When a shape becomes twice as long in every direction, its area does not merely double and its volume does not merely double. The change happens in two dimensions for area and three dimensions for volume. That is the central idea connecting similarity, scale factors and mensuration.
This Secondary 3 Mathematics Learning Guide explains why corresponding lengths scale by k, corresponding areas by k² and corresponding volumes by k³. It then applies that reasoning to triangles, maps, reverse-scale questions, composite shapes and truncated solids. The harder questions are not solved by collecting more formulae; they are solved by deciding which dimensions correspond and which boundaries actually count.
The official 2027 SEC G3 Mathematics syllabus, sections G2 and G5, includes similarity, area and volume ratios, and mensuration of composite figures and solids. This is a G3-oriented learning guide; use the examples appropriate to your school’s sequence.
This article develops the earlier Geometry, Trigonometry and Multi-Step Reasoning guide within the Secondary Mathematics Hub. Choose the diagnostic, scaling laws, composite shapes, practice, or explained answers. The questions and dimensions are original teaching examples.
Similarity Preserves Shape, Not Necessarily Size
Similar figures have equal corresponding angles and proportional corresponding lengths. Congruent figures have the same shape and the same size, so their corresponding lengths are equal. Congruence is therefore the scale-factor-one case of similarity, allowing for a change of position or orientation.
A rotation does not destroy similarity. Nor does placing one triangle upside down on the page. What matters is the correspondence of vertices and sides, not their visual location. A side drawn horizontally in one triangle may correspond to a sloping side in another.
Equal area alone does not establish similarity. A 2 cm by 6 cm rectangle and a 3 cm by 4 cm rectangle both have area 12 cm², but their side proportions differ. There is no single length multiplier that changes one into the other. Before using area or volume scale laws, establish that uniform scaling is actually justified.
Five Questions to Check Before Starting
A triangle has sides 6, 8 and 10 cm. A similar triangle has corresponding sides 9, 12 and 15 cm. What is the enlargement factor? What is the area multiplier? What would the volume multiplier be for similar solids with that same length factor? Convert 1 m² into cm². Finally, explain whether two shapes of equal perimeter must be similar.
The answers are 3/2; 9/4; 27/8; 10,000 cm²; and no. The first three answers should be connected, not memorised independently. The fourth checks dimensional conversion. The fifth checks whether the learner understands the condition required for similarity rather than treating any shared measurement as proof.
If the length factor is correct but the area multiplier is not, focus on dimensions. If the three pairs of sides produce inconsistent factors, focus on correspondence. If the formula is correct but the units are not, focus on conversion. These are different mathematical jobs and deserve different repairs.
Match Corresponding Parts Before Writing a Ratio
If triangle ABC is similar to triangle PQR in that stated order, then A corresponds to P, B to Q and C to R. Therefore AB corresponds to PQ, BC to QR and CA to RP. The vertex order is a compact record of the matching.
Write one direction consistently. To scale from ABC to PQR, use PQ/AB = QR/BC = RP/CA. To scale backwards, use the reciprocal ratios. Mixing “large over small” in one fraction with “small over large” in another changes the relationship unless the factor happens to be one.
A useful diagram habit is to mark corresponding angles before pairing sides. Then describe each side by its endpoints or the angle opposite it. This is more dependable than pairing the two sides that look most alike in a rough drawing.
How Can Similarity Be Established?
For triangles, two equal corresponding angles establish similarity because the third pair is then equal too. Another route is to show that all three pairs of corresponding sides are proportional. A third route uses two pairs of proportional sides and the equal included angle between them.
The word “included” matters. Two side ratios and an unrelated angle do not automatically establish the required shape. Likewise, equal-looking angles in a sketch are not given facts. State the reason for equality, such as corresponding angles formed by parallel lines, or use the angle information explicitly supplied.
For example, triangles with angles 50°, 70° and 60° are similar regardless of their sizes. Triangles with side lengths 6, 8, 10 and 9, 12, 15 are similar because every corresponding side is multiplied by 3/2. The examples use different evidence to establish the same relationship.
