2. 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
(E) 16
(C) 24
(D) 30
2. (C)
$$
\begin{array}{l}
\because M A=M B=M C=5, \\
\therefore \angle A C B=90^{\circ} .
\end{array}
$$

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