Question

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find the area of the triangle.

Answer 1

Let ABC be an isosceles triangle with perimeter 32 cm.

We have, ratio of equal side to its base is 3 : 2.

Let sides of triangle be AB = AC = 3x, BC = 2x

∵ Perimeter of a triangle = 32 m

Now, 3x + 3x + 2x = 32

`\implies` 8x = 32

`\implies` x = 4

∴ AB = AC = 3 × 4 = 12 cm

And BC = 2x = 2 × 4 = 8 cm

The sides of a triangle are a = 12 cm, b = 12 cm and c = 8 cm.

∴ Semi-perimeter of an isosceles triangle,

`s = (a + b + c)/2`

= `(12 + 12 + 8)/2`

= `32/2`

= 16 cm

∴ Area of an isosceles ΔABC

= `sqrt(s(s - a)(s - b)(s - c))`  ...[By Heron’s formula]

= `sqrt(16(16 - 12)(16 - 12)(16 - 8))`  

= `sqrt(16 xx 4 xx 4 xx 8)`

= `4 xx 4 xx 2sqrt(2)  cm^2`

= `32sqrt(2)  cm^2` 

Hence, the area of an isosceles triangle is `32sqrt(2)  cm^2`.