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).
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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.
The fraction of the rectangle covered by n identical tangent circles in a row is always constant at π / 4 approximately 11/14 (or 78.5%) regardless of whether n equals 10, 20 or 50.
Each circle of diameter d sits in a square cell of area d². The ratio of circle area to each cell is (π / 4)d² / d² = π / 4. Both figures have covered fraction π / 4 approximately ...
Yes, they are congruent. Same perimeter gives the same sum of sides and same area gives the same product. They share the identical quadratic equation, producing the exact same side lengths.