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In the figure below, square ABCD is inscribed in circle O. I

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In the figure below, square ABCD is inscribed in circle O. I [#permalink] New post 02 Feb 2019, 02:27
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In the figure below, square ABCD is inscribed in circle O. If the perimeter of ABCD is 24, what is the area of the shaded region?

Attachment:
#GREpracticequestion In the figure below, square ABCD.jpg
#GREpracticequestion In the figure below, square ABCD.jpg [ 13.34 KiB | Viewed 642 times ]


A) \(18 \pi - 36\)

B) \(18 \pi - 24\)

C) \(12 \pi - 36\)

D) \(9 \pi - 36\)

E) \(9 \pi - 24\)
[Reveal] Spoiler: OA

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Re: In the figure below, square ABCD is inscribed in circle O. I [#permalink] New post 02 Feb 2019, 04:15
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Carcass wrote:
In the figure below, square ABCD is inscribed in circle O. If the perimeter of ABCD is 24, what is the area of the shaded region?

Attachment:
#GREpracticequestion In the figure below, square ABCD.jpg


A) \(18 \pi - 36\)

B) \(18 \pi - 24\)

C) \(12 \pi - 36\)

D) \(9 \pi - 36\)

E) \(9 \pi - 24\)


Explanation::

Perimeter of the square = 24

i.e. 4*Side of square = 24

or side of the square = 6

Area of square = \(6^2 = 36\)

Now Area of square = \(\frac{{diagonal}^2}{2}\)

or \(36 * 2 = diagonal^2\)

or diagonal = diameter of the circle = \(6\sqrt{2}\)

Area of the circle = \(\pi * radius^2 = \pi * (3\sqrt{2})^2 = 18\pi\)

Required area = Area of circle - Area of square

= \(18\pi − 36\)
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Re: In the figure below, square ABCD is inscribed in circle O. I   [#permalink] 02 Feb 2019, 04:15
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In the figure below, square ABCD is inscribed in circle O. I

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