Class 8: Chapter 28 - Exercise 28A
1. In the adjourning figure, name:
i.
The point are 𝑀, 𝑁, 𝑃, 𝑄 , 𝑋, 𝑌
̅̅̅̅̅ MP,
̅̅̅̅̅ ̅̅̅̅
̅̅̅̅, MN
̅̅̅̅̅, PQ
̅̅̅̅
ii.
Five line segments are XM,
YN, NQ
⃗⃗⃗⃗
iii.
Four Rays are ⃗⃗⃗⃗⃗
PB, ⃗⃗⃗⃗⃗
QD, ⃗⃗⃗⃗⃗
XA, YC
iv.
v.
⃡⃗⃗⃗⃗ , CD
⃡⃗⃗⃗ , EF
⃡⃗⃗⃗ , HG
⃡⃗⃗⃗⃗
Four lines are AB
Four Collinear points are
𝐴, 𝑋, 𝑀, 𝑃 𝑜𝑟 𝐶, 𝑌, 𝑁, 𝑄 𝑜𝑟 𝑋, 𝑀, 𝑃, 𝐵
2. In the adjoining figure, name :
i.
Two points of intersecting lines are
⃡⃗⃗⃗⃗ , 𝐺𝐻
⃡⃗⃗⃗⃗ 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑛𝑔 𝑎𝑡 𝑅
𝐸𝐹
ii.
iii.
iv.
⃡⃗⃗⃗⃗
𝐶𝐷 , ⃡⃗⃗⃗⃗
𝐺𝐻 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑛𝑔 𝑎𝑡 𝑄.
Three concurrent lines and their point of concurrence.
⃡⃗⃗⃗⃗
𝐴𝐵 , ⃡⃗⃗⃗⃗
𝐸𝐹 , ⃡⃗⃗⃗⃗
𝐺𝐻 and point of concurrence is R
⃗⃗⃗⃗⃗⃗ , 𝑃𝐵
⃗⃗⃗⃗⃗ , 𝑃𝐷
⃗⃗⃗⃗⃗ some other rays
Three rays are, 𝑄𝐻
⃗⃗⃗⃗⃗⃗ , etc.
are⃗⃗⃗⃗⃗⃗
𝑅𝐴, ⃗⃗⃗⃗⃗
𝑅𝐸 , 𝑅𝐺
Two line segments are ̅̅̅̅
𝑄𝑅 , ̅̅̅̅
𝑃𝑄 , ̅̅̅̅
𝑅𝑃
3. State whether the following statements are true or false :
i.
A ray has no end Point: False
ii.
A line AB is the same as line BA: True
iii.
A ray AB is the same as BA: False
iv.
A line has a definite length: False
v.
Two planes always meet in a line: True
vi.
A plane has length and breadth but no thickness: True
vii.
Two distinct points always determine a unique line: True
viii.
Two lines may intersect in two points: False
ix.
Two intersecting lines cannot be both parallel to the same line: True
4. 𝑇𝑤𝑜 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑎𝑛𝑔𝑙𝑒𝑠 𝑜𝑛 𝑎 𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑙𝑖𝑛𝑒 𝑎𝑟𝑒 𝑥° 𝑎𝑛𝑑 (2𝑥 − 21) ° 𝐹𝑖𝑛𝑑
(𝑖) 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 𝑖𝑖) 𝑡ℎ𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑎𝑛𝑔𝑙𝑒
i.
∠𝐴𝑂𝐵 + ∠𝐶𝑂𝐵 = 180°
2𝑥 − 21 + 𝑥 = 180°
3𝑥 = 201
𝑥 = 67°
ii.
Hence ∠𝐶𝑂𝐵 = 67° 𝑎𝑛𝑑 ∠𝐴𝑂𝐵 = 113°
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5. 𝑇𝑤𝑜 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑎𝑛𝑔𝑙𝑒𝑠 𝑜𝑛 𝑎 𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑙𝑖𝑛𝑒 𝑎𝑟𝑒 (3𝑥 − 2)° 𝑎𝑛𝑑 4(𝑥 + 7)° − 𝐹𝑖𝑛𝑑 ∶
(𝑖) 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 𝑖𝑖) 𝑡ℎ𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑎𝑛𝑔𝑙𝑒
i.
3𝑥 − 2 + 4(𝑥 + 7) = 180°
3𝑥 + 4𝑥 + 26 = 180°
7𝑥 = 154
𝑜𝑟 𝑥 = 22
ii.
The measure of angles
Angle1 = 3 × 22 − 2 = 64°
Angle 2 = 4 × (22 + 7) = 116°
6. Two adjacent angles on a straight line are in the ratio 3 : 2. Find the measure of each angle:
The ratio of angles = 3:2
Therefore the angles are 3𝑥 𝑎𝑛𝑑 2𝑥
3𝑥 + 2𝑥 = 180°
5𝑥 = 180°
𝑥 = 36°
The two angles are 108° and 72°
7. 𝐼𝑛 𝑡ℎ𝑒 𝑎𝑑𝑗𝑜𝑖𝑛𝑖𝑛𝑔 𝑓𝑖𝑔𝑢𝑟𝑒, 𝐴𝑂𝐵 𝑖𝑠 𝑎 𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑙𝑖𝑛𝑒. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑥.
