NCERT Solutions for Class 9th Maths Chapter 6 C/D's Adventurous Journey: From Ancient Approximations to the Exact Formula of Mādhava — In-text Questions
Book page 121 Updated on2026-09-08
Q1.
Fig. 6.6 The Mesopotamian Hexagon-to-Circle comparison. Can you see why this shows that π > 3?
Answer
Yes. The hexagon inscribed in a circle of radius 1 has perimeter exactly 6, while the circle that contains it is longer.
Each side of the inscribed regular hexagon equals the radius, so its perimeter is 6.
∠ at the centre for one side = 360° ÷ 6 = 60° the two radii are equal, so the triangle is isosceles with a 60° apex the other two angles are 60° each ∴ the triangle is equilateral, so each side = radius = 1
Why it happens: going from one vertex of the hexagon to the next along the straight side is the shortest route; going along the arc is a detour. Six detours make the circle longer than the hexagon. This is the simplest rigorous statement anyone can make about π, and the Mesopotamians used it around 1900 BCE to reject the crude value 3 and adopt 3 + 1/8 = 3.125.
Q2.
Fig. 6.7: Archimedes' method utilising inscribed and circumscribed polygons. 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.)
Answer
The inscribed hexagon gives the lower bound (previous question). The circumscribed hexagon gives the upper bound, and that is where the Baudhāyana–Pythagoras theorem is needed.
Let the circle have radius 1. For the circumscribed hexagon, each side touches the circle at its midpoint. Join the centre O to that point of contact M and to one endpoint A of the side. Then OM = 1 (a radius, perpendicular to the side), and ∠AOM = half of 60° = 30°.
In right triangle OMA: ∠OMA = 90°, ∠AOM = 30°, so ∠OAM = 60° OA = 2 × AM (side opposite 30° is half the hypotenuse) By Baudhāyana–Pythagoras: OA² = OM² + AM² (2·AM)² = 1² + AM² 4AM² − AM² = 1 ⟹ AM² = 1/3 ⟹ AM = 1/√3
full side = 2AM = 2/√3 perimeter = 6 × 2/√3 = 12/√3 = 4√3
circumference < perimeter of the outer hexagon 2π < 4√3 ∴ π < 2√3 ≈ 3.46
Together with π > 3: 3 < π < 2√3
Why it happens: the circle is squeezed. It contains the inner hexagon, so it is longer than it; it is contained in the outer hexagon, so it is shorter than that. Every time Archimedes doubled the number of sides — 12, 24, 48, 96 — both polygons hugged the circle more tightly and the gap between the bounds shrank, until he could assert 3 10⁄71 < π < 3 1⁄7.
Tip: notice that this argument never measures anything. It is a proof, not an experiment — which is what makes it so much more powerful than wrapping a thread round a reel.