Annulus: area, width, circumference, formulas and worked examples
An annulus is the region between two concentric circles. If you have ever looked at a washer, gasket or the circular cross-section of a hollow pipe, you have already seen this geometry. The central idea is simple: start with the large circle and remove the smaller circle from its center.
The basic idea behind annulus area
For a circle, area is π times the square of its radius. An annulus uses the same formula twice: once for the outer circle and once for the inner circle. Subtracting the second from the first leaves exactly the ring-shaped region.
- A
- area of the annulus
- R
- outer radius
- r
- inner radius
Start with the outer disk
Its area is πR². This includes both the ring and the hole.
Remove the inner disk
The hole has area πr² and must be subtracted.
What remains is the ring
That remainder is π(R² − r²), the annulus area.
Annulus formulas at a glance
The two reference sources use the same notation and core relationships: inner radius r, outer radius R, ring width w, inner circumference c, outer circumference C, inner area a, outer area A, and annulus area AA.
| Quantity | Formula | What it describes |
|---|---|---|
| Outer circumference | C = 2πR | Distance around the outside edge |
| Inner circumference | c = 2πr | Distance around the hole |
| Outer circle area | A = πR² | Entire large disk |
| Inner circle area | a = πr² | Central hole |
| Annulus area | AA = π(R² − r²) | Material in the ring |
| Ring width | w = R − r | Radial thickness |
Why the difference-of-squares form is useful
The area formula can be factored without changing its value:
This second form gives a useful geometric interpretation. The factor π(R + r) is the circumference associated with the mean radius, while w is the radial width. For a thin ring, it behaves like a long strip that has been wrapped around a circle.
Worked examples
Good geometry is not just memorising a formula. The goal is to identify what the measurements mean, choose the correct equation, keep the units consistent, and then check whether the answer makes geometric sense.
R = 12 cm, r = 5 cm
Subtract the squared radii before multiplying by π.
A = π(144 − 25)
A = 119π
R = 12 cm, r = 5 cm
Width is a length, so do not square the result.
w = 12 − 5
w = 7 cm
D = 100 mm, d = 80 mm
Convert diameters to radii first: R = 50 mm and r = 40 mm.
A = 900π
R = 10 cm, A = 50π cm²
Solve for the missing inner radius by rearranging the area equation.
r = √(100 − 50)
r = √50
Circumference: there are two boundaries
An annulus is bounded by two circles, so there are two circumferences to keep separate. The outer boundary uses R; the inner boundary uses r. If a problem asks for the combined length of both circular edges, add them.
Ring width and mean-radius thinking
The radial width is simply the distance between the two concentric circles:
The mean radius is useful because it sits halfway between the inner and outer boundaries:
When the problem gives diameters
Many real objects are measured by outside and inside diameters rather than radii. Let D be the outer diameter and d the inner diameter. Since a radius is half a diameter, R = D/2 and r = d/2.
For a washer with D = 100 mm and d = 80 mm, the radii are 50 mm and 40 mm, giving an annulus area of 900π ≈ 2,827.43 mm².
Annulus geometry in a hollow pipe
The end face of a hollow cylindrical pipe is an annulus. If the pipe has length L, multiplying the annulus cross-sectional area by L gives the volume of material in the wall:
Common mistakes to avoid
| Wrong approach | Why it fails | Correct approach |
|---|---|---|
| Using diameter as radius | A diameter is twice the radius. | Convert D to R = D/2 first. |
| π(R − r)² | The radii must be squared before subtraction. | Use π(R² − r²). |
| Forgetting the inner boundary | An annulus has two circular boundaries. | Keep outer and inner circumference separate. |
| Mixing units | Squared quantities amplify unit mistakes. | Convert all lengths to one unit before calculating. |
| Allowing r ≥ R | The inner circle cannot be larger than the outer circle for this ring. | Require R > r. |
Frequently asked questions
It is the region between two concentric circles: an outer circle with radius R and an inner circle with radius r.
A = π(R² − r²). It is the area of the outer circle minus the area of the inner circle.
The radial width is w = R − r. It is a length, not an area.
Yes. Convert the diameters to radii, or use A = π/4(D² − d²).
The central hole disappears and the annulus becomes an ordinary circle, giving A = πR².
Annulus formula summary
The whole topic can be reduced to a small set of relationships. Once you know which measurement is outer, which is inner, and which quantity the question asks for, the rest is direct circle geometry.
The core notation and formula set above follows the two reference calculators supplied for this page; the explanations, worked examples and diagrams are expanded for teaching clarity.