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Answer 1

$\large{\underline{\underline{\mathfrak{\green{\bf{Questions:-}}}}}}$.

• If the area of an equilateral triangle is $\:81\sqrt{3}$cm^2, then find its height and Side

$\large{\underline{\underline{\mathfrak{\bf{Given\:Here:-}}}}}$.

• Area of Triangle = $\:81\sqrt{3}$.

$\large{\underline{\underline{\mathfrak{\bf{Find\:here:-}}}}}$.

• height of triangle

• Side of triangle

$\large{\underline{\underline{\mathfrak{\bf{Explanation:-}}}}}$.

★We know that,

$\large\boxed{\:Area\:of\:Equiletateral\:triangle\:=\frac{1}{4}\sqrt{3}a^2}$

Where,

• a is the side of triangle

Therefore ,

$\implies\:Area\:=\frac{1}{4}\sqrt{3}a^2$.

$\implies\:81\sqrt{3}\:=\frac{1}{4}\sqrt{3}a^2$

$\implies\:a^2\:=\frac{81\times\:4\times\sqrt{3}}{\sqrt{3}}$

$\implies\:a^2\;=\:81×4$.

$\implies\:a\:=\sqrt{81×4}$

$\implies\:a\:=\:9×2$

$\huge\boxed{\:a\:=\18}$

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Again, Using Formula of Triangle

$\large\boxed{\:Area\:of\:triangle\:=\frac{1}{2}×base×height}$.

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We , Know Here,

• a = 18, Side of triangle(Base).

So, Keep value of a=18 in area of triangle ,

We Get,

$\implies\:81\sqrt{3}\:=\frac{1}{2}×18×height$

$\implies\:9×height\:=\:81\sqrt{3}$

$\implies\:height\:=\frac{81×\sqrt{3}}{9}$

$\large\boxed{\:hieght\:=\:9\sqrt{3}}$

$\large{\underline{\underline{\mathfrak{\green{\sf{Answer:-}}}}}}$.

• Side (a) = 18

• Height = $\:9\sqrt{3}$

___________________________

$\large{\underline{\underline{\mathfrak{\pink{\sf{\:Answer\:Cheching:- }}}}}}$.

$\large\boxed{\:Area\:of\:triangle\:=\frac{1}{2}×base×height}$

Keep value of Area, base and height .

$\implies\:81\sqrt{3}\:=\frac{1}{2}×18×9\sqrt{3}$

$\implies\:81\sqrt{3}\:=\:9×9\sqrt{3}$

$\large\:Thats\:proved$

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