As can be seen from the meaning of the question, the center of the ball is just on the plane where the BCD triangle is located (refer to Mu Zi's diagram), so it can be concluded that the radius of the ball is the radius of the circumscribed circle of the BCD triangle.

The base of the triangle BCD is 6*√2, and the height is 4 of the triangle in front view.

Bc 2 = CD 2 =18+16 = 34 is calculated by Pythagoras' law.

The formula 2r = BC * CD/4 =17/2r =17/4 is calculated by the circumscribed circle.

The surface area of the circumscribed sphere is S = 4π r 2 = 289/4π.