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?
Age
14 to 16
Challenge level
Problem
Image
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.
Student Solutions
Image
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 $\frac{3}{32}$ = $\frac{3}{8}$.