Republic of the Philippines

Department of Education
Regional Office IX, Zamboanga Peninsula

9 Zest for Progress

Z Peal of artnership

MATHEMATICS
Quarter 3 – Module 5:
Proportion and Its Application

Name of Learner: ___________________________

Grade & Section: ___________________________
Name of School: ___________________________
 What I Need to Know
In this lesson, you will learn to:
• Describe a proportion.
• Use the properties of proportions to find the missing quantity of the given proportion.
• Solve problems involving proportions.
• Apply the fundamental theorems of proportionality to solve problems involving
proportions.

What I Know
Find out how much you already know about this lesson. Encircle the letter of the correct answer. Take note of
the items that you were not able to answer correctly and find out the right answer as you go through this
module.

_____1. Which statement of equality between of two ratios?

a. Rate
b. Proportion
c. Similar
d. Congruent

_____2. Which ratio makes 6:8 = ______ a correct proportion?

a. 3:4
b. 2:4
c. 4:5
d. 5:6

_____3. What is the value of y in the proportion 3:6 = y: 24?

a. 4
b. 8
c. 12
d. 16

_____4. In the Proportion w : x = y : z

Which of the mathematical sentence is true?
a. wx=yz
b. wy = xz
c. x/y =w/z
d. w/y = x/z

1
_____5. In the proportion m:6 = 3m-1:12,
What is the value of m?
a. 1
b. 2
c. 3
d. 4

𝑦𝑦
_____6. What is the value of in the proportion 5𝑦𝑦 − 2𝑠𝑠: 10 = 3𝑦𝑦 − 𝑠𝑠: 7
𝑠𝑠
a. 2/5
b. 4/5
c. 3/5
d. 6/5

_____7. In ∆PQR, ST intersects side PQ and PR.

Which of the following proportions state that ST // QR?

a. PS:SQ = PT:TR
b. PQ:QR=PR:QR
c. PQ:PS = PT:TR
d. QS:SP = RP:TP

_____8. Given: BD bisects ∠ABC of ∆ABC. If m∠ABD = 300,

What is the measure of ∠ABC?
a. 150
b. 600
c. 300
d. 450

______
9. Which proportion justifies that CT bisects ∠ACR?
a. AC:CR = AT:TR
b. AR:AT = AC:CT
c. AC:AT = TR:CR
d. AC:CR = AR:TR

.____10. In the figure below, if AC=9, RC= 6 and AR = 10 applying the Triangle Angle-Bisector Theorem.

What is the Measure of TR?

a. 3
b. 4
c. 5
d. 6

2
 What’s In
Ratio is used to compare two or more quantities of the same kind. When two ratios are equal then they are
proportional.

Activity 1: Complete Me.

Directions: The figures that follow show ratios or rates that are proportional. Study the figures and complete
the table that follows by indicating proportional quantities on the appropriate column. Two or more
proportions can be formed from some of the figures. The first one is given for your guide.

Figure Ratios or Rates Proportional Quantities

A Feet : Inches Sample: 1ft:12in = 3ft :36 in
Answer:____________________________________
B Kg of Mango: Amount paid Sample: 3kg : 240 pesos = 2kg : 160 pesos
Answer: ____________________________________
C Minutes : Meters Sample: 2 min : 100m = 1 min : 50 m
Answer:_____________________________________

3
Let us RECALL the accuracy of the determined proportions by checking the equality of the ratios or rates.

A.) Solution 1: Reduce each fraction

Solution 2 : Cross products

The solutions show that the corresponding quantities are proportional. In short, they form a proportion because
the ratios are equal.

Thus,
Proportion is the equality of two ratios and two ratios are equal when the product of the extremes
is equal to the product of the means.

4
 What’s New
The properties that follow show different ways of rewriting proportions that do not change the meaning of
their values.

