Class 9 Maths · Chapter 6
Short answer:
Class 9 Maths Ganita Manjari Chapter 6 "Measuring Space: Perimeter and Area" mein circumference, arc length, sector, Heron's formula, trapezium, parallelogram, rhombus aur circle ka area cover hota hai — sab step-by-step is page par solve kiya gaya hai.
Class 9 Maths Ganita Manjari Part 1 ka Chapter 6 — "Measuring Space: Perimeter and Area" — is naye NCF-2023-based textbook ka sabse "hands-on" aur history-rich chapter hai. Ye chapter ek simple sawaal se shuru hota hai: race track ke outer lane ke runners ko head-start kyun milta hai? Isi puzzle se aap circumference, π, arc length aur sector jaisi cheezein explore karte ho, phir dheere-dheere Heron's formula, trapezium, parallelogram aur rhombus ke area tak pahunchte ho. Purane rationalised syllabus mein ye topics do-teen alag chapters — "Areas of Parallelograms and Triangles", "Heron's Formula" aur "Circles" ke area-wale hisse — mein bante the; Ganita Manjari ne inhe ek hi unified "measuring space" chapter mein pack kar diya hai, saath mein Āryabhaṭa, Brahmagupta, Mādhava aur Baudhāyana jaisi historical figures ki kahaniyan bhi. Agar aap poori Ganita Manjari Class 9 Maths NCERT solutions series follow kar rahe ho, to ye page Chapter 6 ke saare in-text aur teeno exercise-set (6.1, 6.2, 6.3) questions ke step-by-step solutions deta hai.
Part 1 ke baaki 7 chapters bhi is naye structure ka hissa hain — Chapter 1 "Orienting Yourself: Use of Coordinates", Chapter 2 "Introduction to Linear Polynomials", Chapter 3 "The World of Numbers", Chapter 4 "Exploring Algebraic Identities", Chapter 5 "I'm Up and Down, and Round and Round" (Circles), aur aage Chapter 7 "Probability" aur Chapter 8 "Sequences and Progressions" — dono bilkul naye dedicated chapters hain jo pehle Class 9 mein is form mein nahi the. Ganita Manjari Part 2 abhi (4 August 2026 tak) release nahi hui hai, isliye is series mein filhaal sirf Part 1 ke 8 chapters cover kiye ja rahe hain.
Chapter 6 Summary — 5 Minute Revision
Chapter 6 "Measuring Space: Perimeter and Area" teen major exercise-sets mein organized hai:
- Exercise 6.1 — Perimeter aur Circumference: circle ka circumference (2πr), arc length, sector ka perimeter, aur composite shapes (semicircle+rectangle, wheel revolutions, race-track lanes) ka perimeter.
- Exercise 6.2 — Area of Triangles aur Quadrilaterals: Heron's formula se triangle ka area, trapezium, parallelogram aur rhombus ka area, plus "same base–same parallels" theorem-based proofs (parallelograms/triangles ki area-equality, median-based results).
- Exercise 6.3 — Area of Circle, Sector aur Segment: circle ka area (πr²), sector area ((θ/360°)×πr²), quadrant area, aur clock ki minute-hand jaise real-life sector-sweep problems.
Chapter ke beech-beech mein historical boxes hain — Āryabhaṭa ka π approximation, Brahmagupta ka cyclic quadrilateral formula, Mādhava ki infinite series, aur Baudhāyana Sulba Sutra ki area constructions — jo pure calculation-focused chapter ko conceptual depth dete hain.
In-Text Questions — Solutions
Ek circular running track ki radius 35 m hai. Uska circumference nikaliye. (π = 22/7)
Formula: C = 2πr
C = 2 × 22/7 × 35 = 220 m
Track ka circumference 220 m hai.
Ek triangle ki sides 5 cm, 12 cm aur 13 cm hain. Heron's formula se area nikaliye.
s = (5+12+13)/2 = 15 cm
Area = √[s(s−a)(s−b)(s−c)] = √[15 × 10 × 3 × 2] = √900 = 30 cm²
Chunki 5² + 12² = 13² hai, ye ek right triangle bhi hai — check: ½×5×12 = 30 cm², jo match karta hai.
Ek circle ka circumference 44 cm hai. Uski radius nikaliye. (π = 22/7)
2πr = 44 ⟹ r = 44 × 7 / (2 × 22) = 7 cm
Ek parallelogram ka base 8 cm aur height 5 cm hai. Uska area nikaliye.
Area = base × height = 8 × 5 = 40 cm²
Ek sector ki radius 14 cm hai aur central angle 90° hai. Uska area nikaliye. (π = 22/7)
Area = (θ/360°) × πr² = (90/360) × 22/7 × 14 × 14 = ¼ × 616 = 154 cm²
Ek rhombus ki diagonals 10 cm aur 8 cm hain. Uska area nikaliye.
Area = ½ × d₁ × d₂ = ½ × 10 × 8 = 40 cm²
Exercise Questions — Solutions (Q1–Q20)
Do concentric circular tracks hain — inner track ki radius r aur outer track ki radius (r + d) hai, jahan d lane ki width hai. Dikhaiye ki ek chakkar (lap) mein outer lane runner ko kitni extra distance cover karni padti hai, aur ye extra distance radius r par depend karti hai ya nahi.
Inner lap ki length = 2πr
Outer lap ki length = 2π(r + d)
Extra distance = 2π(r+d) − 2πr = 2πd
Extra distance sirf lane-width d par depend karti hai, radius r par nahi — isiliye race officials outer lane runners ko fixed head-start (staggered start) dete hain, chahe track ki radius kuch bhi ho.
Ek car ke pahiye ka diameter 70 cm hai. 11 km distance cover karne ke liye pahiye kitne revolutions lagayega? (π = 22/7)
Ek revolution mein tay ki gayi distance = circumference
C = πd = 22/7 × 70 = 220 cm = 2.2 m
Total distance = 11 km = 11000 m
Revolutions = 11000 ÷ 2.2 = 5000
Pahiya 5000 revolutions lagayega.
Ek protractor (semicircular plate) ki radius 7 cm hai. Uska perimeter nikaliye. (π = 22/7)
Perimeter = arc (πr) + diameter (2r)
P = 22/7 × 7 + 2 × 7 = 22 + 14 = 36 cm
Do circles ke circumference ka ratio 3 : 4 hai. Unki radii ka ratio kya hoga?
C = 2πr, isliye C₁ : C₂ = r₁ : r₂ (2π dono jagah common hai)
r₁ : r₂ = 3 : 4
Ek flower bed rectangle (20 m × 14 m) ke ek 14 m wale side par semicircle jodkar banaya gaya hai (semicircle ka diameter = 14 m). Poore flower bed ka perimeter nikaliye. (π = 22/7)
Semicircle ki radius = 7 m
Perimeter = 2 lengths + 1 breadth (jo semicircle se replace nahi hui) + semicircular arc
P = 20 + 20 + 14 + (22/7 × 7) = 20 + 20 + 14 + 22 = 76 m
Ek sector ki radius 21 cm hai aur central angle 60° hai. Uska perimeter nikaliye. (π = 22/7)
Arc length = (60/360) × 2 × 22/7 × 21 = (1/6) × 132 = 22 cm
Perimeter = 2r + arc = 42 + 22 = 64 cm
Ek clock ka minute hand 10.5 cm lamba hai. 20 minute mein wo kitni distance cover karega? (π = 22/7)
20 minute mein angle = (20/60) × 360° = 120°
Distance = (120/360) × 2 × 22/7 × 10.5 = (1/3) × 66 = 22 cm
Ek circular flower bed (radius 12 m) ke chaaron taraf 2 m chauda path hai. Path ki outer boundary ka circumference nikaliye. (π = 22/7)
Outer radius = 12 + 2 = 14 m
C = 2 × 22/7 × 14 = 88 m
28 cm radius ke circular wire ko mod kar ek rectangle banaya gaya, jiski length aur breadth ka ratio 3 : 2 hai. Rectangle ki dimensions nikaliye. (π = 22/7)
Circle ka circumference = rectangle ka perimeter
C = 2 × 22/7 × 28 = 176 cm
Length = 3x, breadth = 2x maan lo:
2(3x + 2x) = 176 ⟹ 10x = 176 ⟹ x = 17.6
Length = 3 × 17.6 = 52.8 cm, Breadth = 2 × 17.6 = 35.2 cm
Ek triangle ki sides 18 cm, 24 cm aur 30 cm hain. Heron's formula se uska area nikaliye.
s = (18+24+30)/2 = 36 cm
Area = √[36 × 18 × 12 × 6] = √46656 = 216 cm²
Ek triangular park ki sides ka ratio 3 : 5 : 7 hai aur perimeter 300 m hai. Heron's formula se area nikaliye.
Sides = 3x, 5x, 7x; perimeter = 15x = 300 ⟹ x = 20
Sides = 60 m, 100 m, 140 m; s = 150 m
Area = √[150 × 90 × 50 × 10] = √6750000 = 1500√3 m² ≈ 2598.1 m²
Ek trapezium ki parallel sides 25 cm aur 15 cm hain, aur height 8 cm hai. Uska area nikaliye.
Area = ½ × (25 + 15) × 8 = ½ × 40 × 8 = 160 cm²
Ek rhombus ka area 240 cm² hai aur ek diagonal 16 cm hai. Doosri diagonal nikaliye.
