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 6sin beta)
= (frac{1}{2} (11.305) (6 times 0.9288))
= 31.501 cm(^2)
(approxeq) 32 cm(^2) (to the nearest whole number)