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Scaling in Three Dimensions: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A whole-box resize

A closed rectangular box is 3 cm by 4 cm by 10 cm. A similar image has its longest edge 15 cm long.

Find the image dimensions, total surface area, and volume.

Schematic; use the stated measurements.

Text description of this figure

Two similar closed boxes; original dimensions 3, 4, and 10 cm, and image longest edge 15 cm.
1. Hint 1
  Compare corresponding longest edges to find one factor.
2. Hint 2
  Apply that factor to all three dimensions; surface area and volume use different powers.
Answer

Dimensions 4.5, 6, 15 cm. Total surface area $369\text{ cm}^2$ . Volume $405\text{ cm}^3$ .

Full solution

The longest edges correspond, giving length factor $15/10$ , or $3/2$ .
The image dimensions are 4.5, 6, and 15 cm.

The original total surface area is
$S_o=2(12+30+40).$
Thus $S_o$ is 164 square cm.
The image surface area is
$S_i=164\cdot\frac94.$
It is 369 square cm.

The original volume is 120 cubic cm.
The image volume is
$V_i=120\cdot\frac{27}{8}.$
It is 405 cubic cm.
Direct multiplication of the image dimensions gives the same volume.

Answer

Dimensions 4.5, 6, 15 cm. Total surface area $369\text{ cm}^2$ . Volume $405\text{ cm}^3$ .

Key idea

Whole-solid similarity transfers one length factor to every dimension.
Problem 2 Return from an image

Two solids are similar. The image volume is 216 cubic cm and the original volume is 729 cubic cm. An image diagonal is 14 cm long.

Find the corresponding original diagonal and the image-to-original surface-area factor.
1. Hint 1
  First recover the factor that takes original lengths to image lengths.
2. Hint 2
  The known diagonal belongs to the image, so reverse the length factor to return.
Answer

Original diagonal $21$ cm. Image-to-original surface-area factor $4/9$ .

Full solution

The positive cube root of the image-to-original volume ratio gives
$k=\frac23.$
The image diagonal is two thirds of the original.

Divide the image diagonal by that factor.
$L_o=14\cdot\frac32.$
The original diagonal is 21 cm.

Square the common length factor for surface area.
$k^2=\frac49.$

Answer

Original diagonal $21$ cm. Image-to-original surface-area factor $4/9$ .

Key idea

Name the direction before using or reversing a scale factor.
Problem 3 An area record and a volume record

Similar solids have original total surface area 180 square cm and image total surface area 405 square cm. Their original volume is 160 cubic cm.

Find the image volume and the image-to-original ratio of corresponding heights.
1. Hint 1
  Total surface area reveals the square of the length factor.
2. Hint 2
  Recover the positive factor, then cube it for volume.
Answer

Image volume $540\text{ cm}^3$ . Height ratio image:original is $3:2$ .

Full solution

The image-to-original surface-area factor is $405/180$ , or $9/4$ .
Its positive square root gives
$k=\frac32.$
This is also the corresponding height factor.

The image volume is
$V_i=160\left(\frac32\right)^3.$
It is 540 cubic cm.

Answer

Image volume $540\text{ cm}^3$ . Height ratio image:original is $3:2$ .

Key idea

An area record must pass through a square root before it determines a volume factor.
Problem 4 Two spherical boundaries

A sphere of radius 7 cm bounds a ball. A second sphere has radius 21 cm.

Find each sphere surface area divided by its ball volume, in inverse centimeters. Compare the second ratio with the first.

The spheres share one drawing scale; the dashed outline identifies the second sphere.

Text description of this figure

Two spheres of radii 7 cm and 21 cm, drawn at one common scale. Solid and dashed circles distinguish their outlines. Dotted ellipses indicate hidden equatorial arcs; no computed areas or volumes are shown.
1. Hint 1
  Sphere area and ball volume use the same radius but different powers.
2. Hint 2
  Simplify the quotient before substituting; the constant pi cancels.
Answer

First ratio $3/7\text{ cm}^{-1}$ . Second ratio $1/7\text{ cm}^{-1}$ , one third of the first.

Full solution

For positive radius r, divide the established formulas.
$\frac{4\pi r^2}{(4/3)\pi r^3}=\frac3r.$
The result has inverse-length units.

For the 7 cm radius, the ratio is $3/7$ per centimeter.
For the 21 cm radius, it is $3/21$ , or $1/7$ , per centimeter.

The second radius has length factor 3, so its surface-area-to-volume ratio has factor $1/3$ , agreeing with these values.

Answer

First ratio $3/7\text{ cm}^{-1}$ . Second ratio $1/7\text{ cm}^{-1}$ , one third of the first.

