Area of a Circle : āĻŦ⧃āĻ¤ā§āϤ⧰ āĻ•ā§āώ⧇āĻ¤ā§ā§°āĻĢāϞ


1. Formula: Area of a circle: āĻŦ⧃āĻ¤ā§āϤ⧰ āĻ•ā§āώ⧇āĻ¤ā§ā§°āĻĢāϞ : A = πr2


2. Divide Into Sectors : (āĻ–āĻŖā§āĻĄāϤ āĻŦāĻŋāĻ­āĻžāϜāύ)               Find Square // Rectangle // Circle // Triangle : Click Here



  • Divide the circle into many equal sectors (like pizza slices) : āĻŦ⧃āĻ¤ā§āϤāĻ–āύ āĻŦāĻšā§āϤ⧋ āϏāĻŽāĻžāύ āĻ–āĻŖā§āĻĄāϤ (sector) āĻ­āĻžāĻ— āϕ⧰āĻž āĻšāϝāĻŧ

  • Rearrange these sectors alternately to form a rectangle-like shape : āĻāχ āĻ–āĻŖā§āĻĄāĻŦā§‹ā§°āĻ• āĻĒāĻžāϞ⧇āĻĒāĻžāϞ⧇ āϏāĻžāϜāĻŋāϞ⧇ āĻāϟāĻž āφāϝāĻŧāϤāĻ•ā§āώ⧇āĻ¤ā§ā§° āϏāĻĻ⧃āĻļ āφāĻ•āĻžā§° āĻĒā§‹ā§ąāĻž āϝāĻžāϝāĻŧ

  • As the number of sectors increases, the shape becomes closer to a rectangle : āĻ–āĻŖā§āĻĄā§° āϏāĻ‚āĻ–ā§āϝāĻž āĻŦāĻžāĻĸāĻŧāĻŋāϞ⧇ āφāĻ•āĻžā§°āĻŸā§‹ āφāϝāĻŧāϤāĻ•ā§āώ⧇āĻ¤ā§ā§°ā§° āĻĻ⧰⧇ āĻšāϝāĻŧ

  • The idea is NOT directly related to Pythagoras theorem : āĻāχāĻŸā§‹ āĻĒāĻžāχāĻĨāĻžāĻ—ā§‹ā§°āĻžāĻ› āϏ⧂āĻ¤ā§ā§°ā§° āϏ⧈āϤ⧇ āϏ⧰āĻžāϏ⧰āĻŋ āϏāĻŽā§āĻĒā§°ā§āĻ•āĻŋāϤ āύāĻšāϝāĻŧ

  • It is based on rearrangement of shapes with equal area : āχ āĻšā§ˆāϛ⧇ āĻāϕ⧇āχ āĻ•ā§āώ⧇āĻ¤ā§ā§°āĻĢāϞ āĻĨāĻ•āĻž āφāĻ•ā§ƒāϤāĻŋā§° āĻĒ⧁āύ⧰-āĻŦā§āĻ¯ā§ąāĻ¸ā§āĻĨāĻž (rearrangement)


Important Correction (āϗ⧁⧰⧁āĻ¤ā§āĻŦāĻĒā§‚āĻ°ā§āĻŖ āϏāĻ‚āĻļā§‹āϧāύ)



  • Pythagoras theorem (a² + b² = c²) is used for right triangle, not for circle area derivation

  • Circle area derivation uses cutting & rearranging method


Quick Idea(āĻĻā§ā§°ā§āϤ āϧāĻžā§°āĻŖāĻž)



  • Circle → divide → rearrange → rectangle → Area = πr²

  • āĻŦ⧃āĻ¤ā§āϤ → āĻ–āĻŖā§āĻĄ → āϏāĻžāϜāύāĻŋ → āφāϝāĻŧāϤāĻ•ā§āώ⧇āĻ¤ā§ā§° → āĻ•ā§āώ⧇āĻ¤ā§ā§°āĻĢāϞ = πr²

  • Divide circle → sectors 

  • Rearrange → rectangle

  • Base = πr, Height = r

  • Final Area = πr²


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MCQ


Q1. The circumference of a circle is 44 cm. Find its area. : āĻĒā§ā§°āĻļā§āύ: āĻāϟāĻž āĻŦ⧃āĻ¤ā§āϤ⧰ āĻĒā§°āĻŋāϧāĻŋ 44 cmāĨ¤ āχāϝāĻŧāĻžā§° āĻ•ā§āώ⧇āĻ¤ā§ā§°āĻĢāϞ āĻ•āĻŋāĻŽāĻžāύ ?


Options: (a) 121 cm² (b) 154 cm² (c) 100 cm² (d) 200 cm²


Ans: (b) 154 cm²


ExplanationC = 2πr → 44 = 2 × 22/7 × r → r = 7, Area (āĻ•ā§āώ⧇āĻ¤ā§ā§°āĻĢāϞ) = πr² = 154 cm²


Q2. Area of a circle is 616 cm². Find its circumference. āĻĒā§ā§°āĻļā§āύ: āĻāϟāĻž āĻŦ⧃āĻ¤ā§āϤ⧰ āĻ•ā§āώ⧇āĻ¤ā§ā§°āĻĢāϞ 616 cm²āĨ¤ āĻĒā§°āĻŋāϧāĻŋ āĻ•āĻŋāĻŽāĻžāύ ?


