Question

Find the area of a triangle whose sides are 12 cm , 6 cm & 10cm.

Please dont answer if you dont know ​

Answer 1

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

$\sf \: Sides \: of \: triangle \: are \: 12 \: cm , \: 6 \: cm \: and \: 10 \: cm.$

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

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

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

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

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

$\bf \underline{ Now},$

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

$\sf \implies s \: = \: \dfrac{12 +6 + 10}{2}$

$\sf \implies s \: = \: \cancel\dfrac{28}{2}$

$\sf \implies s \: = \: 14$

$\underline{ \sf \: Thus, \: semi-perimeter = 14\:cm }$

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

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

$\sf Substitute \: the \: values,$

$\sf{Area = \sqrt{14(14 - 12) (14 - 6) (14- 10)}}$

$\sf{Area = \sqrt{14 \times 2 \times 8 \times 4}}$

$\sf{Area = \sqrt{2 \times 7 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}}$

$\sf{Area =2 \times 2 \times 2 \sqrt{2 \times 7 }}$

$\sf{Area =8 \sqrt{14 }} \: cm ^{2}$

$\underline{ \boxed{ \red{ \bf Therefore, \:area \:of\:triangle = 8 \sqrt{14} \: cm^2}}}$

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$\bf \underline{Some\: related\: formulas} :$

$\bf{1.}\: \sf Area \: of \: equilateral \: triangle = \dfrac{ \sqrt{3}}{4} a^{2}$

2. Area of triangle with given base and corresponding altitude :

$\sf \implies \dfrac{1}{2} \times base \times corresponding \: altitude$

3. Area of a right triangle :

$\sf \implies \dfrac{1}{2} \times Product \:of\: two\: sides\: containing\: right\: angle$