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.

Answer 1

Let the sides of the triangle ABC be 4x, 5x and 3x

Let AB=4x, AC =5x and BC =3x

Perimeter =4x+5x+3x=96

$\Rightarrow 12x=96$

$\Rightarrow x=\frac{96}{12}$

$\therefore x=8$

$\therefore$ Sides are

BC=3(8)=24 cm, AB=4 (8)=32 cm,

C=5(8)=40 cm

Since $(AC{)}^2=(AB{)}^2+(BC{)}^2$

$[\because (5x{)}^2=(3x{)}^2+(4x{)}^2]$

$\therefore$ By Pythagorus Theorem, $\angle B={90}^{\circ }$

$\therefore$ Area of $\Delta ABC=\frac{1}{2}\left(BC\right)\left(AB\right)=\frac{1}{2}\left(24\right)\left(32\right)$

$=12\times 32=384c{m}^2$