A square with a side of 6√2 cm is inscribed in a circle, find the side of a regular triangle circumscribed about this circle.

The diagonals of the square are equal, at the point of intersection they are divided in half and intersect at right angles.

Then in a right-angled triangle ABO, AB ^ 2 = AO ^ 2 + BO ^ 2.

(6 * √2) ^ 2 = 2 * OA ^ 2.

36 * 2 = 2 * OA ^ 2.

ОА = 6 cm.

OA is the radius of the circle in which the square is inscribed.

Apply the formula for the radius of a circle inscribed in a regular triangle.

R = OA = KP * √3 / 6.

КР = ОА * 6 / √3 = 6 * 6 / √3 = 36 * √3 / 3 = 12 * √3 cm.

Answer: The length of the side of the triangle is 12 * √3 cm.