Annulus - Definition, Area Formula, and Examples

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 $= \pi R^2$

Area of inner circle $= \pi r^2$

Area of annulus $= \pi R^2 - \pi r^2$

Factor out $\pi$:

$$A = \pi(R^2 - r^2)$$

There is a second useful form. Since $R^2 - r^2$ is a difference of two squares, it factors:

$$A = \pi(R + r)(R - r)$$

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 $\pi \approx 3.1416$ 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 $= 2\pi R$

Inner circumference $= 2\pi r$

$$P = 2\pi R + 2\pi r = 2\pi(R + r)$$

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

Six examples, from a direct area to recovering a radius from a given area.

Example 1

Find the area of an annulus with outer radius 5 cm and inner radius 3 cm. Use $\pi \approx 3.14$.

$A = \pi(R^2 - r^2)$

$A = 3.14 \times (5^2 - 3^2)$

$A = 3.14 \times (25 - 9)$

$A = 3.14 \times 16$

$A = 50.24 \text{ cm}^2$

Final answer: 50.24 cm².

Example 2

An annulus has outer radius 7 cm and inner radius 4 cm. A student computes its area as $\pi(7 - 4)^2 = 9\pi$. 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 $9\pi \approx 28.3$ cm².

That squares the wrong quantity. The formula subtracts the squares of the radii, $R^2 - r^2$, not the square of the difference, $(R - r)^2$. Those are different: $R^2 - r^2 = 49 - 16 = 33$, while $(R - r)^2 = 9$.

The correct method squares each radius first, then subtracts:

$A = \pi(R^2 - r^2)$

$A = \pi(49 - 16)$

$A = 33\pi \approx 103.6 \text{ cm}^2$

Final answer: $33\pi \approx 103.6$ cm². Square each radius before subtracting; do not subtract then square.

Example 3

Find the area of an annulus using the factored form, with $R = 10$ cm and $r = 6$ cm. Use $\pi \approx 3.14$.

Use $A = \pi(R + r)(R - r)$.

$R + r = 10 + 6 = 16$

$R - r = 10 - 6 = 4$

$A = 3.14 \times 16 \times 4$

$A = 3.14 \times 64$

$A = 200.96 \text{ cm}^2$

Final answer: 200.96 cm². The factored form gives the same result as $\pi(R^2 - r^2)$.

Example 4

Find the perimeter of an annulus with outer radius 8 cm and inner radius 5 cm. Use $\pi \approx 3.14$.

$P = 2\pi(R + r)$

$P = 2 \times 3.14 \times (8 + 5)$

$P = 2 \times 3.14 \times 13$

$P = 81.64 \text{ cm}$

Final answer: 81.64 cm. Both circle edges are included.

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 $\pi \approx 3.14$.

The track surface is an annulus.

$A = \pi(R^2 - r^2)$

$A = 3.14 \times (50^2 - 42^2)$

$A = 3.14 \times (2500 - 1764)$

$A = 3.14 \times 736$

$A = 2311.04 \text{ m}^2$

Final answer: 2311.04 m². The difference of squares, $736$, is the ring's true footprint.

Example 6

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

Start from $A = \pi(R^2 - r^2)$ and solve for $R$.

$48\pi = \pi(R^2 - 1^2)$

Divide both sides by $\pi$:

$48 = R^2 - 1$

$R^2 = 49$

$R = 7 \text{ cm}$

Final answer: 7 cm. Working backward from the area recovers the missing radius.

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.

The design of a radio telescope dish with a central receiver leaves an annular collecting surface, and its effective light-gathering area is computed as exactly $\pi(R^2 - r^2)$. Whether the ring is a washer or a telescope, the same subtraction of a hole from a disc does the work, which is why the annulus sits so close to the plain circle in any geometry course.

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 $\pi(R - r)^2$ for the area.

The correct way: The area is $\pi(R^2 - r^2)$ - square each radius, then subtract. $(R - r)^2$ is a different, smaller number (Example 2 shows $9$ versus $33$). The first instinct is to subtract the radii and square; the fix is to square each radius before subtracting, or use the factored form $\pi(R + r)(R - r)$, which builds the subtraction in correctly.

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 $2\pi R$ alone as the annulus perimeter.

The correct way: An annulus has two edges, an outer circle and an inner circle, so its perimeter is $2\pi R + 2\pi r = 2\pi(R + r)$. The memorizer who recalls "circumference $= 2\pi r$" for a single circle lands here; the fix is to picture the hole and remember its rim is a boundary too.

Conclusion

• An annulus is the ring-shaped region between two concentric circles, with outer radius $R$ and inner radius $r$.

• Its area is $\pi(R^2 - r^2)$, equal to $\pi(R + r)(R - r)$ - the big disc minus the hole.

• Its perimeter adds both circle edges: $2\pi(R + r)$.

• The most common slip is squaring $(R - r)$ instead of subtracting $R^2 - r^2$; square each radius first.

Practice these to solidify your understanding

Work through these, then check:

1. Find the area of an annulus with $R = 6$ cm and $r = 4$ cm, using $\pi \approx 3.14$. (Answer to Question 1: 62.8 cm².)

2. Find the perimeter of an annulus with $R = 9$ cm and $r = 3$ cm, using $\pi \approx 3.14$. (Answer to Question 2: 75.36 cm.)

3. An annulus has area $24\pi$ cm² and outer radius 5 cm. Find the inner radius. (Answer to Question 3: 1 cm.)

If Question 1 tripped you, return to Example 2 and the square-then-subtract order. To keep building circle geometry with a teacher, explore Bhanzu's geometry tutor, a middle school math tutor, or math classes online. Want a trainer to walk through ring and circle problems with your child? Book a free demo class.

Read More

• Circumference of a circle - the $2\pi r$ edge length that builds the annulus perimeter.

• Sector of a circle - a pie-slice region, another way of carving up a disc.

• Diameter of a circle - twice the radius, and a quick route to either boundary of the ring.

• Center of a circle - the shared point that makes the two circles concentric.