Let OA = OB = r. Semicircle on diameter AB has radius r/√2 and area (1/4)πr². Subtracting circular segment AFB from both the semicircle and the quarter circle OAB leaves equal areas for both shaded regions.
Discussion Forum Latest Questions
Let the rectangle have dimensions X and Y. The vertical dividing line of height h gives C = (1/2)hx₂. Combining expressions A + C = (1/2)hX and B + C = (1/2)x₂Y yields the rectangle area XY = [2(A + ...
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.
By Pythagoras theorem, leg semicircles sum to the hypotenuse semicircle: Semicircle(a) + Semicircle(b) = Semicircle(c). Subtracting the two unshaded circular segments from both sides shows that the lunes’ area Area(A) + Area(B) equals the triangle’s area Area(C).
Radius OA is perpendicular to tangent BC, bisecting it into segments of length l/2. In right-angled triangle OAB, R² – r² = (l/2)² = l²/4. The ring area is pi(R² – r²) = (1/4)pi l².