Math 7 · 7.G.B.4 · G.GMD.A.1

Area of a circle

A circle has no straight sides to multiply. How much flat surface is inside one?

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Area of composite figures and shaded regionsMath 7 Sectors, segments and their areasGeometry Volume of prisms and cylindersMath 7 Why the circle area formula worksMath 7 Surface area of a sphereGeometry Surface area of prisms and cylindersMath 7 +3 more

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Area of a circle

Arc length Sectors, segments and their areas ▾ Show all 15 (6 more)

Why this exists

The area of a rectangle comes from counting rows of unit squares, and every polygon after that is found by cutting it into rectangles and triangles. A circle defeats both moves. It has no straight side to call a base, and however you cut it, curved edges are left over. Unit squares laid on a circle never fit at its edge: some poke out, some leave gaps.

Yet round things need areas all the time: how much mat a trampoline has, how much paint a round sign takes, how much lawn a sprinkler soaks. The answer is one short formula, and it needs only the radius.

Definition

The area of a circle is the amount of flat surface inside it, measured in square units, the way any area is: square centimeters, square meters, square feet. Strictly, the circle is the curve itself and the region inside it is a disc, but everyday speech says “the area of a circle” for the area of that inside region, and so does this lesson.

For a circle of radius r, the area A is

A = \pi r^2

where \pi is the same number that links a circle’s distance around to its distance across, in the Circumference formula C = \pi d = 2\pi r. Here r^2 means r \times r: square the radius first, then multiply by \pi.

When the diameter d is given instead, the radius is half of it, \displaystyle r = \frac{d}{2}, so

A = \pi\left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}

\pi is about 3.14159, and its decimals never end or repeat, so no decimal is exactly \pi. An answer such as 49\pi\text{ ft}^2 is exact. A decimal such as 153.9\text{ ft}^2 is rounded, and is written with \approx. The exact form keeps every digit; the rounded form tells you how big the area actually is.

Intuition

Take a circle of radius r and draw a square on one radius: a square r long and r wide, with area r^2. Call it a radius square. The formula A = \pi r^2 says the circle holds \pi radius squares, a little more than three of them.

Figure 1 shows why that number has to be between 2 and 4.

Figure 1The circle holds more than the diamond (2r^2) and less than the outer square (4r^2).

The square drawn round the circle is 2r wide and 2r tall, so it is made of four radius squares, area 4r^2. The circle sits inside it and misses its four corners, so the circle’s area is less than 4r^2.

The diamond inside the circle has its corners at the ends of two diameters. Each radius square holds one quarter of the diamond, a right triangle that is exactly half the radius square. So the diamond is four halves, area 2r^2, and the circle, which contains the diamond and more, has area more than 2r^2.

So the circle holds between 2 and 4 radius squares. By eye it fills about three quarters of the outer square, a little more than 3 radius squares, and the exact number turns out to be \pi itself.

Results

Area of a circle. For a circle of radius r or diameter d,

A = \pi r^2 = \frac{\pi d^2}{4}

Radius from area. If a circle has area A, then \displaystyle r^2 = \frac{A}{\pi}, so the radius is the positive number whose square is \displaystyle \frac{A}{\pi}:

r = \sqrt{\frac{A}{\pi}}

Half and quarter circles. A half circle of radius r has area \displaystyle \frac{1}{2}\pi r^2, and a quarter circle has area \displaystyle \frac{1}{4}\pi r^2.

Rings. The ring between a circle of radius R and a smaller circle of radius r inside it has area

\pi R^2 - \pi r^2

Scaling. Multiplying the radius by k multiplies the area by k^2. Twice the radius gives four times the area.

Why it is true

The full argument is its own lesson, Why the circle area formula works. Here is the outline in three steps.

The area is some number times r^2. Every circle is a scaled copy of every other circle: a circle of radius r is a circle of radius 1 with every length multiplied by r. Multiplying every length by r multiplies every area by r^2, because an area is made of two lengths multiplied together (Scale factor, area and volume). So a circle of radius r has r^2 times the area of a circle of radius 1, whatever that area is. Call it k: then A = k r^2.

That number is between 2 and 4. Figure 1 traps a circle of radius 1 between a diamond of area 2 and a square of area 4.

That number is \pi. Cut a circle into many thin wedges, like slices from the center, and lay them side by side, pointing alternately up and down. They make a shape very close to a parallelogram, and the thinner the wedges, the closer it gets to a rectangle. Its height is the radius r. Its length is half the curved edge, since half the wedges point each way, and half the circumference is \displaystyle \frac{1}{2} \times 2\pi r = \pi r. So the area is

A = \pi r \times r = \pi r^2

Worked examples

Examples 1 to 3 belong to Part 1 of the lesson and Examples 4 to 7 to Part 2. Unless an example says otherwise, rounded answers use the calculator’s \pi key and are rounded to one decimal place at the end.

