If a circle is inscribed in an equilateral triangle of side a, then the area of any square (in sq. units) inscribed in this circle is

• A

  $\frac{2 a^2}{3}$
• B

  $\sqrt{3} \frac{a^2}{2}$
• C

  $\frac{a^2}{2 \sqrt{3}}$
• Correct answer
  D

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

Solution

Area of $\triangle A B C(\Delta)=\frac{\sqrt{3}}{4} a^2$, Semi-perimeter of $\triangle A B C(s)=\frac{3 a}{2}$ Radius (r) $=\frac{\Delta}{s}=\frac{a}{2 \sqrt{3}}$ Diagonal of square PQRS $=2 \times \frac{a}{2 \sqrt{3}}=\frac{a}{\sqrt{3}}$ Area of square $=\frac{(\text { diagonal })^2}{2}=\frac{a^2}{6}$

Hence, the correct answer is option: 'D'.