WAEC 2022 · Paper 2 · Q4

(a)
The volume of a cylinder of radius $14\text{ cm}$ is $9240\text{ cm}^3$ . Calculate the curved surface area of the cylinder. $\left[\text{Take }\pi = \frac{22}{7}\right]$
(b)
In the diagram, $PQR$ is a triangle with $\angle Q = n$ ; $y$ is the exterior angle at $P$ and $x$ the exterior angle at $R$ (the base $PR$ is produced both ways). If $x + y = 210^\circ$ , find the value of $n$ .
The paper’s diagram is not drawn to scale; this redraw uses one possible triangle.

Worked solution (try it first)

Find the height from the volume: $\frac{22}{7} \times 14^2 \times h = 9240$ , so $616h = 9240$ and $h = 15$ cm.
Curved surface area $= 2\pi rh$
$= 2 \times \frac{22}{7} \times 14 \times 15$
$= 1320\text{ cm}^2$ .

An exterior angle and its interior angle make $180^\circ$ : the interior angles at $P$ and $R$ are $180^\circ - y$ and $180^\circ - x$ .
The angles of the triangle add up to $180^\circ$ : $(180^\circ - y) + (180^\circ - x) + n = 180^\circ$ .
So $n = x + y - 180^\circ$
$= 210^\circ - 180^\circ$
$= 30^\circ$ .