G3 SEC Mathematics K310 | Learner’s Guide Vol 0145. A uniform scale factor changes lengths once, areas by its square and volumes by its cube. First prove similarity, then choose the correct power.
This original lesson takes similar solids beyond slogans: it connects geometry, unit conversions, exposed surfaces, capacities and real-world constraints with fully explained exercises.
Chapter index
- 1. A model twice as long is not just twice as big
- 2. The official 2027 K310 scope
- 3. First establish that the solids really are similar
- 4. Choose and label one direction for the scale factor
- 5. Why area follows the square
- 6. Why volume follows the cube
- 7. Surface area has the square factor, but count the right faces
- 8. How to work backward from an area ratio
- 9. How to work backward from a volume ratio
- 10. Keep units alongside the scale factor
- 11. Similarity versus merely having the same kind of shape
- 12. A composite-solid example: two cubes joined together
- 13. The same material: why mass often follows volume
- 14. The surface-to-volume effect
- 15. An original container problem from capacity to dimensions
- 16. A model that is smaller: reduction is not a different rule
- 17. The trap in drawings, models and labels
- 18. Which method does a K310 question really ask for?
- 19. Twenty-eight original scale-factor questions
- 20. Repair one misconception instead of repeating twenty pages
- 21. Seven days from scales to confident real-world answers
- 22. Volume answers need units, boundaries and sense
- 23. Thirty original questions with worked answers
1. A model twice as long is not just twice as big
Imagine a cube with a side length of 3 cm. Make a new cube with every edge twice as long, so that each side is 6 cm. A child may say it is twice as big. But “big” could refer to length, exposed surface area or the space inside. The second cube has four times the surface area and eight times the volume of the first. All three answers are correct for different quantities.
G3 SEC Mathematics questions on similar figures and solids reward precisely this distinction. A single scale factor can answer several questions once you know what is changing. Length responds to the first power of the scale factor, area to its square, and volume to its cube. Confusing them turns an otherwise correct ratio into a wrong amount of material or capacity.
This Learner’s Guide Vol 0145 begins with correspondence and shape, then builds toward area and volume ratios, inverse problems, unit conversion, composite solids, and checking a calculation without measuring a diagram. It is related to the map-scale work in an earlier volume, but owns a distinct three-dimensional skill: recognising how a genuinely similar solid changes in all directions.
2. The official 2027 K310 scope
The 2027 SEAB G3 subject list identifies Mathematics as K310. The published K310 syllabus includes congruence and similarity, equal corresponding angles, proportional corresponding sides, area ratios for similar plane figures, and volume ratios for similar solids. It also covers volumes and surface areas of common solids and composite solids.
K310 has two written papers, each lasting 2 hours and 15 minutes and carrying 90 marks, each with 50 percent weighting. Paper 1 has about 26 short-answer questions. Paper 2 has nine or ten questions and concludes with a real-world application problem. Candidates answer all questions, and essential working must be shown.
The usual instruction for non-exact numerical answers is three significant figures, or one decimal place for angles in degrees, unless the question specifies otherwise. Similarity results can often be kept as exact ratios, fractions or multiples of π. Use the requested form and units; do not round an intermediate length factor before cubing it.
A scale factor is not a separate exam paper or a prediction that a particular solid will appear. It is a mathematical idea threaded through geometry, measurement and contextual problems. The useful skill is to identify the quantity the problem asks for and preserve a consistent direction for the comparison.
3. First establish that the solids really are similar
Similar figures have matching shapes even when their sizes differ. Corresponding angles agree, and every pair of corresponding lengths has the same ratio. Similar solids inherit a consistent length scale in three dimensions. A cylinder whose radius doubles but whose height remains unchanged is not generally similar to the original cylinder.
Take two cuboids with dimensions 2 × 3 × 5 and 4 × 6 × 10, in the same unit and corresponding order. Every dimension in the second is twice the corresponding dimension in the first. These cuboids are similar and the length scale from small to large is 2.
Compare 2 × 3 × 5 with 4 × 7 × 10. Two ratios are 2, but the middle is 7/3. The solids are not similar, so applying the cube-of-one-ratio shortcut would be wrong. You can calculate their volumes directly, but a single uniform scale factor does not describe the change.
A drawing can help you match corresponding edges, yet its apparent proportions do not prove similarity. Use the stated lengths, angle conditions or a proven shape relationship. If the question says “similar”, that is a condition. If it asks you to determine whether they are similar, the condition is the target to establish.
