I’m Up and Down, and Round and Round (Main Upar-Neeche, aur Gol-Gol Ghoomta Hoon): Class 9 Maths Chapter 5 Hinglish mein

Circles, chords, arcs aur unse bante angles

Ganita Manjari Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

105 min
3 simulations
50 practice questions
36 MCQs

Is chapter mein aap kya seekhenge

Circle ko ek locus ki tarah samjhao, aur uske centre, radius, chords, diameters aur arcs ke naam batao
Circle ki rotational aur mirror symmetries samjhao, aur folding se centre dhoondho
Dikhao ki do points se anginat circles guzarte hain, lekin teen non-collinear points se theek ek circle guzarta hai, aur circumcircle construct karo
Equal chords, centre par equal angles aur centre se equal distances ke beech ke rishte prove karo aur use karo
Centre se daale gaye perpendicular aur Baudhāyana–Pythagoras theorem se chord lengths, distances aur radii nikaalo
Prove karo ki ek arc centre par circle ke baaki points ke mukaable double angle banata hai, aur iske results use karo
Tay karo ki chaar points kab concyclic hote hain, aur cyclic quadrilaterals ki angle property aur uska converse use karo

NCERT Mathematics Class 9, Chapter 5, pp. 92–117 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

Hamare aas-paas circles

Aas-paas dekho, circles har jagah dikhte hain: steel ki katori ka kinara, kaanch ki choodi, kate hue santre ka face, kaate gaye lakdi ke tane ke andar ke rings, Konark ke Sun Temple ke bade-bade patthar ke pahiye. Jab se insaan ne kuch bhi banana shuru kiya, tab se woh yeh shape banata aa raha hai.

Circle itna common kyun hai, aur yeh itna perfect kyun dikhta hai? Yeh sab ek simple property se aata hai. Is chapter mein hum us property ko jahan tak ho sake, wahan tak follow karenge: kuch diye gaye points se kitne circles guzarte hain, chords kaise behave karti hain, aur ek arc alag-alag points par kitna angle banata hai. Raaste mein tum har result ko sirf naapoge nahi, balki prove bhi karoge.

Shuru karte hain ek maali se jo gol flower bed mark kar rahi hai (Fig. 5.1). Woh mitti mein ek peg gaadti hai aur usse ek rassi baandhti hai. Rassi ke doosre end par ek stick baandh kar, woh rassi ko tight rakhte hue chaaron taraf chalti hai aur chalte-chalte zameen par lakeer kheenchti jaati hai. Har lakeer peg se theek ek rassi-length door hai, aur saari lakeerein milkar ek circle banati hain.

3/3Curve ka har point peg se 1.5 m door hai: peg hi centre hai.

Fig. 5.1Stick jo bhi point mark karti hai, woh peg se ek rassi-length door hota hai.

Upar se dekha gaya ek garden bed, jiske corners mein chhote phool hain. Step 1 mein beech mein ek peg hai aur 1.5 m lambi rassi right side ki ek stick tak tani hui hai. Step 2 mein rassi lagbhag poori ghoom chuki hai, aur stick ne peg ke chaaron taraf lagbhag teen-chauthai tak ek orange curve kheench diya hai. Step 3 mein poora orange circle bana hai; rassi ki paanch aur directions mein dashed copies dikhati hain ki har point peg se 1.5 m door hai, aur ab peg par centre likha hai.

ActivityNCERT Activity · p. 93

Circle ki khoj

Tumhe chahiye

ek notebook aur ek pencil
das minute aur tez nazar

Pehle guess karo

Inmein se kaunsa circle nahi hai?

ek rupaye ke sikke ka kinaramurgi ke ande ki outlinecycle ke pahiye ka rimped ke tane par paint kiya gaya ring, upar se dekhne par

Das minute ghar, school ya park mein gol cheezein dhoondho. Kam se kam aath dhoondhne ki koshish karo.
Unhe do groups mein baanto: nature ki banayi hui, ya insaanon ki banayi hui.
Har cheez ke liye us jagah ki taraf ishara karo jahan tumhe lagta hai uska centre hai.

Table 5.1Meri circle ki khoj

Cheez	Natural ya insaanon ki banayi?	Iska centre kahan hai?

Tumhari entries apne aap save hoti hain.

Warm-up

2 mein se 1

Jo bhi shape gol dikhe, jaise ande ki outline, woh circle hai.

SahiGalat

Answer dekho

Answer Galat

Circle ke liye ek aisa centre chahiye jo kinare ke har point se barabar door ho. Ande ki outline mein aisa koi point nahi hota, isliye woh oval hai, circle nahi.

5.1 Definitions

Is chapter ka har figure ek flat surface, yaani plane, par hai, jaise paper ki sheet ya tumhare classroom ka farsh. Tum pehle se jaante ho ki circle ka har point uske centre se barabar door hota hai. Mathematicians ek kadam aage jaate hain aur isi ek fact ko definition bana dete hain: circle kya hai, yeh theek-theek batane ke liye aur kuch nahi chahiye.

Definition

Circle

Circle (वृत्त) plane ke un saare points se banta hai jinki us plane ke ek fixed point, yaani centre, se distance ek fixed length, yaani radius, ke barabar hoti hai.

ExampleMaali ka peg centre hai aur uski 1.5 m ki rassi radius hai.

Saare points shabdon par dhyaan do. Circle sirf boundary wala curve hai. Uske andar ke region ko disc kehte hain: chapati ek disc hai, aur uska kinara ek circle hai.

Definition

Locus

Locus (बिंदुपथ) sirf aur sirf un points ka set hai jo diye gaye rule ko follow karte hain: rule follow karne wala har point usmein hota hai, aur koi doosra point nahi hota.

ExamplePoint $O$ se 4 cm door wale points ka locus woh circle hai jiska centre $O$ aur radius 4 cm hai.

To circle ek fixed point se fixed distance par pade points ka locus hai. Locus check karne ke hamesha do hisse hote hain: figure par pada har point rule follow kare, aur rule follow karne wala har point figure par ho. Yeh do-taraf wala check humein Section 5.3 mein chahiye hoga.

Ab parts ke naam rakhte hain (Fig. 5.2). Centre $O$ wala ek circle lo. $O$ se circle ke kisi point $P$ tak ka segment ek radius (त्रिज्या) hai; uski length $r$ ko bhi radius kehte hain. Circle ke do points $A$ aur $B$ ko jodne wala segment ek chord (जीवा) hai.

$A$ aur $B$ ko centre se jodo. Angle $∠ A O B$ woh angle hai jo chord $A B$ centre par subtend karti hai, yaani $O$ se chord ko dekhne par jo opening dikhti hai. Centre se guzarne wali chord, jaise $C D$ , ek diameter (व्यास) hai. Yeh do radii se bani hai, isliye iski length $2 r$ hai.

4/4Centre se guzarne wali chord diameter hai, jiski length $2 r$ hai.

Fig. 5.2Circle ke woh parts jo hum har theorem mein use karte hain.NCERT Fig. 5.3 · p. 93

Centre O wala ek circle. Step 1 mein centre O aur length r wala ek green radius OP hai. Step 2 mein circle ke upar ki taraf ek blue chord AB judti hai. Step 3 mein A aur B ko grey radii se O tak joda gaya hai aur chord centre par jo angle AOB subtend karti hai, use shade kiya gaya hai. Step 4 mein O se guzarta ek orange diameter CD judta hai, jis par diameter barabar 2r likha hai.

Quick check

4 mein se 1

Ek bachhda 3 m ki rassi se ek khambe se bandha hai aur rassi tight rakhte hue chakkar lagata hai. Uski positions kaunsa figure banati hain?

3 m radius wali bhari hui disc3 m diameter wala circlekhambe ko centre maan kar 3 m radius wala circle3 m side wala square

Answer dekho

Answer (C) khambe ko centre maan kar 3 m radius wala circle

Rassi tight hone par bachhda hamesha khambe se theek 3 m door hota hai, isliye woh 3 m radius wale circle par hi rehta hai. Disc woh area hai jahan woh rassi dheeli hone par char sakta hai.

5.2 Circle ki Symmetries

Circle itna achha lagta hai, iski ek wajah iski perfect symmetry hai. Table par ek saadi steel ki thali rakho aur aankhein band karo, jab tak ek dost use ghumata hai. Aankhein kholne par kya tum bata sakte ho ki use kitna ghumaya gaya? Nahi bata sakte. Centre ke around kisi bhi angle se ghumane ke baad plate bilkul waisi hi dikhti hai.

Square tile se pata chal jaata: woh sirf quarter turn, half turn ya three-quarter turn ke baad hi badli hui nahi lagti. Hum kehte hain ki circle ke paas uske centre ke around complete rotational symmetry hai (Fig. 5.3).

Circle mein mirror symmetry bhi hai, aur woh bhi square se kahin zyada. Square ki sirf chaar mirror lines hoti hain. Circle ke liye har diameter kaam karta hai: uske dono taraf ke halves ek doosre ki mirror images hain. Paper se yeh jaldi dikh jaata hai. Jo bhi fold disc ke kinare ko bilkul usi par rakh de, uski crease ek diameter ke saath banti hai, aur uski direction tum apni marzi se chun sakte ho.

4/4Circle ka har diameter ek line of symmetry hai.

Fig. 5.3Circle kisi bhi turn ke baad waisa hi dikhta hai, aur use kisi bhi diameter ke saath fold kiya ja sakta hai.

Ek blue square aur ek orange circle aamne-saamne, dono par ek chhota laal sticker. Step 1 mein dono ruke hue hain. Step 2 mein dono 30 degrees ghoomte hain: square saaf tircha ho jaata hai aur alag dikhta hai, jabki circle waisa hi dikhta hai, bas uska sticker khisak gaya hai. Step 3 mein square ki chaar dashed mirror lines aur circle ka ek dashed diameter fold line ki tarah dikhte hain. Step 4 mein circle par kai dashed diameters dikhte hain, aur har ek mirror line hai.

Try karke dekho

Fold karke centre dhoondho

Tumhe chahiye

paper ki ek disc (steel ki plate ya katori ke chaaron taraf line kheench kar kaat lo)
ek pencil aur ek ruler

Pehle guess karo

Do creases, jinmein se har ek aise fold kiya gaya ki kinara usi par aa jaaye, ek point par cross karti hain. Woh point kya hai?

disc ka centrekinare par ek pointyeh is par depend karta hai ki tum kaunse folds banate ho

Disc ko aise fold karo ki uska kinara bilkul usi par aa jaaye. Daba kar crease banao, phir khol do.
Ise phir se isi tarah kisi doosri direction mein fold karo, aur khol do.
Jahan do creases cross karti hain, us point ko mark karo. Wahan se kinare tak kai directions mein naapo.

5.3 Kitne Circles?

Page par do points $A$ aur $B$ mark karo. Dono se kitne circles guzarte hain?

$A$ aur $B$ se guzarne wala sabse chhota circle dhoondhna aasaan hai. Uska centre $A B$ ke midpoint $M$ par rakho. Kyunki $M A = M B$ , centre $M$ aur radius $M A$ wala circle dono points se guzarta hai, aur $A B$ uska ek diameter hai, isliye uska radius $\frac{1}{2} A B$ hai. $A$ aur $B$ se guzarne wala koi bhi circle isse chhota nahi ho sakta: $A B$ har aise circle ki ek chord hai, aur koi bhi chord diameter se lambi nahi hoti (yeh hum Section 5.6.1 mein prove karenge).

Kya aur bhi circles hain? $A$ aur $B$ se guzarne wale kisi bhi circle ke centre $O$ ke liye $O A = O B$ hona chahiye, kyunki dono radii hain. To humein woh saare points jaanne honge jo $A$ aur $B$ se barabar door hain.

LemmaPerpendicular bisector ek locus ki tarah

Do points $A$ aur $B$ se barabar door wale points theek wahi points hain jo $A B$ ke perpendicular bisector par hain.

$A B$ ka midpoint $M$ mark karo, aur maan lo $ℓ$ woh line hai jo $M$ se $A B$ ke right angle par guzarti hai. Hum do-taraf wala check karte hain.

$ℓ$ ka har point $A$ aur $B$ se barabar door hai. $ℓ$ par $M$ ke alawa koi point $P$ lo ( $M$ ke liye to yeh saaf hai).

P A = P B

$A$ aur $B$ se barabar door har point $ℓ$ par hai. Ek point $Q$ lo jiske liye $Q A = Q B$ ho aur jo line $A B$ par na ho ( $A B$ par aisa sirf ek point $M$ hai).

Q M ⊥ A B

Dono hisse sahi hain, isliye perpendicular bisector hi theek-theek locus hai.

To $A$ aur $B$ se guzarne wale circles ke possible centres theek $A B$ ke perpendicular bisector ke points hain, aur yeh infinitely many hain (Fig. 5.4). Jaise-jaise centre bisector ke saath $A B$ se door khiskata hai, radius badhta hai aur circle $A$ aur $B$ ke paas zyada flat dikhta hai.

5/5Har centre bisector par hai, aur bisector ka har point ek centre hai.

