NECO 2024 · Paper 2 · Q8

Solid mensuration

A cone has a radius of $10\text{ cm}$ and a slant height of $26\text{ cm}$ . Find the: $\left[\pi = \frac{22}{7}\right]$

(i)
total surface area, correct to 1 decimal place;
(ii)
volume, correct to 2 decimal places;
(iii)
angle of the sector that formed the cone, correct to the nearest degree.

Worked solution (try it first)

(i)

Total surface area of a solid cone $= \pi r l + \pi r^2$
$= \pi r(l + r)$
$= \frac{22}{7} \times 10 \times 36$
$\approx 1131.4\text{ cm}^2$ .

(ii)

The height comes from Pythagoras: $h = \sqrt{26^2 - 10^2} = \sqrt{576} = 24$ cm.
Volume $= \frac13 \times \frac{22}{7} \times 10^2 \times 24$
$\approx 2514.29\text{ cm}^3$ .

(iii)

The cone was made from a sector of radius 26 cm (the slant height), whose arc became the base circumference: $\frac{\theta}{360} \times 2\pi \times 26 = 2\pi \times 10$ .
So $\theta = \frac{10}{26} \times 360^\circ$
$\approx 138^\circ$ .

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