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The area of an isosceles triangle is 60cm2 and the length of each of its equal sides is 13cm. Find its base.

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10 cm

24 cm

Both 1 and 2

None of the above

answer is C.

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Detailed Solution

Area = \frac{1}{2} \times b \times h

60 = \frac{1}{2} bh

bh = 120 \to eq (1)

a^{2} = {(\frac{b}{2})}^{2} + h^{2}

h^{2} = 13^{2} - \frac{b^{2}}{4} \to eq (2)

Squaring eq (1) \to h^{2} = \frac{120^{2}}{b^{2}}

eq (2) \to \frac{120^{2}}{b^{2}} = 13^{2} - \frac{b^{2}}{4}

{4 \times 120}^{2} = b^{2} [4 \times 169 - b^{2}]

b^{4} - 676 b^{2} + 57600 = 0

b^{4} - 576 b^{2} - 100 b^{2} + 57600 = 0

b^{2} [b^{2} - 576] - 100 [b^{2} - 576] = 0

(b^{2} - 100) (b^{2} - 576) = 0

b^{2} = 100 or b^{2} = 576

b = 10 cm or b = 24 cm

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