2017 AMC 10B Problem 15

The diagonal is $AC = 5$ since $3$-$4$-$5$ is a right triangle. In right triangle $ABC$, the altitude $BE$ to the hypotenuse creates $\triangle ABE \sim \triangle ACB$ (both right, sharing angle $A$). Hence
$$\frac{AE}{AB} = \frac{AB}{AC} \quad\Longrightarrow\quad AE = \frac{AB^2}{AC} = \frac{9}{5}.$$

Now compare $\triangle AED$ with $\triangle ACD$. Both have a vertex at $D$, and their bases $AE$ and $AC$ lie along the same line, so they share the same height from $D$. Their areas are proportional to their bases:
$$[AED] = \frac{AE}{AC} \cdot [ACD] = \frac{9/5}{5} \cdot \frac{3 \cdot 4}{2} = \frac{9}{25} \cdot 6 = \frac{54}{25}.$$

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

An altitude to the hypotenuse splits a right triangle into two triangles similar to the original, giving the standard relation $AE \cdot AC = AB^2$. Once a point on a triangle's side is located as a fraction of that side, the area of a sub-triangle is the same fraction of the whole; no new heights are needed.

Coordinates: $A = (0,0)$, $B = (3,0)$, $C = (3,4)$, $D = (0,4)$. Then $E$ is $\frac{9}{5}$ along the unit direction $(\frac35, \frac45)$, so $E = (\frac{27}{25}, \frac{36}{25})$. Triangle $AED$ has base $AD = 4$ on the $y$-axis and height equal to the $x$-coordinate of $E$: area $\frac12 \cdot 4 \cdot \frac{27}{25} = \frac{54}{25}$.

• Using AE = 3 or computing BE = 12/5 (the altitude) and treating it as AE.
• Trying to find the height from $E$ to $AD$ by a second similar-triangle argument and making an arithmetic slip, when the base-ratio shortcut avoids it entirely.