Areas Related to Circles: Class 10 Maths · Chapter 11

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

40 min
1 simulation
23 practice questions
13 MCQs

What you will learn in Areas Related to Circles

Name the minor and major sectors and segments made by two radii or a chord, and find the angle of a major sector.
Derive the area of a sector and the length of an arc by the unitary method, and use them with π = 22/7 or 3.14.
Find the perimeter of a sector and use area = ½ × arc length × radius.
Find the area of triangle OAB from the perpendicular OM and trigonometric ratios for any central angle.
Find the areas of minor and major segments as differences of areas.
Solve everyday and board-exam problems on clock hands, grazing animals, sprinklers, wheels and designs.

Original companion text aligned to NCERT Mathematics Class 10, Chapter 11, pp. 154–160. Not affiliated with or endorsed by NCERT.

11.1 Areas of Sector and Segment of a Circle

Share a pizza with friends and the first question is always: how big is my slice? The minute hand of a clock sweeps a wedge of the dial every few minutes. A rangoli artist fills the curved strip between a straight chalk line and the rim of a circle. Each of these is a part of a circle, and this chapter is about measuring such parts.

You already have the tools. From Class 9 you know that a circle of radius $r$ has circumference $2 π r$ and area $π r^{2}$ , and you have met arcs and sectors. Here we name every part precisely and build each formula by the unitary method. Then we use the trigonometric ratios of Chapter 8 to find the triangle hidden inside a segment. That last step is the one board questions test most.

Sectors and segments

Draw a circle with centre $O$ and mark two points $A$ and $B$ on it. The radii $O A$ and $O B$ cut the disc into two pieces, just as two cuts from the centre cut a pizza. Each piece is bounded by the two radii and one of the arcs $A B$ , and is called a sector (त्रिज्यखंड). The angle $∠ A O B$ at the centre is the angle of the sector.

The two sectors are usually of different sizes. The smaller one is the minor sector and the larger one is the major sector. The two angles at $O$ together make a full turn. So if the minor sector has angle $θ$ , the major sector has angle $360^{\circ} - θ$ , as steps 1 and 2 of Fig. 11.1 show.

Definition

Sector

The region of a disc bounded by two radii and the arc between their ends. Its angle is the angle between the two radii at the centre.

ExampleOne slice of a round pizza cut from the centre.

Now join $A$ and $B$ by a straight line instead. The chord $A B$ also splits the disc into two pieces. Each piece is bounded by the chord and one of the arcs, and is called a segment (वृत्तखंड). The smaller one is the minor segment and the larger one is the major segment (steps 3 and 4).

Definition

Segment

The region of a disc bounded by a chord and one of the two arcs that the chord cuts off.

ExampleThe piece that breaks off a round papad along a straight crack.

1/4Two radii make the minor sector, with angle $θ$ at $O$ .

Fig. 11.1Two radii make two sectors; one chord makes two segments.NCERT Figs. 11.1–11.2 · p. 154

A circle with centre O and two points A and B on its right side. Step 1 shades the minor sector, the narrow wedge between radii OA and OB, with its angle theta marked at O. Step 2 shades instead the major sector, the large wedge on the left, with the angle 360 degrees minus theta. Step 3 hides the radii, draws the chord AB as a dashed line and shades the thin minor segment between the chord and the short arc. Step 4 shades the major segment, everything on the other side of the chord.

One convention saves words. When a question says just “sector” or “segment”, it means the minor one; the larger piece is always called “major”. Notice also how the boundaries differ: a sector always has a corner at the centre, while the corners of a segment are the two ends of a chord.

Quick check

1 of 3

A round papad cracks along a straight line that does not pass through its centre. What are the two pieces called?

a minor and a major sectora minor and a major segmenttwo equal semicirclestwo arcs of the circle

Show answer

Answer (B) a minor and a major segment

The crack is a chord, and a chord cuts a disc into two segments. Sectors need two radii meeting at the centre. Equal semicircles appear only when the chord is a diameter, and this one misses the centre.

Area of a sector

How much of the disc does a sector hold? In Class 9 you saw why equal angles at the centre give equal slices. The quickest way to turn that into a formula is the unitary method, the one a shopkeeper uses to price a single item: find the area for one degree, then multiply. The starting point is the whole disc, which is a “sector” whose angle is a full turn.

