$20 \cdot 85$ is inscribed in a circle within a rhombus whose diagonals are 10 and 24. The radius of the circle is 
(A)4.
(B) $\frac{58}{13}$.
(C) $\frac{60}{13}$.
(D) 5 .
(E)6.
[Solution] As shown in the figure, let the diagonals of rhombus $ABCD$ be $AC=24$, and $BD=10$. The radius of the inscribed circle $O$ is $r$.
Since $AC \perp BD$, we have
\[
\begin{array}{l}
S_{ABCD}=\frac{1}{2} AC \cdot BD=\frac{1}{2} \times 24 \times 10=120, \\
\text { Also, } S_{ABCO}=4 S_{\triangle AOD}=4 \cdot \frac{1}{2} \sqrt{AO^2+OD^2} \cdot r \\
=2 \sqrt{12^2+5^2} r=26 r .
\end{array}
\]

Thus, $26 r=120$, solving for $r$ gives $r=\frac{60}{13}$. Therefore, the answer is $(C)$.