2016 AMC 10A Problem 11

Put the rectangle on a grid with corners $(0,0)$ and $(8,5)$. From the labels, one slanted segment runs from $(1,5)$ on the top edge to $(7,0)$ on the bottom edge, and the other runs from $(0,4)$ on the left edge to $(8,1)$ on the right edge. The shaded region is a bow-tie: a piece in the top-left corner and a piece in the bottom-right corner, touching where the two segments cross.

1. Find the crossing point. Spin the rectangle half a turn about its center $(4, 2.5)$. The point $(1,5)$ lands on $(7,0)$ and $(0,4)$ lands on $(8,1)$, so each slanted segment lands on itself. A segment that is unchanged by a half turn about a point must pass through that point. So both segments go through the center, and they cross at $M = (4, 2.5)$.
2. The two pieces are equal. The same half turn carries the top-left shaded piece onto the bottom-right one. So we only need the top-left piece and then double it.
3. The top-left piece has corners $(0,4)$, $(0,5)$, $(1,5)$ and $M = (4, 2.5)$. Cut it along the line from the corner $(0,5)$ to $M$ into two triangles:
- Triangle $(0,5), (1,5), M$: its base along the top edge has length $1$, and $M$ is $5 - 2.5 = 2.5$ below that edge. Area $= \tfrac12 \cdot 1 \cdot 2.5 = 1.25$.
- Triangle $(0,5), (0,4), M$: its base along the left edge has length $1$, and $M$ is $4$ to the right of that edge. Area $= \tfrac12 \cdot 1 \cdot 4 = 2$.

So the top-left piece has area $1.25 + 2 = 3.25$.
4. The shaded region is two such pieces: $2 \cdot 3.25 = 6.5 = 6\tfrac12$.

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

The labels $1, 7$ on top and bottom and $1, 4$ on the sides are arranged the same way after a half turn, which tells you the whole figure has that symmetry. Symmetry finds the crossing point for free and cuts the work in half. Splitting an odd shape into triangles whose bases lie along the rectangle's edges makes every height easy to read.

Use a parallelogram. The four endpoints $(0,4)$, $(1,5)$, $(8,1)$, $(7,0)$ form a parallelogram (the short sides both go $1$ right and $1$ up), and the two slanted segments are its diagonals. Its area is the rectangle minus four corner triangles: $40 - 14 - 14 - \tfrac12 - \tfrac12 = 11$. The diagonals cut a parallelogram into four triangles of equal area, and the bow-tie holds two of them, which is $\tfrac{11}{2}$, plus the two small corner triangles of area $\tfrac12$ each. Total: $5.5 + 1 = 6.5$.