The learning goal
When every corresponding length is multiplied by k, area is multiplied by k² and volume by k³. A 2×3×4cm box has surface area 52cm² and volume 24cm³. Doubling all edges produces 4×6×8cm, surface area 208cm² and volume 192cm³. Changing just one edge is a different problem.
Calculate the complete original box first
Take a cuboid 2cm long, 3cm wide and 4cm high. Its three pairs of opposite faces have individual areas 6, 8 and 12cm². Total surface area is therefore 2×(6+8+12)=52cm². Volume is 2×3×4=24cm³. These are original exercise values, not measurements of a real package.
The Math is Fun cuboid reference supplies the surface-area and volume formulas. Sketch the three pairs of faces before substituting numbers. That helps prevent missing a back face or treating one rectangle as the entire surface. Keep units visible: one length uses cm, two multiplied lengths use cm² and three use cm³.
| Quantity | Calculation | Result |
|---|---|---|
| One corresponding edge | 2 | 2cm |
| Total surface area | 2×(2×3+2×4+3×4) | 52cm² |
| Volume | 2×3×4 | 24cm³ |
Each direction contributes another multiplier
The enlarged cuboid is 4×6×8cm. A corresponding face changes from 2×3 to (2×2)×(2×3). Each of its two directions contributes a factor of two, making four together. Every face follows the same rule, so total surface area becomes 208cm². Three volume directions each contribute two, turning 24cm³ into 192cm³.
Algebra makes the condition explicit: new surface area is 2×[(ka)(kb)+(ka)(kc)+(kb)(kc)], which is k² times the original. New volume is (ka)(kb)(kc), or k³ times the original. Explain where each k comes from rather than memorizing a slogan without its assumptions.
Doubling edges does not double every quantity
Start with surface area 52 and volume 24, then write 104 and 48 after doubling.
New edges are 4, 6 and 8; surface area is 2×(24+32+48)=208; volume is 4×6×8=192.
Use the same reasoning when shrinking
Multiply all edges by 0.5 to obtain 1×1.5×2cm. Surface area is 2×(1.5+2+3)=13cm²; volume is 1×1.5×2=3cm³. Relative to the original, the ratios are 13÷52=0.25 and 3÷24=0.125, matching (0.5)² and (0.5)³.
Recalculate using the original formulas before relying on the rule. If the methods disagree, check a missing edge factor, a single face mistaken for total surface, or square and cubic units mixed together. The unit-cancellation and map-scale exercise helps with dimensions, but a length ratio cannot simply become an area ratio.
| Length factor | Dimensions (cm) | Surface area (cm²) | Volume (cm³) |
|---|---|---|---|
| 1 | 2×3×4 | 52 | 24 |
| 2 | 4×6×8 | 208=52×4 | 192=24×8 |
| 0.5 | 1×1.5×2 | 13=52×0.25 | 3=24×0.125 |
Doubling only height changes the question
Keep length 2 and width 3, but change height from 4 to 8. Volume becomes 2×3×8=48cm³, twice the original. Surface area becomes 2×(6+16+24)=92cm², neither twice nor four times 52. The three corresponding lengths did not share one scale factor, so the similar-shape rule does not apply.
Likewise, knowing only that volume doubled does not establish that every edge doubled. Changing height alone already achieves that volume change. To infer a uniform length factor from a volume ratio, you also need similarity and proportional changes in all directions.
Classify the condition before calculating
Every corresponding length uses the same k
Use area k² and volume k³, then verify from the original formulas.
Only one direction changes
Return to the formula and retain the unchanged lengths.
Only an area or volume ratio is supplied
Ask whether the shapes are similar before inferring edge changes.
Check whether you can explain a new case
The OpenStax discussion of cell size uses the different growth of surface area and volume as a cross-subject example. It shows where scaling can matter, but a box exercise alone does not predict every feature of a real organism. A mathematical model has conditions; another subject adds its own evidence and limits.
Finish with an explanation task: tell a classmate why doubling only height does not justify k² for the total surface. Use the worked-example self-explanation method to identify where the condition first changes, then recalculate area and volume. Explaining which edges stayed fixed is more reusable than memorizing 92.
Check the learning outcome
- I can identify length, area and volume units separately.
- I check whether every corresponding edge uses one scale factor.
- I can verify enlargement and shrinking with the original formulas.
- When one edge changes, I can explain why uniform scaling no longer applies.
Common questions
Is “twice the size” always the same as “increased by twice the original”?
No. Wording can be ambiguous. Establish whether the new value is two times the original or the original plus two more copies before calculating.
Does four times the area prove every edge doubled?
Only with similarity and proportional corresponding lengths. An area ratio alone does not uniquely determine each edge.
Can these ratios directly predict material weight or usable capacity?
Not by themselves. Weight also depends on density and structure; usable capacity depends on wall thickness and internal space. This lesson concerns ideal geometric quantities.
