Math 7 · 7.G.B.4 · G.GMD.A.1

Circumference

Measuring across a circle is easy and measuring round it is awkward. Is the distance round always the same multiple of the distance across, and if so, what is that number?

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Multiplying decimalsGrade 5 Radius and diameterMath 7

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Arcs and central anglesGeometry Area of a circleMath 7 Irrational numbersMath 8

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Circumference

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Why this exists

The distance across a round thing is easy to measure: lay a ruler over the middle. The distance round it is not. A ruler will not bend, a tape slips off a curved edge, and nobody can wrap a tape round a pond or a Ferris wheel.

So it would be very useful if the distance round were always the same multiple of the distance across: one easy measurement times one fixed number would give the hard one. That is exactly how it turns out, for every circle there is, and the fixed number is called pi. This lesson is about that number and what it lets you work out, from the length of a fence to the distance a bicycle travels.

The lesson is taught over two days. Part 1 is the number itself and the formula it gives. Part 2 runs the formula backwards and uses it on pieces of circles and on turning wheels.

Definition

The circumference of a circle is the distance once round it: the length of the circle itself. It is a length, measured in centimeters, meters or any other unit of length. It is written C.

The diameter d is the distance across the circle through its center, and the radius r is the distance from the center to the circle, so d = 2r.

Pi, written \pi, is the circumference of a circle divided by its diameter:

\pi = \frac{C}{d}

This defines one number, not one number per circle, because the division gives the same answer for every circle (see Why it is true). Its value begins

\pi = 3.14159\ldots

and its decimal digits never end and never fall into a repeating pattern.

Intuition

Part 1: the same number every time

Wrap a piece of string once round a circle, mark where it meets itself, and lay it straight next to the diameter. Three diameters fit along it, with a short piece left over: about a seventh of a diameter.

Figure 1The string that goes once round a circle, laid straight. Three diameters fit along it, with a little over.

Lowri tried this with four round objects, measuring each one across with a ruler and round with a string, and divided:

Object	Across, d	Round, C	C \div d
a coin	2.6 cm	8.2 cm	3.15
a mug	8.0 cm	25.2 cm	3.15
a dinner plate	26 cm	81.5 cm	3.13
a bicycle wheel	66 cm	207.5 cm	3.14

The objects are very different sizes, and the last column hardly moves. The small differences come from measuring: a string stretches a little and a ruler is read to the nearest millimeter. The true ratio, with perfect measuring, is the same number for all four: \pi.

Part 2: rolling a circle out

A wheel turning once along the ground does what the string did: it lays its rim out along the ground, so it moves forward exactly one circumference (Figure 3). That is why “how far does a wheel go?” is a circumference question.

Results

Circumference of a circle. For every circle, with diameter d and radius r,

C = \pi d \qquad\text{and}\qquad C = 2\pi r

The second form is the first with d = 2r put in.

The size of pi. \pi = 3.14159\ldots is not a fraction and has no exact decimal. The two approximations in everyday use are

3.14 < \pi < \frac{22}{7}

so 3.14 is a little too small and \displaystyle \frac{22}{7} = 3.142857\ldots is a little too big. Neither is exact.

Exact and rounded answers. An answer such as 14\pi m is exact: it is the length itself, with \pi kept as a symbol, and it is called an answer in terms of \pi. Replacing \pi by a decimal gives a rounded answer, written with \approx rather than =. Say which kind of answer you are giving and, for a rounded one, which value of \pi was used. Keep \pi as a symbol until the last step: rounding \pi early carries the error into every later step.

