Area & Volume of Similar Shapes

What Does 'Mathematically Similar' Mean?

Two shapes are mathematically similar if one is an enlargement or reduction of the other.

All corresponding angles are equal.

All corresponding sides are in the same ratio (called the length scale factor).

This scale factor affects more than just length:

Area scale factor is the square of the length scale factor.

Volume scale factor is the cube of the length scale factor.

 

Key Scale Factor Formulas

Let $k$ be the length scale factor:

• Lengths:

$$k = \frac{\text{New length}}{\text{Original length}}$$

• Areas:

$$\text{Area SF} = k^2$$

• Volumes:

$$\text{Volume SF} = k^3$$

To reverse:

• $$k = \sqrt{\text{Area SF}}$$
• $$k = \sqrt[3]{\text{Volume SF}}$$

 

 

 

When Do We Use These?

If you:

• Know how two shapes are similar, and
• Know one or more quantities (e.g. height, volume),

you can find unknown values using scale factors.

If a shape doubles in size ($k = 2$):

• Its area becomes $4 \times \text{bigger. i.e,  } (2^2)$
• Its volume becomes $8 \times \text{bigger, i.e,  } (2^3)$

 

Example: Volume and Height

 

 

Cone A and Cone B are mathematically similar.

The total surface area of cone $A$ is $24 \text{cm}^2$. The total surface area of cone $B$ is $96 \text{cm}^2$. The height of cone $A$ is $4 \text{cm}$

(a) Work out the height of cone $B$

The volume of cone $A$ is $12 \text{cm}^3$ 

(b) Work out the volume of cone $B$.

 

(a) Work out the height of cone $B$

Step 1: Find the area scale factor ($k^2$)

$$\frac{\text{Surface Area A}}{\text{Surface Area B}} = \frac{96}{24} = 4$$

 

Step 2: Find the length scale factor ($k$)

$$k = \sqrt{4} = 2$$

 

Step 3: Use the scale factor to find the height of B

$$\text{Height of B} = 4 \times 2 = 8 \text{ cm}$$

 

(b) Work out the volume of cone $B$

Step 1: Find the volume scale factor ($k^3$)

$$\text{Volume scale factor} = \text{Length scale factor}^3 = 2^3 = 8$$

 

Step 2: Find volume of B 

$$V_B = V_A \times \text{Volume Scale Factor} = 12 \times 8 = 96 \text{cm}^3$$