ANSWER:

[32\sqrt{2}cm^{2} ]

Given\; perimeter\; of\; isosceles\; triangle = 32 cm

We\; have\; ratio\; of\; equal\; side\; and\; base = 3 : 2

Thus,\; let\; the \; sides\; of\; triangle\; be\; AB = AC = 3x\; and\; BC = 2x

\therefore Perimeter \; of\; a\; triangle = 32 cm

3x + 3x + 2x = 32

8x = 32

x=\frac{32}{8}

x = 4 cm

So, AC = AB = 3 \times 4 cm\; and\; BC = 2 \times 4 cm

AC = AB = 12 cm\; and \; BC = 8 cm

Now, a = 8\; cm, b = 12\; cm, c = 12 \; cm

Using Heron’s formula

S= \frac{a+b+c}{2}=\frac{8+12+12}{2}=\frac{32}{2}=16cm

\therefore\; AREA \; of\; isosceles\; triangle\; \Delta ABC =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}

=\sqrt{16\left ( 16-8 \right )\left ( 16-12 \right )\left ( 16-12 \right )}

= \sqrt{16\times 8\times 4\times 4}

= \sqrt{4\times 4\times 4\times 2\times 4\times 4}

=4\times 4 \sqrt{ 2\times 2\times 2}=4\times 4\times 2\sqrt{2}=32\sqrt{2}\; cm^{2}

HENCE,\; the\; area\; of\; isosceles \; triangle\; is\; 32\sqrt{2}\; cm^{2}

Step-by-step explanation: