MATHEMATICS
GRADE 9
DATE: ……………………………………….
TOPIC: Geometry 2D Shapes
CONCEPTS & SKILLS TO BE ACHIEVED:
By the end of the lesson learners should know and be able to:
 Revise 2 D shapes: Triangles and Quadrilaterals
 Solve problems relating the above shapes
RESOURCES:
ONLINE RESOURCES
DAY 1:
DBE Workbook, Sasol-Inzalo book, Textbooks,
https://www.youtube.com/watch?v=mLeNaZcy-hE
https://www.youtube.com/watch?v=xEzm8wBSpAc
https://www.youtube.com/watch?v=K3h-PpaSwQI
INTERIOR ANGLES OF A TRIANGLE
LESSON DEVELOPMENT
Recap our knowledge on Triangles.
A triangle is a closed shape with three sides and three interior angles.
The sum of the interior angles of a triangle add up to 𝟏𝟖𝟎°.
Therefor in any triangle:
We
write it
as….
[ ∠' s of ∆=180°]
𝑏
𝑎
𝑎 + 𝑏 + 𝑐 = 180°
𝑐
80°
40° + 60° + 80° = 180°
40°
60°
𝑥
110° + 𝑥 + 30° = 180°
30°
Grade 9
Geometry of 2D shapes
110°
Page 1 of 13
You need to determine the size of the of the missing angles in each of the
following diagrams.
1. Solve for 𝑥
SOLUTION
𝑎 + 40° + 65° = 180°
𝑎 + 105° = 180°
𝑎 = 180° − 105°
𝑎 = 75°
[ ∠′ 𝑠 𝑜𝑓∆= 180°]
2.
105°
𝑏
55°
3. Detemine the value of 𝑎 in the
diagram below:
4𝑎
3𝑎
2𝑎
105° + 55° + b = ___ °
____ ° + 𝑏 = 180°
𝑏 = 180° − _______
𝑏 = ________
[∠′ 𝑠 𝑜𝑓∆= 180°]
You need to fill
in the missing
information:
3𝑎 + ___ + 2𝑎 = 180°
[________________________]
9𝑎 = 180°
𝑎 = 20°
 Add like
terms
Remember you can check your answer!
 Solve
If 𝑎 = 20° 𝑡ℎ𝑒𝑛 3𝑎 = 3 × 20° = 60°
equation
4𝑎 = 80° and
2𝑎 = 40°
60° + 80° + 40° = 180°
ACTIVITY 1
Grade 9
Geometry of 2D shapes
Page 2 of 13
Do the following exercises, applying what you have learnt today. FIRST ATTEMPT TO DO ALL THE
PROBLEMS BEFORE YOU CHECK YOUR ANSWERS IN THE MEMORANDUM BELOW
Calculate the size of the unknown angles. You need to provide reasons.
1.
2.
𝑥 + 15°
𝑥
3𝑥 − 25°
3. In the figure AB ∥ ED, AĈD = 95°
̂ = 30°.
and D
̂ N=60°.
4. In the figure KL ∥ MO, ML̂N=53° and LM
Determine the size of the following angles with
reasons.
