Question

In the given figure, angle ACP = ∠BDP = 90°, AC = 12 m, BD = 9 m and PA= PB = 15 m. Find:
(i) CP
(ii) PD
(iii) CD

Answer 1

Given : AC = 12 m
BD = 9 m
PA = PB= 15 m

(i) In right angle triangle ACP
(AP)² = (AC)² + (CP)²
15² = 12² + CP²
225 = 144 + CP²
225 – 144 = CP²
81 = CP
`sqrt81` = CP
 ∴ CP = 9 m

(ii) In right angle triangle BPD
(PB)² = (BD)² + (PD)²
(15)² = (9)² + PD²
225 = 81 + PD²
225 – 81 = PD²
144 = PD²
`sqrt144` = PD²
∴ PD = 12 m

(iii) CP = 9 m
PD = 12 m
∴ CD = CP + PD
= 9 + 12
= 21 m