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The square is inscribed within the circle and has a side len

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The square is inscribed within the circle and has a side len [#permalink] New post 25 Jul 2020, 09:40
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The square is inscribed within the circle and has a side length of \(\sqrt{2}\) . What is the area of the shaded portion of the drawing?



A. \(2-\pi\)

B. \(\pi -2\)

C. \(\pi - \sqrt{2}\)

D. \(\pi -4\)

E. \(4-\pi\)
[Reveal] Spoiler: OA

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Re: The square is inscribed within the circle and has a side len [#permalink] New post 25 Jul 2020, 11:10
The square as a side of \sqrt{2}. Now draw a line from the center of the circle to a vertex of the square, and let the radius of the circle be r. By Pythagoras' theorem,

\((\sqrt{2}/2)^2 + (\sqrt{2}/2)^2 = r^2.
1 = r^2 \rightarrow r = 1.\)

Hence the area of the shaded region is pi - 2 (area of the square) => B.
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Re: The square is inscribed within the circle and has a side len [#permalink] New post 27 Jul 2020, 11:55
S_shaded = S_circle - S_square
S_sircle = Pi*r^2
S_square = 2
So the choice B looks like what we need S = something - 2
To find the S_circle we need to find the r.
R is a side of a right triangle inscribed into the square.
Its side is equal to (√/2)^2+(√/2)^2=2 => r = 2/2 = 1
S_circle = Pi
Re: The square is inscribed within the circle and has a side len   [#permalink] 27 Jul 2020, 11:55
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The square is inscribed within the circle and has a side len

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