Volume & Surface Areas Questions and Answers

Answer: A

The curved surface of a hemisphere is half a sphere's, so

\[ \text{CSA} = 2\pi r^{2} = 175 \]

\[ r^{2} = \frac{175}{2\pi} = \frac{175}{6.2832} = 27.85 \]

\[ r = \sqrt{27.85} = 5.28 \text{ cm} \]

Only the curved dome is asked for, so the flat base is excluded; using the total surface \( 3\pi r^{2} \) would give \( r = 4.31 \) cm instead.

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: B

A sphere's surface area is

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

Note that it is exactly four times the area of a great circle (\( \pi r^{2} \)) - the flat disc you would get by slicing the sphere through its centre.

The distractors are all real formulas from elsewhere: \( 2\pi r \) is the circumference of a circle, and \( 2\pi r^{2} \) is the curved surface of a hemisphere. A surface area must be proportional to \( r^{2} \), which rules out \( 2\pi r \) on dimensional grounds alone.

Answer: A

The curved surface needs the slant height, so find it by Pythagoras first.

\[ l = \sqrt{r^{2} + h^{2}} = \sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm} \]

\[ \text{CSA} = \pi r l = 3.1416 \times 5 \times 13 \]

\[ = 204.2 \approx 204 \text{ cm}^{2} \]

5–12–13 is a Pythagorean triple, which is why the slant height comes out whole. Including the base would add \( \pi r^{2} = 78.5 \) cm² and give the total surface area of about 283 cm².

Answer: A

The volume of a sphere is

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

A volume must scale with the cube of a length, so any candidate formula has to carry \( r^{3} \) - that alone eliminates the alternatives without recalling the constant.

The relation to the surface area is worth noting: \( V = \tfrac{1}{3} \times S \times r = \tfrac{1}{3} \times 4\pi r^{2} \times r \), the same one-third that appears for cones and pyramids.

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: D

Find the radius from the volume, then use it for the surface area.

\[ \frac{4}{3}\pi r^{3} = 850 \quad\Longrightarrow\quad r^{3} = \frac{850 \times 3}{4\pi} = \frac{2550}{12.566} = 202.9 \]

\[ r = \sqrt[3]{202.9} = 5.88 \text{ m} \]

\[ S = 4\pi r^{2} = 4 \times 3.1416 \times (5.88)^{2} = 12.566 \times 34.57 \]

\[ S \approx 434 \text{ m}^{2} \]

Only the pair 5.88 m and 434 m² is internally consistent - checking one value against the other is quicker here than computing both from scratch.

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: 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: 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.