Rectangle $ABCD$ has $AB=3$ and $BC=4$. Point $E$ is the foot of the perpendicular from $B$ to diagonal $\overline{AC}$. What is the area of $\triangle AED$? $\textbf{(A)}\ 1\qquad$ $\textbf{(B)}\ \frac{42}{25}\qquad$ $\textbf{(C)}\ \frac{28}{15}\qquad$ $\textbf{(D)}\ 2\qquad$ $\textbf{(E)}\ \frac{54}{25}$
Solution
In the right triangle $ABC$ with altitude $BE$, $AE=\frac{AB^2}{AC}$ and $CE=\frac{BC^2}{AC}$ (theorem 7), so $AE:CE=AB^2:BC^2=9:16$.
Triangles $ADE$ and $ADC$ have the same height from $D$, so $[ADE]=\frac{9}{25}[ACD]=\frac9{25}\cdot6=\frac{54}{25}$.
Rectangle $ABCD$ has $AB=5$ and $BC=12$. Point $E$ is the foot of the perpendicular from $B$ to diagonal $\overline{AC}$. What is the area of $\triangle AED$?
