Class 9 Maths · Chapter 5
Short answer:
Chapter 5 — "I'm Up and Down, and Round and Round" (Circles) — Ganita Manjari Class 9 Part 1 ka geometry-heavy chapter hai jisme circle ki basic definitions se shuru hokar 12 theorems tak (chord properties, angle subtended by arc, cyclic quadrilateral) cover hota hai. Agar aap "ganita manjari class 9 maths ncert solutions" dhundh rahe ho step-by-step proof-style answers ke saath, ye spec construction (Exercise 5.1) se lekar cyclic quadrilateral theorems (5.5–5.6) tak sab deta hai.
Circles wala chapter Class 9 Maths ke naye syllabus me sabse zyada proof-based reasoning maangta hai. Purani NCERT "Circles" chapter se core geometry topic milta-julta hai — chord properties, angle subtended by arc, cyclic quadrilateral — bas naam playful ho gaya hai aur presentation NCF-2023 ke style me "Given → To Show → Why is this true?" structure follow karti hai, jisse students rote-memorised proofs ki jagah genuine reasoning seekhte hain.
Chapter 8 sections me phaila hai — nature me circles dekhne se shuru hokar (raindrop, sunflower, chaand-sooraj ka eclipse) formal definitions (centre, radius, chord, diameter, circumcircle, circumcentre) tak jaata hai, phir 12 theorems aur exercises (5.1 se 5.6) me practice hoti hai. Ye poore "class 9 maths new syllabus 2026-27 chapter list" ka Chapter 5 hai — Part 1 book me isse pehle Chapter 1 "Orienting Yourself — Use of Coordinates", Chapter 2 "Introduction to Linear Polynomials", Chapter 3 "The World of Numbers" aur Chapter 4 "Exploring Algebraic Identities" aate hain, aur baad me Chapter 6 "Measuring Space — Perimeter and Area", Chapter 7 "Probability" aur Chapter 8 "Sequences and Progressions" hain.
Is spec me har construction aur proof ka step-by-step working diya gaya hai — exam-answer style stepwise format me — jo "class 9 maths chapter 5 im up and down and round and round circles important questions" practice karne ke liye kaam aayega.
Chapter 5 Summary — 5 Minute Revision
Chapter 5 circle ke around ek complete logical structure banata hai: pehle circle, centre, radius, chord, diameter jaise terms define hote hain, phir circumcircle construction (perpendicular bisectors se) sikhaya jaata hai. Uske baad chord-related theorems (perpendicular from centre bisects chord, equal chords equidistant from centre) aate hain, phir angle subtended by an arc ka central theorem — arc jo angle centre par banata hai woh circle ke remaining part par bane angle ka double hota hai. Isi se angle-in-semicircle (=90°) aur angles-in-same-segment (equal) nikalte hain. Chapter cyclic quadrilateral ke do bade theorems par khatam hota hai — opposite angles ka sum 180°, aur exterior angle = interior opposite angle — jo aage Class 10 ke tangent-circle chapter ki neev bante hain.
In-Text Questions — Solutions
Ek circle me chord aur diameter me kya fark hai? Kya har diameter chord hota hai aur kya har chord diameter hota hai?
Chord woh line segment hai jiske dono end-points circle par hoon. Diameter bhi ek chord hai — bas woh circle ke centre se hokar guzarta hai, isliye woh sabse badi chord hoti hai. Isliye har diameter chord hota hai, lekin har chord diameter nahi hota — sirf woh chord jo centre se pass kare, diameter kehlata hai.
Ek point se hokar kitne circles draw kiye ja sakte hain? Do points se hokar? Teen non-collinear points se hokar?
Ek point se hokar infinite (anant) circles ja sakte hain — kyunki centre kahin bhi rakh sakte ho. Do points se hokar bhi infinite circles ja sakte hain, kyunki centre un dono points ke perpendicular bisector par kahin bhi ho sakta hai. Teen non-collinear points se hokar exactly ek circle jaata hai — kyunki teeno jodiyon (AB, BC, CA) ke perpendicular bisectors sirf ek hi common point (circumcentre) par milte hain.