Options: (a) 44 cm (b) 88 cm (c) 77 cm (d) 66 cm


Ans: (b) 88 cm


Explanation616 = 22/7 × r² → r = 14, C = 2πr = 88 cm. āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: r = 14 → āĻĒā§°āĻŋāϧāĻŋ = 88 cm


Q3. If the radius of a circle is increased by 50%, find the % increase in area.


āĻĒā§ā§°āĻļā§āύ: āĻ…ā§°ā§āϧāĻŦā§āϝāĻžāϏ 50% āĻŦ⧃āĻĻā§āϧāĻŋ āϕ⧰āĻŋāϞ⧇ āĻ•ā§āώ⧇āĻ¤ā§ā§°āĻĢāϞ āĻ•āĻŋāĻŽāĻžāύ % āĻŦ⧃āĻĻā§āϧāĻŋ āĻĒāĻžāϝāĻŧ ?


Options: (a) 50% (b) 75% (c) 100% (d) 125%


Ans: (d) 125%


ExplanationNew radius = 1.5r → Area ∝ r², New area = (1.5)² = 2.25 → increase = 125%


āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: (1.5)² = 2.25 → āĻŦ⧃āĻĻā§āϧāĻŋ = 125%


Q4. The ratio of radii of two circles is 3:4. Find ratio of their circumferences.


āĻĒā§ā§°āĻļā§āύ: āĻĻ⧁āϟāĻž āĻŦ⧃āĻ¤ā§āϤ⧰ āĻ…ā§°ā§āϧāĻŦā§āϝāĻžāϏ⧰ āĻ…āύ⧁āĻĒāĻžāϤ 3:4āĨ¤ āĻĒā§°āĻŋāϧāĻŋā§° āĻ…āύ⧁āĻĒāĻžāϤ āĻ•āĻŋāĻŽāĻžāύ ?


Options: (a) 9:16 (b) 3:4 (c) 4:3 (d) 16:9


Ans: (b) 3:4


ExplanationCircumference ∝ r → ratio same : āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āĻĒā§°āĻŋāϧāĻŋ ∝ r → āĻ…āύ⧁āĻĒāĻžāϤ āĻāϕ⧇āχ


Q5. The difference between circumference and diameter of a circle is 88 cm. Find radius.


āĻĒā§ā§°āĻļā§āύ: āĻĒā§°āĻŋāϧāĻŋ āφ⧰⧁ āĻŦā§āϝāĻžāϏ⧰ āĻĒāĻžā§°ā§āĻĨāĻ•ā§āϝ 88 cmāĨ¤ āĻ…ā§°ā§āϧāĻŦā§āϝāĻžāϏ āĻ•āĻŋāĻŽāĻžāύ ?


Options: (a) 7 cm (b) 14 cm (c) 21 cm (d) 28 cm


Ans: (b) 14 cm


Explanation:
C − D = 88
2πr − 2r = 88 → 2r(π − 1) = 88
2r(22/7 − 1) = 88 → r = 14


Q6. A wire of length 44 cm is bent to form a circle. Find its area. āĻĒā§ā§°āĻļā§āύ: 44 cm āĻĻā§ˆā§°ā§āĻ˜ā§āϝ⧰ āϤāĻžā§° āĻāϟāĻž āĻŦ⧃āĻ¤ā§āϤāϤ āĻŦāĻžāρāĻ•ā§‹ā§ąāĻž āĻšā§ˆāϛ⧇āĨ¤ āĻ•ā§āώ⧇āĻ¤ā§ā§°āĻĢāϞ āĻ•āĻŋāĻŽāĻžāύ ?


Options: (a) 154 cm² (b) 121 cm² (c) 100 cm² (d) 200 cm²


Ans: (a) 154 cm²


Explanation:
Wire = circumference = 44 → r = 7
Area (āĻ•ā§āώ⧇āĻ¤ā§ā§°āĻĢāϞ ) = 154 cm²


Q7. Find the area of a semicircle of radius 14 cm. āĻĒā§ā§°āĻļā§āύ: 14 cm āĻ…ā§°ā§āϧāĻŦā§āϝāĻžāϏ⧰ āĻ…ā§°ā§āϧāĻŦ⧃āĻ¤ā§āϤ⧰ āĻ•ā§āώ⧇āĻ¤ā§ā§°āĻĢāϞ āĻ•āĻŋāĻŽāĻžāύ ?


Options: (a) 308 cm² (b) 154 cm² (c) 616 cm² (d) 200 cm²


Answer: (a) 308 cm²


ExplanationArea = ½ × πr² = ½ × 616 = 308


Q8. The area of a circle increases by 21 cm² when radius increases by 1 cm. Find original radius.


āĻĒā§ā§°āĻļā§āύ: āĻ…ā§°ā§āϧāĻŦā§āϝāĻžāϏ 1 cm āĻŦ⧃āĻĻā§āϧāĻŋ āϕ⧰āĻŋāϞ⧇ āĻ•ā§āώ⧇āĻ¤ā§ā§°āĻĢāϞ 21 cm² āĻŦ⧃āĻĻā§āϧāĻŋ āĻĒāĻžāϝāĻŧāĨ¤ āφāĻ—ā§° āĻ…ā§°ā§āϧāĻŦā§āϝāĻžāϏ āĻ•āĻŋāĻŽāĻžāύ ?


Options: (a) 3 cm (b) 7 cm (c) 10 cm (d) 14 cm


Ans: (a) 3 cm


Explanation:
π[(r+1)² − r²] = 21
π(2r+1) = 21 → r = 3


SSC Tricks 



  • Area ∝

  • Circumference ∝ r

  • Radius ↑ → Area increases faster

  • Always use 22/7 for exact answers