Circles — Print

Every circle has the same ratio of circumference to diameter. That ratio is π, and it drives both formulas.

What the practice questions cover

These questions use 8 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.

Stand a can upright, mark the table where it touches, and roll it one full turn.

Mark where it touches again. The distance between the marks is the distance around the can, unrolled into a straight line.

roll a can along the table for one full turn one full turn r one turn leaves a straight track — how long is it?

Now measure it, and measure across the can too. Divide the first by the second.

Do it with a mug, a coin, a bicycle wheel. You always get about 3.14.

Not roughly. Always — for every circle that has ever existed or ever will.

That constant ratio is π. It is not a fact about a circle; it is what makes a circle a circle. C = π d is simply that measurement written down.

A circle labelled with radius, diameter, chord and sector radius diameter chord sector C = 2πr — a length, so r once A = πr² — a square measure, so r² π = C ÷ d the same for every circle, ≈ 3.14159 a sector of n° is n/360 of the whole

Where π comes from

Wrap a string round any circular object and measure it, then measure straight across. Divide the first by the second.

You always get about 3.14, whether the circle is a coin or a stadium. That constant ratio is π.

So C ÷ d = π, which rearranges to C = π d, and since d = 2r, also C = 2π r.

A circle rolled through one full turn: the track it leaves is its circumference, just over three diameters long roll a circle through one full turn and measure the track it leaves d one full turn C ddd 0.14d C ÷ d≈ 3.14 C ÷ d = 3.14159… = π, for a coin or a stadium alike so C = πd, and since d = 2r, C = 2πr

Area: why π r2

Cut the circle into thin wedges and lay them alternately point-up and point-down. They interlock into a shape very close to a rectangle.

That rectangle is r tall, and its width is half the circumference, π r.

So the area is π r × r = π r2. The thinner the wedges, the more exact it becomes.

A circle cut into twelve wedges and laid top-to-tail: a near-parallelogram r tall and πr wide cut it into wedges, then lay them alternately point-up and point-down r half of Chalf of C πr πr — half the circumference r area ≈ πr × r = πr² the thinner the wedges, the closer the shape is to a true rectangle

Telling them apart

Circumference is a length, so it uses r once. Area is a square measure, so it uses r2. If your answer is in cm but you squared, something is wrong.

Radius or diameter?

Both formulas want the radius. Problems love to give you the diameter instead.

Using d where r belongs makes a circumference twice too big and an area four times too big.

Write down r = … as your very first line, every time.

A circle of diameter 10: the right area uses r = 5; using 10 as the radius draws a circle four times the area r = 5 — correctd = 10, so halve it first 25 5 d = 10 A = π · 5² = 25π C = 2π · 5 = 10π d used as r — wrongthat is a circle twice as wide 25252525 10 A = π · 10² = 100π 4 × too big C = 2π · 10 = 20π 2 × too big

Halve it first

A circle of diameter 10 has area π(5)2 = 25π, not π(10)2 = 100π.

Parts of a circle

A semicircle is half: arc π r, area 12π r2. A quarter circle is a quarter of each.

In general, a sector of n degrees is n360 of the whole circle — use that fraction for both arc length and sector area.

For a 90° sector, 90360 = 14.

Exact or decimal?

Leaving the answer as 25π is exact. Writing 78.5 uses π ≈ 3.14 and is approximate.

Read the question: "in terms of π" means keep the symbol. "Use π = 3.14" means give a decimal.

When several steps are involved, keep π until the very end — rounding early compounds the error.

Worked examples

From diameter

A circular table has diameter 1.4 m. Find its circumference and area. Use π ≈ 227.

  1. r = 1.4 ÷ 2 = 0.7 m Always the first line.
  2. C = π d = 227 × 1.4 = 4.4 m Using d directly is fine for circumference.
  3. A = π r2 = 227 × 0.49 0.72 = 0.49 — square the radius, not the diameter.
  4. = 1.54 m2 Note the units: m for C, m2 for A.

C = 4.4 m, A = 1.54 m2.

A sector

Find the area of a 60° sector of a circle with radius 6. Give an exact answer.

  1. The sector is 60360 = 16 of the circle. Degrees out of 360.
  2. Whole circle: π(6)2 = 36π Square the radius first.
  3. 16 × 36π = 6π "Exact" means leave π in.

6π

Running track

A shape is a 20 m by 14 m rectangle with a semicircle of diameter 14 m on each short end. Find the perimeter. Use π ≈ 3.14.

