Volume & Surface Areas Questions and Answers

Answer: A

Volume needs the perpendicular height, not the slant height, so find it by Pythagoras first.

\[ h = \sqrt{l^{2} - r^{2}} = \sqrt{5^{2} - 4^{2}} = \sqrt{9} = 3 \text{ cm} \]

\[ V = \frac{1}{3}\pi r^{2} h = \frac{1}{3}\pi (4)^{2}(3) = \frac{1}{3}\pi (16)(3) = 16\pi \text{ cm}^{3} \]

Using the slant height of 5 in place of \( h \) would give \( \tfrac{80}{3}\pi \) - the standard slip in every cone question that supplies \( l \) rather than \( h \).

Answer: A

The total surface area of any solid is the sum of the areas of every exposed face.

For a pyramid that means the base plus all the triangular lateral faces:

\[ \text{TSA} = \text{base area} + \text{lateral surface area} \]

The base area alone is only one part of it, and the slant edge is a length, not an area - a total surface area must have units of length squared.

Answer: A

Each metre is 100 centimetres, so a cubic metre is \( 100^{3} \) cubic centimetres.

\[ 1 \text{ m}^{3} = 100 \times 100 \times 100 = 1{,}000{,}000 \text{ cm}^{3} \]

\[ 0.8 \text{ m}^{3} = 0.8 \times 1{,}000{,}000 = 800{,}000 \text{ cm}^{3} \]

The conversion factor for a volume is the cube of the length factor - using 100 or 10,000 instead of 1,000,000 produces the smaller distractors.

Answer: A

A cuboid's total surface area sums its three pairs of faces.

\[ \text{TSA} = 2(lb + bh + hl) \]

\[ = 2(15 \times 10 + 10 \times 9 + 9 \times 15) \]

\[ = 2(150 + 90 + 135) = 2(375) = 750 \text{ cm}^{2} \]

Each product must be counted twice, once for each of the opposite faces - the single sum of 375 is exactly half the answer.

Answer: C

Melting conserves volume, and every sphere's volume is proportional to \( r^{3} \), so the \( \tfrac{4}{3}\pi \) cancels throughout.

The radii are 1.5, 2 and 2.5 cm:

\[ R^{3} = 1.5^{3} + 2^{3} + 2.5^{3} = 3.375 + 8 + 15.625 = 27 \]

\[ R = \sqrt[3]{27} = 3 \text{ cm} \]

Radii add in the cube, never directly: \( 1.5 + 2 + 2.5 = 6 \) is the first distractor and is far too large.

Answer: C

A pyramid's volume is a third of the prism on the same base:

\[ V = \frac{1}{3} \times \text{base area} \times h \]

\[ \text{base area} = 30 \times 20 = 600 \text{ cm}^{2} \]

\[ 2000 = \frac{1}{3} \times 600 \times h = 200h \]

\[ h = 10 \text{ cm} \]

Omitting the \( \tfrac{1}{3} \) gives 3.33 cm; the same one-third factor appears in the cone formula, and for the same geometric reason.

Answer: C

Rearrange the sphere volume formula for the radius.

\[ \frac{4}{3}\pi r^{3} = 5600 \quad\Longrightarrow\quad r^{3} = \frac{5600 \times 3}{4\pi} = \frac{16800}{12.566} = 1337 \]

\[ r = \sqrt[3]{1337} \approx 11 \text{ cm} \]

Checking against the options is quick: \( 10^{3} = 1000 \) is too small and \( 12^{3} = 1728 \) too large, so 11 is the only candidate that can work.

Answer: B

Write the claim as an equation and solve for the constant \( c \).

\[ V = c \times S \times r \]

\[ \frac{4}{3}\pi r^{3} = c \times 4\pi r^{2} \times r = 4c\pi r^{3} \]

\[ \frac{4}{3} = 4c \quad\Longrightarrow\quad c = \frac{1}{3} \]

The same one-third appears in the cone and pyramid formulas, and for the same reason: a solid built of thin pyramids on a surface has volume \( \tfrac{1}{3} \times \text{surface} \times \text{height} \).

Answer: C

Take the original dimensions as \( 1k, 2k, 3k \).

\[ V_{\text{old}} = 1k \times 2k \times 3k = 6k^{3} \]

An increase of 100% doubles a length, and 200% triples it:

\[ 1k \to 2k, \qquad 2k \to 6k, \qquad 3k \to 9k \]

\[ V_{\text{new}} = 2k \times 6k \times 9k = 108k^{3} \]

\[ V_{\text{new}} - V_{\text{old}} = 102k^{3} = 17 \times 6k^{3} \]

So the old box falls short by 17 times its own volume. Note that a 200% increase means the length becomes three times the original, not two - that single slip turns the answer into 7.

Answer: A

Same material means the same density, so weight is proportional to volume.

\[ V_{\text{solid}} = \frac{4}{3}\pi (3)^{3} = 36\pi \text{ cm}^{3} \;\longrightarrow\; 6 \text{ kg} \]

The hollow sphere has outer radius 9 cm and inner radius 6 cm:

\[ V_{\text{hollow}} = \frac{4}{3}\pi\left(9^{3} - 6^{3}\right) = \frac{4}{3}\pi (729 - 216) = \frac{4}{3}\pi (513) = 684\pi \text{ cm}^{3} \]

\[ \text{weight} = 6 \times \frac{684\pi}{36\pi} = 6 \times 19 = 114 \text{ kg} \]

The cavity must be subtracted; treating the object as a solid 18 cm sphere would give 162 kg.