# Mathematics Questions and Answers – Parallel Lines and Transversal – 3

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This set of Mathematics Online Test for Class 9 focuses on “Parallel Lines and Transversal – 3”.

1. Which of the following pairs are parallel? a) AB and CD
b) AB and PQ
c) EF and CD
d) EF and PQ

Explanation: ∠AZY + ∠ZYC = 62° + 110° = 162°
Since AB and CD are two lines with PQ as transversal, ∠AZY and ∠ZYC are interior angles on the same side of transversal and their sum is not equal to 180°. So, Line AB is not parallel to CD.
Now, ∠EXY = ∠XYZ = 110°
EF and PQ are two lines with CD as transversal, ∠EXY and ∠XYZ are alternate interior angles and are equal. So, Line EF is parallel to PQ.

2. Find the value of ∠x if line l || m and line n || p. a) 90°
b) 75°
c) 110°
d) 120°

Explanation: ∠1 = 75°  (Vertically Opposite Angles)
Line n || p ⇒ ∠1 = ∠2 = 75°  (Alternate Interior Angles)
Also, Line l || m ⇒ ∠2 = x°  (Corresponding Angles)
⇒ ∠x = 75°.

3. Find the value of k if ∠1 = 150 – k and ∠2 = 60 + k and line l || m. a) 80°
b) 75°
c) 45°
d) 150°

Explanation: Since line l || m, ∠1 = ∠2 (Corresponding Angles)
⇒ 150 – k = 60 + k
⇒ 2k = 150 – 60
⇒ k = 45°.

4. Find the value of y if line l || m. a) 60°
b) 75°
c) 55°
d) 80°

Explanation: ∠1 + ∠(60 + y) = 180°  (Linear Pair)
⇒ ∠1 = 180° – 60° – y
⇒ ∠1 = 120° – y
Since line l || m, ∠1 = ∠y = 75°  (Alternate Interior Angles)
⇒ 120° – y = y
⇒ 2y = 120°
⇒ y = 60°.

5. Find the value of x, y and z if y : z = 2 : 3 and AB || CD. a) 40°, 40°, 40°
b) 56°, 70°, 62°
c) 40°, 56°, 84°
d) 32°, 50°, 50°

Explanation: Since line AB || CD, ∠APQ = ∠APC   (Alternate Interior Angles)
⇒ ∠x = 40°
Now, y : z = 2 : 3 ⇒ y = 2k, z = 3k  ————— (1)
Also, ∠AQR + ∠QRC = 180°  (Sum of Interior Angles on the same side of transversal is 180°)
⇒ 40° + y + z = 180°
⇒ 40° + 2k + 3k = 180°  (From equation 1)
⇒ 5k = 140°
⇒ k = 28°
So, y = 2k = 2 x 28 = 56°
and z = 3k = 3 x 28 = 84°

6. Find the value of ∠PQR if AB || CD || EF. a) 70°
b) 120°
c) 110°
d) 80°

Explanation: Line AB || CD || EF
Now, ∠CRS = ∠RSF = 20°   (Alternate Interior Angles, CD || EF)
Also, 120° + ∠QRC = 180°  (Sum of Interior Angles on the same side of transversal is 180°, AB || CD)
⇒ ∠QRC = 180° – 120° = 60°
Now, ∠QRS = ∠CRS + ∠QRC
⇒ ∠QRS = 20° + 60°
⇒ ∠QRS = 80°.

7. Find the value of ∠QRT if PQ || SR and SR || UT. a) 10°
b) 20°
c) 110°
d) 120°

Explanation: Extend Line SR till A Line PQ|| RS ⇒ ∠PQR = ∠QRS = 50°  (Alternate Interior Angles)
Line PQ|| RS ⇒ ∠UTR = ∠TRA = 140°  (Alternate Interior Angles)
Also, ∠QRS + ∠QRA = 180°  (Sum of Interior Angles on the same side of transversal is 180°)
⇒ ∠QRA = 180° – 50° = 130°
Now, ∠QRT = ∠TRA – ∠QRA
⇒ ∠QRT = 140° – 130°
⇒ ∠QRT = 10°. 