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प्रश्न
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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उत्तर
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`.
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