1992 - WAEC Mathematics Past Questions and Answers - page 2

L = θ/360 x 2πR = 240/360 x 2 x 22/7 x 8/1 ... (1)

This must be equal to the circumference of the circle which is 2πr = 44R/7 .... (2)

equate (1) and (2)

r = 16/3 = 51/3

Curved surface area of a cylindrical tin = \(2\pi rh\)

\(\therefore 2\pi rh = 704cm^2\)

\(2 \times \frac{22}{7} \times 8 \times h = 704\)

\(h = \frac{704 \times 7}{2 \times 22 \times 8}\)

\(h = 14cm\)

1/3 πr\(^2\) x 9 = 1848
3 x πr2 = 1848 r2 = 1848/3 x 7/22 r = 14

\(\text{Volume of a cone} = \frac{1}{3} \pi r^2 h\)

\(\frac{1}{3} \times \frac{22}{7} \times r^2 \times 9 = 1848\)

\(r^2 = \frac{1848 \times 7}{22 \times 3}\)

\(r^2 = 196 \therefore r = 14cm\)

Area of parallelogram = base \(\times\) height.

\(513 = base \times 19 \implies base = \frac{513}{19}\)

= 27 cm

Area of \(\Delta OXY\) = \(12\frac{1}{4} cm^2\)

Area of sector OXY = \(\frac{30}{360} \times \frac{22}{7} \times 7 \times 7\)

= \(\frac{77}{6} = 12\frac{5}{6} cm^2\)

Area of the shaded portion = \(12\frac{5}{6} - 12\frac{1}{4}\)

= \(\frac{7}{12} cm^2\)

∠RQT = 180° - (95° + 48°) = 73°
∠OQR = 90° - 37° = 53°
∠QOR = 180° - (53° + 53°) = 74°
QSR = 74°/2 = 37°

Reflex < POR = 360° - 126° = 234°

\(\therefore\) < PQR = \(\frac{1}{2} \times 234°\)

= 117°

∠TRS = ∠RST = 60^o
∠PSR + ∠PQR = 180^o, 120^o + x^o = 180^o
x = 60^o

< QOS = 2 < QPS (angle subtended at the centre)

\(\therefore\) < QOS = 2 x 36° = 72°

Construction: Join PQ.

Then < RSP = 37° = < RQP (angles on the same segment)

But < RPQ = 90° (angle in a semi-circle)

\(\therefore\) < PRQ = 180° - (90° + 37°)

 = 53°