Areas Related to Circles 🔵

Sector, segment, arc length — sab formulas aur worked examples ke saath.

🌐 Circle ke Basic Areas — Quick Revision

🍕 Pizza slice ki tarah socho:

Ek pizza = full circle. Ek slice = sector. Slice ka rounded edge = arc. Slice ko ek straight cut se kaato = triangle aur segment alag ho jaate hain. Yahi concepts hain is chapter mein!

Circle ke basic formulas:
Area of circle = πr²  |  Circumference = 2πr
π = 22/7 (approx) ya 3.14159...
θ Sector A = (θ/360)πr² Segment Minor Segment A = Sector − Triangle

📐 Sector Formula

Sector of angle θ (in degrees)
Area of Sector = (θ/360) × πr²
Length of Arc = (θ/360) × 2πr

Perimeter of Sector = 2r + Arc length = 2r + (θ/360)×2πr
Remember: Full circle (360°) ka area = πr². θ° wala sector ka area = θ/360 wala proportion!
Angle θFraction of circleSector AreaArc Length
360°1 (full circle)πr²2πr
180°1/2 (semicircle)πr²/2πr
90°1/4 (quadrant)πr²/4πr/2
120°1/3πr²/32πr/3
60°1/6πr²/6πr/3
Example 1: Sector of radius 7cm, angle 60°. Area aur arc length nikalo. (π=22/7)
Area = (60/360) × (22/7) × 49 = (1/6) × 154 = 77/3 ≈ 25.67 cm²
Arc = (60/360) × 2 × (22/7) × 7 = (1/6) × 44 = 22/3 ≈ 7.33 cm
Example 2: Sector area = 38.5 cm², angle = 45°. Radius nikalo.
38.5 = (45/360) × (22/7) × r²
38.5 = (1/8) × (22/7) × r²
r² = 38.5 × 8 × 7/22 = 38.5 × 56/22 = 98
r = 7√2 cm

📐 Segment Formula

Minor Segment
Area of Segment = Area of Sector − Area of Triangle
= (θ/360)πr² − (1/2)r²sinθ

Jab θ = 90°: Triangle area = (1/2)r², Sector = πr²/4
Segment = πr²/4 − r²/2 = r²(π/4 − 1/2) = r²(π−2)/4
Triangle area in sector:
Agar O center hai aur angle AOB = θ, to △AOB ka area = (1/2) × r × r × sinθ = r²sinθ/2
(Using formula: Area = (1/2)ab sinC)
Example 3: Radius 10 cm, angle 60°. Minor segment ka area. (π=3.14)
Sector area = (60/360) × 3.14 × 100 = (1/6) × 314 = 52.33 cm²
Triangle: (1/2) × 100 × sin60° = 50 × (√3/2) = 25√3 ≈ 43.30 cm²
Segment = 52.33 − 43.30 = 9.03 cm²
Example 4: Radius 14 cm, angle 90°. Major segment area.
Full circle = π×196 = (22/7)×196 = 616 cm²
Sector (90°) = 616/4 = 154 cm²
Triangle (90°) = (1/2)×14×14 = 98 cm²
Minor segment = 154−98 = 56 cm²
Major segment = 616 − 56 = 560 cm²

🎨 Combined Shape Problems

Bahut saare problems mein shapes combine hote hain — circle mein square, ya triangle par semicircle.
Strategy: Shapes ko alag-alag calculate karo, phir add ya subtract karo.
Example 5: Ek square ke side = 10 cm. Square ke andar inscribed circle ka area? Shaded region area?
r=5 Shaded = Square − Circle Circle radius = 10/2 = 5 cm
Circle area = π×25 = 25π = 78.57 cm²
Square area = 100 cm²
Shaded area = 100 − 78.57 = 21.43 cm²
Example 6: Equilateral triangle of side 14 cm ke teeno corners par circle sectors hain (angle=60° each). Shaded area (inside triangle, outside circles)?
Triangle area = (√3/4) × 196 = 49√3 ≈ 84.87 cm²
Each corner angle = 60°. Each sector area = (60/360)×(22/7)×49 = 154/6 ≈ 25.67 cm²
Total 3 sectors = 77 cm²
But note: 3 × 60° = 180° = half circle (all inside triangle)
Shaded = 84.87 − 77 = 7.87 cm²

📝 Practice Questions

Q1. Sector area = 77 cm², radius = 7 cm. Angle nikalo. (π=22/7)

77 = (θ/360)×(22/7)×49 = (θ/360)×154. θ = 77×360/154 = 180°

Q2. Radius 21 cm, arc length 22 cm. Sector angle aur area nikalo.

22 = (θ/360)×2×(22/7)×21 = (θ/360)×132. θ = 22×360/132 = 60°. Area = (60/360)×(22/7)×441 = 231 cm²

Q3. Clock ghante ki sui 8 cm hai. 12 baje se 3 baje tak mein kitna area cover kiya?

12 se 3 baje = 90°. Area = (90/360)×(22/7)×64 = (1/4)×201.14 = 50.28 cm²

Q4. Radius 6 cm, angle 120°. Minor segment ka area nikalo. (Use sin120°=√3/2)

Sector = (120/360)×π×36 = 12π cm². Triangle = (1/2)×36×sin120° = 18×(√3/2) = 9√3 cm². Segment = 12π−9√3 = 37.7−15.59 = 22.11 cm²
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