Sec 3 · G3 (Express)

Area and volume of similar figures

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1. Area ratio

Ratio	Formula	Example: lengths $2 : 3$
lengths	$\dfrac{l_1}{l_2}$	$\dfrac{2}{3}$
areas	$\dfrac{A_1}{A_2} = \left(\dfrac{l_1}{l_2}\right)^{2}$	$\dfrac{4}{9}$
volumes	$\dfrac{V_1}{V_2} = \left(\dfrac{l_1}{l_2}\right)^{3}$	$\dfrac{8}{27}$
masses (same material)	$\dfrac{m_1}{m_2} = \left(\dfrac{l_1}{l_2}\right)^{3}$	$\dfrac{8}{27}$

Area from lengths: 6 cm and 9 cm, larger area 36 cm²

\displaystyle \dfrac{A}{36} = \left(\dfrac{6}{9}\right)^{2}

e.g.

\displaystyle \dfrac{A}{36} = \left(\dfrac{6}{9}\right)^{2}

\displaystyle A = 16

Length from areas: 8 cm² and 15.68 cm², larger side 7 cm, find the smaller side l

\displaystyle \dfrac{l}{7} = \sqrt{\dfrac{8}{15.68}}

e.g.

\displaystyle \dfrac{l}{7} = \sqrt{\dfrac{8}{15.68}}

\displaystyle l = 5

Areas on a map: square the scale

\displaystyle 1\text{ cm} : 0.4\text{ km}

$1$ cm² $: 0.16$ km²
$18$ cm² $: 2.88$ km²

📖 Worked example

Triangles $ABC$ and $PQR$ are similar, with areas $8$ cm² and $15.68$ cm² respectively. $PQ = 7$ cm. Find $AB$ .

Step 1 of 3

Lengths are in the ratio of the square roots of the areas

\dfrac{l_1}{l_2} = \sqrt{\dfrac{A_1}{A_2}}

📖 Worked example

A map is drawn to a scale of $1 : 40\,000$ . A park has an area of $18$ cm² on the map. Find the actual area, in km², of the park.

Step 1 of 3

Write the scale as a ratio with units (1 km = 100 000 cm)

\mathstrut 1 : 40\,000

\mathstrut 1\text{ cm} : 40\,000\text{ cm}

1\text{ cm} : 0.4\text{ km}

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2. Volume and mass ratio

Volume from lengths: heights 5 cm and 6 cm, smaller 80 cm³

\displaystyle \dfrac{V}{80} = \left(\dfrac{6}{5}\right)^{3}

e.g.

\displaystyle \dfrac{V}{80} = \left(\dfrac{6}{5}\right)^{3}

\displaystyle V = 138.24

Length from volumes: 189 cm³ and 448 cm³, smaller height 6 cm

\displaystyle \dfrac{h}{6} = \sqrt[3]{\dfrac{448}{189}}

e.g.

\displaystyle \dfrac{h}{6} = \sqrt[3]{\dfrac{448}{189}}

\displaystyle h = 8

Mass: same material, heights 4 cm and 6 cm, smaller 108 g

\displaystyle \dfrac{m}{108} = \left(\dfrac{6}{4}\right)^{3}

e.g.

\displaystyle \dfrac{m}{108} = \left(\dfrac{6}{4}\right)^{3}

\displaystyle m = 364.5

Model at $1 : n$	Real object	Units
length	$\times n$	$1$ m $= 100$ cm
area	$\times n^2$	$1$ m² $= 10\,000$ cm²
volume	$\times n^3$	$1$ m³ $= 1\,000\,000$ cm³

📖 Worked example

A model of a building is made to a scale of $1 : 20$ . The volume of the model is $64$ cm³. Find the volume, in m³, of the real building.

Step 1 of 2

Volumes: the ratio of lengths cubed

Volume ratio

= 1 : 20^3

= 1 : 8000

📖 Worked example

Two similar cylinders have volumes $189$ cm³ and $448$ cm³. The height of the smaller cylinder is $6$ cm. Find the height of the larger cylinder.

Step 1 of 3

Lengths are in the ratio of the cube roots of the volumes

\dfrac{h_1}{h_2} = \sqrt[3]{\dfrac{V_1}{V_2}}

📖 Worked example

An inverted cone of height $12$ cm has a capacity of $750$ cm³. It contains $93.75$ cm³ of water. Find the depth of the water.

