Area and Volume Ratios
Discover why doubling every length makes the area four times larger and the volume eight times larger. Master the power rule: linear : area : volume = k : k² : k³.
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A small square has side length 2 cm. A larger similar square has side length 6 cm. Without calculating, guess: how many times larger is the area? How many times larger would the volume be if these were cubes instead of squares? What pattern do you notice?
When a shape is enlarged or reduced by a linear scale factor $k$, its area changes by $k^2$ and its volume changes by $k^3$. This is one of the most powerful rules in geometry -- and one of the most commonly tested in exams.
The relationship is linear : area : volume = $k$ : $k^2$ : $k^3$. If you double every length ($k = 2$), the area becomes $2^2 = 4$ times larger and the volume becomes $2^3 = 8$ times larger. If you halve every length ($k = \frac{1}{2}$), the area becomes $(\frac{1}{2})^2 = \frac{1}{4}$ and the volume becomes $(\frac{1}{2})^3 = \frac{1}{8}$.
Know
- Linear : area : volume ratios follow $k : k^2 : k^3$
- How to calculate scale factors from given ratios
Understand
- Why area scales as the square and volume as the cube of the linear scale factor
- How to work backwards from area or volume to find the linear scale factor
Can Do
- Find unknown areas and volumes using scale factors
- Solve model and map scale problems involving area and volume
- Apply the power rule to cylinders, spheres and composite shapes
Wrong: If the linear scale factor is $k$, then the area scale factor is also $k$.
Right: The area scale factor is $k^2$, and the volume scale factor is $k^3$. Length, area and volume scale differently.
Wrong: Area and volume scale factors are added, not multiplied.
Right: Area scale factor = $(\text{linear scale factor})^2$. Volume scale factor = $(\text{linear scale factor})^3$. They are powers, not additions.
When every length in a shape is multiplied by $k$, the area is multiplied by $k^2$. This happens because area is a two-dimensional measurement -- it depends on two lengths multiplied together.
Consider a rectangle with length $l$ and width $w$. Its area is $A = l \times w$. If we scale by factor $k$, the new dimensions are $kl$ and $kw$. The new area is $kl \times kw = k^2 lw = k^2 A$. The same logic applies to every shape -- triangles, circles, trapeziums -- because every area formula involves multiplying two lengths.
When every length in a solid is multiplied by $k$, the volume is multiplied by $k^3$. Volume is a three-dimensional measurement -- it depends on three lengths multiplied together.
Consider a box with dimensions $l \times w \times h$. Its volume is $V = lwh$. After scaling by $k$, the new volume is $kl \times kw \times kh = k^3 lwh = k^3 V$. This applies to every solid -- spheres, cylinders, cones -- because volume formulas all involve multiplying three lengths (or equivalent).
Most exam questions give you one piece of information and ask for another. You might know the linear scale factor and need the volume, or know the area ratio and need the linear scale factor. The key is identifying what you have and what you need, then applying the correct power.
Forward: Given $k$, find area ratio ($k^2$) or volume ratio ($k^3$).
Backward: Given area ratio, find $k = \sqrt{\text{area ratio}}$. Given volume ratio, find $k = \sqrt[3]{\text{volume ratio}}$.
Map scales are a classic application: a map scale of 1:50,000 means $k = \frac{1}{50,000}$. A 4 cm by 3 cm field on the map has actual dimensions $4 \times 50,000$ cm by $3 \times 50,000$ cm.
Watch Me Solve It · 3 examples
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1Identify the linear scale factorScale 1:24 means $k = \frac{1}{24}$ (model / real)The model is smaller than the real car, so $k < 1$.
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2Find the model lengthModel length = $4.8 \div 24 = 0.2$ m = $20$ cmLinear measurements scale by $k$.
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3Find the volume scale factorVolume scale factor = $k^3 = (\frac{1}{24})^3 = \frac{1}{13,824}$Capacity (volume) scales as the cube of the linear factor.
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4Find the model fuel tank capacityModel capacity = $48 \div 13,824 \approx 0.00347$ L $\approx 3.47$ mLCheck: the model is much smaller, so the fuel tank should be tiny. 3.47 mL is reasonable.
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1Find actual dimensionsActual length = $4 \times 50,000$ cm = $200,000$ cm = $2,000$ m = $2$ kmActual width = $3 \times 50,000$ cm = $150,000$ cm = $1,500$ m = $1.5$ kmEach map length is multiplied by 50,000 to get the actual length.
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2Calculate actual areaActual area = $2 \times 1.5 = 3$ km²Area = length $\times$ width, using the actual dimensions.
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3Alternative method (area scale factor)Map area = $4 \times 3 = 12$ cm²Linear scale factor = $50,000$, so area scale factor = $(50,000)^2 = 2.5 \times 10^9$Actual area = $12 \times 2.5 \times 10^9$ cm² = $3 \times 10^{10}$ cm² = $3$ km²Both methods give the same answer. Choose whichever you prefer.
