A Slice of a Circle Is Just a Fraction of It, and the Angle Tells You Which

Cut a circular roti into six equal pieces and each piece is one-sixth of the whole. You do not need a new formula for the piece — you need the area of the roti and the fraction $\frac{1}{6}$.

That is the entire idea of this chapter, and the fraction is read off the angle at the centre. A full turn is $360^{\circ }$, so a slice with a central angle of $\theta$ is the fraction

$$\frac{\theta }{360}$$

of the circle. Six equal pieces means $\theta =60^{\circ }$, and $\frac{60}{360}=\frac{1}{6}$. The fraction and the picture agree.

Apply that fraction to the circle's area and you get the sector; apply it to the circumference and you get the arc. Two formulas, one idea.

The vocabulary has to be exact, because the questions use it precisely.

- An arc is a part of the circumference — a curved piece of the boundary
- A sector is the region between two radii and an arc, shaped like a slice
- A chord joins two points on the circle
- A segment is the region between a chord and an arc, shaped like the piece cut off by a straight knife

The difference between a sector and a segment is where the marks are. A sector is bounded by two straight radii; a segment is bounded by one straight chord. And that gives the whole method for a segment:

$$\text{area of a segment}=\text{area of the sector}-\text{area of the triangle}$$

because removing the triangle formed by the two radii and the chord is exactly what turns a slice into the piece beyond the chord.

**Take $\pi =\frac{22}{7}$ when the radius is a multiple of $7$**, and $\pi =3.14$ otherwise — the numbers in a question are always chosen so that one of the two divides cleanly.

This page covers the CBSE Class 10 Maths chapter on areas related to circles: arc length, the area of a sector, the areas of the minor and major segments, and real-life problems on sectors.

**Learn the three fractions rather than recomputing them: $60^{\circ }$ is one-sixth, $90^{\circ }$ is one-quarter, and $120^{\circ }$ is one-third of the circle.**

$$\frac{60}{360}=\frac{1}{6},\frac{90}{360}=\frac{1}{4},\frac{120}{360}=\frac{1}{3}$$

A few others worth recognising on sight: $30^{\circ }$ is one-twelfth, $45^{\circ }$ is one-eighth, $180^{\circ }$ is one-half, and $270^{\circ }$ is three-quarters.

Worked example 1 — a quarter. Find the area of the sector of a circle of radius $10$ cm with a central angle of $90^{\circ }$. Take $\pi =3.14$.

$$\frac{1}{4}\times 3.14\times 100=\frac{314}{4}=78.5{\text{ cm}}^2$$

Worked example 2 — a third. Find the area of the sector of a circle of radius $12$ cm with a central angle of $120^{\circ }$. Take $\pi =3.14$.

$$\frac{1}{3}\times 3.14\times 144=\frac{452.16}{3}=150.72{\text{ cm}}^2$$

Check: the whole circle is $452.16$ cm², and a third of that is $150.72$ cm². Correct, and doing the whole-circle area first and then taking the fraction is usually the least error-prone order.

Worked example 3 — the major sector. For the same circle of radius $12$ cm, find the area of the major sector corresponding to a central angle of $120^{\circ }$.

The major sector is what is left, so its angle is $360^{\circ }-120^{\circ }=240^{\circ }$:

$$\frac{240}{360}\times 452.16=\frac{2}{3}\times 452.16=301.44{\text{ cm}}^2$$

Check by addition:

$$150.72+301.44=452.16{\text{ cm}}^2$$

which is the whole circle. That addition is the compulsory check whenever a question involves both a minor and a major part.

The distinction between minor and major, stated precisely.

- The minor sector has the smaller central angle, which is the $\theta$ the question usually gives
- The major sector has central angle $360^{\circ }-\theta$
- **When $\theta =180^{\circ }$ the two are equal and each is a semicircle, so neither is minor or major

Worked example 4 — two sectors of different circles compared.** A sector of $60^{\circ }$ is cut from a circle of radius $6$ cm, and a sector of $120^{\circ }$ from a circle of radius $3$ cm. Which has the larger area? Take $\pi =3.14$.

First:

$$\frac{1}{6}\times 3.14\times 36=\frac{113.04}{6}=18.84{\text{ cm}}^2$$

Second:

$$\frac{1}{3}\times 3.14\times 9=\frac{28.26}{3}=9.42{\text{ cm}}^2$$

The first is larger, by exactly a factor of two.