Fundamental Rule of Proportions

𝑤𝑤 𝑦𝑦
If w:x = y:z then = provided x and z are not zero.
𝑥𝑥 𝑧𝑧
Properties of proportion
𝑤𝑤 𝑦𝑦
Cross multiplication property If = 𝑧𝑧 , then wz = xy
𝑥𝑥
𝑤𝑤 𝑦𝑦 𝑤𝑤 𝑥𝑥
Alternation property IF = , then =
𝑥𝑥 𝑧𝑧 𝑦𝑦 𝑧𝑧
𝑤𝑤 𝑦𝑦 𝑥𝑥 𝑧𝑧
Inverse Property IF = , then =
𝑥𝑥 𝑧𝑧 𝑤𝑤 𝑦𝑦
𝑤𝑤 𝑦𝑦 𝑤𝑤+𝑥𝑥 𝑦𝑦+𝑧𝑧
Addition Property IF = , then =
𝑥𝑥 𝑧𝑧 𝑥𝑥 𝑧𝑧
𝑤𝑤 𝑦𝑦 𝑤𝑤−𝑥𝑥 𝑦𝑦−𝑧𝑧
Subtraction Property IF = , then =
𝑥𝑥 𝑧𝑧 𝑥𝑥 𝑧𝑧
𝑢𝑢 𝑤𝑤 𝑦𝑦 𝑢𝑢 𝑤𝑤 𝑦𝑦 𝑢𝑢+𝑤𝑤+𝑦𝑦
Sum Property If 𝑣𝑣 = 𝑥𝑥 = 𝑧𝑧 , then 𝑣𝑣 = 𝑥𝑥 = 𝑧𝑧 = 𝑣𝑣+𝑥𝑥+𝑧𝑧 = k, where k is
Constant at proportionality and v, x, and z are not zero.

Activity 2: Rewrite Me! and Check Me!

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
If = , then y:3 = a :4 Then ( Product of extremes and
𝑦𝑦 𝑎𝑎
3 4 product of means)
Cross Multiplication 𝑦𝑦 𝑎𝑎
= 4
3
Alternation Property of the original 𝑦𝑦 𝑎𝑎
=
proportion 3 4
Inverse Property of the original 𝑦𝑦 𝑎𝑎
=
proportion 3 4
Addition Property of the original 𝑦𝑦
=
𝑎𝑎
proportion 3 4
Subtraction Property of the original 𝑦𝑦 𝑎𝑎
=
proportion 3 4
Sum property of the original 𝑦𝑦 𝑎𝑎
proportion =
3 4

When k is considered in the sum property of the original proportion, the following proportions can be formed:
𝑦𝑦 𝑎𝑎
=k y=3k and =k a=4k. When we substitute the value of y and a to the original proportion,
3 4
all ratios in the proportion are equal to k, representing the equality of ratios in the proportion.
𝑦𝑦 𝑎𝑎 𝑦𝑦 + 𝑎𝑎
= = = 𝑘𝑘
3 4 7
3𝑘𝑘 4𝑘𝑘 3𝑘𝑘 + 4𝑘𝑘
= = = 𝑘𝑘
3 4 7
3𝑘𝑘 4𝑘𝑘 7𝑘𝑘
= = = 𝑘𝑘
3 4 7
5
 What is it
Study the examples on how to determine the indicated quantities from a given proportion:

Examples:
1. If 9:6 = n :2, find n.
Solution: -The product of the extremes equals the product of the means
- Divide both sides by 6,
6n = 18
n=3
then, n = 3

2. If 2 + d : 5= 9 : 15, what is d?
Solution: 2 +d :5 = 9 :15
30 + 15d = 45 -the product of the extremes equals the product of the means
15d = 45-30
15d = 15 - subtract both sides by 30
d=1 -divide both sides by 15

3. If m:n = 4:3, find 3m-2n:3m+n

Solution:
𝑚𝑚 4 𝟒𝟒𝟒𝟒
= 3𝑚𝑚 = 4𝑛𝑛 𝒎𝒎 =
𝑛𝑛 3 𝟑𝟑
4𝑛𝑛
𝟒𝟒𝟒𝟒 3𝑚𝑚−2𝑛𝑛 3� �−2𝑛𝑛 4𝑛𝑛−2𝑛𝑛 2𝑛𝑛 2
Using 𝒎𝒎 = , = 3
4𝑛𝑛 = = =
𝟑𝟑 3𝑚𝑚+𝑛𝑛 3� �+𝑛𝑛 4𝑛𝑛+𝑛𝑛 5𝑛𝑛 5
3

Therefore, 3m-2n: 3m+ n = 2:5

4. If e and b represent two non-zero numbers find the ratio e : b if 2e2 + eb - 3b2 = 0
Solution:
2e2 + eb - 3b2 = 0
(2e +3b ) (e -b) = 0 - Factoring
2e +3b = 0 or e-b =0 - Equate each factor to 0
2e = -3b e=b
2𝑒𝑒 −3𝑏𝑏 𝑒𝑒 𝑏𝑏
= =
2𝑏𝑏 2𝑏𝑏 𝑏𝑏 𝑏𝑏
𝑒𝑒 −3 𝑒𝑒 1
= =
𝑏𝑏 2 𝑏𝑏 1