4. Choose and label one direction for the scale factor
Suppose a small model has a corresponding height of 4 cm and a larger model has height 10 cm. The scale factor from small to large is 10/4 = 2.5. Write that direction in words. Reversing the ratio gives 0.4, which is the correct factor for travelling from large to small, not from small to large.
Ratios are often written as small:large = 2:5. This does not mean the large object is multiplied by 2/5 to obtain an even larger object. It means the large object’s corresponding lengths are 5/2 times the small object’s. Carry the direction label into the calculation rather than depending on which number appears first in a sentence.
A good exam line is “linear scale, small to large, k = 5/2”. Once that is secure, small-to-large area ratio becomes 1:k², and the volume ratio becomes 1:k³. You can calculate those accurately without guessing whether to square a 2 or a 5.
When reversing direction, use the reciprocal: if small to large is 5/2, large to small is 2/5. Squaring and cubing preserve this reciprocal relationship. The direction must remain consistent for every linked quantity.
5. Why area follows the square
Area is built from two perpendicular lengths. A rectangle with sides a and b has area ab. If each side is multiplied by k, the new area is (ka)(kb) = k²ab. The result is not a trick for memorising; it is what happens when two dimensions both change.
A rectangle 3 cm by 4 cm has area 12 cm². Enlarge every length by 3 to make it 9 cm by 12 cm. Its area is 108 cm², exactly nine times the original. The length scale is 3, and the area scale is 3² = 9.
The same principle applies to all genuinely similar plane regions, not just rectangles. A triangle’s base and height both scale by k. A circle’s radius scales by k, so πr² scales by k². Even irregular outlines preserve the area-square relationship when the entire shape is uniformly scaled.
Do not confuse this with doubling only one dimension. A rectangle whose width is doubled while its length stays fixed doubles its area, but it is usually not similar to the first. Before using k², establish that the same k applies to both directions.
6. Why volume follows the cube
Volume measures three-dimensional space. If a cuboid has dimensions a, b and c, its volume is abc. Enlarging every corresponding length by k gives (ka)(kb)(kc) = k³abc. The cube power is the mathematical record of all three spatial dimensions changing together.
For a cube of side 3 cm, volume is 27 cm³. A similar cube with side 6 cm has volume 216 cm³. The linear factor is 2 and the volume factor is 8. The larger cube holds the volume of eight smaller cubes, although each edge is only twice as long.
A cylinder has volume πr²h. If both radius and height scale by k, its new volume is π(kr)²(kh) = k³πr²h. A sphere has volume proportional to r³ and follows the same relationship under uniform enlargement. The solid’s name changes the specific volume formula, not the general scaling rule for similar solids.
If only the radius changes while the height does not, direct calculation is required: doubling a cylinder’s radius with fixed height makes its volume four times larger, not eight. Similarity is the gate that permits the general cube rule.
7. Surface area has the square factor, but count the right faces
The exposed surface area of a similar solid scales with the square of the linear factor. That is because every face is a two-dimensional region. If every length doubles, the area of each corresponding face is multiplied by 4; the total corresponding surface area is multiplied by 4 as well.
For a cube of side 3 cm, total surface area is 6 × 3² = 54 cm². For a similar cube of side 6 cm, it is 6 × 6² = 216 cm², four times as large. The volume ratio for the same pair is eight. This contrast is central to questions about packaging, paint and capacity.
When solids are joined or faces are hidden, the task may ask for external surface area rather than the sum of every separate face. Two identical cubes joined along one face have ten exposed square faces, not twelve. If the whole composite shape is uniformly enlarged, its external surface area still scales by k², but the original surface area must first be found correctly.
Distinguish “total surface area”, “curved surface area”, “painted area” and “cross-sectional area”. They are all areas but may refer to different boundaries. A correct scale factor does not rescue a calculation performed on the wrong set of faces.
8. How to work backward from an area ratio
If the areas of similar plane figures are in ratio 9:25, the corresponding positive length ratio is √9:√25 = 3:5. The square root reverses the area-square rule. Do not report 9:25 as the length ratio merely because it appears prominently in the question.
Suppose two similar rectangular signs have area 36 cm² and 100 cm². Their small-to-large area factor is 100/36 = 25/9. The length factor is √(25/9) = 5/3. If the small sign is 12 cm wide, the larger is 12 × 5/3 = 20 cm wide.
