Annulus - Definition, Area Formula, and Examples

TL;DR

An annulus is the flat ring-shaped region between two concentric circles - circles sharing a centre but with different radii. Its area is π(R²−r²), where R is the outer radius and r the inner radius. This article defines the annulus, derives its area and perimeter, works through examples, and clears up the mistakes students make.

What Is An Annulus?

An annulus is the region between two concentric circles - two circles that share the same centre but have different radii. It is a flat, ring-shaped figure: take a large disc and remove a smaller disc from its centre, and the material left behind is the annulus. The word is Latin for "little ring."

Two lengths describe it. The outer radius R reaches from the shared centre to the outer circle; the inner radius r reaches to the inner circle. Because both circles share a centre, the ring has a constant width w=R−r all the way around. Each boundary is a full circle, so an annulus is built entirely from the radius of a circle and the circle ideas that go with it.

Where Does The Area Formula π(R² − r²) Come From?

The area formula is not memorised blind; it is subtraction, and seeing that keeps the two radii in the right places.

The annulus is a big disc with a small disc removed. So its area is the big circle's area minus the small circle's area:

Area of outer circle = πR²

Area of inner circle = πr²

Area of annulus = πR²−πr²

Factor out π:

A=π(R²−r²)

There is a second useful form. Since R²−r² is a difference of two squares, it factors:

A=π(R+r)(R−r)

Both forms give the same number. The second is handy when the ring's width w=R−r is known, because (R−r) is already sitting there. Each symbol has a job: R is the outer radius, r the inner radius, and π≈3.1416 is the area of a circle constant. The whole formula relies on both circles sharing one centre - otherwise the region is not an annulus.

How Do You Find The Perimeter Of An Annulus?

An annulus has two boundaries, so its perimeter is the sum of both circumferences.

Outer circumference = 2πR

Inner circumference = 2πr

P=2πR+2πr=2π(R+r)

Both edges count, because the ring is bounded on the outside and the inside. A common slip is to include only the outer circle; the hole has an edge too.

Examples Of Annulus

Example 1

Find the area of an annulus with outer radius 5 cm and inner radius 3 cm. Use π≈3.14.

A=π(R²−r²)

A=3.14×(5²−3²)

A=3.14×(25−9)

A=3.14×16

A=50.24 cm².

Example 2

An annulus has outer radius 7 cm and inner radius 4 cm. A student computes its area as π(7−4)²=9π. What went wrong?

The tempting move is to subtract the radii first and then square, writing (R−r)². It looks like a clean shortcut and gives 9π≈28.39 cm².

That squares the wrong quantity. The formula subtracts the squares of the radii, R²−r², not the square of the difference, (R−r)².

The correct method squares each radius first, then subtracts:

A=π(R²−r²)

A=π(49−16)

A=33π≈103.6 cm².

Example 3

Find the area of an annulus using the factored form, with R=10 cm and r=6 cm. Use π≈3.14.

Use A=π(R+r)(R−r).

R+r=10+6=16

R−r=10−6=4

A=3.14×16×4

A=200.96 cm².

Example 4

Find the perimeter of an annulus with outer radius 8 cm and inner radius 5 cm. Use π≈3.14.

P=2π(R+r)

P=2×3.14×(8+5)

P=81.64 cm.

Example 5

A circular running track has an outer radius of 50 m and an inner radius of 42 m. Find the area of the track surface. Use π≈3.14.

The track surface is an annulus.

A=π(R²−r²)

A=3.14×(50²−42²)

A=3.14×736

A=2311.04 m².

Example 6

The area of an annulus is 48π cm² and its inner radius is 1 cm. Find the outer radius.

Start from A=π(R²−r²) and solve for R.

48π=π(R²−1²)

Divide both sides by π:

48=R²−1

R²=49

R=7 cm.

Why The Annulus Matters: Rings That Do Real Work

The annulus turns up wherever something round has a round hole, and its area answers a concrete question every time: how much material, or surface, is in the ring itself.

• Engineering washers and pipes. A washer's flat face is an annulus, and the cross-section of a pipe wall (outer circle minus the bore) is an annulus too. The area tells you the load-bearing material.

• Sports and paving. A circular track lane, a garden path around a pond, or a ring of paving is an annulus, and its area is exactly the surface to be laid.

• Astronomy and optics. Planetary rings, the light-gathering aperture of some telescopes with a central obstruction, and the diffraction "airy ring" are annular regions where the ring area governs how much light or material is present.

What Are The Most Common Mistakes With An Annulus?

Two errors cause most wrong answers, and both mishandle the two radii.

Mistake 1: Squaring the difference instead of subtracting the squares

Where it slips in: The instant a student sees two radii and reaches to combine them, subtracting first feels natural.

Don't do this: Writing π(R−r)² for the area.

The correct way: The area is π(R²−r²) - square each radius, then subtract.

Mistake 2: Counting only one boundary for the perimeter

Where it slips in: On perimeter questions, where "the circle's circumference" is the reflex answer and the inner edge gets forgotten.

Don't do this: Giving 2πR alone as the annulus perimeter.

The correct way: An annulus has two edges, an outer circle and an inner circle, so its perimeter is 2πR+2πr.

Conclusion

• An annulus is the ring-shaped region between two concentric circles, with outer radius R and inner radius r.
• Its area is π(R²−r²), equal to π(R+r)(R−r) - the big disc minus the hole.
• Its perimeter adds both circle edges: 2π(R+r).
• The most common slip is squaring (R−r) instead of subtracting R²−r²; square each radius first.

Practice these to solidify your understanding

1. Find the area of an annulus with R=6 cm and r=4 cm, using π≈3.14. (Answer to Question 1: 62.8 cm².)
2. Find the perimeter of an annulus with R=9 cm and r=3 cm, using π≈3.14. (Answer to Question 2: 75.36 cm.)
3. An annulus has area 24π cm² and outer radius 5 cm. Find the inner radius. (Answer to Question 3: 1 cm.)