Scale and Similar Solids
Theory
For similar solids with length scale factor \(k\), areas are multiplied by \(k^2\) and volumes by \(k^3\). This Year 10A maths page covers scaling of objects: finding \(k\), working back from an area or volume ratio with square and cube roots, scale models, capacity, mass, paint and cost problems, percentage changes in size, and cutting a cone or pyramid. Curriculum: VC2M10M04 / AC9M10M05.
Two solids are similar if one is an exact enlargement of the other: the same shape, with every length multiplied by the same scale factor \(k\).
If the lengths are in the ratio \(a : b\), the surface areas are in the ratio \(a^2 : b^2\) and the volumes in the ratio \(a^3 : b^3\). Area is length \(\times\) length, so it picks up two factors of \(k\); volume picks up three.
Anything that depends on area (paint, fabric, a label, steel for a tank) scales by \(k^2\). Anything that depends on volume (capacity, the mass of a solid object, the material inside) scales by \(k^3\).
A scale of \(1 : n\) means real lengths are \(n\) times the model's, so \(k = n\).
To check that two solids are similar, test every pair of matching lengths: they must all give the same ratio.
| Quantity | Changes by | Examples |
|---|---|---|
| Lengths | \(k\) | height, radius, perimeter, edge |
| Areas | \(k^2\) | surface area, paint, label, steel, fabric |
| Volumes | \(k^3\) | capacity, mass of a solid object, amount of material inside |
How to solve a similar-solids problem
- Find the length scale factor \(k\) from a pair of matching lengths (or from an area or volume ratio, using a root).
- Decide what is being asked: a length (\(k\)), an area (\(k^2\)) or a volume (\(k^3\)).
- Multiply (larger) or divide (smaller), and convert units at the end.
Other questions
- Ratio form: simplify, such as \(4.5 : 3 = 3 : 2\), then cube for volumes: \(27 : 8\).
- Exact volumes: find the volume of the given solid with its formula, then multiply by \(k^3\) as a fraction.
- Value for money: compare cost per unit of volume, since the amount of contents scales by \(k^3\).
- Percentage change in every length: \(20\%\) longer means \(k = 1.2\), so volume \(\times\, 1.2^3\).
- Required scale factor: to multiply the volume by \(m\), solve \(k^3 = m\), so \(k = \sqrt[3]{m}\).
- Unknown lengths: write the length ratio as a fraction equation, such as \(\dfrac{x}{x + 3} = \dfrac{2}{3}\), and solve.
- Cutting parallel to the base: the small top cone or pyramid is similar to the whole; frustum = whole \(-\) top.
(i) the capacity of the large jar
(ii) the area of its label.
Scale factor for lengths, small to large:
| \(k\) | \(=\) | \(\dfrac{15}{9} = \dfrac{5}{3}\) |
(i) Capacity is a volume, so multiply by \(k^3\):
| \(V\) | \(=\) | \(270 \times \left(\dfrac{5}{3}\right)^3 = 270 \times \dfrac{125}{27}\) |
| \(\) | \(=\) | \(\mathbf{1250}\text{ mL}\) |
(ii) A label covers an area, so multiply by \(k^2\):
| \(A\) | \(=\) | \(54 \times \left(\dfrac{5}{3}\right)^2 = 54 \times \dfrac{25}{9}\) |
| \(\) | \(=\) | \(\mathbf{150}\text{ cm}^2\) |
(i) the ratio of their lengths
(ii) the surface area of the larger solid
(iii) the height of the smaller solid.
(i) Lengths are in the ratio of the cube roots of the volumes:
| \(\sqrt[3]{125} : \sqrt[3]{343}\) | \(=\) | \(\mathbf{5 : 7}\) |
(ii) Areas are in the ratio of the squares, \(25 : 49\):
| \(A\) | \(=\) | \(100 \times \dfrac{49}{25}\) |
| \(\) | \(=\) | \(\mathbf{196}\text{ cm}^2\) |
(iii) Going from large to small, multiply lengths by \(\dfrac{5}{7}\):
| \(h\) | \(=\) | \(28 \times \dfrac{5}{7}\) |
| \(\) | \(=\) | \(\mathbf{20}\text{ cm}\) |
(i) the volume in cubic metres
(ii) the roof area in square metres.
Every real length is \(50\) times the model length, so \(k = 50\).
(i)
| \(V\) | \(=\) | \(2400 \times 50^3 = 2400 \times 125\,000\) |
| \(\) | \(=\) | \(300\,000\,000\text{ cm}^3\) |
| \(\) | \(=\) | \(300\,000\,000 \div 1\,000\,000\) |
| \(\) | \(=\) | \(\mathbf{300}\text{ m}^3\) |
(ii)
| \(A\) | \(=\) | \(150 \times 50^2 = 375\,000\text{ cm}^2\) |
| \(\) | \(=\) | \(375\,000 \div 10\,000\) |
| \(\) | \(=\) | \(\mathbf{37.5}\text{ m}^2\) |
\(1\text{ m}^3 = 100^3\text{ cm}^3\) and \(1\text{ m}^2 = 100^2\text{ cm}^2\).
The small top cone is similar to the whole cone, with \(k = \dfrac{4}{10} = 0.4\).
| \(\dfrac{V_{\text{top}}}{V_{\text{whole}}}\) | \(=\) | \(0.4^3 = 0.064\) |
| \(\text{frustum}\) | \(=\) | \(1 - 0.064 = 0.936\) |
| \(\) | \(=\) | \(\mathbf{93.6\%}\text{ of the volume}\) |
The top cone has \(40\%\) of the height but only \(6.4\%\) of the volume.
Common pitfalls
Frequently asked questions
What is the ratio of the volumes of similar solids?
If the lengths are in the ratio \(a : b\), the volumes are in the ratio \(a^3 : b^3\). With scale factor \(k\), the new volume is \(k^3\) times the old.
What is the ratio of the surface areas of similar solids?
The square of the length ratio. If every length is multiplied by \(k\), every area is multiplied by \(k^2\).
How do you find the scale factor from a volume ratio?
Take the cube root of each part of the ratio. For example, \(8 : 125\) gives a length ratio of \(2 : 5\).
If you make every length of a solid model 3 times as long, what happens to its mass?
If it is solid and made of the same material, the mass follows the volume, so it is multiplied by 3 cubed, which is 27.
How do you know if two solids are similar?
Every pair of matching lengths must be in the same ratio, and the solids must have the same shape. Two cylinders are similar only if their radii and heights have the same ratio.