Worked Example 1: Similarity Inside One Triangle
In triangle ABC, D lies on AB and E lies on AC, with DE parallel to BC. Given AD = 6 cm, DB = 3 cm, AE = 8 cm and BC = 15 cm, find AC and DE. Draw the large triangle, place D and E on their stated sides, and mark the parallel segments. The small triangle is ADE; the large triangle is ABC.
The triangles share angle A, and the parallel lines give equal corresponding angles at D and B. Hence triangle ADE is similar to triangle ABC. The whole side AB is 6 + 3 = 9 cm, not 3 cm. The small-to-large length ratio is AD/AB = 6/9 = 2/3.
Thus AE/AC = 2/3, so 8/AC = 2/3 and AC = 12 cm. Also DE/BC = 2/3, so DE = (2/3)(15) = 10 cm. The length EC would be 12 − 8 = 4 cm.
The frequent error 6/3 uses a small segment and the leftover segment, rather than corresponding sides of the two similar triangles. The equation must compare AD with AB. Similarity does not make every visible segment a corresponding length.
Why Area Scales by the Square of the Length Factor
Take a rectangle with length L and width W. Its area is LW. Under a uniform enlargement by positive factor k, the dimensions become kL and kW. The new area is (kL)(kW) = k²LW. Two independent length directions have each contributed one factor of k.
The same reasoning applies to a triangle’s base and perpendicular height. The new area is 1/2 × kb × kh = k²(1/2 bh). The factor 1/2 does not change because the shape type has not changed. For a circle, replacing r with kr gives π(kr)² = k²πr².
Perimeter, by contrast, is a sum of lengths. If each side is multiplied by k, their sum is multiplied by k. Thus perimeter follows k while area follows k². The distinction comes from what is being measured, not from which formula happens to be written first.
Why Volume Scales by the Cube
A cuboid has volume LWH. Enlarging every length by k gives (kL)(kW)(kH) = k³LWH. For a cylinder, both radius and height scale: π(kr)²(kh) = k³πr²h. A cone follows the same relationship because its unchanged factor of 1/3 does not alter the three length factors.
Uniformity is essential. Doubling a cylinder’s radius while leaving its height unchanged multiplies its volume by four, not eight. That change does not produce a similar cylinder because the radius-to-height proportion changes. The phrase “all corresponding lengths” is therefore part of the rule.
| Measurement under uniform scaling | Multiplier |
|---|---|
| Corresponding length, height, radius or perimeter | k |
| Corresponding area or total surface area | k² |
| Corresponding volume | k³ |
Worked Example 2: Area of the Part Left Over
In a new triangle ABC, D lies on AB and E lies on AC, with DE parallel to BC and AD:AB = 2:3. The smaller triangle ADE has area 48 cm². Find the area of triangle ABC and the remaining quadrilateral DBCE. This example uses a separate set of dimensions; use the ratio and area given here, not the side lengths from Worked Example 1.
The small-to-large area ratio is 4:9. Therefore the large area is 48 × 9/4 = 108 cm². The remaining region has area 108 − 48 = 60 cm².
The area ratio applies to the two similar triangles, not directly to the small triangle and the leftover quadrilateral. Once the whole area is known, subtraction finds the remainder. Equivalently, the remainder is 1 − 4/9 = 5/9 of the whole, not 1/3 of the whole. A linear ratio cannot be substituted for an area fraction.
Worked Example 3: Recover a Length Factor From an Area Ratio
Two similar plane figures have areas in the ratio 25:49. A corresponding side of the smaller figure is 15 cm. Find the side of the larger figure. The length ratio is the positive square root of the area ratio, giving 5:7.
The required side is 15 × 7/5 = 21 cm. Multiplying 15 by 49/25 would use an area multiplier on a length and would be incorrect. If the smaller area were 75 cm², however, the larger area would be 75 × 49/25 = 147 cm². The same ratio information supports different multipliers depending on the requested measurement.
