A right triangle ABC with hypotenuse AB, has side AC=15, altitude CH divide AB into segments AH and HB =16. The area of $Δ A B C$ is

120

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144

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150

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216

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Solution

The correct option is B $150$
$∠ H = ∠ C = 90^{\circ}$

$∠ A = ∠ A$ (common)

$A H C \sim A C B$ by $A A$

$∴ \frac{A H}{A C} = \frac{A C}{A H + 16}$

$A C = 15$

$\frac{A H}{15} = \frac{15}{A H + 16}$

$A H (A H + 16) = 225$

solving the above equation, we get, $A H = 9$

$A H C \sim C H B$

$∴ \frac{A H}{H C} = \frac{H C}{H B}$

$H B = 16$

$\frac{9}{H C} = \frac{H C}{16}$

$H C^{2} = 144 \Rightarrow H C = 12$

Area of $△ A B C = \frac{1}{2} (A B) (H C) = \frac{1}{2} (25) (12) = 150$

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