ICSE-VIII-Mathematics
17: Special Types of Quadrilaterals Class 8 Maths
- #5ABCD is a rectangle, if ∠BPC = 124°
Calculate: (i) ∠BAP (ii) ∠ADP
Ans : 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 = 1/2 (180° - ∠APB)
= 1/2 ×(180°- 56°)= 1/2 ×124° = 62°
Hence (i) ∠BAP = 62° (ii) ∠ADP = 28°