Step 1 of 3

The water forms a cone similar to the whole cone

\dfrac{d}{H} = \sqrt[3]{\dfrac{\text{volume of water}}{\text{volume of cone}}}

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Practice questions by skill

1. Area ratio

Ratio	Formula	Example: lengths $2 : 3$
lengths	$\dfrac{l_1}{l_2}$	$\dfrac{2}{3}$
areas	$\dfrac{A_1}{A_2} = \left(\dfrac{l_1}{l_2}\right)^{2}$	$\dfrac{4}{9}$
volumes	$\dfrac{V_1}{V_2} = \left(\dfrac{l_1}{l_2}\right)^{3}$	$\dfrac{8}{27}$
masses (same material)	$\dfrac{m_1}{m_2} = \left(\dfrac{l_1}{l_2}\right)^{3}$	$\dfrac{8}{27}$

Area from lengths: 6 cm and 9 cm, larger area 36 cm²

\displaystyle \dfrac{A}{36} = \left(\dfrac{6}{9}\right)^{2}

e.g.

\displaystyle \dfrac{A}{36} = \left(\dfrac{6}{9}\right)^{2}

\displaystyle A = 16

Length from areas: 8 cm² and 15.68 cm², larger side 7 cm, find the smaller side l

\displaystyle \dfrac{l}{7} = \sqrt{\dfrac{8}{15.68}}

e.g.

\displaystyle \dfrac{l}{7} = \sqrt{\dfrac{8}{15.68}}

\displaystyle l = 5

Areas on a map: square the scale

\displaystyle 1\text{ cm} : 0.4\text{ km}

$1$ cm² $: 0.16$ km²
$18$ cm² $: 2.88$ km²

📖 Worked example

Triangles $ABC$ and $PQR$ are similar, with areas $8$ cm² and $15.68$ cm² respectively. $PQ = 7$ cm. Find $AB$ .

Step 1 of 3

Lengths are in the ratio of the square roots of the areas

\dfrac{l_1}{l_2} = \sqrt{\dfrac{A_1}{A_2}}

📖 Worked example

A map is drawn to a scale of $1 : 40\,000$ . A park has an area of $18$ cm² on the map. Find the actual area, in km², of the park.

Step 1 of 3

Write the scale as a ratio with units (1 km = 100 000 cm)

\mathstrut 1 : 40\,000

\mathstrut 1\text{ cm} : 40\,000\text{ cm}

1\text{ cm} : 0.4\text{ km}

Practise this →

2. Volume and mass ratio

Volume from lengths: heights 5 cm and 6 cm, smaller 80 cm³

\displaystyle \dfrac{V}{80} = \left(\dfrac{6}{5}\right)^{3}

e.g.

\displaystyle \dfrac{V}{80} = \left(\dfrac{6}{5}\right)^{3}

\displaystyle V = 138.24

Length from volumes: 189 cm³ and 448 cm³, smaller height 6 cm

\displaystyle \dfrac{h}{6} = \sqrt[3]{\dfrac{448}{189}}

e.g.

\displaystyle \dfrac{h}{6} = \sqrt[3]{\dfrac{448}{189}}

\displaystyle h = 8

Mass: same material, heights 4 cm and 6 cm, smaller 108 g

\displaystyle \dfrac{m}{108} = \left(\dfrac{6}{4}\right)^{3}

e.g.

\displaystyle \dfrac{m}{108} = \left(\dfrac{6}{4}\right)^{3}

\displaystyle m = 364.5

Model at $1 : n$	Real object	Units
length	$\times n$	$1$ m $= 100$ cm
area	$\times n^2$	$1$ m² $= 10\,000$ cm²
volume	$\times n^3$	$1$ m³ $= 1\,000\,000$ cm³

📖 Worked example

A model of a building is made to a scale of $1 : 20$ . The volume of the model is $64$ cm³. Find the volume, in m³, of the real building.

Step 1 of 2

Volumes: the ratio of lengths cubed

Volume ratio

= 1 : 20^3

= 1 : 8000

📖 Worked example

Two similar cylinders have volumes $189$ cm³ and $448$ cm³. The height of the smaller cylinder is $6$ cm. Find the height of the larger cylinder.

Step 1 of 3

Lengths are in the ratio of the cube roots of the volumes

\dfrac{h_1}{h_2} = \sqrt[3]{\dfrac{V_1}{V_2}}

📖 Worked example

An inverted cone of height $12$ cm has a capacity of $750$ cm³. It contains $93.75$ cm³ of water. Find the depth of the water.

Step 1 of 3

The water forms a cone similar to the whole cone

\dfrac{d}{H} = \sqrt[3]{\dfrac{\text{volume of water}}{\text{volume of cone}}}

Practise this →