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Lesson 20 · Circles
One constant, one trap
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One constant, one trap

Circles bring one number and one trap. The number is π\pi — the distance around any circle divided by the distance across it, the same for every circle ever drawn. On this paper, π=3.14\pi = 3.14. Space inside: A=πr2A = \pi r^2 (A equals pi r squared). Distance around: C=2πrC = 2 \pi r (C equals two pi r) — that one is the circle's perimeter wearing its own name, the circumference.

The trap: both formulas want the radius, centre to edge — but problems love to hand you the diameter, edge to edge through the middle. r=d2r = \tfrac{d}{2}; always halve first. Feed a diameter into πr2\pi r^2 unhalved and the square doubles your doubling: the answer lands FOUR times too big. That slip never retires — meet it now, on purpose.

rA

A=πr2A = \pi r^{2}

  • AA= Area (area)
  • rr= Radius (length)
Area of a Circle solver →
rC

C=2πrC = 2 \pi r

  • CC= Circumference
  • rr= Radius

each variable a length

Circumference of a Circle solver →
abP

P≈π[3(a+b)−(3a+b)(a+3b)]P \approx \pi \left[ 3(a+b) - \sqrt{(3a+b)(a+3b)} \right]

  • PP= Perimeter
  • aa= Semi-major axis
  • bb= Semi-minor axis

each variable a length

Ellipse Perimeter (Ramanujan Approximation) solver →