**Download NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 – Lines and Angles. This Exercise contains 6 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 9 for Maths NCERT solutions for Chapter 6 or other Chapters, you can click the link at the end of this Note.**

**Q.1 In given figure, find the values of x and y and then show that AB || CD.**

**Solution:**

x + 50˚ = 180˚ [ Linear Pair of Angles ]

x = 180˚ – 50˚

x = 130˚

y = 130˚ [ Vertically Opposite Angles ]

**x = y = 130****˚
**x and y are Alternate interior angle for Line AB and CD

**AB || CD**

**Q.2 In given figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.**

**Given :**

AB || CD

CD || EF

i.e, AB || CD || EF

y : z = 3 : 7

**To Find : **x = ?

**Solution:**

x = z [ Alternate Interior Angles ]

y : z = 3 : 7

Let, y = 3a and z = 7a

x + y = 180˚ [ Co interior Angles ]

z + y = 180˚ [ x = z from above]

7a + 3a = 180˚

10 a = 180˚

a = 180˚/10

a = 18˚

x = z = 7a = 7 x 18˚ = 126˚

**x =126****˚**

**Q.3 In given figure, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE**

**Given :**

AB || CD

EF ⊥ CD

∠ GED = 126°

**To Find :**

∠ AGE = ?

∠ GEF = ?

∠ FGE = ?

**Solution:**

**∠ AGE = ∠ GED = 126° **[ Alternate Interior Angles ]

∠ GED = 126° [ Given ]

∠ GEF + ∠ DEF = 126°

∠ GEF + 90° = 126°

∠ GEF = 126° – 90°

**∠ GEF = 36°
**∠ AGE + ∠ FGE = 180° [ Linear Pair of Angles ]

126° + ∠ FGE = 180°

∠ FGE = 180° – 126°

**∠ FGE = 54°**

**Q. 4 In given figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS.**

**Given :**

PQ || ST

∠ PQR = 110°

∠ RST = 130°

**To Find : **∠ QRS = ?

**Solution:**

Let us draw XY || ST passing through point R

∠ PQR + ∠ XRQ = 180° [ Co interior Angles ]

110° + ∠ XRQ = 180°

∠ XRQ = 180° – 110° = 70°

∠ XRQ + ∠ QRS = 130° [ Co interior Angles ]

70° + ∠ QRS = 130°

QRS = 130° – 70°

**QRS = 60°**

**Q.5 In given figure, if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, find x and y.**

**Given :**

AB || CD

∠ APQ = 50°

∠ PRD = 127°

**To Find :**

x = ?

y = ?

**Solution:**

∠ APQ = ∠ PQR = 50° [ Alternate interior Angles]

**x = 50°
**∠APR = ∠ PRD = 127° [ Alternate interior Angles]

∠APQ + ∠ QPR = 127°

50° + ∠ QPR = 127°

∠ QPR = 127° – 50°

**∠ QPR = 77°**

**Q.6 In given 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**

**Solution:**

Let us draw, BM ⊥ PQ, CN ⊥ RS

PQ || RS [ Given ]

Therefore, BM || CN

∠ 2 = ∠ 3 [ Alternate interior Angles]

∠ 1 = ∠ 2, ∠ 3 = ∠ 4 [By Laws of Reflection]

∠1 = ∠ 2 = ∠3 = ∠ 4

Also, ∠1 + ∠ 2 = ∠3 + ∠ 4

∠ ABC = ∠ CDB

∠ ABC & ∠ CDB are Alternate interior angles

**AB || CD**

**NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 – Lines and Angles, has been designed by the NCERT to test the knowledge of the student on the following topics:-**

- Parallel Lines and a Transversal
- Lines Parallel to the Same Line

**The next Exercise for** **NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.3 – Lines and Angles can be accessed by clicking here.**

**Download NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 – Lines and Angles**