imc

Since quadrilateral $UVCM$ has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that $△AUV$ has $\frac{1}{4}$ the area of triangle $AMC$ by similarity, so $[UVCM]=\frac{3}{4}\cdot [AMC].$ Thus, $$\frac{1}{2}\cdot 12\cdot 12=\frac{3}{4}\cdot [AMC]$$ $$72=\frac{3}{4}\cdot [AMC]$$ $$[AMC]=96\to \boxed{96}.$$ Note: We know that since $AU$ is the median $AU$ divided by $AM$ is $1/2$. Since triangle $AUV$ is similar to triangle $AMC$, and by a factor of $1/2$, then every sidelength of $AMC$ must also be scaled by a factor of $1/2$ to get $AUV$. The base and height are all scaled by $1/2$, so then the ratio is $1/2\cdot 1/2$.