Changing Dimensions

If you double the size of a pizza, do you get twice as much pizza? It turns out you get a lot more than that. When you stretch a shape or an object, its perimeter, area and volume don’t all grow at the same rate. This page shows you exactly how each one changes, so you can predict the effect of a change before you make it, whether you’re choosing a pizza, designing a box or planning a garden.

Key ideas

Formulas you’ll use

Shape or object	Measure	Formula
Rectangle, length $l$ and width $w$	perimeter	$P = 2l + 2w$
	area	$A = lw$
Circle, radius $r$	circumference	$C = 2\pi r$
	area	$A = \pi r^2$
Rectangular prism (box), $l$ by $w$ by $h$	surface area	$SA = 2lw + 2lh + 2wh$
	volume	$V = lwh$
Cylinder, radius $r$ and height $h$	volume	$V = \pi r^2 h$

You’ll see more about surface area and volume on their own pages.

Changing just one dimension

If you change only one dimension, look at the formula to see what happens.

• Area and volume are products. If one factor doubles and the rest stay the same, the area or volume doubles. For a rectangle, $A = lw$; double $l$ and you get $(2l)w = 2lw$.
• Perimeter is a sum, so it does not double. For a $5$ m by $3$ m rectangle, $P = 16$ m. Double the length to $10$ m and $P = 26$ m, not $32$ m.

Changing all dimensions: the scale factor

If you multiply every length of a shape or object by the same number $k$, the new shape is a scaled copy, and $k$ is the scale factor. If $k \gt 1$ it’s an enlargement, and if $0 \lt k \lt 1$ it’s a reduction.

What you measure	Units	Multiplied by	When $k = 2$	When $k = 3$	When $k = \frac{1}{2}$
Lengths: sides, perimeter, circumference, height	cm	$k$	$\times 2$	$\times 3$	$\times \frac{1}{2}$
Areas: area, surface area	cm²	$k^2$	$\times 4$	$\times 9$	$\times \frac{1}{4}$
Volume	cm³	$k^3$	$\times 8$	$\times 27$	$\times \frac{1}{8}$

The units are a good memory trick. Area is in square units, so the factor is squared; volume is in cubic units, so the factor is cubed.

Doubling every length fits $4$ copies of a flat shape, but $8$ copies of a solid.

The best rectangle

Two classic questions:

• Fixed perimeter. Of all the rectangles with the same perimeter, the square has the largest area.
• Fixed area. Of all the rectangles with the same area, the square has the smallest perimeter.

The same idea holds in 3-D: of all the boxes with the same volume, the cube has the smallest surface area. That’s one reason packaging designers like boxes that are close to cube-shaped: they use less cardboard. You can explore all of these with a table of values (as in Example 4) or with a spreadsheet or graphing tool, which makes it easy to try many shapes at once.

Worked examples

Example 1: A garden, one dimension vs both

A rectangular garden is $5$ m long and $3$ m wide.

• (a) Find its perimeter and area.
• (b) The length is doubled and the width stays the same. Find the new perimeter and area.
• (c) Instead, both the length and the width are doubled. Find the new perimeter and area.

Solution.

(a) $P = 2(5) + 2(3) = 16$ m and $A = 5 \times 3 = 15 \text{ m}^2$.

(b) The garden is now $10$ m by $3$ m:

$$P = 2(10) + 2(3) = 26 \text{ m} \qquad A = 10 \times 3 = 30 \text{ m}^2$$

The area doubled, but the perimeter didn’t (it grew from $16$ m to $26$ m).

(c) The garden is now $10$ m by $6$ m:

$$P = 2(10) + 2(6) = 32 \text{ m} \qquad A = 10 \times 6 = 60 \text{ m}^2$$

This is a scale factor of $k = 2$. The perimeter doubled ($16 \times 2 = 32$) and the area was multiplied by $2^2 = 4$ ($15 \times 4 = 60$). ✓

Example 2: Which pizza is the better deal?

A medium pizza is $30$ cm across and costs $12. A large pizza is $40$ cm across and costs $18.

• (a) How many times as much pizza is the large?
• (b) Which pizza gives you more pizza per dollar?

Solution.

(a) The radii are $15$ cm and $20$ cm. The areas are

$$A_{\text{medium}} = \pi (15)^2 \approx 706.9 \text{ cm}^2 \qquad A_{\text{large}} = \pi (20)^2 \approx 1256.6 \text{ cm}^2$$

Or use the scale factor: $k = \dfrac{40}{30} = \dfrac{4}{3}$, so the area is multiplied by $k^2 = \dfrac{16}{9} \approx 1.78$. The large has about $1.78$ times as much pizza, even though it’s only $\dfrac{4}{3} \approx 1.33$ times as wide.

