Question

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

Answer 1

Given, Area of triangle $ABC=48sq.cm$

Height $AD=8cm$

Area of triangle $ABC$,

$\Rightarrow 48=\frac{1}{2}\times BC\times AD=\frac{1}{2}\times BC\times 8$

$\Rightarrow BC=\frac{48\times 2}{8}=12cm$

Now, in an isosceles triangle height is also median, so $D$ is midpoint of $BC$

$\Rightarrow BD=DC=6cm$

By, Pythagoras theorem in triangle $ADB$.

$\Rightarrow A{B}^2=B{D}^2+A{D}^2$

$\Rightarrow A{B}^2=6^2+8^2$

$\Rightarrow A{B}^2=36+64=100$

$\Rightarrow AB=10cm$

$\therefore AC=AB=10cm$

Now, perimeter of triangle $=AB+AC+BC$

$=10+10+12$

$=32cm$