Question

the area of an equilateral triangle is 36√3 cm^2 Then it's perimeter is​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that side of an equilateral triangle is x cm.

As it is given that,

$\rm \: Area_{(equilateral\:\triangle)} = 36 \sqrt{3}$

We know,

$\boxed{ \rm{ \:Area_{(equilateral\:\triangle)} = \dfrac{ \sqrt{3} }{4} {(side)}^{2} \: }}$

So, using this result, we get

$\rm \: \dfrac{ \sqrt{3} }{4} {x}^{2} = 36 \sqrt{3}$

$\rm \: {x}^{2} = 36 \times 4$

$\rm \: x = \sqrt{36 \times 4}$

$\rm \: x = 6 \times 2$

$\rm\implies \:x = 12 \: cm$

Now,

$\rm \:Perimeter_{(equilateral\:\triangle)} = 3x$

$\rm \:Perimeter_{(equilateral\:\triangle)} = 3 \times 12$

$\rm\implies \:Perimeter_{(equilateral\:\triangle)} = 36 \: cm$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$

Answer 2

Question :-

• The Area of Equilateral Triangle is 36√3 cm² . Then, find it's Perimeter.

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Answer :-

• Perimeter of Triangle is 36 cm .

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Explanation :-

• Here, Area of Triangle is given 36√3 cm² . And, we have to calculate Perimeter .

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● First, we will calculate the Side :-

$\Longrightarrow \: \sf{Area \: _{Equilateral\:Traingle} = \frac{ \sqrt{3} }{4}(side){}^{2} }$

$\Longrightarrow \: \: \sf{ \frac{ \sqrt{3} }{4} \: {x}^{2} \: = \: 36 \: \sqrt{3} }$

$\Longrightarrow \: \: \sf{ \frac{ \cancel {\sqrt{3}} }{4} \: {x}^{2} \: = \: 36 \: \: \cancel{\sqrt{3}} }$

$\Longrightarrow \: \: \sf{ {x}^{2} \: = \: 36 \times 4 }$

$\Longrightarrow \: \: \sf{x \: = \: \sqrt{36 \: \times \: 4} }$

$\Longrightarrow \: \: \sf{x \: = \: 6 \: \times \: 2}$

$\Longrightarrow \: \:\sf{x\: = \: 12 \: cm}$

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● Now, let's calculate the Perimeter :-

$\Longrightarrow \: \: \sf { Perimeter \: _{Equilateral\:Traingle} = 3 \: x}$

$\Longrightarrow \: \: \sf{Perimeter \: _{Equilateral\:Traingle} = 3 \times 12}$

$\Longrightarrow \: \: \sf{Perimeter \: _{Equilateral\:Traingle} = 36 \: cm}$

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Hence :-

• Perimeter = 36 cm .

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$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c} \\ \underline{ { \pmb {\sf \red{ \dag \: \: More \: Formulas \: \: \dag}}}} \\ \\ \\ \footnotesize \bigstar \: \sf{Area \: of \: Square = Side \times Side} \\ \\ \\ \footnotesize\bigstar \: \sf{Area \: of \: Rectangle = Lenght \times Height} \\ \\ \\ \footnotesize \bigstar \: \sf{Area \: of \: Triangle = \frac{1}{2} \times Base \times Height } \\ \\ \\ \footnotesize \bigstar \: \sf{Area \: of \: Parallelogram = Base \times Height} \\ \\ \\ \footnotesize \bigstar \: \sf{Area \: of \: Trapezium = \frac{1}{2} \times [ \: A + B \: ] \times Height } \\ \\ \\ \footnotesize \bigstar \: \sf {Area \: of \: Rhombus = \frac{1}{2} \times Diagonal \: 1 \times Diagonal \: 2}\end{array}}\end{gathered}\end{gathered}$