Area Contained in a Circle

This article presents multiple methods for finding the equation for the area contained within a circle. It is heavily inspired by integral calculus and is intended as a potential first encounter with the concept of a limit. It does not assume any previous knowledge of calculus though.

Introduction

Terminology

You will notice that I am taking great pains to say “area contained within a circle.” It would be simpler (and just as correct) to say “area of a disk.” The more common phrase “area of a circle” is incorrect. The circle refers only to the boundary, not the interior. When we say the area of a rectangle, that is fine because the word rectangle can refer to the whole shape.

Rectangles are defined as a special case of polygons, which have no agreed-upon definition. Three common definitions contradict each other: polygon refers only to the interior; polygon refers only to the boundary; polygon refers to both the interior and boundary. This is before dealing with the possibility of edges that cross and thus the possibility that only some interior parts are filled.

If we were to say “the area of the edges of a rectangle,” that would obviously be ridiculous. The edges of the rectangle are lines and thus infinitely thin. They have no area. A circle is the same. It is the edge surrounding a disk and is also made of an infinitely thin curve. It also cannot have an area.

I won’t be upset if you continue to use “area of a circle.” It is the most common way to refer to this concept. It is taught this way from as early as preschool (where rectangles, triangles, and circles are often treated as the same sort of thing even though they are often defined to be different creatures altogether). However, it is important that you know it is an incorrect shorthand. The distinction does make a difference in certain contexts, particularly if you want to distinguish between the interior (disk) and boundary (circle).

Finding the Area Contained Within a Circle: Your First Encounter

To start, I’d like to invite you to try to come up with a solution on your own. Take a few minutes and think about how you have found equations for the area contained within other two-dimensional shapes (squares, rectangles, parallelograms, triangles, etc.). Can you think of any way to reuse or extend one of those techniques? Can you create your own method? Please don’t work too hard. If you find yourself getting frustrated, take a break or move on to the next section. There are some nice pointers to get you started. Please stop often and reassess if you have new ideas about how to solve the problem. The fewer hints you use, the stronger your creative and problem-solving skills will become. Remember that there are lots of ways to solve these problems. If you invent one for yourself that is not covered here, that is awesome!

The Method of Boxes (Cartesian Integral)

Setup

Have a look at this app. Leave the hint buttons alone for now. You can use the full-screen button if you like. Try setting the white and red number dials. What do you think they represent? How might this help you discover the area contained within the circle? Please be patient when you have both dials set to numbers above 50. The app may take a moment to process the data for you, especially on an older or handheld device. At the settings of 100 and 100, it is handling 400 million boxes for you!

Digital Manipulative: Area Contained in a Circle (Method of Boxes)

Hint 0 (Introduction to the Tools)

If you have had time to explore the tools, perhaps you have discovered that the red number dial represents the length of the radius in arbitrary units (miles, feet, kilometers, meters, or whatever you choose). The white number dial represents the number of subdivisions in each unit. How might these grid patterns aid you in finding the area contained within a circle?

Hint 1

Feel free to activate the first (leftmost) hint button. What do the blue and green represent? How might this help you? What does this new white number represent?

Hint 2

The blue represents the boxes that are wholly within the circle and the green represents the boxes that are intersected by the circle. They are counted for you in the corresponding color. The new white number is the total number of boxes in the square surrounding the circle. Feel free to check a few cases (the numbers get big fast, so try to keep both dials set to only 1 or 2). How does this bring you closer to a solution? How should you proceed next?

Hint 3

There are several potential avenues to explore at once, so this hint will be split into sub-hints.

Sub-Hint A

Try setting the red and white dials to a variety of values. Which ones seem to do the best job approximating the area inside the circle?

Sub-Hint B

Are there any particular values you might set the dials to get useful information?

Sub-Hint C

Are there any particular patterns in the numbers of colored boxes? Any particular digits that appear? Perhaps for specific values set on the dial?

