Question

Find the area of an isosceles triangle, each of whose equal sides measure $13cm$ and whose base measures$20cm.$

Answer 1

Finding the area of the triangle:

Let$a,bandc$ be the sides of a triangle and $s$ be the semi-perimeter.
Then its area A , by Heron's formula is given by

$\begin{array}{c}A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\\ a=13cm,b=13cm,c=20cm\\ s=\frac{a+b+c}{2}\\ \Rightarrow s=\frac{13+13+20}{2}\\ s=\frac{46}{2}\\ s=23cm\\ \\ \Rightarrow A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\\ =\sqrt{23\left(23-13\right)\left(23-13\right)\left(23-20\right)}\\ =\sqrt{23\left(10\right)\left(10\right)\left(3\right)}\\ =10\sqrt{23\times 3}(\because \sqrt{63}=8.307...)\\ =83.07c{m}^2\end{array}$

Hence, the area of the triangle is $83.07c{m}^2$