Ek right-angled triangle ka circumcentre kahan hoga? Reasoning dijiye.
Right-angled triangle ka circumcentre uske hypotenuse ke midpoint par hota hai. Reasoning: angle-in-semicircle theorem ke converse se, agar koi angle 90° hai to jis chord ne woh angle banaya hai woh diameter hai. Toh hypotenuse hi circumcircle ka diameter ban jaata hai, aur diameter ka midpoint hi centre hota hai.
Agar teen points collinear hain, to unse hoke circumcircle kyun nahi banaya ja sakta?
Agar A, B, C collinear hain, to AB aur BC ke perpendicular bisectors dono ek hi line (AB wali line) ke perpendicular honge, matlab woh parallel ho jaayenge aur kabhi mile hi nahi (ya ek hi line ban jaayein). Isliye koi unique intersection point (circumcentre) nahi milta, aur teeno collinear points se hokar koi circle nahi ja sakta.
Kya ek circle ke andar do unequal chords ho sakte hain jo centre se equal distance par hoon?
Nahi. Theorem ke according, ek hi circle me equal distance se centre wale chords hamesha equal length ke hote hain (aur converse bhi sach hai). Isliye agar do chords centre se equal distance par hain, to woh length me bhi equal hi honge — unequal nahi ho sakte.
AB ek diameter hai. C circle par koi point hai. Angle ACB kitna hoga, chahe C kahin bhi ho?
∠ACB hamesha 90° hoga, chahe C circle par kahin bhi ho (A aur B ke alawa). Kyunki AB diameter hai, isliye AB dwara centre par bana angle 180° (straight line) hota hai, aur angle-subtended theorem se remaining part par bana angle uska half, yaani 90°, hota hai — position se independent.
Kya ek square hamesha cyclic quadrilateral hota hai? Kaaran dijiye.
Haan. Square ke sabhi angles 90° hote hain, isliye koi bhi do opposite angles ka sum 90°+90°=180° hota hai, jo cyclic quadrilateral ki condition poori karta hai. Square ka circumcentre uske diagonals ke intersection point par hota hai, aur circumradius = half diagonal.
Do circles maximum kitne points par intersect kar sakte hain?
Do circles maximum 2 points par intersect kar sakte hain. Agar woh 3 ya usse zyada points par milein, to un teeno common points se sirf ek hi unique circle ja sakti hai (jaisa upar dekha), isliye dono circles same ban jaayenge — do alag circles ke liye max 2 intersection points hi possible hain.
Exercise Questions — Solutions (Q1–Q16)
Exercise 5.1: ∆ABC banaiye jisme AB = 5 cm, ∠A = 70°, ∠B = 60° ho. Iska circumcircle banaiye. Bataiye circumcentre triangle ke andar hai ya bahar.
Construction steps: (1) AB = 5 cm ka line segment banayiye. (2) A par 70° ka angle aur B par 60° ka angle banaiye — dono rays ka intersection C hai. (3) Kisi bhi do sides (jaise AB aur BC) ke perpendicular bisectors draw kijiye — unka intersection point O circumcentre hai. (4) OA (= OB = OC) ko radius lekar circle draw kijiye — yehi circumcircle hai.
Reasoning: ∠C = 180° − 70° − 60° = 50°. Teeno angles (70°, 60°, 50°) 90° se kam hain, isliye ∆ABC acute-angled hai — aur acute triangle ka circumcentre hamesha triangle ke andar padta hai.
Exercise 5.1: ∆ABC banaiye jisme AB = 5 cm, ∠A = 100°, AC = 4 cm ho. Circumcircle banaiye aur circumcentre ki position bataiye.
Construction steps: (1) AB = 5 cm banayiye. (2) A par 100° ka angle banakar us ray par AC = 4 cm cut kijiye — C mil gaya. (3) BC join kijiye taaki ∆ABC poora ho. (4) Do sides ke perpendicular bisectors se O (circumcentre) nikaliye aur OA radius lekar circle draw kijiye.
