Math chapter 6 exercise 6.2 Question 3 and question 4 Solutions

Question 3. 

 In the figure, AB || CD, EF ⊥ CD And ∠ GED = 126°, Find, 

∠ AGE, ∠ GEF And ∠ FGE.

Solution. AB  CD and GE is transversal.

∴ ∠AGE = ∠GED [Alternate angles]

⇒ AGE = 126°.

Further, ∠GED = ∠GEF + ∠FED

⇒ 126° = ∠GEF + 90° [.

°. EF ⊥ CD]

⇒ ∠GEF = 126° – 90°  =  036°

Again, AB  CD and GE is transversal.

∴ ∠FGE + ∠GED = 180° [Sum of interior angles on

the same side of transversal is 180°.]

⇒ ∠FGE + 126° = 180°

Then, ⇒ ∠FGE = 180° – 126° = 054°

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Question 4.

 In Figure, If PQ || ST, ∠ PQR = 110° And ∠ RST = 130°,

Find ∠ QRS.

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

Solution. Construction: Through R draw a line XRY parallel to

PQ.

Proof: PQ  XRY and QR is transversal.

∴ ∠PQR = ∠QRY = 110° ...(i) [Alternate angles]

Also, PQ  ST [Given]

and PQ  RY [Construction]

∴ ST  XRY and SR is transversal.

∴ ∠TSR + ∠SRY = 180°

[Sum of interior angles on the same side of the transversal]

⇒ 130° + ∠SRY = 180°

⇒ ∠SRY = 180° – 130° = 050° ....(ii)

Also, ∠QRY = ∠QRS + ∠SRY

⇒ 110° = ∠QRS + 50° [From (i) and (ii)]

⇒ ∠QRS = 110° – 50° = 60°.