Question

If in a right angled isosceles triangle area of $△=32$sq.cm.Find the sides of the triangle.

Answer 1

Area of triangle$=\frac{1}{2}\times base\times altitude$

In a right angled isosceles triangle, base$=$altitude

Let the base$=$altitude$=x$

$\Rightarrow \frac{1}{2}\times x\times x=32$

$\Rightarrow x^2=64$

$\Rightarrow x=8$cm

$\therefore$ two sides containing right angle are equal to $8$cm

Third side of the triangle$=\sqrt{8^2+8^2}=8\sqrt{2}$cm