Question

Area of an isosceles triangle is $48{\text{cm}}^2$. If the altitudes corresponding to the base of the triangle is $8$ cm, find the perimeter of the triangle.

Answer 1

We know,

Area of triangle = $\frac{1}{2}\times$ base $\times$ height

$\therefore 48=\frac{1}{2}\times b\times 8$

$\Rightarrow 48=4b$

$\Rightarrow$ Base $=12$ cm

$\therefore DC=\frac{base}{2}=6cm$

In $\Delta ADC,\angle D={90}^{\circ }$

$\therefore$ By Pythagoras theorem,

$\therefore 8^2+6^2=A{C}^2$

$\Rightarrow 64+36=A{C}^2$

$\Rightarrow 100=A{C}^2$

$\Rightarrow AC=\sqrt{100}$

$\Rightarrow AC=10$

$\therefore AB=AC$

$\therefore$ perimeter of $\Delta ABC$ $=AB+BC+AC$

$=10+12+10$

$=32$ cm