Worked Example 4: Recover a Length Factor From a Volume Ratio
Two similar solids have volumes in the ratio 64:125. The larger has height 20 cm. Find the smaller height and the ratio of their surface areas. Taking cube roots gives a small-to-large length ratio of 4:5.
The smaller height is 20 × 4/5 = 16 cm. Squaring the length ratio gives a surface-area ratio of 16:25. Do not square the volume ratio; first return to the underlying length factor, then move to the required dimension.
This suggests a useful central route for reverse problems: supplied ratio → length factor → required ratio. Moving through the length factor reduces the chance of taking the wrong root or power.
Percentage Changes Are Multipliers, Not Labels
An increase of 20% in every length means k = 1.20. The area multiplier is 1.20² = 1.44, an increase of 44%. The volume multiplier is 1.20³ = 1.728, an increase of 72.8%. Neither result is obtained by simply doubling or tripling the percentage.
To reverse a volume change, take a cube root. A similar container with 50% greater geometric capacity needs a length factor of 1.51/3, approximately 1.1447. Its corresponding dimensions increase by about 14.5%, not 50%. This assumes the stated capacity follows geometrically similar internal dimensions; a real container’s wall thickness and usable fill level may require a more detailed model.
The index notation is developed in Indices, Standard Form and Estimation. Here it represents a geometric question: which one-dimensional change produces the required three-dimensional change?
Worked Example 5: Map Scale for Distance and Area
A map uses scale 1:25,000. A route measures 3.6 cm on the map, and a region has map area 2.4 cm². Find the actual route length and region area. The scale compares lengths in the same unit. One centimetre on the map represents 25,000 cm, or 250 m, in reality.
The route length is 3.6 × 250 = 900 m. For area, one square centimetre represents 250 m × 250 m = 62,500 m². Therefore the region’s actual area is 2.4 × 62,500 = 150,000 m², or 0.15 km².
The map area is not multiplied by 25,000 alone. It is multiplied by the square of the length scale, with compatible units. Converting the one-centimetre length scale to 250 m before squaring makes the unit relationship especially clear.
Dimensional Conversion Before Mensuration
Since 1 m = 100 cm, a square metre contains 100 × 100 = 10,000 square centimetres. A cubic metre contains 100 × 100 × 100 = 1,000,000 cubic centimetres. These are not separate arbitrary conversion facts; they follow from applying the length conversion in each dimension.
Consequently, 0.36 m² = 3,600 cm², and 125,000 cm³ = 0.125 m³. In volume contexts, 1,000 cm³ equals 1 litre. State whether a question asks for geometric volume, liquid capacity, surface area or perimeter before selecting a conversion.
When a diagram mixes centimetres and metres, convert to one length unit before applying a formula unless there is a clear reason not to. Multiplying 2 m by 30 cm as if the numbers had the same unit produces neither a valid area in m² nor a valid area in cm².
Composite Shapes: The Boundary and the Region Are Different Objects
Area measures the region covered. Perimeter measures the boundary around that region. Adding or removing a piece affects these two quantities differently. A line used to divide a composite shape for area calculation may lie inside the final figure and therefore contribute nothing to its perimeter.
Before calculating, describe the shape in words: a rectangle with a semicircular notch removed, a cylinder capped by a hemisphere, or a cone with its top cut off parallel to the base. Then identify which dimensions are known, which must be derived, and which surfaces remain exposed.
Worked Example 6: A Semicircular Notch
A rectangle is 12 cm long and 8 cm wide. A semicircular notch of radius 4 cm is cut inward from one 8 cm side, with that side as the semicircle’s diameter. Find the remaining area and perimeter.
The original rectangular area is 96 cm². The removed semicircle has area 1/2 × π × 4² = 8π cm². The remaining area is therefore 96 − 8π cm².
