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

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

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New post 03 Dec 2019, 23:41
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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
ABCD Square.JPG [ 46.58 KiB | Viewed 249 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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New post 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
ABCD Square.JPG [ 60.46 KiB | Viewed 203 times ]

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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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