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.
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)