Part 2: working backwards, pieces of circles, wheels

From the circumference back. Dividing both sides of C = \pi d by \pi, and both sides of C = 2\pi r by 2\pi,

d = \frac{C}{\pi} \qquad\text{and}\qquad r = \frac{C}{2\pi}

Half and quarter circles. The curved edge of a half circle is half the circumference, \displaystyle \frac{1}{2}\pi d. Its perimeter, the whole distance round the shape, adds the straight edge:

\text{half circle: } P = \tfrac{1}{2}\pi d + d

The curved edge of a quarter circle is a quarter of the circumference, \displaystyle \frac{1}{4} \cdot 2\pi r = \frac{1}{2}\pi r, and its perimeter adds two radii:

\text{quarter circle: } P = \tfrac{1}{2}\pi r + 2r

Turning wheels. A wheel that rolls without slipping moves forward one circumference for each full turn:

\text{distance} = \text{number of turns} \times C

Why it is true

Every circle gives the same ratio

Scale copies already showed why C \div d cannot depend on the circle: any circle is a scale copy of any other, and a scale copy multiplies every length, the curved ones included, by the same scale factor. Here is what that fact gives next.

Take the circle with diameter 1. Its circumference is some definite length, and that length is what \pi names: \pi = C \div 1. A circle with diameter d is a copy of that circle at scale factor d, so its circumference is d times as long too:

C = \pi \times d

That is the formula, and it needed no measuring at all. Measuring only tells you how big \pi is. Since d = 2r, the same formula reads C = 2\pi r.

Pi is between 3 and 4

Figure 2The circle is longer than the hexagon inside it and shorter than the square round it.

Draw a regular hexagon inside a circle of radius r, with its six corners on the circle, and join the center to two neighboring corners as in Figure 2. The six angles at the center share the full turn equally, so each is 360^\circ \div 6 = 60^\circ. The triangle has two sides equal to r, so its other two angles are equal, and by the triangle angle sum they share 180^\circ - 60^\circ = 120^\circ, which makes each of them 60^\circ. All three angles are 60^\circ, so all three sides are equal, and the side of the hexagon is r.

Between two neighboring corners, the hexagon goes straight and the circle goes along a curve. A straight segment is the shortest path between two points, so each of the six pieces of circle is longer than r, and

C > 6r = 3d

Now draw a square round the circle, touching it at the middle of each side. Its sides are as long as the diameter, so its perimeter is 4d. Each quarter of the circle runs from the middle of one side to the middle of the next and cuts across the corner, while the square goes all the way out to the corner and back. Cutting the corner looks like the shorter route, and it is, though proving it for a curve takes more than this lesson has; take it here as a convincing idea rather than a full proof. So

C < 4d

Together, 3d < C < 4d, which says 3 < \pi < 4. With polygons of many more sides the gap closes. Archimedes, using polygons with 96 sides inside and outside a circle, showed that

3\tfrac{10}{71} < \pi < 3\tfrac{1}{7}

His upper bound, \displaystyle 3\frac{1}{7} = \frac{22}{7}, is still the fraction people use for \pi, but no fraction is exact: \pi is irrational, which was proved long after Archimedes.

Why a wheel moves one circumference per turn

Figure 3In one full turn, every point of the rim touches the ground once, so the wheel moves forward one circumference.

Mark the point of the rim touching the ground. As the wheel rolls without slipping, each point of the rim comes down onto a different spot of ground in turn. When the marked point is back at the bottom, the whole rim has been laid along the ground once, so the stretch covered is as long as the rim: one circumference, \pi d.

Worked examples

Examples 1 to 3 belong to Part 1 and Examples 4 to 7 to Part 2.

Example 1

Fencing a vegetable plot

A circular vegetable plot has a diameter of 21 m. Find the length of fence needed to go once round its edge: exactly, then rounded using a calculator’s \pi, using 3.14, and using \displaystyle \frac{22}{7}.

Solution

The diameter is given, so use C = \pi d with d = 21:

C = \pi \times 21 = 21\pi \text{ m}

That is the exact answer, in terms of \pi. For a number you can buy fence by, replace \pi:

with a calculator’s \pi: 21\pi \approx 65.97 m, to two decimal places;
with 3.14: 21 \times 3.14 = 65.94 m;
with \displaystyle \frac{22}{7}: \displaystyle 21 \times \frac{22}{7} = 66 m.

