Class 10 Maths · Chapter 11
Short answer:
Is chapter me 14 questions cover kiye hain — circle ke perimeter/area ka quick recap, phir sector aur segment ke area (clock hands, chords, umbrella ribs, grazing horse, brooch design), aur end me combination of plane figures (square + circle, triangle + semicircle, square + 4 circles, equilateral triangle + incircle). Dono π values — 22/7 aur 3.14 — use hue hain, jahan jo question specify kare.
Circle ka poora area hamesha nahi chahiye hota — kabhi sirf ek slice (sector) chahiye, kabhi slice minus triangle (segment). Ye chapter bas itna hi hai: circle ke chhote-chhote pieces ka area nikalna, aur phir circle ko doosre shapes (square, triangle) ke saath combine karke shaded region nikalna. Formula ek hi hai baar baar ghoomta hai — (θ/360°) × poora circle ka area — bas angle theek se pehchano.
Chapter 11 Summary — 5 Minute Revision
1. Perimeter aur Area of a Circle — Recap
- Circumference (perimeter) = 2πr
- Area = πr²
- π ka value question specify karega — 22/7 (jab radius 7 ka multiple ho) ya 3.14 (baaki cases me). Jo question bole wahi use karo, apni marzi se mat badlo.
2. Sector of a Circle
Sector = circle ka "pizza slice" — do radii aur unke beech ka arc. Sector ka area poore circle ka fraction hota hai, bas dekhna hai ki angle θ, 360° ka kitna fraction hai.
| Quantity | Formula |
|---|---|
| Arc length | (θ/360°) × 2πr |
| Area of sector | (θ/360°) × πr² |
↔ Table ko side me swipe karein
Minor sector = chhota angle wala slice. Major sector = poora circle − minor sector.
3. Segment of a Circle
Segment = chord aur arc ke beech ka region. Isme sabse common mistake yahi hoti hai: log segment ko sector samajh lete hain. Yaad rakho —
Segment area = Sector area − Triangle area (triangle jo do radii aur chord se banta hai)
Minor segment = sector − triangle. Major segment = poora circle − minor segment.
4. Areas of Combination of Plane Figures
Real exam questions mostly yahi hote hain — circle kisi doosre shape (square, rectangle, triangle) ke andar ya bahar fit hota hai, aur shaded (chhaya wala) region poochha jaata hai. Strategy hamesha same: bada shape ka area − chhote shape(s) ka area (ya vice versa).
- Square/triangle me inscribed circle → incircle radius formula yaad rakho (equilateral triangle: r = a/(2√3))
- Semicircle pe right-angled triangle → diameter hi hypotenuse hota hai (Thales' theorem se radius nikaalo)
- Peg/rope wale grazing questions → rope ki length hi radius hai, angle field ke corner ka angle hota hai
Poore Class 10 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q14)
Q1. Find the area of a sector of a circle with radius 6 cm if the angle of the sector is 60°. (Use π = 22/7)
Radius r = 6 cm, angle θ = 60°.
Area of sector = (θ/360°) × πr² = (60/360) × (22/7) × 6²
= (1/6) × (22/7) × 36 = (22 × 36) / (7 × 6) = 792/42 = 132/7
Area of sector = 132/7 cm² = 18.86 cm² (approx)
Q2. Find the area of a quadrant of a circle whose circumference is 22 cm.
Circumference 2πr = 22 cm. Using π = 22/7:
2 × (22/7) × r = 22 ⇒ r = 22 × 7 / (2 × 22) = 7/2 = 3.5 cm
A quadrant is a sector with θ = 90°, i.e. 1/4 of the circle.
Area of circle = πr² = (22/7) × 3.5 × 3.5 = 269.5/7 = 38.5 cm²
Area of quadrant = 38.5 / 4 = 9.625 cm²
Area of quadrant = 9.625 cm²
Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. (Use π = 22/7)
Minute hand 60 minutes me 360° ghoomti hai, isliye 1 minute me 6°.
Angle swept in 5 minutes θ = 5 × 6° = 30°
Minute hand khud radius hai: r = 14 cm.
Area swept = (θ/360°) × πr² = (30/360) × (22/7) × 14²
= (1/12) × (22/7) × 196 = (22 × 196)/(7 × 12) = 4312/84 = 154/3
Area swept = 154/3 cm² = 51.33 cm² (approx)
Q4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. (Use π = 3.14)
r = 10 cm, θ = 90°.
