Area of a Sector and Length of an Arc for Class 10

Sec­tors and arcs#

Def­i­n­i­tions#

Let $O$ be the cen­tre of a cir­cle and let $A$ and $B$ be two points on the cir­cle.

• An arc is a part of the cir­cle between two points. The shorter one is the minor arc $AB$; the longer one is the major arc $AB$.
• A sec­tor is the region bounded by two radii $OA$, $OB$ and the arc between them. The sec­tor on the minor arc is the minor sec­tor; the rest of the cir­cle is the major sec­tor.
• The angle of the sec­tor is the angle $\angle AOB = \theta$ at the cen­tre. If the minor sec­tor has angle $\theta$, the major sec­tor has angle $360^\circ - \theta$.

In this les­son "sec­tor" means the minor sec­tor unless the ques­tion says oth­er­wise.

The two key for­mu­las#

State­ment#

For a sec­tor of angle $\theta$ (in degrees) in a cir­cle of radius $r$:

$$\displaystyle \text{Length of arc} \; l = \frac{\theta}{360^\circ} \times 2\pi r$$

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

Why they are true: rea­son­ing by pro­por­tion#

Think of the whole cir­cle as a sec­tor with angle $360^\circ$. Its arc is the full cir­cum­fer­ence $2\pi r$ and its area is $\pi r^2$.

Now imag­ine the angle at the cen­tre divided into $360$ equal parts of $1^\circ$ each. The radii through those divi­sions cut the cir­cle into $360$ iden­ti­cal sec­tors, because turn­ing the cir­cle about $O$ car­ries each one exactly onto the next. Iden­ti­cal sec­tors have equal arcs and equal areas, so:

$$\displaystyle \begin{aligned} \text{arc of a } 1^\circ \text{ sector} &= \frac{2\pi r}{360^\circ} \\ \text{area of a } 1^\circ \text{ sector} &= \frac{\pi r^2}{360^\circ} \end{aligned}$$

A sec­tor of angle $\theta$ is made of $\theta$ of these small sec­tors (for a whole num­ber $\theta$; the same pro­por­tion holds for any angle). Mul­ti­ply­ing by $\theta$ gives the two for­mu­las. In short, arc length and sec­tor area are both pro­por­tional to the angle at the cen­tre, and the frac­tion of the cir­cle you have is $\displaystyle \frac{\theta}{360^\circ}$.

A use­ful link between the two#

Divide the area for­mula by the arc for­mula:

$$\displaystyle \frac{\text{Area of sector}}{l} = \frac{\frac{\theta}{360^\circ}\pi r^2}{\frac{\theta}{360^\circ}\, 2\pi r} = \frac{r}{2}$$

So $\displaystyle \text{Area of sector} = \frac{1}{2}\, l\, r$. This is handy when the arc length is given and the angle is not.

The major sec­tor#

You can find the major sec­tor in either of two ways; both give the same answer.

$$\displaystyle \begin{aligned} \text{Area of major sector} &= \pi r^2 - \text{area of minor sector} \\ &= \frac{360^\circ - \theta}{360^\circ} \times \pi r^2 \end{aligned}$$

Like­wise, the major arc has length $2\pi r - l$.

Method: find­ing an arc or a sec­tor#

1. Read off the radius $r$ and the angle $\theta$ at the cen­tre. If a time or a frac­tion of a turn is given, con­vert it into degrees first.
2. Write the frac­tion $\displaystyle \frac{\theta}{360^\circ}$ and sim­plify it (for exam­ple $\displaystyle \frac{60^\circ}{360^\circ} = \frac{1}{6}$).
3. Choose the for­mula: $2\pi r$ for a length, $\pi r^2$ for an area.
4. Sub­sti­tute the stated value of $\pi$ and can­cel before mul­ti­ply­ing.
5. Write the unit: cm or m for lengths, cm² or m² for areas.
6. If the major sec­tor or major arc is asked, sub­tract from the whole cir­cle.

Worked exam­ples#

Exam­ple 1: a basic sec­tor#

A cir­cle has radius $21$ cm. Find the length of the arc and the area of the sec­tor whose angle is $60^\circ$. Use $\displaystyle \pi = \frac{22}{7}$.

