Math, asked by agaur9013, 3 months ago

The sides of a triangle are 3 cm, 5 cm and 6 cm, the area is :: Answer will be in square root​

Answers

Answered by ShírIey
118

Given that, the sides of the triangle are 3 cm, 5 cm & 6 cm. We've to find out the area.

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⠀⠀⠀\sf\underline{\bigstar\;Using\; Heron's\: formula\;:}\\ \\

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

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\sf Here \begin{cases}\sf{a = \bf{3\:}} \\\sf{b = \bf{5}} \\\sf{c = \bf{6}} \end{cases}\\ \\

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:\implies\sf S_{(\: Semiperimeter)} = \dfrac{a + b + c}{2} \\\\\\:\implies\sf s_{(\: Semiperimeter)} = \dfrac{3 + 5 + 6}{2} \\\\\\:\implies\sf s_{(\: Semiperimeter)} = \dfrac{\cancel{14}}{\cancel{\:2}}\\\\\\:\implies{\underline{\boxed{\frak{\pink{S_{(\: Semiperimeter)} = 7}}}}}

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

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:\implies\sf Area_{\triangle} = \sqrt{7(7 -3) (7 - 5) (7 - 6)}\\\\\\:\implies\sf Area_{\triangle} = \sqrt{7 \times 4 \times 2 \times 1} \\\\\\:\implies{\underline{\boxed{\frak{\purple{Area_{\triangle} = 2\sqrt{14} cm^2}}}}}

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\therefore\:{\underline{\sf{Area\:of\:triangle\:is\: \bf{2 \sqrt{14}\:cm^2}.}}}

Answered by VinCus
102

Given:-

\bigstarThe sides of a triangle are 3 cm , 5 cm and 6 cm.

To Find:-

\bigstarArea of triangle..

Solution:-

\bigstarTo find the area of triangle, we area using herons formula,

Let,

  • a = 3

  • b = 5

  • c = 6

\bigstarFirst we have known the semi perimeter to Substitute the value given in the herons formula.

 \longrightarrow{ \underline{ \boxed{ \sf{Semi \: perimeter \:  =  \frac{a + b + c}{2} }}}}

 \\  \longrightarrow \sf \: Semi \: perimeter =  \frac{3 + 5 + 6}{2}

 \\  \longrightarrow \sf \: Semi \: perimeter =  \frac{14}{2}

 \\\longrightarrow \sf \pink \: Semi \: perimeter = 7

\bigstarUsing herons formula,

 \longrightarrow{ \underline{ \boxed{ \sf{Herons \: formula =  \sqrt{s \times(s - a) \:  (s - b) \: (s - c)} }}}}

 \\\longrightarrow \: \sf \:  =  \sqrt{7  \times  (7 - 3) \: (7 - 5) \: (7 - 6)}

 \\\longrightarrow \: \sf \:  =  \sqrt{7  \times  (4) \: (2) \: (1)}

\\ \longrightarrow \: \sf \:  =  \sqrt{7  \times  8 \times 1}

 \\\longrightarrow \: \sf \:  =  \sqrt{7 \times 2 \times 2 \times 2 \times 1}

 \\\longrightarrow \: \sf \:  =  \sqrt[2] {7 \times 2}

\\ \longrightarrow \:  \sf \:  \sqrt[2]{14}  \:  {cm}^{2}

 \bigstar{ \bold{ \: Area \: of \: the \: triangle \: is \:  \sqrt[2]{14}  \: cm {}^{2} }}

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