; RT
̅̅̅̅ ≅ AS
̅̅̅̅ 2. f.definition of supplementary
angles
̅̅̅̅ ∥ TS
3.From R, draw RE ̅̅̅ where E 3.
̅̅̅̅. g.Same Side Interior Angles
lies on AS
4.REST is a parallelogram 4.Definition of parallelogram Theorem
h.angle addition postulate
̅̅̅ ≅ ̅̅̅̅
5.TS RE 5. i.interior angle sum theorem
̅̅̅̅ ≅ ̅̅̅̅
6.AR RE 6.Transitive Property on triangle
j.definition of isosceles
7.∆ARE is an isosceles triangle. 7.
triangle
8.∠3 ≅ ∠A 8. k.parallel postulate

21
 9.m∠1 + m∠3 + m∠A 9.
10.∠3 ≅ ∠2 10.Alternate interior angles are congruent
11.∠A ≅ ∠S 11.
12.m∠1 + m∠2 + m∠S 12.Substitution (statement 9,10&11)
13.∠1 + ∠2 = ∠ART 13.
14.m∠ART + m∠S 14.
15.m∠S + m∠T 15.
16.m∠A + m∠T 16.Substitution
18. ∠ART and ∠S are supplementary 17.
∠A and ∠T are supplementary

Let’s Do Some Exercise V.3!

Directions: Let’s apply the theorem on isosceles trapezoid by answering the following questions.

1. In isosceles trapezoid TRAP, the degree measure

of ∠P= 60 degrees. What is the measure of ∠R?
2.In isosceles trapezoid TRAP, if the degree measure
of ∠T = 120 degrees. What is the measure of ∠A?
1. Find the measure of x in the given figure below at the right.
4.In degrees, find the measure of the sum of angles x and y in the given figure below.

ACTIVITY 4: Show Me!

Theorem 3: The diagonals of an isosceles trapezoid are congruent.

Given: Isosceles Trapezoid ROMA

̅̅̅̅̅ ≅ AO
Prove: RM ̅̅̅̅

Proof:
Statements Reasons CHOICES
Statements:
1. 1.Given
a.Isosceles Trapezoid ROMA
̅̅̅̅ ≅ MA
2.OR ̅̅̅̅̅ 2. b.∆ROM ≅ ∆AMO
Reasons:
3.∠ROM ≅ ∠AMO 3.
a. Definition of isosceles
̅̅̅̅̅ ≅ ̅̅̅̅̅
4.OM MO 4. trapezoid
b. CPCTC
5. 5. SAS Congruence Postulate
c. base angles of an isosceles
̅̅̅̅̅ ≅ AO
6.RM ̅̅̅̅ 6. triangle are congruent
d. reflexive property

22
 Let’s Do Some Exercise V.4!

Directions: Let’s apply the theorem on isosceles trapezoid by answering the following questions.
Refer to the given figure at the right.

1. In isosceles trapezoid ROMA, what is the measure of ̅̅̅̅

AO
̅̅̅̅̅ measures 20cm? ______
if RM

2. What is measure of ̅̅̅̅

AE, if ̅̅̅̅̅
RM measures 20cm and ̅̅̅̅
OE
measures 15cm? ______
3. What is the measure of ̅̅̅̅̅
RM, if ̅̅̅̅
AE measures 8cm and ̅̅̅̅
OE
measures 12cm? ______
̅̅̅̅, if RM
4. What is measure of OE ̅̅̅̅̅ measures 22cm and ̅̅̅̅
AE
measures 9cm? ______

Congratulations! You’ve completed proving the theorems on trapezoids.

References
Reflection
Bryant, M.L., Bulalayao, L. E., Callanta, M.M.,Cruz, J.D.,De Vera, R.F.,
Garcia, G.T., Javier, S.E.,Lazaro, R.A., Mesterio, B.J. &Saladino,
Write a short lyrics from a certain
R.H.A,(2014), Mathematics Learner’s Material 9, FEF Printing Corp.
song that represents your feelings
while answering the activities. https://www.wyzant.com/resources/lessons/math/geometry/quad
rilaterals/trapezoids_and_kites
https://www.onlinemathlearning.com/trapezoid-median.html
https://www.ck12.org/geometry/properties-of-equality-and-
congruence/lesson/Properties-of-Equality-and-Congruence-BSC-
GEOM/

4. 13cm 4.120° 4. 24cm 4. ∠A https://mathbitsnotebook.com/Geometry/Quadrilaterals/QDTrapKi

3. 20cm 3. 108° 3. 30cm 3. 35° te.html#:~:text=DEFINITION%3A%20An%20isosceles%20trapezoid%
2. 5cm 2. 60° 2. 11cm 2. 115° 20is,it%20is%20an%20isosceles%20trapezoid.
1. 20cm 1. 120° 1. 25cm 1. 60° https://sciencestruck.com/geometry-terms
V.4 V.3 V.1! V.2
https://www.nagwa.com/en/worksheets/723150624896/#:~:text=
Let’s Do Some Exercise
True%20or%20False%3A%20The%20midsegment,two%20bases%2
Reasons: 2.a; 3.c; 4.d; 6.b 0of%20the%20trapezoid.&text=%F0%9D%90%B4%20%F0%9D%90
Statements: 1.a;5.b %B5%20%F0%9D%90%B6%20%F0%9D%90%B7%20is,%F0%9D%90
Activity 4:
%B4%20%F0%9D%90%B7%20and%20%F0%9D%90%B5%20%F0%9
15.g; 17.f
D%90%B6%20.
2.a; 3.k; 5.b; 7.j; 8.c; 9.i; 11.d; 13.h; 14.e;
Activity 3: https://www.freemathhelp.com/feliz-trapezoids.html
2. f; 5. e; 8. d; 9. c; 11. b;12. a; 14. g
Activity 2:
4.a; 5. b; 6. c; 8. d; 10. e
Activity 1:
Answer Key

Prepared by:
CRISTINE R. OPIANA

23
 MATHEMATICS 9
Name: ___________________________________Grade&Section:_____________Date: __________
LEARNING ACTIVITY SHEET
PROVING THEOREMS ON KITE
I. Background Information for Learners
A kite is also a quadrilateral with two distinct pairs of adjacent sides that are congruent.
Kites have two pairs of congruent sides that meet at two different points.

In kite ABCD,
Segment AB is adjacent and congruent to segment BC.
Segments AD and CD are also adjacent and congruent.

Kites also have a couple of properties that will help us identify them from other quadrilaterals

(1) The diagonals of a kite meet at a right angle.

(2) Kites have exactly one pair of opposite angles that are congruent.

̅ and ̅̅̅̅̅
In kite JKLM, the diagonals 𝐽𝐿 𝐾𝑀 are perpendicular.

And, the two opposite angles that are congruent are ∠JKL and ∠JML

Now, let’s familiarize first the two theorems on kite that we are going to prove later.

Theorem 1: In a kite, the perpendicular bisector of at least

one diagonal is the other diagonal.

̅̅̅̅ ≅
In kite ABCD, 𝐵𝐷 bisects 𝐴𝐶. Therefore, 𝐴𝑋
̅̅̅̅
𝐶𝑋

Theorem 2: The area of a kite is half the product of the

lengths of its diagonals (𝐷1 𝑎𝑛𝑑 𝐷2 ).

