Using ruler and a pair of compasses only, (a) construct \(\Delta PQR\) such that |PQ| = 7 cm, |PR| = 6 cm and < PQR = 60°. (b) locate point M, the mid-point...

Answer Details

This question tests accurate ruler-and-compasses construction and the ability to read a required measurement from the finished figure. The three given facts, \(|PQ| = 7\) cm, \(\angle PQR = 60^\circ\) and \(|PR| = 6\) cm, fix the triangle, and the interesting result is the size of \(\angle RMQ\).

Construction steps (ruler and compasses only):

1. Draw a base line and mark off \(|PQ| = 7\) cm.
2. At \(Q\), construct an angle of \(60^\circ\) using the equilateral-triangle arc method (an arc from \(Q\) cutting \(QP\), then the same radius stepped once round it), and draw the ray \(QR\).
3. With centre \(P\) and radius \(6\) cm, draw an arc to cut the \(60^\circ\) ray at \(R\). Join \(PR\) and \(QR\) to complete \(\Delta PQR\).
4. (b) Bisect \(PQ\) with equal arcs from \(P\) and \(Q\) above and below the line; the bisector meets \(PQ\) at its mid-point \(M\), so \(|PM| = |MQ| = 3.5\) cm. Join \(RM\).
5. (c) Place the protractor at \(M\) and read \(\angle RMQ\).

The accurately drawn figure:

(c) Measurement: \(\angle RMQ = 60^\circ\).

Why the answer is exactly \(60^\circ\). When the \(60^\circ\) ray at \(Q\) is drawn and an arc of radius \(6\) cm is swung from \(P\), the point \(R\) lands where \(PR\) is perpendicular to \(QR\), so \(\angle PRQ = 90^\circ\). In a right-angled triangle the mid-point of the hypotenuse is equidistant from all three vertices, so

Since \(|MQ| = |MR|\), triangle \(MQR\) is isosceles and \(\angle MRQ = \angle MQR = \angle PQR = 60^\circ\). The three angles of triangle \(MQR\) must sum to \(180^\circ\), so

(In fact \(|QR| = 7\cos 60^\circ = 3.5\) cm as well, so \(MQR\) is equilateral, which confirms \(\angle RMQ = 60^\circ\).)

Examination reminder: the marks are awarded for the visible construction arcs (the \(60^\circ\) at \(Q\) and the perpendicular bisector of \(PQ\)) and for a measured value within about \(\pm 1^\circ\) of the true figure. A reading of \(90^\circ\) is a sign that \(RM\) was mistaken for the perpendicular bisector rather than the line to \(R\); measure the angle between \(MR\) and \(MQ\) at \(M\), and the correct value is \(60^\circ\).