Question

30. in. AABC, AB = 15 c.m. + BC = 13c.m CA-14c.m., Find the area of AABC.​

Answer 1

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Answer 2

$area \: of \: triangle = \sqrt{s(s-a)(-b)(s-c)}$

$where \: is \: the \: semi-perimeter$

$Here's = \frac{13 + 14 + 15}{2} = 21$

$Therefore,area$

$Therefore,area = \sqrt{21(6)(7)(8)} =$

$\sqrt{7050} = 84 {units}^{2}$

$using \: area = \frac{1}{2} \times base \times height$

$84 = \frac{1}{2} \times h = >h = 12$

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