Example: If RP = 8 and QS = 6 , then by using theorem,

1
Area of Kite PQRS = 2 [(D1 )(D2 )]

1
Area of Kite PQRS = 2 [(8)(6)]

1
Area of Kite PQRS = 2 [(48)]
1
Area of Kite = 2 [(𝐷1 )(𝐷2 )]
Area of Kite PQRS = 24

24
 The following definition of some important terms may be of help to you:

Adjacent sides - are sides of a polygon that have a common

vertex.
Reflexive Property - Any algebraic or geometric item is equal in
value to itself. ̅̅̅̅
𝑨𝑩 ≅ ̅̅̅̅
𝑨𝑩 or ∠B ≅∠B SSS Congruence
SSS (Side-Side-Side) Congruence Postulate - states that if the Postulate:

three sides of one triangle are congruent to the three sides of

another triangle, then the triangles are congruent.
of one triangle are congruent to the corresponding two sides
and the included angle of another triangle, then the triangles are Area Addition
Area of Figure=
Postulate:
congruent Area A+AreaB+Area C

Area Addition Postulate - states that if we have two shapes that

do not overlap, the total area equals the sum of the areas of the
individual shapes
(You may also review definition
Associative Property - (a + b) + c = a + (b + c) in addition, and (a of other terms, algebraic
properties, postulates, and
• b) •c = (a • b) •c in multiplication theorems that may help you
accomplish the given activities.)

II. Learning Competency: Proves the theorems on trapezoids and kites. (M9GE-IIId-2)

ACTIVITY 1: Show Me!

Theorem 1: In a kite, the perpendicular bisector of at least one diagonal is the other diagonal.

Given: Kite JKLM

̅ and 𝐾𝑀
diagonals 𝐽𝐿 ̅̅̅̅̅ are perpendicular.
̅ is the perpendicular bisector of 𝐾𝑀
Prove: 𝐽𝐿 ̅̅̅̅̅ .

Proof:
Statements Reasons
1.Kite JKLM 1.Given CHOICES
̅̅̅ ≅ 𝐽𝑀
2.𝐽𝐾 ̅̅̅̅ ; ̅̅̅̅
𝐾𝐿 ≅ ̅̅̅̅
𝑀𝐿 2. a. Reflexive Property
3..𝐽𝐿 ≅ 𝐽𝐿 ; 𝐽𝑁̅ ≅ ̅𝐽𝑁
̅ ̅ ̅̅̅ ̅̅̅ 3. b. Definition of Kite
4.∆JML ≅ ∆JKL 4 c. SSS (Side-Side-Side)
5.∠MJL ≅ ∠KJL 5.CPCTC Congruence Postulate
6.∆JMN ≅ ∆JKN 6. d. SAS (Side-Angle-
̅̅̅̅̅ ≅ ̅̅̅̅̅
7.𝑀𝑁 𝐾𝑁 7. Side) Congruence
̅ is the perpendicular
8.𝐽𝐿 8.A segment bisector forms two Postulate
bisector of ̅̅̅̅̅
𝐾𝑀 . congruent segments.
e. CPCTC

25
 ACTIVITY 2: Show Me!

Theorem 2: The area of a kite is half the product of the lengths of its diagonals (𝐷1 𝑎𝑛𝑑 𝐷2 ).
Given: Kite PQRS
1
Prove: Area of kite PQRS = (QS)(PR)
2

Proof:
Statements Reasons
1.Kite PQRS 1. CHOICES
2.QS ⊥ RP 2. a. Substitution
3.Area of kite PQRS = Area of ∆QRS + Area of ∆QPS 3.Area Addition b. Associative
Postulate
1 4.Area Formula for ∆s
Property
4.Area of ∆QRS = 2 (QS)(RP) c. SAP (Segment
1
Area of ∆QPS = 2 (QS)(RP) Addition
1 1 5.Substitution Postulate)
5.Area of kite PQRS = (QS)(RO) + (QS)(OP)
2 2 d. Given
1 6.
6.Area of kite PQRS = 2 (QS)(RO + OP) e.The diagonals of
7.RO + OP = RP 7. a kite are
1 8.
8.Area of kite PQRS = 2 (QS)(RP) perpendicular to
Congratulations! You’ve just completed the proof of the theorems each other.

on kite. This theorem can be applied to solve problems. Try the activity that follows.
Activity: Play a Kite
Directions: Consider the figure at the right and answer the given questions.
Given: Quadrilateral PLAY is a kite.

1. Given: PA = 12 cm; LY = 6 cm 2. Given: Area of kite PLAY = 135 cm2 ;

Questions: LY = 9 cm
• What is the area of kite PLAY? Questions:
• What theorem justifies your • How long is PA?
answer? • What theorem justifies your answer?

Reflection References
Bryant, M.L.,Bulalayao, L. E., Callanta, M.M.,Cruz, J.D.,De Vera, R.F.,
What #word best describes your Garcia, G.T., Javier, S.E.,Lazaro, R.A., Mesterio, B.J. &Saladino,
R.H.A,(2014), Mathematics Learner’s Material 9, FEF Printing
feelings while doing the activity?
Corp.

https://www.wyzant.com/resources/lessons/math/geometry/quad
rilaterals/trapezoids_and_kites
3. PA = 30 cm, theorem 2(Kite) https://mathbitsnotebook.com/Geometry/Quadrilaterals/QDTrapK
ite.html#:~:text=DEFINITION%3A%20An%20isosceles%20trapezoid
2. Area of Kite PLAY= 36 cm2 , Theorem 2(KIte)
%20is,it%20is%20an%20isosceles%20trapezoid.
Activity: Play a Kite https://www.wyzant.com/resources/lessons/math/geometry/quad
ACTIVITY 2: Show Me! 1.d; 2.e; 6.b; 7.c; 8.a rilaterals/trapezoids_and_kites
https://byjus.com/area-of-a-kite-
ACTIVITY 1: Show Me! 2.b; 3.a; 4.c; 6.d; 7.e
formula/#:~:text=Formula%20for%20Area%20of%20a,_%7B1%7D
Answer Key D_%7B2%7D
https://slideplayer.com/slide/12361799/
Prepared by:
CRISTINE R. OPIANA

26
 MATHEMATICS 9
Name: ______________________________________ Grade & Section: ________Date: __________
LEARNING ACTIVITY SHEET
Solving Problems Involving Parallelograms, Trapezoids and Kites
I. BACKGROUND INFORMATION
Parallelograms, trapezoids and kites are the different kinds of quadrilaterals. When we say
quadrilateral, it is a closed plane figure consisting of four line segments or sides. These learning activity
sheets will help you in enhancing your problem-solving skills in parallelograms, trapezoids, and kites.
It would be easier for you to solve this problem when you know the different properties and theorems
of each type of quadrilateral.

P R O P E R T I E S A N D T H E O R E M S

PARALLELOGRAMS KITES
Parallelogram ABCD Kite IJKL J
A B
I K

D C L
 The opposite sides ̅̅̅̅
𝐴𝐵 ≅ ̅̅̅̅
𝐷𝐶
are congruent ̅̅̅̅ ≅ ̅̅̅̅
𝐴𝐷 𝐵𝐶
 The diagonals are ̅̅̅̅
𝐽𝐿 ⏊ ̅̅̅
𝐼𝐾
(equal).
perpendicular.

