For example, the perimeter of $\angle S A B C$ 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
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\begin{array}{c}
\because B M=M A= \\
M C=5, \\
\therefore \angle 1=\angle A, \\
\angle 2=\angle B . \\
\text { So } \angle 1+\angle 2 \\
=90^{\circ} . \\
\therefore A C^{2}+B C^{2}=A B^{2}=100 . \\
\text { Also, since } B B+A C+B C=24, \\
\therefore A C+B C=14, \\
(A C+B C)^{2}=14^{2}, \\
2 A C \cdot B C=196-\left(A C^{2}+B C^{2}\right)=96 . \\
\therefore S_{\triangle A B C}=\frac{1}{2} A C \cdot B C=24 .
\end{array}
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Solution: As shown in Figure 1,

The correct choice is (C).
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