\(|TQ|^{2} = 4^{2} + 6^{2} - 2 \times 4 \times 6 \times \cos 30^\circ\) = \(16 + 36 - 48 \times 0.8660\) = \(52 - 41.568 = 10.432\) \(\therefore |TQ| = \sqrt{10.432} \approxeq 3.23 cm\) (a) By the rules of similar triangles, \(\frac{|PT|}{|TQ|} = \frac{|PS|}{|SR|}\) \(\frac{4}{3.23} = \frac{10}{|SR|}\) \(|SR| = \frac{3.23 \times 10}{4}\) = \(8.075 cm \approxeq 8 cm\) (b) Using sine rule, \(\frac{|PQ|}{\sin \alpha} = \frac{|TQ|}{\sin 30}\) \(\frac{6}{\sin \alpha} = \frac{3.23}{\sin 30}\) \(\sin \alpha = \frac{6 \sin 30}{3.23}\) sin \(\alpha\) = 0.9288 From the diagram, \(\alpha\) = \(\beta\) \(\sin \alpha = \sin \beta = 0.9288\) \(\sin \beta = \frac{h}{6}\) h = \(6 \sin \beta\) Hence, area of quadrilateral TQRS = \(\frac{1}{2} (TQ + SR) \times h\) = \(\frac{1}{2} (3.23 + 8.075) \times 6\sin \beta\) = \(\frac{1}{2} (11.305) (6 \times 0.9288)\) = 31.501 cm\(^2\) \(\approxeq\) 32 cm\(^2\) (to the nearest whole number)