A regular hexagon consists of 6 equilateral triangles of side r, giving area 6 x (√3/4)r² = (3√3/2)r². Dividing by circle area πr² gives (3√3) / (2π) ≈ 0.827, which is twice Question 8’s ratio because its area is double.
A hexagon is inscribed in a circle of radius r. Show that the ratio of the area of the hexagon to the area of the circle is equal to (3√3) / (2π) ≈ 0.827. Can you see why the answer is exactly twice the answer to Question 8?
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A regular inscribed hexagon has side equal to the circle’s radius r and consists of 6 equilateral triangles of side r:
Area(hexagon) = 6 x [(√3 / 4) x r²] = (3√3 / 2) x r².
Area of the circle = πr².
Ratio = Area(hexagon) / Area(circle) = [(3√3 / 2)r²] / [πr²] = (3√3) / (2π) ≈ 0.827.
Why it is twice the answer to Question 8:
Connecting alternating vertices of the hexagon forms an inscribed equilateral triangle whose area is exactly half of the hexagon’s area:
(3√3 / 2π) = 2 x [(3√3) / (4π)].
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/