Question

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

Answer 1

$Area=60c{m}^2$

$AD=\sqrt{x^2-5^2}=\sqrt{x^2-25}$ (Since $\Delta$ is isosceles)

Area of triangle is given by

$Area=\frac{1}{2}\times base\times height$

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

But, according to question ,

$Area=60c{m}^2$

So,

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

$\Rightarrow 12=\sqrt{x^2-25}$

$\Rightarrow 144=x^2-25$

$\Rightarrow x^2=169$

$\therefore x=13cm.$