The diameter of the ball is 40. The plane is 16 from the center of the ball. Find the area of the ball’s section by the plane.

Let us draw from the center of the ball to the secant plane the segments OC and OD, which, in length, are equal to the radius of the ball.

OC = OD = AB / 2 = 40/2 = 20 cm.

Consider a right-angled triangle OO1C, whose angle O1 is straight, then, according to the Pythagorean theorem, the leg CO1 will be equal to:

CO1 ^ 2 = OC ^ 2 – OO1 ^ 2 = 20 ^ 2 – 16 ^ 2 = 400 – 256 = 144.

CO1 = 12 cm.

The radius of the section circle is 12 cm, then the section area will be equal to:

S = n * R2 = n * 144 cm2.

Answer: The cross-sectional area is 144 cm2.