2017 AMC 10B Problem 15

Triangle $AED$ has its base $AE$ lying along the diagonal $AC$. So if we know what fraction of $AC$ the piece $AE$ is, we get the area as that fraction of triangle $ACD$.

1. The diagonal. Triangle $ABC$ is a right triangle with legs $3$ and $4$, so $AC = 5$.
2. Find $AE$. Triangles $ABE$ and $ACB$ both have a right angle (at $E$ and at $B$) and both have the angle at $A$. So they have the same shape (they are similar). Matching sides:
$$\frac{AE}{AB} = \frac{AB}{AC}, \qquad \frac{AE}{3} = \frac{3}{5}, \qquad AE = \frac{9}{5}.$$
3. What fraction of $AC$? $\frac{AE}{AC} = \frac{9/5}{5} = \frac{9}{25}$.
4. Compare areas. Triangles $AED$ and $ACD$ both have their third corner at $D$, and their bases $AE$ and $AC$ lie on the same line. So they have the same height from $D$, and their areas are in the same ratio as their bases.
5. Triangle $ACD$ is half the rectangle: $\frac12 \cdot 3 \cdot 4 = 6$.
6. So the area of $AED$ is $\frac{9}{25} \cdot 6 = \frac{54}{25}$.

The answer is $\boxed{\textbf{(E)}\ \frac{54}{25}}$.

Dropping a perpendicular from the right-angle corner to the long side of a right triangle makes two smaller triangles with the same shape as the big one, which gives lengths like $AE$ quickly. Then, when two triangles share a corner and their bases lie on one line, their areas compare just like their bases, so no new height has to be found.

Use coordinates: $A = (0, 0)$, $B = (3, 0)$, $C = (3, 4)$, $D = (0, 4)$. The point $E$ is on $AC$ at distance $\frac95$ from $A$, which is $\frac{9}{25}$ of the way to $C$: $E = \left(\frac{9}{25} \cdot 3,\ \frac{9}{25} \cdot 4\right) = \left(\frac{27}{25}, \frac{36}{25}\right)$. Triangle $AED$ has base $AD = 4$ along the $y$-axis, and its height is how far $E$ is from that axis, $\frac{27}{25}$. Area: $\frac12 \cdot 4 \cdot \frac{27}{25} = \frac{54}{25}$.