Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2

Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2 Textbook Questions and Answers.

BSEB Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2

Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2

Question 1.
In figure, find the values of x and y and then show that AB || CD.
Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2 1
Solution:
Since AB || CD and transversal PQ intersects them at R and S respectively.
Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2 2
∴ ∠ARS = ∠RSD [Alternate angles]
⇒ x = y
But ∠RSD = ∠CSQ
[ Vertically opp. angles]
⇒ y = 130° [∵∠CSQ = 130°]
Hence, x = y = 130°.

Question 2.
In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.
Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2 3
Solution:
Since CD || EF and transversal PQ intersects them at S and T respectively.
∠CST = ∠STF [Alternate Angles]
⇒ 180° – y = z
[∵ ∠y + ∠CST = 180° being linear pair]
⇒ y + z = 180°
Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2 4
Given y : z = 3 : 7.
So, the sum of ratios = 3 + 7 = 10
∴ y = \(\frac { 3 }{ 10 }\) x 180°
= 3 x 18° = 54°
and, z = \(\frac { 7 }{ 10 }\) x 180°
= 7 x 18° = 126°
Since AB || CD and transversal PQ intersects them at R and S respectively.
∴ ∠ARS + ∠RSC = 180° [Consecutive interior angles are supplementary]
⇒ x + y = 180°.
⇒ x = 180° – y
= 180° – 54° = 126° [∵ y = 54°]
Hence x = 126°.

Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2

Question 3.
In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2 5
Solution:
Since AB || CD and transversal GE cuts them at G and E respectively.
∠AGE = ∠GED [Alternate angles]
∠AGE = 126° [∵ ∠GED = 1260 (given)]
∠GEF = ∠GED – ∠FED = 126° – 90° = 36°
∠FGE = ∠GEC [Alternate angles]
∠FGE = 90° – ∠GEF
= 90° – 36° = 54°
Hence, ∠AGE = 126°, ∠GEF = 36° and ∠FGE = 54°.

Question 4.
In figure, if PQ || ST, ∠PQR = 110° and ∠RST =130°, find ∠QRS.
 Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2 6
Solution:
Produce PQ to intersect SR in a point M.
Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2 7
Now, PM || ST and transversal SM intersects them at M and R respectively.
∴ ∠SMQ = ∠TSM [Alternate angles]
⇒ ∠SMQ = 130°
⇒ ∠QMR = 180° – 130° = 50° [∵ ∠SMQ + ∠QMR = 180° linear pairs]
Since, ray RQ stands at Q on PM.
∴ ∠PQR + ∠RQM = 180°
⇒ 110° + ∠RQM = 180°
⇒ ∠RQM = 70°
∴ ∠QRS = 180° – (70° + 50°) = 60°
[∵ Sum of the angles of a triangle is 180°]

Question 5.
In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2 8
Solution:
∵ AB || CD and transversal PQ intersects them at P and Q respectively.
∴ ∠PQR = ∠APQ [Alternate angles]
⇒ x = 50° [∵ ∠APQ = 50° (given)]
∵ AB || CD and transversal PR intersects them at P and R respectively.
∴ ∠APR = ∠PRD [Alternate angles]
⇒ ∠APQ + ∠QPR = 127° [∵ ∠PRD = 127°]
⇒ 50° + y = 127° [∵∠APQ = 50°]
⇒ y = 127° – 50° = 77°
Hence, x = 50° and y = 11°.

Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2

Question 6.
In figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.
Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2 9
Solution:
Two-plane mirrors PQ and RS are placed parallel to each other i.e., PQ || RS. An incident ray AB after reflections takes the path BC and CD.
Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2 10
BN and CM are the normals to the plane mirrors PQ and RS respective.
Since BN ⊥ PQ, CM ⊥ RS and PQ || RS
∴ BN ⊥ RS ⇒ BN || CM
Thus BN and CM are two parallel lines and a transversal BC cuts them at B and C respectively.
∴ ∠2 = ∠3 [Alternate interior angles]
But, ∠1 = ∠2 and ∠3 = ∠4 [By laws of reflection]
∠1 + ∠2 = ∠2 + ∠2 and ∠3 + ∠4= ∠3 + ∠3
⇒ ∠1 + ∠2 = 2(∠2) and ∠3 + ∠4 = 2(∠3)
⇒ ∠1 + ∠2 = ∠3 + ∠4 [∵ ∠2 = ∠3 ⇒ 2(∠2) = 2(∠3)]
⇒ ∠ABC = ∠BCD
Thus, lines AB and CD are intersected by transversal BC such that
∠ABC = ∠BCD.
i.e., alternate interior angles are equal.
Therefore, AB || CD.