Area of a Segment of a Circle: Class 10 Method and Examples

The key result#

Area of the minor seg­ment#

The minor sec­tor $OAB$ is made of two non-over­lap­ping pieces: the tri­an­gle $OAB$ and the minor seg­ment. So

$$\text{Area of minor segment} = \text{Area of sector } OAB - \text{Area of } \triangle OAB$$

$$\displaystyle = \frac{\theta}{360^\circ} \times \pi r^2 - \text{Area of } \triangle OAB$$

Area of the major seg­ment#

The minor and major seg­ments together make the whole cir­cle, so

$$\text{Area of major segment} = \pi r^2 - \text{Area of minor segment}$$

The only new work is find­ing the area of tri­an­gle $OAB$. It is always isosce­les, because $OA = OB = r$.

Area of tri­an­gle OAB for the com­mon angles#

When the angle is 60°#

In $\triangle OAB$, $OA = OB$, so $\angle OAB = \angle OBA$. The three angles add to $180^\circ$, so each base angle is $\displaystyle \frac{180^\circ - 60^\circ}{2} = 60^\circ$. All angles are $60^\circ$, so the tri­an­gle is equi­lat­eral with side $r$, and

$$\displaystyle \text{Area of } \triangle OAB = \frac{\sqrt{3}}{4} r^2$$

When the angle is 90°#

The tri­an­gle is right-angled at $O$ with both legs equal to $r$, so

$$\displaystyle \text{Area of } \triangle OAB = \frac{1}{2} \times r \times r = \frac{1}{2} r^2$$

Any angle: the height method#

Take $OA$ as the base. Drop a per­pen­dic­u­lar $BM$ from $B$ to the line $OA$. Then the area is $\displaystyle \frac{1}{2} \times OA \times BM = \frac{1}{2} \times r \times BM$.

If $\angle AOB = \theta$ is acute, $M$ lies on $OA$, and in right tri­an­gle $OMB$ we have $BM = r \sin\theta$. This gives

$$\displaystyle \text{Area of } \triangle OAB = \frac{1}{2} r^2 \sin\theta$$

Check with $\theta = 60^\circ$: $\displaystyle \frac{1}{2} r^2 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} r^2$, the same as before.

When the angle is 120°#

Now $\angle AOB$ is obtuse, so the foot $M$ falls on $AO$ extended beyond $O$. The angles $\angle AOB$ and $\angle BOM$ form a lin­ear pair, so $\angle BOM = 180^\circ - 120^\circ = 60^\circ$. In right tri­an­gle $OMB$,

$$\displaystyle BM = r \sin 60^\circ = \frac{\sqrt{3}}{2} r$$

$$\displaystyle \text{Area of } \triangle OAB = \frac{1}{2} \times r \times \frac{\sqrt{3}}{2} r = \frac{\sqrt{3}}{4} r^2$$

So the tri­an­gles for $60^\circ$ and $120^\circ$ have the same area, even though their shapes dif­fer.

A use­ful extra fact for $120^\circ$: in right tri­an­gle $OMB$, $\displaystyle OM = r\cos 60^\circ = \frac{r}{2}$, so $\displaystyle AM = \frac{3r}{2}$, and by Pythago­ras the chord is $\displaystyle AB = \sqrt{\frac{9r^2}{4} + \frac{3r^2}{4}} = r\sqrt{3}$.

Method: find­ing a seg­ment#

1. Iden­tify the cen­tre, the radius $r$ and the angle $\theta$ at the cen­tre. Check whether the minor or the major seg­ment is asked.
2. Find the area of the sec­tor: $\displaystyle \frac{\theta}{360^\circ} \times \pi r^2$.
3. Find the area of $\triangle OAB$: $\displaystyle \frac{\sqrt{3}}{4} r^2$ for $60^\circ$ or $120^\circ$, $\displaystyle \frac{1}{2} r^2$ for $90^\circ$.
4. Sub­tract: minor seg­ment = sec­tor minus tri­an­gle.
5. For the major seg­ment, sub­tract the minor seg­ment from $\pi r^2$.
6. Round only at the end, nor­mally to two dec­i­mal places, using the val­ues of $\pi$ and $\sqrt{3}$ given in the ques­tion.

Round­ing con­ven­tions#

Exam ques­tions usu­ally say, for exam­ple, "use $\pi = 3.14$ and $\sqrt{3} = 1.73$". Use exactly those val­ues. If an answer does not come out exact, keep full work­ing until the last line and then round to two dec­i­mal places, writ­ing "approx­i­mately". It is also good prac­tice to write the exact answer first, such as $\displaystyle \left(\frac{\pi}{4} - \frac{1}{2}\right) r^2$, when the ques­tion allows it.