Example 1

Area from the radius

A round trampoline has a jumping mat of radius 7 ft. How much mat is there? Give the area exactly and rounded.

Solution

The amount of mat is the circle’s area. With r = 7,

A = \pi r^2 = \pi \times 7^2 = 49\pi

So the exact area is 49\pi\text{ ft}^2. As a decimal, 49\pi \approx 153.938, so the mat is about 153.9\text{ ft}^2. Feet times feet gives square feet, which is why the unit is squared.

If \pi is replaced by 3.14 instead, 49 \times 3.14 = 153.86, which is close but slightly low, because 3.14 is a little less than \pi.

Check. The mat sits inside a square 14 ft wide, area 196\text{ ft}^2, and fills about three quarters of it: 0.75 \times 196 = 147. The answer is a little more than that, as it should be.

Example 2

Area from the diameter

A round wooden sign is 1.2 m across. Its face is given two coats of paint. How much surface does the paint cover in total?

Solution

The width across is the diameter, so the radius is half of it: r = 1.2 \div 2 = 0.6 m. One coat covers the area of the face:

A = \pi \times 0.6^2 = 0.36\pi

The same answer comes straight from the diameter, using \displaystyle A = \frac{\pi d^2}{4}:

\frac{\pi \times 1.2^2}{4} = \frac{1.44\pi}{4} = 0.36\pi

Two coats cover the face twice:

2 \times 0.36\pi = 0.72\pi \approx 2.26

The paint covers 0.72\pi\text{ m}^2, about 2.26\text{ m}^2 in total. Here two decimal places are kept, because the area is small and a tenth of a square meter is a large part of it.

Example 3

How area grows with the radius

Olusegun can buy one sprinkler that waters a circle of radius 12 m, or two sprinklers that each water a circle of radius 6 m. Which choice waters more ground, if the circles do not overlap?

Solution

The large sprinkler waters

\pi \times 12^2 = 144\pi \approx 452.4\text{ m}^2

Each small one waters \pi \times 6^2 = 36\pi, so the two together water

2 \times 36\pi = 72\pi \approx 226.2\text{ m}^2

The single large sprinkler waters twice as much as the two small ones together, and four times as much as one of them.

The reason is the square in the formula. Doubling the radius doubles both the width and the length of everything inside the circle, so the area is multiplied by 2 \times 2 = 4. In general, a circle of radius kr has area

\pi(kr)^2 = k^2 \pi r^2

which is k^2 times the area of a circle of radius r.

Example 4

Finding the radius from the area

(a) A round window has area 1600\pi\text{ cm}^2. Find its radius and its diameter.

(b) Another round window has area 2000\text{ cm}^2. Find its radius and diameter to one decimal place.

Solution

(a) The area formula still holds; this time A is known and r is not:

1600\pi = \pi r^2

Divide both sides by \pi, which keeps them equal:

r^2 = 1600

The radius is the number whose square is 1600. A radius is a length, so it is positive, and 40 \times 40 = 1600, so \displaystyle r = \sqrt{1600} = 40. The radius is 40 cm and the diameter is twice that, 80 cm.

Check. \pi \times 40^2 = 1600\pi.

(b) The same two moves, with no \pi to cancel this time:

\begin{aligned} \pi r^2 &= 2000 \\ r^2 &= \frac{2000}{\pi} \approx 636.62 \end{aligned}

The radius is the positive number whose square is 636.62. Since 25^2 = 625 is a little too small and 26^2 = 676 is too big, r is between 25 and 26, close to 25. A calculator’s square root key gives r \approx 25.23.

Check by squaring. 25.23^2 \approx 636.55, very close to 636.62.

The radius is about 25.2 cm. The diameter is twice the unrounded radius, 2 \times 25.231\ldots \approx 50.5 cm. Doubling the rounded 25.2 would give 50.4, which is why rounding is left to the very end.

Example 5

Half and quarter circles

(a) A doormat shaped as a half circle has a straight edge 90 cm long, as in the left of Figure 2. Find its area.

(b) A corner of a playground is covered with soft matting in a quarter circle of radius 9 m, as on the right of Figure 2. Find the area of the matting.

Figure 2A half circle is cut off by a diameter, a quarter circle by two radii at a right angle.

