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In the figure below, square ABCD is inscribed in circle O. I
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02 Feb 2019, 02:27
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?
#GREpracticequestion In the figure below, square ABCD.jpg [ 13.34 KiB | Viewed 8497 times ]
A) \(18 \pi - 36\)
B) \(18 \pi - 24\)
C) \(12 \pi - 36\)
D) \(9 \pi - 36\)
E) \(9 \pi - 24\)
Attachment:
#GREpracticequestion In the figure below, square ABCD.jpg [ 13.34 KiB | Viewed 8497 times ]
A) \(18 \pi - 36\)
B) \(18 \pi - 24\)
C) \(12 \pi - 36\)
D) \(9 \pi - 36\)
E) \(9 \pi - 24\)
Re: In the figure below, square ABCD is inscribed in circle O. I
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02 Feb 2019, 04:15
02 Feb 2019, 04:15
3
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?
A) \(18 \pi - 36\)
B) \(18 \pi - 24\)
C) \(12 \pi - 36\)
D) \(9 \pi - 36\)
E) \(9 \pi - 24\)
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
= \(18\pi − 36\)
Re: In the figure below, square ABCD is inscribed in circle O. I
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04 Jan 2022, 21:43
04 Jan 2022, 21:43
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