Fig. 5.4

A B

ke perpendicular bisector ka har point

A

aur

B

se guzarne wale ek circle ka centre hai.NCERT Fig. 5.4 · p. 95

Points A aur B, segment AB right ki taraf upar ko jhuka hai, aur AB ka dashed blue perpendicular bisector use right angle par cross karta hai. Step 1 mein sirf A, B, barabar halves ke marks aur bisector hain. Step 2 mein sabse chhota circle judta hai, jiska centre midpoint M hai aur AB diameter hai. Step 3 mein P centre wala bada circle hai, jo bisector par aur aage hai. Step 4 mein Q centre wala circle hai, jo AB ki doosri taraf hai. Step 5 mein yeh saare circles ek saath dikhte hain, saath mein R centre wala ek bahut bada circle bhi; har centre bisector par hai.

Ab teen points $A$ , $B$ aur $C$ lo. Agar woh collinear (संरेख) hain, yaani ek hi seedhi line par hain, to koi bhi circle teeno se nahi guzarta. $A B$ aur $B C$ ke perpendicular bisectors dono us line ke perpendicular hain, isliye woh parallel hain aur kabhi nahi milte. Koi bhi point teeno se barabar door nahi hai. Isse yeh bhi dikhta hai ki ek seedhi line circle ko zyada se zyada do points par milti hai.

Agar teeno points collinear nahi hain, to kahani poori tarah badal jaati hai.

Theorem 1Teen points se guzarta circle

Kinhi bhi teen points se, jo ek seedhi line par na hon, theek ek circle guzarta hai.

Maan lo points $A$ , $B$ aur $C$ hain, jo ek line par nahi hain (Fig. 5.5). Maan lo ek circle teeno se guzarta hai, aur uska centre $O$ hai. Hum nikaalte hain ki $O$ kahan hona chahiye.

O = do perpendicular bisectors ka crossing point

To sirf ek hi possible centre hai. Aur yeh kaam karta hai: crossing point $A$ aur $B$ se barabar door hai, aur $A$ aur $C$ se bhi, isliye us centre aur radius $O A$ wala circle teeno points se guzarta hai. Ek centre aur ek radius se theek ek circle banta hai.

4/4Yeh $O$ par cross karti hain, jo teeno points se barabar door hai. Teesra bisector bhi $O$ se guzarta hai.

Fig. 5.5Do perpendicular bisectors ek aur sirf ek centre ki jagah batate hain.NCERT Fig. 5.5 · p. 96

Teen points A, B aur C ek dashed triangle se jude hain. Step 1 mein sirf points hain. Step 2 mein AB ka blue perpendicular bisector banta hai, barabar halves ke marks aur right angle ke saath. Step 3 mein AC ka purple perpendicular bisector judta hai, jo pehli line ko cross karta hai. Step 4 mein crossing point O mark hota hai, A, B aur C se guzarta orange circle barabar radii ke saath banta hai, aur dikhta hai ki BC ka dashed green bisector bhi O se guzarta hai.

Definition

Circumcircle aur circumcentre

Triangle ke teeno vertices se guzarne wala circle uska circumcircle (परिवृत्त) hai, aur uska centre circumcentre (परिकेंद्र) hai. Hum kehte hain ki circle triangle ko circumscribe karta hai, aur triangle circle mein inscribed hai.

ExampleTheorem 1 se, har triangle ka theek ek circumcircle hota hai.

Crossing point $O$ , $B$ aur $C$ se bhi barabar door hai, isliye woh teesre perpendicular bisector par bhi hai. Triangle ke teeno perpendicular bisectors hamesha ek point par milte hain, jo circumcentre hai. Yeh point kahan hoga, yeh triangle ki shape par depend karta hai (Fig. 5.6).

2/3 $∠ A = 90^{\circ}$ : $O$ hypotenuse $B C$ ka midpoint hai.

Fig. 5.6Circumcentre kahan hai, isse pata chalta hai ki triangle kis tarah ka hai.NCERT Figs. 5.5–5.7 · p. 97

Side BC fixed rehti hai aur vertex A khisakta hai; har step mein blue triangle, uska orange circumcircle aur dashed radii ke saath circumcentre O dikhta hai. Step 1: ek acute triangle, O uske andar hai. Step 2: angle A right angle hai aur O theek BC, yaani hypotenuse, ke midpoint par hai. Step 3: angle A obtuse hai aur O, BC ke neeche, triangle ke bahar hai.

Right-angled wala case yaad rakhne layak hai. Iska circumcentre hypotenuse ka midpoint hai, isliye hypotenuse circumcircle ka ek diameter hai. Aisa kyun hai, yeh tum Section 5.7.1 mein dekhoge.

ActivityNCERT Exercise Set 5.1 · p. 98

Circumcircle construct karo

Tumhe chahiye

ek ruler
ek pair of compasses
ek protractor
ek nukeeli pencil

Pehle guess karo

Ek triangle ke angles

40^{\circ}

30^{\circ}

aur

110^{\circ}

hain. Iska circumcentre kahan hoga?

triangle ke andaruski kisi side partriangle ke bahar

Koi bhi triangle $P Q R$ banao jiski sides 5 cm aur 8 cm ke beech hon.
Compasses ko $P Q$ ke aadhe se zyada khol lo. $P$ aur $Q$ se $P Q$ ke dono taraf arcs banao, aur dono crossing points ko jodo. Yeh line $P Q$ ka perpendicular bisector hai.
Side $Q R$ ke liye bhi yahi karo. Jahan dono bisectors milte hain, us point $O$ ko mark karo.
Compass ki nok $O$ par rakho, use $P$ tak kholo aur circle banao. Kya yeh $Q$ aur $R$ se guzarta hai?
$O P$ , $O Q$ aur $O R$ naapo, aur note karo ki $O$ triangle ke andar hai ya nahi.

Table 5.2Circumcircle ke measurements

Triangle	OP (cm)	OQ (cm)	OR (cm)	O andar hai?
Mera triangle
Dost ka triangle

Tumhari entries apne aap save hoti hain.

Quick check

4 mein se 1

Do points

A

aur

B

ke beech 10 cm ki doori hai. Dono se guzarne wale circle ka sabse chhota possible radius kya hai?

10 cm5 cm2.5 cmkoi sabse chhota radius nahi hai

Answer dekho

Answer (B) 5 cm

Sabse chhote circle mein

A B

diameter hai, isliye uska radius

10 \div 2 = 5

cm hai. Koi sabse bada circle nahi hota, lekin sabse chhota hota hai.

Practice · 6 questions

Practice 5.1: answers ke saath

DiagramEasyNCERT Exercise Set 5.1, Q1
$△ P Q R$ construct karo jismein $Q R = 6$ cm, $∠ Q = 50^{\circ}$ aur $∠ R = 65^{\circ}$ ho. Iska circumcircle banao. Circumcentre triangle ke andar hai ya bahar?
Solution dikhao
AnswerAndar, kyunki teeno angles ( $65^{\circ}$ , $50^{\circ}$ , $65^{\circ}$ ) acute hain.
$Q R = 6$ cm banao, phir $Q$ par $50^{\circ}$ aur $R$ par $65^{\circ}$ ke angles banao. Inki arms $P$ par milti hain.
$∠ P = 180^{\circ} - 50^{\circ} - 65^{\circ} = 65^{\circ}$ , isliye har angle acute hai.
$Q R$ aur $P Q$ ke perpendicular bisectors banao; yeh $O$ par milte hain. Centre $O$ aur radius $O Q$ wala circle $P$ , $Q$ aur $R$ se guzarta hai, aur $O$ triangle ke andar hai.
DiagramEasyNCERT Exercise Set 5.1, Q2
$△ P Q R$ construct karo jismein $P Q = 4.5$ cm, $∠ P = 115^{\circ}$ aur $P R = 5$ cm ho. Iska circumcircle banao. Circumcentre triangle ke andar hai ya bahar?
Solution dikhao
AnswerBahar, side $Q R$ ke us paar, kyunki $∠ P$ obtuse hai.
$P Q = 4.5$ cm banao, $P$ par $∠ Q P R = 115^{\circ}$ banao aur $P R = 5$ cm lekar $R$ mark karo. $Q R$ ko jodo.
$P Q$ aur $P R$ ke perpendicular bisectors banao. Yeh ek point $O$ par milte hain jo $Q R$ ki us taraf hai jo $P$ se door hai.
Centre $O$ wala jo circle $P$ se guzarta hai, woh $Q$ aur $R$ se bhi guzarta hai. Kyunki $∠ P$ obtuse hai, $O$ triangle ke bahar hai.
DiagramMediumNCERT Exercise Set 5.1, Q3
$△ X Y Z$ construct karo jismein $X Y = 5$ cm, $Y Z = 6$ cm aur $Z X = 7$ cm ho. Iska circumcircle banao, circumcentre $O$ mark karo aur $O X$ , $O Y$ aur $O Z$ naapo.
Solution dikhao
AnswerTeeno lagbhag 3.6 cm aati hain.
Teeno sides se triangle construct karo, do sides ke perpendicular bisectors banao aur unka crossing point $O$ mark karo.
$O X$ , $O Y$ aur $O Z$ teeno lagbhag $3.6$ cm aate hain (dhyaan se calculate karne par $3.57$ cm aata hai). Yeh barabar hain kyunki $O$ perpendicular bisectors par hai, isliye woh teeno vertices se barabar door hai.
NumericalEasyNCERT Exercise Set 5.1, Q4
Do points ke beech 7 cm ki doori hai. Dono se guzarne wale circle ka radius kitna chhota ho sakta hai?
cm
Answer dekho
Answer 3.5 cm
Sabse chhote aise circle mein yeh segment diameter hota hai, isliye uska radius $7 \div 2 = 3.5$ cm hai.
5
ProofMediumNCERT Ex. Set 5.1, Think, Draw and Infer Q1
Points $P$ , $Q$ aur $R$ ek line par hain, jahan $P Q = 3$ cm aur $Q R = 4$ cm hai.
1. $P Q$ aur $Q R$ ke perpendicular bisectors banao. Inka aapas mein kya rishta hai?
  Solution dikhao
  AnswerYeh parallel hain, aur inke beech 3.5 cm ki doori hai.
  Dono bisectors line $P R$ ke perpendicular hain, isliye parallel hain. Yeh $P R$ ko $P Q$ aur $Q R$ ke midpoints par cross karte hain, jinke beech $1.5 + 2 = 3.5$ cm ki doori hai.
2. Kya koi aisa point hai jo $P$ , $Q$ aur $R$ se barabar door ho?
  Solution dikhao
  AnswerNahi.
  Aisa point dono bisectors par hona chahiye, lekin parallel lines kabhi nahi milti.
3. Kya ek seedhi line circle ko teen alag points par kaat sakti hai?
  Solution dikhao
  AnswerNahi.
  Agar aisa hota, to woh teen collinear points ek circle par hote, jo part (b) mein galat saabit ho chuka hai.
Short answerMediumNCERT Ex. Set 5.1, Think, Draw and Infer Q2
$△ A B C$ ka circumcircle bana hai. Kya $△ A B C$ ke congruent koi doosra triangle bhi isi circumcircle wala ho sakta hai? Agar haan, to aise kitne triangles hain?
Solution dikhao
AnswerHaan, infinitely many.
$△ A B C$ ko circumcentre $O$ ke around kisi bhi angle se ghumao. Har vertex $O$ se utni hi door rehta hai, isliye ghumaya hua triangle $△ A B C$ ke congruent hai aur uske vertices abhi bhi circle par hain.
Kisi bhi diameter mein reflect karna bhi kaam karta hai. Ghumane ke liye infinitely many angles hain, isliye aise infinitely many triangles hain.

5.4 Chords aur Unke Subtend Kiye Angles

Har chord ke saath ek angle aata hai: woh angle jo woh centre par subtend karti hai. Lambi chords ke saath chaude angles aur chhoti chords ke saath patle angles lagte hain. Is section mein hum is rishte ko exact banayenge.

Deewar ki ghadi dekho. 12 ke mark se 2 ke mark tak ki seedhi line uske rim ki ek chord hai, aur 3 ke mark se 5 ke mark tak ki line bhi. Clock face ka quarter turn pehli chord ko bilkul doosri chord par le aata hai. To dono chords ki length same hai, aur dono centre par same angle, $60^{\circ}$ , banati hain. Centre ke around turn kisi bhi chord ke saath yahi karta hai (Fig. 5.7).

3/3 $A B = C D$ aur $∠ A O B = ∠ C O D$ .

Fig. 5.7Chord ko centre ke around ghumane se uski length aur centre par uska angle, dono same rehte hain.NCERT Figs. 5.8–5.10 · p. 98

Centre O wala ek circle. Step 1 mein neeche left ke paas ek blue chord AB hai, saath mein radii OA aur OB aur unke beech 80 degree ka angle. Step 2 mein chord, uske radii aur uska angle O ke around 150 degrees ghoom kar upar ke paas nayi position CD par aa jaate hain, aur peeche AB ki ek dashed purple copy reh jaati hai. Step 3 mein triangles OAB aur OCD shade hote hain, AB aur CD ko barabar mark kiya jaata hai, aur dono ke liye centre par 80 degree ka angle dikhta hai.

Ghumane se result par bharosa hota hai, lekin proof ko kinhi bhi do equal chords ke liye kaam karna chahiye, unke liye bhi jo humein ghumane se nahi mili. Congruent triangles yeh kaam karte hain.

