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# ABCD is a square of side 'x' enclosed in a circle with its vertices to

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Director
Joined: 07 Mar 2019
Posts: 595
Location: India
GMAT 1: 580 Q43 V27
WE: Sales (Energy and Utilities)
ABCD is a square of side 'x' enclosed in a circle with its vertices to  [#permalink]

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03 Dec 2019, 23:41
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Difficulty:

45% (medium)

Question Stats:

30% (02:22) correct 70% (03:14) wrong based on 10 sessions

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ABCD is a square of side 'x' enclosed in a circle with its vertices touching the circle as shown in diagram.
Find the area of shaded region?

a. $$\frac{x^2}{16} (11 - 8Π)$$

b. $$\frac{x^2}{8} (4Π - 11)$$

c. $$\frac{x^2}{8} (16Π - 11)$$

d. $$\frac{x^2}{16} (8Π - 11)$$

e. $$\frac{x^2}{16}11$$

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File comment: ABCD Square

ABCD Square.JPG [ 46.58 KiB | Viewed 250 times ]

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Re: ABCD is a square of side 'x' enclosed in a circle with its vertices to  [#permalink]

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04 Dec 2019, 05:03
Area of shaded region= Area of circle - area of ABCD + 1/2*(Area of EFGH)+ ( 1/2*(Area of IJKL)

Radius of circle= 1/2* AC= $$\frac{x}{\sqrt{2}}$$

Area of circle= $$Π*\frac{x^2}{2}$$

Area of ABCD= x^2

Diagonal of EFGH= Side of ABCD= x

EF= $$\frac{x}{\sqrt{2}}$$

Area of EFGH= $$\frac{x^2}{2}$$

Similarly we can get

IJ= $$\frac{x}{2\sqrt{2}}$$

Area of IJKL= $$\frac{x^2}{8}$$

Area of shaded region= $$Π*\frac{x^2}{2}$$$$- x^2+ \frac{x^2}{4} + \frac{x^2}{16}$$

= $$\frac{x^2}{2}[ Π - 2 +\frac{1}{2}+ \frac{1}{8}]$$

= $$\frac{x^2}{16} [ 8Π-11]$$

lnm87 wrote:
ABCD is a square of side 'x' enclosed in a circle with its vertices touching the circle as shown in diagram.
Find the area of shaded region?

a. $$\frac{x^2}{16} (11 - 8Π)$$

b. $$\frac{x^2}{8} (4Π - 11)$$

c. $$\frac{x^2}{8} (16Π - 11)$$

d. $$\frac{x^2}{16} (8Π - 11)$$

e. $$\frac{x^2}{16}11$$

Attachments

ABCD Square.JPG [ 60.46 KiB | Viewed 204 times ]

Re: ABCD is a square of side 'x' enclosed in a circle with its vertices to   [#permalink] 04 Dec 2019, 05:03
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