Question

A. 2. In the adjacent figure, find m PDQ 1) 45° 3) 25° 2) 360 4) 30°​

Answer 1

Answer:

1) 45°

Step-by-step explanation:

Since AP = AD, so, ∠APD = ∠ADP = x (let)

Also, QB=BD, so, ∠BQD = ∠BDQ = y (let)

In ΔABC, ∠A + ∠B + ∠C = 180°

∠A + ∠B = 90°...….(i)

Now, in ΔAPD,

∠A + ∠P + ∠D = 180°,

∠A + x + x = 180°

∠A + 2x = 180°...….(ii)

Now, in ΔBQD,

∠B + ∠Q + ∠D = 180°,

∠B + y + y = 180°,

∠B + 2y = 180°...…(iii)

Adding (ii) and (iii),

∠A + 2x + ∠B + 2y = 360°

2x + 2y = 360° - 90° ….from (i)

2(x + y) = 270°

x + y = 135°.

Since, ∠ADP + ∠PDQ + ∠QDB = 180°,

x + ∠PDQ + y = 180°,

∠PDQ = 180° - 135°

∠PDQ = 45°.

Hence, option 1 is the answer.