Question

The base of an isosceles triangle measures $80cm$ and its area is $360sq.cm$. Find the perimeter of the triangle (in cm).

Answer 1

Area of triangle $=\frac{1}{2}\times base\times height=360$ sq cm

$\frac{1}{2}\times 80\times height=360$

$=>$ height $=9$ cm.

Since the height of the triangle, divides it into two right angled triangles,

$a^2=h^2+{\left(\frac{b}{2}\right)}^2=>a^2=9^2+{\left(\frac{80}{2}\right)}^2=1681$

Hence, $a=41$ cm.

Perimeter of the triangle $=$ Sum of all sides $=$ a + a + b $=41+41+80=162cm$