E5.3 - Similar areas and volumes (Extended)

E5.3 - Similar areas and volumes (Extended)

A model is twice as long as another similar model. Does it need twice as much paint, and can it hold twice as much liquid? E5.3 distinguishes length, surface-area and volume scale factors, then works backwards when an area or volume ratio is given.

Area scale factors

Area counts the square units covering a region. A solid's total surface area adds the areas of all its outside surfaces. Imagine a rectangle 3 units long and 2 units wide. Its area is 3×2=63\times2=6 square units. Double both lengths: the new area is 6×4=246\times4=24, four times as large. There are two independent length directions to scale.

[DIAGRAM: asset_name: Length, area and volume factors; asset_slug: similarity_dimensions; caption: Doubling a length gives four times the area and eight times the volume.; description: Exact unit-grid rectangles of three by two units and six by four units show 6 and 24 unit squares. Alongside, a unit cube and a two by two by two cube show 1 and 8 equal unit cubes, with visible subdivision edges and counts labelled. The length multiplier is 2, area multiplier 4 and volume multiplier 8.]
Doubling a length gives four times the area and eight times the volume.

Area and surface-area scale factor

If all corresponding lengths of similar shapes or solids multiply by k>0k>0, then

area factor=k2\text{area factor}=k^2

This applies to plane areas, matching faces and total surface areas of similar solids. Not on the supplied formula list: learn the dimensional relationship.

For a rectangle, (kl)(kw)=k2lw(kl)(kw)=k^2lw. For other similar shapes, the same length scaling acts in two directions. For a solid, every matching face scales in area by k2k^2, so their total surface area does too. Similarity is essential: changing just one dimension is not a uniform scale change.

Worked example 1: a plane area

Question. Two similar shapes have corresponding lengths 4 cm and 6 cm. The smaller has area 20 cm220\text{ cm}^2. Calculate the larger area.

Plan. Find the length factor, square it, then scale the area.

k=6/4=1.5length factork2=1.52=2.25area factorA=20×2.25=45 cm2larger area\begin{aligned} k&=6/4=1.5 &&\text{length factor}\\ k^2&=1.5^2=2.25 &&\text{area factor}\\ A&=20\times2.25=45\text{ cm}^2 &&\text{larger area} \end{aligned}

Answer. 45 cm245\text{ cm}^2.

Check. Reversing the lengths gives factor 2/32/3; 45×(2/3)2=45×4/9=20 cm245\times(2/3)^2=45\times4/9=20\text{ cm}^2.

The next example changes a plane area into a solid's surface area. The exponent stays 2 because surface area is still measured in square units.

Worked example 2: a total surface area

Question. Two similar solids have corresponding lengths 4 cm and 6 cm. The smaller has total surface area 80 cm280\text{ cm}^2. Calculate the larger total surface area.

Plan. Use the same length factor and square it for every face.

k=6/4=1.5length factorS=80×1.52=180 cm2surface-area scaling\begin{aligned} k&=6/4=1.5 &&\text{length factor}\\ S&=80\times1.5^2=180\text{ cm}^2 &&\text{surface-area scaling} \end{aligned}

Answer. 180 cm2180\text{ cm}^2.

Check. The area ratio 180:80180:80 simplifies to 9:49:4, the square of the length ratio 3:23:2.

The same rule works for reductions: a length factor 2/32/3 gives an area factor 4/94/9, which is smaller still.

Watch out

Being a solid does not make every calculation a cube. Paint covers its surface, so paint coverage uses the square of the length factor.

Use the direction specified by the corresponding lengths.

Volume scale factors

Volume measures the three-dimensional space a solid occupies, in cubic units: one cubic unit is a cube of side one unit. A cuboid's volume is its length times width times height. If each is doubled, the factors multiply as 2×2×2=82\times2\times2=8. The larger solid can be filled by eight copies of the original-sized unit cube.

Volume scale factor

If all corresponding lengths of similar solids multiply by k>0k>0, then

volume factor=k3\text{volume factor}=k^3

The three length directions each contribute a factor of kk. Not on the supplied formula list: learn the dimensional relationship.

For a cuboid, (kl)(kw)(kh)=k3lwh(kl)(kw)(kh)=k^3lwh. The same relationship holds for all similar solids. This answers the opening question: doubling all lengths requires four times the paint coverage and gives eight times the volume or capacity.

Worked example 3: doubling every length

Question. A small solid has volume 120 cm3120\text{ cm}^3. A similar solid has every corresponding length twice as large. Calculate its volume.

Plan. Cube the length factor, then multiply the given volume.

k3=23=8volume factorV=120×8=960 cm3larger volume\begin{aligned} k^3&=2^3=8 &&\text{volume factor}\\ V&=120\times8=960\text{ cm}^3 &&\text{larger volume} \end{aligned}

Answer. 960 cm3960\text{ cm}^3.

Check. Halving each length divides volume by 23=82^3=8; 960÷8=120 cm3960\div8=120\text{ cm}^3.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account