Question

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

please give me the correct answer ​

Answer 1

Answer:

Step-by-step explanation:

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

√81 = 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

√144 = PD ‘

∴ PD = 12 m

(iii) CP = 9 m

PD = 12 m

∴ CD = CP + PD

= 9 +12 = 21 m

Answer 2

Answer:

yes he gave the correct answer it's 9m, 12m, 21m