Area of sector (90°) = (90/360) × 3.14 × 10² = 0.25 × 314 = 78.5 cm²
Triangle formed by the two radii and chord is right-angled with both legs = 10 cm:
Area of triangle = 1/2 × 10 × 10 = 50 cm²
(i) Minor segment = Sector area − Triangle area
= 78.5 − 50 = 28.5 cm²
(ii) Major sector = Area of circle − Minor sector
Area of circle = πr² = 3.14 × 100 = 314 cm²
Major sector = 314 − 78.5 = 235.5 cm²
Q5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord. (Use π = 22/7, √3 = 1.732)
r = 21 cm, θ = 60°.
(i) Arc length
= (θ/360°) × 2πr = (60/360) × 2 × (22/7) × 21 = (1/6) × 132 = 22 cm
(ii) Area of sector
= (60/360) × (22/7) × 21² = (1/6) × (22/7) × 441 = (1/6) × 1386 = 231 cm²
(iii) Area of segment = Sector − Triangle
Since θ = 60° and both radii = 21 cm, the triangle is equilateral with side 21 cm:
Area of triangle = (√3/4) × 21² = (√3/4) × 441 = 110.25 × 1.732 = 190.95 cm² (approx)
Area of segment = 231 − 190.95 = 40.05 cm² (approx)
Q6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14, √3 = 1.73)
r = 15 cm, θ = 60°.
Area of sector = (60/360) × 3.14 × 15² = (1/6) × 3.14 × 225 = (1/6) × 706.5 = 117.75 cm²
Triangle is equilateral (side 15 cm, since θ = 60°):
Area of triangle = (√3/4) × 15² = (1.73/4) × 225 = 56.25 × 1.73 = 97.3125 cm²
Minor segment = Sector − Triangle
= 117.75 − 97.3125 = 20.4375 cm² ≈ 20.44 cm²
Major segment = Area of circle − Minor segment
Area of circle = 3.14 × 225 = 706.5 cm²
Major segment = 706.5 − 20.4375 = 686.0625 cm² ≈ 686.06 cm²
Q7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14, √3 = 1.73)
r = 12 cm, θ = 120°.
Area of sector = (120/360) × 3.14 × 12² = (1/3) × 3.14 × 144 = (1/3) × 452.16 = 150.72 cm²
Triangle formed by two radii (12 cm each) with included angle 120° — area is 1/2 × a × b × sin(included angle):
Area of triangle = 1/2 × 12 × 12 × sin120° = 72 × (√3/2) = 36√3 = 36 × 1.73 = 62.28 cm²
Area of segment = Sector − Triangle = 150.72 − 62.28 = 88.44 cm²
Area of segment = 88.44 cm²
Q8. A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by means of a 5 m long rope. Find (i) the area of the field the horse can graze (ii) the increase in grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)
Peg is at a corner of the square, so the corner angle available to the horse is 90°. Since rope length (5 m, then 10 m) is less than the side (15 m), the horse grazes a quarter-circle each time, with no overlap onto adjacent sides.
(i) Rope = 5 m (radius = 5 m)
Grazing area = (90/360) × 3.14 × 5² = 0.25 × 78.5 = 19.625 m²
(ii) Rope = 10 m (radius = 10 m)
Grazing area = (90/360) × 3.14 × 10² = 0.25 × 314 = 78.5 m²
Increase in area = 78.5 − 19.625 = 58.875 m²
Q9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used to make 5 diameters which divide the circle into 10 equal sectors. Find: (i) the total length of the silver wire required (ii) the area of each sector of the brooch. (Use π = 22/7)
Diameter d = 35 mm, so radius r = 17.5 mm. There are 5 diameters (= 10 radii) plus the circumference.
(i) Total wire length = Circumference + 5 × diameter
Circumference = πd = (22/7) × 35 = 110 mm
Total wire = 110 + 5 × 35 = 110 + 175 = 285 mm
(ii) Area of each sector = Total area ÷ 10 (10 equal sectors)
Total area = πr² = (22/7) × 17.5² = (22/7) × 306.25 = 962.5 mm²
Area of each sector = 962.5 / 10 = 96.25 mm²
Q10. An umbrella has 8 ribs which are equally spaced. Assuming the umbrella to be a flat circle of radius 45 cm, find the area between two consecutive ribs. (Use π = 22/7)
8 ribs equally spaced around the centre means the circle is divided into 8 equal sectors.
