In the diagram, |PT| = 4 cm, |TS| = 6 cm, |PQ| = 6 cm and < SPR = 30°. Calculate, correct to the nearest whole number: (a) |SR| ; (b) area of TQRS.

From the diagram, in triangle PSR the point T lies on PS and Q lies on PR, with TQ parallel to SR (shown by the arrows). Given \(|PT| = 4\text{ cm}\), \(|TS| = 6\text{ cm}\), \(|PQ| = 6\text{ cm}\) and \(\angle SPR = 30^\circ\).

Preliminary - use the parallel lines. Because \(TQ \parallel SR\), triangles PTQ and PSR are similar (equiangular). The full side PS is

\[|PS| = |PT| + |TS| = 4 + 6 = 10\text{ cm}.\]

The ratio of similarity is \(\dfrac{PT}{PS} = \dfrac{4}{10} = 0.4\), so

\[\frac{PQ}{PR} = 0.4 \;\Rightarrow\; |PR| = \frac{PQ}{0.4} = \frac{6}{0.4} = 15\text{ cm}.\]

(a) Finding |SR|. In triangle PSR use the Cosine Rule with \(|PS| = 10\), \(|PR| = 15\), \(\angle P = 30^\circ\):

\[|SR|^2 = |PS|^2 + |PR|^2 - 2|PS||PR|\cos 30^\circ.\]

\[|SR|^2 = 10^2 + 15^2 - 2(10)(15)\cos 30^\circ = 100 + 225 - 300(0.8660).\]

\[|SR|^2 = 325 - 259.8 = 65.2.\]

\[|SR| = \sqrt{65.2} = 8.07\ldots \approx 8\text{ cm}.\]

(b) Area of TQRS. TQRS is the trapezium left when the small triangle PTQ is removed from triangle PSR:

\[\text{Area of } TQRS = \text{Area of } \triangle PSR - \text{Area of } \triangle PTQ.\]

Using \(\text{Area} = \tfrac{1}{2}ab\sin C\) with the common angle \(30^\circ\):

\[\text{Area of } \triangle PSR = \tfrac{1}{2}(10)(15)\sin 30^\circ = \tfrac{1}{2}(150)(0.5) = 37.5\text{ cm}^2.\]

\[\text{Area of } \triangle PTQ = \tfrac{1}{2}(4)(6)\sin 30^\circ = \tfrac{1}{2}(24)(0.5) = 6\text{ cm}^2.\]

\[\text{Area of } TQRS = 37.5 - 6 = 31.5 \approx 32\text{ cm}^2.\]

\[\boxed{|SR| \approx 8\text{ cm},\qquad \text{Area of } TQRS \approx 32\text{ cm}^2.}\]