Theorem 2Equal chords se equal angles

Agar circle ki do chords length mein barabar hain, to centre par unke subtend kiye angles barabar hote hain.

Ab sawaal ko ulta karo. Maan lo do chords centre par same angle banati hain. Kya woh barabar honi hi chahiye? Ek haath ke pankhe ke baare mein socho jo apne centre par pin se laga hai. Uski do teeliyon ko ek fixed angle tak kholo aur pankha ghumao: dono teeliyon ke siron ke beech ki doori hamesha same rehti hai.

Theorem 3Equal angles se equal chords

Jab ek circle ki do chords uske centre par equal angles banati hain, to dono chords ki length same hoti hai.

Example 5.1

Equal chords se angles

Centre $O$ wale circle mein chords $P Q$ aur $R S$ barabar hain aur $∠ P O Q = 48^{\circ}$ hai. $∠ R O S$ aur $∠ O R S$ nikaalo.

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

Practice · 2 questions

Practice 5.2: answers ke saath

1
ProofEasyNCERT Exercise Set 5.2, Q1
Centre $O$ wale circle ki chord $A B$ , $∠ A O B = 110^{\circ}$ subtend karti hai.
1. Samjhao ki $△ O A B$ isosceles kyun hai.
  Solution dikhao
  Answer $O A = O B$ .
  $O A$ aur $O B$ dono circle ke radii hain, isliye barabar hain. Do barabar sides wala triangle isosceles hota hai.
2. $∠ O A B$ degrees mein nikaalo.
  °
  Answer dekho
  Answer 35 °
  $∠ O A B = ∠ O B A = (180^{\circ} - 110^{\circ}) \div 2 = 35^{\circ}$ .
ProofMediumNCERT Exercise Set 5.2, Q2
$A B$ aur $C D$ centre $O$ wale circle ki equal chords hain. Dikhao ki $△ O A B ≅ △ O C D$ , aur isliye in dono triangles ke areas barabar hain.
Solution dikhao
$O A = O C$ aur $O B = O D$ (radii), aur $A B = C D$ (diya gaya hai). SSS rule se, $△ O A B ≅ △ O C D$ .
Congruent triangles ek doosre ke upar bilkul fit ho jaate hain, isliye unke areas barabar hote hain.

5.5 Chords ke Midpoints aur Perpendicular Bisectors

Ek chord banao, uska midpoint mark karo aur midpoint ko centre se jodo. Kaunsa angle milta hai? Aur agar hum ulta, centre se chord par daale gaye perpendicular se shuru karein, to kya hota hai?

Theorem 4Centre se midpoint tak

Agar $M$ centre $O$ wale circle ki chord $A B$ (jo diameter nahi hai) ka midpoint hai, to $O M$ , $A B$ ke perpendicular hai.

5/5Seedhi line par do equal angles: har ek $90^{\circ}$ hai.

Fig. 5.8Do congruent halves midpoint par right angle banane par majboor karte hain.NCERT Fig. 5.12 · p. 101

Centre O wala ek circle aur centre ke neeche ek blue chord AB. Step 1 mein sirf chord hai. Step 2 mein uska midpoint M barabar halves ke marks ke saath mark hota hai. Step 3 mein orange segment OM aur grey radii OA aur OB bante hain. Step 4 mein triangles OMA aur OMB do rangon mein matching marks ke saath shade hote hain, aur OA = OB, AM = MB aur OM common likha hai, isliye SSS. Step 5 mein M par right-angle mark judta hai: dono equal angles milkar 180 degrees bante hain, isliye har ek 90 degrees hai.

Converse bhi sahi hai, aur is baar RHS rule kaam karta hai.

Theorem 5Centre se perpendicular

Circle ke centre se kisi chord par daala gaya perpendicular us chord ko do barabar hisson mein baant deta hai.

Dono theorems milkar ek bahut kaam ka fact dete hain: har chord ka perpendicular bisector centre se guzarta hai. $O$ se daala gaya perpendicular chord ko aadha karta hai, isliye wahi chord ka perpendicular bisector hai. Do chords banao aur dono ko bisect karo: bisectors centre par cross karte hain.

Theorem 5 chord ke sawaalon ko right-triangle ke sawaalon mein bhi badal deta hai. Radius $O B$ , distance $O M$ aur half-chord $M B$ ek triangle banate hain jiska right angle $M$ par hai, isliye Baudhāyana–Pythagoras theorem lagta hai.

Half-chord relation

r^{2} = d^{2} + {(\frac{c}{2})}^{2}

jahan

$r$: radius
$d$: centre se chord ki distance
$c$: chord ki length

Example 5.2

Kuein ke upar ek takhta

Ek gol kuein ke mouth ka radius 1.3 m hai. Mouth ke upar ek seedha takhta rakha hai, aur uska kinara centre se 0.5 m door ek chord banata hai. Yeh chord kitni lambi hai?

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

Quick check

4 mein se 1

O

ek circle ka centre hai aur

M

ek aisi chord

A B

ka midpoint hai jo diameter nahi hai. Kaunsa statement hamesha sahi hai?

O M = A M

O M = \frac{1}{2} r

O M ⊥ A B

M

circle par hai

Answer dekho

Answer (C) $O M ⊥ A B$

Theorem 4 se, centre se midpoint tak ka segment chord ke perpendicular hota hai.

O M = A M

sirf ek khaas chord ke liye hota hai.

Practice · 3 questions

Practice 5.3: answers ke saath

ProofMediumNCERT Exercise Set 5.3, Q1
Theorem 5 ka ek doosra proof do, congruent triangles ki jagah Baudhāyana–Pythagoras theorem use karke.
Solution dikhao
Kyunki $O M ⊥ A B$ , triangles $O M A$ aur $O M B$ dono $M$ par right-angled hain.
$A M^{2} = O A^{2} - O M^{2} = O B^{2} - O M^{2} = M B^{2}$
Yahan $O A = O B$ (radii) use hua hai. Lengths positive hoti hain, isliye $A M = M B$ .
ProofMediumNCERT Exercise Set 5.3, Q2
Triangle $P Q R$ centre $O$ wale circle mein inscribed hai, aur $P Q = P R$ hai. Dikhao ki $P$ ko $Q R$ ke midpoint $N$ se jodne wali line $O$ se guzarti hai.
Solution dikhao
$P Q = P R$ , isliye $P$ , $Q$ aur $R$ se barabar door hai aur $Q R$ ke perpendicular bisector par hai.
$O$ centre hai, isliye $O Q = O R$ , aur $O$ bhi usi bisector par hai. Woh bisector $Q R$ ke midpoint $N$ se bhi guzarta hai.
To $P$ , $N$ aur $O$ teeno ek hi line par hain, jo $Q R$ ka perpendicular bisector hai.
NumericalMediumNCERT Exercise Set 5.3, Q3
17 cm radius wale circle ki do parallel chords 16 cm aur 30 cm lambi hain, aur woh centre ki ulti taraf hain. Unke midpoints ke beech kitni doori hai?
cm
Answer dekho
Answer 23 cm
Distances $\sqrt{17^{2} - 8^{2}} = 15$ cm aur $\sqrt{17^{2} - 15^{2}} = 8$ cm hain. Dono midpoints chords ke perpendicular diameter par, $O$ ki ulti taraf hain, isliye unke beech $15 + 8 = 23$ cm ki doori hai.

5.6 Centre se Chords ki Distance

Kisi point se line ki distance ka matlab hai point se line par daale gaye perpendicular ki length, jo unke beech ka sabse chhota raasta hai. To jab hum poochte hain ki chord centre se kitni door hai, to humara matlab length $O M$ se hai, jahan $M$ , $O$ se daale gaye perpendicular ka foot hai. Theorem 5 se, $M$ chord ka midpoint bhi hai.

4/4Creases $M$ par $90^{\circ}$ par milti hain, aur $O M$ centre se $A B$ ki distance hai.

Fig. 5.9Do folds se ek chord aur centre se uska perpendicular banta hai.NCERT Fig. 5.13 · p. 102

Halke rang ki paper disc jiska centre mark kiya hua hai. Step 1 mein saadi disc hai. Step 2 mein ek flap dashed blue crease ke saath fold hua hai, jo chord AB hai. Step 3 mein doosra fold A ko B par le aata hai, centre se guzarti ek purple crease banti hai aur fold kiya hua aadha hissa shade hota hai. Step 4 mein dono creases khuli hain: woh M par right angle par milti hain, M, AB ko aadha karta hai, aur orange segment OM centre se chord ki distance hai.

ActivityNCERT Activity · p. 102

Fold karo, crease banao aur naapo

Tumhe chahiye

paper ki ek disc jiska centre mark kiya ho
ek ruler aur ek protractor
tracing paper ya butter paper
ek pair of compasses

Pehle guess karo

Ek hi circle ki do chords ki length same hai. Centre se unki distances mein kya rishta hoga?

pehle banayi gayi chord zyada paas hogidono barabar hongiyeh is par depend karta hai ki chords kahan hain

Disc ka ek flap andar ki taraf fold karo aur crease banao. Kholo: crease ek chord $A B$ hai.
Disc ko aise fold karo ki $A$ bilkul $B$ par aa jaaye, crease banao aur kholo. Jahan creases cross karti hain, us point ko $M$ kaho.
$A M$ , $M B$ , creases ke beech ka angle, aur $O M$ naapo.
Compasses ko length $A B$ par set karo. Nok ko kinare par kahin bhi rakho aur jahan pencil kinare ko phir se milti hai, wahan mark karo: isse same length ki doosri chord mil jaati hai. Uski bhi $O$ se distance naapo.
Circle, pehli chord aur $O M$ ko tracing paper par trace karo. Dono centres ko ek saath rakho aur tracing ko ghumao: traced chord same length ki doosri chords ke upar aa jaati hai, aur traced $O M$ bhi uske saath ghoomta hai.

Table 5.3Do equal chords

Chord	Length (cm)	Centre se distance (cm)
Pehli chord
Doosri chord

Tumhari entries apne aap save hoti hain.

Abhi nahi kar sakte? Simulation try karo · Fig. 5.11

Kuch chords ko naapna, chahe kitne bhi dhyaan se, sirf unhi chords ke baare mein batata hai. Yeh pakka karne ke liye ki pattern kabhi fail nahi hota, humein aisi wajah chahiye jo equal chords ke har pair par ek saath lagu ho. Right triangles yeh wajah dete hain.

Theorem 6Equal chords centre se barabar door hoti hain

Circle ki equal chords uske centre se barabar distances par hoti hain.

Maan lo $A B$ aur $C D$ centre $O$ wale circle ki equal chords hain, jahan $O M ⊥ A B$ aur $O N ⊥ C D$ . Theorem 5 se, $M$ aur $N$ midpoints hain, isliye $A M = \frac{1}{2} A B = \frac{1}{2} C D = C N$ .

Fig. 5.10Right triangles

O M A

aur

O N C

congruent hain (RHS).NCERT Fig. 5.14 · p. 103

Centre O wala ek circle aur do equal chords, neeche left mein blue chord AB aur upar right mein purple chord CD. Orange perpendiculars OM aur ON chords ko unke midpoints M aur N par right angle par milte hain, aur dashed radii OA aur OC right triangles OMA aur ONC ko poora karte hain.

O M = O N

Ek doosra raasta: triangles $O A B$ aur $O C D$ SSS se congruent hain, aur congruent triangles ki matching sides par heights barabar hoti hain, isliye $O M = O N$ .

Theorem 7Centre se barabar door chords barabar hoti hain

Agar circle ki do chords centre se same distance par hain, to unki length same hoti hai.

SimulationEk circle mein do chords

Centre O wala ek circle aur do chords, blue mein AB aur purple mein CD, jinke chaaron endpoints tum circle par drag kar sakte ho. Perpendiculars OM aur ON har chord ki centre se distance dikhate hain, aur ek panel dono chord lengths, dono distances aur centre par dono angles list karta hai. Ek button CD ko AB ke barabar kar deta hai, aur ek message batata hai ki lengths aur distances mein kya rishta hai.

Fig. 5.11Chaaron points ko drag karo. Panel har chord ki length, centre se uski distance aur centre par uska angle dikhata hai.

Try karo

Make CD = AB dabao. $O M$ ko $O N$ se, aur $∠ A O B$ ko $∠ C O D$ se compare karo.
$D$ ko aise drag karo ki $C D$ , $A B$ se lambi ho jaaye. Ab kaunsi chord $O$ ke zyada paas hai?
Diameter se lambi chord banane ki koshish karo. Tumhe kya rokta hai?

Example 5.3

Har chord kitni door hai?

6.5 cm radius wale circle ki do chords mein se har ek 12 cm lambi hai. Har chord centre se kitni door hai?

