Math 7 · 7.G.B.4 · G.GMD.A.1

Why the circle area formula works

A circle has no straight sides to measure. Why should the space inside it be exactly pi times the radius times the radius?

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Area of a circleMath 7

Where you are 15 in Area and perimeter

Measurement12

Area and perimeter15

▴ 10 above Area of triangles Area of any polygon by cutting it into triangles and rectangles Area of composite figures and shaded regions

Why the circle area formula works

Area of a regular polygon

Why this exists

A rectangle’s area comes from counting rows of unit squares, and a parallelogram or a triangle is cut and rearranged until it is a rectangle, or half of one. All of those shapes have straight sides. A circle has none, so it seems no amount of cutting could turn it into a rectangle. Yet the area formula is exact:

A = \pi r^2

The \pi in it is the same number as in the Circumference formula C = 2\pi r, which counts how many diameters go once round the circle. Nothing obvious links the distance round a circle to the space inside it. This lesson shows the link: a circle can be cut into pieces that come closer and closer to a rectangle whose sides are the radius and half the circumference.

Definition

The statement this lesson explains is:

The area inside a circle equals the area of a rectangle whose height is the radius and whose length is half the circumference.

In symbols, with r the radius and C the circumference,

A = \tfrac12 C \times r

Because C = 2\pi r, half the circumference is \pi r, and the rectangle’s area is \pi r \times r = \pi r^2.

A wedge is a slice cut from the center to the edge, like a slice of a round cake. Its two straight sides are radii, each exactly r long. Its curved edge is its crust, and the crusts of all the wedges together make the whole circumference.

Intuition

Cut a circle into 8 equal wedges and shade the 4 in the top half. Lay the plain wedges in a row, points up and crusts on the table, and drop each shaded wedge, point down, into the gaps between them.

Figure 1Eight wedges interlock into a shape that is nearly a rectangle. The crusts along the bottom are half the circumference.

Neighboring pieces fit along their straight sides, because every straight side is a radius of length r. The result is a bumpy, leaning near rectangle:

Along the bottom lie the 4 plain crusts, the bottom half of the circle’s edge: \tfrac12 C.
Along the top lie the 4 shaded crusts, the other \tfrac12 C.
Each end is a radius, r long. Because the ends lean, the height is a little less than r.

Now use 16 wedges.

Figure 2With 16 wedges the bumps are smaller and the ends lean less. The shape is closer to a rectangle r tall and \tfrac12 C long.

Each crust is half as long, so it bends away from a straight line far less and the bumps shrink. Each wedge is thinner, so the ends stand nearer upright and the height is nearer r. With 32 wedges, or 1000, the bumps keep flattening and the lean keeps straightening. The area never changes, since it is always the same circle in pieces. Only the shape changes, toward a rectangle whose area is

\tfrac12 C \times r = \pi r \times r = \pi r^2

A second picture: rings

Cut the circle instead into thin rings, like the rings of a tree trunk. Cut every ring along one radius and straighten it into a strip.

Figure 3Cut the rings along one radius and straighten them. Stacked, they make a staircase that hugs a triangle with base C and height r.

The outer strip is nearly the whole circumference C. A ring halfway in is about half as long, because its radius is half as big, and the rings near the center are tiny. Stacked longest at the bottom and lined up at one end, the strips shrink steadily from about C to almost nothing, and the stack is r tall, since the rings reach from the edge to the center. That staircase hugs a triangle with base C and height r, and a triangle’s area is half its base times its height:

\tfrac12 \times C \times r = \tfrac12 \times 2\pi r \times r = \pi r^2

Thinner rings make smaller steps and a closer fit. The two pictures agree: the rectangle \tfrac12 C by r and the triangle with base C and height r both have area \tfrac12 C r.

Results

Area from circumference. For every circle, the area is half the circumference times the radius:

A = \tfrac12 C r

The area formula. Since C = 2\pi r,

A = \pi r^2

One shape for every circle. The near rectangle of any circle is \pi r long and r tall, so its length is always \pi times its height. One measurement of it fixes the other.

Why it is true

Moving pieces does not change area. The wedges are moved, not stretched or shrunk. None overlaps another and no gap is left, because neighbors meet along a full radius. Area adds over pieces, so the bumpy shape has exactly the circle’s area, whether it has 8 wedges or 8000. It is the same move as the cut and slide that turns a parallelogram into a rectangle.

