The area of a segment

The segment

Cut a round cake with one straight cut that does not pass through the middle. You get a small piece and a large piece. Each of them is bounded by a chord and an arc. Each piece is a segment of the circle.

• The smaller piece is the minor segment. The larger piece is the major segment.
• When the chord is a diameter, the two segments are equal. Each one is a semicircle.

A segment is not a sector. A sector has two radii as sides. A segment has a chord as its straight side. But they are linked. Join the ends $A$ and $B$ of the chord to the centre $O$. Then:

The area of the triangle

Triangle $OAB$ has two sides equal to $r$ and an angle $\theta$ between them. Its area is $\dfrac{1}{2}\,r\,r\sin\theta$. You know $\sin\theta$ for the special angles from the chapter on trigonometry:

• $\sin 30^\circ = \sin 150^\circ = \dfrac{1}{2}$
• $\sin 45^\circ = \dfrac{\sqrt{2}}{2}$
• $\sin 60^\circ = \sin 120^\circ = \dfrac{\sqrt{3}}{2}$
• $\sin 90^\circ = 1$

For $60^\circ$ the triangle is equilateral with side $r$, so its area is $\dfrac{\sqrt{3}}{4}r^2$, which agrees with $\dfrac{1}{2}r^2 \times \dfrac{\sqrt{3}}{2}$. For $90^\circ$ the triangle has a right angle at $O$ and its area is $\dfrac{1}{2}r^2$.

Put the two parts together:

In radians this becomes $\dfrac{1}{2}r^2(\theta - \sin\theta)$, which is the form A-Level and IB use.

The major segment is the rest of the circle:

Example 1. A chord subtends $90^\circ$ at the centre of a circle of radius $14\ \text{cm}$. Find the areas of the minor and the major segments. Take $\pi = \dfrac{22}{7}$.

1. The sector. $\dfrac{1}{4} \times 616 = 154\ \text{cm}^2$.
2. The triangle. $\dfrac{1}{2} \times 14 \times 14 = 98\ \text{cm}^2$.
3. The minor segment. $154 - 98 = 56\ \text{cm}^2$.
4. The major segment. $616 - 56 = 560\ \text{cm}^2$.

Example 2. A chord subtends $60^\circ$ at the centre of a circle of radius $12\ \text{cm}$. Find the area of the minor segment in terms of $\pi$ and $\sqrt{3}$. Then give a number.

1. The sector. $\dfrac{60}{360} \times \pi \times 144 = 24\pi\ \text{cm}^2$.
2. The triangle. $\dfrac{\sqrt{3}}{4} \times 144 = 36\sqrt{3}\ \text{cm}^2$.
3. The segment. $(24\pi - 36\sqrt{3})\ \text{cm}^2$.
4. With the book values $\pi = 3.14$ and $\sqrt{3} = 1.73$: $24 \times 3.14 = 75.36$ and $36 \times 1.73 = 62.28$, so the segment is $13.08\ \text{cm}^2$.
5. With a calculator the answer is $13.04\ \text{cm}^2$. The small difference comes from the rounded values $3.14$ and $1.73$.

Always say which values of $\pi$ and $\sqrt{3}$ you used.

Example 3, a segment of a larger angle. A chord subtends $120^\circ$ at the centre of a circle of radius $6\ \text{cm}$. The sector is $\dfrac{1}{3} \times \pi \times 36 = 12\pi$. The triangle is $\dfrac{1}{2} \times 36 \times \dfrac{\sqrt{3}}{2} = 9\sqrt{3}$. The minor segment is $(12\pi - 9\sqrt{3})\ \text{cm}^2$.

The length of the chord. The perpendicular from the centre bisects the chord. So the chord is $2r\sin\dfrac{\theta}{2}$. For $\theta = 90^\circ$ and $r = 14\ \text{cm}$, the chord is $2 \times 14 \times \dfrac{\sqrt{2}}{2} = 14\sqrt{2}\ \text{cm}$.

Practise with new numbers

Try it in Python

This program finds the sector, the triangle and the segment for five angles in a circle of radius $12\ \text{cm}$. It uses the exact $\pi$ and the exact sine.

Common mistakes

• Giving the sector as the segment. The segment is the sector minus the triangle.
• Taking the triangle as half of the chord times the radius. Use $\dfrac{1}{2}r^2\sin\theta$, or the base and the height of the triangle.
• Giving the minor segment when the question asks for the major one.
• Mixing the book values. If the question says $\pi = 3.14$ and $\sqrt{3} = 1.73$, use them throughout.

Remember these

• Minor segment $= \dfrac{\theta}{360}\pi r^2 - \dfrac{1}{2}r^2\sin\theta$.
• Major segment $= \pi r^2 - \text{minor segment}$.
• For $\theta = 90^\circ$ the triangle is $\dfrac{1}{2}r^2$. For $\theta = 60^\circ$ it is $\dfrac{\sqrt{3}}{4}r^2$.
• In radians, minor segment $= \dfrac{1}{2}r^2(\theta - \sin\theta)$.

Check yourself

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

1. A chord subtends $90^\circ$ at the centre of a circle of radius $7\ \text{cm}$. Take $\pi = \dfrac{22}{7}$. Find the area of the minor segment.

2. A circle of radius $10\ \text{cm}$ has a minor segment of area $20\ \text{cm}^2$. What is the area of the major segment? Take $\pi = 3.14$.

3. A chord is as long as the radius of its circle. What angle does it subtend at the centre?

4. For $\theta = 60^\circ$ and $r = 6\ \text{cm}$, write the minor segment in terms of $\pi$ and $\sqrt{3}$.

5. Why is the minor segment zero when $\theta = 0^\circ$ and the half circle when $\theta = 180^\circ$?