$24 \cdot 8$ When a lake froze, a ball was floating on it. After the ball was removed (without breaking the ice), a hole with a top diameter of $24 \mathrm{~cm}$ and a depth of $8 \mathrm{~cm}$ was left on the ice surface. What is the radius of the ball?
(A) $8 \mathrm{~cm}$.
(B) $12 \mathrm{~cm}$.
(C) $13 \mathrm{~cm}$.
(D) $8 \sqrt{3} \mathrm{~cm}$.
(E) $6 \sqrt{6} \mathrm{~cm}$.
(38th American High School Mathematics Examination, 1987)
[Solution] Let the radius of the sphere be $R$. On the great circle of the sphere, using the intersecting chords theorem, we have $(2 R-8) \cdot 8=12 \cdot 12$, solving for $R$ gives $R=13$. Therefore, the answer is $(C)$.