A regular hexagon and a square are inscribed in a circle. The area of the hexagon is 24 √3. Find the area of the square.

The large diagonals of a regular hexagon divide it into six equal, equilateral triangles. Then Swam = S6 / 6 = 24 * √3 / 6 = 4 * √3 cm2.

Through the area of a regular triangle, we determine the length of its side.

Swam = ОВ ^ 2 * √3 / 4.

ОВ ^ 2 = 4 * Svom / √3 = 4 * 4 * √3 / √3 = 16.

ОВ = √16 = 4 cm.

In a right-angled triangle AOB, AO = OB = 4 cm, then AB ^ 2 = OA ^ 2 + OB ^ 2 = 16 + 16 = 32.

AB = 4 * √2 cm.

Then Savsd = (4 * √2) ^ 2 = 32 cm2.

Answer: The area of the square is 32 cm2.