Question

One of the equal sides of an isosceles triangle is 13 cm and its perimeter is 50 cm. Find the area of the triangle in cm${}^2$

Answer 1

Let the third side of the iscosceles triangle $=a$

Perimeter of the triangle $=$ Sum of all sides $=13+13+a=50=>a=24cm$

Let the side of length $24cm$, be the base of the triangle.

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

${13}^2=h^2+{12}^2$
$h^2=169-144=25$
$h=\sqrt{25}=5$ cm

Now, area of triangle $=\frac{1}{2}\times base\times height=\frac{1}{2}\times 24\times 5=60$ sq cm