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.
For Fig. 6.43, sides are divided into halves and thirds; subtracting unshaded triangles leaves 7/18 of the total area. For Fig. 6.44, midpoints of the square sides produce the classic inner square of area 1/5.
Scaling linear dimensions by factor k increases area by k²: 2² = 4 and 3² = 9. Yes, in each case exactly 4 or 9 congruent copies tile and perfectly fit into the scaled figure without gaps.
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.