Archimedes’ method utilising inscribed and circumscribed polygons (see the figure). Can you see why this diagram of an inscribed and circumscribed hexagon tells us that π is between 3 and 2√3? (Hint: Use the Baudhāyana–Pythagoras Theorem.)
Step-by-step solution
Idea: The circle lies between the two hexagons, so its circumference is more than the inner perimeter and less than the outer perimeter. The outer hexagon’s side comes from the Baudhāyana–Pythagoras theorem.
- Lower limit: the inner hexagon has side equal to the radius, 1, so its perimeter is 6. The circle is longer, so 2π > 6, i.e. π > 3.1 mark
- Outer hexagon: joining O to its corners makes 6 equilateral triangles of side a. The radius to the point where the circle touches a side is the height of one triangle, and it meets the side at right angles at its midpoint. So height = 1 and half the side = a/2.1 mark
- Baudhāyana–Pythagoras: a2 = 12 + (a/2)2 ⇒ 3a24 = 1 ⇒ a2 = 43 ⇒ a = 2√3. Perimeter = 6a = 12√3 = 4√3.1 mark
- The circle lies inside the outer hexagon, so its circumference is shorter: 2π < 4√3, i.e. π < 2√3 ≈ 3.46. Together: 3 < π < 2√3.1 mark
Check: 2√3 ≈ 3.464, and π ≈ 3.1416 does lie between 3 and 3.464 ✓.
Answer to write in the exam
Inscribed hexagon: side = 1 ⇒ perimeter = 6
Circumscribed hexagon: side a, height of each equilateral triangle = r = 1
a2 = 12 + (a/2)2 (Baudhāyana–Pythagoras)
3a24 = 1 ⇒ a = 2√3
Perimeter = 6a = 4√3
6 < 2π < 4√3
∴ 3 < π < 2√3
Common mistakes that cost marks
- Taking the outer hexagon’s side as 1 as well. Its height (from O) is 1, not its side.
- Using a2 = 1 + a2: the right-angled triangle has half the side, a/2.
- Forgetting to divide the perimeters by 2 (the circumference is 2π, not π).
How this can come in the exam
A regular hexagon is drawn around a circle of radius √3 cm so that each side touches the circle. The side of the hexagon is
- 1 cm
- √3 cm
- 2 cm
- 2√3 cm
Show answer
(C) 2 cm
Height of each equilateral triangle = √3 = (√3/2)a, so a = 2 cm.
A circle has radius 3 cm. Find the perimeters of the regular hexagons drawn inside it and around it, and use them to write limits for its circumference.
Show answer
Inside: side 3, perimeter 18 cm. Outside: side 2√3 × 3 = 2√3, perimeter 12√3 ≈ 20.78 cm (1 mark). So 18 < circumference < 20.78 cm; the true value 6π ≈ 18.85 cm fits (1 mark).Try one yourself
For a circle of radius 1, find the perimeter of a circumscribed square, and the upper limit for π it gives.
Show answer
Square side = 2, perimeter 8 > 2π, so π < 4 (weaker than π < 2√3).
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