(b) Divide the area by the price to find the pizza per dollar:

$$\text{medium: } \frac{706.9}{12} \approx 58.9 \text{ cm}^2 \text{ per dollar} \qquad \text{large: } \frac{1256.6}{18} \approx 69.8 \text{ cm}^2 \text{ per dollar}$$

The large gives you more pizza for your money. The price went up $1.5$ times, but the amount of pizza went up about $1.78$ times.

Example 3: Scaling a box

A box is $10$ cm by $8$ cm by $5$ cm. A company makes a larger box with every dimension $3$ times as long. Find the surface area and volume of both boxes.

Solution. Original box:

$$\begin{aligned} SA &= 2(10)(8) + 2(10)(5) + 2(8)(5) = 160 + 100 + 80 = 340 \text{ cm}^2 \\ V &= 10 \times 8 \times 5 = 400 \text{ cm}^3 \end{aligned}$$

With $k = 3$, surface area is multiplied by $3^2 = 9$ and volume by $3^3 = 27$:

$$SA = 340 \times 9 = 3060 \text{ cm}^2 \qquad V = 400 \times 27 = 10\,800 \text{ cm}^3$$

Check directly: the new box is $30$ by $24$ by $15$ cm.

$$\begin{aligned} SA &= 2(30)(24) + 2(30)(15) + 2(24)(15) = 1440 + 900 + 720 = 3060 \text{ cm}^2 \\ V &= 30 \times 24 \times 15 = 10\,800 \text{ cm}^3 \end{aligned}$$

Both match. ✓ The big box holds $27$ times as much but needs only $9$ times as much cardboard.

Example 4: Fixed perimeter

You have $20$ m of fencing to make a rectangular vegetable garden. What dimensions give the largest area?

Solution. Half the fence goes on one length and one width, so $l + w = 10$. Make a table:

Width (m)	Length (m)	Perimeter (m)	Area (m²)
$1$	$9$	$20$	$9$
$2$	$8$	$20$	$16$
$3$	$7$	$20$	$21$
$4$	$6$	$20$	$24$
$5$	$5$	$20$	$25$

Every rectangle uses all $20$ m of fence, but the areas are very different. The largest area is $25 \text{ m}^2$, from a $5$ m by $5$ m square. (Widths past $5$ just repeat the table in reverse: $6$ by $4$ is the same as $4$ by $6$.)

Common mistakes

Assuming area doubles when you double every length. If all the lengths double, the area is multiplied by $2^2 = 4$ and the volume by $2^3 = 8$. Only lengths (like perimeter) double.

Assuming perimeter doubles when you double one side. Perimeter is a sum. Doubling just the length of a $5$ m by $3$ m rectangle takes the perimeter from $16$ m to $26$ m, not $32$ m.

Using $k$ for everything. Match the factor to the units: $k$ for lengths (cm), $k^2$ for areas (cm²), $k^3$ for volumes (cm³).

Comparing pizzas (or circles) by diameter. A $40$ cm pizza isn’t “a third bigger” than a $30$ cm pizza; it has about $78\%$ more area. Always compare areas when you care about how much there is.

Working backwards the wrong way. If the area of a scaled shape is $9$ times as big, the lengths are $\sqrt{9} = 3$ times as long, not $9$ times. If the volume is $8$ times as big, the lengths are $2$ times as long, since $2^3 = 8$.

Practice

1. (Warm-up) A square has sides of $4$ cm. Every side is tripled. Find the old and new perimeter and area.

Solution

Old: $P = 4 \times 4 = 16$ cm, $A = 4^2 = 16 \text{ cm}^2$.

New side: $12$ cm. $P = 4 \times 12 = 48$ cm (that’s $\times 3$) and $A = 12^2 = 144 \text{ cm}^2$ (that’s $\times 9$).

2. (Warm-up) A circle has a radius of $5$ cm. The radius is doubled. How do the circumference and area change? Find both before and after, to one decimal place.

Solution

Before: $C = 2\pi(5) = 10\pi \approx 31.4$ cm and $A = \pi(5)^2 = 25\pi \approx 78.5 \text{ cm}^2$.

After ($r = 10$): $C = 20\pi \approx 62.8$ cm and $A = 100\pi \approx 314.2 \text{ cm}^2$.

The circumference doubles ($k = 2$), and the area is multiplied by $4$ ($k^2 = 4$).

