Circles

Learning goals

Name the radius, diameter and center, and use $d = 2r$
Define $\pi$ as circumference over diameter, the same for every circle
Find a circumference with $C = 2\pi r$
Derive $A = \pi r^2$ by rearranging thin sectors into a parallelogram
Separate a semicircle's area from its perimeter, which adds the diameter
Split a composite figure into circles and polygons, then add

The parts of a circle

Every measurement of a circle is built from a few named parts, so it is worth learning the vocabulary first.

The center is the fixed point that every point on the circle is equidistant from. A circle is often named by its center, like circle $O$ .
A radius is a straight segment from the center to any point on the circle. Because every point on the circle is the same distance from the center, every radius of a circle has the same length. We call that length $r$ .
A diameter is a straight segment that passes through the center and joins two points on the circle. It is a radius extended straight across to the far side, so a diameter is made of two radii laid end to end. Its length is written $d$ .
A chord is any straight segment joining two points on the circle. A diameter is the special, longest chord, the one that happens to pass through the center; a chord in general does not.
An arc is a piece of the circle itself, a curved section of the boundary between two points.
The circumference is the distance all the way around the circle. It is the circle’s version of a perimeter: the total length of its curved boundary.

A circle with center O. The radius r runs from the center to the edge; the diameter d crosses through the center; the dashed segment is a chord joining two edge points without passing through the center.

The most useful relationship here is the simplest one. Since a diameter is exactly two radii in a straight line, the diameter is always twice the radius:

d = 2r \qquad \text{and equivalently} \qquad r = \frac{d}{2}.

Keep this conversion ready, because some facts about circles are stated with the radius and others with the diameter. Slipping between the two is the source of most circle mistakes.

What pi really is

Here is the question that unlocks everything. Take any circle and walk all the way around it: that distance is the circumference $C$ . Now measure straight across it through the center: that is the diameter $d$ . How do those two lengths compare?

The answer is the most important fact in this lesson. For every circle, no matter how big or small, the circumference is the same number of times the diameter. Roll a coin and a dinner plate each through one full turn. In both cases the distance covered is just over three times the width you rolled across. That fixed ratio of circumference to diameter is the number called pi, written with the Greek letter $\pi$ :

\pi = \frac{C}{d}.

The value of $\pi$ is about $3.14159$ , a little more than $3$ . It is not a fraction of two whole numbers, and its decimals never end and never repeat. So in practice we use a rounded value: $\pi \approx 3.14$ , or sometimes the fraction $\tfrac{22}{7}$ , which is close. Because we round $\pi$ , any number we compute from it is approximate. That is why circle answers are usually written with the " $\approx$ " sign rather than " $=$ ".

Unrolling a circle's boundary into a straight bar shows the diameter laid off along it. The diameter fits three whole times plus a short leftover, about 0.14 of a diameter, so the circumference is about 3.14 diameters. That ratio is pi.

Why the ratio is the same for every circle

It would be a strange coincidence if a coin and a planet happened to share the same ratio by luck. They do not: the ratio is forced to be constant, because all circles have the same shape. A small circle is just a scaled-down copy of a large one, and scaling is the key.

Why $C \div d$ is the same constant for every circle#

Take any circle and imagine enlarging it by some factor $k$ , the way a photocopier blows up an image. Every length in the figure is multiplied by the same $k$ . The radius becomes $k$ times as long, and the diameter becomes $k$ times as long. The whole boundary is scaled uniformly as well, so the circumference also becomes $k$ times as long. Scaling stretches every single part of the picture by the same amount. So scaling cannot stretch the boundary by a different factor than it stretches the width across.

Now look at the ratio of circumference to diameter for the enlarged circle. The circumference is $k \times C$ and the diameter is $k \times d$ , so their ratio is

\frac{k \times C}{k \times d} = \frac{C}{d}.

The factor $k$ cancels top and bottom. The enlarged circle has exactly the same circumference-to-diameter ratio as the one you started with. Since every circle is a scaled copy of every other circle, they must all share that one ratio. That single shared number is what we name $\pi$ .

So $\pi$ is not a property of one particular circle; it is a property of circles as a shape. That is exactly why a formula written with $\pi$ works for every circle at once.

Circumference: the distance around

The definition of $\pi$ does the work for us. Starting from $\pi = \dfrac{C}{d}$ and multiplying both sides by $d$ gives the circumference directly:

C = \pi \, d.

The distance around a circle is $\pi$ times its diameter. And since the diameter is twice the radius ( $d = 2r$ ), we can write the same formula with the radius instead:

C = \pi \, d = \pi (2r) = 2 \pi r.

Both forms say the same thing; use whichever matches what you are given. If you know the diameter, use $C = \pi d$ . If you know the radius, use $C = 2\pi r$ .

Worked example 1 Circumference from a radius

A circular pond has a radius of $7$ m. How far is it around the pond? Use $\pi \approx 3.14$ .

The radius is given, so use $C = 2\pi r$ :

C = 2 \times 3.14 \times 7.

