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

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

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

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

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

• Ratio of their areas

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

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

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

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

Side of triangle = 2x

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

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

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

❛❛Ratio of their areas❜❜ :-

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

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

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

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

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

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

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

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

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

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

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

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

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

• Dear user if you are not able to see the diagrams from app. Please see it from the the site (brainly.in). It will be correctly displayed there.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

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