Sub-Hint D

Perhaps it would help to convert the number of boxes of each color to the equivalent number of square units?

Hint 4

There are several potential avenues to explore at once, so this hint will be split into sub-hints.

Sub-Hint A

Having a finer mesh seems to cause the colored boxes to match the circular boundary more accurately. Higher values of the red and white numbers might be useful.

Sub-Hint B

Every combination of 1, 10, and 100 seems to cause interesting things to happen.

Sub-Hint C

Again, setting the red and white numbers to 1, 10, and 100 in every combination might help. Do you see a famous progression of digits appearing as the dials are set to higher powers of 10?

Sub-Hint D

If you need a refresher on converting square units, check out LESSON FORTHCOMING. I recommend that you take the time to understand how to convert units. However, if you would like to proceed anyway, please use the second (middle) hint button to get the app to do the conversions for you. The new red number is the area of the large square surrounding the circle in the arbitrary square units you chose. The new blue number shows the area covered by the squares that are entirely inside the circle, measured in square units. The green number gives the area covered by the squares that the circle intersects, measured in square units.

Hint 5

How could you use these numbers to get an estimate for the area contained within a circle?

Hint 6

How does the blue area compare to the actual area within the circle? How can you use the green area?

Hint 7

The blue area is always less than the area contained in the circle, but as the mesh gets finer, the approximation gets better. There are always some green pieces missing. If you add the green number though, you will certainly add too much. Some of the green boxes are partially outside the circle. Thus, the blue plus green will always be bigger than the area in the circle, but as the mesh gets finer, the overestimate becomes less and less. How should you proceed?

Hint 8

You could choose to use just the blue number or the blue plus the green. Since both seem to be converging to the same result, you may also choose to take the sum of the blue and half of the green. It would always be sandwiched between the two other approaches. It also seems reasonable to guess that the green boxes are placed roughly randomly along the edge of the circle. For a large enough number of green boxes, you might expect them to appear halfway inside the circle on average. That makes taking half of the green seem like a good choice. This is especially obvious when you take many small boxes. As the boxes get small, the curvature of the circle becomes harder to notice (within the boxes). For small enough boxes, the part of the circle passing through them will look like a straight line. Boxes placed randomly across a line will (on average) be half on one side and half on the other.

Hint 9

Turn on the third (rightmost) hint. It adds the blue area and half of the green area. The sum is given in teal.

Hint 10

Again, setting the red and white numbers to 1, 10, and 100 in every combination might help. Do you see a famous progression of digits appearing as the dials are set to higher powers of 10? Look at the teal sum.

Hint 11

When setting the dials to powers of 10, you should see the digits of pi (3.14159…) starting to appear. More digits will be visible for higher powers of ten. They are the digits of pi, but there is something slightly changed (other than the fact that they stop being the digits of pi after the first few).

Hint 12

Why does the decimal place move around in the digits of pi? What could cause that? What does that have to do with changing the dials?

Hint 13

Try making a table of values for the red and white dial numbers along with the teal result (you could include blue numbers, green numbers, and blue plus green numbers if you like). Choosing powers of ten as your dial inputs again may be helpful.

Hint 14

Having the white dial set to 100 the whole time will help you get precise results. Leave it at 100 going forward.

Hint 15

Try red dial values of 1, 10, and 100. What happens to the decimal in the teal value? Does that help you make a conjecture?

Hint 16

Try red dial values of 1, 2, 3, 4, and 5. Does this match your conjecture (if you made one)?

Hint 17

If you do not have a working conjecture, try adding the red area number from hint button 2 to your table entries for Hints 14 and 15. Do you see a pattern in the red area number?

Hint 18

Try adding the square of the red dial number to your table. How does it compare to the red area number?

Hint 19

The area number seems to always be four times the square of the red dial number. Why is that? Use this to inspire you with the teal number and pi.

Hint 20

Compare the square of the red dial number to the teal number. How do they compare? How is pi involved? Can you make a conjecture as to why this is?