 The opposite angles ∠A ≅ ∠C

are congruent. ∠B ≅ ∠D
 The area of a kite
 The non-opposite or ∠A+∠B=180°
1
adjacent angles are ∠B+∠C=180° is half the product Area=2 (𝐼𝐾)(𝐽𝐿)
supplementary angles ∠C+∠D=180°
(equal to 180°). of the lengths of
∠D+∠A=180°
its diagonals.
 The diagonals bisect ̅̅̅̅
𝐴𝑂 ≅ ̅̅̅̅
𝑂𝐶
each other. ̅̅̅̅ ≅ 𝑂𝐵
̅̅̅̅
𝐷𝑂

 Each diagonal divides ∆𝐷𝐴𝐵 ≅ ∆𝐵𝐶𝐷

a parallelogram into ∆𝐴𝐵𝐶 ≅ ∆𝐶𝐷𝐴
two congruent ∆𝐷𝑂𝐶 ≅ ∆𝐵𝑂𝐴
triangles. ∆𝐴𝑂𝐷 ≅ ∆𝐶𝑂𝐵

27
 P R O P E R T I E S A N D T H E O R E M S

ISOSCELES TRAPEZOIDS TRAPEZIUMS

Trapezoid EFGH Trapezium MNOP
E F M N

H G P O
 The base angles are ∠E ≅ ∠F  Only one pair of ̅̅̅̅̅
𝑀𝑁 ‖𝑃𝑂̅̅̅̅
congruent. ∠H ≅ ∠G opposite sides are
parallel to each
 The diagonals are ̅̅̅̅
𝐸𝐺 ≅ ̅̅̅̅
𝐻𝐹 other.
congruent.
 The median of a  The area of the 1
1 Area=2(h)(MN+PO)
trapezoid is parallel 𝑀𝑁 = (𝐸𝐹 + 𝐻𝐺) trapezium is equal
2
to the bases and its
to the half of the
measure is half the
sum of the product of its
measures of the height and sum of
bases. lengths of its
parallel sides.

 The area of a 1
𝐴𝑟𝑒𝑎 = ℎ(𝐸𝐹 + 𝐻𝐺)
2  The perimeter of Perimeter=MN+NO+OP+
trapezoid is equal to
the trapeziums is
half the product of PM
the height and the equal to the sum
sum of the two of lengths of all
bases. the sides.

II. LEARNING COMPETENCY: Solves problems involving parallelograms, trapezoids and

kites. (M9GE-IIIe-1)

ILLUSTRATIVE EXAMPLES:
1. Given the parallelogram FACT, if m∠F = 40° what is m∠A?
Solution: The non-opposite or adjacent angles are supplementary angles (equal to 180°). This
tells us that m∠F + m∠A = 180°
F A
40° + m∠A = 180°
m∠A = 180° - 40°

T C m∠A = 140° Therefore, the m∠A is 140°.

28
 2. Quadrilateral MATH is an isosceles trapezoid with bases 𝑀𝐴 and 𝐻𝑇, 𝑆𝑉 is the median. If the
measure of 𝑀𝐴 is 7 cm and 𝐻𝑇 is 10 cm, how long is the median?
Solution:
M 7 cm A
1 17 𝑐𝑚
SVmedian = (𝑀𝐴 + 𝐻𝑇) SV =
2 2
1
S V 𝑆𝑉 = 2 (7𝑐𝑚 + 10𝑐𝑚) SV = 8.5 𝑐𝑚
1
SV = (17 𝑐𝑚) Therefore, the median is 8.5 cm long.
2
H 10 cm T
3. Quadrilateral SAFE is a kite. What is its area if the measure of 𝑆𝐹 is 15 cm and 𝐸𝐴 is 12 cm?
Solution:
𝟏
Area =𝟐 (𝑺𝑭)(𝑬𝑨)
A
S
𝟏
Area = (𝟏𝟓 𝒄𝒎)(𝟏𝟐 𝒄𝒎)
𝟐
𝟏
Area = (𝟏𝟖𝟎𝒄𝒎²)
𝟐
E F
Area =𝟗𝟎 𝒄𝒎²
Therefore, the area of quadrilateral SAFE is 𝒄𝒎².
III. ACTIVITY 1: PARALLELOGRAM, TRAPEZOID and KITE SQUAD!
Directions: Solve each problem on a clean sheet of paper. You can use the different properties and
theorems in solving problems involving parallelograms, trapezoids
and kites. Show your solution. L A
1. Mang Tonyo is a farmer in Libon, Albay who owns a field in a
shape of parallelogram named LAND as shown in the figure.
If the farmer is standing at point L and m∠A = 52°, find the
D N
m∠L where the man is standing.

I
W
2. In Grade 9 – St. Dominic class, the math teacher let her
N
D
students make a kite and named it as kite WIND with the
following measurements: 𝑾𝑵 = 10 inches and 𝑰𝑫 = 17 inches.
Using the given measurement, what is the area of kite WIND?

H O
3. Ela bought a Wi-Fi to be used in her online class and she
named it as trapezoid HOME. If the measure of ̅̅̅̅
𝐻𝑂 is 2x – 4,
̅̅̅̅̅
𝐸𝑀 is 3x + 2 and the median is 2x + 4. What is the value of x?

E M
29
 30
Writer
IVY A. MORCO
Prepared by:
3. Given: 𝐻𝑂= 2x-4
𝐸𝑀= 3x+2
Median = 2x+4 2. Given: 𝑊𝑁 =10 inches
1. Given: m∠A = 52°
Find: x 𝐼𝐷 = 17 inches
Solution: Use the formula of median Find: Area of kite WIND Find: m∠L
to find the value of x. Solution:
𝟏 Solution: Since ∠L and ∠A are
𝟏 Area = 𝟐 (𝑾𝑵)(𝑰𝑫)
𝑴𝒆𝒅𝒊𝒂𝒏 = (𝑯𝑶 + 𝑬𝑴) 𝟏 adjacent angles; therefore they are
𝟐 Area = 𝟐 (𝟏𝟎 𝒊𝒏)(𝟏𝟕 𝒊𝒏)
𝟏
𝟐𝒙 + 𝟒 = (𝟐𝒙 − 𝟒 + 𝟑𝒙 + 𝟐) 𝟏 supplementary. Then, we have
𝟐 Area = (𝟏𝟕𝟎 𝒊𝒏²)
𝟐
𝟏 (𝟏𝟕𝟎 𝒊𝒏²) m∠L + m∠A = 180°
𝟐𝒙 + 𝟒 = (𝟓𝒙 − 𝟐) Area = 𝟐
𝟐
𝟓𝒙 − 𝟐 Area = 𝟖𝟓 𝒊𝒏² m∠L + 52° = 180°
𝟐𝒙 + 𝟒 = Therefore, the area of kite WIND is
𝟐 m∠L = 180° - 52°
𝟐(𝟐𝒙 + 𝟒) = 𝟓𝒙 − 𝟐 𝟖𝟓 𝒊𝒏².
𝟒𝒙 + 𝟖 = 𝟓𝒙 − 𝟐 m∠L = 128°
𝟖 + 𝟐 = 𝟓𝒙 − 𝟒𝒙
𝒙 = 𝟏𝟎
Therefore, the value of x is 10.
ANSWER KEY (ACTIVITY 1: PARALLELOGRAMS, TRAPEZOIDS AND KITES SQUAD!)
 Kites Examples (shmoop.com)
(lumenlearning.com)
 Using the Properties of Trapezoids to Solve Problems | Prealgebra
LearnZillion
 Use triangle and parallelogram theorems to solve a real-world problem |
Engaged Learning Grade 9.
 Lim, Y. F., Nocon, R. C., Nocon, E. G. & Ruivivar, L. A. (2014) Math for
 Curriculum Guide in Grade 9 Mathematics
 Learner’s Materials in Grade 9 Mathematics
REFERENCES IV.
Not So Good Okay
Good Great!
Drop and share your emotion here!
How do you feel about what you learned this week?
My Reflections!
 Mathematics 9

Name: _________________________________Grade & Section: ______________Date: __________

LEARNING ACTIVITY SHEET

Describing a Proportion

I. Background Information for Learners:

Ratio is used to compare two or more quantities. Quantities involved in a ratio are of the same
kind so that ratio does not make use of units. However, when quantities are of different kinds, the
comparison of the quantities that consider the units is called rate. Proportions are statements of
equality involving ratios.
The table show ratios or rates that are proportional. Study the table that follows.

