Volume & Surface Areas Questions and Answers

Answer: C

Spin a right-angled triangle about one of its perpendicular sides. That side stays fixed as the axis, the other perpendicular side sweeps out a flat circular disc at the bottom, and the hypotenuse sweeps out a slanted surface meeting at a point on the axis.

The result is a right circular cone, with the height equal to the axis side, the base radius equal to the other perpendicular side, and the slant height equal to the hypotenuse.

Revolving a rectangle instead would give a cylinder, and revolving about the hypotenuse would give two cones joined base to base.

Answer: C

Form the ratio and let the \( \pi \) cancel.

\[ \frac{S}{V} = \frac{4\pi r^{2}}{\tfrac{4}{3}\pi r^{3}} = \frac{3}{r} \]

\[ \frac{3}{r} = \frac{1}{18} \quad\Longrightarrow\quad r = 54 \text{ cm} \]

The ratio has units of inverse length, which is why the answer 1/18 cm gives a radius in centimetres. Reading the relation as \( r/3 \) instead of \( 3/r \) gives 6 cm - the second distractor.

Answer: A

A pyramid is called a right pyramid when its apex sits directly above the centre of its base, so the axis meets the base at a right angle.

When the apex is displaced sideways, the axis leans and the pyramid is oblique.

The other options name pyramids by the shape of their base - triangular, pentagonal, hexagonal - which is a separate classification altogether. A hexagon-based pyramid can be either right or oblique.

Answer: C

Only the water is wanted, so use its own height of 11 cm rather than the tank's 18 cm.

\[ V = 21 \times 13 \times 11 = 3003 \text{ cm}^{3} \]

Since \( 1000 \text{ cm}^{3} = 1 \) litre:

\[ V = \frac{3003}{1000} = 3.003 \text{ litres} \]

Using the full 18 cm depth would give 4.914 litres - the tank's capacity, not the water it holds.

Answer: D

Solve for the radius, taking \( \pi = \tfrac{22}{7} \).

\[ \frac{4}{3}\pi r^{3} = 4851 \]

\[ r^{3} = \frac{4851 \times 3 \times 7}{4 \times 22} = \frac{101871}{88} = 1157.625 \]

\[ r = \sqrt[3]{1157.625} = 10.5 \text{ cm} \]

\[ d = 2r = 21 \text{ cm} \]

The question asks for the diameter, so the radius of 10.5 cm must be doubled - stopping at the radius is the intended slip.

Answer: C

Unrolled flat, a cone's curved surface is a sector of a circle of radius \( l \) whose arc is the base circumference \( 2\pi r \). Its area is

\[ \text{CSA} = \frac{1}{2} \times \text{arc} \times \text{radius} = \frac{1}{2} \times 2\pi r \times l = \pi r l \]

The total surface area adds the circular base:

\[ \text{TSA} = \pi r l + \pi r^{2} = \pi r (l + r) \]

Note that the slant height \( l \) is used here, not the perpendicular height \( h \); the two are related by \( l^{2} = r^{2} + h^{2} \).

Answer: A

Melting conserves volume, and the two shapes share the same radius \( r = 3 \) cm.

\[ V_{\text{sphere}} = \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi (27) = 36\pi \text{ cm}^{3} \]

\[ V_{\text{cylinder}} = \pi r^{2} h = 9\pi h \]

\[ 9\pi h = 36\pi \quad\Longrightarrow\quad h = 4 \text{ cm} \]

Whenever a sphere is recast as a cylinder of the same radius, the height is always \( \tfrac{4}{3}r \) - here \( \tfrac{4}{3} \times 3 = 4 \).

Answer: D

A solid hemisphere has two surfaces: the curved dome and the flat circular base.

\[ \text{curved surface} = 2\pi r^{2}, \qquad \text{flat base} = \pi r^{2} \]

\[ \text{total surface area} = 2\pi r^{2} + \pi r^{2} = 3\pi r^{2} \]

The \( 2\pi r^{2} \) option is the curved surface alone - correct for a hollow bowl, but not for a solid hemisphere, which still has its flat face.

Answer: D

The cubes must fit a whole number of times along each edge, so their edge length must divide 6, 12 and 15. To make the fewest cubes, take the largest such edge - the highest common factor.

\[ \text{HCF}(6, 12, 15) = 3 \]

\[ n = \frac{6}{3} \times \frac{12}{3} \times \frac{15}{3} = 2 \times 4 \times 5 = 40 \]

Equivalently \( \tfrac{6 \times 12 \times 15}{3^{3}} = \tfrac{1080}{27} = 40 \). Fewest cubes means largest cubes - the two requirements pull in opposite directions.

Answer: B

With the same base radius and the same height:

\[ V_{\text{cylinder}} = \pi r^{2} h, \qquad V_{\text{cone}} = \frac{1}{3}\pi r^{2} h \]

\[ \frac{V_{\text{cylinder}}}{V_{\text{cone}}} = \frac{\pi r^{2} h}{\tfrac{1}{3}\pi r^{2} h} = 3 \quad\Longrightarrow\quad 3 : 1 \]

Three coneloads of water exactly fill the matching cylinder - the standard classroom demonstration of where the \( \tfrac{1}{3} \) comes from.