Question

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

Answer 1

✒ Correct Question:- what is the ratio of the areas of a circle and an equilateral triangle whose dianeter and a side are respctively equal.

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✒ Required answer: $\sf{\pi\:\ratio\:\sqrt{3}}$

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✒ Given: Diameter of circle = side of equilateral triangle.

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✒ To find: Ratio of the areas of a circle and equilateral triangle.

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✎ Step by step solution:

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Let Diameter of circle = $\large\sf D$

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side of triangle = $\large\sf D$

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Radius of circle $\large\sf{ = \dfrac{D}{2}}$

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✒ Diagrams:-

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$\large\bf{ 1. \big\rangle \;\; Circle }$

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$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3)\put(0,0){\line(1,0){2.3}}\put(0.5,0.3){\bf\large D/2}\end{picture}$

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$\large\bf{ 2. \big\rangle \;\; Triangle }$

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$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){ \bf A }\put(0.5,-0.3){ \bf C }\put(5.2,-0.3){ \bf B }\end{picture}$

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side of equilateral triangle $\large\rm =$ $\large\sf D$

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area of equilateral triangle $\large\sf{ = \dfrac{\sqrt{3}}{4} (s)^2}$

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$\large\sf { Ratio \: = \: \Bigg ( \dfrac{ A(\bigcirc)}{A( Eq. \triangle)} \Bigg )}$

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$\large\sf{ = \Bigg ( \dfrac{\pi r^{2}}{\frac{\sqrt{3}}{4} (s)^2} \Bigg )}$

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$\large\sf{ = \dfrac{\pi \times ( \frac{d}{2} )^2}{\frac{\sqrt{3}}{4} \times d^2}}$

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$\large\sf{ = \dfrac{\frac{\pi \times d^2}{4}}{\frac{\sqrt{3}}{4} \times d^2}}$

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$\large\sf{ \dfrac{\pi \times d \times d \times 4}{\sqrt{3} \times d \times d \times 4}}$

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$\large\sf{ = \dfrac{\pi \times \cancel{d^2} \times \cancel{4}}{\sqrt{3} \times \cancel{d^2} \times \cancel{4}}}$

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$\large\sf{ = \dfrac{\pi}{\sqrt{3}}}$

$\large\rm \therefore$ ratio $\large\rm{ = \pi : \sqrt{3}}$

Answer 2

Step-by-step explanation:

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