Reasoning: ∠A = 100° obtuse hai. Obtuse-angled triangle ka circumcentre hamesha us angle ke saamne wali side (yahaan BC) ke bahar padta hai — isliye O triangle ke bahar hoga.
Exercise 5.1: ∆ABC banaiye jisme AB = 6 cm, BC = 7 cm, CA = 7 cm ho (isosceles). Circumcircle banaiye — circumcentre kis line par padega aur kyun?
Construction steps: (1) AB = 6 cm banayiye. (2) A se 7 cm aur B se 7 cm ke arcs draw karke C nikaliye (SSS construction). (3) Perpendicular bisectors se circumcentre O nikaliye aur circle draw kijiye.
Reasoning: Yahaan CA = CB = 7 cm, isliye triangle AB ke perpendicular bisector ke around symmetric hai. Isliye circumcentre O hamesha AB ke perpendicular bisector par hi padega — yeh triangle ki symmetry axis bhi hai.
Exercise 5.2: Prove kijiye ki centre se kisi chord par draw kiya gaya perpendicular, us chord ko bisect karta hai.
Given: Circle ka centre O, chord AB, aur OM ⊥ AB.
To show: AM = MB.
Proof: OA aur OB join kijiye. ∆OMA aur ∆OMB me: OA = OB (radii), OM = OM (common), ∠OMA = ∠OMB = 90°. Isliye RHS congruence rule se ∆OMA ≅ ∆OMB.
⟹ AM = MB (CPCT)
Hence proved — perpendicular chord ko bisect karta hai.
Exercise 5.2: Ek circle ki do chords AB aur CD equal hain (AB = CD). Prove kijiye ki ye centre se equal distance par hain.
Given: Circle ka centre O, AB = CD. OM ⊥ AB aur ON ⊥ CD.
To show: OM = ON.
Proof: Perpendicular chord ko bisect karta hai, isliye AM = AB/2 aur CN = CD/2; chunki AB = CD, isliye AM = CN. OA aur OC join kijiye — dono radii hain, OA = OC. Right triangles OMA aur ONC me Pythagoras se:
OM² = OA² − AM² aur ON² = OC² − CN²
Chunki OA = OC aur AM = CN, isliye OM² = ON², yaani OM = ON. Hence proved.
Exercise 5.2: Ek circle me ek chord ki length 8 cm hai aur woh centre se 3 cm door hai. Circle ka radius nikaliye.
Chord AB = 8 cm, OM ⊥ AB, OM = 3 cm. Perpendicular chord ko bisect karta hai, isliye AM = 4 cm. Right triangle OMA me Pythagoras theorem se:
OA² = OM² + AM² = 3² + 4² = 9 + 16 = 25
OA = 5 cm
Isliye circle ka radius = 5 cm.
Exercise 5.3: Prove kijiye ki circle ke kisi arc dwara centre par banaya gaya angle, us arc dwara circle ke baaki hisse ke kisi point par banaye gaye angle ka double hota hai.
Given: Arc AB, centre O par ∠AOB banata hai, aur remaining part par point P par ∠APB banata hai.
To show: ∠AOB = 2∠APB.
Proof (main case): P se hokar O se guzarti hui line PQ khaincho (PQ ko O ke aage extend kijiye). ∆OAP me OA = OP (radii), isliye yeh isosceles hai, isliye ∠OAP = ∠OPA. Exterior angle theorem se ∠AOQ = ∠OAP + ∠OPA = 2∠OPA. Isi tarah ∆OBP se ∠BOQ = 2∠OPB. In dono ko jodne par: ∠AOB = ∠AOQ + ∠BOQ = 2∠OPA + 2∠OPB = 2(∠OPA + ∠OPB) = 2∠APB.
Hence proved (baaki cases — P major/minor arc par — isi logic se, sirf addition/subtraction badalta hai).
Exercise 5.3: Ek circle me chord AB centre par 70° ka angle banati hai. Same chord major arc ke kisi point C par kitna angle banayegi?
Angle subtended by arc theorem se: angle at centre = 2 × angle at remaining part of circle.
