Note that $\Delta OPU \sim \Delta OQT$. Hence, we have
$\begin{array}{rcl}
\dfrac{\text{the area of $\Delta OPU$}}{\text{the area of $\Delta OQT$}} & = & \left(\dfrac{OP}{OQ}\right)^2 \\
\dfrac{\text{the area of $\Delta OPU$}}{\text{the area of $\Delta OQT$}} & = & \left(\dfrac{1}{2}\right)^2 \\
\dfrac{\text{the area of $\Delta OPU$}}{\text{the area of $\Delta OQT$}} & = & \dfrac{1}{4}
\end{array}$
Note that $\Delta OPU \sim \Delta ORS$. Hence, we have
$\begin{array}{rcl}
\dfrac{\text{the area of $\Delta OPU$}}{\text{the area of $\Delta ORS$}} & = & \left(\dfrac{OP}{OR}\right)^2 \\
\dfrac{\text{the area of $\Delta OPU$}}{\text{the area of $\Delta ORS$}} & = & \left(\dfrac{1}{3}\right)^2 \\
\dfrac{\text{the area of $\Delta OPU$}}{\text{the area of $\Delta ORS$}} & = & \dfrac{1}{9}
\end{array}$
Therefore, we have $\text{the area of $\Delta OPU$}:\text{the area of $\Delta OQT$} : \text{the area of $\Delta ORS$} = 1 : 4 : 9$.
Let $\text{the area of $\Delta OPU$} = k$, $\text{the area of $\Delta OQT$} = 4k$ and $\text{the area of $\Delta ORS$} = 9k$, where $k \neq 0$.
$\begin{array}{rcl}
\dfrac{\text{the area of trapezium $PQTU$}}{\text{the area of trapezium $QRST$}} & = & \dfrac{\text{the area of $\Delta OQT$} – \text{the area of $\Delta OPU$}}{\text{the area of $\Delta ORS$} – \text{the area of $\Delta OQT$}} \\
\dfrac{\text{the area of trapezium $PQTU$}}{\text{the area of trapezium $QRST$}} & = & \dfrac{4k – k}{9k – 4k} \\
\dfrac{\text{the area of trapezium $PQTU$}}{\text{the area of trapezium $QRST$}} & = & \dfrac{3}{5} \\
\end{array}$
