Question

Find area of triangle whose two sides are 13 cm and third side is 10

Answer 1

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$\underline{SOLUTION}$ ➡

$Let, \: \: \\ \\ ABC \: \: is \: the \: \: triangle \: \: where ,\\ \\ AB = 13 \: cm\\ \\ AC=13 \: cm\\ \\ BC = 10 \: cm$

$ABC \: \: is \: \: a \: \: isosceles \: \: triangle \: \: whose \\ \\ equal \: \: sides \: are \: ,\: a\: = AB = AC = 13 \: \: cm\\ \\ and \: \: \: base\: ,\: b \: BC = 10 \: \: cm$

$We \: \: know \: \: that \\ \\ Area \: \: of \: \: an\: \: isosceles \: \: triangle \\ \\ = \frac{b}{4} \sqrt{4a {}^{2} - b {}^{2} } \\ \\ =[ \frac{10}{4} \sqrt{4 \times (13) {}^{2} - (10) {}^{2} } ]\: \: sq. \: cm \\ \\ = [\frac{5}{2} \sqrt{4 \times 169 - 100} ]\: \: sq. \: cm\\ \\ = [\frac{5}{2} \sqrt{676 - 100} ]\: \: sq. \: cm\\ \\ =[ \frac{5}{2} \sqrt{576}] \: \: sq. \: cm\\ \\ =[ \frac{5}{2} \sqrt{(24) {}^{2} } ]\: \: sq. \: cm \\ \\ = ( \frac{5}{2} \times 24) \: \: sq. \: cm\\ \\ =( 5 \times1 2)\: \: sq. \: cm \\ \\ = 60 \: \: sq. \: cm$

$So , \: Area \: \: of \: \: the \: \: triangle \: = 60 \: \: sq.cm$

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