Question

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

Answer 1

Let the sides of triangle be $a,b$ and $c$ and $2s$ be its perimeter such that $AB\left(a\right)=15cm,BC\left(b\right)=13cm$ and $AC\left(c\right)=14cm$

$s=\frac{a+b+c}{2}=\frac{15+13+14}{2}=\frac{42}{2}=21$

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

$=\sqrt{21(21-15)(21-13)(21-14)}$

$=\sqrt{21\times 6\times 8\times 7}$

$=\sqrt{7\times 3\times 3\times 2\times 2\times 2\times 2\times 7}$

$=84c{m}^2$

Let the altitude on $AC$ be $h$ cm

Now, area of triangle $=\frac{1}{2}\times \text{Base}\times \text{Height}$

$\frac{1}{2}\times AC\times h=84$

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

$h=12cm$

Hence, altitude is $12cm$.