The ball is crossed by a plane 8 cm from its center. The area of the resulting section is 125 pi cm ^ 2

The ball is crossed by a plane 8 cm from its center. The area of the resulting section is 125 pi cm ^ 2. Find the radius of the ball.

Knowing the cross-sectional area of the ball, we determine the radius of the ball cross-section.

Ssection = n * r ^ 2.

125 * n = n * r ^ 2.

r ^ 2 = 125.

r = 5 * √5 cm2.

From the center of the ball we draw segments OA and OB, which are equal in length to the radius of the ball OA = OB = R.

The segment OO1 is the distance between the centers of the ball and the circle in the section and is equal, by condition, 8 cm.

Triangle AOO1 is rectangular, then by the Pythagorean theorem we determine the radius of the ball.

ОВ ^ 2 = R ^ 2 = OO1 ^ 2 + r ^ 2 = 8 ^ 2 + (5 * √5) ^ 2 = 64 + 125 = 189.

R = √189 = 3 * √21 cm.

Answer: The radius of the ball is 3 * √21 cm.