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

Chapter 6. Lines and Angles

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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