(a) In the diagram, PQSR and SRYZ are parallelograms and PQYZ is a straight line. If /QY/ = 2cm and /RS/ = 3cm, find /PZ/. (b) P and Q are two towns on the ...

(a) Finding /PZ/

The diagram shows the parallelograms PQSR and SRYZ sharing the common side SR, with the points P, Q, Y, Z lying on one straight line in that order.

In a parallelogram opposite sides are equal, so:

• In parallelogram PQSR, side \(PQ\) is opposite side \(SR\), hence \(PQ = SR = 3\text{ cm}\).
• In parallelogram SRYZ, side \(ZY\) is opposite side \(SR\), hence \(ZY = SR = 3\text{ cm}\).

Since \(P, Q, Y, Z\) are collinear in that order:

\[ PZ = PQ + QY + YZ \]

\[ PZ = 3 + 2 + 3 = 8\text{ cm} \]

/PZ/ = 8 cm.

(b) Distance PQ along the parallel of latitude 56\(^{\circ}\)N

Both towns lie on latitude \(56^{\circ}\)N. Their longitudes are \(25^{\circ}\)E and \(95^{\circ}\)E, so the difference in longitude is:

\[ \theta = 95^{\circ} - 25^{\circ} = 70^{\circ} \]

The radius of the parallel of latitude is \(r = R\cos 56^{\circ}\), and the required distance is an arc of that parallel:

\[ PQ = \frac{\theta}{360^{\circ}} \times 2\pi R\cos 56^{\circ} \]

Substituting \(R = 6400\text{ km}\), \(\pi = \frac{22}{7}\) and \(\cos 56^{\circ} = 0.5592\):

\[ PQ = \frac{70}{360} \times 2 \times \frac{22}{7} \times 6400 \times 0.5592 \]

\[ PQ = \frac{70}{360} \times 40228.57 \times 0.5592 \]

\[ PQ = 0.19444 \times 40228.57 \times 0.5592 \approx 4374\text{ km} \]

Distance PQ \(\approx\) 4374 km (to the nearest km).