Question

The base of an isosceles triangle measures 80 cm and its area is 360 $c{m}^2$ Find the perimeter of the triangle.

Answer 1

$\text{Area of triangle}=\frac{1}{2}\times base\times height360=\frac{1}{2}\times 80\times h720=80\times hh=9cm\text{Now in}△ADB,x^2={40}^2+h^2={40}^2+9^{2}1600+811681\therefore x=\sqrt{1681}=41Perimeter=x+x+80=41+41+80=162cm$