Geometry

Area of Circles

Last updated: March 2026 · Beginner
Before you start

You should be comfortable with:

Real-world applications
🌡️
HVAC

Refrigerant charging, airflow, system sizing

Circles show up everywhere — pipes, ducts, wheels, tanks, and round openings. To work with circles, you need to understand a few key parts and one essential formula: A=πr2A = \pi r^2.

Parts of a Circle

  • Center: The point in the exact middle of the circle.
  • Radius (rr): The distance from the center to any point on the circle’s edge.
  • Diameter (dd): The distance across the circle through the center. The diameter is always twice the radius.

d=2ror equivalentlyr=d2d = 2r \quad \text{or equivalently} \quad r = \frac{d}{2}

Parts of a Circle

centerrd = 2r
  • Pi (π\pi): The ratio of every circle’s circumference to its diameter. It is approximately 3.141593.14159, and for most calculations you can use π3.14\pi \approx 3.14 or the π\pi button on your calculator.

Area of a Circle

The area of a circle with radius rr is:

A=πr2A = \pi r^2

This means: multiply pi by the radius squared. Squaring the radius (r2=r×rr^2 = r \times r) comes first, then multiply by π\pi.

Example 1: Find the area of a circle with radius 5 in.

A=πr2=π(5)2=25π25×3.14=78.5 in2A = \pi r^2 = \pi (5)^2 = 25\pi \approx 25 \times 3.14 = 78.5 \text{ in}^2

Answer: The area is approximately 78.5 square inches.

Example 2: Find the area of a circle with radius 3.5 m.

A=π(3.5)2=π(12.25)=12.25π38.48 m2A = \pi (3.5)^2 = \pi (12.25) = 12.25\pi \approx 38.48 \text{ m}^2

Answer: The area is approximately 38.48 square meters.

Working with Diameter Instead of Radius

Many real-world measurements give you the diameter rather than the radius. Before using the area formula, convert diameter to radius by dividing by 2.

Example 3: A circular table top has a diameter of 4 ft. Find its area.

Step 1: Find the radius.

r=d2=42=2 ftr = \frac{d}{2} = \frac{4}{2} = 2 \text{ ft}

Step 2: Apply the area formula.

A=π(2)2=4π12.57 ft2A = \pi (2)^2 = 4\pi \approx 12.57 \text{ ft}^2

Answer: The area is approximately 12.57 square feet.

You can also substitute r=d2r = \frac{d}{2} directly into the formula to get an equivalent version:

A=π(d2)2=πd24A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}

Both formulas give the same result. Use whichever feels more natural.

Area of a Semicircle

A semicircle is half a circle. Its area is half the area of the full circle:

Asemicircle=12πr2A_{\text{semicircle}} = \frac{1}{2}\pi r^2

Example 4: A semicircular window has a diameter of 30 in. Find its area.

Step 1: Find the radius: r=302=15 inr = \frac{30}{2} = 15 \text{ in}

Step 2: Calculate the semicircle area.

A=12π(15)2=12π(225)=112.5π353.43 in2A = \frac{1}{2}\pi (15)^2 = \frac{1}{2}\pi (225) = 112.5\pi \approx 353.43 \text{ in}^2

Answer: The area is approximately 353.43 square inches.

Area of a Sector

A sector is a “pie slice” of a circle — the region between two radii and the arc connecting them. If the central angle of the sector is θ\theta (in degrees):

Asector=θ360×πr2A_{\text{sector}} = \frac{\theta}{360} \times \pi r^2

This formula takes the fraction of the full circle (θ360\frac{\theta}{360}) and multiplies it by the full circle’s area.

Example 5: A sprinkler covers a sector with radius 20 ft and a central angle of 90°90\degree. What area does it water?

A=90360×π(20)2=14×400π=100π314.16 ft2A = \frac{90}{360} \times \pi (20)^2 = \frac{1}{4} \times 400\pi = 100\pi \approx 314.16 \text{ ft}^2

Answer: The sprinkler waters approximately 314.16 square feet.

Notice that 90°90\degree is one-quarter of 360°360\degree, so the sector area is one-quarter of the full circle area. This makes intuitive sense.

Circle Area Formula Reference

ShapeFormulaVariables
Full circle (radius)A=πr2A = \pi r^2rr = radius
Full circle (diameter)A=πd24A = \frac{\pi d^2}{4}dd = diameter
SemicircleA=12πr2A = \frac{1}{2}\pi r^2rr = radius
SectorA=θ360πr2A = \frac{\theta}{360}\pi r^2θ\theta = central angle in degrees, rr = radius

Real-World Application: HVAC — Sizing Round Ductwork

HVAC technicians use the cross-sectional area of round ducts to determine how much air can flow through them. Airflow capacity is directly related to the duct’s area.

Problem: A technician needs to compare two round ducts: one with a 6-inch diameter and one with an 8-inch diameter. How much more area does the larger duct have?