The new boundary contains the two 12 cm sides, the opposite 8 cm side, and the semicircular arc. The arc length is half of 2πr, so it is 4π cm. The perimeter is 32 + 4π cm. The removed diameter is not part of the new boundary and must not be counted again.
This example shows why adding the perimeters of component shapes usually fails. Component boundaries can disappear when shapes join or when material is removed. Trace the outside of the final region instead of adding formula results mechanically.
Worked Example 7: A Cylinder With a Hemispherical Top
A solid consists of a cylinder of radius 3 cm and height 8 cm, capped by a hemisphere of the same radius. Its circular bottom is closed. Find the volume and total external surface area. The cylinder and hemisphere share a circular interface, which is internal to the assembled solid.
The cylinder’s volume is π × 3² × 8 = 72π cm³. The hemisphere’s volume is half of 4πr³/3, giving 18π cm³. Total volume is 90π cm³. There is no overlap of their interior volumes, so addition is appropriate.
The exposed cylinder curve has area 2πrh = 48π cm². The hemisphere’s curved area is 2πr² = 18π cm². The bottom circle contributes 9π cm². Total external surface area is therefore 75π cm².
Neither of the two circular faces at the join is exposed. Adding the separate total surface areas would count both of those internal faces. For surface-area questions, a correct formula is not enough; the correct inventory of exposed surfaces is also necessary.
Truncated Solids: Recover the Missing Similar Shape
A frustum is the remaining part of a cone or pyramid after a smaller similar top portion is removed by a cut parallel to the base. The two full pointed shapes share a vertex before the cut, which makes their heights and radii or base lengths proportional.
The similarity comparison is between the original full cone and the removed small cone. The frustum itself is not similar to either full cone. This distinction is important when deciding which heights can be compared and which volumes can be subtracted.
Worked Example 8: Frustum Volume From Similarity
A cone has base radius 6 cm and perpendicular height 18 cm. Its top is removed by a plane parallel to the base, 6 cm below the vertex. Find the remaining volume. The removed cone has height 6 cm, so its length factor relative to the full cone is 6/18 = 1/3.
The removed radius is therefore 6/3 = 2 cm. Full-cone volume is 1/3 × π × 6² × 18 = 216π cm³. Removed volume is 1/3 × π × 2² × 6 = 8π cm³. The frustum volume is 208π cm³.
A second route uses the volume ratio 1:27. The removed cone occupies 1/27 of the full volume, so the remaining fraction is 26/27. Multiplying 216π by 26/27 again gives 208π. The two routes independently connect the similarity and the volume calculation.
The frustum height is 18 − 6 = 12 cm, but using 12/18 as the small-cone factor would be wrong. That ratio compares the leftover height with the whole height, not corresponding dimensions of the two similar cones.
Worked Example 9: Frustum Surface Area Requires a New Face
For the same frustum, find the total external surface area, including both circular ends. The full cone’s slant height is √(18² + 6²) = √360 = 6√10 cm. By similarity, the removed cone’s slant height is 2√10 cm.
The full curved area is π × 6 × 6√10 = 36π√10 cm². The removed curved area is π × 2 × 2√10 = 4π√10 cm². Their difference is 32π√10 cm².
The exposed bottom circle contributes 36π cm², and the new top circle created by the cut contributes 4π cm². Total external area is therefore 32π√10 + 40π cm².
Subtracting the small cone’s entire surface area from the large cone’s entire surface area would remove a circular face when the cut actually creates one. Volume subtraction and surface-area subtraction do not follow identical bookkeeping. For area, name every exposed face after the change.
Worked Example 10: Design a Similar Cuboid From a Target Volume
A cuboid model has dimensions 10 cm, 8 cm and 6 cm. A geometrically similar solid must have volume 1,620 cm³. Find its dimensions and total surface area. The original volume is 10 × 8 × 6 = 480 cm³.
The volume multiplier is 1,620/480 = 27/8. Therefore the length multiplier is 3/2. The new dimensions are 15 cm, 12 cm and 9 cm. Their product is 1,620 cm³, which checks the target.
