Question

Question 4
The perimeter of an isosceles triangle is 32 cm. The ratio of the equal to its base is 3:2. Find the area of the triangle.

Answer 1

Let ABC be an isosceles triangle with perimeter 32cm.
We have, the ratio of equal side to its base is 3:2.
Let the sides of the triangle be AB = AC = 3x, BC = 2x
$\because$ Perimeter of a triangle = 32m
Now, 3x+3x+2x = 32
$\Rightarrow$ 8x = 32
$\Rightarrow$ x = 4
$\therefore$ AB = AC = 3 $\times$ 4 = 12 cm
And BC = 2x = 2 $\times$ 4 = 8cm
The sides of the triangle are a = 12cm, b = 12cm
And c = 8cm.
$\therefore$ Semi-perimeter of an isosceles triangle,
$s=\frac{a+b+c}{2}$
$=\frac{12+12+8}{2}=\frac{32}{2}=16m$
$\therefore$ Area of an isosceles $\Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron’s form]
$=\sqrt{16(16-12)(16-12)(16-8)}=\sqrt{16\times 4\times 4\times 8}$
$\Rightarrow =4\times 4\times 2\sqrt{2}c{m}^2=32\sqrt{2}c{m}^2$
Hence, the area of an isosceles triangle is $32\sqrt{2}c{m}^2$