In rhombus $ABCD$, point $P$ lies on segment $\overline{AD}$ so that $\overline{BP}$ $\perp$ $\overline{AD}$, $AP = 3$, and $PD = 2$. What is the area of $ABCD$? (Note: The figure is not drawn to scale.)

$\textbf{(A) }3\sqrt 5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }6\sqrt 5 \qquad \textbf{(D) }20\qquad \textbf{(E) }25$
\[\textbf{Figure redrawn to scale.}\]
$AD = AP + PD = 3 + 2 = 5$.
$ABCD$ is a rhombus, so $AB = AD = 5$.
$\bigtriangleup APB$ is a $3-4-5$ right triangle, hence $BP = 4$.
The area of the rhombus is base times height: $bh = (AD)(BP) = 5 \cdot 4 = \boxed{\textbf{(D) }20}$.
~richiedelgado