(a)
(b)
(c) L̂3
̂1
N
L̂2
Determine with reasons:
(a) Ĉ1
̂
(b) E
̂
(c) A
Memo: Activity 1
1.
52° + 79° + 𝑥 = 180° [∠′ 𝑠 of ∆= 180°]
𝑥 + 131° = 180°
𝑥 = 180° − 131°
𝑥 = 49°
2.
(𝑥 + 15°) + 𝑥 + (3𝑥 − 25°) = 180° [∠′ 𝑠 of ∆= 180°]
5𝑥 − 10° = 180°
5𝑥 = 180° + 10°
5𝑥 = 190°
𝑥 = 38°
3.
(a)
4.
̂1 = 180° − (60° + 53°) [∠′ 𝑠 𝑜𝑓∆= 180°]
(a) 𝑁
̂ = 67°
𝑁
̂ [ 𝑎𝑙𝑡 ∠′ 𝑠; 𝐾𝐿 ∥ 𝑀𝑂]
(b) 𝐿̂2 = 𝑀
𝐿̂2 = 60°
̂1 [ 𝑐𝑜𝑟𝑟𝑒𝑠 ∠′ 𝑠; 𝐾𝐿 ∥ 𝑀𝑂]
(c)
𝐿̂3 = 𝑁
𝐿̂3 = 67°
𝐶̂1 = 180° − 95° [ Str line]
= 85°
(b)
𝐸̂ + 85° + 30° = 180°
[∠′ 𝑠 𝑜𝑓∆]
𝐸̂ = 180° − 115° = 65°
(c) 𝐴̂ + 𝐸̂ = 180° [ co − int ∠′ s; AB ∥ DE]
𝐴̂ = 180° − 65° = 115°
Grade 9
Geometry of 2D shapes
Page 3 of 13
DAY 2:
Types of Triangles
Types of Triangles. Do you remember?
 Isosceles
All angles are
equal
All sides are equal
Each angle= 60°
Two angles equal
Two sides are the
same length
Scalene
Different lengths
90° angle
All sides may have
different lengths
• All angles may be
different
Right-Angled 
Equilateral 
Let’s apply this knowledge to some problems.
1. Calculate 𝑥 in the follwing
diagram:
A
64°
B
𝑥
C
Solution:
Complete where needed.
𝐴̂ = 𝐶̂
[∠′ 𝑠 opposite = sides]
∴ 𝐴̂ = 𝐶̂ = 𝑥
𝑥 + 𝑥 + 64° = ______
[∠' s of ∆=180°]
2𝑥 + 64° = 180°
2𝑥 = 180° − 64°
2𝑥 = ________
𝑥 = 58°
2. ABCD is a paralellogram.
Determine with reasons the size of:
̂ = 55°
(a) B
[∠′ s opposite = sides]
̂ + 55° + ___ = 180°
(b) A
̂ = 180° − 110°°
A
̂ = 70°
A
(a)
(b)
(c)
̂1
B
̂
A
̂2
B
(c) 𝐵̂2 = ____
[∠' s of ∆=180°]
[ corres ∠′ 𝑠; AE ∥ BF]
ACTIVITY 2
1. Name each of these triangles.
Grade 9
Geometry of 2D shapes
Page 4 of 13
2.
Determine the size of the unknown
angles.
3.
46°
𝑦
𝑥
𝑥
65°
𝑦
MEMORANDUM: ACTIVITY 2:
1.
(a) scalene
(b) isosceles
(c) right-angled triangle
2.
𝑥 = 60° [∠' s opposite=sides]
𝑦 = 180° − 60° [str line]
3.
2𝑥 = 180° − 46° [ ∠' s opposite=sides]
2𝑥 = 134°
𝑥 = 67°
𝑦 = 180° − 90° − 65° [∠′ 𝑠 𝑜𝑓∆= 180°]
𝑦 = 35°
Grade 9
Geometry of 2D shapes
Page 5 of 13
DAY 3:
The exterior angle of a triangle
Let’s study the diagrams below:
Diagram 1
Diagram 2
70°
(a)
50°
42°
40°
𝑎
𝑏
Diagram 3
𝑎
𝑏
𝑎
𝑏
Complete:
𝑎 = 180° − ( 70° + 40°)
𝑎 = 180° − 110°
𝑎 = 70°
𝑏 = 180° − 70° [∠’s on str line]
[∠’s on str line]
𝑏=
𝑎 = 180° − (50° + ______)
𝑎 = 180° − (____ + 42°)
𝑎 = 180° − 100°
𝑎 = 180° − 132°
𝑎 = 80°
𝑎 = 48°
𝑏 = 180 − 80° [__________]
𝑏 = _____________
𝑏=
𝑏=
This is the
exterior angle
b
a
𝐚+𝐛
The exterior angle of a triangle is equal to the SUM of the opposite interior angles.
[ ext ∠ of ∆]
In the above sketches
Use this for your reason
Diagram 1: 𝑏 = 70° + 40° = 110°
Diagram 2: 𝑏 = 50° + 50° = 100°
Diagram 3: 𝑏 = 90° + 42° = 132°
`1.
Determine the values of 𝑥 and 𝑦.
Solution:
𝑥 = _______________ [ext ∠ of ∆]
= ______________
𝑦
18°
̂ = ________ [ ∠’ s opp=sides ]
C
𝑦 = 180° − 2𝑥 [ ___________________ ]
𝑦 = ___________
𝑥
59°
Grade 9
Geometry of 2D shapes
Page 6 of 13
2.
Determine the size of the letters in
the sketch below:
N
D
𝑎 + 10°
J
136° 𝑏
K
D
46°
M
𝑐
F
Solution:
Fill in the gaps
̂ N=a+10°+46° [ ______________ ]
JK
136° = 𝑎 + 56°
𝑎 = 136° − 56°
𝑎 = 80°
̂ =80°+10°=90°
∴N
𝑏 = 180° − (90° + ____) [ ____________ ]
𝑏 = 44°
𝑐 = 44° + 90°
[ _______________ ]
𝑐 = 134°
ACTIVITY 3
̂ C.
2. Find the value of BA
1. Determine the values of 𝑥 and 𝑦.
3.
In ∆PQR, QR is extended to S. Determine the size of 𝑄̂ .
4. Study the figure below:
(a)
(b)
Caculate the size 𝑄̂2
Hence calculate the size of 𝑀𝑂̂𝑁.
Grade 9
Geometry of 2D shapes
2
Page 7 of 13
MEMORANDUM: ACTIVITY 3:
1. 𝑦 = 56° + 32°
[ext ∠ of ∆]
𝑦 = 88°
𝑥 = 56° + (32° + 12°) [ext ∠ of ∆]
𝑥 = 56° + 44°
𝑥 = 100°
2. 3𝑦 + 𝐵𝐴̂𝐶 = 5𝑦
[ext ∠ of ∆]
𝐵𝐴̂𝐶 = 5𝑦 − 3𝑦
𝐵𝐴̂𝐶 = 2𝑦
But 4𝑦 + 5𝑦 = 180° [Str line]
9𝑦 = 180°
𝑦 = 20°
𝐵𝐴̂𝐶 = 2(20°)
= 40°
̂
[𝑐𝑜 − 𝑖𝑛𝑡 ∠ ; 𝑃𝑄 ∥ 𝑀𝑁]
4. (a) 𝑄2 + 55° = 180°
̂
𝑄2 = 125°
3. 2𝑥 + 7𝑥 = 180° [Str line]
9𝑥 = 180°
𝑥 = 20°
(b)
[ext ∠ of ∆]
𝑄̂ + 4𝑥 = 7𝑥
̂
𝑄 = 3𝑥
𝑄̂ = 2(20°) = 40°
2𝑥 − 5° + 3𝑥 + 40 = 125° [ext ∠ of ∆]
5𝑥 − 35° = 140°
5𝑥 = 175°
𝑥 = 35°
𝑀𝑂̂𝑁 = 2(35°) − 5°
= 70° − 5°
= 65°
IT IS IMPORTANT TO REMEMBER:

The sum of the interior angles of a triangle add
up to 𝟏𝟖𝟎°.

The exterior angle of a triangle is equal to the
sum of the opposite interior angles

Apply the properties of the different triangles at all times.
Grade 9
Geometry of 2D shapes
Page 8 of 13
DAY 4 and 5:
Quadrilaterals
A quadrilateral is a closed shape with four sides.
Important Fact
The sum of the interior angles adds up to 𝟑𝟔𝟎°
Types of Quadrilaterals
Special quadrilaterals
Parallelogram




Properties
Opposite sides are equal
Opposite angles are parallel
Opposite angles are equal
Diagonals bisect each other
Rectangle





Opposite sides are equal
Opposite sides are parallel
Angles = 90°
Diagonals are equal
Diagonals bisect each other
Square





All sides are equal
Opposite sides are parallel
Angles = 90°
Diagonals are equal
Diagonals bisect each other
Diagonals bisect each other at right
angles
Diagonals bisect the interior angles

Rhombus
Grade 9
Geometry of 2D shapes
Page 9 of 13
All sides are equal
Opposite sides are parallel
Opposite angles are equal
Diagonals bisect each other at
right angles.
Diagonals bisect the interior
angles
Diagonals are equal
Diagonals bisect each other







Trapesium

One pair of opposite sides are
parallel.


Two pairs of adjacents sides equal.
Opposite angles between unequal
sides are equal.
One diagonal of the kite bisects the
other diagonal at 90°.
One diagonal bisects the interior
angles.
Kite


Application
Determine the value of the unknown
angles in the shape.
1.
2.
Steps
Identify the shape – paralellogram
Apply properties - opposite angles
are equal
𝒅 = 𝟒𝟓°
[ 𝒐𝒑𝒑 ∠ 𝒐𝒇 𝒑𝒂𝒓𝒎]
3. Identify parts – parallel lines
𝒅 + 𝒆 = 𝟏𝟖𝟎° [𝒄𝒐 − 𝒊𝒏𝒕 ∠; ∥ 𝒍𝒊𝒏𝒆𝒔
𝒆 = 𝟏𝟖𝟎° − 𝟒𝟓°
𝒆 = 𝟏𝟑𝟓°
Grade 9
Geometry of 2D shapes
Page 10 of 13
ACTIVITY 4
Do the following exercises, applying what you have learnt today. FIRST ATTEMPT TO
DO ALL THE PROBLEMS BEFORE YOU CHECK YOUR ANSWERS IN THE MEMORANDUM
BELOW.
̂C = 𝑦
1. ABCD is a parallelogram with EA
̂ B = 𝑥.
and DA
Calculate:
̂1
(a)
D
(b)
𝑦
(c)
̂1
A
2. MNPQ is a trapesium with MQ=MN and
NQ⊥QP. Determine, with reasons, the
sizes of 𝑎, 𝑏, 𝑐 and 𝑑.
3. Calculate, with reasons the value of 𝑥.
Grade 9
Geometry of 2D shapes
Page 11 of 13
Previous examination paper.
Memorandum: Activity 4
1. (a)
(b)
(c)
2.
̂ 1 = 75°
D
[alt ∠' s; AD ∥ BC]
𝑦 = 48° + 75° [ext ∠ of ∆]
𝑦 = 123°
𝐴̂1 = 180° − 123°
𝐴̂1 = 57°
[str line]
𝑎 = 180° − (90° + 65°) [int ∠ of ∆]
𝑎 = 25°
𝑎=𝑏
[alt ∠' s; MQ ∥ NP]
𝑏 = 25°
𝑏=𝑐
[ equal ∠' s opposite equal sides]
𝑐 = 25°
3.
𝐸̂1 = 2𝑥 + 15° [ equal ∠′ 𝑠 opposite = sides]
̂1 = 𝑥
𝐻
[ diagonal bisects angle]
2(2𝑥 + 15°) + 2𝑥 = 180° [int ∠ of ∆]
4𝑥 + 30° + 2𝑥 = 180°
6𝑥 = 150°
𝑥 = 25°
𝑑 = 180° − 50° [int ∠'s of ∆]
𝑑 = 130°
4.
𝑥 = 85° [alt ∠' s; AC ∥ HF]
𝑦 = 180° − 85°
[co-interior ∠' s; BF ∥ CD]
𝑦 = 95°
Grade 9
Geometry of 2D shapes
Page 12 of 13
CONSOLIDATION
IT IS IMPORTANT TO REMEMBER:

You need to identify the shape

You need to know the properties of the

shape
Show all markings on your sketch.
Grade 9
Geometry of 2D shapes
Page 13 of 13