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For the one triangle BXC :guys i need help with question #9(b) on this paper...........
please help!!!!!!!!!!!!!!!!!!!!
The area of segment formula u found out in a(ii) use that to find the area of segment BC with r = 4 and alpha = pi/6 = 8pi/3 - 4sqrt(3)
The area of triangle BXC = {1/2 * 2 * 2sqrt(3)/3 } * 2 = 4sqrt(3)/3
The shaded region BXC = area of triangle BXC - Area of segment BC = 16sqrt(3)/3 - 8pi/3
The shaded area BXC = shaded area AXC = shaded area AXB
Therefore total shaded area = 3 * shahded region BXC = 16sqrt(3) - 8pi