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In the given figure, $AB = 9, BC = 12$ and $AC = 15$ , $BP ⊥ AC$ , Find $BP$ and $AP$ respectively.

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8.4, 4.3

7.2, 5.4

6.9, 3.8

5.5, 5.8

answer is B.

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Detailed Solution

AB = 9, BC = 12, AC = 15

{BP}^{2} = AP \times PC

— (|) [Property of right angle triangle]

{AB}^{2} = AP \times PC

— (||)
Substituting the values in eq (||),

\Rightarrow 9^{2} = AP \times 15

\Rightarrow 81 = 15 AP

\Rightarrow \frac{81}{15} = AP

∴ AP = 5.4

According to given diagram

\Rightarrow AC = AP + PC

\Rightarrow 15 = 5.4 + PC

\Rightarrow PC = 9.6

Putting the values in eq (||)

\Rightarrow {BP}^{2} = 5.4 \times 9.6

\Rightarrow {BP}^{2} = 51.84

\Rightarrow BP = \sqrt{51.84}

\Rightarrow BP = 7.2

Therefore,

AP = 5.4; BP = 7.2

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