Question

The sides of a triangle are $21cm,17cm$ and $10cm$. Find its area.

Answer 1

We have

$AB=10cm$

$BC=21cm$

$CA=17cm$

Now,

$s=\frac{21+17+10}{2}=\frac{48}{2}=24$

Then

Area of $\Delta ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

$=\sqrt{24\left(24-21\right)\left(24-17\right)\left(24-10\right)}$

$=\sqrt{24\times 3\times 7\times 14}$

$=84c{m}^2$