Construction of Triangle PQR and the Required Loci

Given: Construct a triangle PQR such that

• PQ = 8 cm,
• PR = 7 cm,
• ∠QPR = 105°.

(a)(i) Construction of ΔPQR

1. Using a ruler, draw a line segment PQ = 8 cm.
2. At point P, construct an angle of 105°:
   • Construct a 90° angle at P by drawing a perpendicular to PQ.
   • Construct a 120° ray at P using equal arcs with the compasses.
   • Bisect the angle between the 90° ray and the 120° ray. The bisected angle is 105°.
3. On the 105° ray, use the compasses to mark a point R such that PR = 7 cm.
4. Join R to Q. This gives the required triangle ΔPQR.

(a)(ii) Locus L₁ of points equidistant from P and Q

The locus of points equidistant from P and Q is the perpendicular bisector of PQ.

1. With the compasses opened to more than half of PQ, draw arcs centred at P above and below PQ.
2. Without changing the compass width, draw arcs centred at Q to intersect the first arcs.
3. Join the two arc-intersection points with a ruler.

This straight line is labelled L₁.

(a)(iii) Locus l₂ of points equidistant from Q and R

The locus of points equidistant from Q and R is the perpendicular bisector of QR.

1. Draw equal arcs centred at Q and R, using a radius greater than half of QR.
2. Join the two points where the arcs intersect.

This straight line is labelled l₂.

(b)(i) Point of Intersection

Label the point where L₁ and l₂ intersect as T.

Conclusion: The intersection point T is the centre of the circle passing through P, Q and R.