The intersection of the circles consists of two 120° sectors minus two equilateral triangles of side r, giving overlapping area (2π/3 – √3/2)r². The total union area equals 2πr² minus this overlap: (4π/3 + √3/2)r².
Fig. 6.53 shows two circles passing through each other’s centres. Find the area of the region enclosed by the two circles in terms of the common radius r.
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Let the centres be A and B with distance AB = r. The circles intersect at C and D, forming equilateral triangles ABC and ABD of side r, each having area (√3/4)r².
The central angle subtended at each centre is ∠CAD = 60° + 60° = 120°.
Area of 120° sector = (120/360) x πr² = (1/3)πr².
Area of rhombus ACBD = 2 x (√3/4)r² = (√3/2)r².
Area of intersection (shaded region) = 2 x Area(Sector) – Area(Rhombus)
= 2 x (1/3)πr² – (√3/2)r² = (2π/3 – √3/2)r².
Total area enclosed by the two circles (their union):
Area(Union) = Area(Circle 1) + Area(Circle 2) – Area(Intersection)
= 2πr² – [(2π/3 – √3/2)r²]
= (4π/3 + √3/2)r².
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/