Module 5: Quadrilaterals
Lesson 1: Parallelograms
Learning Competency 29: Identifies quadrilaterals that are parallelogram
I – OBJECTIVES
1. Recall the different kinds of quadrilaterals.
2. Identify quadrilaterals that are parallelogram.
3. Appreciate the importance of quadrilateral in real life.
II – SUBJECT MATTER
Topic: Quadrilaterals
Sub-topic: Parallelograms
Materials: laptop and monitor, activity sheet
References: Grade 9 Teacher’s Guide pp.305-306
Grade 9 Learner’s Material pp.305-307
Dilao, Soledad J, et.al (2002). Geometry (New Trends in Math
Series), pp. 140-143
III – PROCEDURE
A.
Preliminary
Showing the following pictures on the TV screen, let each learner study the
features of each picture.
Questions:
1. What have you observed on the pictures shown in the TV screen?
2. Which part of the pictures represent polygon?
3. What kind of polygon did you see in the picture?
B.
Lesson Proper
1. Teaching/Modeling
Activity 1: REFRESH YOUR MIND!
Direction: Complete the table by recalling the definition of each
quadrilateral. Write it on your activity sheet.
Figure
Trapezoid
Parallelogram
Rectangle
Rhombus
Square
Kite
Definition
Activity 2: WE ARE FAMILY!
Study the schematic diagram of quadrilateral and answer the
the questions that follow.
2. Analysis
a. How are quadrilaterals related to each other?
b. Which quadrilaterals are parallelograms?
3. Guided Practice
Study the figure below and name as many parallelograms as you
can. If the parallelogram is special, tell whether if it is a rectangle,
rhombus or square.
A
B
E
F
K
L
C
G
Examples:
Rhombus BCHF
Parallelogram GIML
H
M
D
I
J
N
4. Independent Practice
Using the Cartesian plane, plot each sets of points and connect
consecutively to form a quadrilateral. Identify whether the figure is a
parallelogram or not.
1. A (-1,2), B (-1,0), C (1,0), D (1, 2)
2. E (1, 0), F (3, 0), G (0, -2), H (3, -2)
3. I (-4, -2), J (-4, -4), K (0, -2), L (0, -4)
4. M (3, 4), N (2, 2), O (3, 0), P (4, 2)
5. Q (-4, 2); R (-5,1), S (-3, 1), T (-4, -2)
5. Generalization
Definition:
• Quadrilateral is a closed figure with four sides.
• Parallelogram is a quadrilateral with two pairs of opposite sides
parallel
• Rectangle is a parallelogram with all angles are right
• Rhombus is a parallelogram with all sides congruent
• Square is a parallelogram with all sides congruent and with all
angles are right.
6. Application
In the picture is the temporary ICT room of Kaong National High
School while the construction of the new building is on-going.
Name all the objects that represent parallelogram in this room.
7. Assessment
Using the schematic diagram of quadrilaterals, classify each
statement as true or false.
a. Every rectangle is a quadrilateral.
b. Every rectangle is a parallelogram.
c. Every square is a rectangle
d. Every parallelogram is a square
e. A square is both a rectangle and a rhombus.
IV – ASSIGNMENT
1. Follow-up
Name 5 things inside your house that represent parallelogram.
2. Study
The conditions that guarantee a quadrilateral a parallelogram (Properties
of a parallelogram).
Reference: Mathematics Learner’s Materials (Grade 9), pp. 309-310
Electronic Sources:
www.regentsprep.org/regents/math/geometry/gp9/lparallelogram.htm
https://www.google.com.ph/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=
8&ved=0ahUKEwjvi9Sp0P3PAhWFJJQKHWZaD4EQjRwIBw&url=http%3A%2F%2Fslidepla
yer.com%2Fslide%2F4491674%2F&bvm=bv.136811127,d.dGo&psig=AFQjCNEQYsdp6a6Mhvh0SN2hGk4d32KdQ&ust=1477747017165639
Answer Key
Guided Practice:
1. Square ADNK
2. Rectangle ADJE
3. Rhombus CDIH
4. Rectangle EJNK
5. Parallelogram FGLK
6. Parallelogram BDMK
7. Rhombus FIMK
Independent Practice
1. Parallelogram
2. Not parallelogram
3. Parallelogram
4. Parallelogram
5. Not Parallelogram
Application:
1. Monitor
2. Keyboard
3. Mouse pad
4. Top of the table
5. Key/keys on the key board
Assessment
1. True
2. True
3. True
4. False
5. true
Module 5: Quadrilaterals
Lesson 1: Parallelogram
Learning Competency 30: Determines the conditions that guarantee a
quadrilateral a parallelogram
I – OBJECTIVES
1. Determine the conditions that make a quadrilateral a parallelogram.
2. Identify if the quadrilateral is a parallelogram using the given condition.
3. Appreciate the importance of quadrilaterals in real life.
II – SUBJECT MATTER
Topic: Parallelogram
Sub-topic: Quadrilaterals that are Parallelogram
Materials: Activity sheets, laptop and monitor
References: Grade 9 Teacher’s Guide pp. 210-212
Grade 9 Learner’s Material pp. 309-313
III – PROCEDURE
A. Preliminary
In the table that follows, write T in the second column if your guess on the
statement is true; otherwise, write F. You are to revisit the same table later
on and respond to your guesses by writing R if you were right or W if wrong
under the third column.
STATEMENT
My guess is
I was (R or W)
(T or F)
1. A quadrilateral is a parallelogram if
both pairs of opposite sides are
parallel.
2. A quadrilateral is a parallelogram if
both pairs of opposite sides are ≅.
3. A quadrilateral is a parallelogram if
both pairs of opposite angles are ≅
4. A quadrilateral is a parallelogram if
any two consecutive angles are
complementary.
5. A quadrilateral is a parallelogram if
exactly one pair of adjacent sides
is perpendicular
6. A quadrilateral is a parallelogram if
one pair of opposite sides are both
congruent and parallel
B. Lesson Proper
1. Teaching/Modeling
Illustrative Example # 1:
P
In quadrilateral PQRS,
PQ‖ SR and PS‖ QR
PQ ≅ SR and PS≅ QR
∴ 𝑃𝑄𝑅𝑆 𝑖𝑠 𝑎 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚
Q
S
R
Illustrative Example # 2:
L
In quadrilateral LOVE, ∠𝐿 ≅ ∠𝑉
∠𝑂 ≅ ∠𝐸
∴ 𝐿𝑂𝑉𝐸 𝑖𝑠 𝑎 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚
O
E
V
Illustrative Example # 3:
H
O
In quadrilateral HOME, HT≅ MT
and ET≅ OT
∴ 𝐻𝑂𝑀𝐸 𝑖𝑠 𝑎 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚
T
T
E
Illustrative Example # 4:
A
C
Illustrative Example # 5
H
E
M
R
m∠𝐶 + m∠𝐴 = 180º
m∠𝐴 + m∠𝑅 = 180º
m∠𝑅 + m∠𝐸 = 180º
m∠𝐸 + m∠𝐶 = 180
∴ 𝐶𝐴𝑅𝐸 𝑖𝑠 𝑎 𝑝𝑎𝑟𝑎𝑙𝑙𝑜𝑙𝑜𝑔𝑟𝑎𝑚
O
∆ 𝐻𝐸𝑂 ≅ ∆𝑃𝑂𝐸
∴ 𝐻𝑂𝑃𝐸 𝑖𝑠 𝑎 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚
E
P
2. Analysis
a. What are the conditions that guarantee that a quadrilateral is a
parallelogram?
3. Guided Practice
Given: Parallelogram MATH
1.
2.
c.
d.
e.
MA ≅
∆MAH ≅
MS≅
∆THM ≅
∠ATH ≅
M
A
S
H
T
4. Independent Practice
Complete the table by illustrating the parallelogram being described.
Put markings on the figure to represent the property.
Properties of Parallelogram
Opposite
sides
are parallel
Opposite
Opposite
Consecutive
Diagonals
sides
angles are
angles are
bisect each
are congruent
congruent
supplementary
other
5. Generalization
Conditions that guarantee a quadrilateral a parallelogram:
1. Opposite sides of a parallelogram are congruent
2. Opposite angles of a parallelogram are congruent
3. Consecutive angles of a parallelogram are supplementary
4. The diagonals of a parallelogram bisect each other
5. Either diagonal of a parallelogram forms two congruent
triangles
6. Application
Alex uses parallelogram DEFG in completing the table.
REASON
a. EF ≅
b.∠E ≅
c. FG ≅
d.∆DEF ≅
e. m∠D + m∠G =
.
7. Assessment
A. Identify the following:
a. Quadrilateral BRAD
b. BD and RA
c. ∠B and ∠R
d. ∠D and ∠R
e. DR and BA
B. Complete the statement:
f. ̅B̅T̅ ≅
g. ∠B ≅
h.
≅ ̅D̅A̅
i. m∠B + m∠R =
j. DT ≅
T
IV – ASSIGNMENT
1. Follow-up
Find the values of x, y, and z in each parallelogram.
a.
y⁰
z⁰
x⁰
61⁰
z
b.
53⁰
y
7.5 cm
x⁰
9 cm
Electronic Sources:
www.regentsprep.org/regents/math/geometry/gp9/lparallelogram.htm
Answer Key:
Guided Practice (Let’s Do These!)
a. HT
b. ∆HTA
c. TS
d. ∆MAT
e .∠ HMA
Independent Practice (I Can Do These!)
a.
b.
c.
d
e.
Application (Let’s Do More!)
a. DG; opposite sides of a parallelogram are congruent
b. ∠G; opposite angles of a parallelogram are congruent
c. ED; opposite sides of a parallelogram are congruent
d. ∆FEG; diagonal of a parallelogram forms two ≅ triangles
e. 180; consecutive angles of a parallelogram are supplementary
Assessment (Challenge Yourself)
a. Parallelogram
b. Opposite sides
c. Consecutive angles
d. Opposite angles
e. Diagonals
f. AT
g. ∠A
h. BR
i. 180
j. RT
Module 5: QUADRILATERALS
Lesson 5: Quadrilaterals
Learning Competency 31: Use properties to find measures of angles, sides and
other quantities involving parallelogram
I – OBJECTIVES
1. Describe the angles, sides and other quantities involving parallelogram
2. Find the measure of angles, sides and other quantities involving
parallelogram
3. Show patience in finding the measure of angles and sides of a
parallelogram
II – SUBJECT MATTER
Topic: Quadrilaterals
Sub-Topic: Parallelogram
Materials: Activity Sheets, , laptop and TV monitor
References: Grade 9 Teaching Guide pp. 209-2012
Grade 9 Learning Materials pp. 309 – 324
III – PROCEDURE
A.
A
Preliminary
Write C for Congruent, S for Supplementary ,P for Parallel , B for Bisect
each other to describe the angles , sides and diagonals of a parallelogram
B
1.Opposite Angles:
2.Consecutive Angles:
3.Opposite Sides:
&
1.Opposite Angles:
2.Consecutive Angles:
3.Opposite Sides:
&
4. Diagonals are
4. Diagonals are
5. Sum of consecutive angles:
5. Sum of consecutive angles:
D
E
1.Opposite Angles:
2.Consecutive Angles:
3.Opposite Sides:
&
1.Opposite Angles:
2.Consecutive Angles:
3.Opposite Sides:
&
4. Diagonals are
4. Diagonals are
5. Sum of consecutive angles:
5. Sum of consecutive angles:
Consecutive:
Consecutive:
B.
Lesson Proper
1. Teaching/Modeling
Illustrative Example 1:
Follow the procedure below to find the measure of angles, sides and
other quantities involving parallelogram, trapezoids and kites
a. ∠𝑀 measures 60°, Extend the rays
of 𝑀 , and draw parallelogram
𝑀𝑁𝑂𝑃 on the grid paper,
remember opposite angles are
congruent m∠N=120, m∠M=60,
m∠O =
and
m∠P=
b. 𝑊𝑋𝑌𝑍 is a parallelogram ,Using what
you know about parallelograms,
̅𝑊
̅ ‖̅𝑌̅𝑍̅, m ̅𝑊
𝑋
𝑋̅ = m̅𝑌̅𝑍̅
̅𝑋̅𝑌̅‖̅𝑊
̅𝑍̅, m ̅𝑋̅𝑌̅= m̅𝑊
̅𝑍̅
What is 𝑚𝑋𝑌 and 𝑚𝑌𝑍
?
M
𝑋
3 cm
𝑌
𝑀
𝑊
𝑍
6 cm
𝑚∠𝑊𝑋𝑌 = 113°. Use what you know
about angles in a parallelogram to find the measure of the other
angles.
𝑚∠𝑋𝑌𝑍 =
° 𝑚∠𝑌𝑍𝑊 =
° 𝑚∠𝑍𝑊𝑋 =
°
c. Juan measured some segments in Problem 2.
̅𝑌̅= 8
̅𝑍̅= 3 cm.
He found that ̅𝑊
cm and ̅𝑀
Give the lengths of the following segments: 𝑊𝑀 =
cm 𝑀𝑌 =
cm 𝑋𝑀 =
cm
𝑋𝑍 =
cm
d. Diagonals JL and MK bisect each other.
a) The midpoint of JL is P.
b) Another line MK passing through P so that MP ≅ KP.
c) What quadrilateral did you form using MJKL.
d) If MP=4 What is PK?
2.Analysis
a. What property of a parallelogram is involve letter a, b.c and d
b. How do you find the measure of each angle in the parallelogram?
3.Guided Practice
A .Find the measure of angles and sides in each parallelogram.
e.g
1.
A
9
D
P
I
50
112
B
C
̅ 𝐵̅
note: ̅𝐴̅𝐵̅̅= 7 , 𝐴
̅ ̅ ≅ ̅𝐷̅̅̅𝐶̅, ∴ ̅𝐷̅̅̅𝐶̅= 7
̅𝐴𝐷̅̅
̅
̅
̅ ̅ = 9 , 𝐴̅𝐷̅̅̅ ≅ 𝐵̅𝐶
̅̅
∴ ̅𝐵̅𝐶
̅ ̅ =9
𝑚∠𝐶 = 112, 𝑚∠𝐶 = 𝑚∠𝐴 ∴ 𝑚∠𝐴 = 112
Consecutive angles are supplementary
2.
N
A
𝑚∠𝑁𝐴𝐼 =
𝑚∠𝑃𝑁𝐴 =
𝑚∠𝑃𝐼𝐴 =
3
0
0
X
y
H
O
7
105o
y0
75
m∠x=
,m∠y=
E
,m∠z=_
4.
14
HE =
∠H=
M
, OM =
, ∠O =
5.
a
x
y
b
c
z
60
m∠𝑎=
m∠𝑏=
m∠𝑐=
x=
y=
z=
100
Questions
1. Measure ∠NAI and ∠PNA, What did you find?
2. What can you say about the consecutive angles in each figure?
3. Whenyou draw diagonals in each figure. What did you find?
4. Does quadrilateral HOME appear to be a parallelogram? Why?
5. What specific parallelogram does it represent?
4. Independent Practice
Activity 6.2: Defense! Defense!
Study the following parallelograms below and answer the questions .
Figure
Questions
Answer
1.
A
6
a. if AD = 7
What is BC?
If AB = 6
What is DC?
b. Why did you say
so?
B
7
D
C
2.
L
O
115
0
65 0
V
E
3.
L
a.If m∠ELO=115
Find m∠ LOV
m∠EVO
b. Why did you say
so?
a.PI = 4 and LP = 6
What is KP & PE
b. Why?
I
P
K
E
S
a. If SU = 20
U What is ER?
b. Why?
E
R
4.
5.Generalization
The opposite sides of a parallelogram are congruent and parallel
The opposite angles of a parallelogram are congruent.
Consecutive angles of a square/ rectangle are congruent and
supplementary
Diagonals bisect each other.
6. Application
Do the procedures below and answer the questions that follow.
Materials Needed: bond paper, protractor, ruler, pencil, and compass
Procedure:
1. Mark two points O and P that are 10 cm apart.
2. Draw parallel segments from O and P which are 6 cm each, on the
same side of OP and are perpendicular to OP.
3. Name the endpoints from O and P as H and E, respectively, and
draw HE.
4. Draw the diagonals of the figure formed.
Questions:
1. Measure ∠OHE and ∠PEH. What did you find?
2. What can you say about the four angles of the figure?
3. Measure the diagonals. What did you find?
4. Does quadrilateral HOPE appear to be a parallelogram? Why?
5. What specific parallelogram does it represent
7..Assessment:
Choose the letter of the best answer
1. What is the measure of ∠2 in rhombus
HOME?
a. 75°
b. 90°
c. 105°
d. 180°
2.. Two consecutive angles of a
parallelogram have measures (x + 30)° and
[2(x – 30)]°. What
is the measure of the smaller angle?
a. 30° c. 100°
b. 80° d 140°
3.. In rhombus RHOM, what is the measure of
∠ROH?
a. 35°
b. 45°
c. 55°
d. 90°
4. In rectangle KAYE, YO = 18 cm. Find the length of diagonal AE.
a. 6 cm
b. 9 cm
c. 18 cm
d. 36 cm
IV – ASSIGNMENT
3. Follow-up
Show More What You’ve Got!
Solve each problem completely and accurately on a clean sheet of paper.
Show your solution and write the theorems or properties you applied to
justify each step in the solution process. You may illustrate each given, to
serve as your guide. Be sure to box your final answer.
A. Given: Quadrilateral WISH is a parallelogram.
a. If m ∠∠W = x + 15 and m∠ ∠S = 2x + 5, what is m ∠∠W?
b. If WI = 3y + 3 and HS = y + 13, how long is HS?
c. ❏WISH is a rectangle and its perimeter is 56 cm. One side is 5 cm
less than twice the other side. What are its dimensions and how large
is its area?
d. What is the perimeter and the area of the largest square that can be
formed from rectangle WISH in 1.c.?
B. Given: Quadrilateral POST is an isosceles trapezoid with OS || PT. ER
is its median.
a. If OS = 3x – 2, PT = 2x + 10 and ER = 14, how long is each base?
b. If m ∠P = 2x + 5 and m ∠O = 3x – 10, what is m ∠T?
c. One base is twice the other and ER is 6 cm long. If its perimeter is
27 cm, how long is each leg?
d. ER is 8.5 in long and one leg measures 9 in. What is its perimeter if
one of the bases is 3 in more than the other?
