In triangle ABC,
$A B = A C = x;$ $B C = 10$ cm and the area of the trianle is $60 c m^{2}$ . Find $x$

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Solution

Drop a perpendicular from A to side BC and mark it as D

$\Rightarrow A D = \sqrt{x^{2} - 5^{2}}$

$\Rightarrow a r e a = \frac{1}{2} \times A D \times B C$

$\Rightarrow 60 = \frac{1}{2} \times \sqrt{x^{2} - 25} \times 10$

$\Rightarrow 144 = x^{2} - 25$

$\Rightarrow x^{2} = 169$

$\Rightarrow x = 13 c m$

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