And the reason is worth extracting. Doubling the angle doubles the area, but doubling the radius multiplies the area by four, since the radius is squared. The first sector has half the angle and twice the radius, so its area is $\frac{1}{2}\times 4=2$ times the second's. Radius matters much more than angle, and that is the single most useful intuition in this chapter.

Find the sector, find the triangle formed by the two radii and the chord, and subtract.

$$\text{minor segment}=\text{sector}-\text{triangle}$$
$$\text{major segment}=\text{area of the circle}-\text{minor segment}$$

The triangle is isosceles, with both equal sides equal to the radius and the angle between them equal to $\theta$. Its area is

$$\frac{1}{2}{r}^2\sin \theta$$

and for the three angles this chapter uses, that becomes:

- **$\theta =90^{\circ }$** — the triangle is right-angled, so its area is simply $\frac{1}{2}r\times r=\frac{1}{2}{r}^2$
- **$\theta =60^{\circ }$ — the triangle is equilateral**, since two sides are equal and the included angle is $60^{\circ }$, so its area is $\frac{\sqrt{3}}{4}{r}^2$
- **$\theta =120^{\circ }$** — the area is $\frac{1}{2}{r}^2\sin 120^{\circ }=\frac{\sqrt{3}}{4}{r}^2$, the same expression, since $\sin 120^{\circ }=\sin 60^{\circ }$

Worked example 1 — a right angle at the centre. A chord of a circle of radius $10$ cm subtends a right angle at the centre. Find the areas of the minor and major segments. Take $\pi =3.14$.

The sector, from the previous section:

$$\frac{1}{4}\times 3.14\times 100=78.5{\text{ cm}}^2$$

The triangle, right-angled with both legs equal to the radius:

$$\frac{1}{2}\times 10\times 10=50{\text{ cm}}^2$$

The minor segment:

$$78.5-50=28.5{\text{ cm}}^2$$

The major segment, as the rest of the circle:

$$3.14\times 100-28.5=314-28.5=285.5{\text{ cm}}^2$$

Check by addition: $28.5+285.5=314$ cm², the whole circle. Correct.

Notice how much larger the major segment is. The chord cuts off a thin sliver and leaves almost the whole disc, which is what a sketch would show and which is worth confirming before writing the answer down. **If your minor segment comes out bigger than your major one for an angle under $180^{\circ }$, the subtraction went the wrong way.

Worked example 2 — an obtuse angle.** A chord of a circle of radius $12$ cm subtends an angle of $120^{\circ }$ at the centre. Find the area of the corresponding minor segment. Take $\pi =3.14$ and $\sqrt{3}=1.73$.

The sector:

$$\frac{1}{3}\times 3.14\times 144=150.72{\text{ cm}}^2$$

The triangle, using $\frac{\sqrt{3}}{4}{r}^2$:

$$\frac{1.73}{4}\times 144=1.73\times 36=62.28{\text{ cm}}^2$$

The minor segment:

$$150.72-62.28=88.44{\text{ cm}}^2$$

Check the triangle a second way. Drop the perpendicular from the centre to the chord. It bisects both the chord and the $120^{\circ }$ angle, giving two right triangles with a $60^{\circ }$ angle at the centre. So the half-chord is $12\sin 60^{\circ }=6\sqrt{3}$ and the perpendicular is $12\cos 60^{\circ }=6$. Then

$$\text{triangle}=\frac{1}{2}\times (2\times 6\sqrt{3})\times 6=36\sqrt{3}\approx 36\times 1.73=62.28{\text{ cm}}^2$$

The same value, and that construction is the one to use if you cannot remember the formula.

**Worked example 3 — a $60^{\circ }$ chord.** A chord of a circle of radius $14$ cm subtends $60^{\circ }$ at the centre. Find the area of the minor segment. Take $\pi =\frac{22}{7}$ and $\sqrt{3}=1.73$.

The sector:

$$\frac{1}{6}\times \frac{22}{7}\times 196=\frac{1}{6}\times 616=102.67{\text{ cm}}^2$$

The triangle is equilateral with side $14$ cm:

$$\frac{\sqrt{3}}{4}\times 196=1.73\times 49=84.77{\text{ cm}}^2$$

The minor segment:

$$102.67-84.77=17.90{\text{ cm}}^2$$

A very thin sliver, as it should be — at $60^{\circ }$ the chord is close to the arc, so the segment between them is small. The sector and the triangle come out close together, which is a good sign rather than a worrying one.