3. (Core) A $20$ cm by $15$ cm photo is printed at half size (scale factor $\frac{1}{2}$). Find the area of the original and of the smaller print. What fraction of the original area is the print?

Solution

Original: $20 \times 15 = 300 \text{ cm}^2$.

The print is $10$ cm by $7.5$ cm: $10 \times 7.5 = 75 \text{ cm}^2$.

$\dfrac{75}{300} = \dfrac{1}{4}$, which matches $k^2 = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$.

4. (Core) A cylindrical can has radius $3$ cm and height $10$ cm.

• (a) Find its volume, to one decimal place.
• (b) Find the volume if only the radius is doubled.
• (c) Find the volume if only the height is doubled.
• (d) Which change makes the bigger difference, and why?

Solution

(a) $V = \pi(3)^2(10) = 90\pi \approx 282.7 \text{ cm}^3$.

(b) $V = \pi(6)^2(10) = 360\pi \approx 1131.0 \text{ cm}^3$, which is $4$ times as much.

(c) $V = \pi(3)^2(20) = 180\pi \approx 565.5 \text{ cm}^3$, which is $2$ times as much.

(d) Doubling the radius makes a bigger difference. The radius is squared in $V = \pi r^2 h$, so doubling it multiplies the volume by $4$. The height isn’t squared, so doubling it only doubles the volume.

5. (Core) A cube has edges of $2$ cm. A larger cube has edges of $6$ cm.

• (a) What is the scale factor?
• (b) Find the surface area and volume of both cubes, and check them against the scale factor.

Solution

(a) $k = \dfrac{6}{2} = 3$.

(b) Small cube: $SA = 6 \times 2^2 = 24 \text{ cm}^2$ and $V = 2^3 = 8 \text{ cm}^3$.

Large cube: $SA = 6 \times 6^2 = 216 \text{ cm}^2$ and $V = 6^3 = 216 \text{ cm}^3$.

Check: $24 \times 3^2 = 24 \times 9 = 216$ ✓ and $8 \times 3^3 = 8 \times 27 = 216$ ✓.

6. (Core) A flower bed has an area of $18 \text{ m}^2$. A landscaper makes a new bed of the same shape with every length $1.5$ times as long. Find the area of the new bed.

Solution

Area is multiplied by $k^2 = 1.5^2 = 2.25$:

$$18 \times 2.25 = 40.5 \text{ m}^2$$

7. (Core) A rectangular patio must have an area of $36 \text{ m}^2$, with whole-number side lengths. List the possible rectangles and their perimeters. Which one needs the least edging around the outside?

Solution

Width (m)	Length (m)	Area (m²)	Perimeter (m)
$1$	$36$	$36$	$74$
$2$	$18$	$36$	$40$
$3$	$12$	$36$	$30$
$4$	$9$	$36$	$26$
$6$	$6$	$36$	$24$

The $6$ m by $6$ m square has the smallest perimeter, $24$ m, so it needs the least edging.

8. (Challenge) A farmer has $40$ m of fencing to make a rectangular pen against the side of a long barn. The barn wall forms one side, so the fence only goes on the other three sides: two widths of $x$ metres and one length.

• (a) Explain why the length is $40 - 2x$ and the area is $A = x(40 - 2x)$.
• (b) Make a table for $x = 5, 8, 10, 12, 15$. Which dimensions give the largest area?

Solution

(a) The two widths use $2x$ metres of fence, so the length gets what’s left: $40 - 2x$. Area is width times length: $A = x(40 - 2x)$.

(b)

$x$ (m)	Length $40 - 2x$ (m)	Area (m²)
$5$	$30$	$150$
$8$	$24$	$192$
$10$	$20$	$200$
$12$	$16$	$192$
$15$	$10$	$150$

The largest area is $200 \text{ m}^2$, with widths of $10$ m and a length of $20$ m. Notice it’s not a square here: because the barn saves you one side, the best pen is twice as long as it is wide.

9. (Challenge) A shipping company scales up a box so that its volume is $8$ times the original volume.

• (a) By what factor were the lengths multiplied?
• (b) By what factor was the surface area multiplied?
• (c) The original box needed $0.5 \text{ m}^2$ of cardboard. How much does the new box need?

Solution

(a) Volume is multiplied by $k^3$, so $k^3 = 8$. Since $2^3 = 8$, the lengths were multiplied by $k = 2$.

(b) Surface area is multiplied by $k^2 = 2^2 = 4$.

(c) $0.5 \times 4 = 2 \text{ m}^2$ of cardboard.