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In the figure above, ABCD is a square, and each arc shown is a semicir

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In the figure above, ABCD is a square, and each arc shown is a semicir  [#permalink]

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29 Nov 2017, 22:53
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35% (medium)

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69% (00:56) correct 31% (00:50) wrong based on 36 sessions

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In the figure above, ABCD is a square, and each arc shown is a semicircle. If diagonal BD is equal to √2, what is the sum of the areas of the four semicircles?

(A) π/2
(B) π
(C) √2 π
(D) 2π
(E) 2√2 π

Attachment:

2017-11-30_0949.png [ 52.8 KiB | Viewed 481 times ]

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Re: In the figure above, ABCD is a square, and each arc shown is a semicir  [#permalink]

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29 Nov 2017, 22:58
A.
We can find the side of the square as 1. This is the diameter of the circle. Hence the area of the one semicircle is (1/2)*pi*1^2/4
=pi/8
There are four semicircles. Hence total area=pi/8*4=pi/2

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Re: In the figure above, ABCD is a square, and each arc shown is a semicir  [#permalink]

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02 Dec 2017, 14:51
Diagonal of Square= √2
So, Side of Square = Diameter of each semicircle = 1.

Area of Four Semi-circles= Area of two circles= 2*(πr^2)= 2*π/4 = π/2

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Re: In the figure above, ABCD is a square, and each arc shown is a semicir  [#permalink]

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03 Dec 2017, 18:58
Bunuel wrote:

In the figure above, ABCD is a square, and each arc shown is a semicircle. If diagonal BD is equal to √2, what is the sum of the areas of the four semicircles?

(A) π/2
(B) π
(C) √2 π
(D) 2π
(E) 2√2 π

Attachment:
2017-11-30_0949.png

Since the diagonal of a square = side√2:

side√2 = √2, so side = diameter = 1

Since the radius of each semicircle is 1/2, and since 4 semicircles of equal radii make up 2 full circles, the area of the 4 semicircles is:

2 x (1/2)^2 x π = π/2

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Re: In the figure above, ABCD is a square, and each arc shown is a semicir &nbs [#permalink] 03 Dec 2017, 18:58
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