Let square side be s. Two semicircles have total area 2 x (1/2) x π(s/2)² = (1/4)πs², identical to the quarter circle’s area (1/4)πs². Subtracting their common overlapping region proves Area(A) = Area(B).
The figure shows a quarter circle in a square. Its centre is at one vertex, and it passes through two adjacent vertices. There are two semicircles on two adjacent sides as diameters. They create the shaded regions A and B. Show that A and B have equal area.
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Let the side of the square be s.
The quarter circle has radius s and centre at the top-left vertex, giving:
Area(quarter circle) = (1/4) x π x s² = (π/4)s².
Each semicircle is constructed on a side of length s as diameter, so each has radius r = s/2:
Area of one semicircle = (1/2) x π x (s/2)² = (1/8)πs².
The sum of the areas of the two semicircles = (1/8)πs² + (1/8)πs² = (π/4)s².
Notice that the sum of the areas of the two semicircles equals the area of the quarter circle.
Let C be the unshaded region within the quarter circle that is covered by the two semicircles.
Area of two semicircles = Area(A) + Area(C).
Area of quarter circle = Area(B) + Area(C).
Since both total areas are (π/4)s²:
Area(A) + Area(C) = Area(B) + Area(C).
Subtracting Area(C) from both sides gives Area(A) = Area(B).
Hence proved.
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/