Question

D
P
5. ABCD is a rectangle. A
If ZBPC = 124°.
Calculate :
(i) ZBAP (ii) ZADP.
124°
B
f​

Answer 1

Answer:

Diagonals of rectangle are equal and bisect each other.

∠PBC = ∠PCB = x (say)

But ∠BPC + ∠PBC + ∠PCB = 180°

124° + x + x = 180°

2x = 180° – 124°

2x = 56°

=> x = 28°

∠PBC = 28°

But ∠PBC = ∠ADP [Alternate ∠s]

∠ADP = 28°

Again ∠APB = 180° – 124° = 56°

Also PA = PB

∠BAP = \frac { 1 }{ 2 } (180° – ∠APB)

= \frac { 1 }{ 2 } x (180°- 56°) = \frac { 1 }{ 2 } x 124° = 62°

Hence (i) ∠BAP = 62° (ii) ∠ADP =28°

Step-by-step explanation:

Hope it helps