(a) Using a ruler and a pair of compasses only, construct a parallelogram PQRS with diagonals |PR| = 9cm and |QS| = 6cm, intersecting at K and < QKR = 60°. ...

(a) Constructing \(PQRS\)

The key fact is that the diagonals of a parallelogram bisect each other. Therefore, at their intersection \(K\),

\[ PK=KR=\frac{9}{2}=4.5\text{ cm},\qquad QK=KS=\frac{6}{2}=3\text{ cm}. \]

1. Draw \(PR=9\text{ cm}\) and construct its midpoint \(K\).
2. At \(K\), construct an angle of \(60^\circ\) with \(KR\) as one arm.
3. Mark \(Q\) on this ray so that \(KQ=3\text{ cm}\).
4. Extend \(QK\) through \(K\) and mark \(S\) so that \(KS=3\text{ cm}\).
5. Join \(P\) to \(Q\), \(Q\) to \(R\), \(R\) to \(S\), and \(S\) to \(P\).

(b) Constructing the equal-area rectangle \(PABS\)

Use \(PS\) as the base of the rectangle. The area of parallelogram \(PQRS\) is

\[ \text{area} = PS \times \text{perpendicular height}. \]

Drop a perpendicular from \(Q\) to \(PS\), meeting \(PS\) at \(H\). The length \(QH\) is the perpendicular height of the parallelogram. Construct a perpendicular to \(PS\) at \(P\), on the same side of \(PS\) as \(Q\), and transfer the length \(QH\) onto it to locate \(A\). Thus, \(PA=QH\). Draw a line through \(A\) parallel to \(PS\), and a line through \(S\) parallel to \(PA\); they meet at \(B\).

This makes

\[ \text{area of }PABS=PS\times PA=PS\times QH=\text{area of }PQRS. \]

For the measurement, the parallelogram area is

\[ \frac{1}{2}(9)(6)\sin 60^\circ =27\sin60^\circ \approx 23.38\text{ cm}^2. \]

Also, in triangle \(PKS\), the angle \(PKS\) is \(60^\circ\), not \(120^\circ\). Therefore,

\[ PS^2=4.5^2+3^2-2(4.5)(3)\cos60^\circ =20.25+9-13.5 =15.75, \]

\[ PS\approx 3.97\text{ cm}. \]

Hence

\[ PA=\frac{23.38}{3.97}\approx 5.89\text{ cm}. \]

\(\boxed{PA\approx 5.9\text{ cm}}\)

The earlier value of \(3.6\text{ cm}\) results from using \(120^\circ\) for \(\angle PKS\). Since \(KP\) and \(KS\) are separated by \(60^\circ\), the correct angle is \(60^\circ\).