Ratios or Rates Proportional Quantities

A Feet : Inches 3 ft : 36in = 4 ft : 48 in

B Minutes : Meters 3 min : 60 m = 6 min : 120 m

C Kilograms of Mango : Amount Paid 1: 80 pesos = 3 : 240 pesos

Let us verify the accuracy of determined proportions by checking the equality of the ratios or
rates. Examples are provided. Please be reminded that the objective is to show that the ratios or rates
are equivalent. Hence, solutions need not be in the simplest form. Four different possible solutions
are shown in example A.

Checking the equality of ratios or rates in the cited

Examples Proportional Quantities
proportions
Solution 1: Simplifying Ratios
Identify the common factors of the
3 ft : 36in = 4 ft : 48 in denominators
Can also be written as a fraction 3 4 3 4 1 1
A
? → ? → =
3 𝑓𝑡 4 𝑓𝑡 36 48 3(12) 4(12) 12 12
=
36 𝑖𝑛 48 𝑖𝑛 Solution 2: Simplifying Cross Multiplied Factors
3 4
? → 3(48) ? 36(4) → 3(4)(12) = 3(4)(12)
36 48

31
 Solution 3: Cross Products
3 4
? → 3(48) ? 36(4) → 144 = 144
36 48
Solution 4: Products of Means and Extremes
3 ft : 36 in = 4 ft : 48 in
144
144
3 6 3 6 3 3
B 3 min : 60 m = 6 min : 120 m ? → ? → =
60 120 60 2(60) 60 60
1 3 1 3 1 1
C 1: 80 pesos = 3 : 240 pesos ? → ? → =
180 240 80 3(80) 80 80

The solution in the table that follows show that corresponding quantities are proportional. In
short, they form a proportion because the ratios are equal.

3 4 3 4 1 1
3 ft : 36in = 4 ft : 48 in ? → ? → =
36 48 3(12) 4(12) 12 12

II. Learning Competency

The learner describes a proportion (M9GE-IIIf-1).

III. Activity 1: Classify Me!

Directions: Read the statements/phrases below. If it illustrates a ratio, write the
𝑎
corresponding ratio in the form, a : b or 𝑏
. If it illustrates a proportion, write the
𝑎 𝑐
corresponding proportion in the form, a : b = c : d or 𝑏
= 𝑑
. Write your answer on the

space provided before the number.

_____ 1. A mixture of one part of cement and three parts of sand.

_____ 2. If 20 m of rope weighs 1kg, then 200 m of rope weighs 10 kgs.
_____ 3. 1 inch is equivalent to 2.5 centimeters.
_____ 4. The number of black cars compared to the total number of cars in the parking lot.
_____ 5. The price of ripe mango per kilogram is Php 85. Then 5 kilograms of ripe mango is
Php 425.

32
Activity 2: Find Me!
Directions: Read the statements/phrases below. Determine the ratios/proportions stated.

1. For every 5 boys in the volleyball team there is 1 girl. What is the ratio of boys to girls?
2. For every 8 shoes Sarah owned she had 4 slippers. What is her ratio of shoes to slippers?
3. If 1.5 liters = 5 glasses. How many glasses are in 15 liters?
4. How many street lights posts are there in a 2-km stretch of road, if every street light
post is placed at 20-m intervals?
5. If there are 5 switches for every 8 light bulbs, how many light bulbs are there in 15
switches?

Activity 3: What’s Missing?

Directions: Solve for the missing proportions. You can use any method shown in example A

4 32 𝑘 90
1. = 5. =
10 𝑥 3 10

Answer: ________ Answer: ________

𝑞 2 15 𝑦
2. = 6. =
3 6 5 4

Answer: ________ Answer: ________

20 4 6 𝑏
3. = 7. =
5 𝑧 12 6

Answer: ________ Answer: ________

𝑐 126 8 𝑤
4. = 8. =
3 18 5 15

Answer: ________ Answer: ________

IV. REFLECTION:
After all the activities and assessment in this module, you will now be reflecting on an
important question below. Good luck!

“How important is the study of proportions in solving relevant real-life problems?”

33
 34
Writer
JAY-AR D. LLAVE
Prepared by:
Answer Key
Activity 1 Activity 2 Activity 3
1 𝑐𝑒𝑚𝑒𝑛𝑡 5
1. 1 cement : 3 sand or 1. 5 : 1 or 1. x = 80
3𝑠𝑎𝑛𝑑 1
8 2. q=1
2. 20 : 1 = 200 : 10 or 2. 8 : 4 or 4
20 200 3. z=1
= 3. 50 glasses
1 10 4. c = 21
1 𝑖𝑛𝑐ℎ 4. 100 street light posts
3. 1 inch = 2.5 cm or 5. k = 27
2.5 𝑐𝑚
5. 24 light bulbs
4. number of black cars : total number of cars; 6. y = 12
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑙𝑎𝑐𝑘 𝑐𝑎𝑟𝑠 7. b=3
or
𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑎𝑟𝑠
8. w = 24
5. 1 : 85 = 5 : 425 or
1 5
=
85 425
https://osky6math.weebly.com/world-6---ratios-rates-and-proportional-reasoning.html 
https://www.mathsisfun.com/algebra/proportions.html 
Mathematics Learner’s Material for Grade 9 pp, 358 
References:
 MATHEMATICS 9

Name: ____________________________ Grade and Section ______________Date: _____________

LEARNING ACTIVITY SHEET

ILLUSTRATING SIMILARITY OF FIGURES
I. Background Information for Learners
This activity sheets will serve as your guide to facilitate learning as it focuses on
mastery on illustrating similarity of figures.
A review of ratio and proportion is important in the study of similarity.
RATIO
A ratio is a comparison of two quantities.

Given two numbers a and b such that b ≠ 0, a ratio of a to b is the quotient a divided
by b. A ratio can be represented in four ways.
𝒂
𝒂 to 𝒃 𝒂:𝒃 𝒂/𝒃
𝒃

PROPORTION
The equality of two ratios is called a proportion. If the ratios 𝒂:𝒃 and 𝒄:𝒅 are equal,
𝒂 𝒄
then we write ratios 𝒂:𝒃 = 𝒄:𝒅 or = (where b and d are nonzeros).
𝒃 𝒅

Theorem:
In a proportion, the product of the extremes is equal to the product of the means.
𝒂 𝒄
𝒃
(bd) = 𝒅 (bd) 𝒂 : 𝒃 = 𝒄 : 𝒅

ad = bc or means

extremes
extremes means

PROPORTIONAL SEGMENTS

Corresponding segments are proportional if the segments of one figure have the
same ratio as the segments of the other.

2 3 8 12

A B C M N P
Segments AC and MP are divided proportionally by points B and N, respectively.
𝐴𝐵 𝑀𝑁 2 8
1. 𝐵𝐶
= 𝑁𝑃
→ 3 = 12
𝐴𝐵 𝑀𝑁 Notice that in all the proportions, the
2. = , that is 2:5 = 8:20
𝐴𝐶 𝑀𝑃
𝐵𝐶 𝑁𝑃 product of the means is equal to the
3. = 𝑀𝑃 , that is 3:5 = 12:20
𝐴𝐶 product of the extremes.
𝐴𝐵 𝐵𝐶
4. = , that is 2:8 = 3:12
𝑀𝑁 𝑁𝑃

35
SIMILARITY BETWEEN TRIANGLES

Triangle ABC is similar to triangle XYZ written as ABC ~ XYZ, under the correspondence A ↔ X,
B ↔ Y, C ↔ Z, if and only if A X
i. All pairs of corresponding angles are congruent.
ii. All pairs of corresponding sides are proportional.

Y Z
B C

II. Learning Competency

Illustrates similarity of figures. (M9GE-IIIg-1)

III. Activities
Activity 1.
Find the unknown terms of each proportion.
1. x : 4 = 5 : 10 5. 4 : x = x : 25
8 4
2. =9 6. 3 : 10.5 = x : 5.25
x

3. 6 : 9 = 4 : x 16 𝑥− 8
7. 𝑥+8
= 5
2 𝑥
4. =
3 12

Activity 2.

