Volume & Surface Areas Questions and Answers

Answer: A

A solid hemisphere has both the curved dome and the flat base, giving a total of \( 3\pi r^{2} \).

\[ \text{TSA} = 3 \times \frac{22}{7} \times 7^{2} = 3 \times \frac{22}{7} \times 49 \]

\[ = 3 \times 22 \times 7 = 462 \text{ cm}^{2} \]

The curved surface alone is \( 2\pi r^{2} = 308 \) cm² and the flat base \( \pi r^{2} = 154 \) cm² - that 154 is offered as a distractor.

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: A

Solve the surface-area formula for \( r \), taking \( \pi = \tfrac{22}{7} \).

\[ 4\pi r^{2} = 616 \]

\[ r^{2} = \frac{616 \times 7}{4 \times 22} = \frac{4312}{88} = 49 \]

\[ r = 7 \text{ cm} \]

Using \( 2\pi r^{2} \) - the hemisphere formula - would give \( r = 7\sqrt{2} \approx 9.9 \) cm instead.

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.

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: B

When a sector of paper is rolled into a cone, the sector's radius becomes the cone's slant height - the straight distance from the apex to the rim is unchanged by the rolling.

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

The semicircle's arc becomes the base circumference, so \( \pi \times 10 = 2\pi r \) gives \( r = 5 \) cm - that 5 cm is the base radius, offered as the first distractor, not the slant height.

Answer: A

For cones of equal height the volume \( \tfrac{1}{3}\pi r^{2}h \) varies as \( r^{2} \), so the height cancels from the ratio.

\[ \frac{V_{1}}{V_{2}} = \frac{r_{1}^{2}}{r_{2}^{2}} = \frac{3^{2}}{5^{2}} = \frac{9}{25} \]

The ratio is \( 9 : 25 \). Carrying the radii through unsquared would give \( 3 : 5 \), which is the shape of the second distractor.

Answer: C

A glass with two different circular ends is a frustum, whose curved surface area is

\[ \text{CSA} = \pi (R + r) l \]

\[ = \frac{22}{7} \times (4 + 3) \times 10 = \frac{22}{7} \times 7 \times 10 \]

\[ = 220 \text{ cm}^{2} \]

It is the sum of the radii that enters here, not the difference - the difference appears only when finding the slant height from the vertical height.

Answer: B

Two hemispheres joined along their flat faces make a complete sphere, and the two flat faces vanish inside it.

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

which is exactly the surface area of a sphere of radius \( r \). The \( 3\pi r^{2} \) option is the total surface of a single solid hemisphere, base included - but that base is no longer exposed once the two are joined.

Answer: B

A bucket is a frustum, and its slant height, vertical height and the difference of the radii form a right triangle:

\[ l^{2} = h^{2} + (R - r)^{2} \]

\[ 13^{2} = h^{2} + (9 - 4)^{2} \quad\Longrightarrow\quad 169 = h^{2} + 25 \]

\[ h^{2} = 144 \quad\Longrightarrow\quad h = 12 \text{ cm} \]

Using the sum \( 9 + 4 = 13 \) instead of the difference would make the height zero - it is the difference that gives the horizontal leg of the slant.