Question

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

• A. \( \frac{2a^2}{3} \)
• B. \( \frac{\sqrt{3}a^2}{2} \)
• C. \( \frac{a^2}{2\sqrt{3}} \)
• D. \( \frac{a^2}{6} \)

Hint:

When solving problems involving circles inscribed in triangles, use the known formula for the radius of the incircle and the relationship between the diagonal of the square and the radius to find the area.

Answer 1

The Correct Option is D

We are given that a circle is inscribed in an equilateral triangle of side \( a \), and we need to find the area of any square inscribed in this circle. ### Step 1: Radius of the inscribed circle. For an equilateral triangle with side length \( a \), the radius \( r \) of the inscribed circle (incircle) is given by the formula: \[ r = \frac{a \sqrt{3}}{6}. \] This formula is derived from the relationship between the area of the equilateral triangle and its semiperimeter. ### Step 2: Area of the inscribed square. Now, we need to find the area of the square inscribed in the circle. The diagonal of the square is equal to the diameter of the circle, which is twice the radius: \[ \text{Diagonal of the square} = 2r = \frac{a \sqrt{3}}{3}. \] For a square, the diagonal \( d \) and the side length \( s \) are related by the Pythagorean theorem: \[ d = s \sqrt{2}. \] Thus, the side length of the square \( s \) is: \[ s = \frac{d}{\sqrt{2}} = \frac{a \sqrt{3}}{3\sqrt{2}} = \frac{a \sqrt{6}}{6}. \] ### Step 3: Area of the square. The area \( A \) of the square is the square of its side length: \[ A = s^2 = \left( \frac{a \sqrt{6}}{6} \right)^2 = \frac{a^2}{6}. \] Thus, the area of the inscribed square is: \[ \boxed{\frac{a^2}{6}}. \]