The frac­tion is $\displaystyle \frac{60^\circ}{360^\circ} = \frac{1}{6}$.

$$\displaystyle \begin{aligned} l &= \frac{1}{6} \times 2 \times \frac{22}{7} \times 21 \\ &= \frac{1}{6} \times 132 = 22 \text{ cm} \end{aligned}$$

$$\displaystyle \begin{aligned} \text{Area} &= \frac{1}{6} \times \frac{22}{7} \times 21 \times 21 \\ &= \frac{1}{6} \times 1386 = 231 \text{ cm}^2 \end{aligned}$$

Check with the link: $\displaystyle \frac{1}{2} \times 22 \times 21 = 231$. The arc is $22$ cm and the sec­tor area is $231$ cm².

Exam­ple 2: the minute hand of a clock#

The minute hand of a wall clock is $14$ cm long. Find the area swept by it in $20$ min­utes, and the dis­tance moved by its tip. Use $\displaystyle \pi = \frac{22}{7}$.

The minute hand turns through $360^\circ$ in $60$ min­utes, so through $6^\circ$ each minute. In $20$ min­utes it turns through $20 \times 6^\circ = 120^\circ$. The frac­tion is $\displaystyle \frac{120^\circ}{360^\circ} = \frac{1}{3}$.

$$\displaystyle \begin{aligned} \text{Area swept} &= \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 \\ &= \frac{1}{3} \times 616 = \frac{616}{3} \approx 205.33 \text{ cm}^2 \end{aligned}$$

The tip moves along the arc:

$$\displaystyle \begin{aligned} l &= \frac{1}{3} \times 2 \times \frac{22}{7} \times 14 \\ &= \frac{1}{3} \times 88 = \frac{88}{3} \approx 29.33 \text{ cm} \end{aligned}$$

The area swept is about $205.33$ cm² and the tip trav­els about $29.33$ cm.

Exam­ple 3: minor and major sec­tors with π = 3.14#

In a cir­cle of radius $10$ cm, a chord $AB$ makes an angle of $72^\circ$ at the cen­tre. Find the areas of the minor and major sec­tors and the length of the major arc. Use $\pi = 3.14$.

The frac­tion is $\displaystyle \frac{72^\circ}{360^\circ} = \frac{1}{5}$.

$$\displaystyle \begin{aligned} \text{Minor sector} &= \frac{1}{5} \times 3.14 \times 100 = 62.8 \text{ cm}^2 \\ \text{Whole circle} &= 3.14 \times 100 = 314 \text{ cm}^2 \\ \text{Major sector} &= 314 - 62.8 = 251.2 \text{ cm}^2 \end{aligned}$$

Check with the major angle: $\displaystyle \frac{288^\circ}{360^\circ} \times 314 = \frac{4}{5} \times 314 = 251.2$ cm². For the major arc:

$$\displaystyle \text{Major arc} = \frac{288^\circ}{360^\circ} \times 2 \times 3.14 \times 10 = \frac{4}{5} \times 62.8 = 50.24 \text{ cm}$$

The minor sec­tor is $62.8$ cm², the major sec­tor is $251.2$ cm² and the major arc is $50.24$ cm.

Exam­ple 4: two wind­screen wipers#

A bus wind­screen is cleaned by two wiper arms whose cleaned regions never meet. Each arm is $21$ cm long, piv­ots at one end and turns through $120^\circ$ in one stroke. Find the total area of glass cleaned in one stroke. Use $\displaystyle \pi = \frac{22}{7}$.

Each arm cleans a sec­tor of radius $21$ cm and angle $120^\circ$.

$$\displaystyle \begin{aligned} \text{One sector} &= \frac{120^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21 \\ &= \frac{1}{3} \times 1386 = 462 \text{ cm}^2 \end{aligned}$$

The two regions do not over­lap, so the total is $2 \times 462 = 924$ cm².

The total area cleaned is $924$ cm².

Exam­ple 5: work­ing back­wards from the arc#

An arc of a cir­cle of radius $28$ cm is $44$ cm long. Find the angle it makes at the cen­tre and the area of the sec­tor. Use $\displaystyle \pi = \frac{22}{7}$.

