Question

Find the area of an isosceles triangle whose base is $16cm$ and length of each of the equal sides is $10cm$.

Answer 1

Given

Base$=16cm$

Lengh of both equal side $=10cm$

We will use Herons's formula to find the area

$a=16cm$

$b=10cm$

$c=10cm$

Therefore,

Halfperimeter $=s=\frac{1}{2}\times (16+10+10)$

$=\frac{36}{2}$

$s=18cm$

using Heron's formula

$\sqrt{s(s-a)(s-b)(s-c)}$

$\begin{array}{c}=\sqrt{18(18-16)(18-10)(18-10)}\\ =\sqrt{18\times 2\times 8\times 8}\\ =\sqrt{2\times 9\times 4\times 2\times 4\times 2\times 2}\\ =3\times 2\times 2\times 2\times 2\\ =48c{m}^2\end{array}$

hence,

The area of given triangle is $48c{m}^2$