Question

find the area scalane triangle whose sides are 10cm 12cm and 14cm​

Answer 1

$\bf \underline{ \underline{\maltese\:Given} }$

$\sf Sides \: of \: the \: scalane \: \triangle \: are \: 10 \: cm, \: 12 \: cm \: and \: 14 \: cm.$

$\bf \underline{ \underline{\maltese\:To \: find } }$

$\sf \implies Area \: of \: that \: triangle = \: ?$

$\bf \underline{ \underline{\maltese\:Solution } }$

$\sf Using \: Heron's \: formula :$

$\underline{\boxed{\sf{Area_{\triangle} = \sqrt{s(s - a)(s - b)(s - c)}}}}$

$\bf \underline{Now},$

$\sf \implies s = \dfrac{S um \: of \: all \: sides}{2}$

$\sf \implies s = \dfrac{a + b + c }{2}$

$\sf \implies s = \dfrac{10 + 12 + 14}{2}$

$\sf \implies s = \cancel \dfrac{36}{2}$

$\sf \implies s = 18$

$\underline{ \sf \: Thus, \: semi-perimeter = 18 \: cm }$

$\bf \underline{Now}, \sf \: area \: of \: triangle :$

$\sf{Area_{\triangle} = \sqrt{s(s - a)(s - b)(s - c)}}$

$\sf{Area = \sqrt{18(18 - 10)(18 - 12)(18 - 14)}}$

$\sf{Area = \sqrt{18 \times 8 \times 6 \times 4}}$

$\sf{Area =\sqrt{2 \times \underbrace{3 \times 3} \times \underbrace{2 \times 2} \times \underbrace{2 \times 2} \times 3 \times \underbrace{2 \times 2}}}$

$\sf{Area = \sqrt{2 \times (3)^{2} \times (2)^{2} \times (2)^{2} \times 3 \times (2)^{2}}}$

$\sf{Area = \sqrt{6}} \times \sqrt{(3)^{2}} \times \sqrt{(2)^{2}} \times \sqrt{(2)^{2}} \times \sqrt{(2)^{2}}$

$\sf{Area = \sqrt{6}} \times 3 \times 2 \times 2 \times 2$

$\sf{Area = \sqrt{6}} \times 24$

$\sf{Area = 24\sqrt{6} \: cm^2}$

$\underline{ \boxed{ \bf \red{ Hence, \: area \: of \: \triangle \: is \: 24\sqrt{6} \: cm^2 }}}$