A full turn of $360^{\circ}$ holds $π r^{2}$ , so one degree holds $\frac{1}{360}$ of it, that is $\frac{π r^{2}}{360}$ . A sector of $θ$ degrees is made of $θ$ such one-degree slices, as Fig. 11.2 shows. Its area is therefore $θ$ times $\frac{π r^{2}}{360}$ .

3/3A sector of $θ^{\circ}$ is $θ$ one-degree slices put together.

Fig. 11.2The unitary method: from the whole turn, to one degree, to

θ

degrees.NCERT Fig. 11.3 · p. 155

A shaded disc with centre O and short tick marks all round its rim, standing for equal slices. On the right, the first line says 360 degrees goes with pi r squared. Step 2 picks out one thin slice, drawn wider than one degree so it can be seen, and a second line says 1 degree goes with pi r squared over 360. Step 3 shades a sector of angle theta made of theta such slices, and a third line says theta degrees goes with theta over 360 times pi r squared.

Area of a sector

A = \frac{θ}{360} \times π r^{2}

(11.1)

where

$A$: area of the sector
$r$: radius of the circle
$θ$: angle of the sector, in degrees

Read the formula as a share: a sector takes the same fraction of the disc as its angle takes of a full turn. A quadrant ( $90^{\circ}$ ) is $\frac{90}{360} = \frac{1}{4}$ of the disc, and a $60^{\circ}$ sector is $\frac{1}{6}$ of it. For the major sector, either subtract the minor sector from the whole disc or put $360 - θ$ in place of $θ$ ; both give the same answer.

Area of the major sector

\begin{aligned} A^{'} & = π r^{2} - A \\ = \frac{360 - θ}{360} \times π r^{2} \end{aligned}

where

$A^{'}$: area of the major sector

Example 11.1NCERT Example 1 · p. 156

A pizza in five slices

A pizza of diameter 20 cm is cut from the centre into 5 equal slices. Find the area of one slice, and the area of pizza left after one slice is eaten. (Use $π = 3.14$ .)

2/2One slice is the minor sector; the rest is the major sector.

Fig. 11.3One slice is a minor sector; what is left is the major sector.NCERT Fig. 11.5 · p. 156

A round pizza with a crust, tomato and capsicum toppings, cut from the centre into five equal slices; labels say 5 equal slices, each 72 degrees, radius 10 cm. In step 2 the slice on the right, with its angle marked, slides outwards away from the rest. Labels now give one slice as 62.8 square centimetres and the rest as 251.2 square centimetres.

Try it yourself first. Then reveal the solution one step at a time.

Length of an arc

The same reasoning measures the curved edge of a sector, its arc (चाप). The whole circle, all $360^{\circ}$ of it, is $2 π r$ long. One degree at the centre gets $\frac{1}{360}$ of that length, and $θ$ degrees get $θ$ times as much. Imagine peeling the edge of the circle off and laying it flat, as in Fig. 11.4: the arc is a piece of that straight line in proportion to its angle.

3/3The arc is $\frac{120}{360} = \frac{1}{3}$ of that line.

Fig. 11.4An arc gets the same share of

2 π r

as its angle gets of

360^{\circ}

.NCERT p. 155

Top left, a circle with centre O and a sector of 120 degrees, its arc l drawn thick. Step 2 lays the whole edge of the circle flat as a straight line 2 pi r long, with ticks marked 0, 90, 180, 270 and 360 degrees. Step 3 thickens the first third of the line, from 0 to 120 degrees, labelled l, and shows the working l equals 120 over 360 times 2 pi r.

Length of an arc

l = \frac{θ}{360} \times 2 π r

(11.2)

where

$l$: length of the arc of a sector of angle $θ$

Board questions often ask for the perimeter of a sector, the length of its whole boundary. Walk round it: out along one radius, along the arc, and back along the other radius. So the perimeter is $2 r + l$ , not $l$ alone.

SimulationArc and angle at the centre

A circle with centre O and two draggable points A and B on it. The sector between them is shaded and its arc is drawn thick. Readouts show the angle AOB, the length of arc AB, the chord AB and the radius in centimetres, with the arc-length formula worked out for the current numbers. Points snap to multiples of 5 degrees, and a grip on the circle changes the radius between 2.5 and 6.5 cm.

Fig. 11.5Drag

A

B

to change

∠ A O B

, or drag the grip on the circle to change

r

. The panel works out the arc with (11.2).

Try this

Make $∠ A O B = 90^{\circ}$ . Is the arc a quarter of $2 π r$ ?
Keep the angle fixed and change the radius. When $r$ grows by half, by how much does the arc grow?
Make the angle $180^{\circ}$ . What is the arc now, in terms of $r$ ?