2. Study how find the measure of angles and sides of trapezoids
Electronic Sources:
9 Math LM_U3.M5.v1.0pdf-adobe Reader Module 6: Quadrilateral pages 16-44
ANSWER KEY
Preliminary
A
B
1.Opposite Angles: C
2.Consecutive Angles: C & S
3.Opposite Sides: C & P
1.Opposite Angles: C
2.Consecutive Angles: S
3.Opposite Sides: C & P
4. Diagonals are B
4. Diagonals are B
5. Sum of consecutive angles:180
5. Sum of consecutive angles: 180
D
E
1.Opposite Angles: C
2.Consecutive Angles: S
3.Opposite Sides: C & P
1.Opposite Angles: C
2.Consecutive Angles: C & S
3.Opposite Sides: C & P
4.Diagonals are B
4.Diagonals are B
5.Sum of consecutive angles180
5.Sum of consecutive angles: 180
Teaching/Modeling
Illustrative Example 1:
Follow the procedure below to find the measure of angles, sides and
other quantities involving parallelogram, trapezoids and kites
𝑎. ∠𝑀 measures 60°, Extend the
rays of 𝑀 , and draw
parallelogram 𝑀𝑁𝑂𝑃 on the grid
paper, remember opposite
angles are congruent
m∠N=120, m∠M=60, m∠O = 60
and m∠P=120
𝑁
O
M
P
𝑏. 𝑊𝑋𝑌𝑍 is a parallelogram ,Using what you know
about parallelograms,
̅𝑊
̅ ‖̅𝑌̅𝑍̅, m 𝑋
̅𝑊
̅ = m̅𝑌̅𝑍̅
a. 𝑋
̅𝑋̅𝑌̅‖̅𝑊
̅𝑍̅, m ̅𝑋̅𝑌̅= m̅𝑊
̅𝑍̅
What is 𝑚𝑋𝑌 = 6𝑐𝑚 and 𝑚𝑌𝑍 = 3𝑐𝑚
𝑋
𝑌
𝑀
3 cm
𝑊
𝑍
6 cm
b. 𝑚∠𝑊𝑋𝑌 = 113°. Use what you know about
angles in a parallelogram to find the measure of the other angles.
𝑚∠𝑋𝑌𝑍 = 67°
𝑚∠𝑌𝑍𝑊 = 113°
𝑚∠𝑍𝑊𝑋 = 67°
̅𝑌̅= 8
c.Juan measured some segments in Problem 2. He found that ̅𝑊
̅𝑍̅= 3 cm. Give the lengths of the following segments:
cm and ̅𝑀
𝑊𝑀 = 4 cm
𝑀𝑌 = 4cm
𝑋𝑀 = 3 cm
𝑋𝑍 = 6cm
d.Diagonals JL and MK bisect each other.
a. The midpoint of JL is P.
b. Another line MK passing through P so that
MP ≅ KP.
c. What quadrilateral did you form using
MJKL.
Parallelogram
d. If MP=4 What is PK? 8
3. Guided Practice
A .Find the missing angles and sides in each parallelogram.
e.g
1.
A
9
D
P
I
50
112
B
C
note: ̅𝐴̅𝐵̅̅= 7 , ̅𝐴𝐵̅
̅ ̅ ≅ ̅𝐷̅̅̅𝐶̅, ∴ ̅𝐷̅̅̅𝐶̅= 7
̅𝐴𝐷̅̅
̅ ̅ = 9 , ̅𝐴̅𝐷̅̅̅ ≅ ̅𝐵̅𝐶
̅̅
∴ ̅𝐵̅𝐶
̅ ̅ =9
𝑚∠𝐶 = 112, 𝑚∠𝐶 = 𝑚∠𝐴 ∴ 𝑚∠𝐴 = 112
Consecutive angles are supplementary
N
𝑚∠𝑁𝐴𝐼 =50
𝑚∠𝑃𝑁𝐴 =130
𝑚∠𝑃𝐼𝐴 =130
A
2.
3
0
0
X
y
H
O
7
105o
E
14
M
y0
75
HE =7, OM =7
∠H=_105, ∠O =75
m∠x=105,m∠y=75 ,m∠z=105
4.
5.
a
x
y
b
c
z
60
100
x=100
y=80
z=80
m∠𝑎=60
m∠𝑏=120
m∠𝑐=120
Questions; answer may vary
Independent Practice
Activity 6.2: Defense! Defense!
Study the following parallelograms below and answer the questions .
Figure
Questions
Answer
1.
a. if AD = 7
A
6
B
7
D
C
What is BC?
If AB = 6
What is DC?
b. Why did you say
so?
6
2.
a. If m∠ELO=115
O
L
115
0
Find m∠ LEV
m∠EVO
m∠ LEV=65 0
m∠EVO= 115
b. Why did you say
E
V
3.
L
I
so?
a. PI = 4 and LP = 6
What is KP & PE
b. Why?
KP = 4 & PE =6
a. If SU = 20
What is ER?
b. Why?
ER = 20
P
K
E
4.
S
E
U
R
Application
Do the procedures below and answer the questions that follow.
Materials Needed: bond paper, protractor, ruler, pencil, and compass
Procedure:
1. Mark two points O and P that are 10 cm apart.
2. Draw parallel segments from O and P which are 6 cm each, on the
same side of OP and are perpendicular to OP.
3. Name the endpoints from O and P as H and E, respectively, and
draw HE.
4. Draw the diagonals of the figure formed.
Questions:
1. Measure ∠OHE and ∠PEH. What did you find?
2. What can you say about the four angles of the figure?
3. Measure the diagonals. What did you find?
4. Does quadrilateral HOPE appear to be a parallelogram? Why?
5. What specific parallelogram does it represent
7..Assessment:
Choose the letter of the best answer
1.What is the measure of ∠2 in rhombus HOME?
a. 75°
b. 90°
c. 105°
d. 180°
2.. Two consecutive angles of a parallelogram have measures (x + 30)° and
[2(x – 30)]°. What
is the measure of the smaller angle?
a. 30°
c. 100°
b. 80°
d 140°
3.. In rhombus RHOM, what is the measure of
∠ROH?
a. 35°
c. 55°
b. 45°
d. 90°
4. In rectangle KAYE, YO = 18 cm. Find the length of diagonal AE.
a. 6 cm
b. 9 cm
c. 18 cm
d. 36 cm
ASSIGNMENT
1. Follow-up
Show More What You’ve Got! Page 44
Solve each problem completely and accurately on a clean sheet of paper.
Show your solution and write the theorems or properties you applied to
justify each step in the solution process. You may illustrate each given, to
serve as your guide. Be sure to box your final answer.
A. Given: Quadrilateral WISH is a parallelogram.
a. If m ∠∠W = x + 15 and m∠ ∠S = 2x + 5, what is m ∠∠W?
b. If WI = 3y + 3 and HS = y + 13, how long is HS?
c. ❏WISH is a rectangle and its perimeter is 56 cm. One side is 5 cm
less than twice the other side. What are its dimensions and how large
is its area?
d. What is the perimeter and the area of the largest square that can be
formed from rectangle WISH in 1.c.?
B. Given: Quadrilateral POST is an isosceles trapezoid with OS || PT. ER
is its median.
a. If OS = 3x – 2, PT = 2x + 10 and ER = 14, how long is each base?
b. If m ∠P = 2x + 5 and m ∠O = 3x – 10, what is m ∠T?
c. One base is twice the other and ER is 6 cm long. If its perimeter is
27 cm, how long is each leg?
d. ER is 8.5 in long and one leg measures 9 in. What is its perimeter if
one of the bases is 3 in more than the other?
Answer Key
A. a. m∠W = 25 ;
b. HS = 18 ;
c. dimensions are 11 cm x 17 cm, area is 187cm2
d. area of the largest square = 121 cm2
B. a. The bases measure 10 and 18. ;
b. ∠T = 79O ;
c. Each leg is 7.5 cm long ;
d. Perimeter = 35 in
Module 5: Quadrilaterals
Lesson 1: Special Parallelograms
Learning Competency 32: Proves theorems on the different kinds of
parallelogram (rectangle, rhombus, square)
I – OBJECTIVES
a. Prove theorem on special parallelogram.
b. Apply theorem on special parallelogram.
c. Value accumulated knowledge as means of new understanding.
II – SUBJECT MATTER
Topic: Quadrilaterals
Sub-Topic: Special Parallelogram
Materials: Activity Sheets, Laptop and monitor
References: Learner’s Material for Mathematics 9 pp. 317-326
Teachers Guide for Mathematics 9 pp. 317-326
III – PROCEDURE
A. Preliminaries
1. Pre - Assessment
Direction: Study the photo below.
Question:
How can you tell whether a given figure is a parallelogram, a rectangle,
a rhombus or a square?
2.. Motivation
Are you familiar with one of the projects of Senator Villar here
in Tagaytay? Have you already seen Tagaytay Crosswinds at
Iruhin? Going there feels like you were in Switzerland.
www.vistalandphilippines.com
Motive Question:
How did the contractors and carpenters apply the
conditions for rectangles to make sure that the frame of a house
has correct shape?
B. Lesson Proper
1. Teaching/Modeling
Theorem
Example
a. Parallelogram: If one angle of a parallelogram is a right angle,
then the parallelogram is a rectangle.
b. Parallelogram with diagonals are congruent is a
rectangle.
c. Parallelogram with one pair of consecutive sides are
congruent is a rhombus.
d. Parallelogram with diagonals are perpendicular is a
rhombus.
e. Parallelogram with diagonals bisecting opposite angles is
a rhombus.
f. Parallelogram with diagonals is perpendicular and has
four right angles and four congruent sides is a square.
2. Analysis
Rectangle, rhombus and square are special parallelograms because of
the distinct characteristics of their diagonals. How the diagonals of these
parallelograms differ from one another?
3. Guided Practice
Work in Pairs. Fill in the blanks of the correct answer
1. A contractor of City Land near Olivarez Plaza, built a wood frame for the
side of a house so that 𝑋𝑌≌𝑊𝑍 and 𝑋𝑊≌𝑌𝑍. Using a tape measure, the
̅𝑌̅. Why must the frame be a rectangle?
contractor found that ̅𝑋̅𝑍̅=̅𝑊
Reason: Both pairs of opposite sides of WXYZ are
, so WXYZ is a
̅
parallelogram, since ̅𝑋̅𝑍̅=̅𝑊
𝑌̅, the
of
WXYZ are
,
Therefore the frame is a rectangle by theorem b.
2. Given: PARK is parallelogram
𝑅𝑃 bisects ∠KRA and ∠KPA
Prove: JKLM is a rhombus
Statements
Reasons
1.
2. ∠1≌∠2, ∠3≌∠4
1. Given
2.
3. 𝐽𝐿≌𝐽𝐿
4. ∆ ≅ ∆ _
3.
4. ASA Congruence Postulate
5. 𝐽𝐾≌𝐽𝑀
5.
6.
6.
with one pair of consecutive
sides are ≌
2. Independent Practice
Answer the following using the figure at the right.
1. If QUAD is a rectangle,
a) 𝑈𝐸=21 𝐸𝐷̅̅ =
b) 𝑄𝑈=13.5,𝐷̅̅𝐴=
c) m∠QUA =
d) m∠DQA=38°,m∠UQA=
e) m∠QAU=25°, m∠AEU=
2.If QUAD is a square
a. 𝑄𝐴=17.5 𝐷̅̅𝑈=
b. 𝐷̅̅𝐴=37,𝑄𝑈=
c. m∠QUA=
d. m∠QED=
e. m∠UDA=
Generalization
Things to Remember:
1. Diagonals of rhombus are perpendicular.
2. Each diagonal of a rhombus bisects a pair of opposite
angles.
3. Diagonals of a rectangle are congruent.
4. Diagonals of a square are perpendicular and congruent.
3. Application
Solve the problem below.
Luis of Sonya’s garden is making a frame to provide shelter to his
planted tomatoes; He stretched plastic cover a wooden frame. Each wall of
the frame is a rectangle, with diagonal braces added for support, as shown.
If the brace connecting points A and C has length of 73 inches, how long is
the brace connecting points B and D? What is the length from the point of
intersection to pt A?
4. Assessment
If RHOM is a rhombus, 𝑅𝐻 = 6𝑦 + 4, 𝐻𝑂 = 5𝑦 + 8, Find y, RH, HO, OM
IV- ASSIGNMENT
1. Follow-up
If ABCD is a square, 𝐴𝐵̅=7x-3 and ̅𝐵̅
𝐶̅= 4𝑥 + 9, find the perimeter of
ABCD.
2, Study: State the Midline Theorem .
Answer Key
.
Guided Practice (Let’s Do This!)
1. congruent
diagonals
congruent
2.
1. RP bisects ∠ KRA ∠ KPA
2. given
3. reflexive property
4. ∆ PRK ≅ ∆ PRA
5. reflexive property
6. JKL is a rhombus
3. Independent Practice (I Can Do This!)
The length of the brace connecting points B and D is also 73 inches.
The length from the point of intersection to point A is 36.5 inches.
4. Application (Let’s Do More)
6𝑦 + 4 = 5𝑦 + 8
RH = 6𝑦 + 4
6𝑦 − 5𝑦 = 8 − 4
= 6(4) + 4
𝑦=4
= 28
𝑂𝑀 = 28
𝑅𝑀 = 28
HO = 5(4) + 8
= 28
Assessment (Challenge Yourself!)
1.
2.
a. ED = 21
b. DA = 13.5
c. 90
d. 𝑚∠𝑄𝑈𝐴 = 52
e. 130
a. 17.5
b. 37
c. 90
d. 90
e. 45
Follow-Up
1.rectangle has 4 right angles and 2 pairs of congruent parallel sides
rhombus has 4 congruent sides
square has 4 congruent sides and congruent angles
2. x = 4 perimeter of ABCD = 100
MODULE 5: Quadrilaterals
LESSON 1: Midline Theorem
LEARNING COMPETENCY 33: Proves the Midline
Theorem
I – OBJECTIVES
a.
b.
c.
Proves the Midline Theorem
Apply the Midline Theorem
Show camaraderie in doing the activity
II – SUBJECT MATTER
Topic: Quadrilaterals
Sub-Topic: Midline Theorem
Materials: Activity Sheets, Laptop and monitor
References: Learner’s Material for Mathematics 9 pp.306-308
Teachers Guide for Mathematics 9
Math Time p 2
Geometry by Holt
III – PROCEDURE
A.
Preliminaries
Motivation:
Activity: It’s Paperellelogram!
Form a group of four members and require each member to have the
materials needed. Follow the given procedure.
Materials: 4 pieces of short bond paper, pencil, ruler, adhesive
tape, protractor, and pair of scissors
Procedure:
1. Each member of the group 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.
• Does the segment drawn look parallel to the third side of the triangle you
chose?
3. Measure the segment drawn and the third side you chose.
• Compare the lengths of the segments drawn and the third side you chose.
What did you observe?
4. Cut the triangle along the segment drawn.
• 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.
• After reconnecting the cutouts, what new figure is formed? Why?
• Make a conjecture to justify the new figure formed after doing the above
activity. Explain your answer.
• What can you say about your findings in relation to those of your
classmates?
• Do you think that the findings apply to all kinds of triangles? Why?
B. Lesson Proper
1. Teaching/Modeling
∆HNS, O is the midpoint of
Given:
̅𝐻̅𝑁
, E is the midpoint of ̅𝑁̅𝑆.
Prove:
̅𝑂̅𝐸̅|| ̅𝐻̅𝑆̅, ̅𝑂̅𝐸̅= 1 ̅𝐻̅𝑆
2
Statements
Reasons
1. ∆HNS, O is the midpoint of HN, E is
the midpoint of NS
1. Given
2. ∆HNS, O is the midpoint of HN, E is
the midpoint of NS
2. In a ray, point at a given distance from
the
endpoint of the ray.
3. ̅𝐸̅𝑁̅≅ ̅𝐸̅𝑆
3. Definition of Midpoint
4. ∠2 ≅ ∠3
4. VAT
5. ∆ONE ≅∆TSE ∆ONE ≅ ∆TSE
5. SAS Congruence Postulate
6. ∠1 ≅ ∠4
6. CPCTC
7. HN || ST
7. If AIAC, then the lines are parallel.
8. OH ≅ ON
8. Definition of midpoint
9. ON ≅ TS
9. CPCTC
10. OH ≅ ST
10. Transitive property
11.
Quadrilateral
parallelogram.
HOTS
is
a 11. Definition of parallelogram
12. OE || HS
̅ of HOTS
12. ̅𝑂̅𝐸̅is on the side of ̅𝑇
𝑂
13. OE + ET = OT
13. Segment Addition Postulate
14. OE + OE = 0T
14. Substitution (SN 2)
15. 2OE = OT
15. Addition
16. HS ≅ OT
16. Parallelogram Property1
17. 2OE = HS
17. Substitution
18. OE = 1HS (The segment joining the
18. Substitution (SN 14 & 15)
2
midpoints of two sides of a triangle is half
as long as the third side.)
2.
Analysis
What is midline Theorem?
How do we prove midline Theorem?
3.
Guided Practice
̅𝐴̅𝐶̅, respectively,
Consider ∆ABC, if M and N are midpoints of ̅𝐴̅𝐷̅̅̅ and
̅
̅𝑁is a midline ( or a midsegment). Prove that 𝑁
𝑀‖ ̅𝐷̅̅̅𝐶̅and MN =1DC.
then ̅𝑀
2
Statements
Reasons
⃗⃗⃗⃗𝑂
⃗→ , such that ̅𝑀
̅𝑁‖ ̅𝑁̅𝑂̅
1. Draw ⃗𝑁
1. Construction
2.
2. Given
̅ ≅ ̅𝑀
̅𝐷̅̅̅, ̅𝐴̅𝑁̅≅ ̅𝑁̅𝐶̅
3. ̅𝐴̅𝑀
3.
4.
4.
5. ∆MAN ≅∆OCN
5.
6.
6. CPCTC
7. AD || CO
7. If AIAC, then the lines are parallel.
8. AM ≅ CO
8.
