If you multiply every linear dimension of a solid by 1.5, its surface area becomes 2.25 times the original area, and its volume becomes 3.375 times the original volume. The condition is that the shape's proportions remain constant: length, width and height must all change by the same factor. Stretching a single side is a different problem.
This distinction is useful to calculate in advance when resizing models, packaging, object designs or technical drawings. A "150% size" usually means a linear dimension 1.5 times the original. The volume must be calculated separately from this.
Why square and cube?
Let k be the common multiplier for linear dimensions. The area of a rectangle is a × b. After scaling, (k × a) × (k × b) = k² × a × b. In the volume of a cuboid, k appears in three factors, so the multiplier is k³. For similar shapes, this relationship can be used generally.
The Oak National Academy materials on area scaling and volume scaling discuss the correct square and cubic relationships. The following dimensions are illustrative examples.
Let the dimensions of a solid cuboid be 8 × 6 × 4 cm. Its volume is 8 × 6 × 4 = 192 cm³. After scaling by 1.5 times, you get 12 × 9 × 6 cm, so the new volume is 12 × 9 × 6 = 648 cm³. The ratio is 648 / 192 = 3.375.
The surface area of the original solid is 2 × (8 × 6 + 8 × 4 + 6 × 4) = 208 cm². For the scaled solid, it is 2 × (12 × 9 + 12 × 6 + 9 × 6) = 468 cm². The surface-area ratio relative to the original is 468 / 208 = 2.25.
| Quantity | Multiplier | Increase |
|---|---|---|
| Linear dimension | 1.5 | 50% |
| Surface area | 2.25 | 125% |
| Volume | 3.375 | 237.5% |
Stating "increases to 337.5%" and "increases by 237.5%" describes the same final volume. The first compares the new value to the original, while the second refers only to the added portion.
The direction of scale change in drawings
On a drawing with a scale of 1:50, 4 cm represents a real length of 200 cm. If you want to depict the same object at a scale of 1:40, the drawing length will be 200 / 40 = 5 cm. The required scaling is 5 / 4 = 1.25, or 125%.
You can also calculate directly from the denominators of the two scales: 50 / 40 = 1.25. The corresponding area on the new drawing is 1.25² = 1.5625 times the original drawing area. The actual dimensions of the object remain unchanged; only the representation has become larger.
Use the unit converter to standardise centimetres, metres, square and cubic units. You can also verify powers and the multi-term surface area formula in the graphical calculator (in Hungarian). To determine the scale, you must decide which linear dimensions correspond to each other.
From volume back to linear dimensions
If the goal is merely a volume 20% larger, the volume multiplier is 1.20. The multiplier for length is the cube root of this: k = 1.20^(1/3) ≈ 1.06266. This means an increase in linear dimensions of approximately 6.27% in every direction. A 20% increase in length would result in a much larger volume.
For hollow objects, calculate the material of the wall separately. As an illustration, let the 8 × 6 × 4 cm cuboid be closed, hollow inside, with a 0.2 cm wall thickness on all six sides. Without edge rounding or internal reinforcement, the internal dimensions are 7.6 × 5.6 × 3.6 cm. The material volume is 192 − 153.216 = 38.784 cm³.
Increasing the external dimensions by 1.5 times but keeping the wall thickness at 0.2 cm, the internal dimensions become 11.6 × 8.6 × 5.6 cm. The material volume is 648 − 558.656 = 89.344 cm³, which is approximately 2.3036 times the original. The 3.375 multiplier for external volume would overestimate the material here. The internal and external surfaces together are no longer geometrically similar because the wall was not scaled.
Capacity corresponds to the internal volume. For material requirements, you use the difference between external and internal volume, while the external bounding dimensions determine whether the object fits in its place. These are separate questions, so do not automatically transfer the growth rate of the external volume to the other two.
This is the geometry of the precisely specified shell. In the case of an open lid, reinforcement, edge rounding or partial filling, you must add or subtract the affected volume separately. From the resulting material volume, without further data, printing time, load-bearing capacity or manufacturing cost cannot be inferred.
