Volume & Surface Areas Questions and Answers

Answer: B

The perimeter is 56.56 cm.

The diagonal of a square of side \(a\) is \(a\sqrt{2}\):

\[ \begin{aligned} a &= \frac{20}{\sqrt{2}} = 10\sqrt{2} \approx 14.14\ \text{cm} \\ \text{perimeter} &= 4a \approx 56.56\ \text{cm} \end{aligned} \]

28.28 cm is twice the side, half the perimeter. 141.4 cm is ten times the side, a slip of the decimal point. 40 cm is twice the diagonal, which treats the diagonal as if it were half the perimeter.

Answer: B

The volume is 1200 m³.

Floor plus ceiling is \(2lb\); the four walls are \(2h(l + b)\). Setting them equal:

\[ \begin{aligned} 2 \times 15 \times 12 &= 2h(15 + 12) \\ 360 &= 54h \\ h &= \tfrac{20}{3}\ \text{m} \\ V &= 15 \times 12 \times \tfrac{20}{3} = 1200\ \text{m}^{3} \end{aligned} \]

720 m³ comes from a height of 4 m and 900 m³ from 5 m; both ignore the condition. 840 m³ does not correspond to any height that satisfies it.

Answer: D

The water weighs 1.408 kg.

The cone has radius 8 cm and height 21 cm. With \(\pi = \frac{22}{7}\):

\[ \begin{aligned} V &= \tfrac{1}{3}\pi r^{2} h \\ &= \tfrac{1}{3} \times \tfrac{22}{7} \times 64 \times 21 \\ &= 1408\ \text{cm}^{3} \end{aligned} \]

Water has a density of 1 g/cm³, so 1408 cm³ weighs 1408 g = 1.408 kg.

Using the diameter 16 cm as the radius gives four times as much. Leaving out the \(\frac{1}{3}\) gives 4.224 kg, the volume of a cylinder. Neither matches 2.480 kg or 3.875 kg, which do not follow from the dimensions.

Answer: A

The third ball has a diameter of 2.5 cm.

Volume is conserved when the ball is recast, and a sphere's volume is proportional to the cube of its diameter, so the cubes of the diameters add up:

\[ \begin{aligned} 3^{3} &= 1.5^{3} + 2^{3} + d^{3} \\ 27 &= 3.375 + 8 + d^{3} \\ d^{3} &= 15.625 \\ d &= 2.5\ \text{cm} \end{aligned} \]

Treating diameters as if they simply added would give \(3 - 1.5 - 2 \lt 0\), which is impossible, and squaring instead of cubing gives about 1.66 cm, not offered. 3.6 cm is larger than the original ball, which cannot be.

Answer: B

The total surface area is 622 cm².

A cuboid has three pairs of equal faces:

\[ \begin{aligned} S &= 2(lb + bh + hl) \\ &= 2(7 \times 11 + 11 \times 13 + 13 \times 7) \\ &= 2(77 + 143 + 91) \\ &= 2 \times 311 = 622\ \text{cm}^{2} \end{aligned} \]

311 cm² forgets the factor 2, counting only one face of each pair. 1002 is close to the volume, \(7 \times 11 \times 13 = 1001\) cm³, a different quantity, and 430 cm² does not follow from the dimensions.

Answer: A

The radius of the sphere is 6 cm.

Reshaping keeps the volume of clay the same:

\[ \begin{aligned} \tfrac{1}{3}\pi (6)^{2}(24) &= \tfrac{4}{3}\pi r^{3} \\ 288\pi &= \tfrac{4}{3}\pi r^{3} \\ r^{3} &= 216, \quad r = 6\ \text{cm} \end{aligned} \]

It happens to equal the base radius of the cone. The other radii give far more clay than there is: a sphere of radius 8 cm already has a volume of about \(683\pi\) cm³, against the cone's \(288\pi\).

Answer: C

The total surface area is 214.5 cm².

The hemisphere's height is its radius, 3.5 cm, so the cone's height is \(15.5 - 3.5 = 12\) cm. Its slant height is

\[ \begin{aligned} l &= \sqrt{12^{2} + 3.5^{2}} \\ &= \sqrt{156.25} = 12.5\ \text{cm} \end{aligned} \]

The exposed surface is the cone's curved surface plus the hemisphere's curved surface (the joined circular faces are hidden):

\[ \begin{aligned} \pi r l + 2\pi r^{2} &= \tfrac{22}{7}(3.5)(12.5) + 2 \cdot \tfrac{22}{7}(3.5)^{2} \\ &= 137.5 + 77 = 214.5 \end{aligned} \]

Only the two curved surfaces count; the circular faces where the cone and hemisphere join are hidden inside the toy, and the cone's height is the total height less the hemisphere's radius, not the full 15.5 cm.

Answer: B

An indirect angle measuring tool is the Bevel gauge.

A bevel gauge (sliding bevel) has a blade and stock that can be locked at any angle, but no graduated scale. It is set to the angle of the work and the angle is then transferred to, and read on, a protractor or other measuring instrument, which is why it is an indirect measuring tool.

A bevel protractor and the protractor head of a combination set carry graduated scales and read the angle directly. A marking table (marking-off table) is a flat surface plate for layout work, not an angle-measuring tool. (The paper prints the third option with the letter B as well; options are given here in printed order.)

Answer: B

The ratio is 2:3.

A cylinder circumscribing a hemisphere of radius \(r\) on the same base has radius \(r\) and height \(r\).

\[ \begin{aligned} \text{hemisphere} &= \tfrac{2}{3}\pi r^{3} \\ \text{cylinder} &= \pi r^{2} \times r = \pi r^{3} \\ \text{ratio} &= \tfrac{2}{3} : 1 = 2 : 3 \end{aligned} \]

1:2 would be the ratio for a cylinder twice as tall, and 3:4 and 5:4 do not arise from these volumes. A hemisphere always fills two thirds of its circumscribing cylinder, the same fraction Archimedes found for a sphere in its cylinder.

Answer: C

64000 cubes fit in the box.

Convert to the same unit first: the box edge is 8 cm = 80 mm. Along each edge, \(80 / 2 = 40\) small cubes fit, so

\[ \begin{aligned} \text{cubes} &= 40^{3} \\ &= 64000 \end{aligned} \]

Equivalently, the volume ratio is \(\frac{80^3}{2^3} = \frac{512000}{8} = 64000\). 6400 and 16000 come from slips with the unit conversion or the powers, and 400 is far too few.