Key idea

Sphere surface area divided by ball volume is inversely proportional to radius.
Problem 5 The outside of a resized display

A solid display is to be replaced by a similar solid with every length 30 percent longer. The original exposed surface area is 200 square cm and original volume is 200 cubic cm. The entire boundary remains exposed in both displays.

Find the replacement exposed area and volume.
1. Hint 1
  Convert the length increase to a multiplying factor before using powers.
2. Hint 2
  The area and volume increases are not the same as the length increase.
Answer

Exposed area $338\text{ cm}^2$ . Volume $439.4\text{ cm}^3$ .

Full solution

A 30 percent length increase gives a common positive factor of 1.3.
Similarity and matching exposed boundaries justify squared area scaling.
$S_i=200(1.3)^2.$
The exposed area is 338 square cm.

Volume uses the cube.
$V_i=200(1.3)^3.$
The volume is 439.4 cubic cm.

Answer

Exposed area $338\text{ cm}^2$ . Volume $439.4\text{ cm}^3$ .

Key idea

Translate a length percentage into a factor before finding covering area or enclosed space.
Problem 6 Same target volume, different change

A closed rectangular box is 2 cm by 4 cm by 9 cm.

Plan A. Resize the entire box uniformly so its volume is 8 times the original.

Plan B. Keep the 2 cm by 4 cm base fixed and change only height so volume is 8 times the original.

Find the dimensions and total surface area under each plan. Identify which result is similar to the original.

Schematic; use the stated measurements.

Text description of this figure

Original box with a 2 cm by 4 cm base and perpendicular height 9 cm.
1. Hint 1
  The two plans impose different conditions on corresponding lengths.
2. Hint 2
  For Plan A recover a common length factor from the volume factor.
3. Hint 3
  For Plan B use the fixed base area to find the new height, then count the actual face areas.
Answer

Plan A is similar. Dimensions:
$4,8,18\text{ cm}.$
Surface area:
$496\text{ cm}^2.$
Plan B is not similar. Dimensions:
$2,4,72\text{ cm}.$
Surface area:
$880\text{ cm}^2.$

Full solution

Plan A.
The positive cube root of 8 is 2, so all dimensions double to 4, 8, and 18 cm.
The original surface area is 124 square cm.
The new area is
$S_A=4(124).$
It is 496 square cm.

Plan B.
The fixed base means height must have factor 8.
The dimensions are 2, 4, and 72 cm.
Calculate the actual face areas.
$S_B=2(8+144+288).$
It is 880 square cm.

Plan A is similar because all corresponding lengths use factor 2.
Plan B has factors 1, 1, and 8, so it is not similar.
Both have volume 576 cubic cm, which does not make their shapes the same.

Answer

Plan A is similar. Dimensions:
$4,8,18\text{ cm}.$
Surface area:
$496\text{ cm}^2.$
Plan B is not similar. Dimensions:
$2,4,72\text{ cm}.$
Surface area:
$880\text{ cm}^2.$

Key idea

A shared volume target does not prescribe how each dimension changes.
Problem 7 Derive the ratio change

An original solid has positive total surface area $S$ and positive volume $V$ . A uniform resize uses a positive length factor $k$ and includes corresponding boundaries.

Derive the factor by which surface area divided by volume changes. Explain what this says for an enlargement and a reduction.
1. Hint 1
  Write the new surface area and new volume separately.
2. Hint 2
  Divide them and cancel the common positive factors.
3. Hint 3
  Compare the reciprocal of k with one in each case.
Answer

The ratio has factor $1/k$ . It decreases for an enlargement and increases for a reduction.

Full solution

As a concrete check, length factor 4 gives surface-area factor 16 and volume factor 64.
Their quotient has factor $16/64$ , or $1/4$ .

In general, surface area becomes $k^2S$ and volume becomes $k^3V$ .
Dividing gives
$\frac{k^2S}{k^3V}=\frac1k\cdot\frac SV.$
Cancellation is valid because k and V are positive.

An enlargement has k greater than one, so its reciprocal is less than one and the ratio decreases.
A reduction has k between zero and one, so the reciprocal is greater than one and the ratio increases.
The inverse-length unit stays the same when one common length unit is used.

Answer

The ratio has factor $1/k$ . It decreases for an enlargement and increases for a reduction.

Key idea

The extra length factor in volume becomes a reciprocal factor in surface area per unit volume.
Problem 8 Two unit systems, one solid

A solid has total surface area 0.96 square m and volume 0.064 cubic m.

Find its surface-area-to-volume ratio per meter and per centimeter. Explain why the two numerical values describe the same solid. Use 1 m as 100 cm.
1. Hint 1
  Divide area by volume in the given units first.
2. Hint 2
  The result has one length unit in its denominator; convert that denominator.
Answer

$15\text{ m}^{-1}$ and $0.15\text{ cm}^{-1}$ . The denominator unit changes, not the solid.