9. MD ≅ CO
9.
10. MOCD is a parallelogram
10.
11. MN || DC
11. Definition of parallelogram
12. MO ≅ DC
12.
13. MO = DC
13.
14. MN + NO = MO
14.
15. MN + MN = MO
15.
16. 2MN = MO
16.
17. 2MN =DC
17.
18. MN = 1DC (The segment joining the
18.
2
midpoints of two sides of a triangle is half
as long as the third side.)
4. Independent Practice
Activity: Go for It!
In ∆MCG, A and I are the midpoints of MG and GC, respectively.
Consider each given information and answer the questions that follow.
1. Given: AI = 10.5
Questions:
• What is MC?
•How did you solve for MC?
2. Given: CG = 32
Questions:
• What is GI?
• How did you solve for GI?
3. Given: AG = 7 and CI = 8
Questions:
What is MG + GC?
• How did you solve for the sum?
4. Given: AI = 3x – 2 and MC = 9x – 13
Questions:
• What is the value of x?
• How did you solve for x?
• What is the sum of AI + MC? Why?
5. Given: MG ≅ CG, AG – 2y – 1, IC = y + 5
Questions:
• What is the value of y?
•
• How did you solve for y?
• How long are MG and CG? Why?
5.
Generalization
Midline Theorem
The segment whose endpoints are the midpoints of two sides of
a triangle is parallel to the third side and has a length equal to half the
length of the third side.
The next theorems can be proved by using the midline theorem.
 If the consecutive midpoints of the sides of a y quadrilateral are
joined together, then the quadrilateral formed is a parallelogram.
 If three or more parallel lines cut off congruent segments on one
transversal, then they cut off congruent segments on every
transversal.
6.
Application
Triangle PQR X, Y and Z are the midpoints of PQ, QR and PR
respectively. If PQ = 24, QR = 30 and PR = 18, find perimeter of triangle
XYZ.
7. Assessment
M and N are midpoints of AC and AD respectively. Complete each
statement.
1. If MN = 12 then DC =
2.
3.
4.
5.
If AD = 36, then AN =
If MC = 12.5 then AC =
If DC = 37 then MN =
If MN = x – 2 and DC = x +10, then DC =
IV- ASSIGNMENT
1. Follow-up
Have you been to Sky Ranch Tagaytay where the biggest ferris wheel can
be found?
A lot of kids love to play on the different rides and one of it is the
swing. In the picture below of a swing, the crossbar DE is attached at
the midpoints of the legs BA and BC. The distance AC is 4 ½ feet. The
carpenter has a timber that is 30 inches long. In this timber long
enough to be used as one of the crossbars? Explain.
2. Study:
Describe and state the theorems involving kite.
ANSWER KEY:
Guided Practice:
2. ∆ABC, if M and N are midpoints of ̅𝐴̅𝐷̅̅̅and ̅𝐴̅𝐶̅.
3. Definition of Midpoint
4. ∠ANM ≅ ∠CNO,Vertical Angles are congruent.
5. SAS Congruence Postulate
6. ∠MAN ≅ ∠OCN
7. CPCTC
8. Transitive Property
9. If in quadrilateral, a pair of side is both parallel and congruent, then the
quadrilateral is a parallelogram.
10. Opposite sides of a parallelogram are congruent
11. Definition of congruent segments
12. Segment Addition Postulate
13. Substitution (1)
14, Addition of Like terms
15. Substitution (13)
16. Division Property of Equality
Independent Practice
1. MC = 21
2. GI = 16
3. MG + GC = 30
4. x=3, AI + MC =21
5. y= 6, MG=22 and CG= 22
Application:
Perimeter of triangle XYZ =36
Assessment
1. 24
2. 18
3. 25
4. 18.5
5. 24
Assignment
Follow up
DE = 2.25 feet or 27 inches
Yes, in DE only
Module 5: Quadrilaterals
Lesson 1: Trapezoids
Learning Competency 34: Proves theorems on trapezoids
I – OBJECTIVES
a. Prove theorems on trapezoids.
b. Apply theorems on trapezoids in solving problems.
c. Show camaraderie in doing activities.
II – SUBJECT MATTER
Topic: Quadrilaterals
Sub-Topic: Trapezoid
Materials: Activity Sheets, Laptop and monitor
References: Learner’s Material for Mathematics 9 pp.306-308
Teachers Guide for Mathematics 9 pp.306-308
Math Time p 2
Geometry by Holt
III – PROCEDURE
A. Preliminaries
1. Pre-Assessment
Explain why the figure at right is NOT a parallelogram
2. Motivation
In what places here in Tagaytay can you find big basketball courts?
We have Tagaytay Sports Complex, DAP or Dev’t Academy of the
Philippines and TSAC. The photograph shows the free throw lane of
Tolentino Sports and Activity Center (TSAC)
Motive Question:
1. What is the shape of the free-throw lane above?
2. Describe the angles inside the 4-sided figure
3. Which sides of the figure appear to be parallel? Congruent?
B. Lesson Proper
1. Teaching/Modeling
If ABCD Is an isosceles trapezoid, then
̅𝐴̅𝐷̅̅̅≌ 𝐵̅
̅𝐶̅B e c au se legs of an isosceles trapezoid are ≌
∠𝐴≌∠B because base angles of an isosceles trapezoid ≌
∠D ≌∠C because base angles of an isosceles trapezoid are
≌
d. ̅𝐴̅𝐶̅≌ ̅𝐵̅̅𝐷̅̅̅b e c a u s e diagonals of an isosceles trapezoid are ≌
e. 𝐸𝐹 = ½ (AB+DC) because the median of a trapezoid is ½ the
sum of the bases.
a.
b.
c.
In trapezoid ABCD above, if AB and DC are 15 cm and 23 cm,
respectively, find the length of EF.
Solution:
𝐸𝐹 = ½ (AB+DC)
𝐸𝐹 = ½ (15 +23)
𝐸𝐹 = 19 cm
2.
Analysis
How can be the base angles and diagonals be used to determine if
the trapezoid is isosceles? What is true about median of isosceles
trapezoid?
3. Guided Practice
TRAP is an isosceles trapezoid with median
relation exists between each of the following:
̅𝐸̅𝐷̅̅̅. Determine the
1. ̅𝑅
𝑇̅ and ̅𝑃̅𝐴̅
2. ̅𝑃
𝑇̅ and ̅𝑅̅𝐴̅
3. ̅𝑇̅𝐴̅and ̅𝑅̅𝑃̅
4. ̅𝐸̅𝐷̅̅̅a n d ̅𝑃̅𝐴̅
5. ∠T and ∠R
4. Independent Practice
Many different kinds of ornamental/ medicinal plants are found here
in Tagaytay because of its good climate. Restaurants here like Sonya’s
Garden is popular for their fresh and organic dishes. Considering this pot
of fresh flowers.
Find: HR
5. Generalization
Things to remember
1. Base angles of an isosceles trapezoid are congruent.
2. Diagonals of an isosceles trapezoid are ≅.
3. The median of a trapezoid is parallel to the base and its
length is half the sum of the lengths of the bases.
4. The median of a trapezoid bisects each of the diagonals.
6.
Application
LOVE is an isosceles trapezoid, find the indicated measures.
If LO = 2x -2 , YU = 15 and EV = 3x+2, find x, LO and EV.
7. Assessment
Use isosceles trapezoid TRAP to find the following measure:
TP = 18
2. RP
IV. ASSIGNMENT
1. Follow-up
Compare an isosceles trapezoid to a trapezoid that is not isosceles.
What properties do the figures have in common? What property does one
have that the other does not?
Answer Key
.
Guided Practice (Let’s Do This)
1. they are parallel to each other
2. congruent
3. congruent
4. ED = 1 𝑜𝑓 𝑃𝐴 + 𝑇𝑅
2
5. congruent
Independent Practice (I Can Do This!)
HR = 27
Application (Let’s Do More)
X=6
LO = 10
Assessment (Challenge Yourself!)
1. 9
2. 63 cm
3. 12
4. 69
5. 69
Follow-Up
EV = 20
Module 5: Quadrilaterals
Lesson 1: Kite
Learning Competency 34: Proves theorems on trapezoids and kite.
I – OBJECTIVES
a. Prove theorem on kites.
b. Apply theorem on kites in solving problems.
c. Show camaraderie in doing activities.
II – SUBJECT MATTER
Topic: Quadrilaterals
Sub-Topic: Kite
Materials: Activity Sheets, Laptop and monitor
References: Learner’s Material for Mathematics 9 pp.306-308
Teachers Guide for Mathematics 9
Math Time p 2
Geometry by Holt
III – PROCEDURE
A. Preliminaries
1. Pre-Assessment
ANAGRAM
STIBEC
TIKE
ANALOGID
SALENG
Question:
If you are asked to use these words what certain theorem can you form?
2. Motivation
Have you already been to Picnic Grove? Tourists love to sit on
the grass and dine together with their families. A few more steps is
an overlooking of Taal Volcano. There is also a wide ground for
horseback riding and kite flying where children enjoyed it so much.
Motive Question:
Did you know that a simple kite flew at the Picnic Grove shared
the properties of the geometric figure called a kite?
B. Lesson Proper
1.Teaching/Modeling
Theorem
If a quadrilateral is a kite, then its diagonals are
perpendicular.
If a quadrilateral is a kite, then exactly two
pairs of congruent adjacent sides are ≌.
Example
̅𝐴̅𝐶̅⊥̅𝐵̅̅𝐷̅̅̅
̅𝐴𝐷̅̅
̅ ̅ ≅ ̅𝐴̅𝐵̅̅
̅𝐶̅𝐷̅̅̅≅ ̅𝐶𝐵̅
̅̅
Kite has exactly one pair of opposite angles is ≌
∠𝐵̅ ≅ ∠𝐷̅̅
Kite has exactly one diagonal bisects a pair of opposite angles
∠𝐷̅̅𝐴𝐶 ≅ ∠𝐵̅𝐴𝐶
∠𝐷̅̅𝐶𝐴 ≅ ∠𝐵̅𝐶𝐴
2. Analysis
What is true about the diagonals and adjacent sides of kite?
3. Guided Practice
̅ and
Given: JKLM is a kite with ̅𝐽̅𝐾̅≅ ̅𝐽̅𝑀
Prove: ∠K ≌ ∠M
̅𝐾̅𝐿≅ ̅𝑀
̅𝐿
̅ and 𝐾
̅ ̅𝐿≅ ̅𝑀
̅𝐿. By the
It is
that 𝐽̅ ̅𝐾̅≅ ̅𝐽̅𝑀
𝐽̅𝐿≌𝐽̅𝐿. This means that ΔJKL ≌ ΔJML by
.
Property
So ∠K ≌∠M by
4. Independent Practice
̅𝑅̅and ̅𝐴̅𝐼 complete the following
Given: MARI is a kite with diagonals ̅𝑀
̅𝐼 ≅
1. ̅𝑀
2.
3.
4.
̅𝑀
̅𝐸̅≅
̅𝑅
̅̅
𝑀
∆IER is a
̅𝐴𝐼̅
5. m∠MEA =
6. ∠AMI ≅
7. ∠MAI ≅
8. ∠RIA ≅
̅𝐴̅≅
9. ̅𝑀
10. ∆AMI≅
5. Generalization
Things to Remember
1. The diagonals of a kite are ⊥.
2. Exactly one pair of opposite ∠s is ≌.
3. Exactly one diagonal of a kite bisects a pair of opposite ∠s
6. Application
In KITE PQRS, m∠PQR= 78° and m∠TRS= 59°, Find
a. m∠QRT
b. m∠QPS
c. m∠PSR
7. Assessment
Find the indicated measure if MATH is a
kite
1. AT
2. TH
3. MO
4. AO
5. m ∠MOA
MT =16
6. m ∠MAH
7. m∠MHA
8. m∠ATH
9. m∠MAT
10. m∠M + m∠A + m∠T + m∠H
IV- ASSIGNMENT
1. Follow-up
Find the area of the kite MATH (refer to figure in assessment)
2. Study
a. Define proportion
b. How do we solve for the missing term in a proportion
Answer Key
Preliminaries
BISECTS
KITE
DIAGONAL
ANGLES
Guided Practice (Let’s Do This!)
1. Given
2 Reflexive Property
3. SSS
4. CPCTC
Independent Practice (I Can Do This!)
1. RT
2. ER
3. perpendicular
4. right triangle
5. 90
6. ∠𝐴𝑅𝐼
7. ∠𝑅𝐴𝐼
8. ∠𝑀𝐼𝐴
9. RA
10. ∆𝐴𝑅𝐼
Application (Let’s Do More)
a. 51
b.110
c. 62
Assessment (Challenge Yourself!)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
10
17
8
6
90
52
33
95
104
360
IV. Follow-Up
Area of MATH = 168 square units
Module 5: Quadrilaterals
Lesson 7: Solving Problems Involving Parallelograms, Trapezoids and Kites
Learning Competency 35: Solves problems involving parallelograms,
trapezoids, and kites.
I – OBJECTIVES
a. state the properties of parallelograms, trapezoids, and kites
b. apply the properties of quadrilaterals in solving problems
c. appreciate the importance of the properties of parallelograms, trapezoids,
and kites in solving real – life problems.
II – SUBJECT MATTER
Topic: Quadrilaterals
Sub-topic: Solving Problems Involving Parallelograms, Trapezoids, and Kites
Materials: problem sheet
References: Teacher’s Guide, p. 222
Learner’s Material, p. 341
Geometry by Eunice Lopez, pp. 16 – 19
III – PROCEDURE
A. Preliminaries
1. Pre – Assessment
a. When is a quadrilateral parallelogram?
b. What parallelogram is both equilateral and equiangular?
c. How can you identify that a four – sided polygon is a trapezoid?
d. What are the properties of a kite?
2. Motivation
Which is usually more stable, a three- legged stool or a four-legged one?
Why?
B. Lesson Proper
1. Teaching/Modeling
Illustrative Examples
a. With the measures indicated, form an equation in x, and determine
the numerical degree measure of each angle of the parallelogram.
M
A
x + 40
H 3x – 20
T
Solution: x + 40 + 3x – 20 = 180
4x + 20 = 180
4x = 160
x = 40
Therefore, m∠ M = 80, m∠ A = 100, m∠T = 80, and m∠ H = 100.
b. The diagonal of a kite have lengths 13 cm and 9 cm. Find the area
of the kite.
Solution: A = 𝑑1𝑑2
A =
A=
2
(13)(9)
2
117
2
A = 58.5 cm2
ABCD is an isosceles trapezoid with m∠A = 6x + 10 and
𝑚 ∠B =4x + 50. Find m∠ A.
Solution:
D
6x + 10 = 4x + 50
2x = 40
x = 20
A
B
m∠A = 6(20) + 10 = 130
C
2. Analysis
a. What property of a parallelogram is used in forming an equation in x?
b. How is the measure of each angle obtained?
c. How is the area of a kite determined? What if the length of one
diagonal is missing, what formula is used?
d. What property of a trapezoid is applied in forming the equation?
e. How is m∠A related to m∠ B ?
3. Guided Practice
Solve each problem completely and accurately. Show your solution by
applying the property inside the parenthesis
a. Given: Quadrilateral WISH is a parallelogram.
1. If m∠W = x + 15 and m∠S = 2x + 5, what is m∠W ?
( property on opposite angles)
2. If WI = 3y + 3 and HS = y + 13, how long is ̅𝐻̅𝑆?
( property on opposite sides )
b. Given: Quadrilateral POST is an isosceles trapezoid with OS // ̅𝑃̅𝑇̅, ER
is its median.
1. If OS = 3x – 2 , PT = 2x + 10 and ER = 14, how long is each base?
(property on parallel sides)
2. If m∠𝑃 = 2𝑥 + 5 𝑎𝑛𝑑 m∠𝑂 = 3𝑥 − 10 , 𝑤ℎ𝑎𝑡 𝑖𝑠 m∠𝑇?
( property on consecutive angles)
̅ ≅ ̅𝐸
̅.
c. Given: Quadrilateral LIKE is a kite with ̅𝐿̅𝐼̅ ≅ ̅𝐼̅𝐾̅and ̅𝐿̅𝐸
𝐾
1. LE is twice LI. If its perimeter is 21 cm, how long is ̅𝐿̅𝐸̅?
(property on sides)
2. What is its area if one of the diagonals is 4 more than the other
and IE + LK = 16 in ?
4. Independent Practice
̅𝐶̅
A. Given trapezoid ABCD ,̅𝐴̅𝐵̅̅‖ ̅𝐷̅̅̅𝐶̅; X and Y are midpoints of ̅𝐴̅𝐷̅̅̅and 𝐵̅
respectively.
D
C
1. Find XY if AB = 30 and DC = 18
X
Y Y
2. Find AB if DC = 5 and XY = 8
A
B
B. Given:
ABCD
3. If m∠𝐷̅̅𝐴𝐵̅ = 56, 𝑚∠𝐷̅̅𝐶𝐵̅
4. If DO = 4.5, then DB =
5. If m∠𝐷̅̅𝐶𝐵̅ = 47, 𝑡ℎ𝑒𝑛 𝑚∠𝐶𝐷̅̅𝐴 = ?
B
A
D
C
Q
C. Find the area of the kite QRST
13 cm
6cm
T
R
15 cm
S
5. Generalization
To solve problems on parallelograms, trapezoids, and kites, use or apply
their properties and theorems.
6. Application
Solve the following problems by applying the properties/theorems on
parallelograms, trapezoids, and kites
a. A hectare of land in Amadeo was planted with coffee. Each coffee tree
was allotted 20m2 of land. How many coffee tree were planted on the
said land?
b. A square picture frame made up of bamboo sticks has a side of
12.5 in.
Find its perimeter and area.
c. Find the area of a rectangular garden if its perimeter is 32.8 m and its
length is 10.4 m
d. Find the area of 20 m high trapezoid if the measure of its bases are
16 m and 18 m respectively
e. The area of a kite is 180 cm2 and the length of the diagonal is 36
cm. How long is the other diagonal?
