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GRADE 8 AND 9 GEOMETRY EXERCISES
Exercise 1
 Complete the theorem:
The sum of the angles on a straight line = ___________o
Determine the values of the following angles (𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓, 𝑔 and ℎ) in
the diagram below. Show ALL calculations.
1.1
𝟑𝟓°
𝒂
1.2
𝟑𝟓°
𝟖𝟎°
𝒃
1.3
𝒄
𝟔𝟐°
𝟓𝟒°
𝒅
1.4
𝒆
𝒇 𝟏𝟏𝟎°
𝟒𝟑°
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1.5
𝒈
𝟒𝟏°
𝟏𝟑𝟗° 𝒉
Exercise 2
 Complete the theorem:
When two straight lines intercept (“cut”) each other, the vertically
opposite angels are______________________.
Determine the values of the unknown angels (𝑖, 𝑗, 𝑘, 𝑙, 𝑚, 𝑛) in the diagram
below. Sho ALL calculations.
2.1
𝟏𝟑𝟑°
𝒊
2.2
𝟔𝟏°
𝒋
2.3
𝒌
𝟒𝟏°
𝒍
𝟐𝟖°
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2. 4
𝒎
𝟑𝟕°
𝟗𝟖°
𝒏
Exercise 3
 Complete the theorem:
When two parallel lines are intercepted by a diagonal line, then:

the pair of corresponding angles are _____________________________.

the __________________________ of the co-interior angles on the same
side of the diagonal line = 180o.

the _________________________________ angles are equal.
3.1
In the diagram below, PQ is parallel to RS. AB is a diagonal line,
intercepting PQ and RS at L and M respectively.
A
L1 4
P
2
M1 4
2 3
R
Q
146o
S
B
̂4 , 𝑀
̂2 .
Determine, by giving reasons, the size of 𝐿̂1 , 𝐿̂2 , 𝑀
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3.2
̂ = 43°.
ABCD is a parallelogram. AD || BC, AB || DC, 𝐴𝐹̂ 𝐸 = 36° and 𝐷
E
A
36o
F1
D
43o
2 3
1
B
2
C
Determine, by giving reasons, the size of 𝐹̂3 , 𝐶̂1 , 𝐴̂ en 𝐵̂.
3.3
PQRS is a trapezium. PQ || TU, TU || RS, PR || VZ and 𝑉𝑌̂𝑇 = 62°.
V
X1
2
P
Q
3 4
62o Y1
2 3
T1
2
U1
2
3
2
4
R
S1
Z
Determine, by giving reasons, the size of 𝑇̂1 , 𝑆̂3 , 𝑋̂4 , 𝑃̂, 𝑅̂ and 𝑆̂4 .
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Exercise 4
 Complete the theorem:
o The sum of the interior angles of a triangle = ________________o
o In a(n) ___________________________ triangle, the angles opposite the
equal sides, are equal.
o The exterior angle of a triangle is _________________________ to the
_______________________ of opposite interior angles.
Determine the size of the unknown angles, indicated by variables
𝑎, 𝑏, 𝑐, 𝑥, 𝑦 or 𝑧:
4.1
74o
𝒂
4.2
𝒃
61o
39o
4.3
77o
𝒄
86o
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4.4
70o
55o
𝒙
4.5
68o
𝒛
𝒚
4.6
In the diagram below, AB = AC and DE || BC.
A
𝒄
𝒃 E
𝒙
D 𝒂
B
4.7
65o
C
D
48o
E
𝒚
53o
F
G
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4.8
L
O
52o
𝒛
M
86o
N
4.9
L
𝒂
M
N
71o
42o
O
4.10
𝒂 𝒃
𝒛
𝒙
𝒚
𝟕𝟐°
oooOooo
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The sum of the angles on a straight line = 180o
1.1
𝑎 = 180° − 35°
∴ 𝑎 = 145°
1.2
𝑎 = 180° − (35° + 80°)
𝑎 = 180° − 115°
∴ 𝑎 = 65°
1.3
𝑐 = 90° − 62°
∴ 𝑐 = 28°
𝑑 = 90° − 54°
∴ 𝑑 = 36°
𝑒 = 180° − 43°
∴ 𝑒 = 137°
𝑓 = 180° − 110°
∴ 𝑓 = 70°
1.5
𝑔 = 180° − 41°
∴ 𝑔 = 139°
ℎ = 180° − 139°
∴ ℎ = 41°
When two straight lines intercept (“cut”) each other, the vertically opposite
angels are equal.
2.1
2.3
𝑖 = 133°
𝑘 = 28°;
2.2
2.4
𝑙 = 41°
𝑗 = 61°
𝑚 = 37°;
𝑛 = 98°
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SHORTENED REASONS:
 Vertically Opposite
 Complementary
(angles with a sum of 90o)
 Supplementary
(angles with a sum of 180o)
 Alternate
 Corresponding
 Angles
 Sum of interior angles of a
triangle = 180o
 Exterior angles of a triangle = sum
of opposite interior angles
 In an isoscolese triangle, the angles
opposite the equal sides, are equal.
⟺
⟺
vert. opp.
compl.
⟺
⟺
⟺
⇔
suppl.
alt.
cor.
∠′𝑠
⟺
int. ∠′ 𝑠 𝑜𝑓 ∆
⟺
ext. ∠ 𝑜𝑓 ∆
⟺
∠′ 𝑠 𝑜𝑝𝑝. 𝑒𝑞𝑢𝑎𝑙 𝑠𝑖𝑑𝑒𝑠
When two parallel lines are intercepted by a diagonal line, then:

the pair of corresponding angles are equal.

the sum of the co-interior angles on the same side of the
diagonal line = 180o.