Angle between two consecutive ribs θ = 360°/8 = 45°
Area between two ribs = (45/360) × (22/7) × 45² = (1/8) × (22/7) × 2025
= (22 × 2025) / (7 × 8) = 44550/56 = 795.54 cm² (approx)
Area between two consecutive ribs = ≈ 795.54 cm²
Q11. A square OABC is inscribed in a quadrant OPBQ of a circle, with centre O. If OA = 20 cm, find the area of the shaded region (quadrant minus square). (Use π = 3.14)
Square OABC has side OA = 20 cm. Its diagonal OB is a radius of the quadrant (B lies on the arc).
OB² = OA² + AB² = 20² + 20² = 800 ⇒ OB = radius r, r² = 800
Area of quadrant (θ = 90°, i.e. 1/4 circle):
= (1/4) × π × r² = (1/4) × 3.14 × 800 = 3.14 × 200 = 628 cm²
Area of square = 20 × 20 = 400 cm².
Shaded area = Quadrant area − Square area = 628 − 400 = 228 cm²
Shaded area = 228 cm²
Q12. In a right triangle PQR, right-angled at P, PQ = 24 cm and PR = 7 cm. A semicircle is drawn on QR (the hypotenuse) as diameter. Find the area of the shaded region (semicircle minus triangle). (Use π = 22/7)
Since ∠P = 90°, QR is the hypotenuse:
QR² = PQ² + PR² = 24² + 7² = 576 + 49 = 625 ⇒ QR = 25 cm
QR is the diameter of the semicircle, so radius r = 12.5 cm.
Area of semicircle = (1/2) × πr² = (1/2) × (22/7) × 12.5² = (1/2) × (22/7) × 156.25
= (22 × 156.25) / (7 × 2) = 3437.5/14 = 245.54 cm² (approx)
Area of triangle PQR:
= 1/2 × PQ × PR = 1/2 × 24 × 7 = 84 cm²
Shaded area = Semicircle − Triangle = 245.54 − 84 = 161.54 cm² (approx)
Q13. ABCD is a square of side 14 cm, and inside it are 4 congruent circles arranged in a 2×2 grid, each circle touching two sides of the square and its neighbouring circles. Find the area of the shaded region (square minus the 4 circles). (Use π = 22/7)
Each circle occupies one quarter of the square in a 2×2 arrangement, so each circle's diameter = side/2 = 14/2 = 7 cm, i.e. radius = 3.5 cm.
Area of one circle = πr² = (22/7) × 3.5² = (22/7) × 12.25 = 38.5 cm²
Area of 4 circles = 4 × 38.5 = 154 cm²
Area of square = 14 × 14 = 196 cm².
Shaded area = Square − 4 circles = 196 − 154 = 42 cm²
Shaded area = 42 cm²
Q14. An equilateral triangle ABC of side 12 cm has a circle inscribed in it, touching all three sides. Find the area of the region between the triangle and the circle (shaded region). (Use π = 3.14, √3 = 1.732)
Side a = 12 cm.
Area of equilateral triangle:
= (√3/4) × a² = (1.732/4) × 144 = 0.433 × 144 = 62.352 cm²
Inradius of an equilateral triangle: r = a / (2√3) = 2√3 cm exactly (rationalising), so r² = 12 cm².
r = 12 / (2 × 1.732) = 12/3.464 = 3.464 cm ⇒ r² = 12 cm² (exact)
Area of inscribed circle:
= πr² = 3.14 × 12 = 37.68 cm²
Shaded area = Triangle − Circle = 62.352 − 37.68 = 24.672 cm² ≈ 24.67 cm²
Important Equations — Ek Nazar Me
| Quantity | Formula |
|---|---|
| Circumference of circle | 2πr |
| Area of circle | πr² |
| Arc length of sector (angle θ) | (θ/360°) × 2πr |
| Area of sector (angle θ) | (θ/360°) × πr² |
| Area of minor segment | Area of sector − Area of triangle (formed by the two radii and chord) |
| Area of major segment | Area of circle − Area of minor segment |
| Area of major sector | Area of circle − Area of minor sector |
| Triangle area (two sides a, b, included angle θ) | 1/2 × a × b × sinθ |
| Inradius of equilateral triangle (side a) | a / (2√3) |
| π values used | 22/7 (radius multiple of 7) or 3.14 — always use the value the question specifies |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Diameter ko radius maan lena. Question me diameter diya ho aur seedha usi ko r maan ke πr² laga dena — hamesha pehle r = d/2 nikaalo.
