2018 AMC 10A Problem 9

The trapezoid is what remains of $\triangle ABC$ after removing $\triangle ADE$, so we need the area of $\triangle ADE$.

Along $\overline{DE}$ sit four of the small triangles with their bases on $DE$ (the other three point downward and fill the gaps). Those four bases exactly cover $DE$, so $DE$ is $4$ times the base of a small triangle.

Since $\triangle ADE$ is similar to each small triangle (all triangles in the figure are similar to $\triangle ABC$), its area is the small area scaled by $4^2$:
$$[ADE] = 16 \cdot 1 = 16.$$

Therefore
$$[DBCE] = [ABC] - [ADE] = 40 - 16 = 24.$$

The answer is $\boxed{\textbf{(E)}\ 24}$.

For similar figures, area scales with the square of the length ratio. The only measurement needed is the ratio of $DE$ to a small base, and the picture hands it over: four small triangles stand shoulder to shoulder along $DE$. Everything else (the total area $40$) is given, so the trapezoid is a subtraction.

Count pieces: the seven small triangles form a strip of area $7$ whose top edge is $3$ small bases long. The triangle above the strip is the small triangle scaled by $3$, area $9$. So $[ADE] = 7 + 9 = 16$ and the trapezoid is $40 - 16 = 24$.

• Adding up the seven small triangles and stopping (area 7) or scaling the area by 4 instead of by 4^2.
• Trying to read the ratio $DE : BC$ from the picture, which is not drawn to scale; only the counted bases are reliable.