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Square ABCD is perfectly inscribed in the circle pictured above. If mi

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Square ABCD is perfectly inscribed in the circle pictured above. If mi  [#permalink]

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New post 06 Jun 2017, 11:14
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

70% (02:45) correct 30% (02:41) wrong based on 47 sessions

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Re: Square ABCD is perfectly inscribed in the circle pictured above. If mi  [#permalink]

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New post 06 Jun 2017, 11:41
As Square ABCD is perfectly inscribed in the circle, all the four arcs will be equal

So the total length of all 4 arcs = circumference of the circle = 4 x 2π = 8π

Radius of the circle =\(\frac{8π}{2π}\)= 4

Area of the circle = π X 16 = 16π

Diameter of the Circle = Diagonal of the Square

If each side of the square is a, then 2\(a^2\) = \(8^2\)==> a = \(\frac{8}{\sqrt{2}}\)

Area of Square = \((8/\sqrt{2})^2\) = 32

Area of Shaded Region = Area of Circle - Area of Square = 16π - 32 = 18.2

Answer is D
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Re: Square ABCD is perfectly inscribed in the circle pictured above. If mi  [#permalink]

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New post 25 May 2019, 05:50
Hi, i just wanted to understand why all the arcs are equal? I read the similar concept when the arcs of a circle were circumscribed by an equilateral triangle. is there a concept i am missing out on.
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Re: Square ABCD is perfectly inscribed in the circle pictured above. If mi  [#permalink]

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New post 25 May 2019, 08:42
Each arc is making same angle (90*) at the center.
abhishek31 wrote:
Hi, i just wanted to understand why all the arcs are equal? I read the similar concept when the arcs of a circle were circumscribed by an equilateral triangle. is there a concept i am missing out on.
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Re: Square ABCD is perfectly inscribed in the circle pictured above. If mi   [#permalink] 25 May 2019, 08:42
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Square ABCD is perfectly inscribed in the circle pictured above. If mi

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