The dimensions of similar solids can also be recovered from a surface area ratio, because corresponding exposed surfaces scale by k². The exact surfaces compared must correspond. If one model’s base is counted and the other’s is omitted, the areas do not describe the same surface and cannot be compared by this rule.
Write the inverse process visibly: k² = area factor, so k = √(area factor), taking a positive root for a physical scale. This is much safer than guessing a familiar ratio from the displayed numbers.
9. How to work backward from a volume ratio
If two similar solids have volumes in ratio 27:64, their corresponding positive lengths are in ratio 3:4. The cube root of 27 is 3, and of 64 is 4. Returning to a length requires undoing the power that created the volume.
Imagine a pair of similar storage tanks. The smaller holds 8 litres and the larger holds 27 litres. The small-to-large volume factor is 27/8. The length factor is the cube root, 3/2. A 2 m height on the smaller tank corresponds to 3 m on the larger.
If the problem gives a volume ratio that is not a perfect cube, leave a cube-root expression during working or use the calculator with adequate precision. For example, a volume factor of 5 corresponds to a length factor of cube root of 5, about 1.71. Round only where directed.
This inverse skill becomes especially valuable when a question gives capacity but asks for a missing radius, height or edge. Capacity is a volume measure; you must travel from its cube relation back to the correct linear relation.
10. Keep units alongside the scale factor
A ratio of corresponding lengths is dimensionless when both lengths use the same unit. Before forming it, convert units. If a model edge is 25 cm and the full-size edge is 2 m, convert 2 m to 200 cm; the scale factor is 8, not 2/25.
Area and volume conversions also involve powers. One metre is 100 centimetres. Therefore 1 m² equals 10,000 cm² and 1 m³ equals 1,000,000 cm³. A common error is to apply the linear conversion factor 100 to an area or volume without raising it to the corresponding power.
A box of 3,000,000 cm³ has volume 3 m³. A square panel of 20,000 cm² has area 2 m². These figures are numerically far apart, yet both arise directly from consistent base-unit conversion. Writing the units at each step helps spot a misplaced factor of 100 or 1,000,000.
If a question mixes litres and cubic centimetres, remember that one litre is 1000 cm³. Capacity is measured as volume; a litre figure can therefore be compared to a geometrical volume after conversion. Do not treat a litre as a length because a container’s height is also given.
11. Similarity versus merely having the same kind of shape
Two cylinders are not necessarily similar just because both are cylinders. Their height-to-radius proportions must agree. A cylinder with r = 2 and h = 5 can be similar to one with r = 3 and h = 7.5 because every corresponding length has factor 1.5.
Their volume ratio is 1.5³ = 3.375, which is 27/8. The smaller volume is π × 2² × 5 = 20π. The larger is π × 3² × 7.5 = 67.5π. Dividing confirms the volume ratio. Both routes agree because the cylinders are similar.
Now change only the second height to 6. The radius factor remains 1.5 but the height factor becomes 1.2. These cylinders are not similar. You can find the second volume directly as π × 3² × 6 = 54π, and the volume factor is 54/20 = 2.7. Applying 1.5³ would have been wrong.
This diagnostic helps in unfamiliar word problems. Before asking “Which power?”, ask “Is a common scale factor justified?” The attractive shortcut comes after the relationship is established, never before.
12. A composite-solid example: two cubes joined together
Take two cubes of side 2 cm glued face to face. Each cube has six square faces, but the joined pair has ten faces exposed to the air: the two touching faces are hidden. Its external surface area is 10 × 2² = 40 cm².
If the entire joined shape is enlarged so each little cube has side 5 cm, the length factor is 5/2. The exposed surface area becomes 10 × 5² = 250 cm². The predicted area factor is (5/2)² = 25/4. Indeed, 40 × 25/4 = 250.
The original volume is two cube volumes, or 2 × 2³ = 16 cm³. The larger joined shape has volume 2 × 5³ = 250 cm³. Their volume factor is (5/2)³ = 125/8, and 16 × 125/8 = 250. The area and volume happen to share the number 250 in the larger model, but the units and quantities differ.
This exercise is a useful reminder that identifying which faces are exposed is part of the problem. Scaling the right quantity is necessary, but so is defining that quantity accurately in the first place.
