Question

Find area of an isosceles triangle each of whose equal side is 13 cm and whose base is 24cm

Answer 1

Area of isoceles triangle = 1/2 × base × height
=1/2×24×13
= 156 sq.cm.

Answer 2

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Solution

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$\rm \: here \: each \: equal\: side \: \: a = 13cm \: and \: base = 24cm \\ \\ \therefore \rm \: area \: of \: the \: triangle = \big({ \frac{1}{4} b. \sqrt{4a {}^{2} - b {}^{2} } } \big)cm {}^{2} \\ = \big( \frac{1}{4 } + 24 \times \sqrt{4 \times 169 - 24 \times 24} \big) cm {}^{2} \\ = 60 \: cm {}^{2}$