2007 AMC 10A Problem 18

The cross is made of $4 \times 4$ squares, so a grid makes every point easy to name.

1. Place the points. Put $H$ at $(0, 0)$. From the figure, $A$ is straight above $H$ at $(0, 12)$, $B = (4, 12)$, $C = (4, 8)$ and $G = (4, 0)$. So $A, B, G, H$ are the corners of a $4$ by $12$ rectangle, and $C$ is on its right side.
2. Line $AG$. It goes from $(0, 12)$ down to $(4, 0)$: it drops $12$ while moving $4$ to the right, so it drops $3$ for each $1$ to the right. Its equation is $y = 12 - 3x$.
3. Line $HC$. It goes from $(0, 0)$ to $(4, 8)$: it rises $2$ for each $1$ to the right. Its equation is $y = 2x$.
4. Where they cross. $2x = 12 - 3x$ gives $5x = 12$, so $x = \frac{12}{5}$ and $y = \frac{24}{5}$. That is $M$.
5. Split the area. Triangle $ABG$ is half the rectangle: $\frac12 \times 4 \times 12 = 24$. The quadrilateral $ABCM$ is this triangle with triangle $CMG$ removed.
6. Triangle $CMG$. Its base $CG$ runs along the line $x = 4$ with length $8$. Its height is the sideways distance from $M$ to that line: $4 - \frac{12}{5} = \frac{8}{5}$. So its area is
$$\frac12 \times 8 \times \frac85 = \frac{32}{5}.$$
7. Subtract.
$$24 - \frac{32}{5} = \frac{120}{5} - \frac{32}{5} = \frac{88}{5}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{88}{5}}$.

A shape with an awkward corner like $M$ is easier as "a simple triangle minus a small triangle". The only hard part is locating $M$, and coordinates turn that into solving one small equation. On a figure built from squares, a grid is almost always a good first step.

Use similar triangles instead of coordinates. Segments $AH$ and $CG$ are both vertical, so triangles $AMH$ and $GMC$ have the same shape, one pointing each way through $M$. Their matching sides are $AH = 12$ and $GC = 8$, a ratio of $3 : 2$. So the sideways distances from $M$ to line $AH$ and to line $CG$ are also in ratio $3 : 2$. Those distances add to $4$, so $M$ is $\frac85$ from line $CG$. Then finish with steps 5 to 7 above.