Volume & Surface Areas Questions and Answers

Answer: A

Rearrange the cone volume formula for the height. The \( \pi \) appears on both sides and cancels.

\[ V = \frac{1}{3}\pi r^{2} h \quad\Longrightarrow\quad 84\pi = \frac{1}{3}\pi (6)^{2} h \]

\[ 84 = \frac{36h}{3} = 12h \]

\[ h = 7 \text{ m} \]

Because the volume is quoted as a multiple of \( \pi \), no decimal approximation is needed anywhere.

Answer: D

Water touches the base and the lower part of all four walls, but not the tank's top or the walls above the 11 cm line.

\[ \text{base} = 21 \times 13 = 273 \text{ cm}^{2} \]

\[ \text{two long walls} = 2 \times (21 \times 11) = 462 \text{ cm}^{2} \]

\[ \text{two short walls} = 2 \times (13 \times 11) = 286 \text{ cm}^{2} \]

\[ \text{total} = 273 + 462 + 286 = 1021 \text{ cm}^{2} \]

The tank's own height of 18 cm never enters: only the wetted 11 cm counts, and there is no top surface in contact with the water.

Answer: B

Rearrange the volume formula for the radius, then double it.

\[ V = \pi r^{2} h \quad\Longrightarrow\quad r^{2} = \frac{V}{\pi h} \]

\[ r^{2} = \frac{900}{3.142 \times 20} = \frac{900}{62.84} = 14.32 \]

\[ r = \sqrt{14.32} = 3.785 \text{ cm} \]

\[ d = 2r = 7.57 \text{ cm} \]

The question asks for the diameter, so the square root must be doubled - stopping at the radius leaves 3.79 cm.

Answer: D

The hemisphere of diameter 6 cm just covers the top plate, so each plate is a disc of radius 3 cm and thickness 2 cm.

\[ V_{\text{one plate}} = \pi r^{2} h = \pi (3)^{2}(2) = 18\pi \text{ cm}^{3} \]

\[ V_{\text{14 plates}} = 14 \times 18\pi = 252\pi \text{ cm}^{3} \]

\[ V_{\text{hemisphere}} = \frac{2}{3}\pi r^{3} = \frac{2}{3}\pi (27) = 18\pi \text{ cm}^{3} \]

\[ V_{\text{total}} = 252\pi + 18\pi = 270\pi \text{ cm}^{3} \]

"Just covers" is the clue that fixes the plates' radius at 3 cm - without it the plate dimensions would be unknown.

Answer: B

A solid whose base is a polygon and whose faces are triangles meeting at a single apex is a pyramid.

The base may be any polygon - triangular (a tetrahedron), square, pentagonal and so on - and the name of the pyramid follows the base.

A cone is the analogous solid built on a circular base, so a curved base rules the shape out of this definition.

Answer: C

Halve the diameter to get the base radius, then use the cone volume formula with \( \pi = \tfrac{22}{7} \).

\[ r = \frac{7}{2} = 3.5 \text{ cm}, \qquad h = 14 \text{ cm} \]

\[ V = \frac{1}{3}\pi r^{2} h = \frac{1}{3} \times \frac{22}{7} \times (3.5)^{2} \times 14 \]

\[ = \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 14 = \frac{1}{3} \times 539 \]

\[ V \approx 180 \text{ cm}^{3} \]

Forgetting the \( \tfrac{1}{3} \) would give the volume of the enclosing cylinder, three times too large.

Answer: C

Halve the diameter, then rearrange the volume formula for the height.

\[ r = 2.5 \text{ cm} \]

\[ h = \frac{V}{\pi r^{2}} = \frac{650}{3.142 \times 6.25} = \frac{650}{19.64} \]

\[ h = 33.1 \approx 33 \text{ cm} \]

Using the diameter of 5 instead of the radius would divide by four times as much and give about 8.3 cm.

Answer: C

Let the block measure \( l \times b \times h \). The three adjacent faces are

\[ P = lb, \qquad Q = bh, \qquad R = hl \]

Multiplying all three:

\[ PQR = (lb)(bh)(hl) = l^{2}b^{2}h^{2} = (lbh)^{2} \]

\[ \therefore\; S^{2} = PQR \]

Each dimension appears exactly twice in the product, which is what makes the result the square of the volume. It also gives a quick route to the volume: \( S = \sqrt{PQR} \).

Answer: A

Joining two hemispheres along their bases produces a complete sphere; the two flat circular faces are sealed inside and no longer form part of the surface.

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

which is precisely a sphere's surface area. Adding the two flat faces would give \( 6\pi r^{2} \) - correct for two separate hemispheres, but not for the joined solid.

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 \( h \).