5. In Rt $\triangle A B C$, $\angle B=60^{\circ}, \angle C=$ $90^{\circ}, A B=1$, equilateral triangles $\triangle A B R$, $\triangle B C P$, and $\triangle C A Q$ are constructed outward on $A B$, $B C$, and $C A$ respectively. Line $Q R$ intersects $A B$ at point $T$. Then the area of $\triangle P R T$ is equal to ( ).
(A) $\frac{9 \sqrt{3}}{32}$
(B) $\frac{\sqrt{3}}{4}$
(C) $\frac{1}{2}$
(D) $\frac{\sqrt{3}}{3}$
5. A.

As shown in Figure 4, connect $P Q$. From the given conditions, we have
$$
\begin{array}{l}
B C=\frac{1}{2}, A C=\frac{\sqrt{3}}{2}, \angle Q A T=90^{\circ}, \\
\angle Q C P=150^{\circ}, P, B, R \text { are collinear. }
\end{array}
$$

Since $S_{\triangle A Q T}$
$$
\begin{array}{l}
=\frac{1}{2} A T \cdot A Q \\
=\frac{1}{2} A T \cdot A C \\
=\frac{\sqrt{3}}{4} A T,
\end{array}
$$

and $\frac{S_{\triangle A R T}}{S_{\triangle A R B}}=\frac{A T}{A B}$, therefore,
$$
S_{\triangle A R T}=\frac{\sqrt{3}}{4} A T=S_{\triangle A Q T} .
$$

Thus, $Q T=R T$. Hence,
$$
\begin{array}{l} 
S_{\triangle P R T}=\frac{1}{2} S_{\triangle P Q R} \\
= \frac{1}{2}\left(S_{\triangle A B C}+S_{\triangle A B R}+S_{\triangle B C P}+\right. \\
\left.S_{\triangle C I Q}+S_{\triangle C P Q}-S_{\triangle A Q R}\right) \\
= \frac{1}{2}\left(\frac{\sqrt{3}}{8}+\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{16}+\frac{3 \sqrt{3}}{16}+\frac{\sqrt{3}}{16}-\frac{\sqrt{3}}{8}\right) \\
= \frac{9 \sqrt{3}}{32} .
\end{array}
$$