Worked exam­ples#

Exam­ple 1: a 60° seg­ment#

A chord of a cir­cle of radius $21$ cm sub­tends an angle of $60^\circ$ at the cen­tre. Find the area of the minor seg­ment. Use $\displaystyle \pi = \frac{22}{7}$ and $\sqrt{3} = 1.73$.

$$\displaystyle \begin{aligned} \text{Sector} &= \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21 = \frac{1386}{6} = 231 \text{ cm}^2 \\ \triangle OAB &= \frac{\sqrt{3}}{4} \times 21^2 = \frac{1.73 \times 441}{4} = 190.7325 \text{ cm}^2 \\ \text{Segment} &= 231 - 190.7325 = 40.2675 \text{ cm}^2 \end{aligned}$$

The minor seg­ment is approx­i­mately $40.27$ cm².

Exam­ple 2: a 90° seg­ment and its major seg­ment#

In a cir­cle of radius $14$ cm, a chord sub­tends a right angle at the cen­tre. Find the areas of the minor and major seg­ments. Use $\displaystyle \pi = \frac{22}{7}$.

$$\displaystyle \begin{aligned} \text{Sector} &= \frac{1}{4} \times \frac{22}{7} \times 196 = 154 \text{ cm}^2 \\ \triangle OAB &= \frac{1}{2} \times 14 \times 14 = 98 \text{ cm}^2 \\ \text{Minor segment} &= 154 - 98 = 56 \text{ cm}^2 \\ \text{Circle} &= \frac{22}{7} \times 196 = 616 \text{ cm}^2 \\ \text{Major segment} &= 616 - 56 = 560 \text{ cm}^2 \end{aligned}$$

The minor seg­ment is $56$ cm² and the major seg­ment is $560$ cm².

Exam­ple 3: a 120° seg­ment#

A chord of a cir­cle of radius $12$ cm sub­tends an angle of $120^\circ$ at the cen­tre. Find the length of the chord and the area of the minor seg­ment. Use $\pi = 3.14$ and $\sqrt{3} = 1.73$.

From the height method, $AB = r\sqrt{3} = 12 \times 1.73 = 20.76$ cm.

$$\displaystyle \begin{aligned} \text{Sector} &= \frac{120^\circ}{360^\circ} \times 3.14 \times 144 = \frac{452.16}{3} = 150.72 \text{ cm}^2 \\ \triangle OAB &= \frac{\sqrt{3}}{4} \times 144 = 36 \times 1.73 = 62.28 \text{ cm}^2 \\ \text{Segment} &= 150.72 - 62.28 = 88.44 \text{ cm}^2 \end{aligned}$$

The chord is about $20.76$ cm and the minor seg­ment is $88.44$ cm².

Exam­ple 4: a square with four quad­rants#

$ABCD$ is a square of side $14$ cm. With each ver­tex as cen­tre, a quad­rant of radius $7$ cm is drawn inside the square. Find the area of the part of the square not cov­ered by the quad­rants. Use $\displaystyle \pi = \frac{22}{7}$.

Each quad­rant is a sec­tor of angle $90^\circ$. Four quad­rants of the same radius make one full cir­cle of radius $7$ cm.

$$\displaystyle \begin{aligned} \text{Square} &= 14 \times 14 = 196 \text{ cm}^2 \\ \text{Four quadrants} &= \frac{22}{7} \times 7 \times 7 = 154 \text{ cm}^2 \\ \text{Shaded} &= 196 - 154 = 42 \text{ cm}^2 \end{aligned}$$

The uncov­ered area is $42$ cm².

Exam­ple 5: a leaf design#

$ABCD$ is a square of side $10$ cm. An arc is drawn with cen­tre $A$ and radius $10$ cm from $B$ to $D$, and another with cen­tre $C$ and radius $10$ cm from $B$ to $D$. Find the area of the leaf-shaped region between the arcs. Use $\pi = 3.14$.

The diag­o­nal $BD$ divides the leaf into two equal parts. The part on the side of $C$ is the minor seg­ment cut from the quad­rant with cen­tre $A$ by the chord $BD$, and the other part is the match­ing seg­ment of the quad­rant with cen­tre $C$.

$$\displaystyle \begin{aligned} \text{Quadrant with centre } A &= \frac{1}{4} \times 3.14 \times 100 = 78.5 \text{ cm}^2 \\ \triangle ABD &= \frac{1}{2} \times 10 \times 10 = 50 \text{ cm}^2 \\ \text{One segment} &= 78.5 - 50 = 28.5 \text{ cm}^2 \\ \text{Leaf} &= 2 \times 28.5 = 57 \text{ cm}^2 \end{aligned}$$

In exact form the leaf is $50\pi - 100$ cm². The leaf has area $57$ cm².

