Circumference and Area of a Circle | Grade 7 Math | LevelKite

A circle raises two different measuring questions. How far is it around? The distance once around the circle's curved boundary is its circumference. How much flat surface does it enclose? That amount is its area. Think of a round table: a ribbon wrapped exactly once around its edge is as long as the circumference, and the top you could paint is the area.

Both answers start from a radius or a diameter that you are told. Here are those parts in words:

The centre is the point that is the same distance from every point on the boundary.
A radius is a line segment from the centre to the boundary. Its length is $r$ .
A diameter is a line segment from boundary to boundary that passes through the centre. Its length is $d$ .

A diameter is made of two radii (the plural of radius) end to end, so $d = 2r$ . Both are lengths, so both are positive, and both are measured in the same unit of length, called a linear unit, such as centimetres.

Let $C$ stand for the circumference, measured in the same linear unit as $d$ . Every circle is a scaled copy of every other circle: the same shape, only enlarged or shrunk. Enlarging or shrinking multiplies every length by the same positive number, so circumference and diameter grow together. Divide the circumference of any circle by its diameter and you get the same number. That number is pi, written $\pi$ . It is a little more than $3$ . In symbols, $\pi = \frac{C}{d}$ .

So circumference is proportional to diameter: it is always the same fixed number of diameters long. That fixed number, $\pi$ , is called the constant of proportionality. To turn this into a rule for $C$ , multiply both sides of $\pi = \frac{C}{d}$ by $d$ . A diameter is a positive length, so $d$ is never zero: dividing by it is allowed, and multiplying both sides by it keeps them equal. The result is $C = \pi d$ . Because $d = 2r$ , you can replace $d$ by $2r$ and also write $C = \pi \times 2r = 2\pi r$ .

\pi = \frac{C}{d} \quad\Longrightarrow\quad C = \pi d = 2\pi r \qquad (d = 2r > 0)

Worked example

Circumference from a radius

A circle has a radius of $4$ cm. Find its circumference. Use $\pi \approx 3.14$ .

You are told the radius, 4 cm, which is a positive length. The circumference rule uses the diameter. A diameter is two radii end to end, so double the radius.
$d = 2r = 2 \times 4 = 8 \text{ cm}$
Circumference is pi times the diameter, because pi is the circumference divided by the diameter. Put in the stated value 3.14 for pi and 8 for the diameter.
$C = \pi d \approx 3.14 \times 8$
Split 3.14 into 3 and 0.14 and multiply each part by 8, because multiplying a sum gives the sum of the products: 3 times 8 is 24, and 0.14 times 8 is 1.12. Add the two parts.
$3.14 \times 8 = 3 \times 8 + 0.14 \times 8 = 24 + 1.12 = 25.12$
Attach the unit. A circumference is a length, so it takes the same linear unit as the radius, centimetres. The answer is approximate because 3.14 is a rounded value of pi.
$C \approx 25.12 \text{ cm}$

Answer

The circumference of the circle is approximately $25.12$ cm.

Now for the surface inside. The area of a circle, written $A$ , is the amount of flat surface that the circle encloses. You count area in square units: it tells you how many unit squares, such as squares $1$ cm on each side, it takes to cover that surface, counting parts of squares too. So if $r$ is in centimetres, $A$ is in square centimetres, written $\text{cm}^2$ . A circumference cannot do this job. It is a length, so it only tells you how far it is around.

Here is an informal way to see where the area rule comes from. Imagine the inside of a circle of radius $r$ made of a great many thin rings, one inside the next, like the rings in a slice of onion. Cut every ring along one radius and straighten it into a thin strip. Each ring is itself a thin circle, so its length is $2\pi$ times its own radius. Strip length is therefore proportional to distance from the centre: the strips grow steadily from almost nothing at the centre to $2\pi r$ at the boundary.

Stack the strips in order of length. The thinner the rings, the more closely the stack fills a triangle. Its base is the longest strip, which is the whole circumference, $2\pi r$ . Its height, measured perpendicular (at a right angle) to the base, is the total thickness of all the rings, which is $r$ . A triangle's area is half its base times its perpendicular height. Half of $2\pi r$ is $\pi r$ , and multiplying by the height $r$ gives the area of the circle. Here $r^2$ , read as "r squared", means $r \times r$ :

A = \frac{1}{2} \times 2\pi r \times r = \pi r \times r = \pi r^2 \qquad (r > 0)

Worked example

Area from a radius

A circle has a radius of $7$ m. Find its area. Use $\pi \approx \frac{22}{7}$ .