Hence, e: b = -3/2 or e:b = 1:1

6
 5. 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.
Solution
𝑟𝑟 𝑠𝑠 𝑡𝑡
Let = = = 𝑘𝑘, 𝑘𝑘 ≠ 0
4 3 2
𝑟𝑟
So, = 𝑘𝑘, → r = 4k;
4
𝑠𝑠
= 𝑘𝑘, → s = 3k;
3
𝑡𝑡
= 𝑘𝑘 , → t = 2k
2

Using, r2 - s2 – t2 = 27
(4k)2 – (3k)2 – (2k)2 = 27
16k2 – 9k2 – 4k 2 = 27
3k2 = 27
3𝑘𝑘 2 27
3
=
3

k2 = 9
k = {3 − 3}
Notice that we need to reject -3 since r, s and t are positive numbers.
Then,
r= 4k s = 3k t = 2k
r= 4(3) = 12 s = 3(3) = 9 t = 2(3) = 6

Proportions can be used to solve problems involving two objects that are said to be in
proportion. This means that for ratios comparing the measures of all parts of one object with the
measures of comparable parts of the other object, a true proportion will always exist.

Example 6. A twin jet airplane has a length of 78 m and a wingspan of 90m. A toy model is made in
proportion to the real airplane. If the wingspan of the toy is 36 cm, find the length of the toy.
Solution:
Plane’s length (m) = plane’s wingspan (m)
Model’s length (cm) model’s wingspan(cm)
78 = 90 Substitution
X 36
78(36) = 90(x) Cross product
2808 = 90x Multiply
31.2 = x Divide each side by 90
The length of the model is 31.2 cm

Note: Your knowledge in the process of solving proportions is very helpful as you are going to solve
problems involving fundamental theorems of proportionality.

7
 Triangle Angle Bisector Theorem (TABT)
If a segment bisects an angle of a triangle, then it divides the opposite side into segments
proportional to the other two sides

Proof of the Theorem

Given : ∠1 ≅ ∠2
𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴
Prove : =
𝐷𝐷𝐷𝐷 𝐶𝐶𝐶𝐶

Plan: Draw auxiliary segment BE parallel to CD. Extend AC to point E. Use properties of parallel lines to
conclude that ∠1≅ ∠4 and ∠2 ≅ ∠3, and conclude that CE = CB. Apply the Triangle Proportional Segment
𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴
Theorem to show that = . By substitution = . 𝐷𝐷𝐷𝐷 𝐶𝐶𝐶𝐶 𝐷𝐷𝐷𝐷 𝐶𝐶𝐶𝐶

ILLUSTRATION

𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶
In the figure if AE bisects ∠CAR, then =
𝐸𝐸𝐸𝐸 𝐴𝐴𝐴𝐴
You can also rewrite this proportion in another way by applying the different properties such as:
𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 𝐸𝐸𝐸𝐸 𝐶𝐶𝐶𝐶 𝐸𝐸𝐸𝐸
= ; = ; =
𝐸𝐸𝐸𝐸 𝐴𝐴𝐴𝐴 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴
𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶
= , since CR = CE + ER ; CE corresponds CA and ER corresponds AR
𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶+ 𝐴𝐴𝐴𝐴

𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶+𝐴𝐴𝐴𝐴
=
𝐸𝐸𝐸𝐸 𝐴𝐴𝐴𝐴

Example 1.

Given:
AH bisects ∠MAT
Find MH?
if MA= 8, AT = 5 and HT = 6
Solution:
𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀
=
𝐻𝐻𝐻𝐻 𝐴𝐴𝐴𝐴
𝑀𝑀𝑀𝑀 8
=
6 5

5MH = 48 - cross multiplication

5𝑀𝑀𝑀𝑀 48
= - Solve for MH by dividing both sides by 5
5 5

MH = 9.6

8
Example 2.
Given :
EO bisects ∠ LEV.
If LV = 5, LE = 4, and EV =6 ,
find:
1. LO
2. OV

Solution :
Since LV is given, we will make use of its corresponding segments LE + EV
𝐿𝐿𝐿𝐿 𝐿𝐿𝐿𝐿+𝐸𝐸𝐸𝐸
The proportion could be : =
𝐿𝐿𝐿𝐿 𝐿𝐿𝐿𝐿
5 4+6
=
𝐿𝐿𝐿𝐿 4
5 10
=
𝐿𝐿𝐿𝐿 4

10LO = 20
10𝐿𝐿𝐿𝐿 20
= ║║
10 10

LO = 2
Since LV =5 and LO = 2,
then
OV = LV – LO
OV = 5 - 2
OV = 3

Triangle Proportionality Theorem (TPT)

If a line parallel to one side of a triangle intersects the other two sides, then it divides the sides
proportionally.

