Theory question 109
ConstructionsUsing a ruler and a pair of compasses only, construct (a) triangle QRT with |QR| = 8cm, |RT| = 5cm and |QT| = 4.5cm (b) a quadrilateral QRSP which has a common base QR with ΔQRT such that QTP is a straight line, PQ||SR, |QP| = 9cm and |RS| = 4.5cm (i) Measure |PS| (ii) Find the perpendicular distance between RS and PQ. (iii) What is QRSP?
Show the model answer
(a) Draw QR = 8 cm. With centre Q radius 4.5 cm and centre R radius 5 cm draw arcs meeting at T; join QT and RT.
(b) Produce QT to P with QP = 9 cm. At R construct a line parallel to QP (copy ∠RQP at R on the same side) and mark S on it with RS = 4.5 cm. Join PS.
(i) |PS| ≈ 5.0 cm (TPSR is a parallelogram, so PS = TR = 5 cm).
(ii) Drop a perpendicular from R to QP: about 4.5 cm (calculated 4.55 cm).
(iii) QRSP is a trapezium (one pair of parallel sides, PQ ∥ SR).
Notice TP = 9 − 4.5 = 4.5 cm = RS, which makes TPSR a parallelogram — a useful check on the measurement.