A straight-sided shape that can be measured. Replace every crust by the straight line joining its ends. Each wedge becomes a thin triangle and the whole row becomes an exact parallelogram, whose area is its length times its height. For a circle of radius 1:

Wedges	Length of the bottom	Height	Area of the straight-sided shape
8	3.061	0.924	2.828
16	3.121	0.981	3.061
64	3.140	0.999	3.137
256	3.142	1.000	3.141

These values come from a calculation with trigonometry, met in a later course; here only their trend matters. The length closes in on \pi \approx 3.14159, which is half this circle’s circumference, and the height closes in on 1, the radius. So the parallelogram’s area closes in on \pi \times 1 \times 1, which is \pi r^2 for this circle.

The same shapes close in on the circle’s area. The parallelogram leaves out only the thin slivers between each crust and its straight line. All of them lie in a band along the circle’s edge, and the band gets thinner as the wedges do, so the missing area shrinks toward 0. That is why the last column climbs from below toward \pi; by 256 wedges the slivers together are less than one thousandth of the whole.

From “closer and closer” to “equal”. The parallelogram areas close in on \pi r^2 and also on the circle’s area A. One list of numbers cannot close in on two different values, so the gap between A and \pi r^2 is smaller than any amount you can name, and the only such gap is 0. Therefore

A = \pi r^2

exactly, not approximately. This is the idea of a limit: a value that a process approaches as closely as you like, though no single step of it gets there. It is used informally here. Limits makes it precise, and the ring picture, adding up very many thin strips, grows into the integral as a limit of sums.

Archimedes proved this result in Measurement of a Circle, Proposition 1: a circle has the same area as a right triangle whose two shorter sides are the radius and the circumference.

Worked examples

Example 1

What wide wedges get right and wrong

A circle of radius 18 cm is cut into only 4 wedges, interlocked as in Figure 4. Which measurements of the shape match the rectangle \tfrac12 C by r, and which do not? What is the shape’s area?

Figure 4With 4 wide wedges the pieces still fill the same area, but the shape is a poor rectangle.

Solution

The circumference is C = 2\pi(18) = 36\pi cm, so each of the 4 crusts is 36\pi \div 4 = 9\pi \approx 28.3 cm long.

What still matches. Two crusts lie along the bottom, so the bottom edge, following its bumps, is 2 \times 9\pi = 18\pi \approx 56.5 cm, exactly half the circumference. The top is the same. Each slanted end is a radius, exactly 18 cm. And the area is exactly the circle’s, because the pieces are the circle’s own pieces:

A = \pi (18)^2 = 324\pi \approx 1017.9 \text{ cm}^2

What does not. The top and bottom are deep bumps, not straight lines, and the ends lean steeply, so the shape is not 18 cm tall. With each crust replaced by its straight line, every wedge becomes a right triangle with two 18 cm sides, area \tfrac12 \times 18 \times 18 = 162 cm², and the shape becomes a parallelogram about 50.9 cm long and 12.7 cm tall, with area 4 \times 162 = 648 cm². The four bumps it leaves out hold the other 370 cm² or so.

So the area of 4 wedges is right, but nobody could read it off this shape as length times height. That reading only becomes trustworthy as the wedges get thin.

Example 2

Working backwards from the near rectangle

Tegwen cuts a paper circle into thin wedges and interlocks them. The bottom of her near rectangle measures 59.7 cm. Find the radius of her circle and its area.

Solution

The bottom of the near rectangle is half the circumference, \pi r. So

\pi r = 59.7 \qquad\Longrightarrow\qquad r = \frac{59.7}{\pi} \approx 19.0

The radius is about 19 cm, and that is also the height of the near rectangle. Its area, and so the circle’s, is about

59.7 \times 19 = 1134.3 \text{ cm}^2

Check. The whole circumference would be 2 \times 59.7 = 119.4 cm, and 2\pi(19) \approx 119.4. One measurement was enough, because the length of a near rectangle is always \pi times its height.

Example 3

The ring picture with numbers

A round flower bed has radius 5.2 m. Describe the triangle that the ring picture turns it into, and use it to find the area of the bed.

Solution

The outermost strip is as long as the circumference,

C = 2\pi(5.2) = 10.4\pi \approx 32.7 \text{ m}

and that is the base of the triangle. The rings run from the edge to the center, so the stack is 5.2 m tall, the radius. The triangle’s area is half its base times its height:

\tfrac12 \times 10.4\pi \times 5.2 = 27.04\pi \approx 84.9 \text{ m}^2

Check. The formula gives \pi r^2 = \pi(5.2)^2 = 27.04\pi, the same.

Example 4

Why tripling the radius multiplies the area by nine

Use the near rectangle to explain what happens to a circle’s area when its radius is multiplied by 3. Check with circles of radius 5 mm and 15 mm.