Given:
BD ║AE

Prove:
𝑩𝑩𝑩𝑩 𝑫𝑫𝑫𝑫
=
𝑪𝑪𝑪𝑪 𝑪𝑪𝑪𝑪

Paragraph Proof:

Since BD║AE, ∠4≅∠1 and ∠3≅ ∠2 because they are corresponding angles.
𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶
Then, by AA Similarity, ∆ACE ~∆BCD. From the definition of similar polygons, = .
𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶
By the Segment Addition Postulate, CA=BA+CB and CE=DE+CD.

Substituting for CA and CE in the ratio, we get the following proportion.

9
 𝐵𝐵𝐵𝐵+𝐶𝐶𝐶𝐶 𝐷𝐷𝐷𝐷+𝐶𝐶𝐶𝐶
=
𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶
𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐷𝐷𝐷𝐷 𝐶𝐶𝐶𝐶
+ = + Rewrite as a sum.
𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶
𝐵𝐵𝐵𝐵 𝐷𝐷𝐷𝐷 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶
+1= + 1 then, = 1 and =1
𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶
𝐵𝐵𝐵𝐵 𝐷𝐷𝐷𝐷
= Subtract 1 from each side.
𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶

Illustration

Given : AT//FH Since it is given that AT is parallel to FH ,

one of the sides of the triangle , and it intersects the other two
sides, FI and HI , then it follows that AT divides FI and HI
proportionally.
These are some of the proportions we can derive since
there are plenty of possible proportions available using the
different properties:
𝐹𝐹𝐹𝐹 𝐻𝐻𝐻𝐻 𝐹𝐹𝐹𝐹 𝐻𝐻𝐻𝐻 𝐹𝐹𝐹𝐹 𝐹𝐹𝐹𝐹
= ; = ; =
𝐴𝐴𝐴𝐴 𝑇𝑇𝑇𝑇 𝐴𝐴𝐴𝐴 𝑇𝑇𝑇𝑇 𝐻𝐻𝐻𝐻 𝐻𝐻𝐻𝐻

Example 1.

Determine whether AE // MZ if AY =14, AM=6, EZ =12 and EY = 28.

Solution : Check whether AY : AM = EY : EZ using the given

14 : 6 ? 28 : 12
14(12) ? 6 (28)
168 = 168
Therefore: AE // MZ since the sides are divided proportionally.

Example 2.

10
 What’s More
Let’s find out if you understood the examples provided in applying the properties of proportions. Please
answer the activity.

Activity 3: Your Task

Directions: Read and answer carefully. Write your answer on the space provided.
𝑦𝑦
1. Find if 5y-2s : 10 = 3y - s : 7
𝑠𝑠
Answer  _____________________________
2. If g:h = 4:3, evaluate g + 4h : g + 8h
Answer  _____________________________
3. Solve for the unknown side applying the Triangle Angle-Bisector Theorem.
(Note that the figure is not drawn to scale).

Answer  _____________________________
4. In ∆HKM, HM=15, HN=10, and HJ is twice the length of JK.
Determine whether NJ ║MK.

Answer  _____________________________

5. Given: BC ║ ED.
Find x, AC, and CD if AC = x-3, BE = 20, AB = 16, and CD = x+5.

Answer  _____________________________

11
 What I Have Learned
In solving problems involving proportions we must know how to:
 use the fundamental rule of proportions and apply any of the properties of proportion when
necessary:
• Cross multiplication
• Inverse
• Addition
• Alternation
• Subtraction
• Sum
 apply the fundamental theorems of proportionality
• Triangle Angle Bisector Theorem- (TABT) – If a segment bisects an angle of a
triangle, then it divides the opposite side into segments proportional to the other two
sides.
• Triangle Proportionality Theorem- (TPT)- If a line parallel to one side of a triangle
intersects the other two sides, then it divides the two sides proportionally
•

What I Can Do
Activity 4: Let’s Do It!
Directions: Solve the following problems. Write your answer on the space provide and use extra sheet for
your solution (if necessary).

