Question

An equilateral triangle is circumscribed, a square is inscribed in a circle of radius r. The area of triangle is T and the area of square is S, then $\frac{T}{S}$ is

• A. $\frac{3\sqrt{3}}{2}$
• B. $3\sqrt{3}$
• C. $27$
• D. $\frac{9\sqrt{3}}{2}$

Answer 1

Answer: (A)

$\frac{3\sqrt{3}}{2}$

$\Rightarrow Area△ABC=3\times ar△AOC$

$\frac{\sqrt{3}}{4}{a}^2=3\times \frac{1}{2}\times a\times r$

$a=\frac{3\times 4\times r}{2\sqrt{3}}$

$a=2\sqrt{3}r$

ar $△ABC=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}\times {\left(2\sqrt{3}r\right)}^2=\frac{\sqrt{3}\times 4\times 3\times r^2}{4}$

$T=3\sqrt{3}{r}^2$

Diagonal of square $=$ Dia. of circle $=2r$

Side of square $=d/\sqrt{2}=\frac{2r}{\sqrt{2}}=\sqrt{2}r$

Area of square, $S={\left(\sqrt{2}r\right)}^2={2r}^2$

$\frac{T}{S}=\frac{3\sqrt{3}{r}^2}{2r^2}=\frac{3\sqrt{3}}{2}$