Assume that MUG ~ BAT. Complete the following statements.

U
𝑀𝑈 MG A
1. =
BA ? 14
x 10 x
2. = 21
12 ?
M G 12
10 x
3. =? 10
y
14 21 B T
4. = y
x ?

5. U≅ ?
Activity 3.
O
1. Given: 9 12
BOY ~ DOT with measures of sides
B Y
as indicated in the figure 10
Find: DT, YT and OT 18

D T
36
 B

2. Given:
J
BCF ~ JSD with measures of sides 24 18
a b
as indicated in the figure
S D
Find: the measure of a and b 10
C F
30
3. The sides of a triangle are 8, 15 and 18. The shortest side of a similar triangle is 10.
How long are the other sides?

III. Reflection
Share your experience in performing this lesson. Give at least one reaction by checking
from the given emoticons.

😄______ 😅 _____ 😢 _____ 😰 _____

😂______ 😍 _____ 😭 _____ 😖 _____

Reference:

Bryant, M. L., Bulalayao, L.E., Callanta, M.M., Cruz, J.D., De Vera, R.F., Garcia G.T., Javier, S.E.,
Lazaro, R.A., Mesterio, B.J.& Saladino, R.A. Mathematics Leaner’s Material 9. Vibal Group
Inc.

ANSWER KEY
7. 12
3. 18.75 and 22.5
6. 1.5
b=6 5. O 5. 10
2. a = 8 4. 12 4. 8
CE = 24 3. 12 3. 6
AC = 36 2. Y 2. 18
1. BC = 30 1. BD 1. 2
Activity 3 Activity 2 Activity 1

Prepared by:

FIDELITO M. SAMPER
Writer

37
 MATHEMATICS 9

Name: ____________________________ Grade and Section ______________Date: _____________

LEARNING ACTIVITY SHEET

Proving the Conditions for Similarity Theorems

I. Background Information for Learners

This activity sheets will serve as your guide to facilitate learning as it focuses on mastery on
proving the conditions for similarity of triangles.

Is there a way you can measure difficult-to-obtain lengths without using direct measurement?
How are sizes of objects enlarged or reduced? The concept of similarity of objects will help you solve
the given situations.

SAS SIMILARITY THEOREM

SAS Similarity Theorem

Two triangles are similar if an angle of one triangle is congruent to an angle of

another triangle and the corresponding sides including those angles are in
proportion.

X XY XZ
A =
If: AB AC
X≅ A
R P
B C Then: XYZ ~ ABC
Y Z

STATEMENTS REASONS
1. Let point R be on XY and point P be on
By Construction
XZ such that XR ≅AB and XP ≅ AC
2. Construct RP. Two distinct points determine a line.
3. X≅ A Given

4. XRP ~ ABC SAS Congruence Postulate

XY XZ
5. = Given
AB AC
XY XZ
6. = Substitution
XR XP

38
 7. RP YZ Converse of the Basic Proportionality Theorem
If two parallel lines are cut by a transversal,
8. XRP ≅ XYZ
corresponding angles are congruent.
9. X≅ X Reflexive Property
10. XRP ~ XYZ AA Similarity Theorem
11. XYZ ~ ABC Transitive Property

SSS SIMILARITY THEOREM

SSS Similarity Theorem

Two triangles are similar if the corresponding sides of two triangles are in
proportion.

P
M PR PN RN
If: = =
MT MS TS

X Then: PRN ~ MTS

Y
T S
R N

STATEMENTS REASONS
1. Let X and Y be points of PR and PN,
respectively, such that PX ≅ MT and By construction
PY ≅ MS.
2. Construct XY. Two distinct points determine a line.
3. PX = MT
Definition of congruent segments.
PY = MS
PR PN RN
4. = = Given
MT MS TS
PR PN
5. = Substitution
PX PY
6. P≅ P Reflexive Property
7. PRN ~ PXY SAS Similarity Theorem
XY PX
8. = Definition Of similar triangles
RN PR

39
 XY MT
9. = Substitution
RN PR
MT
10. XY = RN ( )
PR
Multiplication Property of Equality
MT
TS = RN ( )
PR
11. XY = TS Transitive Property
12. PXY ≅ MTS SSS Congruence Postulate
Transitive Property of Equality, Statement 6
13. PXY ~ MTS
and 12

AA SIMILARITY THEOREM

Postulate
AAA Similarity Postulate

Two triangles are similar if and only if their corresponding angles are congruent.

Recall that in two triangles, if two pairs of corresponding angles are congruent, then the third
pair must also be congruent. We then have the following theorem.

AA Similarity Theorem

Two triangles are similar if two angles of one triangle are congruent to two
angles of another triangle.

RIGHT TRIANGLE SIMILARITY THEOREM

Right Triangle Similarity Theorem

If the altitude is drawn to the
hypotenuse of a right triangle, then
the two triangles formed are similar
to the original triangle and to each
other.

40
SPECIAL RIGHT TRIANGLE THEOREMS

45 – 45 – 90 Right Triangle Theorem

In a 45 – 45 – 90 right triangle:
45°
h √2
 Each leg is times the
l 2
hypotenuse; and
45°  The hypotenuse is √2 times
l
each leg.

30 – 60 – 90 Right Triangle Theorem

In a 30 – 60 – 90 right triangle:
30°  The shorter leg s is ½ the
1
h hypotenuse h or times the
l √3
longer leg l
 The longer leg l is √3 times
60° the shorter leg s; and
s  The hypotenuse h is twice the
shorter leg

II. Learning Competency

Proves the conditions for similarity of triangles. (M9GE-111g-h-1)

1.1.1 SAS Similarity Theorem

1.1.2 SSS Similarity Theorem
1.1.3 AA Similarity Theorem
1.1.4 Right Triangle Similarity Theorem
1.1.5 Special Right Triangle Theorem

41
 III. Activities
Activity 1: Write the reasons that are left blank in the proof of Right Triangle Similarity
Theorem.

E Given
MER is a right triangle with MER as the right
angle and MR as the hypotenuse.
EY is an altitude to the hypotenuse of MER.
M R
Y Prove
MER ~ EYR ~ MYE
PROOF
STATEMENTS REASONS
1. MER is a right triangle with MER
as right angle and MR as the
hypotenuse. 1. ______________________________
EY is an altitude to the hypotenuse of
MER.
2. EY MR 2. Definition of ___________________
3. MYE and EYR are right angles 3. Definition of __________________ lines
4. MYE ≅ EYR ≅ MER 4. Definition of _______________ angles
5. YME ≅ EMR ; YRE ≅ ERM 5. __________________ Property
6. MYE ~ MER ; MER ~ EYR 6. _______________ Similarity Theorem

Activity 2: Write the statements that are left blank in the proof of AA Similarity Theorem.
Refer to the hints provided.

L U
W H If: U≅ H; V≅ Y
H Then: LUV ~ WHY

V Y
PROOF
HINTS STATEMENTS REASONS
1 Write all the given. Given
2 Describe the measure of the
Definition of congruent
congruent angles in statement
angles
1.
42
3 Add m V and m Y to Addition Property of
m U and m H respectively. Equality
4 Add the measures of all the The sum of the measures
angles of triangles LUV and of the three angles of a
WHY. triangle is 180.
5 Equate the measures of the
angles of triangles LUV and Transitive Property
WHY from statatement 4.
6 What can you say about m L Addition Property of
and m W? Equality
7 Are angles L and W Definition of congruent
congruent? angles
9 Are triangles LUV and WHY
AAA Similarity Postulate
similar?

Activity 3: Given the figure below, use SAS Similarity Theorem to prove that
R
RAP ~ MAX. The first one is done for you.

