Using ruler and a pair of compasses only, (a) construct a rhombus PQRS of side 7 cm and < PQR = 60°; (b) locate point X such that X lies on the locus of poi...

Answer Details

Construction and loci

1. Draw \(QR=7\text{ cm}\).
2. With centres \(Q\) and \(R\), and radius \(7\text{ cm}\), draw arcs intersecting at \(P\). This gives \(QP=QR=7\text{ cm}\) and \(\angle PQR=60^\circ\).
3. With centres \(P\) and \(R\), again using radius \(7\text{ cm}\), draw arcs intersecting at \(S\) (choose the intersection other than \(Q\)). Join \(PS\) and \(RS\). All four sides are \(7\text{ cm}\), so \(PQRS\) is a rhombus.
4. The locus of points equidistant from the lines \(PQ\) and \(QR\) is the angle bisector of \(\angle PQR\). Construct the internal angle bisector.
5. The locus of points equidistant from points \(Q\) and \(R\) is the perpendicular bisector of \(QR\).
6. The intersection of these loci is \(X\).

Finding \(|XR|\)

The angle bisector makes an angle of \(30^\circ\) with \(QR\). The perpendicular bisector meets \(QR\) at its midpoint, so the horizontal distance from \(X\) to \(R\) is \(3.5\text{ cm}\).

Using coordinates with \(Q=(0,0)\) and \(R=(7,0)\), the perpendicular bisector is \(x=3.5\). At this point on the \(30^\circ\) angle bisector,

\[ y=3.5\tan 30^\circ=\frac{7}{2\sqrt3}. \] \[ |XR|=\sqrt{3.5^2+\left(\frac{7}{2\sqrt3}\right)^2} =\sqrt{\frac{49}{3}} =\frac{7}{\sqrt3} \approx 4.04\text{ cm}. \]

Therefore, \(|XR|\approx 4.0\text{ cm}\) when measured to the nearest millimetre.

The supplied reference answer is inconsistent with the question: it gives \(|XP|=3.8\text{ cm}\), whereas the question asks for \(|XR|\). In the exact construction, \(|XP|=|XR|\approx4.04\text{ cm}\), so \(3.8\text{ cm}\) is not consistent with the stated dimensions.

Examination reminder: “Equidistant from two lines” means an angle bisector; “equidistant from two points” means a perpendicular bisector.