Hint 21

The teal number appears to be roughly equal to pi multiplied by the square of the red dial number. What equation does this suggest for the area contained within a circle?

Hint 22

The teal number is our approximation for the area contained within the circle and the red dial number is the radius of the circle, given in the arbitrary units you chose. What do you think the equation for the area contained within a circle might be?

Hint 23

The data you have seen suggests that a good guess for the equation of the area contained within a circle is pi multiplied by the radius squared or in symbols $A=\pi r^{2}$.

The Method of Segments (Polar Integral: Radius then Angle)

In the last section, you found a good guess for the equation that gives the area contained within a circle. In this section and the next one, you will confirm this guess using different approaches.

Setup

Have a look at the app. Feel free to change the number of segments with the dial and use the full-screen button if you like. Please refrain from using the interact button or the hint button for now though. The app can struggle when showing more than 64 segments (especially on old or handheld devices). Please be patient.

Can you think of a way to find the equation for the area contained within a circle using this approach?

Digital Manipulative: Area Contained in a Circle (Method of Segments)

Hint 1

Try hitting the interact button. What happens?

Hint 2

What happens when you change the number of segments?

Hint 3

What is preserved from the original circle? What is different? Can this be used to find the equation for the area contained in a circle?

Hint 4

Try setting the number of segments to 4, then use the interaction button to change the circle into a strange lumpy shape. Once the strange lumpy shape is created, try increasing the number of segments. What happens?

Hint 5

When you increased the number of segments in the strange lumpy shape, did it start to look like a more familiar simple shape? Which one? How might this help?

Hint 6

The lumpy top and bottom seem to flatten out into straight lines that are of equal length. They seem to be of different length than the sides though.

Hint 7

The corners appear to straighten into 90-degree angles.

Hint 8

The new shape appears to be a rectangle (or a parallelogram that eventually becomes a rectangle if you prefer).

Hint 9

How does the area of this new shape compare to the original circle? How might that help?

Hint 10

The areas are in fact the same. If you can find the area of the new shape, it is equal to the area of the circle. How do you find the area of the new shape?

Hint 11

To find the area of a rectangle, multiply its length by its width. What are its length and width?

Hint 12

Activate the hint button. What do the red, green, and blue represent?

Hint 13

Be sure to try varying the number of segments and alternating between circle and rectangle/lumpy shape to see what the colors mean.

Hint 14

How do the total lengths of these colored lines compare when you use the interact button or change the number of segments?

Hint 15

The total length of these colored lines never changes. How does that help you?

Hint 16

When the diagram is in circle form, the red line has the same length as what measurement of the circle (not a specific number, but the name of a parameter you might be interested in calculating or measuring in a circle)?

Hint 17

When the diagram is in circle form, the red line has the same length as the diameter of the circle. The red line is then split and placed on either end of the rectangle/lumpy shape. How long are each of these pieces (again, not a specific number, but the name of a parameter)?

Hint 18

The red ends of the rectangle/lumpy shape have the same length as the radius of the original circle. The diameter is split evenly (in half) and each part is put on the end.

Hint 19

What about the blue and green lines? How long are they? Do you need to do both separately? Why?

Hint 20

The blue and green lines are of the same length. This is because they are opposite sides of a rectangle. You can also argue this based on the fact that they both have lengths equal to half the distance around the circle.

Hint 21

You can choose to find the length of either the green or blue line. We will assume green here, but you can just as easily substitute blue. How long is that green line?

Hint 22

It is half the distance around the circle, so if you could find the distance that you would travel when traveling around the circle (the perimeter of the circle, also known as its circumference), you could just take half. How do you find the circumference of a circle?

Hint 23

If you don’t know what the equation for the circumference of a circle is, please see LESSON FORTHCOMING for details. It will help to bring you up to speed. Otherwise, the equation for circumference is two times pi times radius. In symbols that is $C=2\pi r$.

Hint 24

The length of the green line is half the circumference. How long is that?