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

Practice · 3 questions

Practice 5.4: answers ke saath

ProofMediumNCERT Exercise Set 5.4, Q1
Theorem 6 ko phir se prove karo, is baar congruent triangles ki jagah Baudhāyana–Pythagoras theorem se.
Solution dikhao
$O M^{2} = r^{2} - A M^{2}, O N^{2} = r^{2} - C N^{2}$
Kyunki $A B = C D$ , half-chords barabar hain: $A M = C N$ . To $O M^{2} = O N^{2}$ , aur kyunki dono distances positive hain, $O M = O N$ .
ProofMediumNCERT Exercise Set 5.4, Q2
Centre $O$ wale circle mein $A B$ aur $C D$ chords hain, jahan $O M ⊥ A B$ , $O N ⊥ C D$ aur $O M = O N$ . Congruent triangles se prove karo ki $A B = C D$ .
Solution dikhao
Triangles $O M A$ aur $O N C$ , $M$ aur $N$ par right-angled hain, inke hypotenuses $O A = O C$ (radii) aur legs $O M = O N$ (diya gaya hai). RHS se yeh congruent hain, isliye $A M = C N$ .
Theorem 5 se, $A B = 2 A M$ aur $C D = 2 C N$ . Isliye $A B = C D$ .
3
NumericalMediumNCERT Exercise Set 5.4, Q3
5 cm radius wale circle ki chords $A B$ aur $C D$ mein se har ek centre se 4 cm door hai.
1. Har chord kitni lambi hai, cm mein?
  cm
  Answer dekho
  Answer 6 cm
  Half-chord $= \sqrt{5^{2} - 4^{2}} = 3$ cm, isliye har chord $6$ cm hai.
2. Baudhāyana–Pythagoras theorem se samjhao ki kisi bhi circle ke centre se same distance $d$ par do chords barabar kyun honi chahiye.
  Solution dikhao
  Har half-chord $\sqrt{r^{2} - d^{2}}$ ke barabar hai, jo sirf $r$ aur $d$ par depend karta hai. Isliye same distance wali do chords ke halves barabar hain, aur unki lengths bhi barabar hain.

5.6.1 Do unequal chords mein se kaunsi centre se zyada door hai?

Equal chords centre se barabar distances par hoti hain. Unequal chords ka kya: lambi chord centre ke zyada paas hoti hai ya zyada door? Abhi apna answer tay karo, phir neeche wali activity mein use test karo.

ActivityNCERT Activity · p. 104

Chords ka survey

Tumhe chahiye

ek pair of compasses
ek ruler
ek set square

Pehle guess karo

Ek circle mein tum 2 cm, 6 cm aur 10 cm lambi chords banate ho. Kaunsi centre ke sabse paas hogi?

2 cm wali chord6 cm wali chord10 cm wali chordsab barabar paas

5 cm radius ka ek circle banao aur uska centre $O$ mark karo.
Compasses ko 2 cm par set karo, nok circle par rakho aur jahan pencil circle ko cross karti hai, wahan mark karo. Dono points ko jodo to 2 cm ki chord mil jaayegi. Isi tarah 4, 6, 8 aur 10 cm ki chords banao, circle par alag-alag jagah.
$O$ se set square ki madad se har chord par perpendicular daalo, aur use naapo.
Apne results table mein likho.

Table 5.4Chord ki length aur centre se distance

Chord ki length	O se distance (cm)
2 cm
4 cm
6 cm
8 cm
10 cm

Tumhari entries apne aap save hoti hain.

Abhi nahi kar sakte? Simulation try karo · Fig. 5.11

5/5Lambi chords centre ke zyada paas hoti hain.

Fig. 5.12Jaise-jaise chord centre se door jaati hai, woh chhoti hoti jaati hai.NCERT Fig. 5.16 · p. 105

5 cm radius ka ek circle, jismein centre O ke neeche ek horizontal chord hai. Step 1: chord O se guzarti hai, isliye uski distance d zero hai aur woh 10 cm ka diameter hai. Step 2: chord O se 3 cm neeche aur 8 cm lambi hai, saath mein distance, aadhi chord aur ek radius se bana right triangle hai. Step 3: O se 4 cm neeche chord 6 cm lambi hai. Step 4: O se 4.8 cm neeche woh sirf 2.8 cm lambi hai. Step 5 mein chaaron chords ek saath dikhti hain, saath mein ek table jo har length ko uski distance ke saath jodti hai.

Theorem 8Lambi chord zyada paas hoti hai

Circle ki do unequal chords mein se lambi chord centre ke zyada paas hoti hai.

Maan lo $A B$ aur $C D$ centre $O$ aur radius $r$ wale circle ki chords hain, jahan $A B > C D$ . $A B$ par perpendicular $O M$ aur $C D$ par $O N$ daalo. Theorem 5 se, $M$ aur $N$ midpoints hain, isliye $A M = \frac{1}{2} A B$ aur $C N = \frac{1}{2} C D$ .

O M < O N

Theorem 8 se Section 5.2 wala sabse lambi chord ka sawaal hal ho jaata hai. Distance kabhi $0$ se kam nahi ho sakti, aur sirf centre se guzarne wali chord hi distance $0$ par hoti hai. Woh chord diameter hai, isliye woh baaki har chord se centre ke zyada paas hai, aur isliye koi bhi chord diameter se lambi nahi hoti. Doosri taraf, kinare ke bahut paas bani chord bahut chhoti hoti hai, jaise Fig. 5.12 ke aakhri steps dikhate hain.

Example 5.4

Chord ki length ka formula

Dikhao ki radius $r$ wale circle ke centre se distance $d$ par chord ki length $2 \sqrt{r^{2} - d^{2}}$ hoti hai. Isse 8.5 cm radius wale circle ke centre se 4 cm door chord ki length nikaalo.

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

Quick check

4 mein se 1

Ek circle mein chord

P Q

7 cm lambi hai aur chord

R S

9 cm lambi hai. Kaunsi centre ke zyada paas hai?

P Q

R S

dono barabar paas haintay nahi kiya ja sakta

Answer dekho

Answer (B) $R S$

Theorem 8 se, lambi chord zyada paas hoti hai. Bahut log ulta guess karte hain, kyunki unhe lagta hai ki lambi chord badi hai, to zyada bahar hogi.

Practice · 3 questions

Practice 5.5: answers ke saath

NumericalEasyNCERT Exercise Set 5.5, Q1
Ek circle ka radius 6.1 cm hai aur ek chord centre se 1.1 cm door hai. Chord kitni lambi hai?
cm
Answer dekho
Answer 12 cm
Half-chord $= \sqrt{{6.1}^{2} - {1.1}^{2}} = \sqrt{37.21 - 1.21} = \sqrt{36} = 6$ cm, isliye chord $12$ cm lambi hai.
2
ProofMediumNCERT Exercise Set 5.5, Q2
Example 5.4 wala chord-length formula $c = 2 \sqrt{r^{2} - d^{2}}$ hai.
1. Ise aise rearrange karo ki distance $d$ , $r$ aur $c$ ke terms mein aa jaaye, aur batao ki figure mein har step ka kya matlab hai.
  Solution dikhao
  Answer $d = \sqrt{r^{2} - \frac{c^{2}}{4}}$
  $\frac{c}{2} = \sqrt{r^{2} - d^{2}}$ , isliye $\frac{c^{2}}{4} = r^{2} - d^{2}$ aur $d = \sqrt{r^{2} - \frac{c^{2}}{4}}$ .
  Yeh wahi right triangle $O M B$ hai, bas ulti taraf se padha gaya: distance ek leg hai, aadhi chord doosri leg hai, aur radius hypotenuse hai.
2. 6.5 cm radius wale circle ke centre se 5 cm ki chord kitni door hai?
  cm
  Answer dekho
  Answer 6 cm
  $d = \sqrt{{6.5}^{2} - {2.5}^{2}} = \sqrt{42.25 - 6.25} = \sqrt{36} = 6$ cm.
ProofHardNCERT Exercise Set 5.5, Q3
5 cm radius wale circle mein chord $P Q$ centre se 4 cm door hai aur chord $R S$ centre se 2 cm door hai, yaani $P Q$ double door hai. Kya $R S$ , $P Q$ se double lambi hai? Kya hum general taur par keh sakte hain ki distance double karne se chord aadhi ho jaati hai?
Solution dikhao
AnswerNahi: $P Q = 6$ cm lekin $R S \approx 9.17$ cm, 12 cm nahi.
$P Q = 2 \sqrt{25 - 16} = 6$ cm aur $R S = 2 \sqrt{25 - 4} = 2 \sqrt{21} \approx 9.17$ cm, isliye $R S$ , $2 \times P Q$ nahi hai.
General taur par hum yeh nateeja nahi nikaal sakte: length $\sqrt{r^{2} - d^{2}}$ par depend karti hai, $d$ ke simple proportion mein nahi. Chords ke ek khaas pair ke liye aisa ho sakta hai (jab paas wali chord distance $r / \sqrt{5}$ par ho), isliye claim hamesha galat nahi hai, lekin hamesha sahi bhi bilkul nahi hai.

5.7 Arc ke Subtend Kiye Angles

Circle par do points $A$ aur $B$ use do tukdon mein kaat dete hain, aur har tukda $A$ se $B$ tak ek arc (चाप) hai. Chhota tukda minor arc hai aur bada tukda major arc. Dono arcs ke end points same hain, isliye hum arc ka naam us par pade ek point ko jod kar rakhte hain: Fig. 5.13 mein $A S B$ minor arc hai aur $A T B$ major arc. Jab $A B$ diameter hota hai, to dono tukde barabar hote hain aur har ek semicircle hota hai.

Definition

Arc

Arc circle ka woh hissa hai jo tum pencil uthaye bina circle ke saath ek point se doosre point tak chalte hue banate ho. Dono points uske end points hain, aur woh hamesha circle ko do arcs mein baant dete hain: ek minor arc aur ek major arc, ya do semicircles.

ExampleClock face par 12 se 3 tak (clockwise) ka rim ek minor arc hai; 3 se aage ghoom kar 12 tak ka rim ek major arc hai.

Arc centre par jo angle subtend karta hai, use naapne ke liye socho ki ek radius $O A$ ke saath shuru hota hai aur $O$ ke around ghoomta hai, uska sira arc ke saath chalta hai, jab tak woh $O B$ ke saath na aa jaaye. Woh jitna ghoomta hai, wahi angle hai. Minor arc ke liye yeh normal angle $∠ A O B$ hai, $180^{\circ}$ se kam. Major arc ke liye radius lamba raasta leta hai, isliye angle reflex hai, $180^{\circ}$ se zyada. Dono angles milkar $360^{\circ}$ ka poora turn banate hain.

4/4Centre par dono angles milkar $360^{\circ}$ hote hain.

Fig. 5.13Points ka har pair circle ko ek minor arc aur ek major arc mein baant deta hai.NCERT Figs. 5.17–5.18 · p. 106

Centre O wala ek circle aur uski right side par do points A aur B; unke beech chhote arc par S hai aur lambe arc par T, jo S ke theek saamne hai. Step 1 mein points aur radii OA aur OB hain. Step 2 mein minor arc ASB blue mein highlight hota hai aur O par uska subtend kiya 110 degree ka angle shade hota hai. Step 3 mein major arc ATB orange mein highlight hota hai aur O par 250 degrees ka reflex angle shade hota hai. Step 4 mein dono arcs dikhte hain aur likha hai ki 110 degrees aur 250 degrees milkar 360 degrees hote hain.

Quick check

3 mein se 1

Circle ka minor arc

A B

centre par

130^{\circ}

subtend karta hai. Major arc

A B

centre par kitna angle subtend karta hai?

Answer dekho

Answer 230 °

Dono angles milkar poora turn banate hain:

360^{\circ} - 130^{\circ} = 230^{\circ}

, ek reflex angle.

5.7.1 Arc ke bahar circle ke kisi point par arc ka subtend kiya angle

Ab circle par ek point $P$ chuno jo us arc $A B$ par nahi hai jise hum padh rahe hain. $P$ ko $A$ aur $B$ se jodne par $∠ A P B$ milta hai, jo angle arc $P$ par subtend karta hai. Ab surprise dekho: $P$ ko circle ke baaki hisse par kahin bhi khiskao, $∠ A P B$ ka size bilkul same rehta hai.

ActivityNCERT Activity · p. 107

Ek arc, kai nazariye

Tumhe chahiye

ek pair of compasses
ek ruler
ek protractor

Pehle guess karo

Arc

A B

centre par

100^{\circ}

subtend karta hai. Circle ke baaki hisse par kisi point

P

par

∠ A P B

kitna hoga?

100^{\circ}

50^{\circ}

200^{\circ}

yeh is par depend karta hai ki

P

kahan hai

Centre $O$ aur 5 cm radius ka ek circle banao. $O$ par protractor rakh kar circle par points $A$ aur $B$ aise mark karo ki $∠ A O B = 80^{\circ}$ ho.
Major arc par door-door chaar points $W$ , $X$ , $Y$ aur $Z$ mark karo.
Har ek ko $A$ aur $B$ se jodo, aur $∠ A W B$ , $∠ A X B$ , $∠ A Y B$ aur $∠ A Z B$ naapo.
$∠ A O B = 140^{\circ}$ lekar sab kuch dohrao.

Table 5.5Ek arc ke subtend kiye angles

Point	Angle (°) jab $∠ A O B = 80^{\circ}$	Angle (°) jab $∠ A O B = 140^{\circ}$
$W$
$X$
$Y$
$Z$

Tumhari entries apne aap save hoti hain.