7. Assessment
Solve the following problems.
a. In rectangle BCDE, if BC = 16.5 and CD = 4.3, what is the perimeter
of the rectangle?
b. Find the length of a diagonal of a kite whose area is 176 sq. cm and
other diagonal is 16 cm long.
c. Quadrilateral BEST is a parallelogram. If m∠𝐵̅ = (𝑥 + 40) and
m∠𝐸 = ( 2𝑥 + 20), what is m∠ 𝐵̅ ?
d. If one of the angles of an isosceles trapezoid is 40, how many degrees
are there in each of the other angles?
e. The lengths of the two sides of a rhombus are 5x – 11 and 2x + 25, find
its perimeter.
IV. Assignment
1. Solve the following problems.
a. One side of a kite 5 cm less than 7times the length of another. If the
perimeter is 86 cm, find the length of each side of the kite.
b. In
CDEF, ̅𝐶̅𝐹̅and ̅𝐷̅̅̅𝐸̅are opposite sides. If the length of ̅𝐶̅𝐷̅̅̅is 3x
and the length of ̅𝐸
𝐹̅ is x + 8, how long is each segment.
2. Study similarity
a. What is a proportion?
b. Define similar polygons
Reference: Learner’s Material, pp. 355 – 365
Electronic Source:
Answer key:
Guided Practice
A.
a. m∠𝑊 = 25
b. HS = 18
B.
a. PT = 18 , OS = 10
b. M∠𝑇 = 101
C.
a. LE = 3.5 cm
b. Area = 30 in2
Independent Practice
A.
1. XY = 24
2. AB = 14
B.
3. m∠𝐷̅̅𝐶𝐵̅ = 56
4. DB = 9
5. m∠𝐶𝐷̅̅𝐴 = 133
C.
Application
a.
b.
c.
d.
e.
Assessment
a.
b.
c.
d.
e.
500 coffee tree
Perimeter = 50 in , Area = 156.25 in2
Area = 62.4 m2
Area = 314 m2
Diagonal is 10 cm
Perimeter = 41.6
Diagonal is 22 cm
M∠𝐵̅ = 80 𝑐𝑚
40° , 40° ,140° , 140°
196
Module 6: Similarity
Lesson 4: Proportion
Learning Competency 36: Describe proportion.
I – OBJECTIVES
a. Describe a proportion
b. Solve for the unknown term in a proportion.
c. Value accumulated knowledge as means of new understanding
II – SUBJECT MATTER
Topic: Similarity
Subtopic: Proportion
References: Learner’s Material for Mathematics 9 pp. 356 – 363
Math Time Grade IX pp. 27 – 31
Materials: Activity Sheets, laptop and monitor
III – PROCEDURE
A. Preliminary
1. Pre-Assessment
Express the following as ratio:
1. 2 meters to 40 centimeters
2. 3 weeks to 6 days
3. 25 minutes to 2 hours
4. 6 years to 1.5 decades
5. a century to a decade
B. Lesson Proper
1. Teaching/Modeling
Illustrative examples:
Find the missing term in a proportion:
a.
x : 8 = 4 : 16
x · 16 = 8 · 4
32
16𝑥
=
16
16
x=2
b. 5: 2 = x : 4
2 ·x = 5· 4
2𝑥
20
=
2
2
x = 10
2. Analysis:
How do you solve for the unknown term in a proportion?
3. Guided Practice
Find the missing term in each of the following proportions.
a.
3
=
4
9
𝑥
Solution: 3 =
4
9
𝑥
3•
=
3𝑥 = 36
𝑥= _
b.
2
3
=
10
3𝑥+3
Solution: 2 =
3
•9
10
3𝑥+3
2(3𝑥 + 3) = 3 •
+ 6 = 30
= 24
𝑥= _
4. Independent Practice
Are the following ratios proportional?
5. G
e
n
5
.
5. Generalization



Ratio is used to compare two or more quantities. Quantities
involved in 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.
Proportion is the equality of two ratios.
Fundamental Rule of Proportion
𝑦
If 𝒘: 𝒙 = 𝒚: 𝒛, then 𝑤 = provided that 𝒙 ≠ 𝟎; 𝒛 ≠ 𝟎.
𝑥
𝑧
6. Application:
A. Tagaytay City is said to be the little Baguio of the Philippines
ecause of its cold and fresh air. It is due to many trees planted in
the said city. Do you know that:
B. Determine if the ratios are proportional or not. Write P or NP.
1. 3 ∶ 5, 9 ∶ 15
2. 7 ∶ 9, 3 ∶ 4
3. 6 ∶ 7, 18 ∶ 21
4. 4 ∶ 𝑥, 12 ∶ 3𝑥
5.
2𝑎 6𝑎
5
,
15
7. Assessment
Apply the fundamental law of proportion by finding the missing variable.
Write the answer on the blank before the number.
1.
2.
3.
4.
5.
3
14
17
8
=
𝑥
13
5
𝑥
𝑥
8
9
=
𝑥
𝑥
64
=
=
24
39
60
84
=9
2
IV – ASSIGNMENT
1. Follow-up
Supply the missing numbers or variables which will make the
statement proportionality.
1.
2.
3.
=
( )
( )
=
3
4
6
2
( )
12
=
=
( )
20
15
18
=(
( )
=
35
35
)
12
66
2. Study: Study about similar polygons.
Electronic Sources: www.analyzemath.com
Answer Key
.
Guided Practice (Let’s Do This!)
a. x, 4
x = 12
b. 10, 6x, 6x,
Independent Practice (I Can Do This!)
Yes
No
No
Yes
Yes
Yes
Application (Let’s Do More)
a. 5
b. 1. P
2. NP
3. P
4. P
5. P
Assessment (Challenge Yourself!)
1. 42
2. 136
3. 8
4. 7
5. 36
Follow-Up
1. 3, 10
2. 5, 18
3. 11, 6.36
Module 6: Similarity
Lesson 6: Proportionality Theorem
Learning Competency 37: Apply the Fundamental Theorems of Proportionality to
Solve Problems Involving Proportions
I – OBJECTIVES
a. Solve problems involving proportions applying the theorems of
proportionality.
II – SUBJECT MATTER
Topic: Similarity
Sub-Topic: Solving Problems Involving Proportion
Materials: Activity Sheets, laptop and monitor
References: Grade 9 Teaching Guide pp.
Grade 9 Learning Module pp. 360 - 361
III – PROCEDURE
C.
Preliminary
1. Below are points A, B, C, and D with their corresponding
coordinates. Use the figure to find the ratio of the following:
a.
b.
c.
d.
AB to BC
DC to AD
BC to CD
AB to AD
A
A
B
C
D
2. The figure in each number are proportional. Find the value of x in
the given figures:
a. How did you find the activity?
b. Were you able to determine the ratio of the first activity?
c. How did you find the value of x on the second activity?
D.
Lesson Proper
1. Teaching/Modeling
Illustrative Example:
Study the examples on how to determine indicated quantities
from a given proportion, then solve the item labeled as Your Task.
Examples
1. If m : n = 4 : 3, find 3m – 2n : 3m + n
Solution
𝑚
4
4𝑛
=
m=
𝑛
3
Your Task
a. Find
𝑦
𝑠
if 5y – 2s : 10 = 3y – s : 7
3
Using m = 4𝑛
3
3( 4𝑛);2𝑛
3𝑚;2𝑛
4𝑛;2𝑛
2𝑛
2
3
=
= 4𝑛+𝑛
= 5𝑛
= 5
4𝑛
3𝑚+𝑛
3( )+𝑛
3
Therefore,
3m – 2n : 3m + n = 2:5
Solution:
5𝑦;2𝑠
10
3𝑦;𝑠
7
=
𝑦
𝑠
4𝑠
5
=
𝑠
7(5y – 2s) = 10(3y – s)
=
35y – 14s = 30y – 10s
= 4𝑠 ·
1
35y – 30y = 14s – 10s
5
𝑠
= 4𝑠
5y = 4s
=4
5𝑦
5
=
y=
4𝑠
4𝑠
5
÷s
5𝑠
5
4𝑠
5
Therefore 𝑦 = 4
𝑠
5
5
2. Analysis
a. How did you find the activity?
b. What property of proportion did you use to obtain
7(5y – 2s) = 10(3y – s)?
c. Why is 7(5y – 2s) = 10(3y – s) become 35y – 14s = 30y –
10s?
d. What did you do to 35y – 14s = 30y – 10s to become
35y – 30y = 14s – 10s?
e. On the second column, explain why
4𝑠
5
𝑠
becomes
4𝑠
5
1
· .
𝑠
3. Guided Practice
Solve the given problem.
a. Find c : r if 3c + 4r : 18 = c + 3r : 9
3𝑐+4𝑟
18
Solution
𝑐; 3𝑟
=
9
9(3c + 4r) = 18(c + 3r)
27c + 36r = 18c + 54r
27c -18c = 54r - 36r
9c = 18r
9𝑐
18𝑟
=
9
9
Write the proportion in fraction
form
Use Cross-multiplication
Property
Use Distributive Multiplication
Use Additive Inverse
Subtraction Property
Multiplicative inverse
C = 2r
𝑐
𝑟
𝑐
𝑟
=
=
2𝑟
𝑟
2
Substitute the value of c
Cancel r
1
b. If e and b represent two non- zero number, find the ratio e : b
if 2e2 + eb – 3b2 = 0.
4. Independent Practice
Use the properties of proportion to solve the given problem.
a. If r, s and t represent three positive numbers such that
r : s : t = 4 : 3 : 2 and r2 – s2 – t2 = 27. Find the values of r, s and t.
b. If g : h = 4 : 3, evaluate 4g + h : 8g + h
5. Generalization
Cross-Multiplication
Property
Alternation Property
If
If
w
𝑥
w
𝑥
𝑦
= , then 𝑤𝑧 = 𝑥𝑦; 𝑥 ≠ 0, 𝑧 ≠ 0
𝑧
𝑦
w
𝑧
𝑦
= , then
w
𝑦
= ,
𝑥
= ; 𝑥 ≠ 0, 𝑦 ≠ 0, 𝑧 ≠ 0
𝑧
𝑥
𝑧
= ; 𝑤 ≠ 0, 𝑥 ≠ 0, 𝑦 ≠
Inverse Property
If
Addition Property
0, 𝑧 ≠ 0
w
𝑦
w+𝑥
𝑦+𝑧
If 𝑥 = 𝑧 , then 𝑥 = 𝑧 ; 𝑥 ≠ 0, 𝑧 ≠ 0
Subtraction Property
If
Sum Property
𝑥
w
𝑥
𝑢
then
𝑧
𝑦
= 𝑧 , then
If =
w
𝑣
𝑥
w;𝑥
𝑦
𝑥
= , then
𝑧
w
=
𝑢
𝑣
𝑦
𝑦;𝑧
=
𝑧
w
𝑥
; 𝑥 ≠ 0, 𝑧 ≠ 0
𝑦
=
𝑧
=
𝑢+w+𝑦
𝑣+𝑥 +𝑧
= 𝑘;
Where 𝑘 is a constant at proportionality and
𝑣 ≠ 0, 𝑥 ≠ 0, 𝑧 ≠ 0.
6. Application:
Use the properties of proportion to solve the given problem.
𝑟
𝑠
5𝑞;6𝑟;7𝑠
a. If 𝑞 = = =
. Find x.
2
3
4
𝑥
b. Find the value of m if
𝑒
1
=
𝑓
2
=
g
=
5𝑒; 6𝑓;2g
3
.
𝑚
7. Assessment
Use the properties of proportion to solve the given problem.
a. If x : y = 7 : 6, find 6x – 5y : 6x + 3y.
b. Evaluate 8a + 4b : 12a + 4b if a : b = 8 : 7
IV – ASSIGNMENT
3. Follow-up
Use the properties of proportion to solve the given problem.
a. Find the value of a, b and c so that 𝑎 = 𝑏 = 6 = 18 .
24
8
16
𝑐
b. Find the value of x, y and z so that 2 =
𝑥
𝑥
𝑦
9
30
c. Find the value of x and y so that =
𝑦
14
=
18
𝑧
=
4
=
.
𝑥
Reference:
Geometry by Eunice Ato-Lopez, MAT, et al, page 64 – 67
Mathematics Learner’s Material, page 360 - 361
12
42
.
Answer Key:
Preliminaries
1. a. 10 : 20 or 1 : 2
b. 5 : 35 or 1 : 7
c. 20 : 5 or 4 : 1
d. 10 : 35 or 2 : 7
2. a. x = 12
b. x = 56
c. x = 90
Independent Practice (I Can Do This!)
𝑠
𝑡
1. let 𝑟 = = = k, k ≠ 0
4
3
2
So, r = 4k; s = 3k; t = 2k
r2 – s2 – t2 = 27
2.
g
4
=
ℎ
3
=𝑘
g = 4k
4(4𝑘)+3𝑘
, h = 3k
8(4𝑘)+3𝑘
(4k)2 – (3k)2 – (2k)2 = 27
16𝑘+3𝑘
32𝑘+3𝑘
16k2 – 9k2 – 4k2 = 27
19𝑘
63𝑘
3k2 = 27
19
63
k = (3 and -3)
therefore:
r = 4k = 4(3) = 12
s = 3k = 3(3) = 9
t = 2k = 2(3) = 6
2. 2e2 + eb – 3b2 = 0
(2e + 3b) (e – b) = 0
2e + 3b = 0 or e – b = 0
therefore
:
2e = - 3b
e=–b
2𝑒
;3𝑏
𝑒
𝑏
= 2𝑏
=𝑏
2𝑏
𝑏
2𝑒
;3𝑏
= 2𝑏
2𝑏
𝑒
;3
= 2
𝑏
𝑒
𝑏
1
=1
Hence, e : b = -3 : 2 or 1 : 1
Application (Let’s Do More!)
Use the properties of proportion to solve the given problem.
𝑟
𝑠
5𝑞;6𝑟;7𝑠
a. let 𝑞 = = =
= k, then
2
3
4
𝑥
q = 2k, r = 3k, s = 4k, and 5q – 6r – 7s = kx
5(2k) – 6(3k) – 7(4k) = kx
10 k – 18k – 28k = kx
-36k = kx
x = 36
b. let
𝑒
1
=
𝑓
2
=
g
3
=
5𝑒; 6𝑓;2g
𝑚
= k, then
e = k, f = 2k, g = 3k, and 5e – 6f – 2g = km
5(k) - 6(2k) – 2(3k) = km
5k – 12k – 6k = km
-13k = km
m = 13
Assessment (Challenge Yourself!)
Use the properties of proportion to solve the given problem.
7
a. 𝑥 =
𝑦
6
7𝑦
6x = 7y
x=
b. .
𝑎
then
6𝑥−5𝑦
6𝑥+3𝑦
6 ( )−5𝑦
= 6 (7𝑦6 )+3𝑦 =
6
7𝑦
=
2𝑦
10𝑦
=
1
5
8
7
7a = 8b
then
8a + 4b
12a + 4b
a=
7𝑥+3𝑦
6
=
𝑏
7𝑦−5𝑦
8𝑏
7
Follow-up
a. a = 9, b = 3, c = 48
b. x = 7, y = 4, z = 63
c. x = 6, y = 20
8𝘣
=
8 ( 7 )+4𝑏
8𝘣
12 ( 7 )+4𝑏
=
64𝘣+28𝘣
7
96𝘣+28𝘣
7
=
92𝑏
124𝑏
=
23
31
Module 6: Similarity
Lesson 6: Fundamental Properties of Proportion
Learning Competency 37: Apply the Fundamental Theorems of Proportionality to
Solve Problems Involving Proportions
I – OBJECTIVES
a. Identify the properties of proportion.
b. Describe a proportion.
c. Use the Theorem of proportionality to solve problems involving proportion.
II – SUBJECT MATTER
Topic: Similarity
Sub – Topic: Fundamental Properties of Proportion
Materials: Activity sheets, laptop and monitor
References: Grade 9 Teaching Guide pp.
Grade 9 Learning Materials pp. 358 – 359
III – PROCEDURE
E.
Preliminaries
Activity 1: My Decisions Now and Then Later
Directions:
1. Replicate the table below on a piece of paper
2. Under the my-decision-now column of the table, write A if you agree
with the statement and D if you don’t.
3. After tackling the whole module, you will be responding to the same
statement under the my-decision-later column.
1
2
3
4
5
6
7
8
9
10
Statement
Proportion is an equality of ratios.
When an altitude is drawn to the hypotenuse of a given
right triangle, the new figure comprises two similar right
triangles.
The Pythagorean Theorem states that the sum of the
squares of the legs of a right triangle is equal to the
square of its hypotenuse.
Polygons are similar if and only if all their corresponding
sides are proportional.
If the scale factor of similar polygons is m : n, the ratios
of their areas and volumes are m2 : n2 and m3 : n3,
respectively.
The set of numbers {8, 15, and 17} is a Pythagorean
triple.
The hypotenuse of a 45-45-90 right triangle is twice the
shorter leg.
Scales are ratios expressed in the form 1 : n.
If a line parallel to one side of a triangle intersects the
other two sides, then it divides those sides proportionally.
Two triangles are similar if two angles of one triangle are
congruent to two angles of another triangle.
My Desicion
Now
Later
Activity 2:
The teacher will show different pictures that are not proportion and the
learners will describe/explain why the picture is not proportion.
F.
Lesson Proper
1. Teaching/Modeling
Illustrative Examples:
1. Express the following ratios:
c. 1 meter to 20 centimeter
Solution: 1 meter = 100 centimeters
100 = 5 𝑜𝑟 5: 1
20
1
d. 5 days to 2 weeks
Solution: 2 weeks = 14 days
5 𝑜𝑟 5: 14
14
e. side of an equilateral triangle to its perimeter
Solution: 𝑃 = 3𝑠
𝑠
1
=
3𝑠
3
2. Find the missing term in each of the following proportions.
9
a. 3 =
4
𝑥
Solution: 3 =
4
9
𝑥
3•𝑥 =4•9
3𝑥 = 36
𝑥 = 12
b.
2
3
10
= 3𝑥+3
Solution: 2 =
3
10
3𝑥+3
2(3𝑥 + 3) = 3 • 10
6𝑥 + 6 = 30
6𝑥 = 24
𝑥=4
3. Complete each statement:
6
a. If 2 = , then (3)(6) = (2)(9)
3
9
3
b. If 𝑎 = , then 3𝑏 = 7𝑎
𝑏
c.
