A circle is inscribed in an equilateral trian | vedclass

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(C) Let $p$ be the altitude of the equilateral triangle. Then $p = a \sin 60^{\circ} = \frac{a\sqrt{3}}{2}$.Since the triangle is equilateral,the cent

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

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(C) Let $p$ be the altitude of the equilateral triangle. Then $p = a \sin 60^{\circ} = \frac{a\sqrt{3}}{2}$.Since the triangle is equilateral,the centroid,orthocentre,circumcentre,and incentre all coincide.Therefore,the radius $r$ of the inscribed circle is $\frac{1}{3}p = \frac{1}{3} 	imes \frac{a\sqrt{3}}{2} = \frac{a}{2\sqrt{3}}$.The diameter of the circle is $D = 2r = \frac{a}{\sqrt{3}}$.Let $x$ be the side length of the square inscribed in the circle. The diagonal of the square is equal to the diameter of the circle.Thus,$x^2 + x^2 = D^2$,which gives $2x^2 = D^2$.Substituting $D = \frac{a}{\sqrt{3}}$,we get $2x^2 = (\frac{a}{\sqrt{3}})^2 = \frac{a^2}{3}$.Therefore,the area of the square is $x^2 = \frac{a^2}{6}$.

$A$ circle is inscribed in an equilateral triangle of side $a$. The area of any square inscribed in the circle is

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