The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$. Altitude $\overline{XC}$ bisects $\overline{YZ}$. The area (in square inches) of the shaded region is

$\textbf{(A)}\ 1\frac{1}2\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2\frac{1}2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 3\frac{1}2$
The shaded region is a right trapezoid. Assume WLOG that $YZ=8$. Then because the area of $\triangle XYZ$ is equal to 8, the height of the triangle $XC=2$. Because the line $AB$ is a midsegment, the top base of the trapezoid is $\frac12 AB = \frac14 YZ = 2$. Also, $AB$ divides $XC$ in two, so the height of the trapezoid is $\frac12 (2) = 1$. The bottom base is $\frac12 YZ = 4$. The area of the shaded region is $\frac12 (2+4)(1) = \boxed{\text{(D)}\ 3}$.