3
5
𝑎
7
= then 8 =
𝑏
5
𝑎+𝑏
𝑏
2. Analysis
a. Can you directly get the ratio of example a and b in illustrative
example #1? Why? If not, what will you do to find its ratio?
𝑎
b. Why is it the ratio of illustrative example #1 letter c is 𝑠 = ?
3𝑠
𝑏
c. What property of proportionality did you use in each of the
following proportion?
3. Guided Practice
Rewrite the given proportions according to the property indicated in
the table and find out if the ratios in the rewritten proportions are still
equal.
a.
Use the cross-multiplication
property to verify that ratios
are equal. Simplify if
necessary. One is done for
you.
𝑦 𝑎
=
Original Proportion
4𝑦 = 3𝑎
3 4
𝑦 3
Alternation Property of the
=
4𝑦 = 3𝑎
original proportion
𝑎 4
3 4
Inverse Property of the
=
4𝑦 = 3𝑎
original proportion
𝑦 𝑎
4y +1 2 = 3a + 12
𝑦+3 𝑎+4
Addition Property of the
=
4y + 12 – 12 = 3a + 12 – 12
original proportion
3
4
4y = 3a
4y - 12 = 3a - 12
𝑦−3 𝑎−4
Subtraction Property of the
=
4y - 12 + 12 = 3a - 12 + 12
original proportion
3
4
4y = 3a
b.
Use the cross-multiplication
property to verify that ratios
are equal. Simplify if
necessary. One is done for
you.
𝑥
𝑦
=
8 12
Original Proportion
12x = 8y
Alternation Property of the
original proportion
Inverse Property of the
original proportion
Addition Property of the
original proportion
Subtraction Property of the
original proportion
4. Independent Practice
Complete each statement.
?
1. If 5M = 8N, then 𝑀 = and
𝑁
2. If 6x = 45, then
𝑥
=
9
𝑦
?
?
and
?
𝑥
5
?
𝑀
= .
8
?
=
?
.
?
?
3. If 𝑥 = , then 𝑥 = .
4
4. If
9
=
𝐾
10
5. If
𝐴
𝐵̅
=
𝑦
?
𝑀
?
, then 𝐾+10 = .
8
10
36
?
?
, then 𝐴−𝐵̅ = .
4
𝐵̅
?
5. Generalization
 Ratio is used to compare two or more quantities. Quantities
involved in 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.
 Proportion is the equality of two ratios.

Fundamental Rule of Proportion
𝑦
If w: 𝑥 = 𝑦: 𝑧, then w = provided that 𝑥 ≠ 0; 𝑧 ≠ 0.
𝑥
𝑧

Properties of Proportion
𝑦
Cross-Multiplication
Property
Alternation Property
If w = , then 𝑤𝑧 = 𝑥𝑦; 𝑥 ≠ 0, 𝑧 ≠ 0
Inverse Property
If
Addition Property
0, 𝑧 ≠ 0
𝑦
If w = , then w+𝑥 =
Subtraction Property
If
w
Sum Property
If
𝑢
𝑥
𝑧
𝑦
𝑥
If w = , then w = ; 𝑥 ≠ 0, 𝑦 ≠ 0, 𝑧 ≠ 0
𝑥
𝑧
w
𝑥
𝑣
𝑧
𝑦
𝑥
𝑧
w
= , then
𝑥
𝑥
𝑦
𝑧
𝑦
𝑥
𝑧
w
𝑥
= , then w−𝑥 =
=
𝑥
𝑦
= , then
𝑧
𝑧
= ; 𝑤 ≠ 0, 𝑥 ≠ 0, 𝑦 ≠
𝑦
𝑦+𝑧
𝑧
𝑦−𝑧
𝑧
𝑢
𝑣
=
; 𝑥 ≠ 0, 𝑧 ≠ 0
; 𝑥 ≠ 0, 𝑧 ≠ 0
w
𝑥
𝑦
𝑢+w+𝑦
𝑧
𝑣+𝑥 +𝑧
= =
=
𝑘;
Where 𝑘 is a constant at proportionality
and 𝑣 ≠ 0, 𝑥 ≠ 0, 𝑧 ≠ 0.
6. Application
Solve the following:
1. Yesterday there were 20 male adults, 177 children and 111
female adults in the Picnic Grove. What was the ratio of
children to male adults?
2. If 3 meters of a pinya cloth cost P200, how much will 9
meters of the same cloth cost?
3. P2,000 is to be distributed among 3 members of a family in
the ratio 5:14:21. How much greater is the largest share
than the smallest share?
4. Alden and Maine share P360 in the ratio 4:5. How much
does each receive?
5. A teacher is 160 cm tall, while a boy is 120 cm. What is the
ratio of the boy to that of the teacher?
7. Assessment
Apply the fundamental law of proportion by finding the missing
variable. Write the answer on the blank before the number.
1. 3 = 9
14
2.
3.
4.
5.
17
8
𝑥
13
5
𝑥
𝑥
8
𝑥
=
=
=
𝑥
64
24
39
60
84
=9
2
IV – Assignment
Follow - up
Supply the missing numbers or variables which will make the statement
proportionality.
1.
2.
3.
4.
5.
=
( )
( )
=
3
4
12
6
2
( )
=
2310
9240
( )
7
20
15
18
=(
( )
=
35
=
=
( )
=
35
)
( )
924
3𝑥
21
=
12
66
=
24
=
=
72
( )
5𝑥
)
=(
( )
=
( )
102
22
( )
=
( )
28
( )
14
References:
Geometry by Eunice – Ato Lopez, et al., page 66
https://www.google.com.ph/search?q=flashlight+clipart&biw=1366&bih=662&source=lnms&tbm=is
ch&sa=X&sqi=2&ved=0ahUKEwjA1c3swf3PAhUJupQKHXBnCzAQ_AUIBigB#tbm=isch&q=animal+with
+two+heads&imgrc=W1dshFcoqtrLIM%3A
https://www.google.com.ph/search?q=flashlight+clipart&biw=1366&bih=662&source=lnms&tbm=is
ch&sa=X&sqi=2&ved=0ahUKEwjA1c3swf3PAhUJupQKHXBnCzAQ_AUIBigB#tbm=isch&q=monkey+wit
h+three+legs&imgrc=S1T-N2DUMIjkPM%3A
https://www.google.com.ph/search?q=flashlight+clipart&biw=1366&bih=662&source=lnms&tbm=is
ch&sa=X&sqi=2&ved=0ahUKEwjA1c3swf3PAhUJupQKHXBnCzAQ_AUIBigB#tbm=isch&q=banana+tre
e&imgdii=6es-3bW9-1GY2M%3A%3B6es-3bW9-1GY2M%3A%3BI_Qw0OhgPLzG-M%3A&imgrc=6es3bW9-1GY2M%3A
https://www.google.com.ph/search?q=flashlight+clipart&biw=1366&bih=662&source=lnms&tbm=is
ch&sa=X&sqi=2&ved=0ahUKEwjA1c3swf3PAhUJupQKHXBnCzAQ_AUIBigB#tbm=isch&q=animal+abn
ormalities&imgdii=HjnGoYN2VNUOOM%3A%3BHjnGoYN2VNUOOM%3A%3BatMK76FaL4jv7M%3A&
imgrc=HjnGoYN2VNUOOM%3A
Answer Key
Preliminaries
Answers may vary.
Teaching / Modeling
Answers may vary.
Analysis
Answers may vary.
Guided Practice
b.
Use the cross-multiplication
property to verify that ratios
are equal. Simplify if
necessary. One is done for
you.
𝑥
𝑦
=
8 12
𝑥
8
=
𝑦 12
8 12
=
𝑥
𝑦
Original Proportion
Alternation Property of the
original proportion
Inverse Property of the
original proportion
Addition Property of the
original proportion
𝑥+8
Subtraction Property of the
original proportion
𝑥−8
Independent Practice
1. 8 and 𝑁
5
5
2.
5
and 9
3.
4
4.
𝑀+8
8
5.
6
6
9
36−4
4
8
8
=
=
𝑦 + 12
12
𝑦 − 12
12
12x = 8y
12x = 8y
12x = 8y
12x + 96 = 8y + 96
12x + 96 – 96 = 8y + 96 –
96
12x = 8y
12x + 96 = 8y + 96
12x + 96 – 96 = 8y + 96 –
96
12x = 8y
Application
1. 177/20
2. X = 600
3. X = 50; 21x = 1050;5x = 250;800php
4. 4x = 160, 5x = 200
5. 3/4
Assessment
1. X = 42
2. X = 136
3. X = 8
4. X = 7
5. X = 36
Follow-up
1. 3 =
4
2.
3.
4.
5.
5
6
=
2
10
9
12
15
18
=
2310
9240
𝑥
7
=
=
=
6.36
35
=
3𝑥
21
15
20
35
42
=
231
924
=
18
=
24
85
=
102
12
=
66
=
72
288
5𝑥
33.33
=
22
121
=
2𝑥
14
7
28
Module: Similarity
Lesson 4: Similarity of Figures
Learning Competency 38: Illustrate similarity of figures
I – OBJECTIVES
a. Illustrate similarity of figures
b. Differentiate similar figures from congruent figures
c. Identify similar figures in real – life situations
II – SUBJECT MATTER
Topic: Quadrilaterals
Sub – Topic: Similarity of Figures
References: Teacher’s Guide pp.235 – 237, Learner’s Material, pp.361 – 365
Sequential Mathematics by Laurence Leff, pp 246 – 248
Materials: Models of similar figures/objects, flat TV screen
III – PROCEDURE
A. Preliminaries
1. Pre – Assessment
Answer each of the following questions.
a. When are two figures congruent?
b. Illustrate a pair of congruent figures
c. If ∆𝐴𝐵̅𝐶 ≅ ∆𝐷̅̅𝐸𝐹, which are the congruent sides? congruent angles?
2. Motivation
a. Which figures seem to be similar? congruent?
D
A
B
C
b. Are congruent figures similar?
c. Are similar figures congruent?
B. Lesson Proper
1. Teaching/ Modeling
Illustrative Examples
Congruent figures have the same shape and the same size, while
similar figures have the same shape but may differ in size, The 1 x 1
school ID picture and the 2 x 2 passport picture are similar.
The two triangles below are also similar.
5
4
I
3
10
8
II
6
Activity 1.
a. Name the corresponding sides of triangles I and II.
b. Find the ratio of the corresponding sides.
c. Determine the perimeter of triangles I and II
2. Analysis
a. How are the corresponding angles of the triangles related?
b. What do you notice about the lengths of the corresponding sides
of triangles I and II?
c. How do you compare the lengths of the corresponding sides of
similar
figures / polygons?
d. When are two or more polygons similar? What are the two
conditions to be met?
e. How do you compare the perimeter of similar polygons to the ratio
of the lengths of any pair of corresponding sides?
3. Guided Practice
Practice 1.
Apply the conditions on similar polygons. Decide whether the
polygons in each pair are similar. Answer yes, no, or not enough
information.
a.
b.
─
─
c.
.
Pracice 2
Tell whether the polygons in each pair are always, sometimes, or
never similar.
a. Two rectangles
b. A regular hexagon and a regular octagon
c. Two squares
4. Generalization
Two or more polygons are similar if their corresponding angles are
congruent and corresponding sides are proportional.
5. Independent Practice
∆MON~ ∆ MON. Complete the following statements.
a. KF = ?
𝑀0
𝑏.
𝑐
K
12
10
F
10
=
?
Y
x
14
Z
10
X
=𝑥
O
?
𝑑 ∠𝑌 ≅ ∠?
e ∠? ≅ ∠𝑀
21
12
N
y
M
6. Application
Give the ratio in simplest form. Use the figure below.
a. AD : AE
A
b. BD : AD
6
4
c. AD : ( AD + DB )
D
d. EC: ( AE + EC )
E
4
3
B
C
7. Assessment
A
5
1
S
3
F
24
B
Z
T
8
4
W
Y
E
C
9
20
X
12
D
Given: Hexagon ABCDEF~ hexagon STWXYZ
a. Find the lengths of the sides of hexagon STWXYZ.
b. Give the ratio of the perimeters of the hexagon
IV – ASSIGNMENT
Given:
1.
FADE~
𝐹𝐸
𝐺𝖶
=
𝐸𝐷̅̅
?
GLOW
F
2. If m∠𝐸 = 60, name two 60° angles in
GLOW
A
D
3. If m∠𝐷̅̅ = 110, then m∠𝑊 =
Study Similarity Theorems
1. What are the similarity theorems?
2. Define AAA, SAS and SSS similarity theorems
Reference: Learner’s Material, pp. 369 – 375
E
G
W
L
O
Answer key:
Guided Practice
Practice 1
a. Yes.
b. Not enough
c. Not enough
Practice 2
a. Sometimes
b. Never
c. Always
Independent Practice
a.
b.
c.
2
3
10
𝑦
𝑥
12
d. ∠O
e. ∠ X
Application
a. 3:2
b. 2:3
c. 3:5
d. 3:7
Assessment
a. ST = 18
TW = 3
WX = 9
XY =15
YZ = 6
SZ = 4
b. 55: 73.3
Module 6: Similarity
Lesson : Similarity of Triangles
Learning Competency 39.1 : Prove the condition for similarity of triangles using
the SAS Similarity Theorem
I – OBJECTIVES
b. Show that the two triangles are similar by SAS Similarity Theorem
c. Prove the condition for similarity of triangles using the SAS theorem
d. Value accumulated knowledge as means of new understanding
II – SUBJECT MATTER
Topic: Similarity of Triangles
Sub-Topic: SAS Similarity Theorem and its Proof
Materials: Activity Sheets, Cartolina, laptop and monitor
References: Grade 9 Teaching Guide pp. 245 – 246
Grade 9 Learning Module pp. 376 – 378
E-Math 9 pp. 315 – 316
III – PROCEDURE
G. Preliminary
Review of Pre-requisite skills
Is either
DOG or
CAT similar to
H
HRS?
16
D
O
18
12
8
12
R
S
28
A
T
12
20
G
C
24
Questions:
1. What similarity theorem did you use in inspecting the three triangles?
2. By inspection, what two triangles are similar with each other? Why?
H. Lesson Proper
2.
Teaching/Modeling
 SAS Similarity Theorem states that if two sides and an included angle
of one triangle are congruent respectively to two sides and an included
angle of another triangle, then the triangles are congruent.
Illustrative Example:
R
Given: RO bisects ∠ MRN
RM ≅ RN
Illustration:
1 2
Prove:
MOR ≅
NOR
M
O
N
Proof
Statements
Reasons
1. RO bisects ∠ MRN
1. Given
2. ∠ 1 ≅ ∠ 2
2. An angle bisector divides an angle
into two congruent angles
3. RO ≅ RO
3. Reflexivity
4. Given
4. RM ≅ RN
MOR ≅
5.
3.
5. SAS Congruence Postulate
NOR
Analysis
a. What is SAS Similarity Theorem?
b. How can you prove that the two triangles are similar using the SAS
Theorem?
4.
Guided Practice
In each pair of given triangles, supply the appropriate condition that makes the
two triangles similar using the SAS similarity theorem. The first one is done for you.
Given:
a.
b.
Figure:
AC
DC
=
BC
D
CE
∠ ACB ≅ ∠ DCE
ABC ~ DEC by SAS
c. Similarity Theorem
Given:
a.
B
C
Figure:
XY
GK
=
b. ∠ Y
c.
A
_
E
G
X
≅
XYZ ~ GKH by SAS
Similarity Theorem
Y
Z
H
K
5.
Independent Practice
Write the statements/reasons that are left blank in the proof of SAS Similarity
Theorem. Refer to the hints provided to help you.
Given:
AB
DE
=
AC
and ∠A
DF
≅ ∠D
E
Prove:
ABC ~
DEF
P
B
Illustration:
A
C
D
Q
Proof:
Statements
Reasons
1. DP ≅ AB
AB
AC
2.
=
DE
DF
.
F
1. Definition of
segments
2.
3.
ABC ≅
DPQ
3.
4.
ABC ~
DPQ
4.
triangles
5.
ABC ~
DEF
5. Triangles similar to the same triangle
are
Postulate
triangles are similar
5. Generalization
 Side-Angle-Side 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.
6. Assessment
Write the reasons that are left blank in the proof of SAS Similarity Theorem. Refer
to the hints provided to help you.
Proof
T
Q
R
P
Given:
QR
TU
=
S
PR
SU
; ∠R ≅ ∠U
Prove:
PQR ~
STU
Proof:
 Construct X on TU
such that XU = QR
 From X, construct
U
XW parallel to TS
intersecting SU at
W
T
X
S
W
U
No.
1
Hints
Which side are parallel
by construction
Write the given related to
sides
Describe angles WXU &
STU and XWU & TSU
based on statement 1
Are WXU and STU
similar?
Write the equal ratios of
similar triangles in
statement 4
What can you say about
triangles PQR and WXU?
2
3
4
5
6
Statements
Reasons
By construction
Given
Corresponding angles
are congruent
AA Similarity Theorem
Definition of Similar
Polygons
SAS Triangle
Congruence Postulate
7. Application
Given the figure use SAS Similarity Theorem to prove that:
R
X
A
M
1
2
3
P
Hints
Write in a proportion the
ratios of two
corresponding
proportional sides
Describe included angles
of the proportional sides
Conclusion based on the
simplified ratios
Statements
Reasons
IV – ASSIGNMENT
1. Follow-up.
Are the two triangles similar? Justify your answer by solving which sides/angles
of the triangles are congruent. The hints are provided to help you.
Solution:
L
R
LV
RY
=?
OV
CY
10
=?
O
m∠ V = ? = m ∠ Y
Therefore,
LOV ~
RCY
30o
12
15
V
Y
30o
18
Similarity Theorem
C
2. Study and research:
a. What is Triangle Angle-Bisector Theorem?
b. How can you prove that two triangles are congruent using the Triangle
Angle-Bisector Theorem
Reference: Grade 9 Learning Module pp. 378 – 380
Answer Key:
Preliminaries
Solution:
 Check the ratios of the corresponding sides in
Shortest Sides
DO
8
2
= =
HR
12
3

HRS and
Longest Sides
OG
16
2
RS
=
Because all the ratios are equal,
24
=
Remaining Sides
DG
12
2
= =
HS
18
3
3
DOG ~
HRS
 Check the ratios of the corresponding sides in
Shortest Sides
CA
12
= =1
RH
12

Longest Sides
AT
28
7
= =
RS
24
6
Since the ratios are not equal,
HRS and
DOG
HRS and
CAT
Remaining Sides
CT
20
10
= =
HS
18
9
CAT are not similar
Guided Practice (Let’s Do This!)