Abhi nahi kar sakte? Simulation try karo

Theorem 9Centre par angle double hota hai

Ek arc centre par jo angle subtend karta hai, woh circle ke baaki hisse ke kisi bhi point par subtend kiye angle ka double hota hai.

Maan lo $A B$ arc hai, $O$ centre hai aur $P$ circle ke baaki hisse par ek point hai (Fig. 5.14). $P O$ ko jodo aur use aage badhao taaki woh circle ko phir se $Q$ par mile. $x = ∠ O P A$ aur $y = ∠ O P B$ likho.

Case 1: $Q$ arc $A B$ par hai. Tab $P Q$ , $P A$ aur $P B$ ke beech se jaati hai, isliye $∠ A P B = x + y$ .

∠ A O B = 2 ∠ A P B

Case 2: $Q$ arc $A B$ ke bahar hai. Ab $P A$ aur $P B$ mein se ek, doosri aur $P Q$ ke beech se jaati hai; maan lo $P B$ , to $∠ A P B = x - y$ . Wahi do steps ab bhi $∠ A O Q = 2 x$ aur $∠ B O Q = 2 y$ dete hain, aur ab $∠ A O B = 2 x - 2 y = 2 (x - y) = 2 ∠ A P B$ .

Yeh argument kaam karta hai chahe arc minor ho ya major. Major arc ke liye $∠ A O B$ reflex angle hai aur $Q$ hamesha arc par aata hai, isliye Case 1 bina badle lagta hai.

3/4 $∠ A O Q = 2 x$ aur $∠ B O Q = 2 y$ , isliye $∠ A O B = 2 (x + y)$ .

Fig. 5.14Do isosceles triangles aur exterior angle theorem se Theorem 9 prove hota hai.NCERT Figs. 5.21–5.22 · p. 108

Centre O wala ek circle. Step 1: neeche arc AB, O par 120 degrees aur upar ek point P par 60 degrees subtend karta hai. Step 2: P se O hokar Q tak diameter banta hai, aur triangles OAP aur OBP shade hote hain, jinke base angles x, x aur y, y hain kyunki har triangle ki do sides radii hain. Step 3: O par exterior angles mark hote hain, angle AOQ barabar 2x aur angle BOQ barabar 2y, isliye angle AOB, 2x plus 2y hai, yaani angle APB ka double. Step 4 mein doosra case hai, P left mein hai taaki Q arc AB ke bahar pade: ab angle AOB, 2x minus 2y hai aur angle APB, x minus y hai, phir bhi theek aadha.

Theorem 9 ka ek zabardast nateeja hai. Circle ke baaki hisse ka har point $P$ arc ko same angle par dekhta hai, jo centre wale angle ka aadha hai. Chord $A B$ aur uske ek arc ke beech ke region ko circle ka segment (वृत्तखंड) kehte hain, isliye yeh fact aksar aise likha jaata hai.

CorollarySame segment mein angles

Ek arc circle ke baaki hisse ke points par jo angles subtend karta hai, woh sab barabar hote hain: same segment mein angles barabar hote hain.

Circle se hatte hi yeh barabari toot jaati hai (Fig. 5.15). Circle ke andar ke point se (figure mein $A B$ ke upar) chord zyada chaudi dikhti hai; bahar ke point se zyada patli. To arc theek woh raasta hai jis par $A B$ ka view same rehta hai, aur is idea ko hum Section 5.8 mein ulta karke use karenge.

1/3 $P$ , $Q$ aur $R$ same arc par hain: har angle $70^{\circ}$ hai.

Fig. 5.15Sirf arc par pade points hi

A B

ko bilkul same angle par dekhte hain.NCERT Fig. 5.25 · p. 110

Centre O wala ek circle aur neeche ke paas ek chord AB, minor arc AB highlight hai. Step 1 mein major arc par teen points P, Q aur R hain, har ek A aur B se juda hai; teeno angles par 70 degrees likha hai. Step 2 mein woh points halke ho jaate hain aur circle ke andar ek point I judta hai, jiska angle AIB lagbhag 93 degrees hai, yaani bada. Step 3 mein circle ke bahar ek point E judta hai, jiska angle AEB lagbhag 38 degrees hai, yaani chhota.

Aise results, jo kisi theorem se lagbhag muft mein nikal aate hain, corollaries (उपप्रमेय) kehlaate hain. Sabse famous corollary tab aati hai jab arc ek semicircle ho.

CorollarySemicircle mein angle

Semicircle mein bana angle right angle hota hai: agar $A B$ diameter hai aur $C$ circle ka koi doosra point hai, to $∠ A C B = 90^{\circ}$ .

Woh arc $A B$ lo jismein $C$ nahi hai. Yeh ek semicircle hai, aur centre par iska subtend kiya angle straight angle $∠ A O B = 180^{\circ}$ hai (Fig. 5.16). Theorem 9 se, $∠ A C B = \frac{1}{2} \times 180^{\circ} = 90^{\circ}$ .

Fig. 5.16Centre ka straight angle aadha hokar

C

par right angle ban jaata hai.NCERT Fig. 5.24 · p. 110

Diameter AB par ek semicircle, AB ke beech mein centre O hai aur O par 180 degrees ka straight angle shade hai. Arc par ek point C blue lines se A aur B se juda hai, aur C par right-angle mark dikhata hai ki angle ACB 90 degrees hai.

Isse Section 5.3 ke right-angled triangles bhi samajh aate hain. Right triangle mein hypotenuse saamne wale vertex par $90^{\circ}$ subtend karta hai, isliye uske us paar ka arc centre par $180^{\circ}$ subtend karta hai. Isliye hypotenuse circumcircle ka diameter hai, aur circumcentre uska midpoint hai.

Example 5.5

Ek chord, do arcs

Points $A$ aur $B$ centre $O$ wale circle par hain, aur minor arc $A B$ , $∠ A O B = 124^{\circ}$ subtend karta hai. $C$ major arc par ek point hai aur $D$ minor arc par ek point hai. $∠ A C B$ aur $∠ A D B$ nikaalo.

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

Example 5.6

Muft mein right angle

$A B$ ek circle ka diameter hai aur $C$ circle par ek point hai jahan $∠ C A B = 34^{\circ}$ . $∠ A C B$ aur $∠ C B A$ nikaalo.

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

Scroll karke explore karo

Ek arc, kai angles

Centre O wala ek circle aur aise points jinhe tum uske upar drag kar sakte ho. Step ke hisaab se stage dikhata hai ki arc centre par kitna angle banata hai, circle ke ek point P par kitna angle banata hai, same segment mein do aise angles, ya semicircle mein bana angle, aur side panel mein live measurements hain.

Centre wale angle se shuru karo. $B$ ko circle par drag karo aur dekho ki $∠ A O B$ aur arc $A B$ ki length saath-saath kaise badalte hain.

Ab circle ke baaki hisse par ek teesra point $P$ hai. $P$ ko arc $A B$ ke bahar kahin bhi drag karo: $∠ A P B$ nahi badalta, aur yeh hamesha $∠ A O B$ ka aadha hota hai.

Yahan $P$ ko $A$ aur $B$ ke beech chhote arc par rakha gaya hai, isliye woh major arc ko dekhta hai. Centre par angle ab reflex hai, $240^{\circ}$ , aur $∠ A P B = 120^{\circ}$ : phir bhi theek aadha.

Same arc par do dekhne wale, $P$ aur $Q$ : tum unhe kaise bhi drag karo, unke angles barabar rehte hain. Ab $Q$ ko doosre arc par le jao. Dono angles ka sum kya aata hai?

Aakhir mein, $A$ aur $B$ ek diameter ke ends hain. $P$ kahin bhi jaaye, $∠ A P B = 90^{\circ}$ : semicircle mein bana angle right angle hota hai.

Practice · 3 questions

Practice 5.6: answers ke saath

1
NumericalEasyNCERT Exercise Set 5.6, Q1
Centre $O$ wale circle mein har case mein chord $P Q$ ki length nikaalo.
1. $∠ P O Q = 60^{\circ}$ aur radius 9.5 cm hai.
  cm
  Answer dekho
  Answer 9.5 cm
  $O P = O Q$ aur unke beech $60^{\circ}$ hai, isliye $△ O P Q$ equilateral hai aur $P Q = 9.5$ cm.
2. $∠ P O Q = 90^{\circ}$ aur radius 5 cm hai.
  cm
  Answer dekho
  Answer 7.07 cm
  $△ O P Q$ , $O$ par right-angled hai, isliye $P Q = \sqrt{5^{2} + 5^{2}} = 5 \sqrt{2} \approx 7.07$ cm.
2
Short answerMediumNCERT Exercise Set 5.6, Q2
$A$ aur $B$ ek circle par do points hain.
1. Maan lo $X$ aur $Y$ dono circle par, $A B$ ki ek hi taraf hain. Kya $∠ A X B$ aur $∠ A Y B$ alag ho sakte hain?
  Solution dikhao
  AnswerNahi.
  $A B$ ki ek taraf ke circle ke points ek hi arc par hote hain, isliye $∠ A X B$ aur $∠ A Y B$ same segment ke angles hain. Har ek centre ke same angle ka aadha hai, isliye dono barabar hain.
2. Circle ke points $X$ aur $Y$ se $∠ A X B = ∠ A Y B$ milta hai. Kya unka $A B$ ki ek hi taraf hona zaroori hai?
  Solution dikhao
  AnswerZaroori nahi.
  Agar $A B$ diameter hai, to circle ka har point, kisi bhi taraf ho, $A B$ ko $90^{\circ}$ par dekhta hai. Kisi aur chord ke liye dono taraf ke angles ka sum $180^{\circ}$ hota hai aur woh unequal hote hain, isliye aisi chord ke liye $X$ aur $Y$ ek hi taraf hone chahiye.
3. Ab $∠ A X B = ∠ A Y B$ hai aur $X$ aur $Y$ , $A B$ ki ek taraf hain, lekin koi nahi kehta ki $Y$ , $A$ , $B$ aur $X$ se guzarne wale circle par hai. Kya woh circle $Y$ se guzarta hai?
  Solution dikhao
  AnswerHaan.
  Yeh theek Section 5.8 ka Theorem 10 hai: agar $A B$ apni ek hi taraf ke do points par equal angles subtend karta hai, to chaaron points concyclic hain.
NumericalEasyNCERT Exercise Set 5.6, Q3
$P Q R S$ ek quadrilateral hai jiske chaaron vertices ek circle par hain, aur $∠ P S R = 112^{\circ}$ . $x = ∠ P Q R$ degrees mein nikaalo.
Fig. 5.18
Ek circle jis par chaar points order mein jud kar quadrilateral PQRS banate hain: P upar ke paas, Q right mein, R neeche ke paas aur S left mein. S par angle 112 degrees mark hai, aur Q par saamne wala angle letter x se mark hai.
°
Answer dekho
Answer 68 °
$∠ P S R$ us angle ka aadha hai jo arc $P Q R$ centre par subtend karta hai, aur $x$ arc $P S R$ ke angle ka aadha hai. Centre par woh do angles milkar $360^{\circ}$ hote hain, isliye $112^{\circ} + x = 180^{\circ}$ aur $x = 68^{\circ}$ .

5.8 Points ki Concyclicity

Teen points jo collinear nahi hain, hamesha ek circle par hote hain (Theorem 1). Chaar points aam taur par nahi hote: chauthe point ko baaki teen se guzarne wale circle par theek-theek aana padta hai. Jo points ek circle par hote hain, unhe concyclic (एकवृत्तीय) kehte hain. Bina koi circle banaye hum chaar points ko kaise test kar sakte hain?

Same-segment property ek test deti hai. Agar $C$ aur $D$ , $A$ aur $B$ se guzarne wale circle par, $A B$ ki ek hi taraf hain, to $∠ A C B = ∠ A D B$ . Kamaal ki baat yeh hai ki converse bhi sahi hai.

Theorem 10Ek taraf equal angles matlab concyclic

Agar segment $A B$ line $A B$ ki ek hi taraf ke do points $C$ aur $D$ par equal angles subtend karta hai, to $A$ , $B$ , $C$ aur $D$ ek circle par hain.

Theorem 1 se humein $A$ , $B$ aur $C$ se guzarne wala ek circle milta hai. Hum dikhate hain ki $D$ bhi us par hona chahiye, baaki har possibility ko galat saabit karke (Fig. 5.19).

Fig. 5.19Agar

D

circle se hat kar hota, to in do pictures mein se ek lagu hoti, aur dono contradiction tak le jaati hain.NCERT Fig. 5.27 · p. 111

Do panels, har ek mein ek circle hai jiski left side par chord AB khadi hai aur right wale arc par ek point C hai, jahan angle ACB mark hai. Left panel par Case 1 likha hai; usmein D circle ke bahar neeche right mein hai; segment AD circle ko E par cross karta hai, aur E par ka angle, jo dashed arc se mark hai, C wale angle ke barabar hai aur D wale angle se saaf bada hai. Right panel par Case 2 likha hai; usmein D circle ke andar hai aur aage badhaya gaya AD circle ko E par milta hai; ab D wala angle E wale angle se bada hai, jo phir se C wale angle ke barabar hai.