Formulas (11.1) and (11.2) share the factor $\frac{θ}{360}$ , and that gives a useful link between them. Multiply the arc length by $\frac{r}{2}$ :

Area of a sector from its arc

A = \frac{1}{2} l r

(11.3)

Fig. 11.6 shows why this makes sense. Cut a sector into many thin slices and fit them together head to tail. They form a shape close to a parallelogram whose height is about $r$ and whose base is about half the arc. Its area is about $\frac{l}{2} \times r$ , and the thinner the slices, the better the fit. The result saves time when a question gives the arc length but not the angle.

3/3Turn every other slice upside down and fit them together: height about $r$ , base about $\frac{l}{2}$ .

Fig. 11.6Thin slices of a sector rearrange into a near-parallelogram: area

\approx \frac{l}{2} \times r

On the left, a sector with centre O, radius r and arc length l. Step 2 cuts it into six thin slices with lines from O. Step 3 shows the six slices on the right, placed side by side with every other slice upside down, so they make a shape close to a parallelogram. Its height is marked r and its base is marked l over 2, and the label says the area is about l over 2 times r.

Example

A level-crossing barrier

At a railway level crossing, the barrier arm is 4.2 m long and turns about a pivot at one end. After a train passes, the arm swings up from horizontal to vertical. (a) How far does its tip travel? (b) What area does the arm sweep?

3/3The arm sweeps a quadrant of radius 4.2 m.

Fig. 11.7The tip moves on an arc; the arm sweeps a quadrant.

A striped barrier arm 4.2 m long lies horizontally from a pivot on a short post with a warning lamp. In step 2 the arm turns up about the pivot until it is vertical, and the path of its tip is drawn as a dashed quarter circle labelled tip moves 6.6 m. In step 3 the quarter circle swept by the arm is shaded and labelled 13.86 square metres.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

A slice of a round cake of radius 7 cm has an angle of

45^{\circ}

at the centre. What is the area of the top of the slice, in cm²?

cm²

Show answer

Answer 19.25 cm²

\frac{45}{360} \times \frac{22}{7} \times 7^{2} = \frac{1}{8} \times 154 = 19.25 c m^{2}

. Using

2 π r

instead of

π r^{2}

gives 5.5, the length of the arc, not an area.

Area of a segment

A segment has no formula of its own: we find it by subtraction. Look at the minor segment cut off by the chord $A B$ in Fig. 11.8. Put the triangle $O A B$ next to it and together they make the whole sector $O A B$ . So the segment is what remains when the triangle is taken out of the sector.

Area of a minor segment

A_{seg} = A_{sec} - A_{△}

(11.4)

where

$A_{seg}$: area of the minor segment
$A_{sec}$: area of the sector, $\frac{θ}{360} π r^{2}$
$A_{△}$: area of triangle $O A B$

3/4Sector minus triangle leaves the minor segment.

Fig. 11.8Segment = sector − triangle; the major segment is the rest of the disc.NCERT Fig. 11.4 · p. 156

A circle with centre O and a chord AB across its upper part, with the angle theta marked at O. Step 1 shades the sector OAB. Step 2 shades the triangle OAB inside it with a dashed outline. Step 3 shades only the thin minor segment between the chord and the arc, with the label sector minus triangle OAB. Step 4 shades the major segment below the chord, labelled pi r squared minus the minor segment.

The major pieces come from the whole disc. The major sector is $π r^{2}$ minus the minor sector, and the major segment is $π r^{2}$ minus the minor segment. Be careful with one tempting shortcut: the major segment is the major sector plus triangle $O A B$ , not minus it, because the triangle lies on the major segment's side of the chord.

Finding the area of triangle OAB

Everything now rests on the triangle. Two of its sides are radii, so $O A = O B = r$ , and the angle between them is $θ$ . For $θ = 90^{\circ}$ it is easy: the radii are perpendicular, so they serve as base and height, and the area is $\frac{1}{2} r^{2}$ . For other angles we find a height using trigonometry.

DerivationArea of triangle OAB for an angle θ below 180°

Draw $O M$ perpendicular to the chord $A B$ (Fig. 11.9). The right triangles $O M A$ and $O M B$ have equal hypotenuses ( $O A = O B$ ) and share the side $O M$ , so they are congruent by RHS. Hence $A M = M B$ , and $O M$ halves the angle at $O$ : $∠ A O M = ∠ B O M = \frac{θ}{2}$ .

In the right triangle $O M A$ , the angle at $O$ is $\frac{θ}{2}$ and the hypotenuse is $r$ . The ratios of Chapter 8 give the two legs:

\begin{aligned} O M & = r \cos \frac{θ}{2} \\ A M & = r \sin \frac{θ}{2} \end{aligned}

Take $A B = 2 A M$ as the base and $O M$ as the height:

Area of triangle OAB

\begin{aligned} A_{△} & = \frac{1}{2} A B \cdot O M \\ = A M \cdot O M \\ = r^{2} \sin \frac{θ}{2} \cos \frac{θ}{2} \end{aligned}

(11.5)

3/3 $O M = r \cos \frac{θ}{2}$ and $A M = r \sin \frac{θ}{2}$ , so the area is $A M \times O M$ .

Fig. 11.9The perpendicular

O M

splits triangle

O A B

into two congruent right triangles.NCERT Fig. 11.7 · p. 157

An isosceles triangle OAB hangs from the centre O at the top, inside a dashed arc of the circle. Step 1 marks OA and OB as r and the angle theta at O. Step 2 draws the perpendicular OM from O to the midpoint M of AB, with a right-angle mark, equal tick marks on AM and MB, and the two halves of the angle marked theta over 2. Step 3 adds the results OM equals r cos theta over 2, AM equals r sin theta over 2, and the area of triangle OAB equals AM times OM.

Table 11.1Triangle

O A B

for the angles that come up most often

$θ$	$O M$	$A M$	Area of $△ O A B$
$60^{\circ}$	$\frac{\sqrt{3}}{2} r$	$\frac{1}{2} r$	$\frac{\sqrt{3}}{4} r^{2}$
$90^{\circ}$	$\frac{1}{\sqrt{2}} r$	$\frac{1}{\sqrt{2}} r$	$\frac{1}{2} r^{2}$
$120^{\circ}$	$\frac{1}{2} r$	$\frac{\sqrt{3}}{2} r$	$\frac{\sqrt{3}}{4} r^{2}$

Two quick routes are worth remembering. At $60^{\circ}$ the triangle is equilateral: $O A = O B$ forces equal base angles, and they must also be $60^{\circ}$ . Its area is $\frac{\sqrt{3}}{4} r^{2}$ straight away. At $90^{\circ}$ the triangle is half of a square drawn on the radius.

Example 11.2NCERT Example 2 · p. 157

Stained glass in a round window

A round window of radius 70 cm has a straight wooden bar $A B$ across it. The bar subtends $120^{\circ}$ at the centre $O$ , and the glass below the bar (the minor segment) is coloured. Find the area of the coloured glass. (Use $π = \frac{22}{7}$ and $\sqrt{3} = 1.732$ .)

2/2Radii $O A$ , $O B$ and the perpendicular $O M$ give the triangle.

Fig. 11.10The coloured glass is the minor segment cut off by the bar.NCERT Fig. 11.6 · p. 157

A round window with a thick frame and a horizontal wooden bar AB across its lower part; the glass below the bar is coloured and patterned, labelled stained glass. Step 2 draws the radii OA and OB with the 120 degree angle at the centre O and the perpendicular OM to the bar with a right-angle mark at M. Labels list OA = 70 cm, angle AOB = 120 degrees and OM = 35 cm.

Try it yourself first. Then reveal the solution one step at a time.

Example

A rangoli with a yellow strip

A circular rangoli has radius 20 cm. A straight line of white powder joins two points $A$ and $B$ on its edge with $∠ A O B = 60^{\circ}$ , and the strip between this line and the edge is filled with yellow. Find the area of (a) the yellow minor segment, (b) the major segment. (Use $π = 3.14$ and $\sqrt{3} = 1.73$ .)

2/3With $∠ A O B = 60^{\circ}$ , triangle $O A B$ has all three sides 20 cm.

Fig. 11.11At

60^{\circ}

the triangle

O A B

is equilateral.

A circular rangoli decorated with eight petals and a ring of dots. A straight line AB near the right edge cuts off a thin yellow strip, labelled minor segment. Step 2 shades the triangle OAB with the 60 degree angle at O and notes that OA, OB and AB all equal 20 cm, so the triangle is equilateral. Step 3 shades everything on the other side of AB, labelled major segment, about 1219.67 square centimetres.

Try it yourself first. Then reveal the solution one step at a time.

InteractiveSector explorer

A circle with centre O, a fixed point A and a draggable point B; dragging B or pressing the arrow keys changes the angle AOB in whole degrees, and buttons jump to 30, 45, 60, 90, 120 and 180 degrees. Four buttons pick the region to shade: the sector of angle theta, the other sector, the segment cut off by chord AB, or the other segment. A slider sets the radius from 1 to 42 cm and a switch chooses pi as 22 over 7 or 3.14. A panel works out the arc length, the sector area, OM, AM and the area of triangle OAB, and the area of the chosen region with the numbers put in, naming it minor or major. A Play button sweeps the angle from 0 to 360 degrees like a clock hand and can be paused.

Fig. 11.12Drag

B

(or use the arrow keys) to set the angle, then pick the region, the radius and the value of

π

. Press Play to sweep the angle like a clock hand.

Practice · 15 questions

Practice 11.1: questions with answers

Use $π = \frac{22}{7}$ unless a question says otherwise.

NumericalEasyNCERT Exercise 11.1, Q1
A sector has radius 9 cm and an angle of $70^{\circ}$ . How much area does it cover?
cm²
Show answer
Answer 49.5 cm²
$\frac{70}{360} \times \frac{22}{7} \times 81 = \frac{22 \times 81}{36} = 49.5 c m^{2}$ .
NumericalMediumNCERT Exercise 11.1, Q2
The border of a circular rangoli is a ribbon 132 cm long. One quadrant of the rangoli is filled with marigold petals. What area do the petals cover?
Show solution
Answer $346.5 c m^{2}$
$2 π r = 132$ , so $r = 132 \times \frac{7}{44} = 21$ cm.
A quadrant is a quarter of the disc: $\frac{1}{4} \times \frac{22}{7} \times 21^{2} = \frac{1}{4} \times 1386 = 346.5 c m^{2}$ .
3
NumericalMediumNCERT Exercise 11.1, Q3
The minute hand of a wall clock is 10.5 cm long. It moves from 10:10 to 10:30.
1. Through what angle does it turn, in degrees?
  degrees
  Show answer
  Answer 120 degrees
  It turns $6^{\circ}$ per minute, and $20 \times 6^{\circ} = 120^{\circ}$ .
2. What area of the dial does it sweep, in cm²?
  cm²
  Show answer
  Answer 115.5 cm²
  $\frac{120}{360} \times \frac{22}{7} \times {10.5}^{2} = \frac{1}{3} \times 346.5 = 115.5 c m^{2}$ .
3. How far does the tip of the hand travel, in cm?
  cm
  Show answer
  Answer 22 cm
  $\frac{1}{3} \times 2 \times \frac{22}{7} \times 10.5 = \frac{1}{3} \times 66 = 22 c m$ .
4
NumericalMediumNCERT Exercise 11.1, Q4
In a circle of radius 8 cm, the radii $O A$ and $O B$ are at right angles, and $A B$ is joined. Use $π = 3.14$ .
1. Find the area of the minor segment, in cm².
  cm²
  Show answer
  Answer 18.24 cm²
  Sector $= \frac{1}{4} \times 3.14 \times 64 = 50.24$ , triangle $= \frac{1}{2} \times 8 \times 8 = 32$ , so the segment is $18.24 c m^{2}$ .
2. Find the area of the major sector, in cm².
  cm²
  Show answer
  Answer 150.72 cm²
  The whole disc is $3.14 \times 64 = 200.96$ , so the major sector is $200.96 - 50.24 = 150.72 c m^{2}$ .
5
NumericalMediumNCERT Exercise 11.1, Q5
An arc of a circle of radius 28 cm makes an angle of $60^{\circ}$ at the centre. Work out:
1. the length of the arc, in cm
  cm
  Show answer
  Answer 29.33 cm
  $\frac{60}{360} \times 2 \times \frac{22}{7} \times 28 = \frac{176}{6} = \frac{88}{3} \approx 29.33 c m$ .
2. the area of the sector, in cm²
  cm²
  Show answer
  Answer 410.67 cm²
  $\frac{1}{6} \times \frac{22}{7} \times 784 = \frac{2464}{6} = \frac{1232}{3} \approx 410.67 c m^{2}$ .
3. the area of the minor segment (exactly, then with $\sqrt{3} = 1.732$ )
  Show solution
  Answer $(\frac{1232}{3} - 196 \sqrt{3}) c m^{2} \approx 71.19 c m^{2}$
  The equilateral triangle has side 28 cm, so its area is $\frac{\sqrt{3}}{4} \times 784 = 196 \sqrt{3} c m^{2}$ .
  Segment $= \frac{1232}{3} - 196 \sqrt{3} \approx 410.667 - 339.472 \approx 71.19 c m^{2}$ .
NumericalMediumNCERT Exercise 11.1, Q6
In a circle of radius 30 cm, the chord $A B$ makes $∠ A O B = 60^{\circ}$ . Work out the areas of both segments it creates, minor and major. (Use $π = 3.14$ and $\sqrt{3} = 1.73$ .)
Show solution
AnswerMinor: $81.75 c m^{2}$ ; major: $2744.25 c m^{2}$
Sector: $\frac{60}{360} \times 3.14 \times 900 = 471 c m^{2}$ .
The triangle is equilateral with side 30 cm: $\frac{1.73}{4} \times 900 = 389.25 c m^{2}$ .
Minor segment: $471 - 389.25 = 81.75 c m^{2}$ .
Disc: $3.14 \times 900 = 2826 c m^{2}$ , so the major segment is $2826 - 81.75 = 2744.25 c m^{2}$ .
NumericalMediumNCERT Exercise 11.1, Q7
A round glass tabletop has radius 60 cm. A gold strip fills the minor segment cut off by a chord $A B$ with $∠ A O B = 120^{\circ}$ . How large is the strip? (Use $π = 3.14$ and $\sqrt{3} = 1.73$ .)
Show solution
Answer $2211 c m^{2}$
Sector: $\frac{120}{360} \times 3.14 \times 3600 = 3768 c m^{2}$ .
$O M = 60 \cos 60^{\circ} = 30$ cm and $A M = 60 \sin 60^{\circ} = 30 \sqrt{3}$ cm.
Triangle: $A M \times O M = 900 \sqrt{3} = 900 \times 1.73 = 1557 c m^{2}$ .
Strip: $3768 - 1557 = 2211 c m^{2}$ .
8
NumericalMediumNCERT Exercise 11.1, Q8
A goat grazes in a rectangular field 20 m by 15 m. Its 6 m rope is fixed to a peg in one corner of the field. (Use $π = 3.14$ .)
Fig. 11.13The corner is a right angle, so the goat grazes a quadrant.NCERT Fig. 11.8 · p. 158
A rectangular field 20 m long and 15 m wide. A peg sits at the bottom left corner with a rope to a shaded quarter circle of radius 6 m, the area the goat can graze. A dashed quarter circle of radius 9 m shows the larger grazing area with a longer rope.
1. What area can the goat graze, in m²?
  m²
  Show answer
  Answer 28.26 m²
  A quadrant of radius 6 m: $\frac{1}{4} \times 3.14 \times 36 = 28.26 m^{2}$ .
2. The owner swaps the rope for a 9 m one. By how much does the grazing area grow, in m²?
  m²
  Show answer
  Answer 35.325 m²
  With 9 m (still shorter than both sides): $\frac{1}{4} \times 3.14 \times 81 = 63.585 m^{2}$ . Increase: $63.585 - 28.26 = 35.325 m^{2}$ .
9
NumericalMediumNCERT Exercise 11.1, Q9
For a school project, Kabir bends brass wire into a model cart wheel: a rim of diameter 42 mm and 4 straight wires that run right across it through the centre, cutting the wheel into 8 equal parts.
1. Find the total length of wire, in mm.
  mm
  Show answer
  Answer 300 mm
  Circumference $= \frac{22}{7} \times 42 = 132$ mm, and four diameters add $4 \times 42 = 168$ mm: $300 m m$ in all.
2. Find the area of each part, in mm².
  mm²
  Show answer
  Answer 173.25 mm²
  Each part is a sector of angle $45^{\circ}$ : $\frac{1}{8} \times \frac{22}{7} \times 21^{2} = \frac{1386}{8} = 173.25 m m^{2}$ .
NumericalEasyNCERT Exercise 11.1, Q10
A prize wheel at a mela is a flat disc of radius 35 cm, divided into prize sections by 10 equally spaced lines from the centre. Find the area of one section.
cm²
Show answer
Answer 385 cm²
Each section is a $36^{\circ}$ sector: $\frac{1}{10} \times \frac{22}{7} \times 35^{2} = \frac{3850}{10} = 385 c m^{2}$ .
NumericalMediumNCERT Exercise 11.1, Q11
Two lawn sprinklers each throw water up to 4.2 m while turning through $100^{\circ}$ . Their wet regions do not overlap. Find the total area watered in one sweep.
m²
Show answer
Answer 30.8 m²
One sector: $\frac{100}{360} \times \frac{22}{7} \times {4.2}^{2} = \frac{5}{18} \times 55.44 = 15.4 m^{2}$ . Two: $30.8 m^{2}$ .
NumericalEasyNCERT Exercise 11.1, Q12
A floodlight on a railway platform throws light 30 m ahead, spread across an angle of $70^{\circ}$ . What area of the platform and track does it light? (Use $π = 3.14$ .)
m²
Show answer
Answer 549.5 m²
$\frac{70}{360} \times 3.14 \times 900 = \frac{7}{36} \times 2826 = 549.5 m^{2}$ .
Long answerHardNCERT Exercise 11.1, Q13
A round brass plate of radius 21 cm has a regular hexagon engraved on it, with its six corners on the rim. The six segments between the hexagon and the rim are enamelled at ₹0.50 per cm². Find the cost. (Use $\sqrt{3} = 1.73$ .)
Fig. 11.14Each side of the hexagon cuts off one segment.NCERT Fig. 11.11 · p. 159
A round brass plate of radius 21 cm with a regular hexagon drawn inside it, its six corners on the rim. The six curved pieces between the sides of the hexagon and the rim are shaded and labelled enamelled segments; a dashed radius is marked 21 cm.
Show solution
AnswerAbout ₹120.80
Joining $O$ to the six corners makes six triangles with $60^{\circ}$ at $O$ , each equilateral with side 21 cm.
One segment: $\frac{1}{6} \times \frac{22}{7} \times 441 - \frac{1.73}{4} \times 441 = 231 - 190.7325 = 40.2675 c m^{2}$ .
Six segments: $6 \times 40.2675 = 241.605 c m^{2}$ .
Cost: $241.605 \times 0.50 = 120.8025$ , that is about ₹120.80.
14
Quick answerMediumNCERT Exercise 11.1, Q14
Which of these give the area of a sector of angle $x^{\circ}$ in a circle of radius $a$ ? Choose all that apply.
$\frac{x}{360} \times π a^{2}$ $\frac{x}{180} \times π a$ $\frac{x}{720} \times 2 π a^{2}$ $\frac{x}{180} \times π a^{2}$ $\frac{1}{2} \times a \times \frac{π a x}{180}$
Select all that apply
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Answer
(A) $\frac{x}{360} \times π a^{2}$
(C) $\frac{x}{720} \times 2 π a^{2}$
(E) $\frac{1}{2} \times a \times \frac{π a x}{180}$
A is (11.1). C is the same, since $\frac{2}{720} = \frac{1}{360}$ . E is $\frac{1}{2} \times r \times l$ with arc $l = \frac{x}{360} \times 2 π a = \frac{π a x}{180}$ . B is the arc length itself, and D is twice the area.
Quick answerMedium
Circle 1 has radius $r$ and an arc of $40^{\circ}$ . Circle 2 has radius $2 r$ and an arc of $20^{\circ}$ . The arc in circle 1 is the longer one.
TrueFalse
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Answer False
By (11.2) the arcs are $\frac{40}{360} \times 2 π r$ and $\frac{20}{360} \times 2 π (2 r)$ , which are equal. Halving the angle and doubling the radius cancel out.

11.2 Summary

Three results carry the whole chapter. Two are a fraction $\frac{θ}{360}$ of a whole-circle quantity, and the third is a difference of two areas.

3/3Segment: sector minus triangle $O A B .$

Fig. 11.15Arc, sector and segment: the three formulas of this chapter.

A circle with centre O and a sector at its upper right. Step 1 thickens the arc and shows the arc-length formula, l equals theta over 360 times 2 pi r. Step 2 shades the sector and adds its area formula, theta over 360 times pi r squared. Step 3 shades only the segment between the chord and the arc and adds the rule: segment equals sector minus triangle OAB.

Recap

Areas Related to Circles: key points and summary

A sector is bounded by two radii and an arc; a segment by a chord and an arc. Each comes as a minor and a major piece, and a lone “sector” or “segment” means the minor one (Sectors and segments).
If the minor sector has angle $θ$ , the major sector has angle $360^{\circ} - θ$ (Sectors and segments).
By the unitary method, the area of a sector is $\frac{θ}{360} \times π r^{2}$ ((11.1)).
The length of an arc is $\frac{θ}{360} \times 2 π r$ , and the perimeter of a sector is $2 r + l$ ((11.2)).
When the arc length is known, the area of a sector is $\frac{1}{2} l r$ ((11.3)).
Area of a minor segment = area of the sector − area of triangle $O A B$ ((11.4)).
Major sector $= π r^{2} -$ minor sector, and major segment $= π r^{2} -$ minor segment (Area of a segment).
With $O M ⊥ A B$ : $O M = r \cos \frac{θ}{2}$ , $A M = r \sin \frac{θ}{2}$ and triangle $O A B = A M \times O M$ ((11.5)).
At $60^{\circ}$ and $120^{\circ}$ the triangle has area $\frac{\sqrt{3}}{4} r^{2}$ ; at $90^{\circ}$ it is $\frac{1}{2} r^{2}$ (Table 11.1).
Use the value of $π$ and $\sqrt{3}$ that the question gives (Key idea).

Flash cards · 8

Formula cards

1 / 8

Quick revision

1 of 2

Match each quantity with its formula.

Tap an item on the left, then its match on the right.

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Answer

Arc length : $\frac{θ}{360} \times 2 π r$
Sector area : $\frac{θ}{360} \times π r^{2}$
Perimeter of a sector : $2 r + l$
Segment area : sector − triangle

Arc length and perimeter are lengths, so they use

r

; sector area uses

r^{2}

; a segment is a difference of two areas.

Board-exam practice

These questions follow the patterns of the CBSE board paper: one-mark MCQs, an assertion–reason item, and 2-, 3-, 4- and 5-mark questions. Show the formula, the substitution and the units in every written answer.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 11.1, Q1	Practice Q1
NCERT Exercise 11.1, Q2	Practice Q2
NCERT Exercise 11.1, Q3	Practice Q3
NCERT Exercise 11.1, Q4	Practice Q4
NCERT Exercise 11.1, Q5	Practice Q5
NCERT Exercise 11.1, Q6	Practice Q6
NCERT Exercise 11.1, Q7	Practice Q7
NCERT Exercise 11.1, Q8	Practice Q8
NCERT Exercise 11.1, Q9	Practice Q9
NCERT Exercise 11.1, Q10	Practice Q10
NCERT Exercise 11.1, Q11	Practice Q11
NCERT Exercise 11.1, Q12	Practice Q12
NCERT Exercise 11.1, Q13	Practice Q13
NCERT Exercise 11.1, Q14	Practice Q14

Quick check: MCQs on Areas Related to Circles

Sectors and segments3 questions
Length of an arc4 questions
Finding the area of triangle OAB4 questions
Quick revision2 questions

Glossary · 9

Key terms (English–Hindi)

arcचापchaap: A part of a circle between two of its points. An arc of angle $θ$ has length $\frac{θ}{360} \times 2 π r$ .First seen in the chapter →

chordजीवाjeeva: A line segment joining two points of a circle.First seen in the chapter →

major sectorदीर्घ त्रिज्यखंडdeergh trijyakhand: The larger of the two sectors made by two radii, with angle $360^{\circ} - θ$ .First seen in the chapter →
major segmentदीर्घ वृत्तखंडdeergh vrittakhand: The larger segment cut off by a chord; its area is $π r^{2}$ minus the minor segment.First seen in the chapter →
minor sectorलघु त्रिज्यखंडlaghu trijyakhand: The smaller of the two sectors made by two radii; its angle $θ$ is less than $180^{\circ}$ .First seen in the chapter →
minor segmentलघु वृत्तखंडlaghu vrittakhand: The smaller segment cut off by a chord; its area is the sector minus triangle $O A B$ .First seen in the chapter →

quadrantचतुर्थांशchaturthansh: A sector of angle $90^{\circ}$ , one quarter of a disc.First seen in the chapter →

sectorत्रिज्यखंडtrijyakhand: The region of a disc between two radii and the arc joining their ends.First seen in the chapter →
segmentवृत्तखंडvrittakhand: The region of a disc between a chord and one of the arcs it cuts off.First seen in the chapter →

Areas Related to Circles: Class 10 Maths · Chapter 11

What you will learn in Areas Related to Circles

11.1 Areas of Sector and Segment of a Circle

Sectors and segments

Area of a sector

Length of an arc

Area of a segment

Finding the area of triangle OAB

Practice 11.1: questions with answers

11.2 Summary

Areas Related to Circles: key points and summary

Board-exam practice

Practice questions with answers (NCERT “Board-exam practice”)

Find your NCERT exercise question

Quick check: MCQs on Areas Related to Circles

Key terms (English–Hindi)

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