The three rounded answers differ by a few centimeters, because the three values of \pi differ in the third decimal place, so 66 m of fence is the sensible order. A diameter that is a multiple of 7 makes \displaystyle \frac{22}{7} especially easy, since the 7 cancels.

Example 2

A ride on a Ferris wheel

A Ferris wheel has a radius of 25 m. How far does a rider travel in one full turn, and in a ride of three full turns? Give each answer exactly and to two decimal places.

Solution

This time the radius is given, so use C = 2\pi r:

C = 2\pi \times 25 = 50\pi \text{ m} \approx 157.08 \text{ m}

The rider sits at the rim, so one turn carries them once round the circle. Three turns carry them three times as far:

3 \times 50\pi = 150\pi \text{ m} \approx 471.24 \text{ m}

Rounding first, 3 \times 157.08 = 471.24, happens to agree here, but each multiplication also multiplies the rounding error, so the safer order is to round at the end.

Example 3

How good are 3.14 and 22/7?

A round running path in a park has a diameter of 350 m. Find the length of one lap using a calculator’s \pi, using 3.14, and using \displaystyle \frac{22}{7}. Which approximation is closer, and by how much is each one off?

Solution

With a calculator’s \pi:

C = 350\pi \approx 1099.56 \text{ m}

With the approximations, 350 \times 3.14 = 1099 m and \displaystyle 350 \times \frac{22}{7} = 50 \times 22 = 1100 m.

The answer with 3.14 is too short by about 1099.56 - 1099 = 0.56 m, and the answer with \displaystyle \frac{22}{7} is too long by about 1100 - 1099.56 = 0.44 m. Here \displaystyle \frac{22}{7} is closer.

In fact it always is: \pi - 3.14 \approx 0.0016 while \displaystyle \frac{22}{7} - \pi \approx 0.0013, and the formula multiplies that error by the diameter. For a coin the errors are far thinner than a pencil line; for a running path they are tens of centimeters.

Example 4

From a label back to the can

(a) A tin can’s paper label goes exactly once round the can and is 50.3 cm long. Find the can’s diameter and radius, to the nearest tenth of a centimeter.

(b) A circle has a circumference of 34\pi cm. Find its radius exactly.

Solution

(a) The label’s length is the can’s circumference, so C = 50.3 and \pi d = 50.3. Dividing both sides by \pi undoes the multiplication by \pi:

d = \frac{50.3}{\pi} \approx 16.0 \text{ cm}

The radius is half the diameter, \displaystyle r = \frac{50.3}{2\pi} \approx 8.0 cm. Check: \pi \times 16 \approx 50.27, which rounds to the 50.3 we started with.

(b) Here the circumference is given in terms of \pi, so the answer can be exact. Use C = 2\pi r and divide both sides by 2\pi:

2\pi r = 34\pi \qquad\Longrightarrow\qquad r = 17

The radius is exactly 17 cm. The \pi cancels, so no rounding is needed at all.

Example 5

A half-circle stage

A stage is a half circle, and its straight back edge is 12 m long (Figure 4, left). Find the length of its curved front edge and the length of a strip of lights that runs all the way round the stage’s edge, exactly and to two decimal places.

Figure 4The stage of Example 5 and the terrace of Example 6. The thick edges are parts of a circumference; the straight edges are not.

Solution

The straight edge goes across the full circle through its center, so it is a diameter: d = 12. The curved edge is half the circumference:

\tfrac{1}{2} \times \pi \times 12 = 6\pi \text{ m} \approx 18.85 \text{ m}

The strip of lights also runs along the straight edge, so its length is the perimeter:

P = 6\pi + 12 \text{ m} \approx 30.85 \text{ m}

The exact answer 6\pi + 12 cannot be tidied into a single term, because 6\pi and 12 are not like terms: one is a number of \pis and the other is not.

