Question

area of equilateral triangle is 81√3 cm square. it's height is a) 9√3 cm b) 6√3 cm c) 18√3 cm d) 9cm ​

Answer 1

Question:

• Area of equilateral triangle is 81√3 cm square. it's height is?

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Given that,

• Area of an equilateral triangle is $\sf\dfrac{81}{3}.$

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⠀⠀⠀⠀$\dag\;{\underline{\frak{As \ We \ know \ that,}}}$⠀

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$\star\:\boxed{\sf{\pink{Area \: of \: equilateral \: \triangle = \dfrac{\sqrt{3}}{\:4} (a)^2 \ cm}}}$

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• Here, a is each side of the equilateral triangle.

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Therefore,

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$:\implies\sf \dfrac{\sqrt{3}}{4} \times (a)^2 = \dfrac{81}{3} \\\\\\:\implies\sf a^2 = \dfrac{81 \sqrt{3} \times 4}{\sqrt{3}} \\\\\\:\implies\sf a^2 = 324 \\\\\\:\implies\sf a = \sqrt{324}\\\\\\:\implies{\underline{\boxed{\frak{\pink{a = 18 \: cm}}}}}\:\bigstar$

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$\therefore\:{\underline{\sf{Hence, \ we \ get \ value \ of \ a \ is \ \bf{18 \ cm}.}}}$⠀⠀⠀

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$:\implies\sf Height = \dfrac{\sqrt{3}}{2} (a) \\\\\\:\implies\sf Height = \dfrac{\sqrt{3}}{\cancel{\: 2}} \times \cancel{18} \\\\\\:\implies{\underline{\boxed{\sf{\purple{ Height = 9\sqrt{3}}}}}}\:\bigstar$

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$\therefore\:{\underline{\sf{Hence, \ Height \ of \ equilateral \ \triangle \ is \ \bf{Option \ a) \ 9 \sqrt{3} \ cm}.}}}$⠀⠀⠀

Answer 2

${\large{\bold{\rm{\underline{Given \; that}}}}}$

• Area of an equilateral triangle = 81√3 cm

${\large{\bold{\rm{\underline{To \; find}}}}}$

• Height of equilateral triangle

${\large{\bold{\rm{\underline{Solution}}}}}$

• Height of equilateral triangle = 9√3 cm (Option a)

${\large{\bold{\rm{\underline{Using \; concept}}}}}$

• Formula to find area of an equilateral triangle.

• Formula to find area of an equilateral triangle.

${\large{\bold{\rm{\underline{Using \; formula}}}}}$

• Area of an equilateral triangle = ${\sf{\dfrac{\sqrt{3}}{4}(a)^{2}}}$

• Height of an equilateral triangle = ${\sf{\dfrac{\sqrt{3}}{2}a}}$

$\; \; \; \; \; \; \; \; \; \; \;{\bf{Where,}}$

• a denotes height

${\large{\bold{\rm{\underline{Full \; Solution}}}}}$

Area of equaliteral triangle = Area of equaliteral triangle that is given.

${\sf{\dfrac{\rightarrow \sqrt{3}}{4}(a)^{2}}}$ = ${\sf{81 \sqrt{3}}}$

${\sf{\green{Let's \: cancel \: square \: roots}}}$

${\sf{\rightarrow \dfrac{1}{4}(a)^{2} \: = \: 81}}$

${\sf{\green{Multiply \: = \: Divide \: ; \: Divide \: = \: Multiply}}}$

${\sf{\rightarrow a^{2} \: = 81 \times 4}}$

${\sf{\green{Let's \: multiply \: 81 \: by \: 4}}}$

${\sf{\rightarrow a^{2} \: = 324}}$

${\sf{\green{Square \: = \: Square \: root}}}$

${\sf{\rightarrow a \: = \: \sqrt{324}}}$

${\sf{\green{Solving \: square \: root}}}$

${\sf{\rightarrow a \: = \: 18 \: cm}}$

${\frak{\red{Henceforth, \: 18 \: cm \: is \: value \: of \: a}}}$

~ Now let's use formula to find height of an equilateral triangle !..

Height of an equilateral triangle = ${\sf{\dfrac{\sqrt{3}}{2}a}}$

${\sf{\green{Put \: 18 \: as \: a}}}$

Height of an equilateral triangle = ${\sf{\dfrac{\sqrt{3}}{2} \times 18}}$

${\sf{\green{Multiplying}}}$

Height of an equilateral triangle = ${\sf{9 \sqrt{3} \: cm}}$

${\frak{\red{Henceforth, \: 9 \sqrt{3} \: cm \: is \: the \: height \: of \: equaliteral \: traingle}}}$

${\large{\bold{\rm{\underline{More \; knowledge}}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cylinder \: = \: \pi r^{2}h}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Surface \: area \: of \: cylinder \: = \: 2 \pi rh + 2 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Lateral \: area \: of \: cylinder \: = \: 2 \pi rh}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Base \: area \: of \: cylinder \: = \: \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Height \: of \: cylinder \: = \: \dfrac{v}{\pi r^{2}}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: cylinder \: = \:\sqrt frac{v}{\pi h}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: square \: = \: Side \times Side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: triangle \: = \: \dfrac{1}{2} \times breadth \times height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: paralloelogram \: = \: Breadth \times Height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle \: = \: \pi b^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: triangle \: = \: (1st \: + \: 2nd \: + 3rd) \: side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: paralloelogram \: = \: 2(a+b)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: cube \: = \: 6(side)^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto LSA \: of \: cube \:= \: 4(side)^{2}}}}$