Question

In a $△ABC,AB=15cm,BC=13cm$ and $AC=14cm$.Find the area of $△ABC$ and hence its altitude on AC ?

Answer 1

area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

where $s$ is the semi-perimeter

here,$s$=$\frac{13+14+15}{2}$=21

therefore,area =$\sqrt{21\left(6\right)\left(7\right)\left(8\right)}$

$=\sqrt{7056}$=84 units${}^2$

using area $\frac{1}{2}\times base\times height$ :

$84\frac{1}{2}\times 14\times h$

$\Rightarrow$ h=12cm