Scaling in Three Dimensions

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Learning goals

Derive length, surface-area, and volume factors for a positive uniform resize
Recover a length factor from surface-area or volume ratios using positive roots
Calculate surface area per unit volume and explain its inverse-length units
Distinguish whole-solid similarity from changing only one dimension using measurements

One factor for the whole solid

Two solids are similar when one is a uniformly resized copy of the other, allowing a turn, reflection, or move. Every corresponding length uses the same positive factor. This includes radii, perpendicular heights, diagonals, and slant heights, not just the edges you can see.

We use original and image to keep the direction clear. The length scale factor is

k=\frac{\text{image length}}{\text{original length}}.

A factor greater than one enlarges; a factor between zero and one reduces. A factor of one preserves size. All solids here have positive volume, and we require $k>0$ so the image does not collapse.

Before reading the values beside the three cubes, predict the surface-area and volume factors for edges twice and three times as long. Each cube has six square faces, and its volume is its edge cubed.

At one drawing scale, the edge factors are 1, 2, and 3. Dotted edges are hidden; dashed outlines distinguish the resized cubes.

The length factors $1$ , $2$ , and $3$ produce area factors $1$ , $4$ , and $9$ , and volume factors $1$ , $8$ , and $27$ . Counting three directions explains the extra multiplication for volume.

Why the powers are two and three

Area and Similarity established that a positive uniform resize multiplies a plane region’s area by the square of the length factor. Every corresponding face of a polyhedron therefore has that squared factor. Adding the face areas gives the same factor for total surface area.

Surface and volume factors for similar solids#

Take a rectangular box $3$ cm by $5$ cm by $8$ cm. Its surface area is $158$ square cm, and its volume is $120$ cubic cm. Enlarge all three lengths by factor $2$ . Each face product contains two factors of $2$ , so the new surface area is $632$ square cm. The volume product contains three factors of $2$ , giving $960$ cubic cm.

The base contributes two length factors and the height contributes the third.

Now use any positive side lengths $a,b,c$ and any positive factor $k$ . Let $S$ denote total surface area and $V$ volume. Subscripts $o$ and $i$ mean original and image.

\begin{aligned} S_o&=2(ab+ac+bc),\\ S_i&=2((ka)(kb)+(ka)(kc)+(kb)(kc)),\\ S_i&=k^2S_o. \end{aligned}

For volume, all three perpendicular lengths change:

\begin{aligned} V_o&=abc,\\ V_i&=(ka)(kb)(kc),\\ V_i&=k^3V_o. \end{aligned}

Nothing depended on the particular box or factor $2$ . For a prism or cylinder, base area contributes $k^2$ and perpendicular height contributes $k$ , giving $k^3$ in $Bh$ . The fixed one-third in a pyramid or cone volume does not change this factor. A sphere’s volume formula contains its radius three times, so it also has factor $k^3$ .

The same reasoning extends to composite solids when every part, opening, and join is resized together. Corresponding added or removed volumes all acquire $k^3$ ; corresponding exposed areas acquire $k^2$ . For curved surfaces, such as a sphere, the area formulas already derived have two length factors. More generally, approximating a curved boundary with finer planar patches preserves this squared factor as the approximation improves. Likewise, finer box approximations preserve the cubic factor for enclosed volume.

Thus, for similar solids with the same corresponding boundary included in both measurements,

\begin{aligned} \frac{L_i}{L_o}&=k,\\ \frac{S_i}{S_o}&=k^2,\\ \frac{V_i}{V_o}&=k^3. \end{aligned}

An open surface must correspond to the same open surface in the image. Comparing a closed original with an image missing its lid changes the boundary as well as the size.

Worked example 1 Transfer one length factor to the outside and inside

Problem. A rectangular box has dimensions $4$ , $6$ , and $9$ cm. A similar image has a $9$ cm side corresponding to the original $6$ cm side. Find the image’s total surface area and volume.