Before changing dimensions, specify whether you are looking for an answer regarding length, surface area, capacity or material quantity. This allows you to decide which ratio is applicable and which requires additional geometric data.
How much scaling fits in the available space?
For the original 8 × 6 × 4 cm solid, there is now at most 11.8 × 9.4 × 6.1 cm of free space available. The position of the solid is fixed: the 8 cm side goes into the 11.8 cm direction, and so on. The dimensions already represent the available internal space.
A common k multiplier must satisfy all three conditions:
- 8k ≤ 11.8, therefore k ≤ 1.475;
- 6k ≤ 9.4, therefore k ≤ 1.5666…;
- 4k ≤ 6.1, therefore k ≤ 1.525.
The smallest upper limit decides: k = 1.475, meaning the linear dimensions can be increased to 147.5% of the original. The resulting solid is 11.8 × 8.85 × 5.9 cm. There is space left in the second and third directions; the first direction limits the common scaling.
Its volume is 11.8 × 8.85 × 5.9 = 616.137 cm³, and its surface area is 208 × 1.475² = 452.53 cm². The volume is 3.209046875 times the original 192 cm³. This limit had to be determined from the linear dimensions, because one side of the solid might still protrude from a space that is sufficient in terms of volume.
If the setting only accepts whole percentages, 147% results in 11.76 × 8.82 × 5.88 cm. The first dimension at 148% is already 11.84 cm, thus exceeding the 11.8 cm limit. Here, rounding the scaling to the nearest whole number would lead to an incorrect decision.
This is the geometric result of the fixed orientation and the specified, precise internal dimensions. If you can rotate the solid, or if you need to leave a fitting clearance for the dimensions, calculate with new conditions. Subtract the required fitting clearance from the specified internal dimensions before scaling. If you start from external dimensions, first determine the internal space based on wall thicknesses; do not subtract this again if you already have internal dimensions.
Final knowledge check
Choose one answer for each question. Each correct answer earns 1 point; an incorrect answer, a skipped question or more than one selected answer earns 0 points. You can score 0–3 points in total. Read the solutions after answering.
1. You scale every linear dimension of a solid by 1.5 times. By how much does the volume increase?
- A) By 237.5%; it changes to 337.5% of the original.
- B) By 50%.
- C) By 150%.
2. You scale the external dimensions of a hollow solid by 1.5 times, leaving the wall thickness unchanged. Can the 3.375 multiplier be automatically transferred to the material volume?
- A) No; the difference between the internal and external solid must be recalculated.
- B) Yes, because wall thickness does not matter.
- C) Yes, every geometric quantity changes cubically.
3. The 8 × 6 × 4 cm solid must fit into at most 11.8 × 9.4 × 6.1 cm in a fixed position. Scaling can only be adjusted in whole percentages. What is the maximum permitted setting?
- A) 150%.
- B) 147%.
- C) 148%.
Solutions and scoring
1. A. The volume multiplier is 1.5³ = 3.375. The increase from this is 3.375 − 1 = 2.375, i.e. 237.5%. For revision: Why square and cube?.
2. A. The unchanged thickness of the wall eliminates the geometric similarity of the entire shell. In the article, the material increases from 38.784 to 89.344 cm³, which is only approximately 2.3036 times. For revision: From volume back to linear dimensions.
3. B. Of the three limits, 11.8/8 = 1.475 is the strictest. 147% fits; at 148%, the first side is 11.84 cm, which is already too large. For revision: How much scaling fits in the available space?.
Each correct answer is worth 1 point; an incorrect or skipped answer is worth 0 points. The total score can be 0–3.
- 0 points: work through the sections marked in the three solutions again, and write down the initial conditions of the examples.
- 1 point: derive the correct answer for the two missed questions; separately note which condition or calculation step was omitted.
- 2 points: return to the section linked to the single missed question, then verify the result again by changing one input.
- 3 points: you correctly applied the three relationships in this task set. Try one of the article's checks with your own data, recording the conditions and units as well.
Translated and adapted from the Hungarian article: Hungarian original.