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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
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Solution
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
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