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A circle is inscribed in an equilateral triangle of side a. what is the area of any square inscribed in this circle?

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\frac{a^{2}}{6}

\frac{a^{2}}{5}

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

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

answer is A.

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Detailed Solution

Given,
A side of equilateral triangle = a
Assume that
In the equilateral

△ ABC

the length of each side = a
Then, radius of the circle = r
Now calculated the area of triangle

△ ABC = \frac{\sqrt{3}}{4} a^{2}

And also, semi perimeter of a triangle is

△ ABC = \frac{3 a}{2}

Now calculated the radius of the circle

r = \frac{Area}{semi - perimeter}

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

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

Now calculate the area of the square
Diagonal of a square PQRS = diameter in circle

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

Thus,

= \frac{(diagonal)^{2}}{2}

= \frac{{(\frac{a}{\sqrt{3}})}^{2}}{2}

= \frac{a^{2}}{6}

Hence, the area of any square inscribed in the circle is

\frac{a^{2}}{6} .

Correct option is 1.

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