vikrantr93 wrote:ABCD is a circle and circles are drawn with AO, CO, DO and OB as diameters. Areas E and F are shaded. E/F is equal to
a. 1
b. 1/2
c. 1/Ï€
d. π/4
e. π/2
Good question! Things to remember here are the areas of sectors and segments of a circle.
Let's take AD = BC = 2 r so that OA = OB = OC = OD = r and hence each included common chord (not named) is r √2/2.
Area of each of the 8 segments = area of sector - area of triangle
= ¼ × π × r^2 - ½ × r √2/2 × r √2/2 = ¼ × r^2 (π - 1)
Hence total shaded F = 8 × ¼ × r^2 (π - 1) =
2 r^2 (Ï€ - 1).
Area of big circle = π × (2 r) ^2 = 4 π r^2
Area inside the big circle EXCEPT the shaded E = area of 4 small circles - total shaded area F
= 4 × π r^2 - 2 r^2 (π - 1)
= 2 π r^2 + 2 r^2 = 2 r^2 (π + 1)
Hence, total shaded area E = 4 π r^2 - (2 π r^2 + 2 r^2)
= 2 π r^2 - 2 r^2 =
2 r^2 (Ï€ - 1).
Hence E/F is equal to
[2 r^2 (Ï€ - 1)]/ [2 r^2 (Ï€ - 1)] = [spoiler]
1
A[/spoiler]
