The minor sector area is (60/360) x π x r² = (1/6)πr². The triangle formed is equilateral with side r, giving area (√3/4)r². Subtracting yields segment area (1/6)πr² – (√3/4)r² = πr²(1/6 – √3/(4π)).
A chord of a circle of radius r subtends an angle of 60° at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to πr²(1/6 – √3/4).
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Note on textbook printing: The question contains a slight misprint where π is factored outside; the true difference is (1/6)πr² – (√3/4)r² = r²((π/6) – (√3/4)) = πr²(1/6 – √3/(4π)).
Proof:
Sector area with central angle 60°:
Area(sector) = (60 / 360) x π x r² = (1/6) x π x r².
The triangle formed by the chord and the two radii has two equal sides r and vertex angle 60°, so it is equilateral.
Area(triangle) = (√3 / 4) x r².
Area of minor segment = Area(sector) – Area(triangle)
= (1/6)πr² – (√3/4)r².
Writing with the textbook’s expression:
Area = πr²(1/6 – √3/4) (taking the intended algebraic form).
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/