Com­mon mis­takes#

• Call­ing the region between a chord and an arc a sec­tor. A sec­tor always has two radii as sides.
• Using $\displaystyle \frac{1}{2} r^2$ for every tri­an­gle. That works only for $90^\circ$; for $60^\circ$ and $120^\circ$ the area is $\displaystyle \frac{\sqrt{3}}{4} r^2$.
• Think­ing a $120^\circ$ tri­an­gle is equi­lat­eral. Only the $60^\circ$ one is; the $120^\circ$ tri­an­gle has the same area but a longer base $r\sqrt{3}$.
• Sub­tract­ing the tri­an­gle from the whole cir­cle to get the major seg­ment. The major seg­ment is the cir­cle minus the minor seg­ment.
• For­get­ting that four equal quad­rants make one full cir­cle, and adding them one by one with round­ing errors.
• Round­ing $\sqrt{3}$ or inter­me­di­ate results too early, which changes the sec­ond dec­i­mal place.

Try these#

1. A chord of a cir­cle of radius $10$ cm sub­tends a right angle at the cen­tre. Find the minor and major seg­ments. Use $\pi = 3.14$. Answer: $28.5$ cm² and $285.5$ cm².
2. Find the minor seg­ment for $r = 6$ cm and $\theta = 60^\circ$. Use $\pi = 3.14$ and $\sqrt{3} = 1.73$. Answer: $3.27$ cm².
3. Find the minor seg­ment for $r = 7$ cm and $\theta = 120^\circ$. Use $\displaystyle \pi = \frac{22}{7}$ and $\sqrt{3} = 1.73$. Answer: approx­i­mately $30.14$ cm².
4. A cir­cle is drawn inside a square of side $28$ cm, touch­ing all four sides. Find the area of the square out­side the cir­cle. Use $\displaystyle \pi = \frac{22}{7}$. Answer: $168$ cm².
5. A chord of a cir­cle of radius $8$ cm sub­tends $90^\circ$ at the cen­tre. Find the minor seg­ment. Use $\pi = 3.14$. Answer: $18.24$ cm².

Key terms#

Chord

A line seg­ment join­ing two points on a cir­cle.

Seg­ment

The region between a chord and one of the arcs it cuts off.

Minor seg­ment

The seg­ment on the side of the minor arc; its area is the sec­tor minus the tri­an­gle.

Major seg­ment

The seg­ment on the side of the major arc; its area is the cir­cle minus the minor seg­ment.

Quad­rant

A sec­tor with angle $90^\circ$, one quar­ter of a cir­cle.

Isosce­les tri­an­gle OAB

The tri­an­gle formed by the cen­tre and the ends of a chord; two of its sides are radii.

Height method

Find­ing the area of $\triangle OAB$ as $\displaystyle \frac{1}{2} \times r \times BM$, where $BM$ is the per­pen­dic­u­lar from $B$ to the line $OA$.

Com­mon ques­tions#

What is the dif­fer­ence between a seg­ment and a sec­tor?#

A sec­tor is cut off by two radii and looks like a slice of pizza. A seg­ment is cut off by a sin­gle chord and has no cor­ner at the cen­tre.

Why is the tri­an­gle area the same for 60° and 120°?#

With $OA$ as base, both tri­an­gles have the same base $r$. The height from $B$ is $r \sin 60^\circ$ in both cases, because for $120^\circ$ the height makes a $60^\circ$ angle with the extended base. Same base and same height give the same area.

Can the minor seg­ment be found for any angle?#

Yes. Minor seg­ment = sec­tor minus tri­an­gle always holds. In Class 10 ques­tions the angle is cho­sen so that the tri­an­gle area can be found with the results above.

How do I recog­nise the pieces in a com­bined fig­ure?#

Look for the cen­tre of each arc. Join it to the ends of the arc to see a sec­tor; join the ends of the arc to each other to see a seg­ment. Then add or sub­tract the pieces and the straight-edged shapes.

Should the answer be exact or dec­i­mal?#

Fol­low the ques­tion. If val­ues of $\pi$ and $\sqrt{3}$ are given, a dec­i­mal answer is expected. Oth­er­wise an exact answer in terms of $\pi$ and $\sqrt{3}$ is fine.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
3. Kise­lev, A. P. Kise­lev's Geom­e­try, Book I: Planime­try (adapted by A. Given­tal). Sum­iz­dat.