Example 6

A quarter-circle terrace

A paved terrace in the corner of a garden is a quarter circle with radius 8 m (Figure 4, right). Find its perimeter, exactly and to two decimal places.

Solution

The curved edge is a quarter of the circumference of a circle with radius 8:

\tfrac{1}{4} \times 2\pi \times 8 = 4\pi \text{ m} \approx 12.57 \text{ m}

The two straight edges are both radii, 8 m each, and they meet at the right angle in the corner. So

P = 4\pi + 16 \text{ m} \approx 28.57 \text{ m}

A quarter circle has two straight edges where a half circle has one.

Example 7

How far a bicycle goes

Gethin’s bicycle wheels have a diameter of 72 cm.

(a) How far does the bicycle travel while the wheels make 200 full turns? Give the answer in meters, exactly and to two decimal places.

(b) How many full turns must the wheels make for the bicycle to travel 1 km?

Solution

(a) In one turn the bicycle moves one circumference, C = 72\pi cm. In 200 turns it moves

200 \times 72\pi = 14400\pi \text{ cm}

There are 100 cm in a meter, so this is 144\pi m \approx 452.39 m.

(b) Now the distance is known and the number of turns is not. 1 km is 100000 cm, and each turn covers 72\pi cm, so the number of turns is

\frac{100000}{72\pi} \approx 442.10

That is not a whole number. After 442 full turns the bicycle has gone 442 \times 72\pi \text{ cm} \approx 999.8 m, just short, and after 443 it has gone about 1002.0 m. So the wheels must make 443 full turns. The answer is rounded up, even though 442.10 is nearer to 442, because the question asks for enough turns to cover the distance.

Boundaries

Only for circles. C = \pi d holds for circles and nothing else. An oval, a rounded square or a running track with straight sides has its own perimeter, worked out piece by piece, and its “across” is not even a single number.

The diameter goes through the center. A chord across a circle that misses the center is shorter than the diameter, and multiplying it by \pi gives a length shorter than the circumference.

A flat circle. A circle drawn on a flat surface has C \div d = \pi. A circle drawn on a curved surface, such as a ring round a ball, measured along that surface, does not: the ratio comes out smaller.

Perimeter of a piece includes its straight edges. The curved edge of a half or quarter circle is a fraction of the circumference; the perimeter of the shape is not, because the straight edges are not part of the circle.

Wheels must roll, not slide. One circumference per turn assumes the rim grips the ground. A wheel spinning on ice, or skidding with the brakes locked, breaks the link between turns and distance.

Where this goes next

The next step is to go only part of the way round. The fraction of a full turn that an angle at the center takes up gives the same fraction of the circumference, which is arc length, and arcs are first named and measured in degrees in arcs and central angles.

The space inside a circle, its area, turns out to involve the same number, and the area of a circle shows why.

And pi is the first number most people meet that is neither a whole number nor a fraction. Seeing why some numbers cannot be written as fractions starts with a simpler case, why the square root of 2 is not a fraction, and leads to the real numbers, the complete number line on which \pi has its place.

Exercises

Part 1: finding the circumference

A circle has a diameter of 17 cm. Find its circumference exactly, and to the nearest tenth of a centimeter.
A circle has a radius of 6 m. Find its circumference exactly, and to the nearest tenth of a meter.
A circle has a radius of 3.2 cm. Find its circumference exactly, and to the nearest tenth of a centimeter.
A car wheel has a diameter of 64 cm. Use 3.14 for \pi to find its circumference.
A round fountain basin has a radius of 3.5 m. Use \displaystyle \frac{22}{7} for \pi to find the distance round its rim.
A round drum has a radius of 9 cm. A strip of tape goes once round it with 2 cm extra to overlap. How long is the tape, exactly and to the nearest tenth of a centimeter?