1. Miguel is using centimeter grid paper to make a scale drawing of his favorite car. The width of the
drawing is11.25 cm. How many feet long is the actual car? (Scale : 1.5cm = 2ft)
Answer  ____________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
2. Alex is standing next to Williams Tower. He is 6 ft tall and the length of his shadow is 8 ft. If the
length of the shadow of the tower is 1,200 ft. how tall is the tower?
Answer  ___________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________

12
 3. Refer to the map below Third avenue and 5th avenue are parallel. If the distance from 5th avenue to
City Mall along State street is 2140 ft, find the distance between 5th avenue and City Mall
along Union St.

Answer  ___________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________

4. The perimeter of triangular lot is 50m. The surveyor’s tape

bisects an angle. Find lengths x and y.

Answer  ___________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________

Assessment
Read the questions carefully. Write the letter of the correct answer on the space provided.
𝐶𝐶 𝐸𝐸
_____1. In proportion . = , What is the form by using cross multiplication property?
𝐷𝐷 𝐹𝐹
𝐶𝐶 𝐸𝐸 𝐶𝐶 𝐷𝐷
a. CE = DF b. CF = DE c. . = d. . =
𝐹𝐹 𝐷𝐷 𝐸𝐸 𝐹𝐹

_____2. Which ratio makes 6:8 = ______ a correct proportion?

a. 2:3 b. 3:4 c. 4:5 d. 5:6
_____3. What is the value of z in the proportion 3 : 5 = z : 15 ?
13
 a. 9 b. 10 c. 11 d. 12

_____4. In the proportion w : x = y : z.

Which mathematical equation is true?
a. wx=yz b. wy = xz c. x/y =w/z d. w/y = x/z

_____5. At a certain time of the day, a post cast a shadow of 3m while a 6m pole cast a shadow of 5m.
What is the height of the post?
a. 5m b. 2.5m c. 3.6m d. 4m

𝑦𝑦
_____6. What is the value of 𝑠𝑠
in the proportion 5𝑦𝑦 − 2𝑠𝑠: 10 = 3𝑦𝑦 − 𝑠𝑠: 7
a. 2/5 b. 4/5 c. 3/5 d. 6/5

_____7. In the figure at the right. ∆ABC, DE intersects side AB and BC.
Which of the following proportion state that DE // AC?
a. AD:DB =CE:EB c. AD:AB = CE:EB
b. DB:BE= EC:DA d. DE:BE =BA:BC

_____8. In the figure at the right. ∆HAT, ER // HT. If HE=6, EA=9, RA = 6.

What is TR?
a.3 c. 4
b. 6 d. 5

______
9. In the figure at the right, CT bisects ∠ACR.
Which proportion is correct?
a. AC:CR = AT:TR c. AC:AT = TR:CR
b. AR:AT = AC:CT d. AC:CR = AR:TR

____10. Using the ∆ABC where AD bisects ∠BAC wher AB=3, AC= 5and BC = 12 .

What is CD?
a. 6 b. 4.5 c. 7.5 d. 5

14
References
Bass, Laurie E., Art Jhonson, Basia Rinesmith Hall, Dorothy F. Wood. Geometry: Tools for a Changing
World:Prentice – Hall, 1998.

Bryant, Merden L., Leonides E. Bulalayao, Melvin M. Callanta, Jerry D. Cruz, et al. Mathematics Teachers Guide 9.
Pasig City: Department of Education, 2014

Bryant, Merden L., Leonides E. Bulalayao, Melvin M. Callanta, Jerry D. Cruz, et al. Learner’s Material Mathematics
9. Pasig City: Department of Education, 2014

Development Team
Writer: Emy Christie P. Roxas
Diplahan National High School

Editor/QA: Eugenio E. Balasabas

Ressme M. Bulay-og
Mary Jane I. Yeban

Reviewer: Gina I. Lihao

EPS-Mathematics

Illustrator:
Layout Artist:
Management Team: Evelyn F. Importante
OIC-CID Chief EPS

Jerry c. Bokingkito
OIC-Assistant SDS

Aurelio A. Santisas
OIC- Assistant SDS

Jenelyn A. Aleman
OIC- Schools Division Superintendent

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