Areas Related to Circles: Class 10 Maths Chapter 11 Hinglish mein

Mathematics · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

40 min
1 simulation
23 practice questions
13 MCQs

Is chapter mein aap kya seekhenge

Do radii ya ek chord se bane minor aur major sectors aur segments ke naam batao, aur major sector ka angle nikaalo.
Unitary method se sector ka area aur arc ki length derive karo, aur unhe π = 22/7 ya 3.14 ke saath use karo.
Sector ka perimeter nikaalo aur area = ½ × arc length × radius use karo.
Kisi bhi central angle ke liye perpendicular OM aur trigonometric ratios se triangle OAB ka area nikaalo.
Minor aur major segments ke areas ko areas ke difference ke roop mein nikaalo.
Clock hands, charte hue jaanwar, sprinklers, wheels aur designs par rozmarra ke aur board-exam wale problems solve karo.

NCERT Mathematics Class 10, Chapter 11, pp. 154–160 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

11.1 Circle ke Sector aur Segment ke Areas

Doston ke saath pizza share karo to pehla sawaal hamesha yahi hota hai: meri slice kitni badi hai? Clock ki minute hand har kuch minute mein dial ka ek wedge ghoom leti hai. Rangoli banane wala artist seedhi chalk line aur circle ke rim ke beech ki curved strip bharta hai. Inmein se har ek circle ka ek hissa hai, aur yeh chapter aise hisson ko naapne ke baare mein hai.

Tools tumhare paas pehle se hain. Class 9 se tum jaante ho ki radius $r$ wale circle ki circumference $2 π r$ aur area $π r^{2}$ hota hai, aur tum arcs aur sectors se mil chuke ho. Yahan hum har hisse ko sahi-sahi naam denge aur har formula unitary method se banayenge. Phir Chapter 8 ke trigonometric ratios use karke segment ke andar chhupa triangle dhoondhenge. Board questions sabse zyada isi aakhri step ko test karte hain.

Sectors aur segments

Centre $O$ wala ek circle banao aur us par do points $A$ aur $B$ mark karo. Radii $O A$ aur $O B$ disc ko do tukdon mein kaat dete hain, bilkul waise jaise centre se do cuts pizza ko kaatte hain. Har tukda do radii aur arcs $A B$ mein se ek se ghira hota hai, aur use sector (त्रिज्यखंड) kehte hain. Centre par angle $∠ A O B$ sector ka angle hai.

Dono sectors aksar alag-alag size ke hote hain. Chhota wala minor sector hai aur bada wala major sector. $O$ par dono angles milkar ek full turn banate hain. Isliye agar minor sector ka angle $θ$ hai, to major sector ka angle $360^{\circ} - θ$ hai, jaise Fig. 11.1 ke steps 1 aur 2 dikhate hain.

Definition

Sector

Disc ka woh region jo do radii aur unke siron ke beech ke arc se ghira ho. Iska angle centre par do radii ke beech ka angle hai.

ExampleGol pizza ki centre se kati ek slice.

Ab $A$ aur $B$ ko iske bajaye ek seedhi line se jodo. Chord $A B$ bhi disc ko do tukdon mein baant deti hai. Har tukda chord aur arcs mein se ek se ghira hota hai, aur use segment (वृत्तखंड) kehte hain. Chhota wala minor segment hai aur bada wala major segment (steps 3 aur 4).

Definition

Segment

Disc ka woh region jo ek chord aur us chord se kate do arcs mein se ek se ghira ho.

ExampleGol papad ka woh tukda jo seedhi daraar ke saath toot kar alag hota hai.

1/4Do radii minor sector banate hain, jiska $O$ par angle $θ$ hai.

Fig. 11.1Do radii do sectors banate hain; ek chord do segments banati hai.NCERT Figs. 11.1–11.2 · p. 154

Centre O wala ek circle, jiske right side par do points A aur B hain. Step 1 minor sector ko shade karta hai, yaani radii OA aur OB ke beech ka patla wedge, jiska angle theta O par mark hai. Step 2 iske bajaye major sector shade karta hai, yaani left ka bada wedge, jiska angle 360 degrees minus theta hai. Step 3 radii ko chhupa deta hai, chord AB ko dashed line se banata hai aur chord aur chhote arc ke beech ka patla minor segment shade karta hai. Step 4 major segment shade karta hai, yaani chord ke doosri taraf ka sab kuch.

Ek convention se shabd bachte hain. Jab question sirf “sector” ya “segment” kahe, to uska matlab minor wala hai; bade tukde ko hamesha “major” kehte hain. Yeh bhi dekho ki boundaries kaise alag hain: sector ka ek corner hamesha centre par hota hai, jabki segment ke corners chord ke do sire hote hain.

