Question

solve the problem... ​

Answer 1

Given:-

• Three sides of a triangle are 5 m, 6 m and 7 m respectively.

To find: -

• Area of triangle?

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$\underline{\bigstar\:\boldsymbol{Using\: Heron's\:Formula\::}}$

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

$:\implies\sf s = semi - perimeter$

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

$\sf Here \begin{cases} & \sf{a = \bf{5\:m}} \\ & \sf{b = \bf{6\:m}} \\ & \sf{c = \bf{7\:m}} \end{cases}$

$:\implies\sf s = \dfrac{5 + 6 + 7}{2}$

$:\implies\sf s = \cancel{\dfrac{18}{2}}$

$:\implies{\underline{\boxed{\frak{\purple{s = 9\:m}}}}}\;\bigstar$

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$\bf{\dag}\;{\underline{\frak{Now,\:Putting\:values\;in\;formula}}}$

$:\implies\sf Area_{\;\triangle} = \sqrt{9(9 - 5)(9 - 6)(9 - 7)}$

$:\implies\sf Area_{\;\triangle} = \sqrt{9 \times 4 \times 3 \times 2}$

$:\implies\sf Area_{\;\triangle} = 3 \times 2 \sqrt{3 \times 2}$

$:\implies{\underline{\boxed{\frak{\purple{Area_{\;\triangle} = 6\sqrt{6}\:m^2}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Area\:of\:triangle\:is\: \bf{6 \sqrt{6}\:cm^2}.}}}$

Answer 2

Answer:

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

$:\implies\sf s = semi - perimeter$

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

$\sf Here \begin{cases} & \sf{a = \bf{5\:m}} \\ & \sf{b = \bf{6\:m}} \\ & \sf{c = \bf{7\:m}} \end{cases}$

$:\implies\sf s = \dfrac{5 + 6 + 7}{2}$

$:\implies\sf s = \cancel{\dfrac{18}{2}}$

$:\implies{\underline{\boxed{\frak{\pink{s = 9\:m}}}}}\;\bigstar$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀

$\bf{\dag}\;{\underline{\frak{Now,\:Putting\:values\;in\;formula}}}$

$:\implies\sf Area_{\;\triangle} = \sqrt{9(9 - 5)(9 - 6)(9 - 7)}$

$:\implies\sf Area_{\;\triangle} = \sqrt{9 \times 4 \times 3 \times 2}$

$:\implies\sf Area_{\;\triangle} = 3 \times 2 \sqrt{3 \times 2}$

$:\implies{\underline{\boxed{\frak{\blue{Area_{\;\triangle} = 6\sqrt{6}\:m^2}}}}}\;\smallstar$

$\therefore\:{\underline{\sf{Area\:of\:triangle\:is\: \bf{6 \sqrt{6}\:cm^2}.}}}$