Math, asked by ritadevi88342, 4 months ago

what is the ratio of the areas of a circle and an equilateral triangle whose dianeter and a side are respctively equal​

Answers

Answered by MяMαgıcıαη
132

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\underline{\underline{\sf{ \color{magenta}{\qquad Given\:: \qquad}  }}}

  • A circle and equilateral triangle whose diameter and side are equal

\underline{\underline{\sf{ \color{magenta}{\qquad To\:Find\:: \qquad}  }}}

  • Ratio of their areas

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Let diameter of circle = 2x

Side of triangle also = 2x

So , Radius of circle = 2x/2 = x

\underline{\underline{\sf{ \color{magenta}{\qquad Diagrams\:: \qquad}  }}}

\bold\red{(1)\:Circle}

Radius of circle = x

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\bold\red{(2)\:Triangle}

Side of triangle = 2x

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\underline{\underline{\sf{ \color{magenta}{\qquad Solution\:: \qquad}  }}}

Ratio of their areas :-

\longmapsto\:\tt{\bigg(\dfrac{(Area\:of\:circle)}{(Area\:of\:equilateral\:triangle)}}\bigg)

\longmapsto\:\tt{\bigg(\dfrac{\pi r^2}{\frac{\sqrt{3}}{4}(side)^2}}\bigg)

\longmapsto\:\tt{\bigg(\dfrac{(\pi)\:\times\:(x)^2}{\frac{\sqrt{3}}{4}\:\times\:(2x)^2}}\bigg)

\longmapsto\:\tt{\bigg(\dfrac{\pi \:\times\:x^2}{\frac{\sqrt{3}}{4}\:\times\:2x\:\times\:2x}}\bigg)

\longmapsto\:\tt{\bigg(\dfrac{\pi\:\times\: x^2}{\frac{\sqrt{3}}{4}\:\times\:4x^2}}\bigg)

\longmapsto\:\tt{\bigg(\dfrac{\pi \:\times\:x^2}{\frac{\sqrt{3}}{\cancel{4}}\:\times\:\cancel{4}\:\times\:x^2}}\bigg)

\longmapsto\:\tt{\bigg(\dfrac{\pi\:\times\: x^2}{\sqrt{3}\:\times\:x^2}\bigg)}

\longmapsto\:\tt{\bigg(\dfrac{\pi\:\times\:\cancel{ x^2}}{\sqrt{3}\:\times\:\cancel{x^2}}\bigg)}

\longmapsto\:\tt{\bigg(\dfrac{\pi}{\sqrt{3}}}\bigg)

\longmapsto\:\boxed{\boxed{\tt{\dfrac{\pi}{\sqrt{3}}}}}\:\orange\bigstar

\underline{\boxed {\frak {\therefore \green {Ratio\:of\:their\:areas\:\leadsto\:\pi\:\ratio\:\sqrt{3}}}}}\:\red\bigstar

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\underline{\underline{\sf{ \color{magenta}{\qquad Note\:: \qquad}  }}}

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Anonymous: Awesome!
Answered by XxRedmanherexX
6

Radius of circle r= \frac{d}{2}  \\And \: side \: of \: triangle \: a=d \\ Ratio= \frac{area \: of \: the \: circle}{area \: of \: the \: triangle}  =  \frac{\pi {r}^{2} }{  \frac{ \sqrt{3} {a}^{2}  }{4}  }\\  =  \frac{ \frac{\pi {d}^{2} }{4} }{ \frac{ \sqrt{3} {d}^{2}  }{4} }  \\  =  \frac{\pi}{ \sqrt{3} }

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