YZ
a.
HK
b. ∠K
Independent Practice (I Can Do This!)
1. Congruent
2. Given
3. SAS Postulate
4. Congruent
5. similar
Application (Let’s Do More!)
Statements
AR AP
1
=
AM AX
∠ RAP ≅ ∠MAX
RAP ~ MAX
2
3
Assessment (Challenge Yourself!)
No.
1
2
3
Statements
XW || ST
QR PR
=
TU SU
∠ WXU ≅ ∠ STU
∠ XWU ≅ ∠ TSU
Reasons
Given
Vertical angles are congruent
SAS Similarity Theorem
4
WXU ~
STU
WX XU WU
=
=
ST TU
SU
5
PQR ≅
6
Assignment
Follow-up
LV
10
2
= or
RY
15
3
OV
CY
=
WXU
12
18
or
2
3
Two pairs of corresponding sides are proportional.
m∠ V = 30 = m∠ Y
Therefore,
LOV ~
RCY
SAS Similarity Theorem
Module 6: Similarity
Lesson 4: Similarity of Triangles
Learning Competency 39.2 & 39.3: Prove the condition for Similarity of
Triangles using the AA and SSS Similarity Theorem.
I – OBJECTIVES
a. Describe AA and SSS Similarity Theorem.
b. prove the condition for similarity of triangles using the AA and SSS
Similarity theorem
c. Value accumulated knowledge as means of new understanding
II – SUBJECT MATTER
Topic: Similarity of Triangles
Sub-Topic: AA and SSS Similarity Theorem and its Proof
Materials: Activity Sheets, Cartolina, laptop and monitor
References: Grade 9 Teaching Guide pp. 9 pp. 368-373
Grade 9 Learning Module pp. 9 pp. 368-373
III – PROCEDURE
A. Preliminary
1. Pre-Assessment
𝐿𝑖𝑡𝑜 𝑠𝑡𝑎𝑛𝑑 𝑖𝑛 𝑓𝑟𝑜𝑛𝑡 𝑜𝑓 𝑎 𝑠𝑢𝑛 𝑚𝑜𝑏𝑖𝑙𝑒 𝑝ℎ𝑜𝑛𝑒 𝑡𝑜𝑤𝑒𝑟 𝑎𝑡 𝐺𝑒𝑛𝑒𝑟𝑎𝑙 𝑀𝑎𝑟𝑖𝑎𝑛𝑜
𝐴𝑙𝑣𝑎𝑟𝑒𝑧 𝐶𝑎𝑣𝑖𝑡𝑒 𝑐𝑎𝑠𝑡𝑖𝑛𝑔 𝑎 𝑠ℎ𝑎𝑑𝑜𝑤 𝑤𝑖𝑡ℎ 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 6𝑚. 𝐻𝑖𝑠 ℎ𝑒𝑖𝑔ℎ𝑡 𝑖𝑠 1.8 𝑚
𝑎𝑛𝑑 ℎ𝑖𝑠 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑤𝑎𝑦 𝑡ℎ𝑒 𝑡𝑜𝑤𝑒𝑟 𝑖𝑠 54 𝑚. 𝐹𝑖𝑛𝑑 𝑥 (ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑜𝑤𝑒𝑟).
𝐺𝑖𝑣𝑒𝑛: ∆𝐴𝐵̅𝐶 𝑎𝑛𝑑 ∆𝐴𝐷̅̅𝐸 𝑎𝑟𝑒 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠.
x
2. Motivation
Determine if the pair of triangles is similar by completing the table
below.
8
6
4
10
5
1. Name congruent angles.
2. Are the sides of the triangles proportion?
3. Based on your answer in number 1 and 2, are the triangles
similar?
Motive Questions:
When do we say that the two triangles are similar?
C. Lesson Proper
1. Teaching/Modeling
Illustrative Examples:
Example 1: Proof of AA Similarity theorem is as follows
A
Given
∠𝐴 ≅ ∠𝐷̅̅
∠𝐵̅ ≅ ∠𝐸
B
C
Prove
D
X
E
Y
F
∆𝐴𝐵̅𝐶~∆𝐷̅̅𝐸𝐹.
Proof
Statements
1. ∠𝐴 ≅ ∠𝐷̅̅
̅ 𝐵̅
̅ ̅ ≤ 𝐷̅̅
̅𝐸
̅ ̅ . 𝑇𝑎𝑘𝑒 a
2. Assume 𝐴
̅𝐸
̅ ̅ so that 𝐷̅̅
̅𝑋
̅̅ =𝐴
̅ ̅𝐵̅̅.
point X on 𝐷̅̅
Through X, draw a line such
that ∠𝐷̅̅𝑋𝑌 = ∠𝐵̅.
3. . ∆𝐴𝐵̅𝐶 = ∆𝐷̅̅𝑋𝑌
Reasons
1. Given
2. Two points determine a line
3. ASA triangle congruence (1, 2)
4. ∠𝐵̅ ≅ ∠𝐸
4. Given
5. ∠𝐷̅̅𝑋𝑌 = ∠𝐸
5. Transitivity (2, 4)
6. ̅𝑋̅𝑌̅‖ ̅𝐸̅𝐹̅.
6. If two lines are cut by a
transversal and a pair of
corresponding
angles
are
congruent, then the lines are
parallel.
7. If two parallel lines are cut by a
transversal, then a pair of
corresponding
angles
is
congruent.
8. CPCTC (3)
7. ∠𝐷̅̅𝑋𝑌 = ∠𝐹
̅
8. ∠𝐶 = ∠𝐷̅̅𝑌𝑋; ̅𝐴̅𝐶̅ = ̅𝐷̅̅̅𝑌̅, ̅𝐶
𝐵̅
= ̅𝑋̅𝑌̅.
9.∠𝐶 ≅ ∠𝐹
10.
11.
𝐷̅̅K
𝐷̅̅𝐸
𝐴𝐵̅
𝐷̅̅𝐸
=
=
𝐷̅̅F
𝐷̅̅𝐹
𝐴𝐶
𝐷̅̅𝐹
=
=
KF
𝐸𝐹
𝐵̅𝐶
9. Substitution (7, 8)
10.
Basic
Theorem
Proportionality
11. Substitution (2, 8, 10)
𝐸𝐹
12. ∆𝐴𝐵̅𝐶~∆𝐷̅̅𝐸𝐹.
12. Definition of Similar Triangles
Example 2: Proof of SSS Similarity theorem is as follows
Given
𝐴𝐵̅
𝐷̅̅𝐸
=
𝐵̅𝐶
𝐸𝐹
=
𝐴𝐶
𝐷̅̅𝐹
Prove
∆𝐴𝐵̅𝐶~∆𝐷̅̅𝐸𝐹.
Proof
Statements
1. On ̅𝐷̅̅̅𝐸̅, let DP =AB
Reasons
1. The Point-PlottingTheorem
2. ̅𝐷̅̅̅𝑃̅≅ ̅𝐴̅𝐵̅̅
2. Definition of congruent segments
3. . Construct ̅𝑃̅𝑄̅‖ ̅𝐸̅𝐹̅
3. Through a given point, there is
exactly one line parallel to given line.
4. Corollary 5.1
4 ∆𝐷̅̅𝑃𝑄~∆𝐷̅̅𝐸𝐹.
5.
𝐷̅̅𝑃
𝐷̅̅𝐸
=
𝑃Q
𝐸𝐹
=
𝐷̅̅Q
𝐷̅̅𝐹
6.
𝐴𝐵̅
𝐷̅̅𝐸
=
𝐵̅𝐶
𝐸𝐹
=
𝐴𝐶
𝐷̅̅𝐹
7.
𝐷̅̅𝑃
𝐷̅̅𝐸
=
𝐴𝐵̅
𝐷̅̅𝐸
8.
𝐷̅̅Q
𝐷̅̅𝐹
=
𝐴𝐶
𝐷̅̅𝐹
5. Definition of similar triangles.
6..Given
7. MPE
;
𝐵̅𝐶
𝐸𝐹
=
𝑃Q
𝐸𝐹
9. ̅𝐷̅̅̅𝑄̅= ̅𝐴̅𝐶̅
̅𝐵̅𝐶
̅ ̅ = ̅𝑃̅𝑄̅
̅
10. 𝐷̅̅̅𝑄̅ ≅ ̅𝐴̅𝐶̅
̅𝐵̅𝐶
̅ ̅ ≅ ̅𝑃̅𝑄̅
8. Transitive Property of Equality
9. Multiplicative Property of Equality
10. Definition of congruent segments
11. ∆𝐴𝐵̅𝐶 ≈ ∆𝐷̅̅𝐸𝐹
11. SSS Postulate
12. ∆𝐴𝐵̅𝐶~∆𝐷̅̅𝐸𝐹.
12. Congruent triangles are similar
2. Analysis
a. How do you find the activity?
b. Is it easy to prove similar triangles?
c. Describe the AA and SSS Similarity Theorem
Given:
Prove:
JK II KH
ΔJGI ~ΔKGH
Statements
1. JK II KH
Reasons
Given
2. ∠𝐽 ≅ ∠ 𝐺𝐾𝐻
Corresponding
Angles
3. ∠𝐼 ≅ ∠ 𝐺𝐻𝐾
Corresponding
Angles
4. ΔJGI ~ΔKGH
AA Similarity
Theorem
𝐴𝐵̅
Example 4: Given:
𝐷̅̅𝐸
=
𝐴𝐶
𝐷̅̅𝐹
=
𝐵̅𝐶
𝐸𝐹
ΔABC ~ΔDEF
Prove:
Statements
1.
𝐴𝐵̅
𝐷̅̅𝐸
=
Reasons
𝐴𝐶
𝐷̅̅𝐹
=
𝐵̅𝐶
Given
𝐸𝐹
2. ΔABC ~ΔDEF
SSS Similarity
Theorem
Guided Questions:
Analyze the example above. Answer the following questions:
a. How are you going to prove similar triangles by AA Similarity
Theorem?
b. How are you going to prove similar triangles by AA Similarity
Theorem?
3. Guided Practice
Activity : COMPLETE ME.
A. Use the AA Similarity Theorem in writing an if-then statement to
describe an illustration or in completing a figure based on an if-then
statement.
R
H
If :
O
Y
Then:
ΔHEY ~
If :
Then:
∠𝐴 ≅ ∠𝑂 , ∠𝐵̅ ≅ ∠𝑇
E
P
Given: ∠𝐸 ≅ ∠𝑅 , ∠𝑌 ≅ ∠𝑂,
∠𝐻 ≅ ∠𝑃
Y
T
A
B
P
O
ΔBAY ~ΔTOP
B. Use the SSS Similarity Theorem in writing an if-then statement
to describe an illustration or in completing a figure based on an
if-then statement.
If :
0F
𝐴𝐹
Then:
=
0𝐽
𝐴𝐿
=
𝐽F
If :
𝐿𝐹
ΔJOY ~ΔLAF
Then:
Activity: ANSWER MO, SHOW MO
Choose the letter of the correct answer.
1. Are the given triangles similar?
a. Yes, they are similar, because they have one set of corresponding
angles of equal measures.
b. No, they are not similar.
c. There is no enough information to tell if they are similar.
d. Yes, they are similar, because they are triangles.
2. Which of the following statements describe similar triangle?
a. Their corresponding angles are equal in measure
b. Their corresponding sides are proportional
c. They have the same shape, but may not be the same size.
d. All of these statements describe similar triangles.
3. According to the AA Similarity Theorem, two triangles are similar if and
only if they have how many corresponding angles with equal measure?
a. Two
b. Four
c. None
d. One
4. According to the SSS Similarity Theorem, two triangles are similar if and
only if the corresponding sides are?
a. Congruent
b. Proportional
c. not equal
d. all of the above
4. Independent Practice
PERFECT TWO
Determine which two of the three given triangles are similar.
5. Generalization
To prove conditions for similarity of triangles, we may use
the following theorems:
AA Similarity Theorem
Two triangles are similar if two angles of one triangle are congruent
to two angles of another triangle.
SSS Similarity Theorem
Two triangles are similar if the corresponding sides of two triangles
are in proportion
6. Application:
Write the statements or reasons that are left blank in the proof of
AA or SSS Similarity Theorem.
a.
Given: 𝐴𝐵̅ ‖ 𝐷̅̅𝐶
Prove: ΔABE ~ΔDCE
Statements
Reasons
a.
Given
b.
Corresponding Angles
e.
∠𝐴 ≅ ∠𝐶𝐷̅̅𝐸
c.
d.
b.
A
W
T
H
C
Statements
a..
b.
Given:
𝐴𝑊
𝐶𝑇
=
𝐴𝑇
𝐶𝐻
=
𝑊𝑇
𝑇𝐻
Prove: ΔATW ~ΔCHT
Reasons
Given
c.
7. Assessment
If the triangles are similar, write a similarity statement between each
pair of triangles.
Statements
Reasons
Statements
Reasons
IV – ASSIGNMENT
4. Follow-up
Find the value of x and y using the similar triangles below.
5. Study:
SAS Similarity Theorem. How is it different from AA and SSS Similarity
Theorem.
Electronic Sources: www.analyzemath.com
Answer Key:
Preliminaries
Answer: x= 18m
Guided Practice (Let’s Do This!)
Activity 1:
A. If ∠𝐸 ≅ ∠𝑅, ∠𝑌 ≅ ∠𝑂, ∠𝐻 ≅ ∠𝑃
Then ΔHEY ~ ΔPRO
B. If
0F
𝐴𝑁
=
𝐽0
𝐽F
𝑀𝑁
= 𝑀𝐴
Then ΔJOY ~ ΔMAN
Activity 2:
1. c
2. d
3. a
4. b
Independent Practice (I Can Do This!)
Perfect Two: ΔDEF ~ ΔGHI
Application (Let’s Do More!)
1.
Statements
a. 𝐴𝐵̅ ‖ 𝐷̅̅𝐶
∠𝐴 ≅ ∠𝐶𝐷̅̅𝐸
c.
∠𝐵̅ ≅ ∠𝐷̅̅𝐶𝐸
d.
2.
ΔABE ~ΔDCE
Statements
𝐴𝑇
𝖶𝑇
a.. : 𝐴𝖶
=
=
𝐶𝑇
𝐶𝐻
𝑇𝐻
Reasons
Given
b. Corresponding Angles
Corresponding Angles
e. AA Similarity Theorem
Reasons
Given
b. ΔATW ~ΔCHT
c. SSS Similarity Theorem
Assessment (Challenge Yourself!)
a.
Statements
Reasons
∠𝐶 ≅ ∠𝐷̅̅
Given
∠𝐴 ≅ ∠𝐸
Given
ΔABC ~ΔEFD
AA Similarity Theorem
b
Statements
𝑅𝑆
𝑉𝑈
=
𝑆𝑇
𝑉𝑇
=
𝑅𝑇
Reasons
Proportionality Theorem
𝑈𝑇
ΔRST ~ΔUVT
SSS Similarity Theorem
Module 6: SIMILARITY
Lesson: Right Triangle Similarity and Its Proof
Learning Competency 39.4: Proves the conditions for similarity of triangles
I – OBJECTIVES
a. Apply properties of similar triangles and proportional theorems to obtain
properties of a right triangle.
b. Prove the conditions for similarity of right triangles.
c. Appreciate right triangle similarity in real life situation.
II – SUBJECT MATTER
Topic: SIMILARITIES
Sub-Topic: Right Triangle Similarity and Its Proof
Materials: Activity Sheets, index card, pencil, ruler and pair of scissors, cut
outs of triangles
References: Grade 9 Teaching Guide pp. 386-387
Grade 9 Learning Module pp. 253-255
https://www.youtube.com/watch?v=kMbC4aTJtAQ
III – PROCEDURE
Preliminary (Similarity on a Right Triangle)
Activity 1: Investigating Similar Right Triangles
1. Cut an index card along one of its diagonals.
2. On one of the right triangles, draw an altitude from the right angle to the
hypotenuse. Cut along the altitude to form two right triangles.
3. You should now have three right triangles. Compare the triangles. What
special property do they share? Explain.
4. Tape your group’s triangles to a piece of paper.
5. What did you observe with your cutouts?
A.
Lesson Proper
1. Teaching/Modeling
Let us watch a video presentation
Illustrative
Example
1:
The Brgy. Chairman of Bulihan would like to
put a bicycle path from the corner of Manga and Bayabas
Streets to Lanzones Street. What is the shortest path if
Manga and Bayabas streets are perpendicular? How long is
the path if BD is 18 meters and AD is 32 meters?
mathematical sentence.
Bayabas St.
Solutions:
The shortest path from the point C to AB is the line segment from
perpendicular to AB. This is the altitude CD of ΔBDC ~ ΔCDA.
So, BD = CD
CD AD
18 = CD
CD 32
CD2 = 18(32)
CD2 = 576
CD = 24 meters
Example 2. Given the illustration separate the new triangles formed from the
original triangle.
s
r
w
u
v
t
t
r
s
w
r
u
w
s
C
v
A
B
Using the definition of Similar Polygon in Right
Triangle
Altitude w is the geometric mean
between u and v.
B and C
Leg r is the geometric mean
between t and u
A and C
Leg s is the geometric mean
between t and v
A and B
w2 = uv
v/w = w/u
t/r = r/u
v/s = s/t
r2 = ut
s2 = vt
r = √ut
s =√ vt
Example 3: Illustration
Given: ΔABC is a right triangle with an altitude of BD and a right ∠ABC
Prove: ΔABC ~ ΔBDC ~ ΔADB
B
A
D
C
w =√uv
Proof:
Statements
1. ΔABC is a right triangle with an altitude
of BD and a right ∠ABC
2. ∠BDA and ∠BDC are right angles
3. ∠BDA ≅ ∠BDC ≅ ∠ABC
4. ∠A ≅ ∠A
∠C ≅ ∠C
5. ΔABC ~ ΔADB
ΔADB ~ ΔBDC
6. ΔABC ~ ΔADB ~ ΔBDC
Reasons
1. Given
2. Definition of altitude
3. Right Angle Theorem
4. RPE
5. AA similarity theorem
6. Triangles similar to the given triangle
are also similar.