You are told the radius, 7 m, a positive length. The area rule is pi times the radius squared, so square the radius first. Squaring means multiplying the radius by itself, and metres times metres gives square metres.
$r^2 = 7 \times 7 = 49 \text{ m}^2$
Multiply the squared radius by pi, using the stated value 22/7.
$A = \pi r^2 \approx \frac{22}{7} \times 49$
Multiplying by 22/7 means multiplying by 22 and dividing by 7, and you may divide first. 49 divided by 7 is 7, so the product becomes 22 times 7. In parts, 20 times 7 is 140 and 2 times 7 is 14.
$\frac{22}{7} \times 49 = 22 \times 7 = 140 + 14 = 154$
Attach the unit. Area is counted in square units, and the radius is in metres, so the area is in square metres. It is approximate because 22/7 is a close value of pi, not an exact one.
$A \approx 154 \text{ m}^2$

Answer

The area of the circle is approximately $154 \text{ m}^2$ (154 square metres).

Key concept

Two rules for two different measures

For a circle with a stated radius $r$ and diameter $d$ , where $d = 2r$ and both are positive lengths in one unit:

The circumference $C$ is the distance once around the circle. It is $\pi$ times the diameter, and it is a length, in linear units such as cm.
The area $A$ is the flat surface the circle encloses. It is $\pi$ times the radius squared, and it is counted in square units such as $\text{cm}^2$ .

Use $\pi$ in the form the question states: keep the symbol for an exact answer, or use the given value and write $\approx$ .

C = \pi d = 2\pi r \qquad A = \pi r^2

One circle with radius 5 cm, measured two ways (using 3.14 for pi)

Measure	What it tells you	What the rule uses	Working	Result
Circumference	Distance once around the circle	Diameter: $d = 2 \times 5 = 10$ cm	$C = \pi d \approx 3.14 \times 10 = 31.4$	About $31.4$ cm, in linear units
Area	Flat surface the circle encloses	Radius squared: $r^2 = 5 \times 5 = 25 \text{ cm}^2$	$A = \pi r^2 \approx 3.14 \times 25 = 75 + 3.5 = 78.5$	About $78.5 \text{ cm}^2$ , in square units

Worked example

Circumference and area from a diameter

A circular flower bed has a diameter of $12$ m. Edging goes exactly once around its boundary, and mulch covers its whole surface. Find the length of the edging and the area that the mulch covers. Leave both answers in terms of $\pi$ .

The edging goes once around the boundary, so its length is the circumference. You are told the diameter, 12 m, a positive length, so multiply pi by it directly. Keep the symbol for pi, as the question asks.
$C = \pi d = \pi \times 12 = 12\pi \text{ m}$
The mulch covers the surface, so you need the area. The area rule uses the radius, not the diameter. The diameter is twice the radius, so the radius is half the diameter.
$r = 12 \div 2 = 6 \text{ m}$
Square the radius by multiplying it by itself. Metres times metres gives square metres.
$r^2 = 6 \times 6 = 36 \text{ m}^2$
Multiply by pi, keeping the symbol again. Both answers are exact because pi was never replaced by a rounded value.
$A = \pi r^2 = \pi \times 36 = 36\pi \text{ m}^2$

Answer

The edging is exactly $12\pi$ m long, and the mulch covers exactly $36\pi \text{ m}^2$ .

The takeaway. A circle has two measures, and each has its own rule. The circumference is the distance around it, a length: $C = \pi d$ , where $\pi$ is the circumference divided by the diameter. The area is the surface inside it, in square units: $A = \pi r^2$ . Start from the positive radius or diameter you are told, use $d = 2r$ to move between them, keep one unit throughout, and write $\pi$ in the form the question states.

Next, you will work backwards from a circumference or an area to find the radius or diameter, and you will find the area and perimeter of semicircles and quarter circles.