Part 1: pi as a ratio

Jakub measures a round mirror. It is 50 cm across and his string round it is 156 cm long. What value of C \div d does he get, to two decimal places? Is it close to \pi?
Three measurements of round objects are recorded as (a) d = 10 cm, C = 31.5 cm; (b) d = 20 cm, C = 62.6 cm; (c) d = 12 cm, C = 75.4 cm. Find C \div d for each, to two decimal places. One was recorded wrongly. Which, and what was the likely mistake?
Circle A has a diameter of 5 cm and circle B a diameter of 20 cm. How many times as long as A’s circumference is B’s? A’s circumference measures 15.7 cm. Find B’s without using \pi.
A circle has a diameter of 49 cm. Find its circumference with a calculator’s \pi (to two decimal places), with 3.14, and with \displaystyle \frac{22}{7}. Which approximation is closer, and by how much is each off, to the nearest hundredth?

Part 1: reasoning

Explain why using 3.14 for \pi always gives a circumference slightly too short, and using \displaystyle \frac{22}{7} always gives one slightly too long.
Lowri doubles the radius of a circle. Show from the formula what happens to the circumference. Then explain the same fact using scale copies.
A regular hexagon with sides of 4 cm has its six corners on a circle. Explain why the circle’s radius is 4 cm and why its circumference must be more than 24 cm. Then find the circumference exactly and to the nearest tenth, and check that it is.

Part 2: back from the circumference

A circle has a circumference of 20\pi cm. Find its diameter and its radius exactly.
A circle has a circumference of 50 m. Find its radius to the nearest tenth of a meter.
A trundle wheel is a wheel on a handle that clicks once each turn, used for measuring distances on the ground. One is made so that each click marks exactly 1 m. Find its diameter to the nearest tenth of a centimeter.

Part 2: half and quarter circles

A half circle has a straight edge of 20 cm. Find the length of its curved edge and its perimeter, exactly and to two decimal places.
A quarter circle has a radius of 10 m. Find the length of its curved edge and its perimeter, exactly and to two decimal places.
A window is a square with sides of 60 cm, with a half circle on top whose straight edge is the top side of the square. Find the distance round the outside of the window, exactly and to two decimal places.
The curved edge of a half circle is 15\pi cm long. Find its straight edge, and its perimeter exactly and to two decimal places.

Part 2: turning wheels

A car’s wheels have a diameter of 60 cm. How far does the car travel while the wheels make 1000 full turns? Give the answer in meters, exactly and to the nearest meter.
Kiri’s bicycle wheels have a radius of 33 cm. Find the distance covered in one turn, to two decimal places. How many full turns must the wheels make to cover a ride of 2 km?
Hedda measures the length of a hall with a trundle wheel of diameter 40 cm. It clicks 25 times. How long is the hall, exactly in centimeters and to the nearest tenth of a meter?

Part 2: reasoning

Ifeoma says the perimeter of a half circle with diameter d is half of \pi d. Explain what she has left out, and show that the perimeter is \displaystyle \left(\frac{\pi}{2} + 1\right)d.
A tricycle’s front wheel has a diameter three times that of each back wheel. As the tricycle rides along, how many turns does a back wheel make for each turn of the front wheel? Explain.

