(a) The area of trapezium PQRS is 60\(cm^{2}\). PQ... | WAEC SSCE - General Mathematics - 2011 (Essay)

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(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?

Answer

(a)(i) Perpendicular height. For a trapezium, area \(=\tfrac{1}{2}(\text{sum of parallel sides})\times h\).

\[60=\tfrac{1}{2}(15+25)h=20h\Rightarrow h=3\text{ cm}\]

Height \(=3\) cm.

(ii) |PS|. \(PS\) is the slant side at \(S\), where \(\angle PSR=60^{\circ}\). The perpendicular height is the vertical component of \(PS\):

\[h=|PS|\sin 60^{\circ}\Rightarrow |PS|=\frac{h}{\sin 60^{\circ}}=\frac{3}{0.8660}\approx 3.46\text{ cm}\]

|PS| \(\approx 3.46\) cm.

(b) Let the total sum be \(T\). Ade takes \(\tfrac{3}{5}T\); remainder \(=\tfrac{2}{5}T\).

Nelly takes \(\tfrac{1}{3}\) of the remainder \(=\tfrac{1}{3}\times\tfrac{2}{5}T=\tfrac{2}{15}T\).

Austin takes the rest of the remainder \(=\tfrac{2}{5}T-\tfrac{2}{15}T=\tfrac{6}{15}T-\tfrac{2}{15}T=\tfrac{4}{15}T\).

Austin's share exceeds Nelly's by \(N3{,}000\):

\[\tfrac{4}{15}T-\tfrac{2}{15}T=\tfrac{2}{15}T=3000\Rightarrow T=3000\times\tfrac{15}{2}=22500\]

Ade's share \(=\tfrac{3}{5}\times22500=N13{,}500\).

Ade got \(N13{,}500.00\).

Read lesson note on Mensuration (JAMB)

Read lesson note on Areas (WAEC)

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