1992 - WAEC Mathematics Past Questions and Answers - page 2
The angle of a sector of a circle of radius 8cm is 240°. This sector is bent to form a cone. Find the radius of the base of the cone.
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
The curved surface area of a cylindrical tin is 704cm\(^2\). Calculate the height when the radius is 8cm. [Take π = 22/7]
Curved surface area of a cylindrical tin = \(2\pi rh\)
\(\therefore 2\pi rh = 704cm^2\)
\(2 \times \frac{22}{7} \times 8 \times h = 704\)
\(h = \frac{704 \times 7}{2 \times 22 \times 8}\)
\(h = 14cm\)
The volume of a cone of height 9cm is 1848cm\(^3\). Find its radius. [Take π = 22/7]
1/3 πr\(^2\) x 9 = 1848
3 x πr2 = 1848 r2 = 1848/3 x 7/22 r = 14
\(\text{Volume of a cone} = \frac{1}{3} \pi r^2 h\)
\(\frac{1}{3} \times \frac{22}{7} \times r^2 \times 9 = 1848\)
\(r^2 = \frac{1848 \times 7}{22 \times 3}\)
\(r^2 = 196 \therefore r = 14cm\)
The area of a parallelogram is 513cm\(^2\) and the height is 19cm. Calculate the base.
Area of parallelogram = base \(\times\) height.
\(513 = base \times 19 \implies base = \frac{513}{19}\)
= 27 cm
The diagram above shows the shaded segment of a circle of radius 7cm. if the area of the triangle OXY is 12\(\frac{1}{4}\)cm\(^2\), calculate the area of the segment
[Take π = 22/7]
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 = \(12\frac{5}{6} - 12\frac{1}{4}\)
= \(\frac{7}{12} cm^2\)
In the diagram above PQT is a tangent to the circle QRS at Q. Angle QTR = 48° and ∠QRT = 95°. Find ∠QRT
∠RQT = 180° - (95° + 48°) = 73°
∠OQR = 90° - 37° = 53°
∠QOR = 180° - (53° + 53°) = 74°
QSR = 74°/2 = 37°
In the diagram above, O is the center of the circle and ∠POR = 126°. Find ∠PQR
Reflex < POR = 360° - 126° = 234°
\(\therefore\) < PQR = \(\frac{1}{2} \times 234°\)
= 117°
∠PSR + ∠PQR = 180o, 120o + xo = 180o
x = 60o
In the diagram above, O is the center of the circle PQRS and ∠QPS = 36°. Find ∠QOS
< QOS = 2 < QPS (angle subtended at the centre)
\(\therefore\) < QOS = 2 x 36° = 72°
In the diagram above, O is the center of the circle QOR is a diameter and ∠PSR is 37°. Find ∠PRQ
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°