∠ACB = ∠AOB ÷ 2 = 70° ÷ 2 = 35°
Isliye major arc ke point C par chord AB 35° ka angle banayegi.
Exercise 5.3: AB ek circle ka diameter hai aur C circle par koi point hai (C ≠ A, B). Prove kijiye ∠ACB = 90°.
Given: AB diameter hai, C circle par koi point.
To show: ∠ACB = 90°.
Proof: Chunki AB diameter hai, A aur O aur B ek straight line par hain, isliye AB dwara centre par bana angle ∠AOB = 180°. Angle-subtended theorem se, remaining part ke point C par bana angle uska half hoga:
∠ACB = 180° ÷ 2 = 90°
Hence proved — yeh 'angle in a semicircle' theorem kehlata hai.
Exercise 5.4: Prove kijiye ki same segment me bane angles equal hote hain.
Given: Chord AB, aur C, D dono same segment (chord ke ek hi side) par circle ke points.
To show: ∠ACB = ∠ADB.
Proof: Angle-subtended theorem se, arc AB (jis side C aur D hain uske opposite arc) dwara centre par bana angle ∠AOB hai. Isliye:
∠ACB = ∠AOB ÷ 2 aur ∠ADB = ∠AOB ÷ 2
Dono ek hi ∠AOB ka half hain, isliye ∠ACB = ∠ADB. Hence proved.
Exercise 5.4: Cyclic quadrilateral ABCD me ∠A = 110° hai. ∠C ka value nikaliye.
Cyclic quadrilateral ke opposite angles ka sum 180° hota hai, isliye:
∠A + ∠C = 180° ⟹ ∠C = 180° − 110° = 70°
Isliye ∠C = 70°.
Exercise 5.4: Cyclic quadrilateral ABCD ki side AB ko E tak extend kiya gaya hai. ∠ADC = 95° hai. Prove karke ∠CBE ka value nikaliye.
Opposite angles ka sum 180° hota hai, isliye ∠ABC + ∠ADC = 180°, ⟹ ∠ABC = 180° − 95° = 85°.
∠ABC aur ∠CBE linear pair hain (AB ek straight line hai jise E tak badhaya gaya):
∠CBE = 180° − ∠ABC = 180° − 85° = 95°
Isliye ∠CBE = 95° = ∠ADC — yehi exterior-angle-equals-interior-opposite-angle theorem hai.
Exercise 5.5: Prove kijiye ki cyclic quadrilateral ke opposite angles ka sum 180° hota hai.
Given: ABCD ek cyclic quadrilateral hai, O centre hai.
To show: ∠ABC + ∠ADC = 180°.
Proof: OA aur OC join kijiye. Arc ADC (jis side B nahi hai) dwara centre par bana reflex angle AOC = 2∠ABC (angle subtended theorem, B remaining part par hai). Arc ABC dwara centre par bana non-reflex angle AOC = 2∠ADC. Reflex aur non-reflex angle AOC milkar 360° banate hain:
2∠ABC + 2∠ADC = 360° ⟹ ∠ABC + ∠ADC = 180°
Hence proved. (Isi tarah ∠BAD + ∠BCD = 180° bhi sabit hota hai.)
Exercise 5.5: Prove kijiye ki cyclic quadrilateral ka exterior angle, interior opposite angle ke barabar hota hai.
Given: ABCD cyclic, AB ko E tak extend kiya gaya.
To show: ∠CBE = ∠ADC.
Proof: ∠ABC aur ∠CBE linear pair hain, isliye ∠ABC + ∠CBE = 180°. Pichhle theorem se ∠ABC + ∠ADC = 180° bhi sach hai. Dono equations ko compare karke:
∠CBE = 180° − ∠ABC = ∠ADC
Hence proved.
Exercise 5.6: Agar quadrilateral ABCD ke opposite angles ka sum 180° hai, to prove kijiye ki A, B, C, D concyclic hain (converse).
Given: ∠ABC + ∠ADC = 180°.
To show: A, B, C, D ek hi circle par hain.