13. The same material: why mass often follows volume
If two similar solid objects are made of the same uniform material, their density is the same. Mass equals density multiplied by volume, so mass scales by k³ along with volume. A model enlarged by a factor of 2 may require eight times as much solid material, not twice as much.
This statement has conditions. If the larger object is hollow while the smaller is solid, or if their materials have different densities, the mass ratio need not match the external volume ratio. A scaled-up shell with fixed wall thickness is not the same geometrical solid as a completely scaled solid block.
For a uniform block with density 2 g/cm³, volume 30 cm³ corresponds to mass 60 g. A truly similar solid block with length factor 3 has volume 810 cm³ and mass 1620 g under the same-density assumption. If the problem asks about cost per gram, this mass scaling can lead to a very different budget from a simple linear price estimate.
State the material and shape assumptions explicitly when applying mass or capacity reasoning. A right numerical factor without the needed condition is a fragile answer.
14. The surface-to-volume effect
When a similar solid becomes larger, surface area grows as k² but volume grows as k³. Therefore surface area divided by volume changes by 1/k. Under a doubling of every length, the surface-to-volume ratio is halved. That relationship is useful for understanding why an object’s size can affect heating, cooling or material demands, although real physical behaviour involves further conditions.
A cube of side a has total surface area 6a² and volume a³, so surface area per unit volume is 6/a. A cube with side 2a has ratio 6/(2a) = 3/a, half the first. The surface has increased, but the space inside has increased even faster.
Do not conclude that doubling size always halves every real cooling rate. Material properties, airflow, temperature differences and shape details also matter. The geometric result is precise; the physical extrapolation requires additional evidence and modelling.
In a Mathematics answer, you can state the clean ratio relation first. In an integrated Science context, keep the geometry separate from any explanation of a measured response. That separation makes the reasoning easier to evaluate.
15. An original container problem from capacity to dimensions
A small display container is similar to a larger one. The small container’s length is 12 cm, width 8 cm and height 5 cm. The larger container’s corresponding length is 18 cm. The linear scale is 18/12 = 3/2, so the larger width is 12 cm and its height is 7.5 cm.
The small container has volume 12 × 8 × 5 = 480 cm³. The volume factor is (3/2)³ = 27/8. Thus the larger container has volume 480 × 27/8 = 1620 cm³. Direct calculation, 18 × 12 × 7.5 = 1620 cm³, confirms the result.
If the question asks how much more it can hold, give 1620 − 480 = 1140 cm³ rather than 1620 cm³. If the context converts capacity to litres, the larger volume is 1.62 litres. The wording “new capacity” and “increase in capacity” asks for different quantities.
A clean solution shows the factor, the power, the calculation and the target. Four purposeful lines are more useful than a long argument written after the wrong quantity has already been chosen.
16. A model that is smaller: reduction is not a different rule
If the new model has length factor 0.5 relative to the original, its area factor is 0.25 and its volume factor is 0.125. The same power rules apply. A reduction by half in every direction leaves one-eighth of the original volume.
Suppose a designer reduces every edge of a solid cube from 20 cm to 10 cm. The original volume is 8000 cm³, and the new volume is 1000 cm³. A pupil may subtract fifty percent because the edge halved, but the correct volume change is 7000 cm³, which is 87.5 percent of the original volume.
A percentage decrease in a length does not transfer unchanged to surface area or capacity. Convert the length percentage into a multiplier first. For example, reducing every edge by 20 percent means k = 0.8, area factor = 0.64 and volume factor = 0.512. The area decreases by 36 percent and the volume by 48.8 percent.
The reverse is equally instructive. An increase of 10 percent in every linear dimension gives volume factor 1.1³ = 1.331, a 33.1 percent increase. This is why dimension changes deserve their own calculation rather than an intuitive guess.
17. The trap in drawings, models and labels
A drawing may show two shapes with different apparent proportions because it is not drawn to scale. Unless the question says the figures are similar or gives enough relationships to establish that fact, treat the picture as a guide to correspondence, not a ruler. The stated measurements control your answer.
When a figure is rotated or reflected, corresponding edges can be difficult to match. Label vertices, mark equal angles and compare relative positions. An incorrect correspondence creates a wrong scale factor even when the power rule is remembered perfectly.
For example, a long edge of one cuboid must be compared with the corresponding long edge of the other, not merely whichever edge happens to lie on the page’s right side. Once correspondence is fixed, check at least one second pair of lengths. If the ratios disagree, similarity may not hold.
