Question

Question 88

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

Answer 1

Given, area of $\Delta ABC=48c{m}^2$ and altitude = 8 cm
$\because \Delta ABC$ is an isosceles trinagle, where AB =AC
$\therefore$ Area of $\Delta ABC=\frac{1}{2}\times BC\times AD=48$
$[\because$ area of triangle = $\frac{1}{2}\times base\times height]$
$\Rightarrow 48=\frac{1}{2}\times BC\times AD$
$\Rightarrow \frac{1}{2}\times BC\times 8=48\Rightarrow BC=\frac{48\times 2}{8}$
BC = 12 cm
Now, in an isosceles triangle, BD = DC = 6 cm $[\because AD\perp BC]$
Applying Pythagoras theorem in right angled $\Delta ADB,$
$A{B}^2=B{D}^2+A{D}^2\Rightarrow A{B}^2=6^2+8^2=36+64\Rightarrow A{B}^2=100\Rightarrow AB=10cm$
Now, perimeter of triangle = AB + AC + BC = AB + AB + BC $[\because AB=AC]$
= 10 + 10 + 12
= 32 cm