The mistake to guard against in all three. Using the chord as the base of a triangle whose height you then guess. The triangle is always the one formed by the two radii and the chord, its two equal sides are radii, and its area comes from $\frac{1}{2}{r}^2\sin \theta$ or from the perpendicular-bisector construction — never from the chord alone.

Anything that sweeps about a fixed point traces a sector. Identify the radius and the angle turned, and the formula does the rest.

Worked example 1 — a clock. The minute hand of a clock is $14$ cm long. Find the area of the face it sweeps in $5$ minutes. Take $\pi =\frac{22}{7}$.

First the angle. The minute hand completes $360^{\circ }$ in $60$ minutes, so in one minute it turns $6^{\circ }$, and in $5$ minutes

$$5\times 6^{\circ }=30^{\circ }$$

Then the sector:

$$\frac{30}{360}\times \frac{22}{7}\times 196=\frac{1}{12}\times 616=51.33{\text{ cm}}^2$$

Check with the fraction. Five minutes is $\frac{5}{60}=\frac{1}{12}$ of an hour, and the hand sweeps $\frac{1}{12}$ of the face — so the answer must be $\frac{1}{12}$ of $616$ cm². The time fraction and the angle fraction are the same fraction, which means you can skip the angle entirely for the minute hand.

The hour hand is different, and it is a favourite trap. The hour hand takes $12$ hours for a full turn, so it moves $30^{\circ }$ per hour and only $0.5^{\circ }$ per minute. **Using $6^{\circ }$ per minute for the hour hand gives an answer twelve times too large, so always ask which hand the question means.

Worked example 2 — a wiper blade.** A car has a wiper blade of length $25$ cm sweeping through an angle of $115^{\circ }$. Find the area cleaned by one sweep of the blade. Take $\pi =\frac{22}{7}$.

$$\frac{115}{360}\times \frac{22}{7}\times 625=\frac{115}{360}\times \frac{13750}{7}$$
$$=\frac{115\times 13750}{2520}=\frac{1581250}{2520}\approx 627.48{\text{ cm}}^2$$

The angle was not one of the tidy ones, so the fraction stayed as it was and the arithmetic had to be carried through. That is normal in applied questions — the tidy angles belong to the pure ones.

And note what the question asked. One blade, one sweep. If a car has two such blades the cleaned area is doubled, and a question that says "two wipers" is testing exactly whether you read it.

Worked example 3 — a grazing animal. A horse is tied to a peg at one corner of a square grassy field of side $15$ m by a rope $5$ m long. Find the area of the field it can graze. Take $\pi =3.14$.

The peg is at a corner, so the horse can swing through a right angle — the two fences block the rest. The grazed region is a quarter circle of radius $5$ m:

$$\frac{1}{4}\times 3.14\times 25=\frac{78.5}{4}=19.625{\text{ m}}^2$$

Worked example 4 — the same field with a longer rope. If the rope were $10$ m long instead, find the increase in the grazing area.

$$\frac{1}{4}\times 3.14\times 100=\frac{314}{4}=78.5{\text{ m}}^2$$
$$\text{increase}=78.5-19.625=58.875{\text{ m}}^2$$

Check the ratio. Doubling the rope should multiply the area by four, and $\frac{78.5}{19.625}=4$. Correct — which is the squared-radius effect from the earlier section showing up in a practical answer.

The boundary case that makes this problem interesting. With a rope of $5$ m in a field of side $15$ m, the horse never reaches a fence, so a clean quarter circle is right. But a rope longer than the side of the field would let it swing past the corners, and the grazed region would then be a quarter circle plus two extra pieces. The quarter-circle answer is only valid while the rope is shorter than the side, and saying so is what distinguishes a complete answer from a lucky one.

Worked example 5 — a shaded region. A quadrant is cut away from a circular tabletop of radius $14$ cm. Find the area of the piece that remains. Take $\pi =\frac{22}{7}$.

The whole circle is

$$\frac{22}{7}\times 196=616{\text{ cm}}^2$$

and the removed quadrant is

$$\frac{1}{4}\times 616=154{\text{ cm}}^2$$

so the remaining area is

$$616-154=462{\text{ cm}}^2$$

Check: the remaining region is a $270^{\circ }$ sector, and $\frac{270}{360}\times 616=\frac{3}{4}\times 616=462$ cm². The two routes agree, and in any shaded-region question it is worth computing the answer both by subtracting and by taking the fraction directly.