In the diagram, $A B C D$ is a rectangle and point $E$ lies on $A B$. Triangle $D E C$ has $\angle D E C=90^{\circ}, D E=3$ and $E C=4$. The length of $A D$ is
(A) 2.6
(D) 1.8
(B) 2.4
(C) 2.8
(E) 3.2

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-118.jpg?height=244&width=371&top_left_y=203&top_left_x=1381)
Solution 1

By the Pythagorean Theorem, $D C^{2}=D E^{2}+E C^{2}$ so $D C=5$.

Draw a line from $E$ to point $P$ on $D C$ so that $E P$ is perpendicular to $D C$. Since $A B C D$ is a rectangle, then $A D=E P$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-230.jpg?height=295&width=436&top_left_y=1278&top_left_x=1430)

Then the area of triangle $D E C$ is equal to $\frac{1}{2}(D E)(E C)$ and also to $\frac{1}{2}(D C)(E P)$.

So

$$
\begin{aligned}
\frac{1}{2}(3)(4) & =\frac{1}{2}(5)(E P) \\
12 & =5(A D) \\
A D & =\frac{12}{5}=2.4
\end{aligned}
$$

## Solution 2

By the Pythagorean Theorem, $D C^{2}=D E^{2}+E C^{2}$

so $D C=5$.

Then $\sin (\angle E D C)=\frac{E C}{D C}=\frac{4}{5}$.

But $\angle A E D=\angle E D C$ since $A B$ and $D C$ are parallel,

so $\frac{4}{5}=\sin (\angle A E D)=\frac{A D}{E D}=\frac{A D}{3}$, and so $A D=\frac{12}{5}=2.4$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-231.jpg?height=290&width=440&top_left_y=294&top_left_x=1409)

Answer: (B)