12. In isosceles $\triangle A M C$, $A M=A C$, the median on the legs $M V=C U=12$, and $M V \perp C U$ at point $P$. Then the area of $\triangle A M C$ is ( ).
(A) 48
(B) 72
(C) 96
(D) 144
(E) 192
12. C.

As shown in Figure 4, connect $U V$. Then
$$
\begin{array}{l}
\triangle A M C \backsim \triangle A U V \\
\Rightarrow \frac{S_{\triangle A U V}}{S_{\triangle A M C}}=\frac{1}{4} \\
\Rightarrow S_{\text {quadrilateral UMCV }} \\
=\frac{3}{4} S_{\triangle A M C} \\
=\frac{1}{2} M V \cdot C U=72 \\
\Rightarrow S_{\triangle A M C}=96 . \\
\end{array}
$$