The ring between two circles

The ring

A washer, a bangle, a CD and the path round a round pond all have the same shape. Two circles share one centre. The part between them is a ring. In geometry it is also called an annulus. Some books call the flat metal one a washer.

• $R$ is the radius of the outer circle and $r$ is the radius of the inner circle.
• The width of the ring is $R - r$.
• The area of the ring is the area of the big disc minus the area of the small disc:

There is a second form: $R^2 - r^2 = (R - r)(R + r)$. So $A = \pi\,(R - r)(R + r)$. This is often quicker, because the width $R - r$ is easy to read.

Example 1. A ring has an outer radius of $14\ \text{cm}$ and an inner radius of $7\ \text{cm}$. Find its width and its area. Take $\pi = \dfrac{22}{7}$.

1. The width. $R - r = 14 - 7 = 7\ \text{cm}$.
2. The area. $A = \dfrac{22}{7}\,(14^2 - 7^2) = \dfrac{22}{7} \times 147 = 462\ \text{cm}^2$.

Check with the second form: $\dfrac{22}{7} \times 7 \times 21 = 462$. The two forms agree.

Example 2, a path round a pond. A circular lawn has a radius of $35\ \text{m}$. A path $3.5\ \text{m}$ wide runs all round the outside. Find the area of the path. Take $\pi = \dfrac{22}{7}$.

1. The outer radius. The path is outside the lawn, so $R = 35 + 3.5 = 38.5\ \text{m}$.
2. The area. $A = \dfrac{22}{7}\,(38.5^2 - 35^2) = \dfrac{22}{7} \times (1482.25 - 1225) = \dfrac{22}{7} \times 257.25 = 808.5\ \text{m}^2$.

If the path were inside the lawn, the inner radius would be $35 - 3.5$ and the outer radius would stay $35$. Draw the picture first, so you know which radius is which.

The chord that touches the inner circle

Here is a classic question. Draw a chord of the big circle that just touches the small circle. It touches at its middle point, and the radius to that point is perpendicular to the chord. So the half chord, the inner radius $r$ and the outer radius $R$ make a right triangle with hypotenuse $R$.

By Pythagoras, $R^2 = r^2 + \left(\dfrac{c}{2}\right)^2$, where $c$ is the chord. So

We can find the area of the ring without knowing $R$ or $r$ at all.

Example 3. A chord of the big circle is $14\ \text{cm}$ long and touches the small circle. Find the area of the ring. Take $\pi = \dfrac{22}{7}$.

1. Half the chord. $7\ \text{cm}$.
2. The area. $A = \dfrac{22}{7} \times 7^2 = 154\ \text{cm}^2$.

Example 4, with the radii. The inner radius is $12\ \text{cm}$ and a chord of the outer circle that touches the inner circle is $10\ \text{cm}$ long. Then $R^2 = 12^2 + 5^2 = 169$, so $R = 13\ \text{cm}$. The area is $\pi\,(169 - 144) = 25\pi\ \text{cm}^2$. Check with the chord: $\pi \times 5^2 = 25\pi$.

Reading a ring from two areas

Sometimes a question gives the two areas. If the big circle has an area of $1386\ \text{cm}^2$ and the small one has an area of $616\ \text{cm}^2$, the ring has an area of $1386 - 616 = 770\ \text{cm}^2$. You never need the radii.

Practise with new numbers

Try it in Python

This program finds the area of a ring from its two radii, and again from the chord that touches the inner circle. The two answers must agree.

Common mistakes

• Subtracting the radii before squaring. The area is $\pi(R^2 - r^2)$, not $\pi(R - r)^2$.
• Adding the width to the wrong radius. A path outside a lawn adds to the radius. A path inside takes away from it.
• Using the whole chord. In the right triangle the leg is half of the chord.
• Mixing up the width and a radius. The width of the ring is $R - r$.

Remember these

• The area of a ring is $\pi(R^2 - r^2) = \pi(R - r)(R + r)$.
• A path outside a circle adds its width to the radius.
• A chord of the big circle that touches the small circle gives the area of the ring directly: $\pi \times (\text{half the chord})^2$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A ring has an outer radius of $10\ \text{cm}$ and an inner radius of $6\ \text{cm}$. Write its area in terms of $\pi$.

2. A circular pond has a radius of $7\ \text{m}$. A path $7\ \text{m}$ wide goes round it. Take $\pi = \dfrac{22}{7}$. Find the area of the path.

3. A chord of the outer circle is $16\ \text{cm}$ long and touches the inner circle. What is the area of the ring in terms of $\pi$?

4. The areas of two circles with the same centre are $154\ \text{cm}^2$ and $616\ \text{cm}^2$. What is the area of the ring between them?

5. Why is the area of a ring not $\pi(R - r)^2$?