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# In the figure above, each of the 4 equal circles touch the square at

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Math Expert
Joined: 02 Sep 2009
Posts: 55271
In the figure above, each of the 4 equal circles touch the square at  [#permalink]

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14 Sep 2018, 02:13
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Question Stats:

61% (02:59) correct 39% (01:53) wrong based on 23 sessions

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In the figure above, each of the 4 equal circles touch the square at exactly 2 points. If each circle touches exactly 2 other circles at one point and has a radius of 4, what is the area of the shaded region?

A. 192 – 48π
B. 192 – 60π
C. 240 – 68π
D. 240 – 60π
E. 256 – 64π

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image003 (1).jpg [ 5.11 KiB | Viewed 563 times ]

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Re: In the figure above, each of the 4 equal circles touch the square at  [#permalink]

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16 Sep 2018, 10:19
OA: A
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Capture.PNG [ 18.47 KiB | Viewed 451 times ]

Area of the shaded portion as marked in above figure $$= 4^2 - \frac{90}{360}* \pi * 4^2 = 16 - 4 \pi$$

There are 12 such shaded areas in the figure.

Total Shaded area $$= 12*(16 - 4 \pi) = 192-48 \pi$$
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In the figure above, each of the 4 equal circles touch the square at  [#permalink]

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16 Sep 2018, 11:00
Area of square = (4 x radius of circle) = 16^ = 256
If you join the center of the circles , you get a smaller square with side = 8. Area = 64
You lose 1/4th of each circle. Area of remaining circle = 3/4 (because you lost 1/4th) * 4 (there are 4 circles) * π * r^2 = 48π
Area of shaded region = 256-64-48π = 192-48π
In the figure above, each of the 4 equal circles touch the square at   [#permalink] 16 Sep 2018, 11:00
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