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SSC CGL Model Paper 7

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Question : 65 of 100

Marks: +1, -0

A circle is inscribed in an equilateral triangle and a square is inscribed in the circle. The ratio of the area of the square to the area of the equilateral triangle is?

$\frac{2\sqrt{3}}{3}$
$\frac{2}{\sqrt{3}}$
$\frac{2}{3\sqrt{3}}$
$\frac{2\sqrt{3}}{\sqrt{7}}$

Solution:

Let the side of the ΔABC be 'a' unit

Inradius,

r = \frac{a}{2\sqrt{3}}

And area of triangles =

\frac{\sqrt{3}}{4a^2}

→ Then diameter of the circle (PR) =

2 \times \frac{a}{2\sqrt{3}} = \frac{a}{\sqrt{3}}

→ Since, PR is the diagonals of the square PQRS

Then, side of the square

= \frac{a}{\sqrt{3} \times \sqrt{2}} = \frac{a}{\sqrt{6}}

Required Ratio

= \frac{\text{Area of square}}{\text{Area of equilateral trianlge}}

= \frac{ \left(\frac{a}{\sqrt{6}}\right)^2 }{ \frac{\sqrt{3}}{4a^2} } = \frac{2}{3\sqrt{3}}

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