Proof (by contradiction): A, B, C se hokar ek unique circle draw kijiye (teen non-collinear points se ek hi circle jaati hai). Maan lijiye D us circle ke andar hai — to BD ko extend karke circle ko D' par milaiye. ABCD' cyclic hai, isliye ∠ABC + ∠AD'C = 180° (theorem se). Diya gaya hai ∠ABC + ∠ADC = 180°, isliye ∠AD'C = ∠ADC. Lekin ∠AD'C, ∆ACD ka exterior angle hai (D circle ke andar hone se), jo ∠ADC se strictly bada hona chahiye — contradiction. Isi tarah D circle ke bahar hone par bhi contradiction aata hai. Isliye D circle par hi hona chahiye — A, B, C, D concyclic hain.
Exercise 5.6: Do circles do points P aur Q par intersect karte hain. Prove kijiye ki dono centres ko milane wali line, common chord PQ ko perpendicularly bisect karti hai.
Given: Do circles centre O₁ aur O₂ ke saath P, Q par intersect karte hain.
To show: O₁O₂, PQ ko perpendicularly bisect karta hai.
Proof: O₁P = O₁Q (dono pehle circle ke radii), isliye O₁, PQ ke perpendicular bisector par hai. O₂P = O₂Q (dono doosre circle ke radii), isliye O₂ bhi PQ ke perpendicular bisector par hai. Chunki O₁ aur O₂ dono PQ ke ek hi perpendicular bisector par hain, isliye line O₁O₂ hi PQ ka perpendicular bisector hai. Hence proved.
Important Equations — Ek Nazar Me
| Theorem / Concept | Statement |
|---|---|
| Perpendicular from centre to chord | Centre se chord par draw kiya perpendicular, chord ko bisect karta hai (aur converse bhi sach hai — chord ka midpoint jodne wali line se centre ka perpendicular banta hai). |
| Equal chords & distance | Ek hi circle ke equal chords centre se equal distance par hote hain; equal distance wale chords equal length ke hote hain (converse). |
| Angle subtended by arc | Arc dwara centre par bana angle = 2 × (arc dwara circle ke remaining part par bana angle). |
| Angle in semicircle | Diameter dwara circle ke kisi bhi point par bana angle 90° hota hai. |
| Angles in same segment | Ek hi segment me, ek hi chord dwara bane angles equal hote hain. |
| Cyclic quadrilateral | Opposite angles ka sum = 180°. |
| Exterior angle (cyclic quad) | Exterior angle = interior opposite angle. |
| Circumcentre | Triangle ke teeno sides ke perpendicular bisectors ka intersection point; O se teeno vertices equidistant (= circumradius). |
| Chord-radius relation | OA² = OM² + AM², jahan OM ⊥ chord AB aur M, AB ka midpoint hai. |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Chord aur diameter ko same samajh lena — diameter khud ek chord hai (sabse badi), lekin har chord diameter nahi hota; sirf woh chord jo centre se guzre.
- Angle-subtended-by-arc theorem me 'double' kis taraf lagana hai ulta kar dena — angle hamesha CENTRE par bada (double) hota hai, remaining part ke point par chota (half) hota hai, ulta nahi.
- Kisi bhi quadrilateral ko bina proof ke cyclic maan lena — opposite angles ka sum 180° verify kiye bina quadrilateral ko circle par maan lena galat hai.
- Perpendicular bisector construction me compass ka radius chord ki half length se kam le lena, jisse dono arcs intersect hi nahi karte aur bisector line galat ban jaati hai.
- Circumcentre ki position triangle ke type se link karna bhool jaana — acute triangle me andar, right-angled me hypotenuse ke midpoint par, obtuse me bahar hota hai; isko yaad na rakhna common galti hai.
- 'Equal chords equidistant from centre' theorem ko do alag circles ke beech directly apply kar dena — yeh theorem sirf ek hi circle (ya equal radii wale congruent circles) ke chords ke liye valid hai, alag-alag radius wale circles ke liye nahi.
Board-Style Important Questions
- Q1. Ek circle ki do chords AB aur CD equal hain. Prove kijiye ki centre se dono chords equidistant hain.
