17-133 The perimeter of $\triangle A B C$ is $24, M$ is the midpoint of $A B, M C=M A=5$. Then the area of $\triangle A B C$ is
(A) 12 .
(B) 16 .
(C) 24 .
(D) 30 .
(China Junior High School Mathematics League, 1999)
$$
\begin{array}{lll}
\text { [Solution] } \because & M A=M B=M C=5, \\
\therefore & & \angle A C B=90^{\circ} .
\end{array}
$$

Given the perimeter is 24, then $A C+B C=14, A C^{2}+$
$$
\begin{array}{l} 
B C^{2}=10^{2} . \\
\therefore \quad 2 A C \cdot B C= \\
=14^{2}-10^{2}=4 \times 24 . \\
\therefore \quad S_{\triangle A B C}=\frac{1}{2} A C \cdot B C=24 .
\end{array}
$$

Therefore, the answer is $(C)$.