A
P X

M
PROOF
HINTS STATEMENTS REASONS
Write in a proportion the
𝑹𝑨 𝑴𝑨
1 ratios of two corresponding = 𝑿𝑨 Given
𝑷𝑨
proportional sides
2 Describe included angles of
the proportional sides
3 Conclusion based on the
simplified ratios

Activity 4: Using the figure below, prove that DAM ~ FAN.

1
M 2 N
A

43
 Activity 5: Triangle TUV is an isosceles right triangle with U as the right angle. (Show your
solution.) U
1. If t = 5 cm, find u.
v t
2. If v = 8 cm, find u.
3. If v = 12.5 cm, find u.
T V
u

Activity 6: In the figure, JA SC Supply the missing length in each row below to show that
triangle JDA is similar to triangle SDC.

SJ JD AC DA
a. 3 2 6 4
D
b. 7 3 5
c. ¾ ½ 4
d. d e f J A
e. m n q
S C
f. 3x 4x 2y

IV. Reflection:

End of Lesson EMOJI

Rate your understanding of today’s lesson by choosing from the emojis below. Draw the
emoji on your answer sheet.

44
 References:

Bryant, M. L., Bulalayao, L.E., Callanta, M.M., Cruz, J.D., De Vera, R.F., Garcia G.T., Javier, S.E.,
Lazaro, R.A., Mesterio, B.J.& Saladino, R.A. Mathematics Leaner’s Material 9. Vibal Group
Inc.

Dilao, S.J. Ed.D., & Bernabe, J.G. Geometry. Alkem Company (S) PTE. Ltd.

ANSWER KEY

3y/2 f.
mq/n e.
ef/d d.
8/3 c.
35/3 b.
4 a.
Activity 6

12.5√2 cm 3.
8√2 cm 2.
5√2 cm 1.
Activity 5

Activity 2
Right Triangle 6.
Reflexive 5.
Activity 4 Right 4.
Perpendicular 3.
Altitude 2.
Given 1.
REASONS
Activity 3 Activity 1

Prepared by:
ANABELLE L. SABAYBAY
Writer

45
 MATHEMATICS 9

Name: _____________________________________ Grade & Section: _____________Date:_______

LEARNING ACTIVITY SHEET

APPLYING THE THEOREMS TO SHOW THAT GIVEN TRIANGLES ARE SIMILAR

I. Background Information

This activity sheet shall serve as your guide to facilitate learning as it focuses on applying the
theorems to show that given triangles are similar.

The following theorems are important to show that triangles are similar.

TRIANGLE SIMILARITY THEOREMS

AA Similarity Theorem SSS Similarity Theorem

If two angles of one triangle are congruent to If the corresponding sides of two triangles are
two angles of another triangle, then the proportional, then the triangles are similar.
triangles are similar.

𝑷𝑸 𝑸𝑹 𝑷𝑹
If ∠𝑼 ≅ ∠𝑯 and ∠𝑽 ≅ ∠𝒀, If = 𝑻𝑼 = ,
𝑺𝑻 𝑺𝑼

Then, ∆𝑳𝑼𝑽~∆𝑾𝑯𝒀 Then, ∆𝑷𝑸𝑹~∆𝑺𝑻𝑼

Example: ∆𝑨𝑩𝑪 ↔ ∆𝑱𝑲𝑳 Example: ∆𝑨𝑺𝑯 ↔ ∆𝑳𝑬𝒀

Since ∠𝑩 ≅ ∠𝑲 and ∠𝑪 ≅ ∠𝑳, 𝐴𝑆 4 𝟏 𝑆𝐻 7 𝟏 𝐴𝐻 9 𝟏

= = ; = = ; = =
Therefore, ∆𝑨𝑩𝑪~∆𝑱𝑲𝑳 𝐿𝐸 8 𝟐 𝐸𝑌 14 𝟐 𝐿𝑌 18 𝟐

𝐴𝑆 𝑆𝐻 𝐴𝐻 𝟏
Since = = =
𝐿𝐸 𝐸𝑌 𝐿𝑌 𝟐
Therefore, ∆𝑨𝑺𝑯 ~ ∆𝑳𝑬𝒀

46
 SAS Similarity Theorem Right Triangle Similarity Theorem
If an angle of one triangle is congruent to an If the altitude is drawn to the hypotenuse of a
angle of another triangle, and the right triangle, then the triangles formed are
corresponding sides including those angles are similar to the original triangle and to each other.
proportional, then the triangles are similar.

If ∆𝑴𝑬𝑹 is a right triangle with ∠𝑴𝑬𝑹 as the

̅̅̅̅̅ as the hypotenuse.
right angle and 𝑀𝑅
If
𝑸𝑹 𝑷𝑹
= 𝑺𝑼 ; ∠𝑹 ≅ ∠𝑼, ̅̅̅̅
𝐸𝑌 is an altitude to the hypotenuse of ∆𝑴𝑬𝑹.
𝑻𝑼 Then, ∆𝑴𝑬𝑹~∆𝑬𝒀𝑹~∆𝑴𝒀𝑬
Then, ∆𝑷𝑸𝑹~∆𝑺𝑻𝑼
The altitude (𝒄) to the hypotenuse of ∆MER is
Example: ∆𝑱𝑴𝑶 ↔ ∆𝑨𝑴𝑹
𝐽𝑀 20 𝟒 the geometric mean between 𝒂 and 𝒃.
= = 𝒂 𝒄
𝐴𝑀 15 𝟑 = ; 𝒄 = √𝒂𝒃
𝒄 𝒃
𝑂𝑀 8 𝟒 The shorter leg (𝒅) of ∆MER is the geometric
= =
𝑅𝑀 6 𝟑 mean between 𝒂 and 𝒇.
𝐽𝑀 𝑂𝑀 𝟒 𝒂 𝒅
= = = ; 𝒅 = √𝒂𝒇
𝐴𝑀 𝑅𝑀 𝟑 𝒅 𝒇
𝐽𝑀 𝑂𝑀
The longer leg (𝑒) of ∆MER is the geometric
Since = 𝑅𝑀 and ∠𝑱𝑴𝑶 ≅ ∠𝑨𝑴𝑹, mean between 𝒃 and 𝒇.
𝐴𝑀
Therefore, ∆𝑱𝑴𝑶 ~ ∆𝑨𝑴𝑹 𝒃 𝒆
= ; 𝒆 = √𝒃𝒇
𝒆 𝒇

II. Learning Competency: Applies the theorems to show that given triangles are similar
(M9GE-IIIi-1)
III. ACTIVITIES
ACTIVITY 1:
Directions: Match the illustrations of similarity concepts to the theorems that justify the similarity.
Write only the number of the figure that corresponds to the similarity concept.

𝐵𝐴 𝐴𝑌 𝐵𝑌 𝐽𝑂 𝐽𝑌
= = = ; ∠𝐽 ≅ ∠𝐹
𝐿𝑂 𝑂𝐺 𝐿𝐺 𝐹𝐼 𝐹𝑇
∆𝐵𝐴𝑌~∆𝐿𝑂𝐺 ∆𝐽𝑂𝑌~∆𝐹𝐼𝑇
① ②

∠𝑋 ≅ ∠𝐴 ;
∆𝑆𝑂𝐿~∆𝑆𝑈𝑂~∆𝑂𝑈𝐿 ∠𝑌 ≅ ∠𝐵
∆𝑋𝑌𝑍~∆𝐴𝐵𝐶
③ ④
Figure No. Similarity Theorem Figure No. Similarity Theorem

Right Triangle Similarity Theorem AA Similarity Theorem

SAS Similarity Theorem SSS Similarity Theorem

47
ACTIVITY 2

Directions: Complete the following statements to justify that the given triangles are similar.