The original surface area is 2(10 × 8 + 10 × 6 + 8 × 6) = 376 cm². Multiply by (3/2)² = 9/4 to obtain 846 cm². Direct calculation gives 2(15 × 12 + 15 × 9 + 12 × 9) = 846 cm² as well.
For an actual hollow package, internal capacity and external surface area may refer to different dimensions because of wall thickness. This exercise treats an ideal solid with exact geometric dimensions. A model should state what it represents before its result is used for a practical decision.
Useful Checks Before Accepting a Mensuration Answer
First check the dimension of the answer. A length needs linear units, an area square units and a volume cubic units. Then check direction: enlarging similar shapes by k > 1 must increase all three measurements, although by different factors. A smaller solid cannot have a larger volume than its similar parent under a factor below one.
Next compare part and whole. A removed region cannot have a negative area, and a remainder cannot exceed the original when material has only been removed. For a composite solid, inspect whether an interface was mistakenly counted as external surface area. These checks target the structural errors that a calculator cannot identify.
Finally preserve exact values, such as multiples of π, until a decimal approximation is requested. If a question specifies accuracy, apply it to the final result rather than rounding each component independently. This is especially useful when two large areas are subtracted to produce a smaller remaining area.
Independent Practice: Decide the Dimension First
For each question, state whether you are using a length, area or volume multiplier before calculating. For composite shapes, list the included and excluded boundaries or surfaces.
1. Similar triangles have corresponding sides 8 cm and 12 cm. Another side of the smaller triangle is 10 cm. Find the corresponding larger side.
2. For the triangles in Question 1, the smaller area is 40 cm². Find the larger area.
3. Similar figures have area ratio 9:16. Their larger perimeter is 52 cm. Find the smaller perimeter.
4. Similar solids have volume ratio 27:64. The larger height is 20 cm. Find the smaller height and the surface-area ratio.
5. Every length of a similar solid is increased by 10%. Find the percentage increase in volume.
6. A map scale is 1:2,000. A line measures 4.5 cm. Find its actual length in metres.
7. On the same map, a region has area 3 cm². Find its actual area in m².
8. Convert 0.36 m² to cm² and 125,000 cm³ to m³.
9. An open-top cylindrical container has radius 4 cm and height 10 cm. Ignoring thickness, find the area of material needed for the curved wall and bottom.
10. A cone has radius 9 cm and height 12 cm. A similar top cone of height 4 cm is removed by a parallel cut. Find the remaining volume.
11. Two similar triangles have length ratio 3:5. The smaller lies inside the larger, and the larger area is 200 cm². Find the area outside the smaller triangle but inside the larger.
12. A cylinder’s radius doubles while its height stays unchanged. By what factor does its volume change, and are the cylinders similar?
13. Explain why two rectangles of equal area need not have the same perimeter.
14. A similar solid has eight times the original volume. What happens to its lengths and total surface area?
Explained Answers
1. The length factor is 12/8 = 3/2. The required side is 10 × 3/2 = 15 cm.
2. The area factor is (3/2)² = 9/4, so the larger area is 40 × 9/4 = 90 cm². Using 3/2 directly would incorrectly scale area as a length.
3. Take the positive square roots of 9:16 to get length ratio 3:4. Perimeter is a length measurement, so the smaller perimeter is 52 × 3/4 = 39 cm.
4. The length ratio is 3:4. The smaller height is 20 × 3/4 = 15 cm. The surface-area ratio is 9:16.
5. The length factor is 1.1, and the volume factor is 1.1³ = 1.331. Subtract 1 to obtain a 33.1% increase.
6. One map centimetre represents 20 m. Therefore the actual length is 4.5 × 20 = 90 m.
7. One square centimetre represents 20² = 400 m². The actual area is 3 × 400 = 1,200 m².
8. Multiply 0.36 by 10,000 to obtain 3,600 cm². Divide 125,000 by 1,000,000 to obtain 0.125 m³.
9. The curved area is 2π(4)(10) = 80π cm². The bottom is π(4²) = 16π cm². Total material area is 96π cm². There is no top circle to include.