Hint 25

If it takes two $\pi r$ to make it all the way around the circle, how many $\pi r$ does it take to make it halfway around?

Hint 26

The green line is of length $\pi r$. This can also be seen by realizing that you will always be dividing the circumference equation by 2 and that will exactly cancel the multiplication by two at the beginning of the circumference equation.

Hint 27

You have a rectangle with length equal to $\pi r$ and width equal to the radius of the original circle. Multiply the length by the width. If you are unsure how to handle this calculation (if algebra is something you learned recently), you may want to use the guess you made in The Box Method to help you.

Hint 28

$A=l\times w$, but you know what the length and width are (in terms of circle-related quantities), so you can replace them.

Hint 29

$A=(\pi r)\times r$. You could stop there, but multiplication can be done in any order and repeated multiplication can be represented with a power, so why not do that too.

Hint 30

Your final expression is $A=\pi r^{2}$.

The Method of Rings (Polar Integral: Angle then Radius)

Can you think of another way to find the area contained within a circle, perhaps one that has to do with rings?

Setup

Take a look at the app. Feel free to change the number of rings with the dial and use the full-screen button if you like. However, please refrain from using the interact or hint buttons for now. For displays that are small or low resolution, strange patterns may occur if you choose to display many thin rings. If you start to see strange patterns (like stretched four-leaf clovers), please lower the number of rings to avoid confusion.

Can you think of a way to find the equation for the area contained within a circle using this approach?

Digital Manipulative: Area Contained in a Circle (Method of Rings)

Hint 1

How might you manipulate this model into a more useful form?

Hint 2

Can you see a way to change this shape into another simple shape, one whose area you can easily calculate?

Hint 3

What if you split the rings and unfurled them? What sort of shape might that make?

Hint 4

Go ahead and hit the interact button. How does this help? Try changing the number of rings. Does this make what is happening clearer?

NOTE

There is a small imperfection in the animation. Can you spot it? It is subtle and doesn’t affect the result, but it may bother your geometric intuition. If you do spot it, how might the animation be improved? We will discuss it in full at the end of this section.

Hint 5

What is the new shape that appears after you use the interact button?

Hint 6

The new shape is a triangle. How does the triangle’s area compare to the original circle?

Hint 7

The areas are the same. How do you find the area of the triangle (and thus the circle)?

Hint 8

The equation for the area contained in a triangle is base times height divided by two, or in symbols $A={\frac {bh}{2}}$.

Hint 9

What are the base and height?

Hint 10

How can you get these values from the model?

Hint 11

How do these values relate back to the original circle?

Hint 12

Try using the hint button in the app. What does it show? How does this help you?

Hint 13

Try switching between circle and triangle using the interact button. What does the green line represent in each case? Does it change?

Hint 14

The green line represents the length of the radius in the circle and the height of the triangle. It doesn’t change. What does this mean?

Hint 15

The height of the triangle is the same as the length of the radius of the circle.

Hint 16

Now, what about the red line? What does it represent in the circle and in the triangle?

Hint 17

The red line represents the length around the outside of the circle (the perimeter, also known as the circumference). It is also the base of the triangle. What does this mean?

NOTE

For those with exceptional geometric intuition, you may have noticed an important imperfection here. The start and end positions of the animation work out exactly right, but during the animation, there is some distortion. See if you can spot the problem and find a way to improve it. We will discuss it shortly.

Hint 18

The base of the triangle has the same length as the circumference of the circle. What is the circumference of the circle?

Hint 19

As mentioned in "The Method of Segments," the circumference of a circle is found with the equation $C=2\pi r$. How can you use this?

Hint 20

You now have substitutions to make into the equation for the area of a triangle. Exchange the base and height for their circle-ish counterparts.

Hint 21

Substituting the equation for the circumference for the base and the radius for the height causes the area equation $A={\frac {bh}{2}}$ to change into $A={\frac {(2\pi r)\times r}{2}}$.

