Understand when changes to lengths or angles produce similar or non-similar figures, and apply the scaling of lengths, areas and volumes by k, k² and k³

Learning goal: Understand how and when changes to the measures of a figure, its lengths or its angles, result in similar figures and when they result in non-similar figures; and, when figures are similar, understand and apply the fact that scaling a figure by a factor k multiplies its lengths by k, its areas by k² and its volumes by k³.

Similar figures, defined with conditions. Two figures are similar when one can be carried onto the other by a similarity transformation: a dilation with some positive scale factor k, possibly combined with rigid motions. The consequences that we use in practice are that corresponding angles are equal and every corresponding length is multiplied by the same k. Both conditions matter. A figure and its image under a dilation are similar by definition; a figure whose angles have been kept but whose lengths have been changed by different factors is not.

Which changes keep similarity. Multiply every length of a figure by the same positive factor k and keep every angle: the result is similar to the original, with scale factor k. That is what a dilation does, whether k is larger than 1 (an enlargement), between 0 and 1 (a reduction) or exactly 1 (a congruent copy). For some families of figures no other change is possible without leaving the family: all circles are similar, all squares are similar, all cubes and all spheres are similar, because a single length determines each one. For triangles, the angles alone fix the shape: two triangles with equal angles are similar (the angle-angle criterion), so a change that alters the set of three angle measures produces a non-similar triangle, and any change that keeps all three angle measures produces a similar one (relabeling the vertices changes no measure, so it changes no shape).

Which changes break similarity. Changing one length but not the others usually breaks it, but the test is whether any correspondence works, not only the one you started with: a 3 by 4 rectangle cut down to 3 by 2.25 is similar to the original after a quarter turn, matching 4 with 3 and 3 with 2.25, both ratios 0.75. A 3 by 4 rectangle stretched to 3 by 8 has all its angles still 90°, yet the ratios of corresponding sides are 3/3 = 1 and 8/4 = 2, not a single factor, and the other pairing (3 with 8, 4 with 3) gives 8/3 and 3/4, so the rectangles are not similar; equal angles are not enough for quadrilaterals. Changing an angle breaks it: a 3 by 4 rectangle pushed into a parallelogram with sides 3 and 4 and angles 60° and 120° has the same side lengths but different angles, so it is not similar to the rectangle. Scaling different lengths by different factors breaks it: the 3-4-5 triangle with only the shortest side doubled becomes a 6-4-5 triangle, which is not even a right triangle, since 4² + 5² = 41 ≠ 36. A cylinder of radius 2 and height 5 whose height alone is doubled becomes a 2 by 10 cylinder, and the ratios 2/2 = 1 and 10/5 = 2 differ, so the cylinders are not similar; a cylinder is similar to another cylinder only when radius and height are scaled by the same factor. The check is always the same: is there a correspondence under which all corresponding angles are equal and all corresponding lengths are in one common ratio? If no correspondence works, the figures are not similar.

Lengths, areas and volumes under a scale factor k. When figures are similar with scale factor k, every length is multiplied by k, every area by k² and every volume by k³. The reason is in the formulas: an area is a product of two lengths, and if each is multiplied by k the product is multiplied by k × k = k²; a volume is a product of three lengths, so it is multiplied by k³. This holds for every shape, not only rectangles and boxes, because any area can be approximated by small squares and any volume by small cubes, and each square or cube scales by k² or k³. Take a box 2 cm by 3 cm by 4 cm: surface area 2(6 + 8 + 12) = 52 cm², volume 24 cm³. Scale by k = 3: the box is 6 by 9 by 12, surface area 2(54 + 72 + 108) = 468 cm² = 52 × 9, volume 648 cm³ = 24 × 27. Scale by k = ½: 1 by 1.5 by 2, surface area 13 cm² = 52 × ¼, volume 3 cm³ = 24 × ⅛. A 3-4-5 triangle has area 6; the similar 6-8-10 triangle has area 24 = 6 × 2². A sphere of radius 2 has surface area 16π and volume 32π/3; the similar sphere of radius 5, with k = 2.5, has surface area 100π = 16π × 6.25 and volume 500π/3 = (32π/3) × 15.625, and 6.25 = 2.5² and 15.625 = 2.5³.

Going backwards. The rule also runs in reverse. If two similar solids have surface areas in the ratio 9 : 25, then k² = 9/25, so k = 3/5, and their volumes are in the ratio k³ = 27/125. If two similar cylinders have volumes 20π and 160π, then k³ = 8, so k = 2, and their radii and heights are in the ratio 1 : 2 and their surface areas in the ratio 1 : 4. A model built at scale 1 : 20 has every area 1/400 of the real object's and every volume 1/8,000 of it: paint needed scales with area, material scales with volume.

When the k² and k³ rule does not apply. The rule is a statement about similar figures. Doubling only the height of a cylinder doubles its volume, not multiplies it by 8, because that change is not a similarity. Doubling only the radius multiplies the volume by 4, since r² appears and h does not change. A student who sees "doubled" and writes "times 8" must first ask whether every length was doubled. Angles, incidentally, are not multiplied by k at all; they are unchanged by a similarity.