Multiply left to right. First $2 \times 3.14 = 6.28$ , then

C = 6.28 \times 7 = 43.96 \text{ m}.

So the distance around the pond is about $44$ m. The answer is approximate because we rounded $\pi$ to $3.14$ ; the true value is a little different.

Check your understanding

A circle has a diameter of $10$ cm. What is its circumference? Use $\pi \approx 3.14$ .

31.4 \text{ cm}

15.7 \text{ cm}

62.8 \text{ cm}

78.5 \text{ cm}

Area: the space inside

Circumference measured the boundary. Area measures the flat space the disk covers, counted in square units just as it was for polygons. The formula is

A = \pi r^2,

read as “pi r squared.” Notice it uses the radius, and the radius is squared, so this is genuinely a measure of square units, not a length. The surprise is that the same constant $\pi$ that governs the boundary also governs the area. We can see exactly why by cutting the disk into pieces and rebuilding it into a shape we already know how to measure.

A disk cut into 12 equal pie-slice sectors. Each sector has two straight edges of length r (the radius) and one curved edge along the circumference.

Slice the disk into many thin equal sectors, like the slices of a pie. Each slice is almost a thin triangle: two straight edges of length $r$ and a slightly curved base along the circle. Now pull the slices apart and lay them in a row, pointing up and down in turn, so they interlock into a long strip.

The same sectors rearranged, pointing up and down in turn, interlock into a strip shaped like a parallelogram. Its height is the radius r, and its base is half the circumference, pi times r.

That strip is nearly a parallelogram, and its dimensions come straight from the circle. The straight side of each sector is the radius $r$ . As the sectors get thinner those sides straighten out, so the perpendicular height of the strip gets closer and closer to $r$ . The strip’s base is made of the curved edges of the slices: half of them point up and form the top. The other half point down and form the bottom, so the bottom edge is exactly half the circumference. The rearranged strip always has exactly the area of the disk. That is because cutting a shape up and moving the pieces never changes how much area it covers. Only the shape changes, sharpening toward a true parallelogram as the slices get thinner.

Why the area of a circle is $\pi r^2$ #

Cut the disk into a large number of equal sectors. Lay them in a row, alternating point-up and point-down. Laid out that way they fit together into a strip. Moving the pieces around never changes how much area they cover in total. So this strip has exactly the area of the disk, no matter how many sectors you use. What does change is the strip’s shape. The more sectors you cut, the thinner each one is, and the closer the strip comes to a true parallelogram. We work out the area of that limiting parallelogram, and since the strip always had the disk’s area, that is the disk’s area too.

The height of the strip is the straight edge of a sector. As the sectors get thinner, those straight edges stand up straighter, so the perpendicular height of the strip approaches the radius $r$ .

Its base is built from the curved edges of the sectors. Half of the sectors point upward and lay their curved edges along the top. The other half point downward and lay their curved edges along the bottom instead. The full circumference is $2\pi r$ , and exactly half of it forms the bottom edge. As the sectors get thinner, each curved edge flattens toward a straight segment, so the bumpy base approaches the straight length

\text{base} = \frac{1}{2}(2\pi r) = \pi r.

In the limit the strip is a parallelogram of base $\pi r$ and perpendicular height $r$ . A parallelogram’s area is base times height, so the area of the disk is

A = \text{base} \times \text{height} = (\pi r)(r) = \pi r^2.

The base contributes one factor of $r$ through $\pi r$ , and the height contributes another factor of $r$ , which is why the radius ends up squared.

So for a circle,

A = \pi r^2 \qquad (r \text{ is the radius}).

If a problem hands you the diameter instead, do not put it where the radius belongs. Halve it first to get the radius, then square. This is the single most common circle error, and the next section comes back to it.

Worked example 2 Area from a radius

A circular tabletop has a radius of $5$ cm. Find its area. Use $\pi \approx 3.14$ .

Use $A = \pi r^2$ . Square the radius first, then multiply by $\pi$ :

r^2 = 5^2 = 25.

Now multiply by $\pi$ :

A = 3.14 \times 25 = 78.5 \text{ cm}^2.

The tabletop covers about $78.5$ square centimetres. The unit is square centimetres because area counts unit squares, and the radius was squared in the formula.

Worked example 3 Area from a diameter (halve first)

A circular plate has a diameter of $12$ cm. Find its area. Use $\pi \approx 3.14$ .

The formula needs the radius, but we were given the diameter, so halve it first:

r = \frac{d}{2} = \frac{12}{2} = 6 \text{ cm}.

Now square the radius and multiply by $\pi$ :

A = \pi r^2 = 3.14 \times 6^2 = 3.14 \times 36 = 113.04 \text{ cm}^2.

So the plate covers about $113 \text{ cm}^2$ . A frequent mistake is to square the diameter by accident, $3.14 \times 12^2 = 452.16$ . That answer is four times too large, because the diameter is twice the radius and squaring doubles that error into a factor of four.

Check your understanding

A circle has a radius of $10$ m. What is its area? Use $\pi \approx 3.14$ .

