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 (π - 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 (π - 1).

Hence E/F is equal to

[2 r^2 (Ï€ - 1)]/ [2 r^2 (Ï€ - 1)] = [spoiler]1



A[/spoiler]

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

Hi,
I dont know if some of the unshaded regions marked as F are mistyped as G. The question states E and F are shaded regions. so, I am taking for granted that the 4 inner shaded regions are F and the four outer shaded regions are E
Let the radius of smaller circles be r. So, the radius of outer circle is 2r
Adding the the areas of 4 smaller circles gives 4.Ï€r^2 = unshaded region + 2.F (as each shaded F is repeated twice)
So, 4.Ï€r^2 = unshaded region + 2F --eqn(1)
The area of outer circle is π(2r)^2 = 4πr^2 = unshaded region+F+E --eqn(2)
From eqns(1) & (@) we get 4.Ï€r^2 = unshaded region + 2F = unshaded region+F+E => E = F.
So, E/F = 1

Hence, A

Cheers!

Isn't there a problem with the question??

cans wrote:Isn't there a problem with the question??

Hi,
I think some of the unshaded regions are mistakenly marked 'F' instead of 'G'. But, in the ques, it is mentioned that E and F are marked regions. So, I have assumed that in my post. Moreover, the options seem to be having 'n'. In fact, it is π(Pie).

Cheers!