Segment area

A segment of a circle is the region cut off by a chord. It is bounded by the chord and one of the two arcs the chord defines. The minor segment uses the minor arc (smaller piece); the major segment uses the major arc (larger piece). Together the two segments make the whole disc.

Don't confuse segment with sector. A sector is the slice cut by two radii; a segment is the slice cut by a chord.

Computing the area of a minor segment

The minor segment is the region bounded by a chord $AB$ and the minor arc $AB$. Two natural ingredients sit inside it: the sector $OAB$ (formed by the two radii and the arc), and the triangle $OAB$ (formed by the two radii and the chord). The relationship is

$\text{(Minor sector $OAB$)} = \text{(Minor segment)} + \text{(Triangle $OAB$)}.$

So

$\boxed{A_{\text{minor seg}} = A_{\text{sector}} - A_{\triangle OAB}.}$

The triangle $OAB$ is isosceles (two sides equal to the radius $r$). If the central angle of the sector is $\theta$, then

$A_{\triangle OAB} = \frac{1}{2} r^2 \sin\theta.$

So the segment area becomes

$A_{\text{minor seg}} = \frac{\theta}{360^\circ} \pi r^2 - \frac{1}{2} r^2 \sin\theta = \frac{r^2}{2}\left(\frac{\pi\theta}{180^\circ} - \sin\theta\right).$

Major segment

The major segment is the rest of the disc. Hence

$A_{\text{major seg}} = \pi r^2 - A_{\text{minor seg}}.$

Equivalently it equals the major sector plus the same triangle $OAB$, because the major sector includes the triangle on its "inside" boundary.

Frequently used angles

For board problems, $\theta$ is almost always one of $60^\circ, 90^\circ, 120^\circ$. Recall:

• $\sin 60^\circ = \sqrt 3/2$.
• $\sin 90^\circ = 1$.
• $\sin 120^\circ = \sqrt 3/2$.

These keep the algebra clean.

Worked examples

Example 1. Find the area of the minor segment of a circle of radius $10$ cm cut by a chord that subtends $90^\circ$ at the centre. (Use $\pi = 3.14$.)

Sector area $= (90/360) \pi r^2 = (1/4)(3.14)(100) = 78.5$ cm$^2$.

Triangle area $= (1/2) r^2 \sin 90^\circ = (1/2)(100)(1) = 50$ cm$^2$.

Minor segment $= 78.5 - 50 = 28.5$ cm$^2$.

Example 2. Find the area of the corresponding major segment in Example 1.

Total $= \pi r^2 = 314$ cm$^2$. Major $= 314 - 28.5 = 285.5$ cm$^2$.

Example 3. A chord of length $r\sqrt 3$ in a circle of radius $r$ subtends what angle at the centre? Find the minor segment area.

Using the cosine rule in $\triangle OAB$: $(r\sqrt 3)^2 = r^2 + r^2 - 2r^2 \cos\theta = 2r^2(1 - \cos\theta)$. So $3r^2 = 2r^2(1 - \cos\theta) \Rightarrow \cos\theta = -1/2 \Rightarrow \theta = 120^\circ$.

Sector $= (120/360) \pi r^2 = \pi r^2/3$.

Triangle $= (1/2) r^2 \sin 120^\circ = (\sqrt 3/4) r^2$.

Minor segment $= \pi r^2/3 - (\sqrt 3/4) r^2 = r^2(\pi/3 - \sqrt 3/4)$.

For $r = 12$: $\approx 144(1.047 - 0.433) \approx 88.4$ cm$^2$.

Example 4. A chord $AB$ of a circle of radius $14$ cm subtends $60^\circ$ at the centre. Find the area of the minor segment. (Use $\pi = 22/7, \sqrt 3 = 1.73$.)

Sector area $= (60/360)(22/7)(196) = (1/6)(616) = 102.67$ cm$^2$.

Triangle area $= (1/2)(196) \sin 60^\circ = 98 \cdot \sqrt 3/2 = 49\sqrt 3 \approx 84.77$ cm$^2$.

Minor segment $\approx 102.67 - 84.77 = 17.9$ cm$^2$.

Example 5. A chord $AB$ of length $24$ in a circle of radius $13$ subtends what angle at the centre? Find the minor segment area.

Half-chord $= 12$, leg of right triangle from the centre. Drop $OM \perp AB$, $OM = \sqrt{169 - 144} = 5$. $\sin(\theta/2) = 12/13$, $\cos(\theta/2) = 5/13$, so $\sin\theta = 2 \cdot 12/13 \cdot 5/13 = 120/169$.

Direct formula: $A_{\triangle OAB} = (1/2)(24)(5) = 60$ (using base-height). Angle: $\theta = 2 \arcsin(12/13) \approx 2(67.38^\circ) = 134.76^\circ$.

Sector $= (134.76/360) \pi (169) \approx (0.374)(530.66) \approx 198.5$.

Minor segment $\approx 198.5 - 60 = 138.5$ (units squared).

(Board problems would typically give $\theta$ as a nice angle.)

Try it yourself

1. A chord of a circle of radius $14$ subtends $60^\circ$ at the centre. Find the minor segment's area.
2. Same as (1), find the major segment's area.
3. A chord subtends $90^\circ$ at the centre of a circle of radius $10$. Find the minor segment area.
4. A chord of length $r\sqrt 2$ in a circle of radius $r$ subtends what angle?
5. Find the area of the region between a $60^\circ$ sector of radius $6$ and the chord.
6. A chord $AB$ of length $10$ in a circle of radius $13$. Find the perpendicular from the centre to the chord.
7. A semicircle of radius $7$. Find its area. (Hint: the chord is the diameter; the segment is the semicircle.)
8. A chord of a circle of radius $21$ subtends $120^\circ$. Find the chord length and the minor segment area.
9. The minor segment of a circle of radius $r$ corresponding to a central angle of $60^\circ$ has area $A$. Find $A$ in terms of $r$.
10. A chord of length $r$ in a circle of radius $r$ subtends what angle? Find the minor segment area in terms of $r$.
11. Show that the segment area is always less than the sector area.
12. A circle of radius $14$ is cut by a chord into two segments. If the chord is a diameter, what is each segment's area?

Pitfalls / Insight

(1) Always subtract the triangle (not the sector) from the sector. The triangle is the isosceles one made by the two radii and the chord.

(2) The triangle area formula $(1/2) r^2 \sin\theta$ is the easiest. Don't compute it via base $\times$ height if $\theta$ is a standard angle.

(3) Note the $\sin$ symmetry: $\sin 60^\circ = \sin 120^\circ = \sqrt 3/2$. Use this to find the triangle area for either acute or obtuse central angle without re-deriving.

(4) When $\theta = 180^\circ$ (chord is a diameter), the segment is half the disc , area $\pi r^2/2$.