Use arrows or a small note such as “small → large”. That three-word annotation prevents many ratio reversals. A few seconds spent defining direction can save several minutes of repairing a final answer.
18. Which method does a K310 question really ask for?
An item about a side length asks for k. An item about shaded area or required sheet material asks for k² once similar surface regions are established. An item about volume, capacity or mass of the same solid material asks for k³ under the stated assumptions. An inverse question gives an area or volume ratio and asks you to use a square root or cube root to recover the corresponding length ratio.
Questions about composite solids may require direct mensuration before a scale factor can help. If the question asks only for the exposed surface of a joined object, count exposed faces correctly first. If it compares dissimilar solids, abandon the uniform-scale shortcut and use their actual dimensions and volume formulae.
A brief self-test is: “What is measured, what corresponds, and which dimension is being changed?” This is the essential method decision. Memorising length-squared-cubed without identifying those three things is not enough for longer real-world questions.
The Vol 0115 map-scale workshop treats the special case of two-dimensional scale maps. Here we apply the wider geometry of similar solids, including volume and capacity, rather than repeating that narrower map problem.
19. Twenty-eight original scale-factor questions
Try these questions without a diagram first, then sketch a corresponding pair of objects if the relationships are confusing. Every quantity needs a unit where appropriate. A few cases deliberately describe non-similar solids; the correct response is to refuse an unjustified shortcut and calculate directly.
Before reading each worked answer, write the ratio in words, such as “length, small to large”. If the question provides an area or volume ratio, ask which root reverses it. If a percentage change is given, convert it into a multiplier before applying a power.
These are original teaching cases, not official SEC items. The purpose is to make an unfamiliar exam setting recognisable through the relationship it describes.
20. Repair one misconception instead of repeating twenty pages
Students who square a length when asked for volume need a three-dimensional contrast, such as a cube whose edges double. Students who cube every radius change need two cylinders with different height behaviour: one truly similar, one with fixed height. Students who reverse ratios need a clear small-to-large arrow.
A mistaken unit conversion calls for a direct demonstration: draw a 1 m square as a 100 cm by 100 cm square, then a 1 m cube as 100 cm on each edge. The factors 10,000 and 1,000,000 become geometric rather than arbitrary. For a composite surface mistake, count visible faces instead of repeating the power rule.
A useful correction record has three columns: the wrong assumption, the evidence that disproved it and one new contrasting example. “Cube the factor because it is a solid” is not precise enough. “Cube k only when all corresponding lengths scale by the same k and the target is volume” is a durable rule.
21. Seven days from scales to confident real-world answers
Day one: determine whether given plane figures and solids are similar. Day two: practise direct length, area and volume ratios. Day three: work backward from area and volume ratios using square roots and cube roots. Day four: compare similar and non-similar cylinders, boxes and spheres.
Day five: practise surface area of composite objects and unit conversions. Day six: attempt a mixed set with one capacity problem, one hidden-face problem, one reduction percentage and one invalid similarity assumption. Day seven: revisit the mistakes you actually made and solve one new transfer problem for each.
Parents and tutors can ask “Did every length change by the same factor?” before suggesting a formula. That question is especially useful for a confident student who calculates rapidly but never verifies similarity. A short pause can prevent a high-speed wrong answer.
The goal is not a collection of memorised model answers. It is an independent decision: recognise the dimension of the target, establish the correct factor and return to the actual context.
22. Volume answers need units, boundaries and sense
For a volume of 250 cm³, do not write 250 cm². For an area of 250 cm², do not call it a capacity. A numerically correct answer with the wrong unit is evidence that the physical quantity has slipped. Write the exponent on the unit from the beginning.
Check plausibility before the final line. If the new object is larger in every corresponding dimension, its volume should be larger. If the factor from small to large is 3, volume factor 27 should be greater than area factor 9. If your result reverses this expectation, inspect the direction and exponent.
Finally, ask what the question requested. A larger container’s full capacity, the added capacity, the percentage increase, and the number of original units that can fit in the new volume are related but different targets. The appropriate final sentence names the actual quantity, complete with units.
23. Thirty original questions with worked answers
Take each problem in order and check your decision about similarity and the target dimension before looking at the solution. All quantities and calculations are original teaching examples.
Question 1. A cube’s edge doubles. What happens to its surface area?
Answer. It becomes 4 times as large.
Reason. Area factor is 2².
Question 2. The same cube’s edge doubles. What happens to its volume?
