# The perimeter of an isosceles triangle is 32 cm.

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.

Solution:

Let $A B C$ be an isosceles triangle with perimeter $32 \mathrm{~cm}$. We have, ratio of equal side to its base is $3: 2$.

Let sides of triangle be $\quad A B=A C=3 x, B C=2 x$

$\because$ Perimeter of a triangle $=32 \mathrm{~m}$

Now, $\quad 3 x+3 x+2 x=32$

$\Rightarrow \quad 8 x=32$

$\Rightarrow \quad x=4$

$\therefore \quad A B=A C=3 \times 4=12 \mathrm{~cm}$

and $\quad B C=2 x=2 \times 4=8 \mathrm{~cm}$

The sides of a triangle are $a=12 \mathrm{~cm}, b=12 \mathrm{~cm}$

and $\quad c=8 \mathrm{~cm}$.

$\therefore$ Semi-perimeter of an isosceles triangle,

$s=\frac{a+b+c}{2}$

$=\frac{12+12+8}{2}=\frac{32}{2}=16 \mathrm{~cm}$

$\therefore$ Area of an isosceles $\triangle A B C=\sqrt{s(s-a)(s-b)(s-c)} \quad$ [by Heron's formula]

$=\sqrt{16(16-12)(16-12)(16-8)}=\sqrt{16 \times 4 \times 4 \times 8}$

$\Rightarrow$ $=4 \times 4 \times 2 \sqrt{2} \mathrm{~cm}^{2}=32 \sqrt{2} \mathrm{~cm}^{2}$

Hence, the area of an isosceles triangle is $32 \sqrt{2} \mathrm{~cm}^{2}$.