Step 1: Find the area of the 6-inch duct.

r1=62=3 inr_1 = \frac{6}{2} = 3 \text{ in}

A1=π(3)2=9π28.27 in2A_1 = \pi (3)^2 = 9\pi \approx 28.27 \text{ in}^2

Step 2: Find the area of the 8-inch duct.

r2=82=4 inr_2 = \frac{8}{2} = 4 \text{ in}

A2=π(4)2=16π50.27 in2A_2 = \pi (4)^2 = 16\pi \approx 50.27 \text{ in}^2

Step 3: Find the difference.

A2A1=16π9π=7π21.99 in2A_2 - A_1 = 16\pi - 9\pi = 7\pi \approx 21.99 \text{ in}^2

Step 4: Find the percentage increase.

A2A1A1×100=7π9π×100=79×10077.8%\frac{A_2 - A_1}{A_1} \times 100 = \frac{7\pi}{9\pi} \times 100 = \frac{7}{9} \times 100 \approx 77.8\%

Answer: The 8-inch duct has approximately 22 more square inches of cross-sectional area — about 78% more than the 6-inch duct. This is a critical insight for HVAC work: increasing a duct’s diameter by just 2 inches nearly doubles its cross-sectional area (and airflow capacity). That is because area depends on the square of the radius, so small changes in diameter produce large changes in area.

Common Mistakes to Avoid

  1. Using diameter instead of radius. The formula is πr2\pi r^2, not πd2\pi d^2. If you are given the diameter, divide by 2 first.
  2. Squaring π\pi instead of rr. Calculate r2r^2 first, then multiply by π\pi. The formula is π×(r2)\pi \times (r^2), not (π×r)2(\pi \times r)^2.
  3. Forgetting units are squared. Area is in square units (in2\text{in}^2, ft2\text{ft}^2), not linear units.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Problem 1: Find the area of a circle with radius 10 cm.

A=π(10)2=100π314.16 cm2A = \pi (10)^2 = 100\pi \approx 314.16 \text{ cm}^2

Answer: Approximately 314.16 cm2314.16 \text{ cm}^2

Problem 2: A circular pond has a diameter of 24 ft. Find its area.

Radius: r=242=12 ftr = \frac{24}{2} = 12 \text{ ft}

A=π(12)2=144π452.39 ft2A = \pi (12)^2 = 144\pi \approx 452.39 \text{ ft}^2

Answer: Approximately 452.39 ft2452.39 \text{ ft}^2

Problem 3: Find the area of a semicircle with radius 7 m.

A=12π(7)2=12(49π)=24.5π76.97 m2A = \frac{1}{2}\pi (7)^2 = \frac{1}{2}(49\pi) = 24.5\pi \approx 76.97 \text{ m}^2

Answer: Approximately 76.97 m276.97 \text{ m}^2

Problem 4: A pizza has a diameter of 16 in. One slice covers a 45°45\degree sector. What is the area of one slice?

Radius: r=162=8 inr = \frac{16}{2} = 8 \text{ in}

A=45360×π(8)2=18×64π=8π25.13 in2A = \frac{45}{360} \times \pi (8)^2 = \frac{1}{8} \times 64\pi = 8\pi \approx 25.13 \text{ in}^2

Answer: Approximately 25.13 in225.13 \text{ in}^2

Problem 5: An HVAC technician needs a round duct with at least 75 in275 \text{ in}^2 of cross-sectional area. What is the minimum duct diameter (to the nearest inch)?

Start with A=πr2A = \pi r^2 and solve for rr:

75=πr275 = \pi r^2

r2=75π753.1423.89r^2 = \frac{75}{\pi} \approx \frac{75}{3.14} \approx 23.89

r23.894.89 inr \approx \sqrt{23.89} \approx 4.89 \text{ in}

Diameter: d=2r9.77 ind = 2r \approx 9.77 \text{ in}

Round up to the nearest inch: d=10 ind = 10 \text{ in}

Check: A=π(5)2=25π78.54 in2>75A = \pi(5)^2 = 25\pi \approx 78.54 \text{ in}^2 > 75 . Confirmed.

Answer: The minimum duct diameter is 10 inches.

Key Takeaways

  • The area of a circle is A=πr2A = \pi r^2 — always use the radius, not the diameter
  • To convert from diameter to radius: r=d2r = \frac{d}{2}
  • Area is in square units (ft2\text{ft}^2, in2\text{in}^2, etc.)
  • A semicircle has half the area of a full circle: A=12πr2A = \frac{1}{2}\pi r^2
  • A sector’s area is a fraction of the full circle: A=θ360πr2A = \frac{\theta}{360}\pi r^2
  • Because area depends on r2r^2, doubling the radius quadruples the area — small changes in size have a big impact

Return to Geometry for more topics in this section.

Last updated: March 28, 2026