𝐻𝑒𝑛𝑐𝑒 , 𝑓𝑖𝑛𝑑, ∠𝐴𝑂𝐶 𝑎𝑛𝑑 ∠𝐵𝑂𝐷
∠𝐴𝑂𝐶 + ∠𝐶𝑂𝐷 + ∠𝐷𝑂𝐵 = 180°
3𝑥 − 5 + 55 + 𝑥 + 20 = 180°
4𝑥 = 110°
⟹ 𝑥 = 27.5°
Therefore ∠𝐴𝑂𝐶 = 3 × 22 − 5 = 77.5°
And ∠𝐵𝑂𝐷 = 22 + 20 = 47.5°
8. 𝐼𝑛 𝑡ℎ𝑒 𝑎𝑑𝑗𝑜𝑖𝑛𝑖𝑛𝑔 𝑓𝑖𝑔𝑢𝑟𝑒, 𝐴𝑂𝐵 𝑖𝑠 𝑎 𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑙𝑖𝑛𝑒. 𝐼𝑓 𝑥 ∶ 𝑦 ∶ 𝑧 = 6 ∶ 5 ∶ 4,
𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑥, 𝑦 𝑎𝑛𝑑 𝑧.
𝑥: 𝑦: 𝑧 = 6: 5: 4
Therefore
6𝑎 + 5𝑎 + 4𝑎 = 180°
Or 𝑎 = 12 therefore 𝑥 = 72°, 𝑦 = 60° & 𝑧 = 48°
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9. In the adjoining figure, what value of 𝑥 make AOB a straight line?
For ⃡⃗⃗⃗⃗
𝐴𝐵 to be a straight line
3𝑥 + 5 + 2𝑥 − 25 = 180°
5𝑥 − 20 = 180°
5𝑥 = 200°
𝑥 = 40°
∠𝐴𝑂𝐶 = 3 × 40 + 5 = 125°
∠𝐵𝑂𝐶 = 2 × 40 − 25 = 55°
10. 𝐼𝑛 𝑡ℎ𝑒 𝑎𝑑𝑗𝑜𝑖𝑛𝑖𝑛𝑔 𝑓𝑖𝑔𝑢𝑟𝑒, 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥.
∠𝐷𝑂𝐴 + ∠𝐴𝑂𝐵 + ∠𝐵𝑂𝐶 + ∠𝐶𝑂𝐷 = 360°
𝑥 + 65 + 90 + 120 = 360
𝑥 = 360° − 275°
𝑥 = 85°
11. 𝐼𝑛 𝑒𝑎𝑐ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑓𝑖𝑔𝑢𝑟𝑒𝑠, 𝑡𝑤𝑜 𝑙𝑖𝑛𝑒𝑠 𝐴𝐵 𝑎𝑛𝑑 𝐶𝐷 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡 𝑎𝑡 𝑎 𝑝𝑜𝑖𝑛𝑡 𝑂.
𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥, 𝑦 𝑎𝑛𝑑 𝑧
i.
∠𝐴𝑂𝐷 = ∠𝐶𝑂𝐵 (vertically opposite angle)
∴ 𝑦 = 75°
∠𝐴𝑂𝐷 + ∠𝐷𝑂𝐵 = 180°
75 + 𝑧 = 180°
Or 𝑧 = 105°
∠𝐴𝑂𝐷 + ∠𝐴𝑂𝐶 = 180°
75 + 𝑥 = 180
𝑥 = 105°
We could have also used
∠𝐷𝑂𝐵 = ∠𝐴𝑂𝐶
𝑥 = 105°
ii.
∠𝐶𝑂𝐵 = ∠𝐴𝑂𝐵 (vertically Opposite angles)
⇒ 𝑦 = 125°
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125 + 𝑧 = 180° (Angles on a straight line are supplementary)
𝑧 = 55°
∠𝐵𝑂𝐷 = ∠𝐶𝑂𝐴 (Vertically opposite angles)
𝑥 = 55°
iii.
∠𝐴𝑂𝐶 = ∠𝐷𝑂𝐵 (Vertically opposite angles)
⟹ 𝑦 = 30°
∠𝐶𝑂𝐵 + ∠𝐵𝑂𝐷 = 180 (Angles on a straight line)
⟹ 𝑧 = 150°
∠COB = ∠AOD (Vertically opposite angles)
⟹ 𝑥 = 150°
iv.
