The diagram shows the top-right-hand portion of the square.

The shaded trapezium is labelled QXYZ and W is the point at which ZY produced meets PQ.

As QXYZ is an isosceles trapezium, ∠QZY = ∠ZQX = 45°.

Also, as YX is parallel to ZQ, ∠XYW = ∠WXY = 45°. So WYX and WZQ are both isosceles right-angled triangles. As ∠ZWQ = 90° and Z is at centre of square PQRS, we deduce that W is the midpoint of PQ. Hence WX = XQ = $\frac{1}{4}$PQ. So the ratio of the side-lengths of similar triangles WYX and WZQ is 1:2 and hence the ratio of their areas 1:4.

Therefore the area of trapezium QXYZ = $\frac{3}{4}$ x area of triangle ZWQ = $\frac{3}{32}$ x area PQRS since triangle ZWQ is one-eighth of PQRS. So the fraction of the square which is shaded is 4 x $\frac{3}{32}$ = $\frac{3}{8}$.

*This problem is taken from the UKMT Mathematical Challenges.*