- Segment ko sector samajh lena. Segment area sirf sector area nahi hota — segment = sector area − triangle area. Ye chapter ki sabse common galti hai.
- Question ka specified π value ignore karna. Agar question 'Use π = 3.14' bole aur tum 22/7 laga do (ya vice versa), answer galat aayega chahe method sahi ho.
- Angle ko degree ki jagah kuch aur maan lena, ya 360° ka fraction ulta le lena. Hamesha (given angle / 360°) hi likho — (360°/given angle) nahi.
- Major sector/segment nikalte waqt poore circle ka area subtract karna bhool jaana. Major sector = poora circle − minor sector, na ki sirf doosra sector alag se calculate karna kisi wrong angle se.
- Combination figures me overlap double-count ya miss kar dena. Jaise square-inscribed-in-circle wale question me square ka area subtract karna bhool jaana, ya galat radius (side ki jagah diagonal, ya diagonal ki jagah side) use kar lena.
Board-Style Important Questions
- 1 mark: Find the area of a sector of a circle of radius 7 cm and central angle 90°.
- 2 marks: Find the area of a quadrant of a circle whose circumference is given.
- 3 marks: Find the area of the minor segment of a circle of given radius when a chord subtends a given angle at the centre.
- 3 marks: A chord of a circle subtends a given angle at the centre. Find the areas of both the minor and major segments.
- 5 marks: Find the area of the shaded region formed by a square (or right triangle) combined with an inscribed or circumscribed circle/semicircle.
Quick Quiz — Score Check Karein
Q1. Class 10 Maths Ch 11 'Areas Related to Circles' — sector ka area nikalne ka formula kya hai (angle θ diya ho)?
Q2. Segment ka area kaise nikalte hain?
Q3. Radius 6 cm wale circle ke 60° sector ka area kya hoga? (π = 22/7)
Q4. Ek circle ki circumference 22 cm hai. Uska radius kya hoga? (π = 22/7)
Q5. Clock ki minute hand 5 minute me kitne degree ghoomti hai?
Q6. Jab chord radius ke saath 60° ka angle banaye, toh do radii aur chord se bana triangle kaisa hota hai?
Q7. Major sector nikalne ka sahi tarika kya hai?
Q8. Equilateral triangle (side a) me inscribed circle ka radius (inradius) kaunsa formula deta hai?
Q9. Do radii (a aur b) aur unke beech included angle θ se bane triangle ka area kaise nikalte hain?
Q10. π ka value 22/7 use karna kab convenient hota hai (jaisa is chapter me bataya gaya hai)?
Aksar Poochhe Jaane Wale Sawaal
Sector aur segment me kya fark hai?
Sector do radii aur unke beech ke arc se bana hota hai (pizza slice jaisa). Segment sirf chord aur arc ke beech ka region hota hai — sector minus us mein bane triangle ka.
Segment ka area seedha kaise nahi nikal sakte?
Segment ka koi direct formula nahi hai jo sirf radius aur angle pe depend kare bina triangle ke — isliye pehle sector nikaalo, phir usme se chord-triangle ka area ghatao.
π = 22/7 kab use karein aur 3.14 kab?
Jo question me likha ho wahi use karo. Aam taur pe jab radius 7 ka multiple ho (7, 14, 21, 35...) tab 22/7 se calculation clean aati hai; baaki cases me aksar 3.14 diya jaata hai.
Major sector aur major segment kaise nikalte hain?
Major sector = poora circle ka area minus minor sector ka area. Major segment = poora circle ka area minus minor segment ka area. Dono me poore circle se minor wala part ghataana hota hai.
Combination of figures wale questions me approach kya honi chahiye?
Pehle poora figure identify karo (bada shape kya hai, andar/bahar kaunsa chhota shape hai), phir bade shape ka area minus (ya plus) chhote shape(s) ka area karo — step by step, koi bhi missing piece na chhoote.
Radians ka use hota hai kya is chapter me?
NCERT syllabus me hamesha degree measure use hota hai, radians nahi. Formula (θ/360°) × ... me θ hamesha degrees me hoga.
Class 10 Maths — Saare Chapters
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