𝐀𝐂 𝐀𝐌
1. If = and ∠CAM ≅ ________, then 2. If ∠T ≅ ∠K and ∠R ≅ ______, then
𝐀𝐒 𝐀𝐘
∆ RST ~ ∆ LJK by AA Similarity Theorem.
∆ CAM ~ ∆ SAY by SAS Similarity Theorem.

3. If ̅̅̅̅
SD is an altitude to the hypotenuse ̅̅̅̅
UT of 4. If
𝐀𝐁
=
𝟏𝟒 𝟐
= ,
𝟐
=
𝐱 𝟐
and =
𝟏𝟎
, then
𝐉𝐎 𝟐𝟏 𝟑 𝟑 𝟏𝟐 𝟑 𝐲
∆UST, then ∆UST ~ ∆UDS, ∆UDS ~ ∆SDT
for what value of 𝒙 and 𝒚 will make
and ∆UST ~ _______.
∆ ABC ~ ∆ JOG by SSS Similarity Theorem?
𝑥 = ___, 𝑦 = _____

IV. Reflection

Choose the emoji that would describe your understanding of the topic.

48
V. References:

Bryant, Merden L., Bulalayao, Leonides E., et al. Mathematics Grade 9 Learner’s Material
Module 6: Similarity, First Edition, (2014), Department of Education, Vibal Group Inc.

Dilao, Soledad J., Bernabe, Julieta G., Orines, Fernando B., Geometry, SD Publications, Inc.,
Gregorio Araneta Avenue, Quezon City, 2009

Primiani, Rose, Caroscio, William, Prentice Hall Comprehensive Review for New York Math
Examination, Pearson Education Inc. 2005

1. ∠SAY, 2. ∠L, 3. ∆SDT, 4. 𝑥 = 8, 𝑦 = 15

Activity 2
Theorem, ①SSS Similarity Theorem
③ Right Triangle Similarity Theorem, ②SAS Similarity Theorem, ④AA Similarity
Activity 1

Answer Key:

Prepared by:

JOMAR A. MONFORTE
Writer

49
 MATHEMATICS 9

Name: ___________________________________Grade and Section: ____________Date: ________

LEARNING ACTIVITY SHEET

Proving Pythagorean Theorem

I. BACKGROUND INFORMATION FOR LEARNERS:

One of the best known and most useful mathematical formulas is the Pythagorean Theorem,
named after the Greek mathematician and philosopher Pythagoras.

The theorem relates the length of the sides of a right triangle. The
side c which is opposite the right angle is the longest side, known as
hypotenuse. The other two sides, a and b are called the legs of the
right triangle.

What is Pythagorean Theorem?

Pythagorean Theorem states that in a right triangle, the square of the length of the
hypotenuse is equal to the sum of the squares of the lengths of the legs.

Let us extend a square on each side of the right triangle as given below.
(Size of each small box in the squares 1, 2 and 3 are same in size)

In square 1, each side is divided into 3 units equally.

Then the side length of square 1, a = 3
In square 2, each side is divided into 4 units equally.
Then the side length of square 2, b = 4
In square 3, each side is divided into 5 units equally.
Then the side length of square 3, c = 5

AREA OF THREE SQUARES

Area of square 1: Area of square 2: Area of square 3:

a = 32 = 9 square units
2
b2 = 42 = 16 square units c2 = 52 = 25 square units

In the areas of above three squares, we can have

Area of square 3 = Area of square 1 + Area of square 2

From the above result, it is clear that the sum of squares of two sides of a right triangle is
equal to the square of the third side.
c2 = a2 + b2

Hence, Pythagorean Theorem is proved.

The Pythagorean Theorem can be used to find the length of any side of a right triangle when
the lengths of the other two sides are given.

50
 C

b a
Two-Column Proof
Given: △ ABC with ∠𝐶 as right angle, CD ⊥ 𝐴𝐵 d c-d
Prove: c² = a² + b² B
A D
c C

STATEMENT REASON

1. 𝐶𝐷 ⊥ 𝐴𝐵 at D 1. Given

2. Let AD = d, then BD = c-d 2. Proportionality Theorem

3. △ ABC~ △ CBD~ △ ACB 3. Similarity in Right Triangles Theorem

𝑐 𝑎 𝑐 𝑏 4. Geometric Mean Theorem
4. 𝑎 = 𝑐−𝑑 𝑎𝑛𝑑 𝑏
= 𝑑

5. c(c-d)=a² and cd=b² 5. Cross Multiplication Property

6. c²-cd=a² and cd=b² 6. Distributive Property

7. c²- cd+ cd = a² + b² 7. Addition Property of Equality

8. c² = a² + b² 8. The square of the length of the hypotenuse is

equal to the sum of the two legs.

II. LEARNING COMPETENCY: The learner proves the Pythagorean Theorem. (M9GE-IIIi-2).

III. ACTIVITIES

Select the statements or reasons from the choices that are left blank in the proof of the
Pythagorean Theorem.
M
Given: LM= r and MN=s as the legs
LN = t as the hypotenuse s
∠𝐿𝑀𝑁 is a right angle r w
Prove: r² + s² = t² v
u
Proof: L N
K
t M

STATEMENT REASON

1. △LMN~ △MKN~ △LKM 1. Right Triangles Similarity Theorem

A. Addition Property
𝑢 𝑟
2. 𝑟 = 𝑡 ∶ 2. Geometric Mean Theorem of Equality

B. r2 = ut, s2 = vt
3. 3. Cross Multiplication Property
𝑣 𝑠
C. =
𝑠 𝑡
4. r²+s²= ut + vt 4.
D. Common
5. r²+s²=t(u + v) 5. Monomial Factoring

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 6. r²+s²=t(t) 6. Segment Addition Postulate

7. Simplify r²+s²=t² 7. Product Law of Exponents or

Pythagorean Theorem

Write the statements or reasons that are left blank in the proof of the Pythagorean Theorem.
E
Given: △ FED with ∠𝐸 as a right angle and EG as altitude of △FED M

2
Prove: e² = f² + d f d
h i
D G F
M
e

STATEMENT REASON

1. h + i = e 1.

2. 2. Geometric Mean Theorem

3. f2 + d2 = he + ce 3.

4. 4. Distributive Property

5. f2 + d2 = e (e) 5.

6. 6. Pythagorean Theorem

Solve the following problems using the Pythagorean Theorem.

1. The size of a TV screen is given by the length of its diagonal. If the dimension of a TV screen is 16
inches by 14 inches, what is the size of the TV screen?

2. A 20-foot ladder is leaning against a vertical wall. If the foot of the ladder is 8 feet from the wall,
how high does the ladder reach? Include an illustration in your solution.

3. The figure of the A-frame of a house is not drawn to scale. Find the lengths GR and OR.
R

5ft
G O
7ft 4ft

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 IV. REFLECTION: Let us summarize what you have learned from the activities by completing the
statements.

The thing I like the most from the The question I still want to ask
The concepts /words I love are _______________________
activities is________________
_________________________________
__________________________ _______________________
_________________________________
V. ANSWER KEY
5. Substitution Property 5. D
RO ≈ 6.40𝑐𝑚 4. f2 + d2 = e (h + i) 4. A
3. GR ≈ 8.60𝑐𝑚 3. Addition Property of Equality 3. B
2. d ≈19.6 feet 2. 𝑓 = 𝑒 ∶ 𝑑 = 𝑒 2. C
ℎ 𝑓 𝑖 𝑑
1. d ≈21.26 inches 1. Segment Addition Postulate
ACTIVITY 3 ACTIVITY 2 ACTIVITY 1

VI. REFERENCES

Bryant, Merden L.,et al. Mathematics Grade 9 Learner’s Material First Edition 2014
Contextualized Detailed Lesson Plans in Mathematics 9 Quarter 3 Region V
De Leon, Cecile M.,et al. Geometry III Textbook
https://www.onlinemath4all.com/proof-of-pythagorean-theorem.html

Prepared by:

Mark Anthony A. Mapula

Writer

53
 MATHEMATICS 9

Name: ______________________________________ Grade & Section: ________Date: __________

LEARNING ACTIVITY SHEET
SOLVING PROBLEMS THAT INVOLVE TRIANGLE SIMILARITY AND RIGHT TRIANGLES

I. Background Information for Learners:

This activity sheet shall serve as your guide to facilitate learning as it focuses in solving problems
that involve triangle similarity and right triangles.
Knowing the key concepts, theorems and properties of triangle similarity will help you solve
problems that involve Triangle Similarity and Right Triangles as shown inside the box.

