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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.

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

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

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

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

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

$3x + 3x + 2x = 32$

$8x = 32$

$x=\frac{32}{8}$

$x = 4 cm$

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

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

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

Using Heron’s formula

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

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

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

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

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

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

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

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