$\frac{5}{12}$

Let $AB$ and $AD$ be of length $b$ and $h$ respectively. Then
the area of $ABCD = bh$ and the area of $$ABQP =
\frac{1}{2}b\left(\frac{h}{2} + \frac{h}{3}\right)=
\frac{5bh}{12}$$

Thus the required ratio is $\frac{5}{12}$.

*This problem is taken from the UKMT Mathematical Challenges.**View the archive of all weekly problems grouped by curriculum topic*

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