2. Analysis
c. How are you going to make or derive a proportion out of the
given formula?
d. What mathematical principles did you use to prove using the
right triangle similarity theorem?
3. Guided Practice:
1. The corresponding sides of the similar
triangles.(Refer to the figure below)
Original
Triangle
Hypotenuse
New
Larger
Triangle
ES
Longer Leg
EZ
Shorter Leg
SZ
Solve for the geometric mean a, b, c
Geometric
Means
New
Smaller
Triangle
Proportion
Altitude a
Shorter leg s
Longer leg b
4. Independent Practice
Answer
Write the statements and reasons that are blank in the proof of
Right Triangle Similarity Theorem.
Given
 MER is a right triangle
E
with angle
 MER as right angle and
as the hypotenuse. EY is an
altitude to the hypotenuse of
ΔMER
Prove
ΔMER  ΔEYR  ΔMYE
M
Y
R
Proof
Statements
1.1 ΔMER is a right triangle with angle
1.
MER as right angle as the
hypotenuse.
1.2 EY is an altitude to the
hypotenuse of ΔMER
2. EY 
3.  MYE and
angles.
2. Definition of altitude
 EYR
are right
3. Definition of
4.  MYE   EYR   MER
5.  YME 
 YRE   ERM
Reasons
Lines
4. Definition of
Angles
;
6. ΔMYE  ΔMER; ΔMER ~ ΔEYR
7. ΔMER  ΔEYR  ΔMYE
5.
Property
6.
Similarity Theorem
7.
Property
5. Generalization
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.
Special Properties of Right Triangles
When the altitude is drawn to the hypotenuse of a right triangle,
1. the length of the altitude is the geometric mean between the segments
of the hypotenuse; and
2. each leg is the geometric mean between the hypotenuse and the
segment of the hypotenuse that is adjacent to the leg.
6. Application:
Answer the following:
a. The school garden of Bulihan NHS is right triangular in shape
as shown. From where his office is, the principal has to walk
through the garden to the canteen. How long is the walk way.
B . Find the missing lengths.
b
12
h
y
x
b. Assessment:
Answer the following:
1.The roof of the Aling Nene’s Food
House is to be built. Suppose AB  AD
and AC  BD. If AB =10m and BC = 8m,
how long is BD?
2.Give the indicated proportions.
a. The altitude of the geometric mean
b. The horizontal leg is a geometric mean
c. The vertical leg is a geometric mean
P
IV – ASSIGNMENT
Q
1. Follow Up
In your respective location search on Google map that describe right
triangle similarity theorem.
2. Study:
a. Define the Pythagorean Theorem
b. Who is the proponent of Pythagorean Theorem?
Reference: Grade 9 Learning Module pp. 389 - 390
Electronic Sources: https://www.youtube.com/watch?v=kMbC4aTJtAQ
https://cdn.kutasoftware.com/Worksheets/Geo/7-Similar%20Right%20Triangles.pdf
R
Answer Key:
Preliminaries
Answers may vary.
Guided Practice (Let’s Do This!)
1. The corresponding sides of the similar triangles
Original
Triangle
New Larger
Triangle
New
Smaller
Triangle
Hypotenuse
ES
EY
SY
Longer Leg
EY
EZ
YZ
Shorter Leg
SY
YZ
SZ
Solve for the geometric mean a, b, c
Geometric Means Proportion
Altitude a
a=√2(8) =
Answer
a=4
√16
Shorter leg s
𝑠 = √2(10)
= √2(2)(5)
𝑠 = 2√5
𝑏 = √8(10)
Longer leg b
= √23(2)(5)
= √24(5)
𝑏 = 4√5
Independent Practice (I Can Do This!)
Given
E
 MER is a right triangle with angle 
MER as right angle and as the
hypotenuse. EY is an altitude to the
hypotenuse of ΔMER
Prove
ΔMER  ΔEYR  ΔMYE
M
Y
R
Proof
Statements
1.1 ΔMER is a right triangle with
1. Given
angle MER as right angle as the
hypotenuse.
1.2 EY is an altitude to the
hypotenuse of ΔMER
2. EY  MR
3.  MYE and  EYR are right
angles.
4.  MYE   EYR   MER
5.  YME   EMR;  YRE  
ERM
6. ΔMYE  ΔMER; ΔMER ~ ΔEYR
7. ΔMER  ΔEYR  ΔMYE
2. Definition of altitude
3. Definition of perpendicular lines
4. Definition of right angles
5. Reflexive Property
6. AA Similarity Theorem
7. Transitive Property
Application (Let’s Do More!)
8
x
1.
=
x
18
x2 = 144
x = √144
x = 12 meters
2.
Assessment (Challenge Yourself!)
1.
BD
10
=
10
8
8BD = 100
BD ≈ 12.5 meters
𝑃𝑆
Q𝑆
2. A.
=
B.
C.
Q𝑆
𝑃𝑅
Q𝑅
𝑃𝑅
𝑃Q
𝑆𝑅
Q𝑅
=
=
𝑆𝑅
𝑃Q
𝑃𝑆
Reasons
Module 6: SIMILARITY
Lesson: Special Triangles
Learning Competency 39.5: Prove the conditions for similarity of triangles
I – OBJECTIVES
a. Illustrate special right triangles.
b. Prove the special right triangles
c. Solve problems involving special right triangles.
II – SUBJECT MATTER
Topic: SIMILARITIES
Sub-Topic:Special Triangles
Materials: Activity Sheets, compass, pencil, protractor, pair of scissors,
powerpoint presentation and AVP of the lesson
References: Grade 9 Teaching Guide pp. 256-257 and Grade 9 Mathematics
Patterns and Practicalities pp. 327-334
Grade 9 Learning Module pp.389-391.
III – PROCEDURE
A. Preliminary
Activity 1. ARRANGE ME!
Arrange the letters to form the word related to the figure.
a. NHPUTYESOE
b-c. EGSL
d. UCNORTEGN
e. RTGIH ALNEG
B. Lesson Proper
1. Teaching/Modeling
The teacher will present the video to discuss the lesson.
2
.
n
2. Analysis:
a. How does the 30º-60º-90º and 45º-45º-90º said to be special?
b.How do you find the unknown sides of right triangle?
c. How does the Pythagorean theorem is necessary in finding the
unknown length of the sides of the special right triangles?
d.How these two special triangles related to each other?
3. Guided Practice:
Clues
1. List down all the
given
Statements
Right with measure of
angle LMK = 60; measure
of angle LKM = 30;
KM =
; LM =
; KL =
2. List down all
constructed angles
and segments and
their measures
measure of angle LKN =
30; measure of angle KNL
= 60;
KN =
and LN =
3. Use Angle Addition
Postulate to angle
LKM and angle MKN
4. What is measure of
angle MKN? Simplify.
5. What do you
observe about
considering its
angles?
6. What conclusion
you make based from
statement 5?
7. With statement 6,
what can you say
about the sides of ?
8. Use segment
Addition Postulate for
measure of angle
=
measure of angle LKM +
measure of angle LKN
measure of angle MKN =
Substitution
is
is
triangle.
Reasons
Definition of
Equiangular
Triangle
.
Equiangular
Triangle is also
equilateral
Definition of
Equilateral
Triangle
Segment
Addition
LN and ML
9. Replace LN, ML
and MN with their
measurements and
simplify.
10. What is the value
of ?
11. Solve for using
statement 9
Postulate
+
=
2
*
Property of
Equality
**
Property of
Equality
Pythagorean
Theorem
Substitution
12. What equation
can you write about ?
13. Use statement 10
in statement 13
14. Simplify the right
side of statement 13
15. Solve for
16. Solve for and
simplify
17. Solving for in
statement 16
=
***
****
Power of a
Product (Law of
Exponent)
Subtraction
Property of
Equality
(Law of
Radicals)
Division Property
of Equality and
Rationalization of
Radicals
4. Independent Practice
Answer the following:
a
b
c
d
e
In an isosceles right triangle with  S as the right angle. Complete the table:
S
SP
PO
1.
2.
4.
16
9
11
P
SO
3.
5.
11
O
5. Generalization
In a 45⁰-45⁰-90⁰ right triangle:
a. Each leg 𝓁 s
√2
2
times the hypotenuse;
b. The hypotenuse is √2 times each leg 𝓁
In a 30-60-90 triangle theorem
a. The shorter leg is
1
the hypotenuse h
2
or
√2
times the
2
b. The longer leg is √3 times the shorter leg ; and
c. The hypotenuse is twice the shorter leg
6. Application
Answer the following problems.
1. The mat used in floor exercise by the Taekwondo
Team of Bulihan NHS for the competition is a square
with a side length of 12 meters. They start at the corner
of the mat and does a kick routine along the diagonal
to the opposite corner. How long is the player’s path?
longer leg
2. Bryan is excited to connect his brand new stereo
system to the television set. The directions say that speakers
should be in line with your television 12 feet apart as shown
A. Find the distance between Bryan and the T.V set.
B. Find the distance between Bryan and each speaker.
6. Assessment:
Find the missing parts of the isosceles right triangle.
A
1.
2.
3.
4.
5.
AT = 13
AC = 25
CT = 10
CT = 30
CT = 36
C
Find AC and CT
Find AT and CT
Find AT and AC
Find AT and AC
Find AT and AC
T
Find the missing parts of the 30-60-90 triangle on the right.
6. DG = 10
Find GO =
OD =
7. DG = 14
Find GO =
OD
=
8. GO = 8
Find DG =
OD
=
9. GO = 17
Find DG =
OD
=
10. OD =
Find DG =
GO=
G
60
30
D
IV – ASSIGNMENT
3. Follow Up
O
Module 6: Similarity
Lesson 6: Similarities in Right Triangle
Learning Competency 40: Apply the Theorems to show that triangles are similar.
I – OBJECTIVES
a. Determine the property to be used in proving the similarity of two
triangles.
b. Apply the Theorems to show the similarity of two triangles.
c. Show patience in proving the similarities of two triangles.
II – SUBJECT MATTER
Topic: Similarities
Sub-Topic: Similarities in Right Triangle
Materials: Activity Sheets, laptop and monitor
References: Grade 9 Teaching Guide pp. 246 - 248
Grade 9 Learning Materials pp.376 – 379
III – PROCEDURE
A. Preliminary
a. Are we similar or congruent?
- Is
ABC similar or congruent to
XYZ ?
b. Use the figure at the right to identify the
1. longer leg of triangle ADB
2. hypotenuse of triangle DEC
3. shorter leg of ADB,
DEC
B. Lesson Proper
1. Teaching/Modeling
Illustrative Example:
Given:
LOV and
ERV are
right triangles
LO II ER
Shows that
LOV ~
O
V
L
ERV
R
E
Answer the following.
1. If triangle LOV and Triangle ERV are right triangles, what are the
right angles?
2. Are the answers in question #1 congruent? Why?
3. What do you mean by parallel?
4. If LO II ER what will be the congruent angles? Why?
5. How many parts of triangle LOV are congruent to triangle ERV?
What are those?
6. Are triangles LOV similar to triangle ERV? Why?
7. What condition/Theorem?
Solution:
1. ∠LVO and / EVR
2. Yes. Because two right angles are congruent.
3. Two lines that do not intersect are parallel lines.
4. ∠L ≅∠ E and O ≅∠R
5. There are three congruent parts. These are∠ LVO ≅ ∠EVR,
∠L ≅∠E and ∠O ≅∠R
6. Yes, triangles LOV is similar to triangle ERV because if three
corresponding angles of two triangles are congruent then the two
triangles are congruent. Congruent triangles are similar.
7. By AAA Similarity Theorem.
2. Analysis
1. How did you find the measure of the required data?
2. What Theorem did you used to prove the similarity of two
triangles?
3. How many parts of triangles to be proved to say that they are
similar?
3. Guided Practice
1. Shows that ABC ~ EDF. Based on the following
conditions: / A ≅ / E and / C ≅ / F.
STATEMENT
REASONS
1. ∠A ≅∠E
1. Given
2. ∠ C ≅ ∠F
2. Given
3. m∠A + m / C + m∠ B = 180
3. The sum of the angles in a triangle is
180.
m ∠ E + m ∠ F + m∠ D =
180
4. m ∠ B = 180 - ( m / A + m /
B)
4. Addition Property of Equality (APE)
m ∠D = 180 - ( m ∠ E + m ∠
F)
5. m ∠ A + m ∠B = m ∠E + m
5. Algebraic sum of statement # 4
∠F
6. m ∠B = m∠D
6. Transitive Property of Equality (TPE)
7. ∠B =∠D
2
8.
A
. BC ~
7. Definition of congruent angles
EDF
8. AAA Similarity Theorem
Given:
̅𝑀
̅𝐸̅and ̅𝑁̅𝑌̅bisect each other at O
Prove: MOY ~ ENO
STATEMENT
1. ME and NY bisect each other
at O
2.
3.
4.
MOY ≅
REASON
1.
2.Definition of line segment
bisector
3.Vertical angles are congruent
4.
ENO
5.
5.Congruent triangles are also
similar
4. Independent Practice
- Complete the column below to prove the similarity of the two
triangles.
1. Given: In DEF ̅𝐶̅𝑆II ̅𝐸̅𝐹̅
Prove: DCS ~ DEF
STATEMENTS
1.
2. ∠DCS ≅ ∠ DEF
3.
4. ∠D ≅ ∠D
5.
DCS ≅
DEF
REASONS
1. Given
2.
3. Same as reason # 2
4.
5.
2. Given: ̅𝐴̅𝐵̅̅II ̅𝐶̅𝐷̅̅̅
𝐵̅𝐸
Prove: 𝐴𝐵̅ =
𝐶𝐷̅̅
𝐶𝐸
STATEMENTS
1.
2.
3. ∠B ≅ / C
REASONS
1. Given
2. If two II lines are cut by a
transversal then alternate interior
angles are congruent.
3.
4. ∠BEA ≅∠CED
5.
𝐴𝐵̅
𝐵̅𝐸
6. 𝐶𝐷̅̅ = 𝐶𝐸
4.
5. AAA Similarity Theorem
6.
5. Generalization
To show that two triangles are similar by AAA Similarity
Theorem prove that their corresponding angles are congruent.
To show that two triangles are similar by SAS Similarity
Theorem prove that two corresponding sides are proportional and
the corresponding angles are congruent.
To show that two triangles are similar by SSS Similarity
Theorem prove that the three corresponding sides are proportional.
6. Application
̅𝐵̅̅
Given: ̅𝑃̅𝑁̅ ⊥ ̅𝑀
/ N ≅ / MRP
Prove: PRM ~
PNB
7. Assessment
Given: The figure
with 𝑃𝑇 = Q𝑇
𝑇𝑆
𝑇𝑅
Prove:
PQT ~ SRT
IV – ASSIGNMENT
1. Follow-up
Answer the following.
a. Prove that any two right isosceles triangles are similar.
b. Prove that all equilateral triangles are similar
2. Study
a. Study Learners’ Material, page
References:
Secondary Mathematics Book III by Antonio G. Tayao, et al., pages 316 - 320
Geometry Book by Eunice Ato - Lopez, MAT, et al., pages 71 - 75
Integrated Mathematics by Elenita R. Rosales, et al., pages 196 - 200
Answer Key:
Preliminaries
A.
ABC is similar to
XYZ.
B. 1. ̅𝐴̅𝐷̅̅̅
̅𝐶
̅̅
2. 𝐸
3. ̅𝐵̅̅𝐷̅̅̅, ̅𝐷̅̅̅𝐶̅
Guided Practice (Let’s Do This!)
2.
STATEMENT
1. ME and NY bisect each other
at O
2. MO ≅ 𝐸𝑂
YO ≅ 𝑁𝑂
3. / MOY ≅ / ENO
4.
MOY ≅ ENO
5.
MOY ~ ENO
Independent Practice (I Can Do This!)
1.
STATEMENTS
̅
̅
̅
̅
̅
1. 𝐶𝑆 ‖
𝐸𝐹
2. / DCS ≅ / DEF
3. / DSC ≅ / DFE
4. / D ≅ / D
DCS ≅
5.
DEF
REASON
1, Given
2.Definition of line segment bisector
3.Vertical angles are congruent
4.SAS Postulate
5.Congruent triangles are also similar
REASONS
1. Given
2. In a transversal, corresponding
angles are ≅.
3. Same as reason # 2
4. Reflexive Property of Equality
(RPE)
5. AAA Similarity Theorem
2.
STATEMENTS
̅𝐶̅𝐷̅̅̅
1. ̅𝐴̅𝐵̅̅ ‖
2. / B ≅ / C
3. / A ≅ / D
4. / BEA ≅ / CED
5.
BEA ~ CED
𝐴𝐵̅
𝐵̅𝐸
6. =
𝐶𝐷̅̅
𝐶𝐸
Application (Let’s Do More!)
STATEMENT
̅
̅
̅
̅
̅
̅
1. 𝑃𝑁 ⊥ 𝑀
𝐵̅
2. / MPR and / BPN are right
angles
3. / MPR ≅ / BPN
4. / N ≅ / MRP
REASONS
1. Given
2. If two II lines are cut by a
transversal then alternate interior
angles are congruent.
3. Same as reason #2.
4. Vertical angles are congruent
5. AAA Similarity Theorem
6. If two triangles are ~ then its
corresponding sides are
proportional.
REASON
1. Given
2. Definition of Perpendicularity
3. Two right angles are ≅
4. Given
5. m / M + m / MPR + m / MRP =
180
6. m / B + m / BPN + m / BNP =
180
7. m / M = 180 – (m / MPR + m /
MRP)
8. m / B = 180 – (m / BPN + m /
BNP)
9. m / M = m / B
10. / m ≅ / B
11. PRM ~ PNB
5. APE
6. Same as reason #5
7. Algebraic sum of #5
8. Algebraic sum of #6
9. Transitivity
10. Definition of congruent angles
11. AAA Similarity Theorem
Assessment (Challenge Yourself!)