Maan lo $D$ circle ke bahar hai, aur segment $A D$ circle ko phir se $A$ aur $D$ ke beech ek point $E$ par milta hai, jaise left picture mein. Tab $E$ aur $C$ same arc par hain, isliye $∠ A E B = ∠ A C B$ . Lekin $∠ A E B$ , $△ E B D$ ka exterior angle hai, isliye woh interior opposite angle $∠ E D B$ , yaani $∠ A D B$ , se bada hai. Isse $∠ A C B > ∠ A D B$ ho jaata hai, jo diye gaye fact ke khilaaf hai. (Agar $D$ aisi jagah hai jahan $A D$ circle ko nahi chhoota, to thodi alag picture se bhi yahi contradiction milta hai.)

Maan lo $D$ circle ke andar hai. $A D$ ko $D$ ke aage badhao taaki woh circle ko $E$ par mile. Ab $∠ A D B$ , $△ D B E$ ka exterior angle hai, isliye $∠ A D B > ∠ A E B = ∠ A C B$ : phir se contradiction.

Koi bhi case possible nahi hai, isliye $D$ , $A$ , $B$ aur $C$ se guzarne wale circle par hai, aur chaaron points concyclic hain.

Jis quadrilateral ke chaaron vertices concyclic hon, use cyclic quadrilateral (चक्रीय चतुर्भुज) kehte hain. Iske angles ek sundar rule follow karte hain (Fig. 5.20).

4/4 $140^{\circ} + 220^{\circ} = 360^{\circ}$ , isliye $∠ A + ∠ C = 180^{\circ}$ .

Fig. 5.20Dono arcs centre par poora turn banate hain, isliye opposite angles aadha turn banate hain.NCERT Fig. 5.28 · p. 112

Centre O wale circle mein inscribed ek quadrilateral ABCD; angle A 70 degrees hai aur angle C 110 degrees. Step 1 mein quadrilateral aur yeh do angles hain. Step 2 mein arc BCD blue mein highlight hota hai aur O par uska subtend kiya 140 degree ka angle shade hota hai; angle A uska aadha hai. Step 3 mein arc DAB pink mein highlight hota hai aur O par uska subtend kiya 220 degree ka reflex angle shade hota hai; angle C uska aadha hai. Step 4 mein dono shaded angles O par poora turn bhar dete hain: 140 plus 220, 360 hai, isliye angle A plus angle C, 180 hai.

Theorem 11Cyclic quadrilateral ke opposite angles

Cyclic quadrilateral mein opposite angles ke har pair ka sum $180^{\circ}$ hota hai.

Maan lo $A B C D$ cyclic hai aur centre $O$ hai. $A$ arc $B C D$ par nahi hai, aur $C$ arc $D A B$ par nahi hai. Yeh dono arcs milkar poora circle banate hain.

∠ A + ∠ C = 180^{\circ}

Arcs $A B C$ aur $C D A$ ke saath wahi argument $∠ B + ∠ D = 180^{\circ}$ deta hai.

SimulationCyclic quadrilateral

Ek quadrilateral ABCD jiske chaaron vertices tum centre O wale circle par drag kar sakte ho, aur woh hamesha order mein rehte hain. Chaaron angles mark hain, aur ek panel har angle ke saath angle A plus angle C aur angle B plus angle D ke sums dikhata hai, jo shape kaise bhi badle, 180 degrees par hi rehte hain.

Fig. 5.21Kisi bhi vertex ko circle ke saath drag karo aur dono sums dekho.

Try karo

$A$ ko circle ke saath dheere-dheere drag karo. Kaunse angles badalte hain, aur kaunse sums fixed rehte hain?
$A B C D$ ko rectangle jaisa banao. Uska har angle kitna hona chahiye?
$B$ ko $A$ ke bahut paas le aao. Kya $∠ B + ∠ D$ abhi bhi $180^{\circ}$ hai?

Activity

Phaado aur jodo

Tumhe chahiye

paper ki ek sheet aur scissors
ek pair of compasses
ek ruler

Pehle guess karo

Tum cyclic quadrilateral ke angles

A

aur

C

phaad kar ek doosre ke bagal mein rakhte ho. Dono milkar kya banate hain?

ek right angleek straight angleek poora turn

Ek bada circle banao aur us par koi bhi chaar points mark karo. Circle ke chaaron taraf order mein unhe $A$ , $B$ , $C$ , $D$ naam do, unhe jod kar quadrilateral banao aur kaat lo.
$A$ aur $C$ ke corners phaad lo aur unhe ek doosre ke bagal mein rakho, aise ki unke vertices chhoo rahe hon.
$B$ aur $D$ ke corners ke saath bhi yahi karo.
Kisi aise dost se compare karo jiske quadrilateral ki shape alag hai.

Abhi nahi kar sakte? Simulation try karo · Fig. 5.21

Example 5.7

Ratio mein opposite angles

Cyclic quadrilateral $A B C D$ mein $∠ A B C$ , $∠ A D C$ ka teen guna hai. Dono angles nikaalo.

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

Theorem 11 ka converse bhi sahi hai, aur yeh chaar concyclic points ka ek jaldi wala test deta hai.

Theorem 12Cyclic quadrilaterals ka ek test

Agar kisi quadrilateral ke opposite angles ke ek pair ka sum $180^{\circ}$ hai, to quadrilateral cyclic hai: ek circle uske chaaron vertices se guzarta hai.

Maan lo $A B C D$ ek quadrilateral hai jismein $∠ A + ∠ C = 180^{\circ}$ . Chaaron angles ka sum $360^{\circ}$ hota hai, isliye $∠ B + ∠ D = 180^{\circ}$ bhi; koi angle reflex nahi hai, aur $A$ aur $C$ diagonal $B D$ ki ulti taraf hain.

$A$ , $B$ aur $D$ se guzarne wala circle banao (Theorem 1). Abhi humein nahi pata ki $C$ us par hai ya nahi (Fig. 5.22).

Fig. 5.22

∠ B E D = 180^{\circ} - ∠ A = ∠ B C D

, isliye Theorem 10,

C

B

D

aur

E

se guzarne wale circle par rakh deta hai.NCERT Fig. 5.29 · p. 113

A, B aur D se guzarta centre O wala ek circle. Quadrilateral ABCD blue shade mein hai, uska chautha vertex C right mein hai aur uske bagal mein question mark hai. Arc BD par A se door ek point E dashed lines se B aur D se juda hai. A, C aur E par angles shade hain, aur ek key mein angle A = 75 degrees, angle C = 105 degrees aur angle BED = 105 degrees likha hai. Ek note mein likha hai: E circle par hai; kya C bhi us par hai?

Arc $B D$ par, jismein $A$ nahi hai, koi bhi point $E$ chuno. Tab $A B E D$ ek cyclic quadrilateral hai, isliye us par Theorem 11 lagta hai.

∠ B E D = ∠ B C D

$E$ aur $C$ dono $B D$ ki us taraf hain jo $A$ se door hai, aur $B D$ un par equal angles subtend karta hai. Theorem 10 se, $B$ , $D$ , $E$ aur $C$ ek circle par hain. Lekin $B$ , $D$ aur $E$ se sirf ek circle guzarta hai: woh circle jo humne banaya. To $C$ us par hai, aur $A B C D$ cyclic hai.

Example 5.8

Kya yeh cyclic hai?

Quadrilateral $P Q R S$ mein $∠ P = 72^{\circ}$ hai. Side $Q R$ ko $R$ ke aage ek point $T$ tak badhaya gaya hai, aur exterior angle $∠ S R T = 72^{\circ}$ hai. Kya $P Q R S$ cyclic hai?

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

Humara safar yahin poora hota hai. Ek simple rule, centre se barabar distance, se shuru karke hum chords, arcs, angles aur concyclic points ke baarah theorems tak pahunch gaye. Class 10 mein tum tangents jodoge, yaani aisi lines jo circle ko theek ek point par chhooti hain, aur dekhoge ki woh bhi inhi jaise saaf-suthre rules follow karti hain.

Recap

Summary aur key points

Circle ka har point, aur plane ka koi doosra point nahi, centre se distance $r$ par hai: circle ek locus hai (Section 5.1).
Circle apne centre ke around kisi bhi turn ke baad waisa hi dikhta hai, aur har diameter ek line of symmetry hai (Section 5.2).
Do diye gaye points se anginat circles guzarte hain, aur unke centres un points ka poora perpendicular bisector banate hain (Section 5.3).
Teen non-collinear points se theek ek circle, yaani circumcircle, guzarta hai; uska centre wahan hai jahan perpendicular bisectors milte hain (Theorem 1).
Barabar length ki chords centre par equal angles banati hain, aur centre par equal angles banane wali chords ki length barabar hoti hai (Theorem 2, Theorem 3).
Centre ko chord ke beech se jodne par right angle milta hai, aur centre se daala gaya perpendicular chord ko aadha kar deta hai (Theorem 4, Theorem 5).
Equal chords centre se barabar door hoti hain, aur centre se barabar door chords barabar hoti hain (Theorem 6, Theorem 7).
Do unequal chords mein se lambi chord centre ke zyada paas hoti hai, isliye diameter sabse lambi chord hai (Theorem 8).
Centre se distance $d$ par chord ki length $2 \sqrt{r^{2} - d^{2}}$ hoti hai (Example 5.4).
Ek arc centre par circle ke baaki hisse ke kisi bhi point ke mukaable double angle subtend karta hai (Theorem 9).
Same segment mein angles barabar hote hain, aur semicircle mein bana angle $90^{\circ}$ hota hai (Section 5.7.1).
Agar $A B$ , $C$ aur $D$ par same angle banata hai, aur dono $A B$ ki ek hi taraf hain, to $A$ , $B$ , $C$ aur $D$ concyclic hain (Theorem 10).
Cyclic quadrilateral ke opposite angles ka sum $180^{\circ}$ hota hai, aur aisa pair wala quadrilateral cyclic hota hai (Theorem 11, Theorem 12).

Flash cards · 13

Theorem flashcards: agar … to …

1 / 13

Chapter ke end ki exercises

Practice · 30 questions

Practice questions answers ke saath (NCERT “Chapter ke end ki exercises”)

