GRADE 1 TO 12
DAILY LESSON
LOG
School
Teacher
Teaching Date and Time
Ramon Magsaysay High School
Jasper Clark F. Abella
April 21, 2021
Grade level
Learning Areas
9
Mathematics
Quarter
Fourth
I. OBJECTIVES
A. Content Standards
The learner demonstrates understanding of key concepts of parallelograms and
triangle similarity.
B. Performance Standards
The learner is able to investigate, analyze, and solve problems involving
parallelograms and triangle similarity through appropriate and accurate
representation.
C. Learning Competencies/
Objectives. Write the LC
code for each.
At the end of the period, the learners, with 80% proficiency, will be able to:
A. Prove the conditions for special right triangles. (M9GE-IIIg-h-1)
 Define the two types of special right triangle
 Find the length of the indicated side using the two special right triangle
II. CONTENT
PROVING CONDITIONS FOR SPECIAL RIGHT TRIANGLES
III. LEARNING RESOURCES
A. References
1. Teacher’s Guide pages
2. Learner’s Material pages
3. Textbook pages
Learning Module for Junior High School Mathematics Grade 9 Q3 Module 13
4. Additional Materials
from Learning
Resource Portal
Grade 9 Learner’s Material pages 394-397
B. Other Learning Resources
IV. PROCEDURE
SYNCHRONOUS (1 hour)
A. Opening prayer and greetings
B. Recap of the last lesson
Special Right Triangles
1. 45°-45°-90° RIGHT TRIANGLE
A. Reviewing previous lessons
or presenting the new
lesson.

In a 45°-45°-90° right triangle:
o Is also known as an isosceles right triangle
o Is a right triangle with 2 equal sides or legs (𝑆𝑖𝑑𝑒1 = 𝑆𝑖𝑑𝑒2),
then the legs are usually labeled 𝑥.
o The hypotenuse is often labeled ℎ.
2. 30°-60°-90° RIGHT TRIANGLE

In a 30°-60°-90° right triangle:
o Is also known as a scalene right triangle
o The side across from the 30° angle is the short side and often
labeled 𝑠.
o The side across from the 60° angle is the long side and often
labeled 𝑙.
o The hypotenuse is often labeled ℎ.
In proving the Special Right Triangles using the Pythagorean theorem.
1. 45°-45°-90° right triangle
𝑐 2 = 𝑎2 + 𝑏 2 or
B. Establishing a purpose for
the lesson.
𝑥 2 + 𝑥 2 = ℎ2
2𝑥 2 = ℎ2
√2 ∙ 𝑥 2 = √ℎ2
𝒙√𝟐 = 𝒉 or 𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 = 𝒍𝒆𝒈 ∙ √𝟐
In finding the leg when the hypotenuse is given
𝑥 2 + 𝑥 2 = ℎ2
2𝑥 2 = ℎ2
ℎ2
𝑥2 =
2
ℎ2
√𝑥 2 = √
2
𝑥=
ℎ
∙
√2
√2 √2
𝒉√𝟐
𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 ∙ √𝟐
𝒙=
𝒐𝒓 𝒍𝒆𝒈 =
𝟐
𝟐
2. 30°-60°-90° right triangle
∆WXZ is equilateral triangle
̅̅̅̅̅ is the perpendicular bisector of ̅̅̅̅
𝑊𝑌
𝑋𝑍
1
2
1
2
Thus, 𝑋𝑌 = 𝑋𝑍 = 𝑋𝑊 , or 𝑋𝑊 = 2𝑋𝑌 = 2𝑥.
Also,
𝑋𝑌 2 + 𝑌𝑊 2 = 𝑋𝑊 2
𝑠 2 + 𝑌𝑊 2 = (2𝑠)2
𝑌𝑊 2 = 4𝑠 2 − 𝑠 2
√𝑌𝑊 2 = √3𝑠 2
𝒀𝑾 = 𝒔√𝟑
Use Pythagorean Theorem
Substitute s for XW and 2s for XW
Substract 𝑠 2 from each side
Simplify
Find the square root of each side
Note : The length of the hypotenuse is twice the length of the shorter leg.
The length of the longer leg is √𝟑 times the length of the shorter leg.
Examples: Find the value of each variable.
C. Presenting examples/
instances of the new
lesson.
