Homothety: scaling from a point
Choose a point O (the centre) and a number k (the ratio). A homothety sends every point P to the point P′ on the ray OP with OP′ = k × OP. If k = 2, every point goes twice as far from O. If k = ½, every point moves halfway towards O. If k is negative, P′ goes to the opposite side of O.
Two things never change: angles and shape. Parallel lines stay parallel. So the image is similar to the original. Two solids are similar when one is the image of the other by a homothety, possibly followed by moving it around (turning or flipping) without changing size.
Lengths, areas and volumes under scaling
If the ratio is k (take k > 0):
- Every length (edge, height, radius, slant height) becomes k times.
- Every surface area becomes k² times, because an area is two lengths multiplied.
- Every volume becomes k³ times, because a volume is three lengths multiplied.
Cube check: edge a → ka. Area 6a² → 6k²a². Volume a³ → k³a³. Notice that you can use ratios even when you do not know the shapes: if two similar solids have lengths in the ratio m : n, their areas are in the ratio m² : n² and their volumes in m³ : n³.
Surface area formulas of common solids
The surface area is the total area of the outside. Use a flattened net to see it.
- Prism: 2 × base area + perimeter of base × height (the sides unrolled form a rectangle).
- Cylinder: lateral 2πrh + two bases 2πr², total 2πr(h + r).
- Cone: lateral πrl (l = slant height = √(r² + h²)) + base πr², total πr(l + r).
- Sphere: 4πr² (exactly four times its great circle).
- Regular pyramid: base area + ½ × perimeter × slant height of a face.
Each formula is made of two lengths multiplied, so under scaling all of them become k² times bigger.
Cutting a cone or pyramid: a similar piece
Cut a cone or pyramid by a plane parallel to its base. The top piece is a smaller cone or pyramid similar to the whole, with centre O at the apex. If the cut is at a fraction k of the height from the apex, then the small solid has all lengths × k, surface area × k², volume × k³. For k = ½: the top has ⅛ of the volume, so the lower part (the frustum) has ⅞ of it.
Solving similarity problems
- Find the ratio of any two matching lengths. That is k.
- For area, use k². For volume, use k³.
- If given areas or volumes, take a square root or cube root to get k back: k = √(area ratio) = ∛(volume ratio).
- Check units: cm → cm² → cm³.
Careful: the ratio of volumes is not the same as the ratio of lengths. Mixing them is the commonest mistake.
Try it: sugar cubes
Take 8 small sugar cubes (or dice). Make a 2 × 2 × 2 big cube. Count the squares on one face (4) and the cubes inside (8). Now predict: for a 3 × 3 × 3 cube, how many squares on a face, and how many cubes? Check with the 3D by sliding k to 3.
Key formulas and definitions
- Homothety: OP′ = k · OP
- Lengths × k, areas × k², volumes × k³
- Ratio of lengths m : n gives areas m² : n² and volumes m³ : n³
- Cylinder: S = 2πr(h + r), V = πr²h
- Cone: S = πr(l + r), l = √(r² + h²), V = ⅓πr²h
- Sphere: S = 4πr², V = (4/3)πr³
- Cut parallel to the base at fraction k of the height from the apex: small volume = k³ × whole
Worked examples
1. A cube has edge 3 cm. It is scaled by k = 2. Find the new surface area and volume.
Original S = 6 × 9 = 54 cm², V = 27 cm³. New S = 54 × 4 = 216 cm². New V = 27 × 8 = 216 cm³.
2. Two similar cones have heights 6 cm and 9 cm. The small one has volume 40 cm³. Find the volume of the larger one.
k = 9/6 = 1.5. Volume × k³ = 40 × 3.375 = 135 cm³.
3. A sphere of radius 3 cm is scaled by k = 2. Find the new surface area.
New radius = 6 cm. S = 4π × 36 = 144π cm². (Check: original 36π × 4 = 144π.)
4. A cone has r = 3 cm and l = 5 cm. Find its total surface area, and the area of a similar cone with scale factor 3.
S = πr(l + r) = π × 3 × 8 = 24π cm². Similar cone: 24π × 9 = 216π cm².
5. A cone of volume 240 cm³ is cut by a plane parallel to its base halfway up the height. Find the volume of the top cone and of the frustum.
k = ½, so top cone = 240 × ⅛ = 30 cm³. Frustum = 240 − 30 = 210 cm³.
6. The volumes of two similar solids are in the ratio 8 : 27. Find the ratio of their surface areas.
k = ∛(8/27) = 2/3. Area ratio = k² = 4/9, so 4 : 9.
7. A 1 : 20 model of a statue needs 0.5 m² of paint. How much paint area does the real statue have, and how many times larger is its volume?
k = 20. Area = 0.5 × 400 = 200 m². Volume is 20³ = 8000 times larger.
Common mistakes
- Multiplying areas by k instead of k². If the sides double, the area becomes four times.
- Using the ratio of lengths for volumes. Volumes use k³.
- Using the cone's height h instead of the slant height l in the lateral surface formula πrl.
- Forgetting to take the cube root when the problem gives volumes and asks for lengths (k = ∛(volume ratio)).