NumericalEasyNCERT End-of-chapter Exercises, Q1
17 cm radius wale circle ki ek chord centre se 8 cm door hai. Chord kitni lambi hai?
cm
Answer dekho
Answer 30 cm
Half-chord $= \sqrt{17^{2} - 8^{2}} = \sqrt{225} = 15$ cm, isliye chord $30$ cm lambi hai.
NumericalEasyNCERT End-of-chapter Exercises, Q2
Ek arc circle ke centre par $154^{\circ}$ subtend karta hai. Circle ke baaki hisse ke kisi point par woh kitna angle, degrees mein, subtend karta hai?
°
Answer dekho
Answer 77 °
Theorem 9 se angle aadha hai: $154^{\circ} \div 2 = 77^{\circ}$ .
NumericalEasyNCERT End-of-chapter Exercises, Q3
Ek circle ka diameter 50 cm hai. 48 cm lambi chord uske centre se kitni door hai?
cm
Answer dekho
Answer 7 cm
$r = 25$ cm aur aadhi chord 24 cm hai, isliye distance $\sqrt{25^{2} - 24^{2}} = \sqrt{49} = 7$ cm hai.
NumericalEasyNCERT End-of-chapter Exercises, Q4
Ek circle ka radius 29 cm hai, aur uski ek chord centre se 20 cm door hai. Chord ki length nikaalo.
cm
Answer dekho
Answer 42 cm
Aadhi chord $\sqrt{29^{2} - 20^{2}} = \sqrt{441} = 21$ cm hai, isliye chord $42$ cm lambi hai.
ProofMediumNCERT End-of-chapter Exercises, Q5
Ek carpenter gol table top ka centre dhoondhna chahta hai. Prove karo ki kisi bhi chord ka perpendicular bisector centre se guzarta hai, aur samjhao ki phir do chords se centre kaise milta hai.
Solution dikhao
Maan lo $A B$ ek chord hai aur $O$ centre hai. $O A = O B$ (radii), isliye $O$ , $A$ aur $B$ se barabar door hai aur isliye $A B$ ke perpendicular bisector par hai (Section 5.3 ka locus wala result).
Do aisi chords banao jo parallel nahi hain aur unke perpendicular bisectors construct karo. Dono $O$ se guzarte hain, aur do lines jo parallel nahi hain, sirf ek point par milti hain, isliye unka crossing point hi centre hai.
6
Short answerEasyNCERT End-of-chapter Exercises, Q6
$P Q$ ek circle ka diameter hai aur $R$ circle par ek point hai jahan $∠ R P Q = 27^{\circ}$ .
1. $∠ P R Q$ degrees mein nikaalo.
  °
  Answer dekho
  Answer 90 °
  $∠ P R Q$ semicircle mein bana angle hai, isliye yeh $90^{\circ}$ hai.
2. $∠ R Q P$ degrees mein nikaalo.
  °
  Answer dekho
  Answer 63 °
  $△ P Q R$ mein angle sum: $180^{\circ} - 90^{\circ} - 27^{\circ} = 63^{\circ}$ .
7
NumericalEasyNCERT End-of-chapter Exercises, Q7
$K L M N$ ek cyclic quadrilateral hai.
1. Agar $∠ K = 58^{\circ}$ hai, to $∠ M$ degrees mein nikaalo.
  °
  Answer dekho
  Answer 122 °
  Opposite angles ka sum $180^{\circ}$ hota hai: $180^{\circ} - 58^{\circ} = 122^{\circ}$ .
2. Agar $∠ L = 96^{\circ}$ hai, to $∠ N$ degrees mein nikaalo.
  °
  Answer dekho
  Answer 84 °
  $180^{\circ} - 96^{\circ} = 84^{\circ}$ .
8
NumericalMediumNCERT End-of-chapter Exercises, Q8
Quadrilateral $W X Y Z$ ek circle mein inscribed hai, jahan $∠ W = (3 y + 5)^{\circ}$ aur $∠ Y = (2 y + 25)^{\circ}$ .
1. $y$ nikaalo.
  Answer dekho
  Answer 30
  $∠ W + ∠ Y = 180^{\circ}$ , isliye $5 y + 30 = 180$ aur $y = 30$ .
2. $∠ W$ degrees mein nikaalo.
  °
  Answer dekho
  Answer 95 °
  $3 \times 30 + 5 = 95$ , isliye $∠ W = 95^{\circ}$ .
3. $∠ Y$ degrees mein nikaalo.
  °
  Answer dekho
  Answer 85 °
  $2 \times 30 + 25 = 85$ , isliye $∠ Y = 85^{\circ}$ . Check: $95^{\circ} + 85^{\circ} = 180^{\circ}$ .
NumericalMediumNCERT End-of-chapter Exercises, Q9
24 cm lambi chord ek circle ke centre se 3.5 cm door hai. Radius nikaalo.
cm
Answer dekho
Answer 12.5 cm
$r = \sqrt{12^{2} + {3.5}^{2}} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5$ cm.
NumericalHardNCERT End-of-chapter Exercises, Q10
Ek cyclic quadrilateral $A B C D$ mein $A B = A D = 8$ units aur $C B = C D = 15$ units hain. Iska area square units mein nikaalo.
sq. units
Solution dikhao
Answer120 square units
Triangles $A B C$ aur $A D C$ congruent hain (SSS: $A B = A D$ , $C B = C D$ aur $A C$ common hai), isliye $∠ B = ∠ D$ .
$A B C D$ cyclic hai, isliye $∠ B + ∠ D = 180^{\circ}$ , jisse $∠ B = ∠ D = 90^{\circ}$ milta hai.
Quadrilateral legs 8 aur 15 wale do right triangles se bana hai, isliye iska area $2 \times \frac{1}{2} \times 8 \times 15 = 120$ square units hai. (Agar sides 8, 15, 8, 15 ke order mein hoti, to shape 8 by 15 ka rectangle hoti, aur area same hota.)
Short answerHardNCERT End-of-chapter Exercises, Q11
Circle banaye bina tum kaise bataoge ki cyclic quadrilateral ke vertices se guzarne wale circle ka centre quadrilateral ke andar hai, uski kisi side par hai, ya uske bahar?
Solution dikhao
AnswerHar side door wale vertex par jo angle subtend karti hai, use naapo: centre tabhi andar hota hai jab yeh chaaron angles acute hon.
Cyclic quadrilateral $A B C D$ ki side $A B$ lo. Yeh vertex $C$ par $∠ A C B$ subtend karti hai.
Agar $∠ A C B < 90^{\circ}$ , to $C$ se door wala arc $A B$ minor arc hai, isliye centre $A B$ ki usi taraf hai jis taraf quadrilateral hai. Agar $∠ A C B = 90^{\circ}$ , to $A B$ diameter hai aur centre us par hai. Agar $∠ A C B > 90^{\circ}$ , to centre $A B$ ke us paar, bahar hai.
To $∠ A C B$ , $∠ B D C$ , $∠ C A D$ aur $∠ D B A$ naapo, jinmein se har ek ek diagonal aur ek side ke beech hai. Centre quadrilateral ke andar tabhi hota hai jab yeh chaaron acute hon.
ProofHardNCERT End-of-chapter Exercises, Q12
Ek circle ki do equal chords ek doosre ko ek point $X$ par cross karti hain. Dikhao ki $X$ ek chord ko jin do tukdon mein baant deta hai, woh doosri chord ke do tukdon ke barabar hain.
Solution dikhao
Maan lo centre $O$ wale circle mein chords $P Q$ aur $R S$ hain, har ek ki length $2 h$ . $O M ⊥ P Q$ aur $O N ⊥ R S$ daalo. Theorems 5 aur 6 se, $M$ aur $N$ midpoints hain aur $O M = O N$ .
Triangles $O M X$ aur $O N X$ , $M$ aur $N$ par right-angled hain, inka hypotenuse $O X$ common hai, aur $O M = O N$ . RHS se yeh congruent hain, isliye $M X = N X$ ; is length ko $k$ kaho.
$X$ har chord ke midpoint se $k$ door hai, isliye $P Q$ ke tukde $h + k$ aur $h - k$ hain, aur $R S$ ke tukde bhi $h + k$ aur $h - k$ hain.
13
DiagramMediumNCERT End-of-chapter Exercises, Q13
Aisa circle construct karo jismein 8 cm lambi chord centre se theek 3 cm door ho.
Solution dikhao
AnswerCircle ka radius 5 cm hai.
$A B = 8$ cm banao aur uska perpendicular bisector construct karo, jo $A B$ ko $M$ par milta hai.
Bisector par $O$ aise mark karo ki $O M = 3$ cm ho.
Centre $O$ aur radius $O A$ wala circle banao.
Construction se $O M ⊥ A B$ aur $O M = 3$ cm hai, isliye $A B$ centre se 3 cm door ek chord hai. Radius $O A = \sqrt{4^{2} + 3^{2}} = 5$ cm hai.
ProofMediumNCERT End-of-chapter Exercises, Q14
Ek parallelogram ke chaaron vertices ek circle par hain. Prove karo ki woh rectangle hi hona chahiye.
Solution dikhao
Parallelogram $A B C D$ mein $∠ A = ∠ C$ . Kyunki $A B C D$ cyclic hai, $∠ A + ∠ C = 180^{\circ}$ (Theorem 11).
To $2 ∠ A = 180^{\circ}$ aur $∠ A = 90^{\circ}$ . Ek right angle wale parallelogram ke chaaron angles right angles hote hain, isliye woh rectangle hai.
ProofMediumNCERT End-of-chapter Exercises, Q15
Ek rectangle $A B C D$ circle mein inscribed hai. Dikhao ki dono diagonals diameters hain, aur isliye woh centre par cross karte hain.
Solution dikhao
$∠ A B C = 90^{\circ}$ , isliye chord $A C$ circle ke point $B$ par $90^{\circ}$ subtend karti hai. Theorem 9 se, $B$ ke bina wala arc $A C$ centre $O$ par $2 \times 90^{\circ} = 180^{\circ}$ subtend karta hai. To $A$ , $O$ aur $C$ ek line par hain, aur $A C$ diameter hai.
Isi tarah $B D$ bhi diameter hai. Do diameters centre par milte hain, isliye diagonals $O$ par cross karte hain.
Short answerHardNCERT End-of-chapter Exercises, Q16
10 cm radius wale circle mein 16 cm lambi saari chords ke baare mein socho. Unke midpoints kaunsa figure banate hain?
Solution dikhao
AnswerSame centre aur 6 cm radius wala circle.
Aisi har chord ka midpoint $M$ , $O$ se daale gaye perpendicular ka foot hai, isliye har ek ke liye $O M = \sqrt{10^{2} - 8^{2}} = 6$ cm.
Ulta, $O$ se 6 cm door har point aisi ek chord ka midpoint hai: us point se radius ke perpendicular chord banao. Isliye midpoints centre $O$ aur radius 6 cm wala circle banate hain.
ProofMediumNCERT End-of-chapter Exercises, Q17
Centre $O$ wale circle ki chords $P Q$ aur $P R$ barabar hain. Prove karo ki $O$ , $∠ Q P R$ ke bisector par hai.
Solution dikhao
$O Q = O R$ (radii), $P Q = P R$ (diya gaya hai) aur $O P$ common hai, isliye $△ O P Q ≅ △ O P R$ (SSS).
Isliye $∠ O P Q = ∠ O P R$ : line $P O$ , $∠ Q P R$ ko bisect karti hai, isliye $O$ bisector par hai.
NumericalHardNCERT End-of-chapter Exercises, Q18
Ek circle ki do parallel chords, 14 cm aur 48 cm lambi, centre ki ek hi taraf hain aur unke beech 17 cm ki doori hai. Radius nikaalo.
cm
Solution dikhao
Half-chords 7 cm aur 24 cm hain. Lambi chord centre ke zyada paas hai (Theorem 8); maan lo uski distance $d$ hai, to chhoti chord centre se $d + 17$ door hai.
$r^{2} = d^{2} + 24^{2} = (d + 17)^{2} + 7^{2}$
Expand karne par: $d^{2} + 576 = d^{2} + 34 d + 289 + 49$ , isliye $34 d = 238$ aur $d = 7$ .
$r = \sqrt{7^{2} + 24^{2}} = \sqrt{625} = 25 c m$
19
NumericalHardNCERT End-of-chapter Exercises, Q19
Ek regular hexagon ke chhe vertices 8 cm radius wale circle par hain.
1. Har side kitni lambi hai?
  cm
  Answer dekho
  Answer 8 cm
  Har side centre par $360^{\circ} \div 6 = 60^{\circ}$ subtend karti hai, isliye centre ke saath bane chhe triangles mein se har ek equilateral hai, aur har side 8 cm hai.
2. Har side centre se kitni door hai?
  cm
  Answer dekho
  Answer 6.93 cm
  Distance $= \sqrt{8^{2} - 4^{2}} = \sqrt{48} = 4 \sqrt{3} \approx 6.93$ cm.
20
Short answerMediumNCERT End-of-chapter Exercises, Q20
Quadrilateral $W X Y Z$ ek circle mein inscribed hai, jahan $W X$ diameter hai aur $∠ X W Z = 38^{\circ}$ . Har answer ki wajah batao.
1. $∠ W Z X$ degrees mein nikaalo.
  °
  Answer dekho
  Answer 90 °
  $∠ W Z X$ ko diameter $W X$ circle ke point $Z$ par subtend karta hai: yeh semicircle mein bana angle hai, isliye $90^{\circ}$ hai.
2. $∠ W X Z$ degrees mein nikaalo.
  °
  Answer dekho
  Answer 52 °
  $△ W X Z$ mein angle sum: $180^{\circ} - 90^{\circ} - 38^{\circ} = 52^{\circ}$ .
3. $∠ W Y Z$ degrees mein nikaalo.
  °
  Solution dikhao
  $∠ W Y Z$ aur $∠ W X Z$ dono ko chord $W Z$ subtend karti hai, points $Y$ aur $X$ par, jo same arc $W X Y Z$ par hain. Same segment mein angles barabar hote hain, isliye $∠ W Y Z = 52^{\circ}$ .
4. $∠ W Y X$ degrees mein nikaalo.
  °
  Answer dekho
  Answer 90 °
  $∠ W Y X$ ko bhi diameter $W X$ subtend karta hai, isliye yeh semicircle mein bana angle hai: $90^{\circ}$ .
ProofMediumNCERT End-of-chapter Exercises, Q21
$P Q R S$ ek cyclic quadrilateral hai, aur side $Q R$ ko $R$ ke aage ek point $T$ tak badhaya gaya hai. Prove karo ki exterior angle $∠ S R T$ interior opposite angle $∠ Q P S$ ke barabar hai.
Solution dikhao
$∠ S R T = 180^{\circ} - ∠ Q R S$ (seedhi line par angles).
$∠ Q P S = 180^{\circ} - ∠ Q R S$ (Theorem 11, kyunki $P$ aur $R$ opposite vertices hain).
To $∠ S R T = ∠ Q P S$ .
ProofMediumNCERT End-of-chapter Exercises, Q22
Is statement ko justify karo: circle ki koi bhi chord diameter se lambi nahi hoti.
Solution dikhao
Maan lo $A B$ ek chord hai aur $O$ centre hai. Agar $O$ , $A B$ par nahi hai, to $△ O A B$ mein triangle inequality se $A B < O A + O B = 2 r$ milta hai.
Agar $O$ , $A B$ par hai, to $A B = O A + O B = 2 r$ , aur chord diameter hai. To har chord zyada se zyada $2 r$ lambi hai. (Yaani $2 \sqrt{r^{2} - d^{2}} \leq 2 r$ , aur barabari sirf tab jab $d = 0$ .)
ProofHardNCERT End-of-chapter Exercises, Q23
$K$ centre $O$ wale circle ke andar ek point hai, aur $K \neq O$ . Dikhao ki $K$ se guzarne wali sabse chhoti chord woh hai jo $O K$ ke perpendicular hai.
Solution dikhao
Maan lo $K$ se guzarne wali ek chord $O$ se distance $d$ par hai, aur $M$ perpendicular ka foot hai. Agar $M \neq K$ , to triangle $O M K$ ka right angle $M$ par hai, isliye hypotenuse $O K$ , $O M$ se lamba hai. To $d \leq O K$ , aur barabari theek tab jab $M = K$ , yaani jab chord $O K$ ke perpendicular ho.
Chord ki length $2 \sqrt{r^{2} - d^{2}}$ hai, jo tab sabse chhoti hoti hai jab $d$ sabse bada ho. Isliye $K$ se guzarne wali sabse chhoti chord woh hai jo centre se distance $O K$ par hai: $O K$ ke perpendicular chord.
ProofMediumNCERT End-of-chapter Exercises, Q24
$A B$ centre $O$ wale circle ka diameter hai, aur $C$ circle par ek aur point hai. $O C$ jodo, aur isosceles triangles $O A C$ aur $O B C$ ke base angles ko $p$ aur $q$ kaho. Inse Theorem 9 use kiye bina dikhao ki $∠ A C B = 90^{\circ}$ .
Solution dikhao
$O A = O C$ , isliye $∠ O A C = ∠ O C A = p$ ; aur $O B = O C$ , isliye $∠ O B C = ∠ O C B = q$ .
$∠ A C B = ∠ O C A + ∠ O C B = p + q$ , isliye $△ A B C$ ke angles $p$ , $q$ aur $p + q$ hain, jinka sum $2 (p + q) = 180^{\circ}$ hai.
Isliye $∠ A C B = p + q = 90^{\circ}$ .
ProofHardNCERT End-of-chapter Exercises, Q25
$A B$ ek circle ka diameter hai. Chords $P P^{'}$ aur $Q Q^{'}$ dono $A B$ ke perpendicular hain. Prove karo ki $P Q$ ke midpoint ko $P^{'} Q^{'}$ ke midpoint se jodne wala segment $A B$ ke perpendicular hai.
Solution dikhao
Diameter $A B$ mein reflection circle ko usi par map karta hai (Section 5.2). $A B$ centre se guzarta hai aur $P P^{'}$ ke perpendicular hai, isliye woh $P P^{'}$ ko bisect karta hai (Theorem 5); isliye reflection $P$ aur $P^{'}$ ki jagah badal deta hai. Isi tarah woh $Q$ aur $Q^{'}$ ki jagah bhi badal deta hai.
To reflection segment $P Q$ ko $P^{'} Q^{'}$ par, aur $P Q$ ke midpoint $M$ ko $P^{'} Q^{'}$ ke midpoint $M^{'}$ par le jaata hai.
Ek point aur uski mirror image ko jodne wala segment mirror line ke perpendicular hota hai, isliye $M M^{'} ⊥ A B$ . (Agar $M$ , $A B$ par hi ho, to $M^{'} = M$ aur prove karne ko kuch nahi bachta.)
ProofMediumNCERT End-of-chapter Exercises, Q26
$A B C D$ ek cyclic quadrilateral hai jiska centre $O$ uske andar hai. $O$ ko chaaron vertices se jodo, aur isosceles triangles $O A B$ , $O B C$ , $O C D$ aur $O D A$ ke base angles ko $w$ , $x$ , $y$ aur $z$ kaho. Inse prove karo ki $∠ A + ∠ C = 180^{\circ}$ .
Solution dikhao
Har vertex angle do base angles mein bant jaata hai: $∠ A = z + w$ , $∠ B = w + x$ , $∠ C = x + y$ aur $∠ D = y + z$ .
Quadrilateral ke angles ka sum $360^{\circ}$ hota hai, isliye $2 (w + x + y + z) = 360^{\circ}$ aur $w + x + y + z = 180^{\circ}$ .
Isliye $∠ A + ∠ C = z + w + x + y = 180^{\circ}$ .
Quick answerEasy
Kaunsa statement galat hai?
Circle ki equal chords uske centre se barabar door hoti hain.Centre se chord par daala gaya perpendicular use aadha kar deta hai.Semicircle mein bana angle right angle hota hai.Lambi chord centre se zyada door hoti hai.
Answer dekho
Answer (D) Lambi chord centre se zyada door hoti hai.
Lambi chord centre ke zyada paas hoti hai (Theorem 8). Baaki teen Theorem 6, Theorem 5 aur semicircle wali corollary hain.
Quick answerEasy
Arc $P Q$ circle ke baaki hisse ke ek point $R$ par $∠ P R Q = 41^{\circ}$ subtend karta hai. Centre par iska subtend kiya angle degrees hai.
Answer dekho
Answer 82
$41^{\circ}$ ka double $82^{\circ}$ hai.
29
Quick answerMedium
5 cm radius wale circle ki in chords ko order mein lagao, centre ke sabse paas se sabse door tak.
1. 8 cm wali chord
2. 6 cm wali chord
3. 10 cm wali chord
4. 2.8 cm wali chord
Answer dekho
Answer
10 cm wali chord
8 cm wali chord
6 cm wali chord
2.8 cm wali chord
Inki distances 0, 3, 4 aur 4.8 cm hain: chord jitni lambi, utni paas.
30
Quick answerMedium
Har theorem ko us congruence rule se match karo jo is chapter mein uske proof mein use hua.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
Theorem 2: equal chords se equal angles : SSS
Theorem 3: equal angles se equal chords : SAS
Theorem 5: perpendicular chord ko bisect karta hai : RHS
Theorem 2 mein sides ke teen pairs milte hain, Theorem 3 mein do sides aur unke beech ka angle, aur Theorem 5 mein ek right angle, ek hypotenuse aur ek side.