  1. The two semicircles make one full circle of diameter 14. Two halves join into a whole — a common shortcut.
  2. That circle contributes C = π × 14 ≈ 43.96 m Use C = π d directly.
  3. The two straight sides contribute 20 + 20 = 40 m The short ends are no longer on the boundary — the curves replaced them.
  4. 43.96 + 40 = 83.96 m Add the curved and straight parts.

About 83.96 m.

Words to know

Radius
Centre to edge. Half the diameter.
Diameter
All the way across through the centre. d = 2r.
Circumference
The distance around: C = 2π r = π d.
Arc / sector
Part of the edge; a "pizza slice" bounded by two radii.
π
Circumference divided by diameter — the same for every circle, about 3.14159.

Mistakes to watch for

Using the diameter in π r2. Halve it first: since r is squared, using d makes the area four times too big, which is not a near miss.

Mixing up circumference and area. C = 2π r is a length in cm; A = π r2 is a covering in cm2. The units on your answer say immediately which one you found.

Including the straight edge that a semicircle replaced. When a semicircle sits on a rectangle, the shared edge is inside the shape — nobody walks along it, so it is not part of the perimeter.

Rounding π too early. Work in terms of π for as long as you can and round once at the end; rounding at every step lets the error grow.

Archimedes trapped π between polygons drawn inside and outside a circle, reaching 96 sides by hand to prove it lies between 31071 and 317.

In 1897 the Indiana state legislature nearly passed a bill that would have implied π = 3.2. A visiting mathematician happened to be in the building and talked them out of it.

Computers have now found over 100 trillion digits. NASA uses about fifteen — enough to navigate the solar system to within a few centimetres.

Problem 1

Find the circumference of a circle with radius 5. Use π ≈ 3.14.

Worked answer

31.4

2 × 3.14 × 5 = 31.4

Problem 2

Find the area of a circle with radius 4. Use π ≈ 3.14.

Worked answer

50.24

42 = 16

3.14 × 16 = 50.24

Problem 3

A circle has diameter 10. Find its radius.

Worked answer

5

10 ÷ 2 = 5

Problem 4

Find the circumference of a circle with radius 7. Use π ≈ 227.

Worked answer

44

2 × 227 × 7 = 44

Problem 5

Find the area of a circle with diameter 14. Use π ≈ 227.

Worked answer

154

r = 7

227 × 49 = 154

Problem 6

A circle has circumference 62.8. Find its radius. Use π ≈ 3.14.

Worked answer

10

62.8 = 6.28r

r = 10

Problem 7

A circle has radius 6. Find its area in terms of π (enter just the number in front of π).

Worked answer

36

62 = 36

A = 36π

Problem 8

Find the area of a semicircle of radius 6, in terms of π (enter just the number in front of π).

Worked answer

18

Full circle: 36π

Half: 18π

Problem 9

A circular table has diameter 1.4 m. Find its area. Use π ≈ 227.

Worked answer

1.54

r = 0.7 m

227 × 0.49 = 1.54 m2

Problem 10

Find the arc length of a quarter circle of radius 8, in terms of π (enter just the number in front of π).

Worked answer

4

Full circumference: 16π

A quarter: 4π

Problem 11

A circle has area 49π. Find its radius.

Worked answer

7

r2 = 49

r = 7

Problem 12

Find the area of a 60° sector of a circle with radius 6, in terms of π (enter just the number in front of π).

Worked answer

6

60360 = 16

Whole circle: 36π

16 × 36π = 6π

Problem 13

A shape is a 20 m by 14 m rectangle with a semicircle of diameter 14 m on each short end. Find the perimeter. Use π ≈ 3.14.

Worked answer

83.96

The two semicircles form one circle: C = 3.14 × 14 = 43.96 m

The two long sides: 20 + 20 = 40 m

43.96 + 40 = 83.96 m

Problem 14

A circle is inscribed in a square of side 12. Find the area inside the square but outside the circle. Use π ≈ 3.14.

Worked answer

30.96

Square: 12 × 12 = 144

Circle: r = 6, so 3.14 × 36 = 113.04

144 - 113.04 = 30.96

Problem 15

Two circles share a centre, with radii 10 and 7. Find the area of the ring between them, in terms of π (enter just the number in front of π).

Worked answer

51

Outer: 100π, inner: 49π

100π - 49π = 51π

Note it is NOT π(10-7)2 = 9π.

Problem 16

Find the arc length of a 90° sector of a circle with radius 10. Use π ≈ 3.14.

Worked answer

15.7

Full circumference: 2 × 3.14 × 10 = 62.8

14 × 62.8 = 15.7

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