Similar Solids (VCE SSCE General Mathematics): Revision Notes

Similar Solids

What are similar solids?

Two three-dimensional objects are called similar solids when they meet two important conditions. First, they must have the same shape. Second, the ratios of their corresponding linear dimensions must be equal. This means that if you compare matching measurements (like length to length, width to width, or height to height), these comparisons should all give you the same ratio.

Understanding similar solids is important because it helps us predict how changes in size affect volume. When you scale up or scale down a 3D object while keeping its shape the same, there are predictable mathematical relationships between the original and the new version.

Understanding scale factors

When working with similar solids, we use something called a scale factor, which we represent with the letter $k$. This is the number by which we multiply all the linear dimensions of one solid to get the corresponding dimensions of the similar solid.

For example, if one cuboid has dimensions that are all three times larger than another similar cuboid, we say the scale factor $k = 3$.

Cuboids

Let's look at a concrete example using cuboids (rectangular prisms). Consider two cuboids: Cuboid A has dimensions of $9$ cm length, $6$ cm width, and $3$ cm height. Cuboid B has dimensions of $3$ cm length, $2$ cm width, and $1$ cm height.

To determine if these cuboids are similar, we need to check both conditions. First, they are both cuboids, so they have the same shape. Second, we need to verify that the ratios of corresponding dimensions are equal.

Now let's examine the volumes. We can calculate the volume scale factor:

$\text{Volume scale factor} = k^3 = \frac{\text{Volume of cuboid B}}{\text{Volume of cuboid A}}$

$= \frac{2 \times 3 \times 1}{6 \times 9 \times 3} = \frac{6}{162} = \frac{1}{27}$

Notice that $\left(\frac{1}{3}\right)^3 = \frac{1}{27}$, which confirms our principle. When the linear dimensions are reduced by a scale factor of $\frac{1}{3}$, the volume is reduced by a scale factor of $\frac{1}{27}$.

Cylinders

The same principles apply to cylinders. Consider two cylinders: Cylinder A has a radius of $2$ cm and height of $6$ cm, while Cylinder B has a radius of $1$ cm and height of $3$ cm.

These cylinders are similar because they are both cylinders (same shape) and their corresponding dimensions have equal ratios:

$\frac{\text{height of cylinder B}}{\text{height of cylinder A}} = \frac{3}{6} = \frac{1}{2}$

$\frac{\text{radius of cylinder B}}{\text{radius of cylinder A}} = \frac{1}{2}$

The scale factor is $k = \frac{1}{2}$.

For the volume comparison, we use the cylinder volume formula $V = \pi r^2 h$:

$\text{Volume scale factor} = k^3 = \frac{\pi \times 1^2 \times 3}{\pi \times 2^2 \times 6} = \frac{3}{24} = \frac{1}{8}$

Again, we can verify: $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$. The pattern holds true for cylinders just as it did for cuboids.

Cones

Let's examine one more example using cones. Cone A has a base radius of $1.5$ cm and height of $5$ cm. Cone B has a base radius of $6$ cm and height of $20$ cm.

These cones are similar because they are both cones and their corresponding dimensions maintain equal ratios:

$\frac{\text{height of cone B}}{\text{height of cone A}} = \frac{20}{5} = 4$

$\frac{\text{radius of cone B}}{\text{radius of cone A}} = \frac{6}{1.5} = 4$

In this case, the scale factor is $k = 4$, meaning Cone B is an enlargement of Cone A.

For the volume comparison, we use the cone volume formula $V = \frac{1}{3}\pi r^2 h$:

$\text{Volume scale factor} = k^3 = \frac{\frac{1}{3} \times \pi \times 6^2 \times 20}{\frac{1}{3} \times \pi \times 1.5^2 \times 5}$

$= \frac{6^2 \times 20}{1.5^2 \times 5} = 64$

We can verify: $4^3 = 64$. Once again, the volume scale factor is the cube of the linear scale factor.

Worked example: comparing volumes of similar solids

Let's apply what we've learned to solve a practical problem.

Practice question

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