Such reasoning suggests that for a circle too, if C = circumference and A = area, the ratio C2 : A must be some fixed constant. But what is this constant?
Answer: C = 2πr and A = πr2, so C2 : A = 4π2r2 : πr2 = 4π ≈ 12.57. (The Babylonians measured it as about 12.)
Step-by-step solution
Idea: Write both C and A in terms of the radius; the r2 cancels, leaving a number that is the same for every circle.
- C = 2πr, so C2 = 4π2r2; and A = πr2.1 mark
- C2A = 4π2r2πr2 = 4π ≈ 12.57. This is why the Babylonian rule A ≈ C212 was close but a little too big.1 mark
For every circle, C² : A = 4π : 1 ≈ 12.57 : 1.
Check: r = 7: C = 44, A = 154; 442 ÷ 154 = 1936 ÷ 154 ≈ 12.57 ✓ (with π = 227, 4π = 887 ≈ 12.57).
Answer to write in the exam
C2 = (2πr)2 = 4π2r2, A = πr2
C2A = 4π2r2πr2 = 4π
∴ C2 : A = 4π ≈ 12.57
Common mistakes that cost marks
- Forgetting to square the 2 in 2πr (getting 2π).
- Thinking the ratio changes with the size of the circle; r2 cancels.
How this can come in the exam
MCQ (1 mark)
The ratio (perimeter)2 : area for every square is
- 4 : 1
- 8 : 1
- 16 : 1
- 4π : 1
Show answer
(C) 16 : 1
(4a)2 : a2 = 16 : 1.
Try one yourself
Find the ratio (perimeter)2 : area for a regular hexagon of side a (area 3√32a2).
Show answer
(6a)2 ÷ 3√32a2 = 723√3 = 8√3 ≈ 13.86, between the circle’s 12.57 and the square’s 16.
More questions like this
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- Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.
- Find the area of a quadrant of a circle whose circumference is 44 cm.
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- A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding: (Use π ≈ 3.14.)