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

In isosceles $\Delta ABC$

AB=AC =13 cm

But perimeter =50 cm

$\therefore$ BC=50 -(13+13)cm

=50-26=24 cm

$AD\perp BC$

$\therefore AD=DC=\frac{24}{2}=12cm.$

In right $\Delta ABD,$

$A{B}^2=A{D}^2+B{D}^2$ (Pythagoras Theorem)

$(13{)}^2=A{D}^2+(12{)}^2$

$\Rightarrow 169=A{D}^2+144$

$\Rightarrow A{D}^2=169-144$

$=25=(5{)}^2$

$\therefore AD=5cm.$

Now area of $\Delta ABC=\frac{1}{2}$ Base $\times$ Altitude

$=\frac{1}{2}\times BC\times AD$

$=\frac{1}{2}\times 24\times 5=60c{m}^2$