Solution

The near rectangle is r tall and \pi r long. Triple the radius and both sides triple: the height becomes 3r and the length \pi(3r) = 3\pi r. Multiplying both sides of a rectangle by 3 multiplies its area by 3 \times 3 = 9:

3\pi r \times 3r = 9 \pi r^2

The circle of radius 5 mm has near rectangle 5\pi by 5, area 25\pi mm², and the circle of radius 15 mm has near rectangle 15\pi by 15, area 225\pi mm². And 225\pi \div 25\pi = 9. The square in \pi r^2 is there because the radius appears in both sides of the rectangle; Scale factor, area and volume treats the same pattern for every shape.

Example 5

The same argument for a slice

A slice cut from the center of a circle has a curved edge 13 cm long and straight sides 12 cm long. Find its area without finding its angle.

Solution

A slice like this is a sector. Cut it into many thin wedges from the center and interlock them. Their crusts together make the curved edge, 13 cm, and half of them lie along the bottom, so the near rectangle is \tfrac12 \times 13 = 6.5 cm long. Its height closes in on the radius, 12 cm. So

A = 6.5 \times 12 = 78 \text{ cm}^2

The argument only needs wedges whose straight sides are radii, so for any sector, area = \tfrac12 \times (\text{curved edge}) \times r.

Check. The full circle of radius 12 cm has circumference 24\pi cm and area 144\pi cm². The slice is \displaystyle \frac{13}{24\pi} of the way round, so its area is \displaystyle \frac{13}{24\pi} \times 144\pi = 78 cm². The Sectors, segments and their areas lesson does this kind of calculation in full.

Boundaries

The argument starts from the circumference. It uses C = 2\pi r, so it takes the Circumference formula as known. What it shows is that the \pi in the area formula is the same number as the \pi in the circumference formula.

Rearranging keeps area only if nothing is stretched. The wedges are moved whole. A ring is different: straightening it stretches its inner edge or squashes its outer edge, so a thick ring does not straighten into an exact rectangle. The ring picture is honest only as the rings get thin.

No finite number of wedges gives a true rectangle. With 8 wedges, or 8 million, there are still bumps and a lean. A drawing is evidence, not proof; the equality A = \pi r^2 comes from the limit step.

It is about circles. An oval (an ellipse) cut from its center gives wedges with straight sides of different lengths, so they do not fit together and there is no single height to read off.

It needs wedges whose straight sides are radii. That is why it extends to a sector (Example 5) and a semicircle, but not to a piece of a circle cut off by a straight line that misses the center.

Where this goes next

The area of a circle lesson uses A = \pi r^2 for round tables, paths and windows, and Example 5 is the start of the area of a sector. The area of a regular polygon uses the same cut from the center with true triangles: half the perimeter times the distance from the center to a side. As the number of sides grows, the perimeter closes in on C and that distance on r, which is this lesson’s argument again.

“Closer and closer, therefore equal” is the seed of Limits, and adding up many thin strips, as in the ring picture, becomes the integral in calculus. Stacking thin slices is also how Cavalieri’s principle and the volume of a cylinder work.

Exercises

Measuring the near rectangle

A circle of radius 14 cm is cut into 28 equal wedges and interlocked. How long is each crust, and how long is the bottom edge of the shape?
A circle has radius 16 m. Give the length and height of the rectangle its wedges close in on, and the circle’s area, exactly and to one decimal place.
A circle has diameter 22 cm. Give the length and height of the rectangle its wedges close in on, and the circle’s area, exactly and to one decimal place.
As the wedges of a circle get thinner, the height of its near rectangle closes in on 23 cm. What does the length close in on, and what is the circle’s area?
The bottom of a near rectangle made from a circle measures 53.4 m. Find the radius of the circle and its area, each approximately.

The ring picture

In the ring picture for a circle of radius 4.2 cm, what are the base and height of the triangle? Use them to find the area.
Wiremu wraps a tape round a round tin lid and measures its circumference as 57.2 cm, and its radius as 9.1 cm. Use A = \tfrac12 C r to find the area without using \pi, then compare with \pi r^2.
A circle of radius 30 cm is cut into 3 rings, each 10 cm thick. Take each strip’s length to be the circumference halfway across its ring. Find the three lengths, the total area of the three strips, and compare it with \pi r^2.

Using the argument elsewhere

A semicircle has radius 26 cm. Cut it into thin wedges from the center and interlock them. How long is the near rectangle, how tall, and what is the semicircle’s area?
A sector has a curved edge 18 cm long and straight sides 17 cm long. Find its area.
A quarter circle has radius 50 cm. Find the length of its curved edge, then use the wedge argument to find its area.
A flat washer is the region between two circles with the same center, of radii 12 cm and 15 cm. Cut it into thin rings and straighten them. What shape do the strips stack into, and what is its area? Check by subtracting two circle areas.

