In the diagram, |PT| = 4 cm, |TS| = 6 cm, |PQ| = 6 cm

Explanation

\(|TQ|^{2} = 4^{2} + 6^{2} - 2 \times 4 \times 6 \times \cos 30°\)

= \(16 + 36 - 48 \times 0.8660\)

= \(52 - 41.568 = 10.432\)

\(\therefore |TQ| = \sqrt{10.432} \approxeq 3.23 cm\)

(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)