Question

The lengths of the sides of a triangle are in the ratio $4:5:3$ and its perimeter is $96$ cm. Find its area (in sq. cm).

Answer 1

Let the sides of the triangle be $4a,5a,3a$.
Perimeter of the triangle $=$ Sum of all sides $=4a+5a+3a=12a$

Given, perimeter of the triangle $=96cm$
$\Rightarrow 12a=96cm$
$a=8cm$

So, the sides of the triangle are $4a=32cm,5a=40cm,3a=24cm$.

Area of triangle with $a,b,c$ and semiperimeter
$s=\frac{a+b+c}{2}$ is $\sqrt{s(s-a)(s-b)(s-c)}$.

For triangle with sides $32cm,40cm$ and $24cm,$ $s=\frac{32+40+24}{2}=48cm$

Area of the triangle with sides $32cm,40cm$ and $24cm=\sqrt{48(48-32)(48-40)(48-24)}$

$=\sqrt{48\times 16\times 8\times 24}$

$=\sqrt{16\times 3\times 16\times 8\times 8\times 3}$

$=16\times 8\times 3=384{cm}^2$