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# In the figure below, square ABCD is inscribed in circle O. If the peri

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In the figure below, square ABCD is inscribed in circle O. If the peri  [#permalink]

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19 Jul 2018, 22:58
00:00

Difficulty:

35% (medium)

Question Stats:

80% (01:42) correct 20% (02:19) wrong based on 25 sessions

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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 shaded region?

A. $$18\pi - 36$$

B. $$18\pi - 24$$

C. $$12\pi - 36$$

D. $$9\pi - 36$$

E. $$9\pi - 24$$

Attachment:

figure 10.jpg [ 17.76 KiB | Viewed 406 times ]

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In the figure below, square ABCD is inscribed in circle O. If the peri  [#permalink]

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19 Jul 2018, 23:07
1
the perimeter of the square ABCD is 24

=> side of the square = $$\frac{24}{4}$$ = 6

=> diagonal of the square = $$\sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}$$

=> radius of the circle = half of square's diagonal = $$\frac{6}{2} * \sqrt{2} = 3\sqrt{2}$$

Required area = Area of circle - Area of square

=> $$\pi (3\sqrt{2})^2 - 6^2$$

=> $$18 \pi - 36$$

Hence option A
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Re: In the figure below, square ABCD is inscribed in circle O. If the peri  [#permalink]

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20 Jul 2018, 00:25
+1 for A.

Perimeter of the inscribed square (P) = 24
4*S = 24, S = Side of square
S = 6

Diagonal = D
D^2 = 6^2 + 6^2
D^2 = 72
D = √72
D = 6√2

Radius = Diagonal / 2 = half of square's diagonal = 6√2/2 = 3√2

Therefore,
Shaded Area = Area of Circle - Area of Square
= π(3√2)^2−6^2
= 18π−36

Hence, A.
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Re: In the figure below, square ABCD is inscribed in circle O. If the peri &nbs [#permalink] 20 Jul 2018, 00:25
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