12. A convex quadrilateral $A B C D$ is such that $A B=14, B C=24, C D=26$ and $D A=16$. Its area is 360. Find the area of the region in the plane inside the quadrilateral consisting of points that are at most 3 units away from its perimeter.
(A) 100
(B) 120
(C) $120+9 \pi$
(D) 156
(E) 200

## Numerical answer problems -5 points
12. The answer is $(\mathbf{E})$. The solution to the problem is based on two key considerations. First, noting that $A B+C D=B C+D A=40$, it follows that the quadrilateral $A B C D$ is circumscribed around a circle. In particular, if $r$ is the radius of this circle, then $r=\frac{360}{40}=9$.

Let $I$ be the center of the circle. Applying a homothety centered at $I$ with a factor of $\frac{2}{3}$, the quadrilateral is mapped to $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, which is circumscribed around a circle with center $I$ and radius 6. We then observe that the sides of $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ are exactly 3 units away from those of $A B C D$, so a point inside $A B C D$ is at most 3 units away from the perimeter if and only if it does not lie inside $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$.

To conclude, $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ has an area of $\left(\frac{2}{3}\right)^{2} \cdot 360=160$, so the answer is $360-160=200$.