$20 \cdot 46$ In $\triangle A B C$, $A C=24, B C=10, A B=26$, then the radius $r$ of the inscribed circle of $\triangle A B C$ is
(A) 26.
(B) 4.
(C) 13.
(D) 0.
(E) None of the above answers is correct.
(1st American High School Mathematics Examination, 1950)
[Solution 1] From the given and $26^{2}=24^{2}+10^{2}$, we know that $\triangle A B C$ is a right triangle. From $A B+2 r=A C+B C$, we get $r=4$.
Therefore, the answer is $(B)$.
[Solution 2] From Solution 1, we know that $\triangle A B C$ is a right triangle. Then
$$
\begin{aligned}
& \frac{1}{2}(A B+B C+C A) \cdot r=\frac{1}{2} A C \cdot B C, \\
\therefore \quad & r=\frac{A C \cdot B C}{A B+B C+C A}=\frac{24 \cdot 10}{26+10+24}=4 .
\end{aligned}
$$

Therefore, the answer is $(B)$.