Question

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3: 2, the sides of the triangle are​

Answer 1

Let ABC be an Isosceles triangle having perimeter 32 cm.

The ratio of the equal side to its base is 3:2.

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☯ Let sides of triangle be in form of 'x'

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Therefore,

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• Equal sides, AB = AC = 3x
• Base of triangle, BC = 2x

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We know that,

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$\star\;{\boxed{\sf{\purple{Perimeter_{\triangle} = Sum\;of\;sides}}}}$

$:\implies\sf 3x + 3x + 2x = 32$

$:\implies\sf 8x = 32$

$:\implies\sf x = \cancel{ \dfrac{32}{8}}$

$:\implies{\boxed{\frak{\pink{x = 4}}}}\;\bigstar$

Therefore, measure of sides are,

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• AB = a = 3 × 4 = 12 cm
• AC = b = 3 × 4 = 12 cm
• BC = c = 2 × 4 = 8 cm

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☯ Also, we can find Area of given Isosceles triangle:

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Semi - Perimeter of an Isosceles triangle,

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$:\implies\sf s = \dfrac{a + b + c}{2}$

$:\implies\sf s = \dfrac{12 + 12 + 8}{2}$

$:\implies\sf s = \dfrac{32}{2}$

$:\implies{\boxed{\frak{s = 16\;cm}}}\;\bigstar$

Now, Finding Area of Isosceles triangle using Heron's Formula,

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$\star\;{\boxed{\sf{\purple{Area = \sqrt{s(s - a)(s - b)(s - c)}}}}}$

$:\implies\sf \sqrt{16(16 - 12)(16 - 12)(16 - 8)}$

$:\implies\sf \sqrt{16 \times 4 \times 4 \times 8}$

$:\implies\sf 4 \times 4 \times 2 \sqrt{2}$

$:\implies{\boxed{\frak{\pink{32 \sqrt{2}\;cm^2}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;the\;area\;of\; Isosceles\;triangle\;is\; \bf{32\sqrt{2}\;cm^2}.}}}$