Quick check

3 mein se 1

Ek gol papad ek seedhi line ke saath tootta hai jo uske centre se nahi guzarti. Dono tukdon ko kya kehte hain?

ek minor aur ek major sectorek minor aur ek major segmentdo barabar semicirclescircle ke do arcs

Answer dekho

Answer (B) ek minor aur ek major segment

Daraar ek chord hai, aur chord disc ko do segments mein kaatti hai. Sectors ke liye centre par milne wale do radii chahiye. Barabar semicircles tabhi bante hain jab chord diameter ho, aur yeh wali centre se nahi guzarti.

Sector ka area

Sector disc ka kitna hissa rakhta hai? Class 9 mein tumne dekha tha ki centre par barabar angles barabar slices kyun dete hain. Ise formula mein badalne ka sabse tez tareeka hai unitary method, wahi jo dukaandaar ek item ki keemat nikaalne ke liye use karta hai: ek degree ka area nikaalo, phir multiply karo. Shuruaat poori disc se hoti hai, jo ek aisa “sector” hai jiska angle full turn hai.

$360^{\circ}$ ka full turn $π r^{2}$ rakhta hai, isliye ek degree uska $\frac{1}{360}$ rakhti hai, yaani $\frac{π r^{2}}{360}$ . $θ$ degrees ka sector aisi $θ$ one-degree slices se bana hai, jaise Fig. 11.2 dikhata hai. Isliye uska area $\frac{π r^{2}}{360}$ ka $θ$ guna hai.

3/3 $θ^{\circ}$ ka sector $θ$ one-degree slices ko jodkar banta hai.

Fig. 11.2Unitary method: poore turn se, ek degree tak, phir

θ

degrees tak.NCERT Fig. 11.3 · p. 155

Centre O wali ek shaded disc, jiske rim par chaaron taraf chhote tick marks hain jo barabar slices dikhate hain. Right mein pehli line kehti hai ki 360 degrees ke saath pi r squared aata hai. Step 2 ek patli slice alag dikhata hai, jo dikhne ke liye ek degree se chaudi banayi gayi hai, aur doosri line kehti hai ki 1 degree ke saath pi r squared over 360 aata hai. Step 3 angle theta ka ek sector shade karta hai jo aisi theta slices se bana hai, aur teesri line kehti hai ki theta degrees ke saath theta over 360 times pi r squared aata hai.

Sector ka area

A = \frac{θ}{360} \times π r^{2}

(11.1)

jahan

$A$: sector ka area
$r$: circle ka radius
$θ$: sector ka angle, degrees mein

Formula ko ek hisse ki tarah padho: sector disc ka utna hi fraction leta hai jitna uska angle full turn ka leta hai. Ek quadrant ( $90^{\circ}$ ) disc ka $\frac{90}{360} = \frac{1}{4}$ hai, aur $60^{\circ}$ ka sector uska $\frac{1}{6}$ hai. Major sector ke liye, ya to poori disc mein se minor sector ghatao ya $θ$ ki jagah $360 - θ$ rakho; dono se same answer aata hai.

Major sector ka area

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

jahan

$A^{'}$: major sector ka area

Example 11.1NCERT Example 1 · p. 156

Paanch slices mein ek pizza

20 cm diameter wale ek pizza ko centre se 5 barabar slices mein kaata gaya hai. Ek slice ka area nikaalo, aur ek slice khane ke baad bache pizza ka area bhi nikaalo. ( $π = 3.14$ use karo.)

2/2Ek slice minor sector hai; baaki major sector hai.

Fig. 11.3Ek slice minor sector hai; jo bacha hai woh major sector hai.NCERT Fig. 11.5 · p. 156

Crust, tomato aur capsicum toppings wala ek gol pizza, centre se paanch barabar slices mein kata hua; labels kehte hain 5 equal slices, sab 72 degrees, radius 10 cm. Step 2 mein right wali slice, jiska angle mark hai, baaki pizza se door bahar ki taraf khisakti hai. Ab labels ek slice ko 62.8 square centimetres aur baaki ko 251.2 square centimetres batate hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Arc ki length

Yahi reasoning sector ke curved edge, yaani uske arc (चाप) ko bhi naapti hai. Poora circle, uske saare $360^{\circ}$ , $2 π r$ lamba hai. Centre par ek degree ko us length ka $\frac{1}{360}$ milta hai, aur $θ$ degrees ko uska $θ$ guna. Socho ki circle ka edge chheel kar seedha bichha diya, jaise Fig. 11.4 mein: arc us seedhi line ka ek tukda hai, apne angle ke proportion mein.

3/3Arc us line ka $\frac{120}{360} = \frac{1}{3}$ hai.