A worked model inside the lesson. A cylindrical can has radius 2 cm and height 5 cm, volume π × 4 × 5 = 20π cm³. A similar can at scale k = ½ has radius 1 cm and height 2.5 cm, volume π × 1 × 2.5 = 2.5π cm³, and 2.5π ÷ 20π = 1/8 = (½)³, as the rule predicts. Its curved surface area went from 2π × 2 × 5 = 20π cm² to 2π × 1 × 2.5 = 5π cm², a quarter, as k² = ¼ predicts. A different can with radius 2 cm and height 10 cm is not similar to the original (ratios 1 and 2), and its volume 40π cm³ is simply double, because only one length changed.

Misconceptions, and why they are wrong. "Equal angles make figures similar" is true for triangles only; a 3 by 4 and a 3 by 8 rectangle are the counterexample. "Doubling a solid doubles its volume" is wrong for similar solids, where volume is multiplied by 8, and right only for a stretch in one direction, such as doubling a prism's or cylinder's height with its base unchanged, which is not a similarity (doubling a cylinder's radius alone multiplies its volume by 4). "Areas scale by k" mistakes a two-dimensional measure for a one-dimensional one; the box's surface area went from 52 to 468, nine times, under k = 3. And "k must be a whole number" is false; k = ½, k = 2.5 and k = 3/5 all appeared above.

Worked similarity check: Decide whether a 6 by 9 by 12 box is similar to a 2 by 3 by 4 box, and whether a 4 by 6 by 10 box is. Corresponding edges of the first pair are in ratios 6/2 = 3, 9/3 = 3 and 12/4 = 3, one common factor, and all angles are right angles, so they are similar with k = 3. The second pair has ratios 4/2 = 2, 6/3 = 2 and 10/4 = 2.5, not a common factor, so the boxes are not similar even though every angle matches. The volume of the 4 by 6 by 10 box, 240 cm³, is 10 times 24 cm³, which would correspond to k = ∛10 ≈ 2.15 if the boxes were similar; since no edge ratio is 2.15, the volume ratio confirms the boxes are not similar. A volume ratio alone can never prove similarity; only the edge ratios can.

Worked forward scaling: A cone has radius 3 cm, height 4 cm, slant height 5 cm, volume ⅓ × π × 9 × 4 = 12π cm³ and curved surface area π × 3 × 5 = 15π cm². A similar cone with k = 2 has radius 6 cm, height 8 cm and slant height 10 cm. Its volume is ⅓ × π × 36 × 8 = 96π cm³, which is 12π × 8, and its curved surface area is π × 6 × 10 = 60π cm², which is 15π × 4. Lengths doubled, area quadrupled, volume multiplied by 8.

Worked reverse scaling: Two similar statues have surface areas 0.36 m² and 2.25 m². Then k² = 2.25/0.36 = 6.25, so k = 2.5. Their heights are in the ratio 1 : 2.5, and their volumes are in the ratio k³ = 15.625. If the small statue has volume 0.008 m³, the large one has 0.008 × 15.625 = 0.125 m³. Taking the square root of the area ratio, not the ratio itself, is the step students skip.

Worked non-similar change: A cylindrical tank of radius 1.5 m and height 4 m holds π × 2.25 × 4 = 9π m³. A second tank keeps the radius and doubles the height to 8 m: volume 18π m³, exactly double, because only one length changed and the tanks are not similar. A third tank doubles both radius and height, to 3 m and 8 m: volume π × 9 × 8 = 72π m³, eight times the original, as k = 2 for a similar solid predicts. The difference between "double" and "eight times" is whether the change was a similarity.

Scaling record: Take the 2 cm by 3 cm by 4 cm box and make a table with rows for k = ½, 1, 2, 3 and 10 and columns for the edge lengths, the surface area 2(ab + bc + ca) and the volume abc. The entries should be 13, 52, 208, 468 and 5,200 cm² for surface area and 3, 24, 192, 648 and 24,000 cm³ for volume. Add two columns showing surface area ÷ 52 and volume ÷ 24, and confirm that they equal k² and k³ in every row. Then add one row for a box that is not similar, 2 by 3 by 8, and explain why its volume ratio 48 ÷ 24 = 2 does not come from a similarity: it would need every edge multiplied by ∛2 ≈ 1.26, but the edge ratios are 1, 1 and 2.

Independent performance: Design a scale model problem. Choose a real solid object (a water tower, a shipping container, a statue, a sports ball) with dimensions you state as hypothetical, choose a scale factor k that is not a whole number, and compute the model's lengths, one area and its volume. Then give one change to the object that is not a similarity (only one dimension altered) and compute the new volume directly. Success criteria: the similarity check (equal angles and one common ratio for all lengths) is written out for the model; the area is multiplied by k² and the volume by k³, with each factor computed from k; the non-similar change is computed from the formula, not from k³; one sentence explains why the k³ rule was correct for the model and wrong for the altered object; and a reverse question is included and solved, such as finding k from a given area ratio by taking a square root.