240 = ½ × 16 × d₂ ⟹ d₂ = 240 × 2 / 16 = 30 cm
Parallelograms ABCD aur ABEF same base AB par hain aur same parallels ke beech hain. Dikhaiye ki dono ka area equal hai.
Dono parallelograms ki height base AB se opposite parallel line tak ki perpendicular distance hai, jo dono ke liye same hai (kyunki dono same do parallel lines ke beech mein hain).
Area(ABCD) = AB × h = Area(ABEF)
Isliye same base aur same parallels ke beech ke parallelograms ka area hamesha equal hota hai.
Triangle ABC mein D, side BC ka midpoint hai. Dikhaiye ki area(△ABD) = area(△ACD).
A se BC par perpendicular AL draw karo.
Area(△ABD) = ½ × BD × AL, Area(△ACD) = ½ × DC × AL
Chunki D midpoint hai, BD = DC, isliye
Area(△ABD) = Area(△ACD)
Yani median ek triangle ko do equal-area triangles mein baant deta hai.
Parallelogram ABCD mein P, side AB ka midpoint hai. Area(△APD) : Area(ABCD) ka ratio nikaliye.
Let height (D se AB tak) = h, AB = base = b. AP = b/2.
Area(△APD) = ½ × (b/2) × h = bh/4
Area(ABCD) = b × h
Ratio = (bh/4) : (bh) = 1 : 4
Ek circle ka circumference 44 cm hai. Uske quadrant ka area nikaliye. (π = 22/7)
2πr = 44 ⟹ r = 7 cm
Quadrant area = πr²/4 = (22/7 × 49)/4 = 154/4 = 38.5 cm²
Ek sector ki radius 21 cm hai aur central angle 150° hai. Uska area nikaliye. (π = 22/7)
Area = (150/360) × 22/7 × 21 × 21 = (5/12) × 1386 = 577.5 cm²
Ek clock ka minute hand 14 cm lamba hai. 10 minute mein wo kitna area sweep karega? (π = 22/7)
10 minute mein angle = (10/60) × 360° = 60°
Area = (60/360) × 22/7 × 14 × 14 = (1/6) × 616 = 102.67 cm² (= 308/3 cm²)
Ek pendulum 30° ka angle banate hue 8.8 cm ka arc describe karta hai. Pendulum ki length nikaliye. (π = 22/7)
Arc length = (θ/360) × 2πr
8.8 = (30/360) × 2 × 22/7 × r = (11/21) × r
r = 8.8 × 21/11 = 16.8 cm
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| Circumference of circle | C = 2πr = πd |
| Arc length | l = (θ/360°) × 2πr |
| Perimeter of sector | P = 2r + l |
| Area of triangle (base–height) | A = ½ × base × height |
| Heron's formula | A = √[s(s−a)(s−b)(s−c)], where s = (a+b+c)/2 |
| Area of parallelogram | A = base × height |
| Area of trapezium | A = ½ × (sum of parallel sides) × height |
| Area of rhombus | A = ½ × d₁ × d₂ |
| Area of circle | A = πr² |
| Area of sector | A = (θ/360°) × πr² |
| Area of quadrant (θ = 90°) | A = πr²/4 |
| Brahmagupta's formula (cyclic quadrilateral) | A = √[(s−a)(s−b)(s−c)(s−d)], s = (a+b+c+d)/2 |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Heron's formula lagate waqt semi-perimeter s = (a+b+c)/2 nikalna bhool jaana aur seedha sides ko formula mein daal dena — isse poora answer galat aa jata hai.
- π ki value ko lekar confusion — jab sides/radius 7 ke multiple hon tab 22/7 use karna chahiye, warna decimal (3.14) — dono ko mix kar dena ek common calculation error hai.
- Sector ka area ya arc length nikalte waqt angle θ ko 360° se divide karna bhool jaana, ya degree aur radian mix kar dena.
- Rhombus ka area nikalte waqt formula mein diagonals ko half karna bhool jaana — galat: d₁ × d₂, sahi: ½ × d₁ × d₂.
- Diameter aur radius mein confuse ho jaana — circumference ya area formula mein diameter ki jagah seedha radius wali value daal dena (ya opposite).
- Composite shapes (jaise rectangle + semicircle) mein final perimeter nikalte waqt us side ki straight length ko bhi count kar lena jo actually arc se replace ho chuki hai — isse extra length add ho jaati hai.
Board-Style Important Questions
- Ek circle ki radius double kar di jaaye to uska circumference kitne guna badhega?
Answer: C = 2πr hai, isliye radius double hone par circumference bhi 2 guna ho jayega. - Ek triangle ki sides 9 cm, 12 cm aur 15 cm hain. Heron's formula se iska area nikaliye.