Solution

(a) The straight edge of a half circle is a diameter of the full circle, so the radius is 90 \div 2 = 45 cm. The full circle would have area \pi \times 45^2 = 2025\pi, and the mat is half of it, because a diameter cuts a circle into two matching halves:

A = \frac{1}{2} \times 2025\pi = 1012.5\pi \approx 3180.9

The doormat has area 1012.5\pi\text{ cm}^2, about 3180.9\text{ cm}^2.

(b) Two radii at a right angle cut off one quarter of the full turn, so the matting is a quarter of a circle of radius 9 m:

A = \frac{1}{4} \times \pi \times 9^2 = \frac{81\pi}{4} = 20.25\pi \approx 63.6

The matting covers 20.25\pi\text{ m}^2, about 63.6\text{ m}^2.

In both parts the radius comes first, then the area of the whole circle, then the fraction of it. A slice that is a third, or any other fraction of the turn, works the same way, and that is the area of a sector.

Example 6

A ring

The floor of a park bandstand is a circle of radius 2 m. Pietro lays a paved border 1 m wide all the way round it, as in Figure 3. Find the area of the border, and say whether it is more or less than the area of the floor.

Figure 3The border is the big circle with the floor taken out of it.

Solution

The border is not a shape the formula covers directly, but it is the difference of two that it does. The floor and the border together make a big circle, whose radius is the floor’s radius plus the border’s width, 2 + 1 = 3 m. Taking the floor out of the big circle leaves the border:

\begin{aligned} \text{border} &= \pi \times 3^2 - \pi \times 2^2 \\ &= 9\pi - 4\pi = 5\pi \end{aligned}

The border has area 5\pi\text{ m}^2, about 15.7\text{ m}^2.

The floor has area 4\pi\text{ m}^2, about 12.6\text{ m}^2, so the border, only 1 m wide, covers more ground than the floor it surrounds. The outer band of a circle holds much of its area, because it runs all the way round on the longest circumference.

Example 7

A length and an area from one circle

Saoirse-Ann has a dartboard 44 cm across. She fits a thin metal band round its rim and covers its face with new cork.

(a) How long is the band, and how much cork does she need?

(b) Is there a size of circle whose circumference and area are the same number?

Solution

(a) The band runs round the edge, so its length is the Circumference, a length:

C = \pi d = 44\pi \approx 138.2\text{ cm}

The cork covers the face, so its amount is the area. The radius is 22 cm:

A = \pi \times 22^2 = 484\pi \approx 1520.5\text{ cm}^2

The band is about 138.2 cm long and the cork about 1520.5\text{ cm}^2. These are different kinds of quantity: one is measured with a tape, the other by covering with squares. A board twice as wide would need twice the band but four times the cork, since the circumference has d in it once and the area has r in it twice.

(b) Setting the two formulas equal and dividing both sides by \pi r (allowed, since r is not zero):

\begin{aligned} 2\pi r &= \pi r^2 \\ 2 &= r \end{aligned}

So a circle of radius 2 m has circumference 4\pi m and area 4\pi\text{ m}^2: the same number. But it is the same circle that has radius 200 cm, and in centimeters its circumference is 400\pi cm and its area is 40000\pi\text{ cm}^2. The numbers no longer match. The match was produced by the choice of unit, not by anything about the circle, and that is exactly what it means for a length and an area to be different kinds of thing: comparing them is like asking whether a distance is bigger than a weight.

Boundaries

The shape must be a true circle. The formula uses one radius, and only a circle has the same radius in every direction. An oval is longer one way than the other and \pi r^2 does not give its area for any choice of r.

The r is the radius. With a diameter, halve it first or use \displaystyle \frac{\pi d^2}{4}; putting d itself into \pi r^2 gives \pi d^2, four times the area.

Every length in the same unit. A radius in meters gives square meters. Changing an area from one unit to another needs the square of the length conversion (Converting square and cubic units).

Decimals are approximations. \pi has no exact decimal, so a rounded area carries \approx, and rounding is left to the last step, as in Example 4.

Only the positive square root. When r^2 = 1600, the number -40 also squares to 1600, but a radius is a length and cannot be negative.

Half and quarter circles are cut through the center. A half circle is cut off by a diameter, and a quarter circle by two radii at a right angle. A piece cut off by a line that misses the center is not half the circle, and its area is a segment, found another way.

A ring is a subtraction. The ring’s area is the big circle’s area minus the small one’s, \pi R^2 - \pi r^2, which is not the area of a circle whose radius is the width R - r. The subtraction works whenever the smaller circle lies wholly inside the larger; the band has the same width all round only when the two share a center.

Where this goes next

The half and quarter circles here are the first cases of a general rule: a slice that is any fraction of the full turn has that fraction of the area, which is the area of a sector. Circles, half circles and rings also join rectangles and triangles in composite figures and shaded regions, where a shape is split into pieces whose areas you know.

In three dimensions, \pi r^2 becomes the floor of a cylinder: stacking copies of that circle up to a height h gives the volume of a cylinder, \pi r^2 h. A sphere’s surface turns out to be exactly four circles of the same radius, the surface area of a sphere.

Exercises

Exercises 1 to 9 go with Part 1 and Exercises 10 to 21 with Part 2. Give each area exactly in terms of \pi and rounded to one decimal place, unless the exercise says otherwise.

Part 1: area from the radius or diameter

A circle has radius 4 cm. Find its area.
A circle has diameter 18 in. Find its area.
A circle has radius 1.5 m. Find its area.
A circle has diameter 11 cm. Find its area.
A circle has radius \displaystyle \frac{3}{4} ft. Find its area.
Morwenna is painting the top of a round concrete step 1.5 m across. She gives it three coats, and one liter of paint covers 10\text{ m}^2 of surface. How much paint does she need, to the nearest hundredth of a liter?
A round tray has radius 8 in and a square tray has sides of 14 in. Which has more surface, and by about how much?

Part 1: how area grows

Circle P has radius 3 cm and circle Q has radius 21 cm. Find both areas exactly. How many times the area of P is the area of Q?
A circle has radius 15 cm. Find the area of the square drawn round it and of the diamond inside it, as in Figure 1. Then find the circle’s area to one decimal place and check that it lies between them.

Part 2: the radius from the area

A circle has area 196\pi\text{ m}^2. Find its radius and diameter.
A circle has area 50\text{ cm}^2. Find its radius to one decimal place, and check by squaring.
Niamh’s sprinkler waters a circle of radius 5 m. What radius would water twice that area? Give it to one decimal place, and check by squaring.

Part 2: half and quarter circles

Find the area of a half circle of radius 8 cm.
A stage shaped as a quarter circle of radius 2.4 m fits into the corner of a hall. Find its area.
A shelf shaped as a half circle has area 1250\text{ cm}^2. Find its radius and the length of its straight edge, each to one decimal place.

Part 2: rings

Find the area of the ring between two circles with the same center, of radii 9 cm and 4 cm.
On an archery target, one scoring ring is the band between two circles with the same center, of radii 16 cm and 17 cm. Find its area.
A round sandpit has radius 3 m, with a strip of rubber matting 1.5 m wide all the way round it. Find the area of the matting.
A ring has area 24\pi\text{ cm}^2, and its inner circle has radius 1 cm. Find the radius of the outer circle and the width of the ring.

Part 2: length or area

A circle has radius 20 cm. Find its circumference and its area, each exactly and to one decimal place, with the right unit on each.
The rim of a round skating rink is 36\pi m long. Find the area of its ice.

Reasoning

Show that every circle has area equal to half its circumference times its radius, \displaystyle A = \frac{1}{2}Cr. Then explain why multiplying the two lengths C and r gives an area.
Rhodri says a border 1 m wide round any round floor needs the same amount of paving, whatever the size of the floor. Write the border’s area for a floor of radius r m, and explain whether he is right.
Which is larger: half of a circle of radius r, or a whole circle of radius \displaystyle \frac{r}{2}? By what factor? Explain.
A circle has radius 1 cm. Compare the numbers in its circumference and its area. Then measure the same circle in millimeters and compare again. What does this show about asking whether a circle’s area is bigger than its circumference?

Answers

16\pi\text{ cm}^2 \approx 50.3\text{ cm}^2. A = \pi \times 4^2 = 16\pi.
81\pi\text{ in}^2 \approx 254.5\text{ in}^2. The radius is 18 \div 2 = 9, and \pi \times 9^2 = 81\pi.
2.25\pi\text{ m}^2 \approx 7.1\text{ m}^2. 1.5^2 = 2.25.
30.25\pi\text{ cm}^2 \approx 95.0\text{ cm}^2. The radius is 5.5, and 5.5^2 = 30.25.
\displaystyle \frac{9}{16}\pi\text{ ft}^2 \approx 1.8\text{ ft}^2. \displaystyle \left(\frac{3}{4}\right)^2 = \frac{9}{16}.
About 0.53 liters. The radius is 0.75 m, so one coat covers \pi \times 0.75^2 = 0.5625\pi, and three coats cover 1.6875\pi \approx 5.30\text{ m}^2. At 10\text{ m}^2 a liter that is 5.30 \div 10 \approx 0.53 liters.
The round tray, by about 5.1\text{ in}^2. The round tray has 64\pi \approx 201.1\text{ in}^2 and the square tray 14 \times 14 = 196\text{ in}^2.
9\pi\text{ cm}^2 and 441\pi\text{ cm}^2, so Q has 49 times the area of P. The radius is 7 times as large, and 7^2 = 49.
The square has area 900\text{ cm}^2 (it is 30 cm wide), the diamond 450\text{ cm}^2 (four right triangles with legs of 15 cm, each 112.5\text{ cm}^2), and the circle 225\pi \approx 706.9\text{ cm}^2, which lies between 450 and 900.
Radius 14 m, diameter 28 m. \pi r^2 = 196\pi gives r^2 = 196, and 14 \times 14 = 196.
About 4.0 cm. \displaystyle r^2 = \frac{50}{\pi} \approx 15.92. Since 4^2 = 16 is just over, r is just under 4: r \approx 3.99, which is 4.0 to one decimal place. Check: 3.99^2 \approx 15.92.
About 7.1 m. Twice the area is 2 \times 25\pi = 50\pi, so r^2 = 50 and r is the number whose square is 50. 7^2 = 49 is just under and 7.1^2 = 50.41 just over; 7.05^2 \approx 49.70 is still under, so r \approx 7.07, which is 7.1. Doubling the radius to 10 m would give four times the area, not twice.
32\pi\text{ cm}^2 \approx 100.5\text{ cm}^2. Half of \pi \times 8^2 = 64\pi.
1.44\pi\text{ m}^2 \approx 4.5\text{ m}^2. A quarter of \pi \times 2.4^2 = 5.76\pi.
Radius about 28.2 cm, straight edge about 56.4 cm. From \displaystyle \frac{1}{2}\pi r^2 = 1250, \displaystyle r^2 = \frac{2500}{\pi} \approx 795.77. Since 28^2 = 784 and 29^2 = 841, r is between 28 and 29; a calculator gives r \approx 28.21, and 28.21^2 \approx 795.80. The straight edge is a diameter, 2r \approx 56.42.
65\pi\text{ cm}^2 \approx 204.2\text{ cm}^2. \pi \times 9^2 - \pi \times 4^2 = 81\pi - 16\pi.
33\pi\text{ cm}^2 \approx 103.7\text{ cm}^2. \pi \times 17^2 - \pi \times 16^2 = 289\pi - 256\pi.
11.25\pi\text{ m}^2 \approx 35.3\text{ m}^2. The outer radius is 3 + 1.5 = 4.5 m, and \pi \times 4.5^2 - \pi \times 3^2 = 20.25\pi - 9\pi.
Outer radius 5 cm, width 4 cm. \pi R^2 - \pi \times 1^2 = 24\pi, so R^2 = 25 and R = 5; the width is 5 - 1 = 4.
C = 40\pi\text{ cm} \approx 125.7\text{ cm} and A = 400\pi\text{ cm}^2 \approx 1256.6\text{ cm}^2. The circumference is a length, in centimeters; the area is in square centimeters.
324\pi\text{ m}^2 \approx 1017.9\text{ m}^2. 2\pi r = 36\pi gives r = 18, and \pi \times 18^2 = 324\pi.
\displaystyle \frac{1}{2}Cr = \frac{1}{2} \times 2\pi r \times r = \pi r^2 = A. Multiplying a length in centimeters by a length in centimeters gives square centimeters, just as length times width does for a rectangle, so the product of two lengths is an area. It is the area of a triangle with base C and height r, which is the picture behind Why the circle area formula works.
He is wrong. The border’s area is \pi(r + 1)^2 - \pi r^2 = \pi(2r + 1), which grows with r: 5\pi for a floor of radius 2 m, 7\pi for radius 3 m. A bigger floor has a longer edge, so a border of the same width round it is longer and needs more paving.
The half circle, by a factor of 2. Half of a circle of radius r has area \displaystyle \frac{1}{2}\pi r^2; a circle of radius \displaystyle \frac{r}{2} has area \displaystyle \pi\left(\frac{r}{2}\right)^2 = \frac{1}{4}\pi r^2. Halving the radius quarters the area, while cutting the circle in half only halves it.
In centimeters, C = 2\pi \approx 6.28 cm and A = \pi \approx 3.14\text{ cm}^2, so the area’s number is smaller. In millimeters the radius is 10 mm, C = 20\pi mm and A = 100\pi\text{ mm}^2, so the area’s number is larger. Which number is bigger depends on the unit chosen, so the question has no answer: a length and an area are different kinds of quantity and cannot be compared.