Reasoning

Explain why the bumpy shape made from the wedges has exactly the same area as the circle, for any number of wedges.
Ulf cuts a circle into 7 equal wedges, an odd number, so they cannot alternate evenly. What shape do they make when interlocked as well as possible, and does the argument still give A = \tfrac12 C r?
The ends of the interlocked shape lean, so its height is less than r. Does it follow that the circle’s area is less than \pi r \times r? Explain.
With a million wedges, is the interlocked shape a rectangle? Explain how the argument reaches an exact answer anyway.
Show that the near rectangle and the ring triangle give the same area for every circle, and explain why they had to.

Answers

Each crust is \pi \approx 3.14 cm and the bottom edge is 14\pi \approx 44.0 cm. The circumference is 2\pi(14) = 28\pi, shared by 28 crusts. Half the wedges point up, so 14 crusts lie along the bottom.
Length 16\pi \approx 50.3 m, height 16 m, area 256\pi \approx 804.2 m². The length is half the circumference, \pi r, and the height is r.
Length 11\pi \approx 34.6 cm, height 11 cm, area 121\pi \approx 380.1 cm². The radius is half the diameter, 11 cm.
The length closes in on 23\pi \approx 72.3 cm, and the area is 529\pi \approx 1661.9 cm². The height is the radius, so r = 23, and the length is \pi r.
Radius about 17.0 m, area about 907.8 m². From \pi r = 53.4, r = 53.4 \div \pi \approx 17.0, and the area is about 53.4 \times 17 = 907.8.
Base 8.4\pi \approx 26.4 cm, height 4.2 cm, area 17.64\pi \approx 55.4 cm². The base is the circumference 2\pi(4.2), and \tfrac12 \times 8.4\pi \times 4.2 = 17.64\pi.
About 260.3 cm², and \pi r^2 \approx 260.2 cm². \tfrac12 \times 57.2 \times 9.1 = 260.26, while \pi(9.1)^2 \approx 260.16. The two agree to within the accuracy of the measurements.
10\pi \approx 31.4, 30\pi \approx 94.2 and 50\pi \approx 157.1 cm; total area 900\pi \approx 2827.4 cm², exactly \pi(30)^2. The halfway radii are 5, 15 and 25 cm, and twice each times \pi gives the lengths. Each strip is 10 cm tall, so the total is (10 + 30 + 50)\pi \times 10 = 900\pi.
Length 13\pi \approx 40.8 cm, height 26 cm, area 338\pi \approx 1061.9 cm². The curved edge is half the circumference, 26\pi, and half of that lies along the bottom. The answer is half of the full circle’s 676\pi.
153 cm². The near rectangle is \tfrac12 \times 18 = 9 cm long and 17 cm tall, and 9 \times 17 = 153.
Curved edge 25\pi \approx 78.54 cm, area 625\pi \approx 1963.5 cm². The curved edge is a quarter of 2\pi(50) = 100\pi. The near rectangle is 12.5\pi by 50, and \tfrac14 \pi (50)^2 = 625\pi agrees.
A trapezoid with parallel sides 24\pi and 30\pi and height 3; area 81\pi \approx 254.5 cm². The inner edge straightens to 2\pi(12), the outer to 2\pi(15), and \tfrac12(24\pi + 30\pi)(3) = 81\pi. Check: 225\pi - 144\pi = 81\pi.
The wedges are only moved: none is stretched, shrunk or cut again, none overlaps another, and no gap is left, since neighbors meet along a whole radius. Area adds over pieces, so the total is the circle’s area, however many wedges there are.
A near trapezoid, and yes. With 4 wedges pointing up and 3 pointing down, the bottom holds \tfrac47 C of crust and the top \tfrac37 C, and both ends lean inwards. As the wedges thin the height closes in on r, and a trapezoid’s area is the average of its parallel sides times its height: \tfrac12\left(\tfrac47 C + \tfrac37 C\right) r = \tfrac12 C r. (Cutting one wedge in half and putting a half at each end also gives the usual near rectangle.)
No. Length times height gives the area of a straight-sided shape, and the interlocked shape is not one. The straight-sided parallelogram is shorter than \pi r and less than r tall, so its area is less than \pi r^2, but it leaves out the bumps above and below it. The bumpy shape, bumps included, has exactly the circle’s area for every number of wedges.
No. A million wedges still leave tiny bumps and a tiny lean. The exact answer comes from the limit: the straight-sided shapes have areas that close in on \pi r^2 and also on the circle’s area A, so A and \pi r^2 differ by less than every positive amount, which forces the difference to be 0.
The rectangle gives \pi r \times r = \pi r^2 and the triangle gives \tfrac12 \times 2\pi r \times r = \pi r^2, the same for every r. They had to agree because both are the same circle cut into pieces and rearranged, so both have the circle’s area.