STATEMENT
𝑃𝑇
Q𝑇
1. Given
1. =
𝑇𝑆
REASON
𝑇𝑅
2. / PTQ ≅ / STR
3.
PQT ~
SRT
2. Vertical angles are congruent
3. SAS Similarity Theorem
Follow-up
(Answers may vary.)
Study:
c. Review your lessons and prepare for a long test.
Reference: Grade 9 Learning Module pp. 394-395
Electronic Sources:
https://www.google.com.ph/search?q=special+right+triangles+worksheet&biw=1366
&bih=662&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjm1dnOtfjPAhWKHZQKHd
DsDGQQ_AUIBigB#imgrc=R_ZB3MObb7erBM%3A
Module 6: SIMILARITY
Lesson: Pythagorean Theorem and Its Proof
Learning Competency 41: Proves the Pythagorean Theorem
I – OBJECTIVES
a. Apply the definition of similar triangles to derive the Pythagorean Theorem.
b. Solve problems involving Pythagorean Theorem
c. Appreciate the application of real life situations of Pythagorean Theorem.
II – SUBJECT MATTER
Topic: SIMILARITIES
Sub-Topic: Pythagorean Theorem and its Proof
Materials: Activity Sheets, powerpoint presentation, cut outs of squares with
different sizes.
References: Grade 9 Teaching Guide pp. 256-257 and Grade 9 Mathematics
Patterns and Practicalities pp. 327-334
Grade 9 Learning Module pp.389-391.
III – PROCEDURE
A. Preliminary
Activity 1. Briefly review the vocabulary below. Ask students to write
definitions or examples for each term on their own. Have volunteers share
their ideas and record/display their definitions. Make them part of a word wall.
 triangle
 right triangle
 hypotenuse
 legs
 square root
Activity 2. Pythagorean Puzzle
Instructions:
1) Cut off the small and medium squares. (Numbered 1-4
and 5)
2) Cut the medium square along the dotted lines.
3) Try to arrange the pieces (1-5) inside the larger, darker square.
B Lesson Proper
1. Teaching/Modeling
Illustrative Example 1:Solving for a missing
hypotenuse.
Remember that the hypotenuse is always the
longest side and the side opposite the right angle.
Note that 8 ft. and 15 ft. must the lengths of the legs since they make up
the right angle. That means that x in this case is the missing
hypotenuse. Plugging those values into the Pythagorean Formula yields
the following:
(8)2 + (15)2 = x2
64 + 225 = x2
289 = x2
√289 = x
17 = x
This means that the missing hypotenuse length is 17 feet.
Illustrative Example 2:
Solving for a missing leg.
First note that it makes no difference
which leg we label as a and which leg we
label as b.
Given the fact that this is a right triangle, we can solve for the missing leg
length, a. Just substitute everything we know into the Pythagorean
Formula. We know that the hypotenuse length, c, is 13 inches and that
the other leg length, b, is 12 inches.
a2 + b2 = c2
a2 + (12)2 = (13)2
a2 + 144 = 169
a2 + 144 -144 = 169 - 144
a2 = 25
a = √25
a=5
The missing leg is 5 inches.
Illustrative Example 3. Give: Right ΔABC with leg lengths a and b, and
hypotenuse length c.
Prove: c2 = a2 + b2
6. c = x+ y
7. c2 = a2 + b2
Reasons
1. Definition of an altitude
2. Leg rule in the similarity on right
triangle theorem
3. Fundamental law of proportion
4. Addition Poperty of equality
5. Distributive property of multiplication
over addition
6. Segment Addition Postulate
7. Substitution
Illustrative Example 4. Keith and Kevin
JM Loyola St.
started bicycling from the
corner of Milograsa St. and JM
Loyola St. in Carmona Town
Proper. At a particular
time,Keith had covered 12
meters and Kevin 5 meters.
How far apart were they at the
time?
Milagrosa St.
Statements
1. Draw altitude CP to the
hypotenuse
x a y b
 ; 
a
c b c
2.
3. ax = a2; cy = b2
4. cx + cy = a2 + b2
5. c(x+y) = a2 + b2
Solution: Since the paths along Milograsa and JM Loyola streets are
perpendicular, the distance between Keith and Kevin at the particular
instance is the hypotenuse of the right triangle formed. By the
Pythagorean theorem,
Keith
c2 = a2 + b2
c2 = 52 + 122
c2= 25 + 144
c = 169
c = 13 meters
C=?
Kevin
12m
5m
Starting point
2. Analysis:
a. How can we use the Pythagorean Theorem to find the missing
length of a right triangle?
b. How does the length of the hypotenuse compare to the length of a
leg?
c. What is the relationship between the sum of the squares of the legs
to t
the square of the hypotenuse?
3. Guided Practice:
A. Use the Pythagorean Theorem to find the unknown side of the given
right triangle if two of its sides are given. Note that these lengths are
known us Pythagorean Triples. The last one is done for you.
B.Apply the concepts of similar right triangles to complete the table.
C
Given : ΔABC is a right triangle with
a
 ACB as a right angle.
Prove: c2 = a2 + b2
A
Proof:
Statements
1. ΔABC is a right triangle
with  ACB as a right angle
2.
3. CD is the altitude of
ΔABC
4.
h
b
x
D
y
Reasons
1. Given
2. Definition of perpendicular line from a
point
3.
4. Geometric Mean
5. a2 + b2 = x(x+y) +
y(x+y)
6.
5.
7. c = x + y
8.
9. a2 + b2 = c2
7. Definition of betweeness
8. Substitution
9.
6.
B
4. Independent Practice
A. Given: Right ∆ABC with ∠B as right angle.
1. If a = 6, c = 8, find b.
2. If a = 21, b = 35, find c.
3. If b = 8, c = 10, find a.
B. Given: Right ΔABC with leg lengths a and b, and hypotenuse length c.
Prove: c2 = a2 +b2
Statements
5. Generalization
Reasons
1.
2.
3.
4.
5.
6.
7.
Pythagorean Theorem
For any right triangle, the sum of the squares of the legs of the triangle is
equal to the square of the hypotenuse, that is a2 + b2 = c2. In words, the
theorem states that: (leg)2 + (leg)2 = (hypotenuse)2 .
Basic steps to solve a Pythagorean Theorem problem.
Step 1: Write the equation.
Step 2: Substitute the length of the hypotenuse for c and any lengths of the
legs for either a or b.
Step 3: Solve the equation for the missing side.
6. Application
Analyze and answer the problem:
1. In preparing for a competition, Mark who is 168 cm
tall, finds that longest string he can use to maintain his
kite flying is 10m. The kite is right above his friend
Kiko, who is six meters away from him. How high is
the kite?
7. Assessment:
Use the Pythagorean Theorem to find x in each triangle.
Answer the following. Show the complete solution.
1. If the diagonal of the square is 18 cm, how long is a side?
2. How high up the wall will a 7 m ladder touch if the foot of the ladder is placed 2 m
from the wall?
3. Maria needed to reach a window that was 3m above ground level. When she
placed a ladder so that it reached the bottom of the window, the base of the ladder
was 1.8 m from the house. How long was the ladder?
IV – ASSIGNMENT
4. Follow Up
A. Follow-up
Given: Right ∆XYZ with right ∠Y, find the missing part.
5. If z = 10 and x = 24, find y.
6. If y = 30, z = 18, find x.
7. If y = 25 and x = 15, how long is z?
Solve. If the width of a picture frame is 2cm less than its length, and the diagonal’s
length is 10cm, find the dimension of the rectangle.
B. Study
1. How to prove the conditions for right triangles using Pythagorean Theorem?
2. How to solve problems applying the Pythagorean Theorem.
3. What are the properties of Isosceles Right Triangle?
4. What is the relationship of its properties to Pythagorean Theorem?
Reference: Learners Manual page 389-392
Electronic Sources:
http://teachers.henrico.k12.va.us/math/HCPSCourse3/8-10/8-10_PythagoreanConstr.pdf
http://www.charleston.k12.il.us/cms/Teachers/math/PreAlgebra/paunit8/L8-1.PDF
Answer Key:
Preliminaries
Activity 1:
(Students answers may vary.)
Activity 2
I can conclude that the area of the small square and the area of the medium
square is the same as the area of the large square in when arranged to form a right
angle triangle
Guided Practice (Let’s Do This!)
A.
A
B
2
2
2
2
f +g =h
f + g2 = h2
2
2
2
3 +g =5
52 + 122 = h2
9 + g2 = 25
25 + 144 = h2
2
g = 25 – 9
h2 = 169
2
g = 16
h = 13
g=4
C
+ g2 = h2
2
f + 242 = 252
f2 + 576= 625
f2 = 625 – 576
f2 = 49
f=7
f2
B.
Statements
1. ΔABC is a right triangle with  ACB as
Reasons
1. Given
a right angle
2. Draw CD ┴ AB
3. CD is the altitude of ΔABC
4. h2 = xy
b2 = x(x+y)
a2 = y(x+y)
5. a2 + b2 = x(x+y) + y(x+y)
6. a2 + b2= (x+y) (x+y)
7. c = x + y
2. Definition of perpendicular line from a
point
3. Definition of an altitude
6. Geometric Mean
5. APE
6. Factoring
7. Definition of betweeness
8. a2 + b2 = c(c)
9. a2 + b2 = c2
8. Substitution
9. Simplifying expression
Independent Practice (I Can Do This!)
A.
1. 28
2. 1,666
3. 36
B.
Statements
1. Draw altitude CD to the hypotenuse
𝑎 𝑦
𝑏
2. 𝑥 = ; =
𝑎
𝑐
𝑏
𝑐
a2
3. cx = ; cy = b2
4. cx + cy = a2 + b2
5. c(x+y) = a2 + b2
6. c = x + y
7. c2 = a2 + b2
Application (Let’s Do More!)
A. 1. No
2. Yes
3. Yes
4. Yes
5. No
6. Yes
Assessment (Challenge Yourself!)
A.
1. 9
2. 5
3. 14.14
4. 17.72
5. 14
B.
1. 12.71
2. 6.71 m
3. 3.5 m
Assignment
Follow-up
1. 26
2. 24
3. 20
Reasons
Definition of an altitude
Leg rule in the similarity on a right
triangle theorem
Fundamental law of proportion
Addition property of equality
Distributive property of multiplication
over addition
Segment addition postulate
Substitution
Module 6: Similarity
Lesson 6: Similarities in Right Triangle
Learning Competency 42: Solves problems that involve triangle similarity and right
triangles.
I – OBJECTIVES
a.Solve problems involving similarities of two triangles.
b.Apply knowledge and skills related to similar triangles to word problems.
c.Show patience in solving the similarities of two triangles.
II – SUBJECT MATTER
Topic: Similarities
Sub-Topic: Similarities in Right Triangle
Materials: Activity Sheets, laptop and monitor
References: Grade 9 Teaching Guide pp.
Grade 9 Learning Module pp. 397 – 399
III – PROCEDURE
A. Preliminary
What Am I ???
Identify what is being referred to in each picture.
1.
2.
4.
a.
b.
c.
d.
3.
5.
How did you find the activity?
Were you able to determine what is being portrayed in the picture?
What did you do to determine what is in the picture?
Do you think the process or procedure in determining the given
pictures are similar in solving word problems? Explain.
2. Recall the procedure in solving word problem.
B. Lesson Proper
1. Teaching/Modeling
Illustrative Example:
Solve the problem, use Polya’s method.
Mr. Kho, the owner of Pulong Saging Bakery, decided to have
renovation of roof of his rest house located at Malabag, Silang,
Cavite near Marco Polo Garden. The braces of the roof is in
triangular form at shown below.
The measure of brace AB is
150 cm, brace AD is 70 cm and brace DC is 140 cm.
1. What theorem can you apply to find the length of brace BD?
2. What is the measure of brace BD?
3. What theorem can you apply to find the length of brace BC?
4. What is the measure of brace BC?
5. If Mr. Kho decided to expand the roof of his rest house and
added 10 cm in each given braces, what will be the length of BD
and BC?
6. What is the ratio of the area of the two triangles?
Solution:
1. 30-60-90 Right Triangle Theorem
2. 70√3 cm
3. Right Triangle Similarity Theorem
4. 171. 46 cm
5. BD = 80√3 cm, BC = 185. 74 cm
2. Analysis
e. How did you find the activity?
f. What procedure did you apply to find the length of BC and BD?
g. Why did you apply such theorem to find BC? BD? Explain.
h. Is similarities in right triangles useful in real-life situations?
3. Guided Practice
The antenna of the Globe telecom located at Buho, Silang, Cavite
casts a shadow of 60 m. At the same time, Penduko, a by – stander
who is 1.8m tall is standing near the antenna casts a shadow of 6m.
How tall is the antenna of the Globe telecom?
A
D
C
E
B
Solution:
- Let ABC be the big triangle.
Let
DEC be the small triangle.
- ABC ~ DEC
-
𝐴𝐵̅
𝐷̅̅𝐸
𝐴𝐵̅
1.8
=
𝐵̅𝐶
=
60
𝐴𝐶
6𝐴𝐵̅
6
=
Determine whether the
triangles are similar.
Write the proportional
Segments
Substitute the value
6
- 6AB = 60(1.8)
-
Assign variables in triangles.
180
6
Apply the property of
proportion (The product of
extreme is equal to the
product of mean)
Apply the multiplicative
Inverse
- AB = 18
Therefore, the height of the antenna of Globe telecom is 18m.
Guided Practice 2:
Light ray are reflected from a mirror at an angle congruent to
that at which they strike it. Anton placed a mirror flat on the ground
and stood where they could see the top of the tree in the mirror. If
Anton’s eyes were 170 cm above ground level, and the mirror was 80
cm from the man and 12 meters from the foot of the tree, how tall is the
tree?
E
A
A
D
4. Independent
Practice
Joy would like to know how
tall the Maliksi building of Malabag
National High School is. In order
to do this, she holds a book
between her eyes so that the top
and bottom of the building are in
line with the edges of the book. If
Joy is 150 cm tall and standing
600 cm from the school building,
how high is the school building?
E
B
5. Generalization
A. Procedures in solving word problems involving similarity:
1. Assign variables in triangles.
2. Determine whether the triangles are similar.
3. Write the proportional segments.
4. Substitute the value.
5. Apply the property of proportion.
6. Apply multiplicative inverse.
B. How can you use the similarities in triangle in solving word
problem?
6. Application
Your classmate is 5 ft. tall and casts a shadow of 4 ft. at the
same time that the flagpole casts a 12 ft. shadow, what is the height of
the flagpole?
5ft
4ft
7. Assessment
Estimate the distance across Taal Lake if CD = 130m, DE = 120m,
and AC = 160m. Given that AB II DE.
B
A
C
D
E
IV – ASSIGNMENT
Follow – up
Caloy is a spider collector. One night he went to out to find one. He saw
a spider hanging on the wall. Find the length of the shadow casted against a
wall if a 12cm web is held between a flashlight and a wall.
6 cm
15cm
45cm
6 cm
References:
Secondary Mathematics Book III by Antonio G. Tayao, et al, pages 333 – 336
Geometry by Soledad Jose-Dilao, et al, pages 178 - 180
Geometry Book by Eunicde Ato-Lopez, MAT, et al,
Resources:
https://www.google.com.ph/search?q=flashlight+clipart&biw=1366&bih=662&source=lnms
&tbm=isch&sa=X&sqi=2&ved=0ahUKEwjA1c3swf3PAhUJupQKHXBnCzAQ_AUIBigB#tbm=isch&q=taal
+lake&imgrc=1mU5-1Qe7PJWcM%3A
https://www.google.com.ph/search?q=flashlight+clipart&biw=1366&bih=662&source=lnms&tbm=is
ch&sa=X&sqi=2&ved=0ahUKEwjA1c3swf3PAhUJupQKHXBnCzAQ_AUIBigB#imgrc=jh5MWRUtHGXKD
M%3A
https://www.google.com.ph/search?q=rest+house+in+tagaytay&biw=1366&bih=662&source=lnms
&tbm=isch&sa=X&ved=0ahUKEwiLzMTD5_7PAhWIlZQKHV7sCCsQ_AUIBigB#tbm=isch&q=problem+a
bout+similarity+&imgrc=jun54JUg0qkxgM%3A
https://www.google.com.ph/search?q=rest+house+in+tagaytay&biw=1366&bih=662&source=lnms
&tbm=isch&sa=X&ved=0ahUKEwiLzMTD5_7PAhWIlZQKHV7sCCsQ_AUIBigB#tbm=isch&q=k0426114.
jpg&imgrc=AlxWLloTJfwqnM%3A
Answer Key:
Preliminaries:
1. Giraffe neck
2. open sea in a periscope
3. ice cream vendor
4. cat climbing on a tree
5. two policemen talking on a road
Modeling:
1. 30-60-90 Right Triangle Theorem
2. 70√3 cm
3. Right Triangle Similarity Theorem
4. 171. 46 cm
5. BD = 80√3 cm, BC = 185. 74 cm
Guided Practice 2: (Let’s Do This!)
𝐴𝐵̅
𝐵̅𝐶
=
𝐸𝐷̅̅
170
𝐸𝐷̅̅
𝐶𝐷̅̅
=
80 𝐸𝐷̅̅
80
80
120
20,400
= 80
ED = 255 cm or 25.5 m
Independent Practice (I Can Do This!)
150
600
=
600
𝑦
150y 360,000
y = 2,400 cm
x = 2400cm + 150 cm
x = 2550 cm
Application (Let’s Do More!)
𝑥
12
=
5
4
4𝑥
4
30
= 60
x = 15 ft
Therefore the height of the flagpole is 15 feet
Assessment (Challenge Yourself!)
𝐷̅̅𝐸
𝐷̅̅𝐶
=
𝐵̅𝐴
120
𝐵̅𝐴
𝐴𝐶
=
130
160
130 𝐵̅𝐴
130
19,200
= 130
ED = 147.69 m
Follow-up
𝐴𝐸
=
𝐴𝐷̅̅
15
60
=
𝐷̅̅𝐶
𝐴𝐶
6
𝐶𝐷̅̅
15 CD = 360
CD = 24 cm
CG = CD + DG
CG = 24 + 24
CG = 48
Therefore the length of the shadow casted against a wall is 48 cm.