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 Set 5.1, Q1	Practice 5.1 · Q1
NCERT Exercise Set 5.1, Q2	Practice 5.1 · Q2
NCERT Exercise Set 5.1, Q3	Practice 5.1 · Q3
NCERT Exercise Set 5.1, Q4	Practice 5.1 · Q4
NCERT Ex. Set 5.1, Think, Draw and Infer Q1	Practice 5.1 · Q5
NCERT Ex. Set 5.1, Think, Draw and Infer Q2	Practice 5.1 · Q6
NCERT Exercise Set 5.2, Q1	Practice 5.2 · Q1
NCERT Exercise Set 5.2, Q2	Practice 5.2 · Q2
NCERT Exercise Set 5.3, Q1	Practice 5.3 · Q1
NCERT Exercise Set 5.3, Q2	Practice 5.3 · Q2
NCERT Exercise Set 5.3, Q3	Practice 5.3 · Q3
NCERT Exercise Set 5.4, Q1	Practice 5.4 · Q1
NCERT Exercise Set 5.4, Q2	Practice 5.4 · Q2
NCERT Exercise Set 5.4, Q3	Practice 5.4 · Q3
NCERT Exercise Set 5.5, Q1	Practice 5.5 · Q1
NCERT Exercise Set 5.5, Q2	Practice 5.5 · Q2
NCERT Exercise Set 5.5, Q3	Practice 5.5 · Q3
NCERT Exercise Set 5.6, Q1	Practice 5.6 · Q1
NCERT Exercise Set 5.6, Q2	Practice 5.6 · Q2
NCERT Exercise Set 5.6, Q3	Practice 5.6 · Q3
NCERT End-of-chapter Exercises, Q1	Chapter ke end ki exercises · Q1
NCERT End-of-chapter Exercises, Q2	Chapter ke end ki exercises · Q2
NCERT End-of-chapter Exercises, Q3	Chapter ke end ki exercises · Q3
NCERT End-of-chapter Exercises, Q4	Chapter ke end ki exercises · Q4
NCERT End-of-chapter Exercises, Q5	Chapter ke end ki exercises · Q5
NCERT End-of-chapter Exercises, Q6	Chapter ke end ki exercises · Q6
NCERT End-of-chapter Exercises, Q7	Chapter ke end ki exercises · Q7
NCERT End-of-chapter Exercises, Q8	Chapter ke end ki exercises · Q8
NCERT End-of-chapter Exercises, Q9	Chapter ke end ki exercises · Q9
NCERT End-of-chapter Exercises, Q10	Chapter ke end ki exercises · Q10
NCERT End-of-chapter Exercises, Q11	Chapter ke end ki exercises · Q11
NCERT End-of-chapter Exercises, Q12	Chapter ke end ki exercises · Q12
NCERT End-of-chapter Exercises, Q13	Chapter ke end ki exercises · Q13
NCERT End-of-chapter Exercises, Q14	Chapter ke end ki exercises · Q14
NCERT End-of-chapter Exercises, Q15	Chapter ke end ki exercises · Q15
NCERT End-of-chapter Exercises, Q16	Chapter ke end ki exercises · Q16
NCERT End-of-chapter Exercises, Q17	Chapter ke end ki exercises · Q17
NCERT End-of-chapter Exercises, Q18	Chapter ke end ki exercises · Q18
NCERT End-of-chapter Exercises, Q19	Chapter ke end ki exercises · Q19
NCERT End-of-chapter Exercises, Q20	Chapter ke end ki exercises · Q20
NCERT End-of-chapter Exercises, Q21	Chapter ke end ki exercises · Q21
NCERT End-of-chapter Exercises, Q22	Chapter ke end ki exercises · Q22
NCERT End-of-chapter Exercises, Q23	Chapter ke end ki exercises · Q23
NCERT End-of-chapter Exercises, Q24	Chapter ke end ki exercises · Q24
NCERT End-of-chapter Exercises, Q25	Chapter ke end ki exercises · Q25
NCERT End-of-chapter Exercises, Q26	Chapter ke end ki exercises · Q26

Quick check: MCQs

Warm-up2 questions
5.1 Definitions4 questions
5.2 Circle ki Symmetries4 questions
5.3 Kitne Circles?4 questions
5.4 Chords aur Unke Subtend Kiye Angles3 questions
5.5 Chords ke Midpoints aur Perpendicular Bisectors4 questions
5.6.1 Do unequal chords mein se kaunsi centre se zyada door hai?4 questions
5.7 Arc ke Subtend Kiye Angles3 questions
5.7.1 Arc ke bahar circle ke kisi point par arc ka subtend kiya angle4 questions
5.8 Points ki Concyclicity4 questions

Glossary · 15

Key terms (English + Hindi)

arcचापchaap: Circle ka woh hissa jo us par ek point se doosre point tak chalne par banta hai. Do arcs mein se chhota minor arc hai, bada major arc.Chapter mein pehli baar yahan aaya →

chordजीवाjeeva: Aisa segment jiske dono ends circle par hon.Chapter mein pehli baar yahan aaya →
circleवृत्तvritt: Ek gol curve jiske saare points uske andar ke ek point, yaani centre, se theek $r$ door hote hain; $r$ radius hai.
circumcentreपरिकेंद्रparikendra: Circumcircle ka centre: jahan triangle ki sides ke perpendicular bisectors milte hain.Chapter mein pehli baar yahan aaya →
circumcircleपरिवृत्तparivritt: Woh ek circle jo triangle ke teeno vertices se guzarta hai.
collinearसंरेखsanrekh: Ek hi seedhi line par pade hue.Chapter mein pehli baar yahan aaya →
concyclicएकवृत्तीयekvrittiya: Points concyclic hote hain jab ek circle un sab se guzarta hai.Chapter mein pehli baar yahan aaya →
corollaryउपप्रमेयupapramey: Aisa result jo abhi prove kiye gaye theorem se lagbhag turant nikal aata hai.Chapter mein pehli baar yahan aaya →
cyclic quadrilateralचक्रीय चतुर्भुजchakriya chaturbhuj: Aisa quadrilateral jiske chaaron vertices ek circle par hon.Chapter mein pehli baar yahan aaya →

diameterव्यासvyaas: Centre se guzarne wali chord; iski length $2 r$ hai aur koi bhi chord isse lambi nahi hoti.Chapter mein pehli baar yahan aaya →

locusबिंदुपथbindupath: Sirf aur sirf un points ka set jo diya gaya rule follow karte hain, aur koi doosra nahi.Chapter mein pehli baar yahan aaya →

perpendicular bisectorलंब समद्विभाजकlamb samadvibhajak: Segment ke midpoint se uske right angle par guzarne wali line; iske points theek wahi hain jo dono ends se barabar door hain.Chapter mein pehli baar yahan aaya →

radiusत्रिज्याtrijya: Centre se circle ke kisi point tak ka segment, ya uski length $r$ .Chapter mein pehli baar yahan aaya →

segment of a circleवृत्तखंडvrittkhand: Ek chord aur uske kaate gaye do arcs mein se ek ke beech ka region.Chapter mein pehli baar yahan aaya →
subtendअंतरित करनाantarit karna: Ek chord ya arc $A B$ point $P$ par $∠ A P B$ subtend karta hai: $P$ ko $A$ aur $B$ se jodne par bana angle.Chapter mein pehli baar yahan aaya →

Hamare aas-paas circles

5.1 Definitions

5.2 Circle ki Symmetries

5.3 Kitne Circles?

5.4 Chords aur Unke Subtend Kiye Angles

5.5 Chords ke Midpoints aur Perpendicular Bisectors

5.6 Centre se Chords ki Distance

5.6.1 Do unequal chords mein se kaunsi centre se zyada door hai?

5.7 Arc ke Subtend Kiye Angles

5.7.1 Arc ke bahar circle ke kisi point par arc ka subtend kiya angle

5.8 Points ki Concyclicity

Chapter ke end ki exercises

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye