# Quarters

Weekly Problem 27 - 2014

Four congruent isosceles trapezia are placed in a square. What fraction of the square is shaded?

Four congruent isosceles trapezia are placed in a square. What fraction of the square is shaded?

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Four congruent isoceles trapeziums are placed so that their longer parallel sides form the diagonals of a square PQRS, as shown. The point X divides PQ in the ratio 3:1. What fraction of the square is shaded?

If you liked this problem, here is an NRICH task that challenges you to use similar mathematical ideas.

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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, âˆ QZY = âˆ ZQX = 45 °.

Also, as YX is parallel to ZQ, âˆ XYW = âˆ WXY = 45 °. So WYX and WZQ are both isosceles right-angled triangles. As âˆ 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 $\frac{3}{32}$ = $\frac{3}{8}$.