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1Identify the linear scale factor$k = \frac{5}{2}$ (larger / smaller) or $k = \frac{2}{5}$ (smaller / larger)Use the ratio that goes from the shape you know to the shape you want.
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2Find the surface area ratioArea ratio = $k^2 = (\frac{5}{2})^2 = \frac{25}{4} = 25:4$Surface area scales as $k^2$.
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3Find the volume ratioVolume ratio = $k^3 = (\frac{5}{2})^3 = \frac{125}{8} = 125:8$Volume scales as $k^3$.
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4Find the larger volume$\frac{V_{\text{large}}}{32} = \frac{125}{8}$$V_{\text{large}} = 32 \times \frac{125}{8} = 4 \times 125 = 500$ cm³The larger cylinder is $(\frac{5}{2})^3 = 15.625$ times the volume of the smaller.
The Power Rule
- Linear : Area : Volume = $k : k^2 : k^3$
- Area ratio = $(\text{linear ratio})^2$
- Volume ratio = $(\text{linear ratio})^3$
Working Forwards
- Given $k$, square for area
- Given $k$, cube for volume
- Model problems: $k = \text{model} / \text{real}$
Working Backwards
- Given area ratio: $k = \sqrt{\text{area ratio}}$
- Given volume ratio: $k = \sqrt[3]{\text{volume ratio}}$
- Always check $k$ makes sense
Map Scales
- Scale 1:n means $k = n$ (real / map)
- Convert linear dims first
- Then calculate area
- Watch unit conversions
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Brain Trainer · 4 problems
Four quick problems on area and volume scale factors. Work each, then reveal the answer.
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1 The linear scale factor between two similar figures is 3. Find the area scale factor and the volume scale factor.
Area scale factor = $3^2 = 9$. Volume scale factor = $3^3 = 27$.Area = 9, Volume = 27 -
2 The linear scale factor is $\frac{1}{2}$. Find the area scale factor and the volume scale factor.
Area scale factor = $(\frac{1}{2})^2 = \frac{1}{4}$. Volume scale factor = $(\frac{1}{2})^3 = \frac{1}{8}$.Area = $\frac{1}{4}$, Volume = $\frac{1}{8}$ -
3 The area scale factor between two similar figures is 36. Find the linear scale factor and the volume scale factor.
Linear scale factor = $\sqrt{36} = 6$. Volume scale factor = $6^3 = 216$.$k = 6$, Volume = 216 -
4 Two similar spheres have radii 2 cm and 6 cm. The smaller sphere has volume 32 cm³. Find the volume of the larger sphere.
$k = \frac{6}{2} = 3$. Volume scale factor = $3^3 = 27$. Larger volume = $32 \times 27 = 864$ cm³.$864$ cm³
Multiple Choice · 5 questions
If linear scale factor $k = 2$, the volume scale factor is:
Two similar cubes have side lengths 2 cm and 6 cm. The volume ratio is:
If area scale factor is 25, the linear scale factor is:
A model car is built with scale factor 1:10. If the real car has volume 4 m³, the model volume is:
For similar cylinders, if radius doubles, volume becomes:
Short Answer · 3 questions
(a) Find the ratio of their surface areas.
(b) Find the ratio of their volumes.
(c) If the smaller cylinder has a volume of 32 cm³, find the volume of the larger cylinder.
(a) Find the linear scale factor from the small mammal to the large mammal.
(b) Find the surface area of the large mammal.
(c) If the small mammal has a mass of 80 g, estimate the mass of the large mammal (assume mass is proportional to volume).
(d) Explain why the large mammal might need a different body shape or behaviour to stay cool, using your answers from (b) and (c).
Reveal solution
(a) $k = \frac{50}{10} = 5$.
(b) Surface area scales as $k^2 = 25$. Large mammal surface area = $200 \times 25 = 5,000$ cm².
(c) Volume (and mass) scales as $k^3 = 125$. Large mammal mass = $80 \times 125 = 10,000$ g = 10 kg.
(d) The large mammal's volume (heat production) is 125 times greater, but its surface area (heat loss) is only 25 times greater. The ratio of surface area to volume drops from $\frac{200}{80} = 2.5$ cm²/g to $\frac{5000}{10000} = 0.5$ cm²/g. It has less skin per unit of body mass to lose heat through, so it needs adaptations like larger ears, thinner limbs, or behavioural changes to cool down.
$k : k^2 : k^3$
Linear : Area : Volume
Area
Square the linear factor
Volume
Cube the linear factor
Backwards
$k = \sqrt{A}$ or $\sqrt[3]{V}$
Maps
Convert dims, then multiply
Check
Does the answer make sense?
Interactive: Tangent Explorer
Solve problems involving tangents, chords and the alternate segment theorem. Test your geometric reasoning with randomly generated problems.
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