ℎ = 𝑥√2
𝒉 = 𝟓√𝟐
Checking:
𝑐 2 = 𝑎2 + 𝑏 2
2
(5√2) = 52 + 52
25 ∙ 2 = 25 + 25
𝟓𝟎 = 𝟓𝟎
ℎ √2
𝑥=
2
𝟑√𝟐
𝒙=
𝟐
Checking:
𝑐 2 = 𝑎2 + 𝑏 2
2
3√2
)
2
32 = (
9×2
2
3√2
)
2
+(
9×2
9= 4 + 4
9 = 4.5 + 4.5
𝟗=𝟗
𝑙 = 𝑠√3
𝑙
√3
=
𝑠=
𝒔=
𝑙 = 𝑠√3
𝑠√3
𝒍 = 𝟓√𝟑
√3
𝑙√3
ℎ = 2𝑠
3
𝟕√𝟑
ℎ = 2(5)
𝟑
𝒉 = 𝟏𝟎
ℎ = 2𝑠
7√3
ℎ = 2(
)
3
𝟏𝟒√𝟑
𝒉=
𝟑
Checking:
Checking:
𝑐 2 = 𝑎2 + 𝑏 2
(
2
14√3
) =(
3
196×3
9
=
588
9
7√3
=
147
9
2
) + 72
102 = 52 + (5√3)
+ 49
100 = 25 + (25 × 3)
3
49×3
9
𝑐 2 = 𝑎2 + 𝑏 2
2
+ 49
65.33 = 16.33 + 49
𝟔𝟓. 𝟑𝟑 = 𝟔𝟓. 𝟑𝟑
100 = 25 + 75
𝟏𝟎𝟎 = 𝟏𝟎𝟎
Solve for the following problems.
1. An escalator lifts people to the second floor, 25 ft. above the first floor. The
escalator rises at a 30° angle. How far does a person travel from the bottom
to the top of the escalator?
Solution:
Let hypotenuse
𝒉 − be the length of the escalator
𝟐𝟓 𝒇𝒕. − be the length of the shorter leg opposite the 30° angle
D. Discussing new concept
and practicing skills #1.
ℎ = 2𝑠
ℎ = 2(25 𝑓𝑡. )
𝒉 = 𝟓𝟎 𝒇𝒕.
Answer:
A person travel from the top of the escalator is about 50 ft.
2. Road Signs. The warning sign at the left is an equilateral triangle. The height
of the sign is 1m. Find the length s of each side of the sign to the nearest
tenth of a meter.
Solution:
1
𝑙√3
𝑠=
2
3
1
1√3
𝑠=
2
3
1
𝑠 = 2 ( 𝑠)
2
1√3
𝑠 = 2(
)
3
2√3
𝑠=
3
𝑠 = 1.155
𝒔 = 𝟏. 𝟐 𝒎𝒆𝒕𝒆𝒓
Activity 1 : Complete the measure of the dimensions each triangle.
2.
E. Developing mastery (Leads
to Formative Assessment)
Answer:
1. 𝑥 = 20 , 𝑦 = 20√3
2. 𝑥 = 24 , 𝑦 = 12√3
3. 𝑥 = 5 , 𝑦 = 5√3
4. 𝑥 = 9 , 𝑦 = 18
5. 𝑥=√3 , 𝑦 = 3
4.
F. Finding practical
applications of concepts
and skills
in daily living.
Activity 2: Answer the following questions.
1. A square-shaped handkerchief measures 16 inches on each side. You fold it
along its diagonal so you can tie it around your neck. How long is this tie?
2. A cake is triangular in shape. Each side measures 1 foot. If the cake is
subdivided equally into two by slicing from one corner perpendicular to the
opposite side, how long is that edge where the cake is sliced?
3. A ladder leaning against a wall makes an angle of 30 degrees with the
ground. If the length of the ladder is 9 m, find the height of the wall.
G. Making generalization and
abstractions about the
lesson.
What are the special right triangles?
ASYNCHRONOUS TASK (1 hour)
Answer the quiz in google form :
I.
Evaluating Learning
https://forms.gle/uhwNFNz3Y5dgkNa38
J. Additional Activities
application/remediation
A. Do Additional Activities on Math 9 Quarter 3 Module 13 page 17
V. REMARKS
VI. REFLECTION
SECTIONS
TIME
ROOM
A. No. of learners who earned 80% in the evaluation.
B. No. of learners who require additional activities for
remediation.
C. Did the remedial lessons work? No. of learners who caught
up with the lesson.
D. No. of learners who continue to require remediation.
E. Which of my teaching strategies worked well? Why did
these work?
F. What difficulties did I encounter which my
principal/supervisor can help me solve?
G. What innovation or localized materials did I use/discover
which I wish to share with other teachers?
Einstein
7:00-9:00
29/25
Darwin
Edison
9:15-11:15
12:00-2:00
2:15-4:15