Question

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

Answer 1

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

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

$\Rightarrow area=\frac{1}{2}\times AD\times BC$

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

$\Rightarrow 144=x^2-25$

$\Rightarrow x^2=169$

$\Rightarrow x=13cm$