Put the known val­ues into the arc for­mula:

$$\displaystyle \begin{aligned} 44 &= \frac{\theta}{360^\circ} \times 2 \times \frac{22}{7} \times 28 \\ 44 &= \frac{\theta}{360^\circ} \times 176 \\ \frac{\theta}{360^\circ} &= \frac{44}{176} = \frac{1}{4} \\ \theta &= 90^\circ \end{aligned}$$

Now the area, using the link for­mula:

$$\displaystyle \text{Area} = \frac{1}{2} \times l \times r = \frac{1}{2} \times 44 \times 28 = 616 \text{ cm}^2$$

Check: $\displaystyle \frac{1}{4} \times \frac{22}{7} \times 28 \times 28 = \frac{1}{4} \times 2464 = 616$. The angle is $90^\circ$ and the sec­tor area is $616$ cm².

Com­mon mis­takes#

• Using $\pi r^2$ when a length is asked, or $2\pi r$ when an area is asked. Lengths use $2\pi r$; areas use $\pi r^2$.
• For­get­ting to con­vert min­utes of a clock into degrees. The minute hand turns $6^\circ$ per minute; the hour hand turns $30^\circ$ per hour, which is $0.5^\circ$ per minute.
• Using the diam­e­ter in place of the radius. If a diam­e­ter of $28$ cm is given, the radius is $14$ cm.
• Giv­ing the minor sec­tor when the major sec­tor is asked. Read which region is shaded.
• Mix­ing the two val­ues of $\pi$ in one ques­tion, or using a value other than the one stated.
• Writ­ing cm instead of cm² for an area, or leav­ing out units alto­gether.
• Round­ing too early. Keep frac­tions such as $\displaystyle \frac{616}{3}$ until the last step, then round to two dec­i­mal places.

Try these#

1. Find the arc length and the sec­tor area for $r = 7$ cm and $\theta = 45^\circ$. Use $\displaystyle \pi = \frac{22}{7}$. Answer: $5.5$ cm and $19.25$ cm².
2. The minute hand of a clock is $21$ cm long. Find the area it sweeps in $5$ min­utes. Use $\displaystyle \pi = \frac{22}{7}$. Answer: $115.5$ cm².
3. Find the area of the major sec­tor of a cir­cle of radius $10$ cm when the minor sec­tor has angle $36^\circ$. Use $\pi = 3.14$. Answer: $282.6$ cm².
4. A sec­tor of a cir­cle of radius $7$ cm has area $77$ cm². Find its angle. Use $\displaystyle \pi = \frac{22}{7}$. Answer: $180^\circ$ (a semi­cir­cle).
5. A gar­den sprin­kler waters up to $14$ m away while turn­ing through $45^\circ$. Find the area watered. Use $\displaystyle \pi = \frac{22}{7}$. Answer: $77$ m².

Key terms#

Arc

A part of a cir­cle between two of its points.

Minor arc

The shorter of the two arcs join­ing two points on a cir­cle.

Major arc

The longer of the two arcs join­ing two points on a cir­cle.

Sec­tor

The region bounded by two radii and the arc between them.

Angle of a sec­tor

The angle between the two bound­ing radii, mea­sured at the cen­tre.

Major sec­tor

The sec­tor with angle $360^\circ - \theta$ that remains when the minor sec­tor is removed.

Cir­cum­fer­ence

The length of the whole cir­cle, equal to $2\pi r$.

Com­mon ques­tions#

Why do we divide by 360?#

A full turn is $360^\circ$. The ratio $\displaystyle \frac{\theta}{360^\circ}$ tells you what frac­tion of the whole cir­cle the sec­tor is, and the arc and area are that same frac­tion of the full cir­cum­fer­ence and full area.

Should I use 22/7 or 3.14?#

Use whichever the ques­tion states. If it states nei­ther, $\displaystyle \frac{22}{7}$ usu­ally suits radii that are mul­ti­ples of 7 and $3.14$ suits radii that are mul­ti­ples of 10. Men­tion your choice in the answer.

Is the perime­ter of a sec­tor the same as its arc length?#

No. The perime­ter of a sec­tor is the arc plus the two radii: $l + 2r$. The arc alone is only the curved part.

How do I find the area if only the arc length and radius are given?#

Use $\displaystyle \text{Area} = \frac{1}{2} l r$. It fol­lows directly from the two sec­tor for­mu­las, so you do not need the angle.

Does the hour hand sweep the same way?#

Yes, the method is the same, but the hour hand is slower: it turns $30^\circ$ each hour. Con­vert the time into degrees and use the sec­tor for­mula with the length of the hour hand as the radius.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.
3. Kise­lev, A. P. Kise­lev's Geom­e­try, Book I: Planime­try (adapted by A. Given­tal). Sum­iz­dat.