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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]  02 Feb 2019, 02:27
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Question Stats:

57% (01:25) correct 42% (01:34) wrong based on 14 sessions
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 [ 13.34 KiB | Viewed 1603 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]  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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