WAEC 2009 · Paper 2 · Q4✱

Solid mensuration

A sector of angle $135^\circ$ is cut from a thin circular metal sheet of radius 40 cm. The sector is then folded, with its straight edges coinciding, to form a right circular cone. [Take $\pi = \frac{22}{7}$ ]

(a)
Calculate the base radius of the cone, correct to two decimal places.
(b)
Calculate the greatest volume of liquid which the cone can hold, correct to the nearest $\text{cm}^3$ .

Worked solution (try it first)

(a)

The arc of the sector becomes the circumference of the base, and the radius of the sheet becomes the slant height $l = 40$ cm.
Arc length $= \frac{135}{360} \times 2\pi \times 40$
$= 30\pi$ cm.
Set it equal to $2\pi r$ : $2\pi r = 30\pi$ , so $r = 15.00$ cm.

(b)

Height by Pythagoras: $h = \sqrt{40^2 - 15^2}$
$= \sqrt{1375}$
$= 37.081$ cm.
Volume $= \frac13\pi r^2 h$
$= \frac13 \times \frac{22}{7} \times 15^2 \times 37.081$ .
$= 8740.5$ , so the cone holds $8741\text{ cm}^3$ to the nearest $\text{cm}^3$ .

Report a problem with this question