2016 AMC 10A Problem 11

Place the rectangle with corners $(0,0)$ and $(8,5)$. Reading the labels, the shaded region is bounded by the segment from $(1,5)$ to $(7,0)$, the segment from $(0,4)$ to $(8,1)$, and the two short $1$-unit pieces of the rectangle's boundary at the top-left and bottom-right corners.

Look at the quadrilateral $W = (0,4),\,(1,5),\,(8,1),\,(7,0)$. Its sides $(0,4)\to(1,5)$ and $(7,0)\to(8,1)$ are both the vector $(1,1)$, so $W$ is a parallelogram, and the two slanted segments are exactly its diagonals. The diagonals of a parallelogram split it into four triangles of equal area, and the shaded region contains two opposite ones, i.e. half of $W$, together with the two small right triangles at the corners $(0,5)$ and $(8,0)$, each with legs $1$ and $1$.

Area of $W$: subtract the four unshaded corner triangles from the $8\times 5$ rectangle:
$$40 - \tfrac12\cdot 7\cdot 4 - \tfrac12\cdot 7\cdot 4 - \tfrac12\cdot 1\cdot 1 - \tfrac12\cdot 1\cdot 1 = 40 - 29 = 11.$$

Shaded area $= \tfrac12\cdot 11 + \tfrac12 + \tfrac12 = 6\tfrac12$.

The answer is $\boxed{\textbf{(D)}\ 6\dfrac{1}{2}}$.

A region bounded by two crossing segments is a bow-tie, not a simple polygon, so decompose it rather than applying a formula blindly. Recognizing the two segments as diagonals of a parallelogram converts the awkward shape into "half a parallelogram," and the parallelogram itself is a rectangle minus four right triangles. Always check whether slanted segments in a figure are parallel or equal; that structure is usually the intended shortcut.

Coordinates: the segments $y = 5 - \tfrac56(x-1)$ and $y = 4 - \tfrac38 x$ meet at $(4, 2.5)$, the center of the rectangle. Shoelace on the quadrilateral $(4,2.5),(0,4),(0,5),(1,5)$ gives $3.25$; the other half is congruent by the $180^\circ$ symmetry about the center, so the total is $6.5$.

• Treating the shaded figure as one simple hexagon and running shoelace around its six listed corners, which gives 0 because the boundary crosses itself.
• Forgetting the two small $1\times 1$ corner triangles and answering $5\tfrac12$, or including the whole parallelogram and getting $12$.