Find the cross-sectional area of a ball with a radius of 41 cm by a plane drawn at a distance of 29 cm

Find the cross-sectional area of a ball with a radius of 41 cm by a plane drawn at a distance of 29 cm from the center of the ball.

O is the center of the circle, AB is the diameter of the resulting section. OH is the distance from the center O to the section, OH is perpendicular to AB.

Consider a triangle ОВН: angle Н = 90 °, ОН = 29 cm, ОВ = 41 (since this is the radius). We calculate the length of the НВ using the Pythagorean theorem:

HB = √ (BO² – OH²) = √ (41² – 29²) = √ (1681 – 841) = √840 cm.

Find the cross-sectional area (this is a circle with a radius of √840 cm).

S = nR² = 840p.