Answer. It becomes 8 times as large.
Reason. Volume factor is 2³.
Question 3. A similar solid has length factor 3. Give its volume factor.
Answer. 27.
Reason. Three spatial dimensions scale, so use 3³.
Question 4. Similar plane areas are in ratio 9:25. Find their corresponding lengths.
Answer. 3:5.
Reason. Take the positive square roots.
Question 5. Similar-solid volumes are in ratio 27:64. Find length ratio.
Answer. 3:4.
Reason. Take the positive cube roots.
Question 6. A small cube has side 4 cm; a large cube has side 10 cm. Find k small to large.
Answer. 2.5.
Reason. Divide 10 by 4 and label the direction.
Question 7. For that cube pair, what is the volume factor?
Answer. 15.625.
Reason. (2.5)³ = 15.625; do not square a volume factor.
Question 8. A 12 cm side becomes 18 cm. Find the area factor for similar plane shapes.
Answer. 2.25.
Reason. k=1.5, so area factor is 1.5².
Question 9. A 12 cm side becomes 18 cm. Find the volume factor for similar solids.
Answer. 3.375.
Reason. k=1.5, so volume factor is 1.5³.
Question 10. Volume of a similar smaller tank is 8 L and the larger is 27 L. Find k.
Answer. 1.5.
Reason. Cube root of 27/8 is 3/2.
Question 11. The small tank is 2 m high. What is the large similar tank’s height?
Answer. 3 m.
Reason. Use the length factor 1.5, not the volume factor.
Question 12. A model is 25 cm high and original is 2 m high. Find model-to-original scale.
Answer. 1:8.
Reason. Convert 2 m to 200 cm before division.
Question 13. Convert 20,000 cm² to m².
Answer. 2 m².
Reason. One square metre is 10,000 cm².
Question 14. Convert 3,000,000 cm³ to m³.
Answer. 3 m³.
Reason. One cubic metre is 1,000,000 cm³.
Question 15. A cylinder’s radius doubles but height is unchanged. Volume factor?
Answer. 4.
Reason. πr²h scales by 2² when only r doubles.
Question 16. A cylinder’s radius and height both double. Volume factor?
Answer. 8.
Reason. Now it is a similar cylinder, so k³ applies.
Question 17. A cube’s side increases by 10 percent. Volume increase?
Answer. 33.1 percent.
Reason. 1.1³ = 1.331, so the increase is 0.331.
Question 18. A solid’s every length falls by 20 percent. Volume factor?
Answer. 0.512.
Reason. 0.8³ = 0.512.
Question 19. What percentage volume decrease is that?
Answer. 48.8 percent.
Reason. 1 − 0.512 = 0.488.
Question 20. Two cubes of side 2 cm are joined face to face. Exposed surface area?
Answer. 40 cm².
Reason. Ten faces remain exposed, each 4 cm².
Question 21. Each little joined cube becomes side 5 cm. Exposed area?
Answer. 250 cm².
Reason. Still ten exposed faces, each area 25 cm².
Question 22. The two original joined cubes have volume?
Answer. 16 cm³.
Reason. 2 × 2³ = 16.
Question 23. The joined large cubes of side 5 cm have total volume?
Answer. 250 cm³.
Reason. 2 × 5³ = 250.
Question 24. Similar signs have areas 36 and 100 cm². Find their linear factor small to large.
Answer. 5/3.
Reason. Square root of 100/36.
Question 25. Small sign width 12 cm. Find corresponding large width.
Answer. 20 cm.
Reason. 12 × 5/3 = 20.
Question 26. Two cuboids are 2×3×5 and 4×7×10. Are they similar?
Answer. No.
Reason. The corresponding length ratios are 2, 7/3 and 2.
Question 27. A uniform solid model has mass 60 g. Every dimension triples, same material. New mass?
Answer. 1620 g.
Reason. Mass scales with volume for the same density: 60×27.
Question 28. A small container holds 480 cm³; its similar copy has k=1.5. New capacity?
Answer. 1620 cm³.
Reason. 480 × 1.5³ = 1620.
Question 29. How much additional capacity is that?
Answer. 1140 cm³.
Reason. Subtract the original 480 cm³ from 1620 cm³.
Question 30. For a scaled similar solid, how does surface-area/volume change when k=2?
Answer. It is halved.
Reason. The ratio gains factor k²/k³ = 1/k.
