Finding the area of a triangle using fractions? 
 To find the area of the triangle do you use Pythagorean theorem from what you have? Could this use similar triangles.
 A: Draw the line $QT$. Note that the area of $\triangle PQT$ is $\frac{3}{4}$ times the area of the big triangle $PQR$: same height, base $\frac{3}{4}$ as big.
Note also that the area of $\triangle PST$ is $\frac{2}{3}$ times the area of $\triangle PQT$. The argument is essentially the same. The two triangles have the same base, and the height of $\triangle PST$ is $\frac{2}{3}$ times the height of $\triangle PQT$.   
Thus the desired ratio of areas is $\frac{3}{4}\cdot \frac{2}{3}=\frac{1}{2}$.
Remark: A more detailed analysis shows that we do not need to know that the line $QV$ is perpendicular to $PR$. The only advantage of perpendicularity is that it makes the fact that the height of $\triangle PST$ is $\frac{2}{3}$ times the height of $\triangle PQT$ obvious. But the relationship holds even without perpendicularity. 
A: The area of $\triangle PST$ is the sum of the areas of $\triangle PSV$ and $\triangle VST$.
$$\begin{align}
\text{The area of }\triangle PSV &= \frac12 PV\cdot SV\\
&=\frac12 PV\cdot\frac23 QV\\
&=\frac13 PV\cdot QV,
\end{align}$$
$$\begin{align}
\text{and the area of }\triangle VST &= \frac12VT\cdot SV\\
&=\frac12 VT\cdot \frac23QV\\
&=\frac13VT\cdot QV.
\end{align}$$
\begin{align}
\text{So the area of }\triangle PST &= \frac13QV(PV+VT)\\
&=\frac13QV\cdot PT\\
&=\frac13QV\cdot\frac34PR\\
&=\frac14QV\cdot PR.
\end{align}
The area of $\triangle PQR = \frac12QV\cdot PR$.
\begin{align}
\text{So the ratio }\frac{\triangle PST}{\triangle PQR} &= \frac{\frac14 QV\cdot PR}{\frac12QV\cdot PR}\\
&=\frac{\frac14}{\frac12}\\
&=\frac12.
\end{align}