Similarity is supplied; these panels use independent drawing scales.

Solution. The image-over-original length factor is $\tfrac32$ . First calculate the original measurements.

\begin{aligned} S_o&=2(4\cdot6+4\cdot9+6\cdot9),\\ S_o&=228\text{ cm}^2,\\ V_o&=4\cdot6\cdot9,\\ V_o&=216\text{ cm}^3. \end{aligned}

Use the square for the surface and the cube for the enclosed space.

\begin{aligned} S_i&=\frac94\cdot228,\\ S_i&=513\text{ cm}^2,\\ V_i&=\frac{27}{8}\cdot216,\\ V_i&=729\text{ cm}^3. \end{aligned}

The image dimensions $6$ , $9$ , and $13.5$ cm give the same results directly.

Check your understanding

Every length of a solid is reduced to one third of its original value. Which pair gives the image-to-original surface-area factor followed by the volume factor?

1/3

and

1/3

1/9

and

1/27

1/27

and

1/9

1/6

and

1/9

Recover a length factor with a root

If a similar solid has eight times the volume, its lengths are twice as long, because $2^3=8$ . A cube root reverses cubing. Use the positive cube root of image volume divided by original volume.

k=\sqrt[3]{\frac{V_i}{V_o}}.

Worked example 2 Volume first, then a length

Problem. Similar solids have original volume $375$ cubic cm and image volume $648$ cubic cm. A length of $10$ cm in the original corresponds to which image length?

Schematic panels give the known volumes without encoding the unknown length factor.

Solution. The volume factor is not the length factor. Reverse cubing.

\begin{aligned} k^3&=\frac{648}{375},\\ k^3&=\frac{216}{125},\\ k&=\frac65. \end{aligned}

Apply the recovered factor to the corresponding length.

\begin{aligned} L_i&=\frac65\cdot10,\\ L_i&=12\text{ cm}. \end{aligned}

Cubing $\tfrac65$ returns the given volume ratio, which checks the factor and its direction.

Check your understanding

Two similar solids have volume ratio original:image of 343:125. What factor takes original lengths to image lengths?

7/5

125/343

25/49

5/7

If the given measurement is surface area, reverse squaring instead. For example, an area factor of $2.25$ gives a positive length factor of $1.5$ .

k=\sqrt{\frac{S_i}{S_o}}.

Worked example 3 Surface area first, then height and volume

Problem. Similar solids have original total surface area $600$ square cm and image total surface area $294$ square cm. The original height is $10$ cm and its volume is $1000$ cubic cm. Find the image height and volume.

The same boundary is included in both surface-area measurements.

Solution. Take the positive square root of the surface-area factor.

\begin{aligned} k^2&=\frac{294}{600},\\ k&=\frac7{10}. \end{aligned}

Height is a length, so it uses this factor once.

\begin{aligned} h_i&=\frac7{10}\cdot10,\\ h_i&=7\text{ cm}. \end{aligned}

Volume uses its cube.

\begin{aligned} V_i&=\left(\frac7{10}\right)^3\cdot1000,\\ V_i&=343\text{ cm}^3. \end{aligned}

Check your understanding

Similar solids have image surface area 144 square cm and original surface area 64 square cm. What is the image-to-original volume factor?

9/4

3/2

27/8

8/27

Similarity must be given or established before either root links the measurements. A volume ratio alone does not establish shape or the ratios of corresponding lengths. Also use one length unit consistently before forming an area or volume ratio.

Surface area per unit volume

The surface-area-to-volume ratio is total surface area divided by enclosed volume. It measures how much surface there is per unit of enclosed space. A cube with edge $2$ cm has $24$ square cm of surface and $8$ cubic cm of volume, so its ratio is $3$ square cm per cubic cm.

\frac{24\text{ cm}^2}{8\text{ cm}^3}=3\text{ cm}^{-1}.