- Q2. Cyclic quadrilateral ABCD me ∠A : ∠C = 2 : 3 hai. Dono angles ka value nikaliye.
- Q3. ∆ABC banaiye jisme BC = 6 cm, ∠B = 45°, ∠C = 75° ho, aur iska circumcircle construct kijiye. Circumcentre ki position bhi bataiye.
- Q4. Prove kijiye ki circle ke kisi arc dwara centre par bana angle, us arc dwara remaining part of circle par bane angle ka double hota hai.
- Q5. Prove kijiye ki cyclic quadrilateral ke opposite angles ka sum 180° hota hai, aur isi se yeh bhi dikhaiye ki quadrilateral ka exterior angle interior opposite angle ke barabar hota hai.
Aksar Poochhe Jaane Wale Sawaal
Class 9 Maths ka naya syllabus 2026-27 me kitne chapters hain aur Circles chapter kaunse number par aata hai?
Ganita Manjari Part 1 me total 8 chapters hain. 'I'm Up and Down, and Round and Round (Circles)' Chapter 5 hai — isse pehle Chapter 1 'Orienting Yourself — Use of Coordinates', Chapter 2 'Introduction to Linear Polynomials', Chapter 3 'The World of Numbers' aur Chapter 4 'Exploring Algebraic Identities' aate hain, aur baad me Chapter 6 'Measuring Space — Perimeter and Area', Chapter 7 'Probability' aur Chapter 8 'Sequences and Progressions' hain. Poori 'class 9 maths new syllabus 2026-27 chapter list' NCERT ki official site (ncert.nic.in) par available hai.
Ganita Manjari Part 2 kab aayega? Kya usme bhi koi geometry chapter hoga?
4 Aug 2026 tak ki status ke hisaab se, Ganita Manjari Part 2 abhi release nahi hui hai aur koi official release date bhi announce nahi hui. Kuch secondary sources ke hisaab se Part 2 me purane-syllabus-jaisi topics (Linear Equations in Two Variables, Euclid's Geometry, Lines and Angles, Triangles, Quadrilaterals, Surface Area and Volume, Statistics) expected hain — lekin yeh sirf 'expected' list hai, koi official NCERT confirmation nahi hai, isliye Part 2 ke exact content ya numbering par abhi kuch confirm nahi kaha ja sakta.
Is chapter me kitne exercises aur theorems hain?
Chapter 5 me total 8 sections aur 12 theorems hain, aur exercises 5.1 se 5.6 tak hain — Exercise 5.1 circumcircle construction se shuru hoti hai aur 5.5–5.6 cyclic quadrilateral theorems tak jaati hai.
Kya purani NCERT 'Circles' chapter aur naya 'I'm Up and Down, and Round and Round' chapter same hai?
Core geometry topic same hai (chord properties, angle subtended by arc, cyclic quadrilateral) — sirf naam playful banaya gaya hai NCF-2023 ke style me. Presentation thodi different hai: har theorem 'Given → To Show → Why is this true?' structure me diya gaya hai, jo purani textbook ke formal proof-only style se thoda alag approach hai.
Class 9 Maths Chapter 1 'Orienting Yourself — Use of Coordinates' aur Chapter 3 'The World of Numbers' ke solutions kahan milenge?
In dono chapters ke 'ncert solutions' is hi tarah ke alag chapter-spec documents me cover hote hain — Chapter 1 coordinate geometry ke basics (plotting points, quadrants) leta hai, aur Chapter 3 number systems ka historical aur decimal-expansion-focused version hai (purani 'Number Systems' chapter se scope thoda alag hai).
Circles ke baad Class 9 Maths me Chapter 6, 7, 8 me kya aata hai?
Chapter 6 'Measuring Space — Perimeter and Area' mensuration cover karta hai, Chapter 7 'Probability' (Class 9 ke liye ab expanded-scope wapas aaya hua chapter) aur Chapter 8 'Sequences and Progressions' (Class 9 ke liye bilkul naya topic hai, pehle nahi tha). In teeno chapters ke important questions aur ncert solutions bhi is hi series me alag se available hain.
Class 9 Maths — Saare Chapters
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