Fig. 11.4Arc ko

2 π r

ka utna hi hissa milta hai jitna uske angle ko

360^{\circ}

ka milta hai.NCERT p. 155

Upar left mein centre O wala ek circle aur 120 degrees ka ek sector, jiska arc l mota banaya gaya hai. Step 2 circle ke poore edge ko 2 pi r lambi seedhi line ki tarah bichha deta hai, jis par 0, 90, 180, 270 aur 360 degrees ke ticks hain. Step 3 line ka pehla teesra hissa, 0 se 120 degrees tak, mota karta hai, jis par l likha hai, aur working dikhata hai: l equals 120 over 360 times 2 pi r.

Arc ki length

l = \frac{θ}{360} \times 2 π r

(11.2)

jahan

$l$: angle $θ$ wale sector ke arc ki length

Board questions aksar sector ka perimeter poochte hain, yaani uski poori boundary ki length. Uske chaaron taraf chalo: ek radius ke saath bahar, arc ke saath, aur doosre radius ke saath wapas. Isliye perimeter $2 r + l$ hai, sirf $l$ nahi.

SimulationArc aur centre par angle

Centre O wala ek circle jis par do draggable points A aur B hain. Unke beech ka sector shaded hai aur uska arc mota banaya gaya hai. Readouts angle AOB, arc AB ki length, chord AB aur radius centimetres mein dikhate hain, aur current numbers ke liye arc-length formula ki working bhi. Points 5 degrees ke multiples par snap hote hain, aur circle par ek grip radius ko 2.5 se 6.5 cm ke beech badalti hai.

Fig. 11.5

∠ A O B

badalne ke liye

A

B

ko drag karo, ya

r

badalne ke liye circle par grip ko drag karo. Panel (11.2) se arc calculate karta hai.

Try karo

$∠ A O B = 90^{\circ}$ banao. Kya arc $2 π r$ ka ek quarter hai?
Angle fix rakho aur radius badlo. Jab $r$ aadha aur badhta hai, to arc kitna badhta hai?
Angle $180^{\circ}$ banao. Ab arc $r$ ke terms mein kya hai?

Formulas (11.1) aur (11.2) mein factor $\frac{θ}{360}$ common hai, aur isse dono ke beech ek kaam ka link milta hai. Arc length ko $\frac{r}{2}$ se multiply karo:

Arc se sector ka area

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

(11.3)

Fig. 11.6 dikhata hai ki yeh kyun sahi baithta hai. Ek sector ko bahut saari patli slices mein kaato aur unhe ulta-seedha jod do. Woh lagbhag ek parallelogram jaisa shape banati hain jiski height lagbhag $r$ aur base arc ka lagbhag aadha hai. Uska area lagbhag $\frac{l}{2} \times r$ hai, aur slices jitni patli, fit utna behtar. Jab question arc length de par angle na de, tab yeh result time bachata hai.

3/3Har doosri slice ko ulta karo aur sabko jod do: height lagbhag $r$ , base lagbhag $\frac{l}{2}$ .

Fig. 11.6Sector ki patli slices ko rearrange karke lagbhag parallelogram banta hai: area

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

Left mein centre O, radius r aur arc length l wala ek sector. Step 2 use O se lines kheench kar chhe patli slices mein kaatta hai. Step 3 right mein chhe slices ko ek ke baad ek rakhta hai, har doosri slice ulti, taaki woh lagbhag parallelogram jaisa shape banayein. Uski height r mark hai aur base l over 2, aur label kehta hai ki area lagbhag l over 2 times r hai.

Example

Level-crossing ka barrier

Railway level crossing par barrier arm 4.2 m lamba hai aur ek sire par pivot ke around ghoomta hai. Train guzarne ke baad arm horizontal se vertical tak upar uthta hai. (a) Uski tip kitni door chalti hai? (b) Arm kitna area sweep karta hai?

3/3Arm 4.2 m radius ka ek quadrant sweep karta hai.

Fig. 11.7Tip ek arc par chalti hai; arm ek quadrant sweep karta hai.

4.2 m lamba dhaariyon wala barrier arm ek chhote post par lage pivot se horizontal pada hai, post par warning lamp hai. Step 2 mein arm pivot ke around upar ghoomta hai jab tak vertical na ho jaaye, aur uski tip ka raasta dashed quarter circle se dikhaya gaya hai jis par tip 6.6 m chali likha hai. Step 3 mein arm se sweep hua quarter circle shaded hai aur us par 13.86 square metres likha hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

7 cm radius wale gol cake ki ek slice ka centre par angle

45^{\circ}

hai. Slice ke top ka area kitna hai, cm² mein?

cm²

Answer dekho

Answer 19.25 cm²

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

π r^{2}

ki jagah

2 π r

use karne se 5.5 aata hai, jo arc ki length hai, area nahi.

Segment ka area

Segment ka apna koi formula nahi hai: hum use subtraction se nikaalte hain. Fig. 11.8 mein chord $A B$ se kata minor segment dekho. Uske saath triangle $O A B$ rakho aur dono milkar poora sector $O A B$ banate hain. Isliye segment woh hai jo sector mein se triangle nikaalne ke baad bachta hai.

Minor segment ka area

A_{seg} = A_{sec} - A_{△}

(11.4)

jahan

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

3/4Sector minus triangle se minor segment bachta hai.

Fig. 11.8Segment = sector − triangle; major segment baaki disc hai.NCERT Fig. 11.4 · p. 156

Centre O wala ek circle jiske upar wale hisse mein chord AB hai, aur O par angle theta mark hai. Step 1 sector OAB shade karta hai. Step 2 uske andar triangle OAB ko dashed outline ke saath shade karta hai. Step 3 sirf chord aur arc ke beech ka patla minor segment shade karta hai, jis par label hai sector minus triangle OAB. Step 4 chord ke neeche ka major segment shade karta hai, jis par likha hai pi r squared minus the minor segment.

Major tukde poori disc se aate hain. Major sector $π r^{2}$ minus minor sector hai, aur major segment $π r^{2}$ minus minor segment. Ek tempting shortcut se saavdhaan raho: major segment major sector plus triangle $O A B$ hai, minus nahi, kyunki triangle chord ke usi taraf hai jidhar major segment hai.

Triangle OAB ka area nikaalna

Ab sab kuch triangle par tika hai. Uski do sides radii hain, isliye $O A = O B = r$ , aur unke beech ka angle $θ$ hai. $θ = 90^{\circ}$ ke liye yeh aasaan hai: radii perpendicular hain, isliye woh base aur height ka kaam karte hain, aur area $\frac{1}{2} r^{2}$ hai. Baaki angles ke liye hum trigonometry se height nikaalte hain.

Derivation180° se chhote angle θ ke liye triangle OAB ka area

Chord $A B$ par $O M$ perpendicular banao (Fig. 11.9). Right triangles $O M A$ aur $O M B$ ke hypotenuse barabar hain ( $O A = O B$ ) aur side $O M$ common hai, isliye woh RHS se congruent hain. Isliye $A M = M B$ , aur $O M$ , $O$ par angle ko aadha karta hai: $∠ A O M = ∠ B O M = \frac{θ}{2}$ .

Right triangle $O M A$ mein, $O$ par angle $\frac{θ}{2}$ hai aur hypotenuse $r$ hai. Chapter 8 ke ratios do legs dete hain:

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

$A B = 2 A M$ ko base aur $O M$ ko height lo:

Triangle OAB ka area

\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 $O M = r \cos \frac{θ}{2}$ aur $A M = r \sin \frac{θ}{2}$ , isliye area $A M \times O M$ hai.

Fig. 11.9Perpendicular

O M

triangle

O A B

ko do congruent right triangles mein baant deta hai.NCERT Fig. 11.7 · p. 157

Ek isosceles triangle OAB upar centre O se latka hai, circle ke dashed arc ke andar. Step 1 OA aur OB ko r aur O par angle theta mark karta hai. Step 2 O se AB ke midpoint M tak perpendicular OM banata hai, right-angle mark ke saath, AM aur MB par barabar tick marks, aur angle ke dono aadhe hisse theta over 2 mark hain. Step 3 results jodta hai: OM equals r cos theta over 2, AM equals r sin theta over 2, aur triangle OAB ka area equals AM times OM.

Table 11.1Sabse zyada aane wale angles ke liye triangle

O A B

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

Do shortcut raaste yaad rakhne layak hain. $60^{\circ}$ par triangle equilateral hai: $O A = O B$ se base angles barabar hote hain, aur woh bhi $60^{\circ}$ hi hone chahiye. Uska area seedha $\frac{\sqrt{3}}{4} r^{2}$ hai. $90^{\circ}$ par triangle radius par bane square ka aadha hai.

Example 11.2NCERT Example 2 · p. 157

Gol khidki mein stained glass

70 cm radius ki ek gol khidki mein ek seedhi wooden bar $A B$ lagi hai. Bar centre $O$ par $120^{\circ}$ subtend karti hai, aur bar ke neeche ka glass (minor segment) rangeen hai. Rangeen glass ka area nikaalo. ( $π = \frac{22}{7}$ aur $\sqrt{3} = 1.732$ use karo.)

2/2Radii $O A$ , $O B$ aur perpendicular $O M$ se triangle milta hai.

Fig. 11.10Rangeen glass bar se kata minor segment hai.NCERT Fig. 11.6 · p. 157

Mote frame wali ek gol khidki jiske neeche wale hisse mein horizontal wooden bar AB lagi hai; bar ke neeche ka glass rangeen aur patterned hai, jis par rangeen glass likha hai. Step 2 radii OA aur OB banata hai, centre O par 120 degree ka angle, aur bar par perpendicular OM, M par right-angle mark ke saath. Labels mein OA = 70 cm, angle AOB = 120 degrees aur OM = 35 cm likha hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Peeli strip wali rangoli

Ek circular rangoli ka radius 20 cm hai. Safed powder ki ek seedhi line uske edge par do points $A$ aur $B$ ko jodti hai, jahan $∠ A O B = 60^{\circ}$ hai, aur is line aur edge ke beech ki strip peele rang se bhari hai. (a) Peele minor segment, (b) major segment ka area nikaalo. ( $π = 3.14$ aur $\sqrt{3} = 1.73$ use karo.)

2/3 $∠ A O B = 60^{\circ}$ hone par triangle $O A B$ ki teeno sides 20 cm hain.

Fig. 11.11

60^{\circ}

par triangle

O A B

equilateral hai.

Aath pankhudiyon aur dots ki ek ring se saji circular rangoli. Right edge ke paas ek seedhi line AB ek patli peeli strip kaat kar alag karti hai, jis par minor segment likha hai. Step 2 triangle OAB shade karta hai, O par 60 degree ka angle hai, aur batata hai ki OA, OB aur AB teeno 20 cm ke barabar hain, isliye triangle equilateral hai. Step 3 AB ke doosri taraf ka sab kuch shade karta hai, jis par major segment likha hai, lagbhag 1219.67 square centimetres.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

InteractiveSector explorer

Centre O wala ek circle, ek fixed point A aur ek draggable point B; B ko drag karne ya arrow keys dabane se angle AOB poore degrees mein badalta hai, aur buttons 30, 45, 60, 90, 120 aur 180 degrees par le jaate hain. Chaar buttons shade karne ka region chunte hain: angle theta ka sector, doosra sector, chord AB se kata segment, ya doosra segment. Ek slider radius ko 1 se 42 cm tak set karta hai aur ek switch pi ko 22 over 7 ya 3.14 chunta hai. Ek panel arc length, sector area, OM, AM aur triangle OAB ka area, aur chune hue region ka area numbers rakh kar nikaalta hai, aur batata hai ki woh minor hai ya major. Play button angle ko clock hand ki tarah 0 se 360 degrees tak ghumata hai aur use pause bhi kiya ja sakta hai.

Fig. 11.12Angle set karne ke liye

B

ko drag karo (ya arrow keys use karo), phir region, radius aur

π

ki value chuno. Angle ko clock hand ki tarah ghumane ke liye Play dabao.

Practice · 15 questions

Practice 11.1: answers ke saath

$π = \frac{22}{7}$ use karo, jab tak question kuch aur na kahe.

NumericalEasyNCERT Exercise 11.1, Q1
Ek sector ka radius 9 cm aur angle $70^{\circ}$ hai. Woh kitna area cover karta hai?
cm²
Answer dekho
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
Ek circular rangoli ka border 132 cm lamba ribbon hai. Rangoli ka ek quadrant genda ki pankhudiyon se bhara hai. Pankhudiyan kitna area cover karti hain?
Solution dikhao
Answer $346.5 c m^{2}$
$2 π r = 132$ , isliye $r = 132 \times \frac{7}{44} = 21$ cm.
Quadrant disc ka ek quarter hai: $\frac{1}{4} \times \frac{22}{7} \times 21^{2} = \frac{1}{4} \times 1386 = 346.5 c m^{2}$ .
3
NumericalMediumNCERT Exercise 11.1, Q3
Ek wall clock ki minute hand 10.5 cm lambi hai. Woh 10:10 se 10:30 tak chalti hai.
1. Woh kitne degrees ke angle se ghoomti hai?
  degrees
  Answer dekho
  Answer 120 degrees
  Woh har minute $6^{\circ}$ ghoomti hai, aur $20 \times 6^{\circ} = 120^{\circ}$ .
2. Woh dial ka kitna area sweep karti hai, cm² mein?
  cm²
  Answer dekho
  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. Hand ki tip kitni door chalti hai, cm mein?
  cm
  Answer dekho
  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
8 cm radius wale circle mein radii $O A$ aur $O B$ right angle par hain, aur $A B$ ko joda gaya hai. $π = 3.14$ use karo.
1. Minor segment ka area nikaalo, cm² mein.
  cm²
  Answer dekho
  Answer 18.24 cm²
  Sector $= \frac{1}{4} \times 3.14 \times 64 = 50.24$ , triangle $= \frac{1}{2} \times 8 \times 8 = 32$ , isliye segment $18.24 c m^{2}$ hai.
2. Major sector ka area nikaalo, cm² mein.
  cm²
  Answer dekho
  Answer 150.72 cm²
  Poori disc $3.14 \times 64 = 200.96$ hai, isliye major sector $200.96 - 50.24 = 150.72 c m^{2}$ hai.
5
NumericalMediumNCERT Exercise 11.1, Q5
28 cm radius wale circle ka ek arc centre par $60^{\circ}$ ka angle banata hai. Nikaalo:
1. arc ki length, cm mein
  cm
  Answer dekho
  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. sector ka area, cm² mein
  cm²
  Answer dekho
  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. minor segment ka area (pehle exact, phir $\sqrt{3} = 1.732$ ke saath)
  Solution dikhao
  Answer $(\frac{1232}{3} - 196 \sqrt{3}) c m^{2} \approx 71.19 c m^{2}$
  Equilateral triangle ki side 28 cm hai, isliye uska area $\frac{\sqrt{3}}{4} \times 784 = 196 \sqrt{3} c m^{2}$ hai.
  Segment $= \frac{1232}{3} - 196 \sqrt{3} \approx 410.667 - 339.472 \approx 71.19 c m^{2}$ .
NumericalMediumNCERT Exercise 11.1, Q6
30 cm radius wale circle mein chord $A B$ se $∠ A O B = 60^{\circ}$ banta hai. Isse bane dono segments, minor aur major, ke areas nikaalo. ( $π = 3.14$ aur $\sqrt{3} = 1.73$ use karo.)
Solution dikhao
AnswerMinor: $81.75 c m^{2}$ ; major: $2744.25 c m^{2}$
Sector: $\frac{60}{360} \times 3.14 \times 900 = 471 c m^{2}$ .
Triangle equilateral hai, 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: $3.14 \times 900 = 2826 c m^{2}$ , isliye major segment $2826 - 81.75 = 2744.25 c m^{2}$ hai.
NumericalMediumNCERT Exercise 11.1, Q7
Ek gol glass tabletop ka radius 60 cm hai. Ek gold strip us minor segment ko bharti hai jo chord $A B$ se kata hai, jahan $∠ A O B = 120^{\circ}$ . Strip kitni badi hai? ( $π = 3.14$ aur $\sqrt{3} = 1.73$ use karo.)
Solution dikhao
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 aur $A M = 60 \sin 60^{\circ} = 30 \sqrt{3}$ 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
Ek bakri 20 m by 15 m ke rectangular field mein charti hai. Uski 6 m ki rassi field ke ek corner mein gadi khoonti (peg) se bandhi hai. ( $π = 3.14$ use karo.)
Fig. 11.13Corner ek right angle hai, isliye bakri ek quadrant mein charti hai.NCERT Fig. 11.8 · p. 158
20 m lamba aur 15 m chauda ek rectangular field. Neeche left corner par ek khoonti (peg) hai jisse rassi 6 m radius ke shaded quarter circle tak jaati hai, yahi woh area hai jo bakri char sakti hai. 9 m radius ka ek dashed quarter circle lambi rassi ke saath bada charne ka area dikhata hai.
1. Bakri kitna area char sakti hai, m² mein?
  m²
  Answer dekho
  Answer 28.26 m²
  6 m radius ka ek quadrant: $\frac{1}{4} \times 3.14 \times 36 = 28.26 m^{2}$ .
2. Maalik rassi badal kar 9 m ki laga deta hai. Charne ka area kitna badhta hai, m² mein?
  m²
  Answer dekho
  Answer 35.325 m²
  9 m ke saath (jo ab bhi dono sides se chhoti hai): $\frac{1}{4} \times 3.14 \times 81 = 63.585 m^{2}$ . Badhat: $63.585 - 28.26 = 35.325 m^{2}$ .
9
NumericalMediumNCERT Exercise 11.1, Q9
School project ke liye Kabir brass wire mod kar ek model cart wheel banata hai: 42 mm diameter ka rim aur 4 seedhe wires jo centre se hote hue uske ek sire se doosre sire tak jaate hain, aur wheel ko 8 barabar hisson mein kaat dete hain.
1. Wire ki total length nikaalo, mm mein.
  mm
  Answer dekho
  Answer 300 mm
  Circumference $= \frac{22}{7} \times 42 = 132$ mm, aur chaar diameters $4 \times 42 = 168$ mm jodte hain: kul $300 m m$ .
2. Har hisse ka area nikaalo, mm² mein.
  mm²
  Answer dekho
  Answer 173.25 mm²
  Har hissa $45^{\circ}$ angle ka sector hai: $\frac{1}{8} \times \frac{22}{7} \times 21^{2} = \frac{1386}{8} = 173.25 m m^{2}$ .
NumericalEasyNCERT Exercise 11.1, Q10
Mele mein ek prize wheel 35 cm radius ki flat disc hai, jo centre se nikalti 10 barabar doori wali lines se prize sections mein bati hui hai. Ek section ka area nikaalo.
cm²
Answer dekho
Answer 385 cm²
Har section $36^{\circ}$ ka sector hai: $\frac{1}{10} \times \frac{22}{7} \times 35^{2} = \frac{3850}{10} = 385 c m^{2}$ .
NumericalMediumNCERT Exercise 11.1, Q11
Do lawn sprinklers mein se har ek $100^{\circ}$ ghoomte hue 4.2 m tak paani phenkta hai. Unke geele regions overlap nahi karte. Ek sweep mein kul kitna area seencha jaata hai, nikaalo.
m²
Answer dekho
Answer 30.8 m²
Ek sector: $\frac{100}{360} \times \frac{22}{7} \times {4.2}^{2} = \frac{5}{18} \times 55.44 = 15.4 m^{2}$ . Do: $30.8 m^{2}$ .
NumericalEasyNCERT Exercise 11.1, Q12
Railway platform par ek floodlight 30 m aage tak roshni phenkti hai, jo $70^{\circ}$ ke angle mein phaili hai. Woh platform aur track ka kitna area roshan karti hai? ( $π = 3.14$ use karo.)
m²
Answer dekho
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
21 cm radius ki ek gol brass plate par ek regular hexagon khuda hai, jiske chhe corners rim par hain. Hexagon aur rim ke beech ke chhe segments par ₹0.50 per cm² ki dar se enamel kiya jaata hai. Kharcha nikaalo. ( $\sqrt{3} = 1.73$ use karo.)
Fig. 11.14Hexagon ki har side ek segment kaat kar alag karti hai.NCERT Fig. 11.11 · p. 159
21 cm radius ki ek gol brass plate jiske andar ek regular hexagon bana hai, uske chhe corners rim par hain. Hexagon ki sides aur rim ke beech ke chhe curved tukde shaded hain aur un par enamelled segments likha hai; ek dashed radius par 21 cm mark hai.
Solution dikhao
AnswerLagbhag ₹120.80
$O$ ko chhe corners se jodne par chhe triangles bante hain jinka $O$ par angle $60^{\circ}$ hai, har ek 21 cm side ka equilateral triangle.
Ek segment: $\frac{1}{6} \times \frac{22}{7} \times 441 - \frac{1.73}{4} \times 441 = 231 - 190.7325 = 40.2675 c m^{2}$ .
Chhe segments: $6 \times 40.2675 = 241.605 c m^{2}$ .
Kharcha: $241.605 \times 0.50 = 120.8025$ , yaani lagbhag ₹120.80.
14
Quick answerMediumNCERT Exercise 11.1, Q14
Radius $a$ wale circle mein angle $x^{\circ}$ ke sector ka area inmein se kaun dete hain? Jitne sahi hain sab chuno.
$\frac{x}{360} \times π a^{2}$ $\frac{x}{180} \times π a$ $\frac{x}{720} \times 2 π a^{2}$ $\frac{x}{180} \times π a^{2}$ $\frac{1}{2} \times a \times \frac{π a x}{180}$
Saare sahi options chuno
Answer dekho
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 (11.1) hai. C bhi wahi hai, kyunki $\frac{2}{720} = \frac{1}{360}$ . E, $\frac{1}{2} \times r \times l$ hai jahan arc $l = \frac{x}{360} \times 2 π a = \frac{π a x}{180}$ . B khud arc length hai, aur D area ka double hai.
Quick answerMedium
Circle 1 ka radius $r$ hai aur usmein $40^{\circ}$ ka arc hai. Circle 2 ka radius $2 r$ hai aur usmein $20^{\circ}$ ka arc hai. Circle 1 wala arc zyada lamba hai.
SahiGalat
Answer dekho
Answer Galat
(11.2) se arcs $\frac{40}{360} \times 2 π r$ aur $\frac{20}{360} \times 2 π (2 r)$ hain, jo barabar hain. Angle aadha karna aur radius double karna ek doosre ko cancel kar dete hain.

11.2 Summary

Poora chapter teen results par tika hai. Do results poore circle ki kisi quantity ka $\frac{θ}{360}$ fraction hain, aur teesra do areas ka difference hai.

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

Fig. 11.15Arc, sector aur segment: is chapter ke teen formulas.

Centre O wala ek circle jiske upar right mein ek sector hai. Step 1 arc ko mota karta hai aur arc-length formula dikhata hai, l equals theta over 360 times 2 pi r. Step 2 sector shade karta hai aur uska area formula jodta hai, theta over 360 times pi r squared. Step 3 sirf chord aur arc ke beech ka segment shade karta hai aur rule jodta hai: segment equals sector minus triangle OAB.

Recap

Summary aur key points

Sector do radii aur ek arc se ghira hota hai; segment ek chord aur ek arc se. Har ek ka ek minor aur ek major tukda hota hai, aur akela “sector” ya “segment” ka matlab minor wala hai (Sectors aur segments).
Agar minor sector ka angle $θ$ hai, to major sector ka angle $360^{\circ} - θ$ hai (Sectors aur segments).
Unitary method se sector ka area $\frac{θ}{360} \times π r^{2}$ hai ((11.1)).
Arc ki length $\frac{θ}{360} \times 2 π r$ hai, aur sector ka perimeter $2 r + l$ hai ((11.2)).
Jab arc length pata ho, to sector ka area $\frac{1}{2} l r$ hai ((11.3)).
Minor segment ka area = sector ka area − triangle $O A B$ ka area ((11.4)).
Major sector $= π r^{2} -$ minor sector, aur major segment $= π r^{2} -$ minor segment (Segment ka area).
$O M ⊥ A B$ ke saath: $O M = r \cos \frac{θ}{2}$ , $A M = r \sin \frac{θ}{2}$ aur triangle $O A B = A M \times O M$ ((11.5)).
$60^{\circ}$ aur $120^{\circ}$ par triangle ka area $\frac{\sqrt{3}}{4} r^{2}$ hai; $90^{\circ}$ par yeh $\frac{1}{2} r^{2}$ hai (Table 11.1).
$π$ aur $\sqrt{3}$ ki wahi value use karo jo question deta hai (Key idea).

Flash cards · 8

Formula cards

1 / 8

Quick revision

2 mein se 1

Har quantity ko uske formula se match karo.

Left mein koi item tap karo, phir right mein uska match.

Answer dekho

Answer

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

Arc length aur perimeter lengths hain, isliye unmein

r

use hota hai; sector area mein

r^{2}

; segment do areas ka difference hai.

Board-exam practice

Yeh questions CBSE board paper ke patterns par hain: one-mark MCQs, ek assertion–reason item, aur 2, 3, 4 aur 5 marks ke questions. Har likhe hue answer mein formula, substitution aur units dikhao.

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
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

Sectors aur segments3 questions
Arc ki length4 questions
Triangle OAB ka area nikaalna4 questions
Quick revision2 questions

Glossary · 9

Key terms (English + Hindi)

arcचापchaap: Circle ke do points ke beech ka hissa. Angle $θ$ wale arc ki length $\frac{θ}{360} \times 2 π r$ hoti hai.Chapter mein pehli baar yahan aaya →

chordजीवाjeeva: Circle ke do points ko jodne wala line segment.Chapter mein pehli baar yahan aaya →

major sectorदीर्घ त्रिज्यखंडdeergh trijyakhand: Do radii se bane do sectors mein bada wala, jiska angle $360^{\circ} - θ$ hota hai.Chapter mein pehli baar yahan aaya →
major segmentदीर्घ वृत्तखंडdeergh vrittakhand: Chord se kata bada segment; iska area $π r^{2}$ minus minor segment hai.Chapter mein pehli baar yahan aaya →
minor sectorलघु त्रिज्यखंडlaghu trijyakhand: Do radii se bane do sectors mein chhota wala; iska angle $θ$ , $180^{\circ}$ se kam hota hai.Chapter mein pehli baar yahan aaya →
minor segmentलघु वृत्तखंडlaghu vrittakhand: Chord se kata chhota segment; iska area sector minus triangle $O A B$ hai.Chapter mein pehli baar yahan aaya →

quadrantचतुर्थांशchaturthansh: $90^{\circ}$ angle ka sector, disc ka ek quarter.Chapter mein pehli baar yahan aaya →

sectorत्रिज्यखंडtrijyakhand: Disc ka woh region jo do radii aur unke siron ko jodne wale arc ke beech ho.Chapter mein pehli baar yahan aaya →
segmentवृत्तखंडvrittakhand: Disc ka woh region jo ek chord aur us chord se kate arcs mein se ek ke beech ho.Chapter mein pehli baar yahan aaya →

Areas Related to Circles: Class 10 Maths Chapter 11 Hinglish mein

Is chapter mein aap kya seekhenge

11.1 Circle ke Sector aur Segment ke Areas

Sectors aur segments

Sector ka area

Arc ki length

Segment ka area

Triangle OAB ka area nikaalna

Practice 11.1: answers ke saath

11.2 Summary

Summary aur key points

Board-exam practice

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

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye