Area and Circumference of a Circle — Formulas and Examples

Grade: 7-8 | Topic: Geometry

What You Will Learn

After this lesson you will be able to calculate the area and circumference of any circle when given its radius or diameter. You will understand the role of $\pi$, know when to use each formula, and confidently solve problems involving semicircles, quarter-circles, and real-world circular shapes.

Theory

Key vocabulary

Every circle problem starts with three measurements:

• Radius ($r$) — the distance from the center of the circle to any point on its edge.
• Diameter ($d$) — the distance across the circle through the center. It is always exactly twice the radius: $d = 2r$.
• Pi ($\pi$) — a mathematical constant approximately equal to $3.14159$. It is the ratio of any circle's circumference to its diameter.

Circumference — the distance around the circle

The circumference is the circle's "perimeter." There are two equivalent forms of the formula:

$C = 2\pi r \qquad \text{or} \qquad C = \pi d$

Use the first form when you know the radius. Use the second when you know the diameter. They give the same result because $d = 2r$.

For example, a circle with radius 5 cm has circumference:

$C = 2\pi(5) = 10\pi \approx 31.42 \text{ cm}$

Area — the space inside the circle

The area formula tells you how much flat surface the circle covers:

$A = \pi r^2$

This means: square the radius, then multiply by $\pi$. Always remember to square the radius first, not multiply $\pi$ by $r$ and then try to square.

For the same circle with $r = 5$ cm:

$A = \pi(5)^2 = 25\pi \approx 78.54 \text{ cm}^2$

Semicircles and quarter-circles

For a semicircle (half a circle):

$A_{\text{semi}} = \frac{1}{2}\pi r^2 \qquad C_{\text{semi}} = \pi r + 2r$

The circumference of a semicircle includes the curved part ($\pi r$) plus the straight diameter ($2r$).

For a quarter-circle:

$A_{\text{quarter}} = \frac{1}{4}\pi r^2 \qquad C_{\text{quarter}} = \frac{1}{2}\pi r + 2r$

The perimeter includes the curved arc ($\frac{1}{2}\pi r$) plus two radii ($2r$).

Worked Examples

Example 1: Finding circumference and area from the radius (easy)

Problem: A circle has a radius of 8 cm. Find its circumference and area. Use $\pi \approx 3.14$.

Step 1: Calculate the circumference. $C = 2\pi r = 2 \times 3.14 \times 8 = 50.24 \text{ cm}$

Step 2: Calculate the area. $A = \pi r^2 = 3.14 \times 8^2 = 3.14 \times 64 = 200.96 \text{ cm}^2$

Answer: Circumference = 50.24 cm, Area = 200.96 cm$^2$

Example 2: Given the diameter instead of the radius (easy)

Problem: A circular table has a diameter of 1.2 m. Find the area of the tabletop. Use $\pi \approx 3.14$.

Step 1: Find the radius. $r = \frac{d}{2} = \frac{1.2}{2} = 0.6 \text{ m}$

Step 2: Calculate the area. $A = \pi r^2 = 3.14 \times (0.6)^2 = 3.14 \times 0.36 = 1.1304 \text{ m}^2$

Answer: $\approx$ 1.13 m$^2$

Example 3: Finding the radius from the circumference (medium)

Problem: A bicycle wheel has a circumference of 2.2 m. What is its radius? Use $\pi \approx 3.14$.

Step 1: Start from the circumference formula and solve for $r$. $C = 2\pi r \implies r = \frac{C}{2\pi}$

Step 2: Substitute. $r = \frac{2.2}{2 \times 3.14} = \frac{2.2}{6.28} \approx 0.35 \text{ m}$

Answer: $\approx$ 0.35 m (or about 35 cm)

Example 4: Area of a semicircular window (medium)

Problem: A window is shaped as a semicircle with a diameter of 80 cm. Find the area of the glass. Use $\pi \approx 3.14$.