Answers

17\pi cm \approx 53.4 cm. C = \pi d = \pi \times 17.
12\pi m \approx 37.7 m. C = 2\pi r = 2\pi \times 6.
6.4\pi cm \approx 20.1 cm. C = 2\pi \times 3.2.
200.96 cm. C \approx 3.14 \times 64. It is approximate, because 3.14 is.
About 22 m. The diameter is 7 m, and \displaystyle \frac{22}{7} \times 7 = 22.
18\pi + 2 cm \approx 58.5 cm. The circumference is 2\pi \times 9 = 18\pi cm, and the overlap adds 2 cm.
3.12. 156 \div 50 = 3.12. That is close to \pi, off by about 0.02, which a slightly loose string or a ruler read to the nearest centimeter easily explains.
(a) 3.15, (b) 3.13, (c) 6.28. Record (c) is wrong: its ratio is about twice \pi, so the 12 cm was almost certainly the radius, measured from the center. With d = 24 cm the ratio is 75.4 \div 24 \approx 3.14.
4 times, and B’s circumference is 62.8 cm. B is a scale copy of A at scale factor 20 \div 5 = 4, so its circumference is 4 \times 15.7.
153.94 cm, 153.86 cm and 154 cm. The answer with 3.14 is off by about 153.94 - 153.86 = 0.08 cm and the one with \displaystyle \frac{22}{7} by about 154 - 153.94 = 0.06 cm, so \displaystyle \frac{22}{7} is closer.
3.14 < \pi, so 3.14 \times d is less than \pi \times d for every positive d: always too short. \displaystyle \frac{22}{7} = 3.1428\ldots is more than \pi = 3.1415\ldots, so \displaystyle \frac{22}{7} \times d is always more than \pi d: always too long.
It doubles. With radius 2r the circumference is 2\pi(2r) = 2(2\pi r), twice the old one. A circle with twice the radius is a scale copy at scale factor 2, and a scale copy multiplies every length, the circumference included, by 2.
8\pi cm \approx 25.1 cm, which is more than 24 cm. Joining the center to two neighboring corners makes a triangle with a 60^\circ angle at the center and two equal sides, so it is equilateral and the radius equals the side, 4 cm. Between neighboring corners the circle curves while the hexagon goes straight, and the straight path is shorter, so C is more than the hexagon’s perimeter, 6 \times 4 = 24 cm. Then C = 2\pi \times 4 = 8\pi.
d = 20 cm and r = 10 cm. From \pi d = 20\pi, dividing by \pi gives d = 20; the radius is half of it.
8.0 m. \displaystyle r = \frac{50}{2\pi} = 7.957\ldots.
31.8 cm. One turn must cover 100 cm, so \pi d = 100 and \displaystyle d = \frac{100}{\pi} = 31.83\ldots.
Curved edge 10\pi cm \approx 31.42 cm; perimeter 10\pi + 20 cm \approx 51.42 cm. The curved edge is \displaystyle \frac{1}{2} \times \pi \times 20, and the perimeter adds the straight edge.
Curved edge 5\pi m \approx 15.71 m; perimeter 5\pi + 20 m \approx 35.71 m. The curved edge is \displaystyle \frac{1}{4} \times 2\pi \times 10, and the perimeter adds two radii of 10 m.
180 + 30\pi cm \approx 274.25 cm. Three sides of the square are on the outside, 3 \times 60 = 180; the fourth is covered by the half circle, whose curved edge is \displaystyle \frac{1}{2} \times \pi \times 60 = 30\pi.
Straight edge 30 cm; perimeter 15\pi + 30 cm \approx 77.12 cm. From \displaystyle \frac{1}{2}\pi d = 15\pi, d = 30.
600\pi m \approx 1885 m. One turn is 60\pi cm, so 1000 turns are 60000\pi cm, which is 600\pi m.
66\pi \approx 207.35 cm per turn, and 965 full turns. 2 km is 200000 cm, and 200000 \div 66\pi \approx 964.6. After 964 turns the bicycle has gone about 1998.8 m, short of 2 km, so it takes 965 full turns to cover the ride.
1000\pi cm \approx 31.4 m. Each click is one turn, 40\pi cm, and 25 \times 40\pi = 1000\pi.
She has left out the straight edge. Half of \pi d is only the curved edge; the perimeter of the shape also runs along the diameter, so it is \displaystyle \frac{1}{2}\pi d + d = \left(\frac{\pi}{2} + 1\right)d, which is d more than half the circumference.
3 turns. All three wheels roll along the same ground, so they cover the same distance. The front wheel’s circumference is \pi(3d) = 3\pi d, three times a back wheel’s \pi d, so while the front wheel turns once a back wheel must turn 3\pi d \div \pi d = 3 times.