To solve the problem, we compute the area of regular pentagon $ABCDE$ in two different ways. First, we can divide regular pentagon $ABCDE$ into five congruent triangles.

If $s$ is the side length of the regular pentagon, then each of the triangles $AOB$, $BOC$, $COD$, $DOE$, and $EOA$ has base $s$ and height 1, so the area of regular pentagon $ABCDE$ is $5s/2$.
Next, we divide regular pentagon $ABCDE$ into triangles $ABC$, $ACD$, and $ADE$.

Triangle $ACD$ has base $s$ and height $AP = AO + 1$. Triangle $ABC$ has base $s$ and height $AQ$. Triangle $ADE$ has base $s$ and height $AR$. Therefore, the area of regular pentagon $ABCDE$ is also
\[\frac{s}{2} (AO + AQ + AR + 1).\]
Hence,
\[\frac{s}{2} (AO + AQ + AR + 1) = \frac{5s}{2},\]
which means $AO + AQ + AR + 1 = 5$, or $AO + AQ + AR = \boxed{4}$. The answer is $\boxed{(C)}$.
$ABCDE$ is a regular pentagon. $AP, AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD, CB$ extended and $DE$ extended, 
respectively. Let $O$ be the center of the pentagon. If $OP = 1$, then $AO + AQ + AR$ equals

$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 1 + \sqrt{5}\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 2 + \sqrt{5}\qquad \textbf{(E)}\ 5$