10. The removed cone has length factor 4/12 = 1/3, hence radius 3 cm. Full volume is 324π cm³ and removed volume is 12π cm³. The remaining volume is 312π cm³.
11. The small-to-large area ratio is 9:25. The smaller area is 200 × 9/25 = 72 cm². The requested remaining area is 200 − 72 = 128 cm².
12. Volume becomes π(2r)²h = 4πr²h, so it is multiplied by four. The cylinders are not similar because the radius has changed by factor two while the height has changed by factor one.
13. A 2 cm by 6 cm rectangle and a 3 cm by 4 cm rectangle both have area 12 cm². Their perimeters are 16 cm and 14 cm. Equal area does not determine either the proportions or the perimeter.
14. The length factor is the cube root of eight, which is two. Total surface area is multiplied by 2² = 4. The original similarity condition is essential to this conclusion.
From an Error to a Better Diagram
When a solution is wrong, redraw only the relationships needed to find the first error. If the problem concerns correspondence, separate the two triangles and label matched vertices. If it concerns a truncated cone, draw the full cone and mark the removed and remaining heights. If it concerns exposed area, list each surface rather than beginning with a formula.
This targeted redraw is different from producing a neat copy of the original picture. It is a new representation built to answer a specific question about the reasoning. A simple labelled sketch that makes the missing relationship visible is more useful than a polished diagram that repeats the original ambiguity.
A Suggested Three-Stage Practice Route
Stage one: recognise. Identify pairs of similar shapes, match corresponding dimensions and reject examples that have only equal area or one matching angle. Explain why each scale factor is valid before using it.
Stage two: transfer. Move from length ratios to area ratios, from volume ratios back to lengths, and from percentages to multipliers. Mix the requested dimensions so the learner must decide which power or root is appropriate.
Stage three: model. Combine similarity with subtraction, unit conversion, Pythagoras or surface accounting. The frustum examples show how several familiar ideas can form one longer route. Keep intermediate quantities labelled so that a radius is not accidentally used as a diameter or a perpendicular height as a slant height.
Teacher and Parent Questions That Help
Ask “Which two shapes are similar?” before asking for a ratio. Ask “Is the question asking for length, area or volume?” before suggesting a power. For a composite shape, ask “Can you trace the boundary or name the exposed surfaces?” These prompts make the choice visible instead of supplying the next formula.
For a learner needing support, use clearly separated similar rectangles and triangles before nested diagrams. For a learner ready for extension, ask how surface-area-to-volume ratio changes under uniform scaling. Since area gains k² and volume gains k³, their ratio is divided by k. This is a geometric conclusion; applying it to a physical process requires additional assumptions about that process.
Questions Students Often Ask
Does a scale factor have to be greater than one? No. A positive factor below one describes reduction. The factors used here are positive ratios of corresponding lengths; signed coordinate-enlargement conventions are a separate representation question.
Why take a square root of an area ratio? Because the area ratio already contains two copies of the length factor. The positive square root returns the one-dimensional ratio.
Can I add the surface areas of solids that have been joined? Only after removing faces that become internal. The cylinder-and-hemisphere example shows why the join must be inspected.
Should I use the whole height or the leftover height? Use the heights belonging to the two similar shapes. For a cut cone, compare the full cone with the removed cone, then use subtraction for the frustum.
Continue the Secondary 3 Learning Route
Read Quadratic Equations and Word Problems when unknown dimensions create squared relationships; Algebraic Fractions and Formula Rearrangement when a mensuration formula needs a different subject; and Indices, Standard Form and Estimation for roots, powers, units and numerical checking.
Good mensuration begins by deciding what is being measured and which relationships justify the calculation. Match the shapes, choose the dimension, identify the boundary, and only then calculate. Return to the Secondary Mathematics Hub for the complete learning route.