Step 1: Find the radius. $r = \frac{80}{2} = 40 \text{ cm}$

Step 2: Calculate the area of the full circle and take half. $A = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.14 \times 40^2 = \frac{1}{2} \times 3.14 \times 1600 = \frac{5024}{2} = 2512 \text{ cm}^2$

Answer: 2 512 cm$^2$ (or about 0.25 m$^2$)

Example 5: Composite shape — rectangle with semicircles (challenging)

Problem: A running track is shaped like a rectangle (100 m by 60 m) with a semicircle on each short end. Find the total area enclosed by the track. Use $\pi \approx 3.14$.

Step 1: The two semicircles together form one full circle. The diameter of each semicircle equals the short side: $d = 60$ m, so $r = 30$ m.

Step 2: Calculate the area of the rectangle. $A_{\text{rect}} = 100 \times 60 = 6000 \text{ m}^2$

Step 3: Calculate the area of the full circle (two semicircles). $A_{\text{circle}} = \pi r^2 = 3.14 \times 30^2 = 3.14 \times 900 = 2826 \text{ m}^2$

Step 4: Add the two areas. $A_{\text{total}} = 6000 + 2826 = 8826 \text{ m}^2$

Answer: 8 826 m$^2$

Common Mistakes

Mistake 1: Forgetting to square the radius in the area formula

❌ $A = \pi r = 3.14 \times 5 = 15.7$

✅ $A = \pi r^2 = 3.14 \times 25 = 78.5$

Why this matters: Leaving out the exponent gives you half the circumference instead of the area. The answer is not just a little wrong — it is an entirely different measurement. Always write $r^2$ to remind yourself to square first.

Mistake 2: Using the diameter directly in the area formula

❌ Given $d = 10$: $A = \pi d^2 = 3.14 \times 100 = 314$

✅ First find $r = 5$: $A = \pi r^2 = 3.14 \times 25 = 78.5$

Why this matters: The area formula requires the radius, not the diameter. Plugging in the diameter without halving it quadruples the area (because $(2r)^2 = 4r^2$). Always check whether the problem gives you $r$ or $d$.

Mistake 3: Confusing circumference and area formulas

❌ "Area $= 2\pi r$" (this is the circumference)

✅ Area $= \pi r^2$, Circumference $= 2\pi r$

Why this matters: Circumference is a length measured in units (cm, m). Area is a surface measured in square units (cm$^2$, m$^2$). A quick unit check reveals which formula you need: if the answer should be in cm$^2$, you need the formula that squares something.

Practice Problems

Try these on your own before checking the answers:

1. A circle has a radius of 6 cm. Find its circumference and area. (Use $\pi \approx 3.14$.)
2. A circular pond has a diameter of 14 m. Find its area.
3. A pizza has a circumference of 40 cm. What is its radius?
4. Find the area of a quarter-circle with radius 10 cm.
5. A rectangular yard (20 m by 12 m) has a circular fountain (diameter 4 m) in the center. Find the area of the yard that is NOT covered by the fountain.

Click to see answers

1. $C = 2 \times 3.14 \times 6 = 37.68$ cm. $A = 3.14 \times 36 = 113.04$ cm$^2$.
2. $r = 7$ m. $A = 3.14 \times 49 = 153.86$ m$^2$.
3. $r = \frac{40}{2 \times 3.14} = \frac{40}{6.28} \approx 6.37$ cm.
4. $A = \frac{1}{4} \times 3.14 \times 100 = 78.5$ cm$^2$.
5. Yard area: $20 \times 12 = 240$ m$^2$. Fountain: $r = 2$, $A = 3.14 \times 4 = 12.56$ m$^2$. Remaining: $240 - 12.56 = 227.44$ m$^2$.

Summary

• The circumference of a circle is $C = 2\pi r = \pi d$ — it measures the distance around the circle.
• The area of a circle is $A = \pi r^2$ — always use the radius, and remember to square it.
• If you are given the diameter, divide by 2 to get the radius before using the area formula.
• For semicircles and quarter-circles, take the corresponding fraction of the full circle's area, and add the straight edges for the perimeter.
• Check your units: circumference is in linear units (cm, m), area is in square units (cm$^2$, m$^2$).

Related Topics

• Area and Perimeter — Formulas, Examples, and Practice
• Area of a Triangle — Formula and Examples
• Surface Area of 3D Shapes — Prisms, Cylinders, and More

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