Full solution

Divide surface area by volume.
$\frac{0.96\text{ m}^2}{0.064\text{ m}^3}=15\text{ m}^{-1}.$
This is 15 per meter.

One meter is 100 cm, so
$\frac{15}{100\text{ cm}}=0.15\text{ cm}^{-1}.$
Equivalently, the area is 9600 square cm and volume is 64000 cubic cm; their quotient is also 0.15 per centimeter.

The numerical values differ because the denominator unit changes.
These are two expressions of the same surface area per unit volume.

Answer

$15\text{ m}^{-1}$ and $0.15\text{ cm}^{-1}$ . The denominator unit changes, not the solid.

Key idea

An inverse-length measurement changes numerical value when its denominator unit changes.
Problem 9 Test two possible explanations

An original closed rectangular box has dimensions 3 cm by 7 cm by 8 cm. A record for a changed box gives volume 336 cubic cm and total surface area 362 square cm.

Explanation A. The changed box is a uniformly enlarged copy of the original.

Explanation B. Only the 8 cm height was doubled while the 3 cm by 7 cm base stayed fixed.

Decide whether each explanation fits all the recorded measurements. Show checks for both volume and surface area.

Only the original box is shown; the two explanations must be tested against the records.

Text description of this figure

The original closed rectangular box has dimensions 3 cm by 7 cm by 8 cm. The changed box is not drawn. Dotted edges are hidden; use the stated dimensions.
1. Hint 1
  Calculate the original measurements before testing either explanation.
2. Hint 2
  Explanation A must use a cube root to match volume and then its squared factor for area.
3. Hint 3
  Explanation B keeps the base faces unchanged; calculate its remaining faces directly.
Answer

Explanation A fails the area record:
$S_A=202\sqrt[3]{4}\text{ cm}^2.$
Explanation B fits both records, with dimensions
$3,7,16\text{ cm}.$

Full solution

The original volume is 168 cubic cm and total surface area is 202 square cm.

Explanation A.
The recorded volume has factor 2, requiring length factor $\sqrt[3]{2}$ .
The area would be
$S_A=202\sqrt[3]{4}.$
This is about 320.66 square cm, not 362.
Thus A matches volume but fails surface area.

Explanation B.
The dimensions would be 3, 7, and 16 cm.
Their volume is
$V_B=3\cdot7\cdot16.$
It is 336 cubic cm, as recorded.

Explanation B.
The surface area is
$S_B=2(21+48+112).$
It is 362 square cm, as recorded.
Thus B fits both measurements.

Answer

Explanation A fails the area record:
$S_A=202\sqrt[3]{4}\text{ cm}^2.$
Explanation B fits both records, with dimensions
$3,7,16\text{ cm}.$

Key idea

Test every linked measurement; fitting volume alone does not establish similarity.
Problem 10 Recover two linked measurements

Two solids are similar and have corresponding boundaries included. Their surface-area-to-volume ratios are $1.2\text{ cm}^{-1}$ for the original and $0.8\text{ cm}^{-1}$ for the image. The original volume is 100 cubic cm.

Find the common length factor, the image volume, and the image total surface area. Verify the given image ratio.
1. Hint 1
  The quotient of the two surface-area-to-volume ratios is the reciprocal length factor.
2. Hint 2
  Once the length factor is known, cube it for image volume.
3. Hint 3
  Multiply the image ratio by the image volume to recover surface area.
Answer

Length factor $3/2$ . Image volume $337.5\text{ cm}^3$ . Image total surface area $270\text{ cm}^2$ . Verified ratio $0.8\text{ cm}^{-1}$ .

Full solution

The image-to-original ratio of the two given quotients is $0.8/1.2$ , or $2/3$ .
It equals the reciprocal length factor, so
$k=\frac32.$
This agrees with the image having the smaller ratio.

The image volume is
$V_i=100\left(\frac32\right)^3.$
It is 337.5 cubic cm.

The given image ratio multiplied by this volume gives
$S_i=0.8(337.5).$
The surface area is 270 square cm.
The inverse-centimeter unit multiplied by cubic centimeters produces square centimeters.

Checking, 270 divided by 337.5 is 0.8, with units per centimeter.
Also, the original area is 120 square cm and multiplying by $9/4$ gives the same 270 square cm.

Answer

Length factor $3/2$ . Image volume $337.5\text{ cm}^3$ . Image total surface area $270\text{ cm}^2$ . Verified ratio $0.8\text{ cm}^{-1}$ .

Key idea

A surface-area-to-volume ratio can recover the length scale when similarity is already known.

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