SAS Triangle Similarity Theorem – Two triangles are similar if an angle of one triangle is congruent
to an angle of another triangle and the corresponding sides including those angles are in
proportion
SSS Triangle Similarity Theorem – Two triangles are similar if the corresponding sides of two
triangles are in proportion.
The AA Similarity theorem – Two triangles are similar if two angles of one triangle are congruent
to two angles of another triangle.

45° √2
In 𝟒𝟓° 𝟒𝟓° 𝟗𝟎° right triangle theorem - The leg is times
2
√2 ℎ = √2𝑙
𝑙= ℎ the hypotenuse.; and the hypotenuse is √2 times each leg
2
𝑙.
45°

√2
𝑙= ℎ
2 1 60°
In 𝟑𝟎° 𝟔𝟎° 𝟗𝟎° right triangle theorem The shorter leg is 𝑠 = ℎ 𝑜𝑟 ℎ = 2𝑠
2
1 √3
the hypotenuse ℎ or times the longer leg.; √3
2 3 𝑠= 𝑙
3 30°
The longer leg 𝑙 is √3 times the shorter leg s.; and
𝑙 = √3𝑠
The hypotenuse ℎ is twice the shorter leg.
Pythagorean Theorem – The square of the hypotenuse of a right triangle is equal to the sum of the
squares of the legs. It is called “Pythagoras’ Theorem” and can be written in one short
equation a² + b² = c².

a² + b² = c² c²
a²
b²
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 Right Triangle Similarity Theorem (RTST) -if the altitude is drawn to the hypotenuse of a right
triangle, then the two triangles formed are similar to the original triangle and to each
other.
Figure Description Proportion

Y s S The altitude of YES, is the 𝑚 𝑎

= → 𝑎 = √𝑚𝑛
geometric mean between m and 𝑎 𝑛
a m n.
b The shorter leg s is the geometric 𝑚 𝑠
= → 𝑠 = √𝑚ℎ
h mean between m and h. 𝑠 ℎ
n
The longer leg b is the geometric 𝑛 𝑏
E = → 𝑏 = √𝑛ℎ
mean between n and h. 𝑏 ℎ
Y Illustrative Example:
1. The corresponding sides of the similar triangles

Y s
S Original New Larger New Smaller
2 Triangle Triangle Triangle
a Hypotenuse ES EY SY
z Longer Leg EY EZ YZ
Shorter Leg SY YZ SZ
8
2. Solve for the geometric mean a, b, and s.
E Geometric Means Proportion Answer
Y Altitude a 𝑎 = √2(8) = √16 𝑎=4

Shorter leg s 𝑠 = √2(10) = √2(2)(5) 𝑠 = 2√5

Longer leg b 𝑏 = √8(10) 𝑏 = 22 √5

= √23 (2)(5) = √24 (5) = 4√5

II. Learning Competency:

Solves Problems that involve triangle similarity and right triangles. (M9GE-IIIj-1)

III. Exercises/Activities:
Direction: Analyze the situation below and supply the missing value inside the box in each
problem.
Illustrative Example #1. “Mang Nilo wants to renovate the shed for his goats by replacing its beams.
The width of the shed is 5 m., while the height of the roof is 1.2 m. What is the length of the roof
beams needed?”

55
 Solution: 𝑐²=𝑎²+𝑏²
Let c = be the length of the roof beam 𝑐 ²= + (5)²
a= height of the roof 𝑐² = 1.44 +
b= width of the shed 𝑐 ² = 26.44
Using the Pythagorean Theorem 𝑐=

The length of the roof beam is 5.14m.

Illustrative Example #2. To estimate the height of the Jin Mao Tower in Shanghai, a tourist sights the
top of the tower in a mirror that is 87.6 meters from the tower. The mirror is on the ground and faces
upward. The tourist is 0.4 meter from the mirror, and the distance from his eyes to the ground is about
1.92 meters. How tall is the tower?

1. Solving for x :
0.4
=
𝑥 87.6
0.4𝑥 = (1.92)
168.192
𝑥=

𝑥=

Therefore, the height of the tower is 420.48 m.

Illustrative Example #3. “An escalator lifts people to the second floor of a building, 25 ft above the
first floor. The escalator rises at 30° angle. To the nearest foot, how far does a person travel from the
bottom to the top of the escalator?

Solution: The hypotenuse h is twice the shorter leg.

Therefore:
Let s= be the shorter leg which is 25ft
h= 2(s)
h= be the hypotenuse
h= 2
By using the In 𝟑𝟎° 𝟔𝟎° 𝟗𝟎° right triangle
theorem: h=

Therefore 50 ft does person travel from the

bottom to the top of the escalator.

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 IV. Assessment:
Direction: Solve the following problems by using the special properties of right triangles and
the right triangle similarity theorem. Show your solution on a clean sheet of paper

1. A 12-meter fire truck ladder is leaning on

a vertical wall of a class room. If the base
of the ladder is 5 meters away from the
wall, how high does the ladder reach?

Solution:

Let c= be the unknown

a= be the 12 m fire truck ladder

b= be the base of the ladder which is 5 m
By using the Pythagorean Theorem c² = a²+ b²

c²= ² + (5)²

c²= 144 +

c²= 169

c=

Therefore the ladder reached high.

2. Similar triangles are drawn with right
angles at R and Q. S is located by sighting
from P to T. By direct measurement, RS =
50 m, PQ= 8 m and QS = 10 m. Find the
distance from one side of the river bank to
a tree on the opposite side.

Solution:

RST ̴ QSP
𝑇𝑅 𝑅𝑆
=
𝑃𝑄 𝑄𝑆
𝑇𝑅 00
=
8 10
50(8) 400
𝑇𝑅 = = = 40𝑚
10 10
Therefore the distance from one side of the river
bank to a tree on the opposite side is m.

57
 This is your chance to write a letter to your teacher
to tell them everything they need to know about
your feelings after answering the activities.

Dear Ma’am/Sir:

Love:

Answer Key:
Therefore the ladder reached 13 m high.
10 10
𝑇𝑅 = = = 40𝑚
c= 13 m 50(8) 400
c²= 169 8 10
c²= 144 + 25 =
𝑇𝑅 50
c²= (12)² + (5)²
b² 𝑃𝑄 𝑄𝑆
=
By using the Pythagorean theorem c² = a²+ 𝑇𝑅 𝑅𝑆
b= be the base of the ladder which is 5 m QSP RST ̴
a= be the 12 m fire truck ladder
Solution: 1.
Let c= be the unknown
2. Solution:

V. References
Learner’s Materials in Grade 9 Mathematics
Curriculum Guide in Grade 9 Mathematics
Geometry III Textbook for Third Year by Cecile M. De Leon, Soledad Jose-Dilao, Ed.D.
Julieta G. Bernabe
Dynamic Math III Textbook for Mathematics for Third Year by Priscila C. De Sagun
Pearson Texas Geometry, Student Text and Homework Helper by Stuart J. Murphy
www.geometryonline.com/self_check_quiz/sol

Prepared by:

JEFFREY P. BOBIS
Writer

58