A ball with a radius of 8 cm crossed by a plane. The distance from the center of the ball to this plane is 7 cm

A ball with a radius of 8 cm crossed by a plane. The distance from the center of the ball to this plane is 7 cm. Find the cross-sectional area of the ball by this plane.

Consider a triangle OAB. In the triangle OA = OB = 8 cm as the radii of the ball, then the triangle AOB is isosceles, and its height OO1 is the median of the triangle, and then O1A = O1B.

From the right-angled triangle OAO1, according to the Pythagorean theorem, we determine the length of the leg O1A.

O1A ^ 2 = OA ^ 2 – OO1 ^ 2 = 8 ^ 2 – 7 ^ 2 = 64 – 49 = 15.

О1А = √15 cm.

Determine the cross-sectional area.

S = n * R^2 = n * OA1^2 = n * 15 cm2.

Answer: The cross-sectional area is equal to n * 15 cm2.




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