Question

An isosceles triangle has perimeter $30cm$ and each of its equal sides is $12cm.$ Find the area of the triangle.

Answer 1

Let the base of the isosceles triangle be $xcm.$
Equal side of the given isosceles triangle $=12cm$
Perimeter $=30cm$ [Given]
or, 1$2+12+x=30$
or, $24+x=30$
or, $x=6cm$

Now, s $=$ semi-perimeter $=\frac{30}{2}=15$

$a=12cm$, $b=12cm$ and $c=6cm$

Apply Heron's Formula, we have
Area of the given isosceles triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{15(15-12)(15-12)(15-6)}$

$=\sqrt{15\left(3\right)\left(3\right)\left(9\right)}$

$=\sqrt{15\times \underset{–}{3\times 3}\times \underset{–}{3\times 3}}$

$=9\sqrt{15}c{m}^2$