Answer: s = 18, Area = √[18×9×6×3] = √2916 = 54 cm². - Ek sector ki radius 7 cm hai aur central angle 120° hai. Iska area aur perimeter dono nikaliye. (π = 22/7)
Answer: Arc = (120/360)×2×22/7×7 = 44/3 cm ≈ 14.67 cm; Area = (120/360)×22/7×49 = 154/3 cm² ≈ 51.33 cm²; Perimeter = 2×7 + 44/3 = 14 + 44/3 = 86/3 cm ≈ 28.67 cm. - Ek park rectangle (length 36 m, breadth 24.5 m) ke ek breadth-side par ek semicircle jodkar bana hai (semicircle ka diameter = breadth). Poore park ka perimeter aur area dono nikaliye. (π = 22/7)
Answer: radius = 12.25 m. Perimeter = 36+36+24.5+(22/7×12.25) = 36+36+24.5+38.5 = 135 m. Area = rectangle area + semicircle area = (36×24.5) + (½×22/7×12.25²) = 882 + 235.81 ≈ 1117.81 m². - Ek rhombus-shaped field ki diagonals 48 m aur 20 m hain. Field ka area nikaliye, aur agar seeds ki cost ₹4 per m² hai to total cost nikaliye.
Answer: Area = ½×48×20 = 480 m²; Cost = 480×4 = ₹1920.
Aksar Poochhe Jaane Wale Sawaal
Class 9 Maths Chapter 6 'Measuring Space: Perimeter and Area' ke important questions kahan se practice karein?
Is chapter ke teeno exercise-sets (6.1 — perimeter/circumference, 6.2 — Heron's formula aur quadrilaterals ka area, 6.3 — sector/circle ka area) se banaye class 9 maths chapter 6 measuring space perimeter and area important questions is page ke exercise section mein step-by-step solve kiye gaye hain — Heron's formula aur composite-shape perimeter wale questions par extra focus rakhein, kyunki yahi sabse zyada exam mein aate hain.
Ganita Manjari Class 9 Maths NCERT solutions poori book (saare 8 chapters) ke liye kahan milegi?
Part 1 ke 8 chapters — Coordinates, Linear Polynomials, World of Numbers, Algebraic Identities, Circles, Measuring Space, Probability aur Sequences & Progressions — sabki solutions is series mein chapter-wise cover hoti hain. Ek jagah se saare ganita manjari class 9 maths ncert solutions, saath hi class 9 maths ganita manjari all chapters pdf download options, chapter-index page par milte hain.
Class 9 Maths Ganita Manjari Part 2 kab aayega?
4 August 2026 tak Ganita Manjari Part 2 officially release nahi hui hai, aur NCERT ne koi release date bhi announce nahi ki hai. Multiple secondary sources ke hisaab se expected topics mein Linear Equations in Two Variables, Euclid's Geometry, Lines and Angles, Triangles, Quadrilaterals, Surface Area & Volume aur Statistics shaamil ho sakte hain — lekin ye abhi tak sirf anticipated list hai, koi official confirmation nahi hai, isliye class 9 maths ganita manjari part 2 kab aayega ka pakka jawaab abhi nahi diya ja sakta.
Class 9 Maths ka new syllabus 2026-27 mein total kitne chapters hain aur chapter list PDF kahan milegi?
Session 2026-27 se Class 9 Maths ki nayi book 'Ganita Manjari' hai, jiska sirf Part 1 (8 chapters) abhi published aur taught ho raha hai. Official class 9 maths new syllabus 2026-27 chapter list pdf NCERT ki website (ncert.nic.in) ke textbook-listing section se download ki ja sakti hai.
Kya Chapter 6 mein purana 'Areas of Parallelograms and Triangles' chapter poora cover hota hai?
Nahi, poora nahi — us purane chapter ke kuch core results (same base–same parallels theorem, median-based area equality) Exercise 6.2 ke chhote hisse ke roop mein is unified 'Measuring Space' chapter mein fold kar diye gaye hain, standalone chapter jitna detail nahi hai.
Chapter 6 ke baad aane wale Chapter 7 (Probability) aur Chapter 8 (Sequences and Progressions) kaise alag hain?
Ye dono bilkul naye dedicated chapters hain jo NCF-2023 ke saath add hue hain — class 9 maths chapter 7 probability ncert solutions aur class 9 maths chapter 8 sequences and progressions ncert solutions is series mein separately cover honge. Inka koi direct overlap Chapter 6 ke perimeter-area concepts se nahi hai, isliye inhe independent topics ki tarah prepare karein.
Class 9 Maths — Saare Chapters
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