WAEC 2020 · Paper 2 · Q4

Solid mensuration

The total surface area of a cone of slant height $l\text{ cm}$ and base radius $r\text{ cm}$ is $224\pi\text{ cm}^2$ . If $r : l = 2 : 5$ , find: $\left[\text{Take }\pi = \frac{22}{7}\right]$

(a)
correct to one decimal place, the value of $r$ ;
(b)
correct to the nearest whole number, the volume of the cone.

Worked solution (try it first)

(a)

$r : l = 2 : 5$ , so $l = \frac52 r$ .
Total surface area $= \pi r(r + l)$
$= \pi r\left(r + \frac52 r\right)$
$= \frac72\pi r^2$ .
So $\frac72\pi r^2 = 224\pi$ .
The $\pi$ cancels: $r^2 = 64$ and $r = 8.0$ cm.

(b)

$l = \frac52 \times 8 = 20$ cm, so $h = \sqrt{20^2 - 8^2}$
$= \sqrt{336}$
$\approx 18.33$ cm.
Volume $= \frac13 \times \frac{22}{7} \times 8^2 \times 18.33$
$\approx 1229\text{ cm}^3$ .

Report a problem with this question