the alternate angles are equal.
3.1
𝐿̂1 = 146° … (vert. opp. )
𝐿̂2 = 180° − 146° … (𝑠𝑢𝑝𝑝𝑙. ∠′ 𝑠)
= 34°
̂4 = 180° − 146° … (𝑐𝑜 − 𝑖𝑛𝑡. ∠′ 𝑠, 𝑃𝑄||𝑅𝑆)
𝑀
= 34°
̂2 = 𝐿̂2 = 34° … (cor. ∠′ 𝑠, 𝑃𝑄||𝑅𝑆)
𝑀
3.2
𝐹̂3 = 180° − 36° … (𝑠𝑢𝑝𝑝𝑙. ∠′ 𝑒)
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= 144°
𝐶̂1 = 36° … (cor. ∠′ 𝑠, 𝐴𝐷||𝐵𝐶)
̂
𝐴̂ = 180° − 𝐷
= 180° − 43°
= 137°
… (𝑐𝑜 − 𝑖𝑛𝑡. ∠′ 𝑠, 𝐴𝐵||𝐷𝐶)
𝐵̂ = 180° − 𝐴̂
… (𝑐𝑜 − 𝑖𝑛𝑡. ∠′ 𝑠, 𝐴𝐷||𝐵𝐶)
= 180° − 137°
= 43°
3.3
𝑇̂1 = 62°
… (𝑐𝑜 − 𝑖𝑛𝑡. ∠′ 𝑠, 𝑃𝑅||𝑉𝑍)
𝑆̂3 = 62°
… (cor. ∠′ 𝑠, 𝑇𝑈||𝑅𝑆)
𝑋̂4 = 𝑆̂3
= 62°
… (alt. ∠′ 𝑒, 𝑃𝑄||𝑅𝑆)
𝑃̂ = 𝑋̂4
= 62°
… (cor. ∠′ 𝑠, 𝑃𝑅||𝑉𝑍)
𝑅̂ = 180° − 𝑃̂
= 180° − 62°
= 118°
… (𝑐𝑜 − 𝑖𝑛𝑡. ∠′ 𝑠, 𝑃𝑄||𝑅𝑆)
𝑆̂4 = 180° − 𝑆̂3
= 180° − 62°
= 118°
… (𝑠𝑢𝑝𝑝𝑙. ∠′ 𝑠)
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o The sum of the interior angles of a triangle = 180o
o In a(n) isosceles triangle, the angles opposite the equal sides, are
eqaul.
o The exterior angle of a triangle is equal to the sum of opposite interior
angles.
4.1
𝑎 = 180° − (90° + 74°)
= 180° − 164°
= 16°
… (int. ∠′ 𝑠 𝑜𝑓 ∆)
4.2
𝑏 = 180° − (39° + 61°)
= 180° − 100°
= 80°
… (int. ∠′ 𝑠 𝑜𝑓 ∆)
4.3
𝑐 = 180° − (77° + 86°)
= 180° − (163°)
= 17°
… (int. ∠′ 𝑠 𝑜𝑓 ∆)
4.4
𝑥 = 55°
… (∠′ 𝑠 𝑜𝑝𝑝. 𝑒𝑞𝑢𝑎𝑙 𝑠𝑖𝑑𝑒𝑠)
4.5
𝑦 = 68°
… (∠′ 𝑠 𝑜𝑝𝑝. 𝑒𝑞𝑢𝑎𝑙 𝑠𝑖𝑑𝑒𝑠)
∴ 𝑧 = 180° − 2(68°)
= 180° − 136°
= 44°
… (int. ∠′ 𝑠 𝑜𝑓 ∆)
𝑎 = 𝐵̂
= 65°
… (𝑐𝑜𝑟. ∠′ 𝑠, 𝐷𝐸||𝐵𝐶)
𝑏 = 𝑎 = 65°
… (∠′ 𝑠 𝑜𝑝𝑝. 𝑒𝑞𝑢𝑎𝑙 𝑠𝑖𝑑𝑒𝑠)
∴ 𝑐 = 180° − 2(65°)
= 180° − 130°
= 50°
… (int. ∠′ 𝑠 𝑜𝑓 ∆)
𝑥 = 180° − 𝑏
= 180° − 65°
= 115°
… (𝑠𝑢𝑝𝑝𝑙. ∠′ 𝑠)
4.6
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4.7
̂ + 𝐸̂
𝑦=𝐷
= 48° + 53°
= 101°
… (ext. ∠ 𝑜𝑓 ∆)
4.8
̂
𝑧 = 𝐿̂ + 𝑁
= 86° + 52°
= 138°
… (ext. ∠ 𝑜𝑓 ∆)
4.9
̂𝑁
𝐿̂ + 𝑂̂ = 𝑂𝑀
∴ 𝑎 + 42° = 71°
𝑎 = 71° − 42°
= 29°
… (ext. ∠ 𝑜𝑓 ∆)
4.10 𝑎 = 72°
… (∠′ 𝑠 𝑜𝑝𝑝. 𝑒𝑞𝑢𝑎𝑙 𝑠𝑖𝑑𝑒𝑠)
𝑏 = 180° − 𝑎
= 180° − 72°
= 108°
… (𝑠𝑢𝑝𝑝𝑙. ∠′ 𝑠)
𝑥 = 180° − 2(72°)
= 180° − 144°
= 36°
… (int. ∠′ 𝑠 𝑜𝑓 ∆)
𝑦 = 90° − 72°
= 18°
… (compl. ∠′ 𝑒)
𝑧 = 180° − (90° + 𝑦)
= 180° − (90° + 18°)
= 180° − 108°
= 72°
… (int. ∠′ 𝑠 𝑜𝑓 ∆)
oooOooo
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