Question

What is the ratio of an area of a circle and equilateral triangle when diametre and height are equal

Answer 1

Solution :-

Given :

Diameter of a circle = Height of the equilateral triangle

⇒ d = (√3a)/2

Radius r = d/2 = (√3a)/4

⇒ r = (√3a)/4

⇒ r/a = √3/4

Ratio of area of a circle and equilateral triangle = Area of circle : Area of equilateral triangle

$= \dfrac{\pi {r}^{2} }{ \dfrac{ \sqrt{3} {a}^{2} }{4} }$

$= \dfrac{4\pi {r}^{2} }{\sqrt{3} {a}^{2}}$

$= \dfrac{4\pi }{\sqrt{3} } \times \dfrac{ {r}^{2} }{ {a}^{2} }$

$= \dfrac{4\pi }{\sqrt{3} } \times \bigg(\dfrac{r }{ a } \bigg)^{2}$

$= \dfrac{4\pi }{\sqrt{3} } \times \bigg(\dfrac{ \sqrt{3} }{4} \bigg)^{2}$

$= \dfrac{4\pi }{\sqrt{3} } \times \dfrac{ {( \sqrt{3}) }^{2} }{4^{2} }$

$= \dfrac{ \sqrt{3} \pi }{4 }$

= √3 π : 4

Hence, the ratio of an area of a circle and equilateral triangle is √3 π : 4

Answer 2

${\large\bf{\mid{\overline{\underline{Given:-}}}\mid}}$

• Diameter of circle = Height of Equaliteral ∆

.

$\Large\underline\mathfrak{Question}$

To Find ratio of Area of circle and Equaliteral ∆ .

$\Large\bold\star\underline{\underline\textbf{Formula\:used}}$

• Radius of circle = Diameter /2
• Height of Equaliteral ∆ with side a is = √3a/2
• Area of circle = πr² (where r is radius)
• Area of Equaliteral ∆ = √3a²/4 .

$\Large\underline{\underline{\sf{Solution}:}}$

• $since \: it \: is \: given \: that \\ \\ \textbf{Diameter of circle = Height of Equaliteral triangle} \\ \\ \red\leadsto \: 2r = \frac{ \sqrt{3} a}{2} \\ \\ \red\leadsto \: \large\boxed{\bold{r = \frac{ \sqrt{3} a}{4} }} \\ \\ now \: required \: ratio:--- \: \\ \\ \pi {r}^{2} : \: \frac{ \sqrt{3} }{4} {a}^{2} \\ \\ \red\leadsto \: \pi( \frac{ \sqrt{3} a}{4} ) ^{2} : \: \frac{ \sqrt{3} }{4} {a}^{2} \\ \\ \red\leadsto \: \pi \frac{ \sqrt{3} \times \cancel{\sqrt{3} {a}^{2}} }{4 \times \cancel{4}} \: : \: \frac{ \cancel{\sqrt{3} }}{ \cancel4} \cancel{{a}^{2}} \\ \\ \red\leadsto \: \frac{\pi \sqrt{3} }{4} :1 \\ \\ \red\leadsto \: \pink{\large\boxed{\boxed{\bold{\pi \sqrt{3}:4 }}}}$

Hence the required ratio of Area of circle and Equaliteral ∆ is =√3π:4..

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$\large\underline\textbf{Hope it Helps You.}$