Question

Find the area of an isosceles triangle each of whose equal side measures 13cm20 cm

Answer 1

$k = b \sqrt{4a {}^{2 } - b {}^{2} } \div 4$
$k = 20 \sqrt{4 \times 13 \times 13 - 20 \times 20} \div 4$
$k = 20 \sqrt{16.6132477528} \div 4$
$k = 332.26465 \div 4$
$k = 83.066238629181$
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Answer 2

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$\rm \: here \: each \: equal\: side \: \: a = 13cm \: and \: base = 20cm \\ \\ \therefore \rm \: area \: of \: the \: triangle = \big({ \frac{1}{4} b. \sqrt{4a {}^{2} - b {}^{2} } } \big)cm {}^{2} \\ = \big( \frac{1}{4 } + 20 \times \sqrt{4 \times 169 - 20 \times 20} \big) cm {}^{2} \\ = 83.07 \: cm {}^{2}$