Hint 22

This equation can get the job done, but you will be using it a lot in the future. It would be helpful to simplify it so you don’t have to work so hard later. Can you see a way to simplify the equation?

Hint 23

Look at the equation you found in the past two sections. Does that give you inspiration?

Hint 24

Look back at how you simplified the equation in ‘The Method of Segments’ section. Does that give you any ideas?

Hint 25

You have several multiplications in the numerator. You can choose to do them in any order you want. Why not choose to multiply like this: $A={\frac {(2\pi )r\times r}{2}}$. Can you simplify further?

Hint 26

Multiplying a number by itself is just squaring. Make that simplification: $A={\frac {2\pi r^{2}}{2}}$. Can you simplify further?

Hint 27

There is one more simplification to make. It is similar to what was done in hint 25 of "The Method of Segments."

Hint 28

If you have two $\pi r^{2}$, how many $\pi r^{2}$ would be half that amount.

Hint 29

The multiplication by two then division by two cancel each other out, leaving $A=\pi r^{2}$. Which matches the previous methods.

Discussion About the Imperfection

If you haven’t taken the time to try to spot the imperfection, please try now. If you run the animation, it is perfect at the beginning and end, but in between, there is an imperfection. The imperfection is at its worst at about halfway through the animation. Do you see it?

The transformation used does preserve area, but it doesn’t preserve lengths. While all the lengths are returned to their proper values at the end of the animation, in between, the length of the circumference shrinks by about a quarter and the widths of the ribbons (at their ends) widen to compensate. I could have fixed this, but I thought it was an interesting discussion to have and gives students a chance to really test their geometric intuition.

How do you think a perfect animation would have looked?

A perfect animation would have had the ribbons not change widths or lengths. This would have caused there to be gaps between the ribbons while they were unfurling.

For those who are interested, the transformation used was $\theta ={\frac {2\pi }{t}}$ and also the horizontal direction was scaled up by $t$. $\theta$ was the amount of the circle which was present (as a function of angle). It ranged from the full $2\pi$ radians to only 1% that amount. $t$ was a scale factor that went between 1 and 100 in a nonlinear fashion.

The Method of Inscribed Regular Polygons (Your Turn Again)

Digital Manipulative: Area Contained in a Circle (Method of Polygons)

It’s your turn again. Have a look at this app. It inscribes regular polygons in a circle. Don’t use the hint buttons unless you really need to. If you do use the hint button, use the first (left) one and try to figure it out. If you still need help, try the second (right) hint button.

Can you find a way to confirm the equation for the area contained in a circle using this approach? Can you come up with more than one way? Would your approach change if you instead inscribed the circle in successive regular polygons? How would the diagram change in this case?

For Teachers and Parents

What is this Hint System

The hint system is designed to allow students to engage with the problems with little to no support (Discovery/Inquiry-Based Math). The hints gradually give more and more information until we end up with a full explanation (Traditional Lecture). Since Math is not a ‘spectator sport,’ it is hoped that repeated use of this hint system in multiple lessons will encourage students to use fewer and fewer hints and learn to solve problems creatively for themselves. This is, of course, only one possible arrangement of these hints. An instructor will be better able to adapt the presentation of the hints to the progress a specific student has made. This is beneficial to the student until they learn to use the system and self-regulate. If the student needs hints, they can choose to set aside their particular approach temporarily while filling in the gaps in the order the text presents them. Later they can analyze how the approach they took compares to that of the text.

Isn't this Basically Integral Calculus?

Yes! I am using this lesson as a Trojan Horse. Exposing students to these ideas early so they have horizon knowledge about what is coming down the track later is important. It gives their subconscious extra years to digest the ideas before they become everyday companions.

Why no Rigor?

Why no Rigor? Sorry, the treatment is not rigorous. To my Pure Math friends, sometimes rigor can get in the way of clarity when first encountering a concept. I promise a rigorous treatment of calculus, straight ‘outta’ Real Analysis, will come in future years.