314 \text{ m}^2

62.8 \text{ m}^2

31.4 \text{ m}^2

100 \text{ m}^2

Semicircles

A semicircle is half a circle, cut along a diameter. Its area is simply half the area of the full circle:

A_{\text{semicircle}} = \frac{1}{2}\pi r^2.

Its boundary needs more care. The curved part is half the circumference, $\tfrac{1}{2}(2\pi r) = \pi r$ , but the boundary of a semicircle is not only the curve. To close the shape you also travel back along the straight diameter, which has length $d = 2r$ . So the full distance around a semicircle is the curved half plus the straight diameter:

P_{\text{semicircle}} = \pi r + 2r.

Forgetting the straight edge is a classic slip: a semicircle’s perimeter is not just half the circle’s circumference. That is because cutting a circle in half adds a new straight side that was not there before.

Worked example 4 Perimeter of a semicircle

A semicircle has a radius of $4$ cm. Find the distance around it. Use $\pi \approx 3.14$ .

The boundary is the curved half plus the straight diameter. The curved half is $\pi r$ :

\text{curve} = 3.14 \times 4 = 12.56 \text{ cm}.

The straight edge is the diameter $d = 2r = 8$ cm. Add the two parts:

P = 12.56 + 8 = 20.56 \text{ cm}.

So the distance around the semicircle is about $20.56$ cm. Leaving out the straight diameter would give only $12.56$ cm, which traces just the curved part and never closes the shape.

Composite figures

Real shapes often combine a circle or semicircle with the polygons from the previous lesson. The method is the same one you used for composite polygons. Break the figure into parts you know, find each part’s area, and add them (or subtract a part that has been cut away). Area is additive, so the whole is the sum of its non-overlapping pieces.

For example, take a running track shaped like a rectangle with a semicircle capping each end. That track’s area is equal to the rectangle plus two semicircular ends. Two semicircles of the same radius together make one full circle, so the total area is the rectangle’s area plus a single circle’s area. Keep the two kinds of measurement straight throughout. Lengths and circumferences are in linear units, while areas are in square units, exactly as in the perimeter and area lesson.

Worked example 5 A rectangle with a semicircle on top

A window is a $4$ m wide rectangle standing $5$ m tall, topped by a semicircle whose diameter is the $4$ m width. Find the total area of the window. Use $\pi \approx 3.14$ .

Split the window into the rectangle and the semicircle, find each area, then add.

The rectangle is $4$ m wide and $5$ m tall:

A_{\text{rect}} = 4 \times 5 = 20 \text{ m}^2.

The semicircle sits on the $4$ m width, so that width is its diameter, and the radius is half of it:

r = \frac{4}{2} = 2 \text{ m}.

A semicircle is half a circle, so its area is

A_{\text{semi}} = \frac{1}{2}\pi r^2 = \frac{1}{2} \times 3.14 \times 2^2 = \frac{1}{2} \times 3.14 \times 4 = 6.28 \text{ m}^2.

Add the two non-overlapping pieces:

A = 20 + 6.28 = 26.28 \text{ m}^2.

So the window covers about $26.28$ square metres. The radius for the semicircle was the half-width $2$ m, not the full $4$ m width, which is the same halve-the-diameter step from before.

Common mistakes

Confusing radius and diameter. The diameter is twice the radius ( $d = 2r$ ). Both area and the $2\pi r$ form of circumference use the radius, so when a problem gives the diameter, halve it before substituting.
Squaring the diameter in the area formula. $A = \pi r^2$ squares the radius. Using the diameter by mistake makes the answer four times too big. That is because squaring turns the factor of $2$ between $d$ and $r$ into a factor of $4$ .
Mixing up circumference and area. Circumference $C = 2\pi r$ is a length, in linear units like cm. Area $A = \pi r^2$ is a count of unit squares, in square units like $\text{cm}^2$ . One has $r$ to the first power, the other has $r$ squared.
Treating a rounded $\pi$ as exact. Since $\pi \approx 3.14$ is rounded, any answer built from it is approximate. Write it with " $\approx$ ", and do not round $\pi$ partway and then claim an exact result.
Forgetting the straight edge of a semicircle. A semicircle’s perimeter is the curved half plus the diameter ( $\pi r + 2r$ ), because slicing a circle in half exposes a new straight side.

Key takeaways

A circle is all points a fixed distance from a center. A radius $r$ reaches from the center to the edge; a diameter $d$ crosses through the center, and $d = 2r$ .
Pi is the ratio of any circle’s circumference to its diameter, $\pi = \dfrac{C}{d} \approx 3.14$ . It is the same constant for every circle because all circles are scaled copies of one another.
Circumference: $C = \pi d = 2\pi r$ , straight from the definition of $\pi$ . It is a length.
Area: $A = \pi r^2$ , derived by slicing the disk into thin sectors that rearrange into a parallelogram with base $\pi r$ and height $r$ . It is a count of square units.
A semicircle has area $\tfrac{1}{2}\pi r^2$ and perimeter $\pi r + 2r$ (the curve plus the diameter).
For a composite figure, split it into circles, semicircles, and polygons, find each area, and add (or subtract a removed piece).