# 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](/content/math/geometry/radius/index.html) 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](/content/math/geometry/area-of-a-circle/index.html) 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.)_