Our ncert solutions for Exercise 6.2 Class 9 maths 6. Lines and Angles - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Exercise 6.2 Class 9 maths 6. Lines and Angles - ncert solutions - Toppers Study

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## Exercise 6.2 Class 9 maths 6. Lines and Angles - ncert solutions - Toppers Study

CBSE board students who preparing for **class 9 ncert solutions maths and Mathematics** solved exercise **chapter 6. Lines and Angles** available and this helps in upcoming exams
2023-2024.

### You can Find Mathematics solution Class 9 Chapter 6. Lines and Angles

- All Chapter review quick revision notes for chapter 6. Lines and Angles Class 9
- NCERT Solutions And Textual questions Answers Class 9 Mathematics
- Extra NCERT Book questions Answers Class 9 Mathematics
- Importatnt key points with additional Assignment and questions bank solved.

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### Exercise 6.2 class 9 Mathematics Chapter 6. Lines and Angles

Sure! The following topics will be covered in this article

- Exercise 6.2 Class 9 Maths 6. Lines And Angles - Ncert Solutions - Toppers Study
- Class 9 Ncert Solutions
- Solution Chapter 6. Lines And Angles Class 9
- Solutions Class 9
- Chapter 6. Lines And Angles Exercise 6.2 Class 9

## 6. Lines and Angles

### | Exercise 6.2 |

## Exercise 6.2 Class 9 maths 6. Lines and Angles - ncert solutions - Toppers Study

**Chapter 6. Lines and Angles**

**Exercise 6.2**

**Q1. In Fig. 6.28, find the values of x and y and then show that AB || CD.**

**Solution: **

x + 50° = 180° (linear pair)

x = 180° - 50°

x = 130° ............. (1)

y = 130° (Vertically oposite angle) ....... (2)

From equation (1) and (2)

x = y = 130° (Alternate Interior Angle )

∴ AB || CD

(If alternate interior angle is equal then a pair of lines are parallel.)

**Q2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.**

**Solution:**

AB || CD (Given) ............. (1)

CD || EF (Given) ............. (2)

From equation (1) and (2)

AB || EF

∴ x = z .......... (3) (Alternate Interior Angle)

y : z = 3 : 7 (Given)

Let y = 3k, z = 7k

∴ x = z = 7k From equ. (3)

AB || CD (Given)

Now, x + y = 180° (Sum of interior adjacent angle is 180°)

⇒ 7k + 3k = 180°

⇒ 10k = 180°

⇒ k = 18°

x = 7k

= 7 × 18°

x = 126°

**Q3. In Fig. 6.30, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE.**

**Solution: **

∠GED = 126°

AB || CD and GE is a transversal. (Given)

∴ ∠AGE = ∠GED (Alternate Interior Angle)

∴ ∠AGE = 126°

EF ⊥ CD (Given)

∴ ∠FED = 90° ............ (1)

Now, ∠GED = 126°

Or, ∠GEF + ∠FED = 126°

⇒ ∠GEF + 90° = 126° From eqa (1)

⇒ ∠GEF = 126° - 90°

⇒ ∠GEF = 36°

∠AGE + ∠FGE = 180° (linear pair)

⇒ 126° + ∠FGE = 180°

⇒ ∠FGE = 180° - 126°

⇒ ∠FGE = 154°

∠AGE = 126°, ∠GEF = 36°and ∠FGE = 154°

**Q4. In Fig. 6.31, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS.**

**[Hint : Draw a line parallel to ST through point R.]**

**Solution: **

Construction: Draw PQ || XY from point R.

∠ PQR = 110° and ∠ RST = 130°

PQ || ST .............. (1) (Given)

PQ || XY .................(2) By construction.

From equa.(1) and (2) we get

ST || XY and SR is a transversal.

∴ ∠ RST + ∠ SRY = 180°

(Sum of interior Adjacent Angle)

Or, 130° + ∠ SRY = 180°

⇒ ∠ SRY = 180° - 130°

⇒ ∠ SRY = 50°

PQ || XY and QR is a transversal

∴ ∠ PQR = ∠ QRY (Alternate Interior Angle)

Or, ∠ PQR = ∠ QRS + ∠ SRY

⇒ 110° = ∠ QRS + 50°

⇒ ∠ QRS = 110° - 50°

⇒ ∠ QRS = 60°

**Q5. In Fig. 6.32, if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, find x and y.**

**Solution: **

Given that:

∠ APQ = 50° and ∠ PRD = 127°

AB || CD and PQ is a transversal.

∴ ∠ PQR = ∠ APQ (Alternate Interior Angle)

Or, x = 50°

Similarily,

∠ APR = ∠ PRD (Alternate Interior Angle)

50° + y = 127°

⇒ y = 127° - 50°

⇒ y = 77°

x = 50°, y = 77°

**Q6. In Fig. 6.33, 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: **

**Given:** PQ || RS and AB is incident ray, CD is reflected ray.

**To prove:** AB || CD

**Construction:**

Draw BM ⊥ PQ and CN ⊥ RS

Proof:

BM ⊥ PQ and CN ⊥ RS

∴ BM || CM and BC is a transverasal line

∴ ∠2 = ∠ 3 ............ (1) (Alternate Interior Angle)

While we know that

Angle of incidence = Angle of reflection, where BM and CN are normal.

∴ ∠1 = ∠ 2 .............. (2)

Similarily,

∴ ∠3 = ∠ 4 .............. (3)

Using (1) (2) and (3) we get

∠1 = ∠ 4 ................ (4)

Adding equa (1) and (4)

∠1 + ∠2 = ∠ 3 + ∠ 4

∠ABC = ∠ BCD (Alternate Interior Angle)

Therefore, AB || CD proved

##### Other Pages of this Chapter: 6. Lines and Angles

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