Chapter 4 Lines and Angles Ex 4.2

Question 1.
In figure, find the values of x and y and then show that AB || CD.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 1
Solution:
∵ x + 50° = 180° (Linear pair)
⇒ x = 130°
∴ y = 130° (Vertically opposite angle)
Here, ∠x = ∠COD = 130°
These are corresponding angles for lines AB and CD.
Hence, AB || CD

Question 2.
In figure, if AB || CD, CD || EF and y: z = 3:7, find x.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 2
Solution:
Given
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 3
⇒ Let y = 3k, z = 7k
x = ∠CHG (Corresponding angles)…(i)
∠CHG = z (Alternate angles)…(ii)
From Eqs. (i) and (ii), we get
x = z …….(iii)
Now, x+y = 180°
(Internal angles on the same side of the transversal)
⇒ z+y = 180° [From Eq. (iii)]
7k + 3k = 180°
⇒ 10k = 180°
⇒ k = 18
∴ y = 3 x 18° = 54°
and z = 7x 18°= 126°
∴ x = z
Now, x + y = 180°
(Internal angles on the same side of the transversal)
⇒ z + y = 180° [From Eq. (iii)]
7k + 3k = 180°
⇒ 10k = 180°
⇒ k = 18
∴ y = 3 x 18° = 54°
and z = 7x 18°= 126°
x = z
⇒ x = 126°

Question 3.
In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 4
Solution:
∵ ∠AGE = ∠GED (Alternate interior angles)
But ∠GED = 126°
⇒ ∠AGE = 126° ….(i)
∴ ∠GEF + ∠FED= 126°
⇒ ∠GEF + 90° =126° (∵ EF ⊥ CD)
⇒ ∠GEF = 36°
Also, ∠AGE + ∠FGE = 180° (Linear pair axiom)
⇒ 126° + ∠FGE =180°
⇒ ∠FGE = 54°

Question 4.
In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 5
Solution:
Drawing a tine parallel to ST through R.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 6
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 7

Question 5.
In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 8
Solution:
We have, AB || CD
⇒ ∠APQ = ∠PQR (Alternate interior angles)
⇒ 50° = x
⇒ x = 50°
Now, ∠PQR + ∠QPR = 127°
(Exterior angle is equal to sum of interior opposite angles of a triangle)
⇒ 50°+ ∠QPR = 127°
⇒ y = 77°.

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.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 9
Solution:
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 10