Mathematics Questions and Answers – Parallel Lines and Transversal – 4

This set of Mathematics Exam Questions and Answers for Class 9 focuses on “Parallel Lines and Transversal – 4”.

1. Find the value of k in line l || m?

a) 90°
b) 83°
c) 132°
d) 41°
View Answer

Answer: b
Explanation: line l || m

From figure, ∠1 = ∠(2x + 15)  (Vertically Opposite Angles)
∠1 + ∠(3x – 40) = 180°  (Sum of Interior Angles on the same side of transversal is 180°, l || m)
⇒ 2x + 15 + 3x – 40 = 180°
⇒ 5x = 180° + 25°
⇒ 5x = 205°
⇒ x = 41°
Now, ∠k = ∠(3x – 40) = 75°  (Alternate Interior Angles)
⇒ k = 3 x 41° – 40°
⇒ k = 83°

2. Find the value of ∠1 and ∠2 if line p || q and ∠1 = 5x – 26°, ∠2 = 2x + 10° .

a) 90°, 90°
b) 45°, 90°
c) 41°, 41°
d) 44°, 34°
View Answer

Answer: d
Explanation:

∠1 = ∠3  (Vertically Opposite Angles)
⇒ ∠3 = 5x – 26°
Also, ∠2 = ∠3   (Corresponding Angles, p || q)
⇒ 2x + 10° = 5x – 26°
⇒ 3x = 36°
⇒ x = 12°
∠1 = 5x – 26° = 5 x 12 – 26 = 44°
and ∠2 = 2x + 10° = 2 x 12 + 10 = 34°

3. What will be the values of x and y if line AB || CD?

a) 60°, 70°
b) 75°, 75°
c) 70°, 60°
d) 60°, 120°
View Answer

Answer: a
Explanation:

Since line AB || CD , ∠(2x – y) = 60° (Vertically Opposite Angles)
⇒ 2x – y = 60°  ——- (i)
Also, ∠(x + y) + 50° = 180°  (Sum of Interior Angles on the same side of transversal is 180°)
⇒ x + y = 130°   ——- (ii)
Adding equation (i) and (ii),
2x – y + x + y = 60° + 130°
⇒ 3x = 180°
⇒ x = 60°
Substituting value of x in equation (ii),
60° + y = 130°
⇒ y = 70°

4. Find the value of ∠CAB if PQ || RS and AB ⊥ RS.

a) 60°
b) 75°
c) 45°
d) 80°
View Answer

Answer: c
Explanation:

AB ⊥ RS ⇒ ∠ABC = 90°
Since line PQ || RS and AB is transversal, ∠QAB = ∠ABC = 90°   (Alternate Interior Angles)
Since line PQ || RS and AC is transversal, ∠RCA = ∠QAC = 135°   (Alternate Interior Angles)
Now, ∠CAB = ∠QAC – ∠QAB
⇒ ∠CAB = 135° – 90°
⇒ ∠CAB = 45°

5. Find the value of y if AB || CD.

a) 60°
b) 80°
c) 132°
d) 280°
View Answer

Answer: d
Explanation: Draw a line POQ parallel to AB and CD

Since line PQ || AB and OB is transversal, ∠ABO = ∠BOQ = 120°  (Alternate Interior Angles)
Since line PQ || CD and OC is transversal,
∠OCD + ∠COQ = 180°  (Sum of Interior Angles on the same side of transversal is 180°)
⇒ ∠COQ = 180° – 140°
⇒ ∠COQ = 40°
Now, ∠BOC = ∠BOQ – ∠COQ
⇒ ∠BOC = 120° – 40°
⇒ ∠BOC = 80°
Now, y = 360° – ∠BOC   (Sum of Angles around a point is 360°)
⇒ y = 360° – 80°
⇒ y = 280°

6. Find the value of x if BF || DE, AB ⊥ BF and ∠BAC : ∠ACB = 2 : 3.

a) 70°
b) 126°
c) 110°
d) 80°
View Answer

7. Find the value of y if PR || AC.

a) 10°
b) 25°
c) 265°
d) 120°
View Answer

Answer: c
Explanation: Draw Line XY parallel to AC and PR

Line PR || XY ⇒ ∠OQR = ∠QOX = 35°  (Alternate Interior Angles)
Line XY || AC ⇒ ∠XOA + ∠ABO = 180°  (Sum of Interior Angles on the same side of transversal)
⇒ ∠XOA + 120° = 180°
⇒ ∠XOA = 60°
Now, ∠QOX + ∠XOA + ∠QOB = 360° (Sum of Angles around a point is 360°)
⇒ 35° + 60° + y = 360°
⇒ y = 265°.

8. Find the value of x if PQ || RS, CD || RB, ∠1 : ∠2 = 3 : 4 and ∠2 = 64°.

a) 43°
b) 54°
c) 68°
d) 72°
View Answer

9. Find the value of k if line l || m || n.

a) 60°
b) 54°
c) 75°
d) 120°
View Answer

Answer: a
Explanation:

From figure, ∠1 = 120°  (Vertically Opposite Angles)
Also, ∠1 + ∠2 = 180°  (Sum of Interior Angles on the same side of transversal is 180°, Line l || m )
⇒ ∠2 + 120° = 180°
⇒ ∠2 = 60°
Now, ∠k = ∠2  (Corresponding Angles, Line m || n)
⇒ k = 60°

10. What is the value of (a + b) if Line l || m?

a) c
b) 90°
c) b + d
d) d
View Answer

Answer: c
Explanation:

From figure, ∠b = ∠d  (Corresponding Angles, Line l || m and p is transversal)
∠1 = ∠a  (Vertically Opposite Angles)
Also, ∠1 = ∠d   (Corresponding Angles, Line l || m and p is transversal)
Hence, a + b = c + d.

11. Find the value of x and y if line AB || CD.

a) 60°, 75°
b) 90°, 90°
c) 36°, 63°
d) 36°, 144°
View Answer

Sanfoundry Global Education & Learning Series – Mathematics – Class 9.

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