∠𝐴𝑂𝐶 + ∠𝐶𝑂𝐵 + ∠𝐵𝑂𝐷 + ∠𝐷𝑂𝐴 = 360°
3𝑥 − 20 + 𝑥 + 𝑧 + 𝑦 = 360° … … … … … … … (𝑖)
𝑥 = 𝑦 (Vertically opposite angles)
3𝑥 − 20 = 𝑧 (Vertically opposite angles)
𝑧 + 𝑦 = 180° … … … … … … … (𝑖𝑖)
𝑥 + 𝑧 = 180° … … … … … … … (𝑖𝑖𝑖)
Substituting (ii) in (i)
3𝑥 − 20 + 𝑥 + 180 = 360
4𝑥 = 180 + 20
𝑥 = 50°
∠𝐴𝑂𝐷 = 50° & ∠𝐴𝑂𝐶 = 130°
From iii) 𝑧 = 180 − 50 = 130°
12. Prove that the bisectors of two adjacent supplementary angles include a right angle.
Let ∠𝐴𝑂𝐶 = 180 − 𝑥
∠𝐵𝑂𝐶 = 𝑥
Bisector of ∠𝐵𝑂𝐶 =
𝑥
2
1
= ∠𝐷𝑂𝐶
𝑥
Bisector of ∠𝐴𝑂𝐶 = 2 (180 − 𝑥 ) = 90 − 2 = ∠𝐶𝑂𝐸
Therefore
𝑥
𝑥
2
2
∠𝐶𝑂𝐸 + ∠𝐷𝑂𝐶 = 90 − + = 90°
Hence
∠𝐷𝑂𝐸 = 90° = 𝑅𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒
13. Find the measure of an angle which is (l) equal to its complement (ii) equal to its supplement.
i.
If the ∠1 = 𝑥, 𝑖𝑡𝑠 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 ∠2 = 90 − 𝑥
If 𝑥 = 90 − 𝑥
⟹ 𝑥 = 45°
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ii.
𝐼𝑓 𝑡ℎ𝑒 ∠1 = 𝑥, 𝑖𝑡𝑠 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡 = 180 − 𝑥
𝑥 = 180 − 𝑥
⟹ 𝑥 = 90°
14. Find the angle which is 84° more than its complement.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 = 𝑥
𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 = 90 − 𝑥
𝐺𝑖𝑣𝑒𝑛 𝑥 = (90 − 𝑥) + 34
2𝑥 = 124
𝑜𝑟 𝑥 = 62°
15. Find the angle which is 16° less than its complement.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 = 𝑥
𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 = 90 − 𝑥
Given
𝑥 + 16 = 90 − 𝑥
2𝑥 = 74
𝑥 = 37°
16. Find the angle which is 26° more than its supplement.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 = 𝑥
𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡 = 180 − 𝑥
𝐺𝑖𝑣𝑒𝑛
𝑥 = (180 − 𝑥) + 26
2𝑥 = 206
𝑂𝑟 𝑥 = 103°
17. Find the angle which is 32° less than its supplement.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 = 𝑥
𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡 = 180 − 𝑥
𝐺𝑖𝑣𝑒𝑛,
𝑥 + 32 = 180 − 𝑥
2𝑥 = 148°
𝑜𝑟 𝑥 = 74°
18. Find the angle which is four times its complement.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 = 𝑥
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𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 = 90 − 𝑥
𝐺𝑖𝑣𝑒𝑛
𝑥 = 4(90 − 𝑥 )
5𝑥 = 360
⟹ 𝑥 = 72°
19. Find the angle which is five times its supplement.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 = 𝑥
𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 = 180 − 𝑥
𝐺𝑖𝑣𝑒𝑛
𝑥 = 5(180 − 𝑥 )
6𝑥 = 5 × 180
𝑂𝑟 𝑥 = 150°
20. 𝐹ind the angle whose supplement is four times its complement.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 = 𝑥
𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 = 90 − 𝑥
𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡 = 180 − 𝑥
𝐺𝑖𝑣𝑒𝑛,
180 − 𝑥 = 4(90 − 𝑥 )
3𝑥 = 180
𝑥 = 60°
21. 𝐹ind the angle whose complement is one third of its supplement.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 = 𝑥
𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 = 90 − 𝑥
𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡 = 180 − 𝑥
𝐺𝑖𝑣𝑒𝑛
1
90 − 𝑥 = (180 − 𝑥 )
3
270 − 3𝑥 = 180 − 𝑥
2𝑥 = 90
⟹ 𝑥 = 45°
22. Two complementary angles are in the ratio 7: 11. Find the angles.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 = 𝑥
𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 = 90 − 𝑥
Given
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𝑥
7
=
90 − 𝑥 11
or 18𝑥 = 630 𝑜𝑟 𝑥 = 35°
The complement = 35°
Hence the angles are 35° 𝑎𝑛𝑑 55°
23. Two supplementary angles are in the ratio 7 ∶ 8. Find the angles.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 = 𝑥
𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡 = 180 − 𝑥
𝐺𝑖𝑣𝑒𝑛
𝑥
7
=
180 − 𝑥 8
8𝑥 = 1260 − 7𝑥
𝑥 = 84° 𝑠𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡 = 96°
𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 84° 𝑎𝑛𝑑 96°
24. Find the measure of an angle, if seven times its complement is 10° less than three
times its supplement.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 = 𝑥
𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 = 90 − 𝑥
𝑆𝑢𝑝𝑝𝑙𝑒𝑚𝑒𝑛𝑡 = 180 − 𝑥
Given
7(90 − 𝑥 ) + 10 = 3(180 − 𝑥 )
630 − 7𝑥 + 10 = 540 − 3𝑥
4𝑥 = 100
𝑥 = 25°
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