Circles: Area, Circumference, and Finding Errors | Grade 8 Math | LevelKite

You can already find the perimeter and area of shapes with straight sides. A circle has no sides to add up, so it needs its own formulas.

A circle is the set of all points on a flat surface that are the same distance from one point, called the center. Four measurements describe it:

The radius, $r$ , is the distance from the center to any point on the circle.
The diameter, $d$ , is the distance across the circle through the center. It is two radius lengths end to end, so $d = 2r$ .
The circumference, $C$ , is the distance around the circle. It is the circle's perimeter.
The area, $A$ , is the amount of flat surface inside the circle.

Both $r$ and $d$ are distances on a real circle, so both are positive: $r > 0$ and $d > 0$ . In each problem in this lesson, $r$ , $d$ and $C$ are measured in one linear unit, such as centimeters, and $A$ is measured in the square of that same unit, such as square centimeters.

Start with the distance around. For every circle, large or small, the circumference divided by the diameter, both in the same unit, is the same number. That number is called pi and is written $\pi$ .

The value of $\pi$ is a little more than $3$ , so the distance around any circle is a little more than $3$ times the distance across it. The number $\pi$ is irrational: its decimal form, $3.14159\ldots$ , never ends and never repeats.

This fact gives you a formula. Write it as $\frac{C}{d} = \pi$ . Because $d > 0$ , you can multiply both sides by $d$ and the two sides stay equal. Then, because $d = 2r$ , you can replace $d$ with $2r$ :

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

Now the surface inside. Think of cutting a circle into many thin wedges, like slices of pizza, and laying them side by side with their points alternating up and down. The thinner the wedges, the closer the arrangement comes to a rectangle:

Its height, measured perpendicular to its base (at a right angle to it), is the radius $r$ , because each wedge reaches from the center to the edge.
Its base is half of the circumference, because half of the curved edges lie along the bottom and half along the top. Half of $2\pi r$ is $\pi r$ .

Rearranging the wedges does not change how much surface they cover, so the circle's area matches the rectangle's area, base times height: $\pi r \cdot r = \pi r^2$ . Here $r^2$ means $r \times r$ . Multiplying a length by a length is why area is measured in square units.

Key concept

One radius gives both measurements

For any circle with radius $r > 0$ and diameter $d = 2r$ , both measured in one linear unit:

The circumference is the distance around: $C = 2\pi r$ , which equals $\pi d$ . Its unit is that linear unit.
The area is the surface inside: $A = \pi r^2$ , where $r^2 = r \times r$ . Its unit is the square of that linear unit.

When you are given the diameter, find the radius with $r = \frac{d}{2}$ before you use a formula that needs $r$ .

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

Worked example

Starting from the radius

A circle has a radius of $5$ cm. Find its circumference and its area. Give each answer exactly and as an approximation, using $3.14$ for $\pi$ .

Name what is stated. The length given is the radius. It is positive and it is in centimeters, so you can use it in both formulas as it is.
$r = 5 \text{ cm}$
Find the circumference, the distance around. Replace r with 5 in C = 2πr. Factors can be multiplied in any order, so multiply 2 by 5 and keep π as a symbol. This is the exact value, in the linear unit.
$C = 2\pi r = 2 \cdot \pi \cdot 5 = 10\pi \text{ cm}$
Approximate the circumference. Replacing π with the rounded value 3.14 changes the value slightly, so the sign becomes ≈ at that step.
$C = 10\pi \approx 10 \times 3.14 = 31.4 \text{ cm}$
Find the area, the surface inside. Replace r with 5 in A = πr². Square before you multiply by π: 5² means 5 × 5, which is 25. A length times a length gives the square unit.
$A = \pi r^2 = \pi \cdot 5^2 = \pi \cdot 25 = 25\pi \text{ cm}^2$
Approximate the area. To multiply 25 by 3.14, split 3.14 into 3 and 0.14: 25 × 3 = 75 and 25 × 0.14 = 3.5. Then add the two parts.
$A = 25\pi \approx 25 \times 3.14 = 75 + 3.5 = 78.5 \text{ cm}^2$

Answer

The circumference is exactly $10\pi$ cm, which is about $31.4$ cm. The area is exactly $25\pi \text{ cm}^2$ , which is about $78.5 \text{ cm}^2$ .

Worked example

Starting from the diameter

A circular flower bed has a diameter of $6$ m. You will put edging around its border and spread mulch over its whole surface. How many meters of edging do you need, and how many square meters will the mulch cover? Use $3.14$ for $\pi$ .

Match each question to a measurement. Edging runs around the border, so it is the circumference, a length in meters. Mulch covers the surface inside, so it is the area, in square meters.
Find the radius, because the area formula needs it. You are given the diameter, and d = 2r. Dividing both sides by 2, which is not zero, keeps the two sides equal.
$r = \frac{d}{2} = \frac{6}{2} = 3 \text{ m}$
Find the circumference. Use C = πd with the stated diameter. Using C = 2πr with r = 3 gives the same product, because 2 × 3 = 6. Then replace π with 3.14: 6 × 3 = 18 and 6 × 0.14 = 0.84.
$C = \pi d = \pi \cdot 6 = 6\pi \approx 6 \times 3.14 = 18.84 \text{ m}$
Find the area with the radius, not the diameter. Square first: 3² = 3 × 3 = 9. For the approximation, 9 × 3 = 27 and 9 × 0.14 = 1.26, and 27 + 1.26 = 28.26.
$A = \pi r^2 = \pi \cdot 3^2 = \pi \cdot 9 = 9\pi \approx 9 \times 3.14 = 28.26 \text{ m}^2$

Answer

You need about $18.84$ m of edging (exactly $6\pi$ m), and the mulch will cover about $28.26 \text{ m}^2$ (exactly $9\pi \text{ m}^2$ ).

Now use these ideas to check work. Error analysis means reading a finished calculation step by step, finding the first step that is not valid, explaining why, and then redoing the work correctly from that step on. Keep reading after you find one error, because one error does not rule out a second.

A quick estimate tells you when to be suspicious. Since $\pi$ is a little more than $3$ , every circle with $r > 0$ has a circumference a little more than $3d$ and an area a little more than $3r^2$ . A result that is below its estimate, or far above it, signals an error somewhere.

The table lists five checks, each with a mistaken calculation, the reason it is not valid, and the correction. The first four are about circles. The last one comes from shapes you already know: a triangle's height is measured perpendicular to its base, so a side that is not perpendicular to the base is not the height.

Five checks for area and perimeter work. Every length is positive and each problem uses one linear unit.

Check	Mistaken work	Why it is not valid	Corrected work
Radius or diameter?	Diameter $10$ cm: $C = 2\pi \cdot 10 = 20\pi$ cm	$C = 2\pi r$ needs the radius, and $r = \frac{10}{2} = 5$ cm.	$C = 2\pi \cdot 5 = 10\pi$ cm
Around or inside?	Area for radius $3$ m: $A = 2\pi \cdot 3 = 6\pi \text{ m}^2$	$2\pi r$ gives the distance around, not the surface inside.	$A = \pi \cdot 3^2 = 9\pi \text{ m}^2$
Linear or square unit?	Radius $7$ cm: $A = \pi \cdot 7^2 = 49\pi$ cm	Area multiplies a length by a length, so its unit is squared.	$A = 49\pi \text{ cm}^2$
Exact or approximate?	Radius $5$ cm: $C = 10\pi = 31.4$ cm	$3.14$ is a rounded value of $\pi$ , so $10\pi$ and $31.4$ are close but not equal.	$C = 10\pi \approx 31.4$ cm
Is the height perpendicular?	Triangle: base $10$ cm, slanted side $13$ cm, height $12$ cm. $A = \frac{1}{2}(10)(13) = 65 \text{ cm}^2$	The height must be perpendicular to the base. The slanted side is not.	$A = \frac{1}{2}(10)(12) = 60 \text{ cm}^2$

Worked example

Find and correct the errors

A circle has a diameter of $12$ cm. A student finds its area this way: $A = \pi \cdot 12^2 = 144\pi \approx 452.16$ cm. Find each error, explain why it is an error, and give the correct area.

Estimate first. You are given the diameter, and d = 2r. Dividing both sides by 2, which is not zero, gives the radius, 6 cm. The area should be a little more than 3 × 6² = 3 × 36 = 108 square centimeters. The student's 452.16 is about four times that, so expect an error.
$r = \frac{d}{2} = \frac{12}{2} = 6 \text{ cm}$
Read the student's work from the left. Area is the surface inside, so A = πr² is the right formula. But the first expression, π · 12², puts the diameter where the radius belongs. That is the first step that is not valid.
$r = 6 \text{ cm}, \text{ not } 12 \text{ cm}$
Redo the work from that step with the radius. Square first: 6² = 6 × 6 = 36. Keeping π as a symbol gives the exact value.
$A = \pi r^2 = \pi \cdot 6^2 = \pi \cdot 36 = 36\pi$
Approximate with 3.14 for π: 36 × 3 = 108 and 36 × 0.14 = 5.04. The result is a little more than the estimate of 108, as it should be.
$36\pi \approx 36 \times 3.14 = 108 + 5.04 = 113.04$
Keep checking the rest of the work. The student squared 12 and multiplied by 3.14 correctly, and used ≈ in the right place, but the unit is a second error. The student wrote cm, a linear unit. Area multiplies a length by a length, so its unit is the square centimeter.
$A = 36\pi \text{ cm}^2 \approx 113.04 \text{ cm}^2$
Explain the size of the first error. Using 12 in place of 6 doubles the number that gets squared, and squaring a doubled number multiplies the result by 2 × 2 = 4. So the student's 144π is exactly 4 times the correct 36π.
$12^2 = (2 \cdot 6)(2 \cdot 6) = (2 \cdot 2)(6 \cdot 6) = 4 \cdot 36 = 144$

Answer

The student used the diameter in place of the radius and wrote a linear unit for an area. The correct area is exactly $36\pi \text{ cm}^2$ , which is about $113.04 \text{ cm}^2$ .

Takeaway. One length, the radius $r > 0$ , gives you both circle measurements, and $d = 2r$ connects it to the diameter. The circumference, $C = 2\pi r = \pi d$ , is the distance around and takes the linear unit. The area, $A = \pi r^2$ , is the surface inside and takes the square of that unit. Keep $\pi$ for an exact answer, and write $\approx$ once you replace it with $3.14$ .

To check any calculation, read it step by step and ask which length is stated, whether the question is about around or inside, and whether the unit and the size make sense. Next, you will find the area of composite figures, which are shapes made by joining simpler shapes.