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

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Founder
Joined: 18 Apr 2015
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The square is inscribed within the circle and has a side len [#permalink]  25 Jul 2020, 09:40
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100% (00:53) correct 0% (00:00) wrong based on 8 sessions
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GRE The square is inscribed within the circle and has a side length.jpg [ 10.99 KiB | Viewed 158 times ]

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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Intern
Joined: 25 Jul 2020
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Kudos [?]: 14 [0], given: 2

Re: The square is inscribed within the circle and has a side len [#permalink]  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.
Intern
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Kudos [?]: 14 [0], given: 7

Re: The square is inscribed within the circle and has a side len [#permalink]  27 Jul 2020, 11:55
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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