The unit $\text{cm}^{-1}$ means per centimeter. Two powers of centimeters cancel, leaving one in the denominator. This ratio has inverse-length units; it is not a unitless fraction or a percentage.

Under a similar enlargement by factor $2$ , surface area is multiplied by $4$ and volume by $8$ . Their quotient is multiplied by $\tfrac12$ . In general, for positive volume and positive $k$ ,

\begin{aligned} \frac{S_i}{V_i}&=\frac{k^2S_o}{k^3V_o},\\ \frac{S_i}{V_i}&=\frac1k\cdot\frac{S_o}{V_o}. \end{aligned}

A larger similar solid has more total surface but less surface per unit volume. A reduction reverses that comparison.

Worked example 4 More surface, smaller ratio

Problem. Compare cubes with edges $3$ cm and $9$ cm. Find each surface-area-to-volume ratio, including units. Which cube has the greater total surface area, and which has the greater ratio?

The edges have length ratio 1:3; the cubes share one drawing scale.

Solution. For a cube of positive edge length $s$ , its surface area is $6s^2$ and volume is $s^3$ , so their quotient is $6/s$ .

For the $3$ cm cube:

\begin{aligned} S&=54\text{ cm}^2,\\ V&=27\text{ cm}^3,\\ S/V&=2\text{ cm}^{-1}. \end{aligned}

For the $9$ cm cube:

\begin{aligned} S&=486\text{ cm}^2,\\ V&=729\text{ cm}^3,\\ S/V&=\frac23\text{ cm}^{-1}. \end{aligned}

The $9$ cm cube has more surface in total. The $3$ cm cube has the greater surface-area-to-volume ratio. The length factor $3$ divides the ratio by $3$ , agreeing with the calculation.

Check your understanding

A solid has surface-area-to-volume ratio 1.8 per centimeter. A similar image has every length five times as large. What is the image ratio?

0.36\text{ cm}^{-1}

9\text{ cm}^{-1}

0.072\text{ cm}^{-1}

1.8\text{ cm}^{-1}

A stretch changes a different set of lengths

Making a box taller while keeping its base unchanged does not resize the whole box uniformly. Some lengths change and others do not, so one common length factor cannot describe it.

Worked example 5 Compare a height change with a uniform resize

Problem. A closed rectangular box is $2$ cm by $5$ cm by $7$ cm. Compare these two changes.

Case A. Double only the $7$ cm height, keeping the base $2$ cm by $5$ cm.

Case B. Double every length of the original box.

Find the new volume and total surface area in each case. Decide which image is similar to the original.

Only height doubles here. The base stays fixed, so this is not a uniform resize.

Solution. The original volume is $70$ cubic cm and total surface area is $118$ square cm.

Case A. Only height doubles. The base area remains $10$ square cm, so volume doubles. The two base faces keep their areas, while the four side faces double theirs.

\begin{aligned} V_A&=2\cdot5\cdot14,\\ V_A&=140\text{ cm}^3,\\ S_A&=2(2\cdot5+2\cdot14+5\cdot14),\\ S_A&=216\text{ cm}^2. \end{aligned}

This box is not similar to the original. Its base lengths use factor $1$ while its height uses factor $2$ .

Case B. All lengths use factor $2$ , so similarity is preserved.

\begin{aligned} V_B&=8\cdot70,\\ V_B&=560\text{ cm}^3,\\ S_B&=4\cdot118,\\ S_B&=472\text{ cm}^2. \end{aligned}

Check your understanding

A right cylinder's height doubles while its radius stays fixed. Which statement follows?

A It is a similar enlargement with length factor 2. B Its volume has factor 8. C Its volume has factor 2, but it is not similar to the original. D Its total surface area has factor 4.

Scaling in Three Dimensions

Learning goals

One factor for the whole solid

Why the powers are two and three

Recover a length factor with a root

Surface area per unit volume

A stretch changes a different set of lengths

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

Your progress