GK 2015 Q22
22. In trapezium \(PQRS\), the sides \(PQ\) and \(SR\) are parallel. Angle \(RSP\) is 120° and \(PS = SR = \tfrac{1}{3}PQ\). What is the size of angle \(PQR\)?
- A 15°
- B 22.5°
- C 25°
- D 30°
- E 45°
Grey Kangaroo 2015, Question 22. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D 30°
The diagram shows the trapezium described with points \(X\) and \(Y\) added on \(PQ\) so that \(PX = XY = YQ = PS = SR\). Since angle \(RSP = 120°\), angle \(SPQ = 60°\) using co-interior angles and so triangle \(SPX\) is equilateral. Similarly, it can easily be shown that triangles \(SXR\) and \(RXY\) are also equilateral. In triangle \(RYQ\) we then have \(RY = YQ\) and angle \(RYQ = 120°\) using angles on a straight line adding to 180°. Hence triangle \(RYQ\) is isosceles and so angle \(YQR = \tfrac{1}{2}(180° - 120°) = 30°\). Therefore angle \(PQR\) is 30°.
Solution © UK Mathematics Trust, from the GK 2015 solutions.