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(a) The area of trapezium PQRS is 60\(cm^{2}\). PQ // RS, /PQ/ = 15 cm, /RS/ = 25 cm ...

(a) The area of trapezium PQRS is 60\(cm^{2}\). PQ // RS, /PQ/ = 15 cm, /RS/ = 25 cm and < PSR = 60°. Calculate the : (i) perpendicular height of PQRS ; (ii) |PS|.
(b) Ade received \(\frac{3}{5}\) of a sum of money, Nelly \(\frac{1}{3}\) of the remainder while Austin took the rest. If Austin's share is greater than Nelly's share by N3,000, how much did Ade get?
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    Correct Answer: Option n
    Explanation:



    (a)
    \(\theta = \frac{60°}{2} = 30°\)
    Area of trapezium = \(\frac{a + b}{2} h\)
    \(60 = \frac{15 + 25}{2} h \implies 60 = 20h\)
    \(h = \frac{60}{20} = 3 cm\)
    (ii) Considering \(\Delta PSR\),
    \(\sin \theta = \frac{Opp}{Hyp}\)
    \(\sin 30 = \frac{3}{|PS|}\)
    \(|PS| = \frac{3}{\sin 30}\)
    = \(\frac{3}{0.5} = 6 cm\)
    (b) Let the total amount to be shared = k.
    Ade's share = \(\frac{3}{5} k \)
    Remainder = \(k - \frac{3}{5}k \)
    = \(\frac{2}{5}k\)
    Nelly's share : \(\frac{1}{3} \times \frac{2}{5}k\)
    = \(\frac{2}{15}k\)
    Austin's share : \(\frac{2}{5}k - \frac{2}{15}k\)
    = \(\frac{4}{15}k\)
    \(\frac{4}{15}k - \frac{2}{15}k = \frac{2}{15}k\)
    \(\frac{2}{15}k = N3,000\)
    \(\implies k = \frac{3,000 \times 15}{2}\)
    = \(N22,500\)
    Ade's share = \(\frac{3}{5}k\)
    = \(\frac{3}{5} \times N22,500 = N13,500\).

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