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ABCD is a square inscribed in a circle. If the length of BC = 8cm. Fin

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ABCD is a square inscribed in a circle. If the length of BC = 8cm. Fin  [#permalink]

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New post 02 Jun 2020, 03:14
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ABCD is a square inscribed in a circle. If the length of BC = 8cm. Find the area of the shaded region.

A. 32(2- π) cm^2
B. 32(π – 3) cm^2
C. 32(π – 2) cm^2
D. 64(π – 2) cm^2
E. 64(π – 1) cm^2



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Re: ABCD is a square inscribed in a circle. If the length of BC = 8cm. Fin  [#permalink]

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New post 02 Jun 2020, 03:25
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Bunuel wrote:
Image
ABCD is a square inscribed in a circle. If the length of BC = 8cm. Find the area of the shaded region.

A. 32(2- π) cm^2
B. 32(π – 3) cm^2
C. 32(π – 2) cm^2
D. 64(π – 2) cm^2
E. 64(π – 1) cm^2



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IMO C,

Area of the square=8^2=64,
Area of circle=Diameter/2 or (8 root 2)/2 so the area of circle=32Pie

are of shaded region=32Pie-64 or C.
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Re: ABCD is a square inscribed in a circle. If the length of BC = 8cm. Fin  [#permalink]

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New post 02 Jun 2020, 03:34
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Bunuel wrote:
Image
ABCD is a square inscribed in a circle. If the length of BC = 8cm. Find the area of the shaded region.

A. 32(2- π) cm^2
B. 32(π – 3) cm^2
C. 32(π – 2) cm^2
D. 64(π – 2) cm^2
E. 64(π – 1) cm^2


BC = 8

Diagonal of Square = Diameter of Circle = 8√2

SHaded Area = Area of Circle - Area of Square \(= π(4√2)^2 - 8^2 = 32π - 64 = 32(π-2)\)

Answer: Option C
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Re: ABCD is a square inscribed in a circle. If the length of BC = 8cm. Fin  [#permalink]

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New post 02 Jun 2020, 03:59
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Bunuel wrote:
Image
ABCD is a square inscribed in a circle. If the length of BC = 8cm. Find the area of the shaded region.

A. 32(2- π) cm^2
B. 32(π – 3) cm^2
C. 32(π – 2) cm^2
D. 64(π – 2) cm^2
E. 64(π – 1) cm^2



IMO C.

Radius f circle is 4 (root 2) .
Area of shaded region = [π (4 (root 2) ) ^2- 64 ] = 32 π - 64 = 32 (π-2 ) cm ^2 .

Ans C
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Re: ABCD is a square inscribed in a circle. If the length of BC = 8cm. Fin  [#permalink]

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New post 02 Jun 2020, 05:46
1
Diagonal of a square = Side*√2
Also, Diagonal= Diameter of Circle
8√2=Diameter
4√2 = radius

Area of circle = πr^2 => π(4√2)^2 => 32π
Area of Square = S^2 => 8*8 => 64

Area of shaded region = Area of circle - Area of Square
=> 32π-64 => 32(π-2)

Answer is C

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ABCD is a square inscribed in a circle. If the length of BC = 8cm. Fin  [#permalink]

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New post 02 Jun 2020, 09:13
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Bunuel wrote:
Image
ABCD is a square inscribed in a circle. If the length of BC = 8cm. Find the area of the shaded region.

A. 32(2- π) cm^2
B. 32(π – 3) cm^2
C. 32(π – 2) cm^2
D. 64(π – 2) cm^2
E. 64(π – 1) cm^2



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Attachment:
2.png


AC = Diameter of Circle = Diagonal of ABCD

Or, AC = Diameter of Circle = Diagonal of ABCD = \(8\sqrt{2}\)

Area of Shadded region = Area of Circle - Area of Square

Or, Area of Shadded region = \( π*4\sqrt{2}*\sqrt{2}- 8^2\)

Or, Area of Shadded region = \(π*32 - 64\) => \(32(π - 2)\), Answer must be (C)
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Re: ABCD is a square inscribed in a circle. If the length of BC = 8cm. Fin  [#permalink]

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New post 02 Jun 2020, 09:28
Abhishek009 wrote:
Bunuel wrote:
Image
ABCD is a square inscribed in a circle. If the length of BC = 8cm. Find the area of the shaded region.

A. 32(2- π) cm^2
B. 32(π – 3) cm^2
C. 32(π – 2) cm^2
D. 64(π – 2) cm^2
E. 64(π – 1) cm^2



Project PS Butler


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Attachment:
2.png


AC = Diameter of Circle = Diagonal of ABCD

Or, AC = Diameter of Circle = Diagonal of ABCD = \(8\sqrt{2}\)

Area of Shadded region = Area of Square - Area of Circle

Or, Area of Shadded region = \(8^2 - π*4\sqrt{2}*\sqrt{2}\)

Or, Area of Shadded region = \(64- π*32\) => \(32(2- π)\), Answer must be (A)

Shouldn't we minus square from circle as square is inscribed inside the circle?

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Re: ABCD is a square inscribed in a circle. If the length of BC = 8cm. Fin  [#permalink]

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New post 02 Jun 2020, 09:41
yashikaaggarwal wrote:
Abhishek009 wrote:
Bunuel wrote:
Image
ABCD is a square inscribed in a circle. If the length of BC = 8cm. Find the area of the shaded region.

A. 32(2- π) cm^2
B. 32(π – 3) cm^2
C. 32(π – 2) cm^2
D. 64(π – 2) cm^2
E. 64(π – 1) cm^2



Project PS Butler


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Attachment:
2.png


AC = Diameter of Circle = Diagonal of ABCD

Or, AC = Diameter of Circle = Diagonal of ABCD = \(8\sqrt{2}\)

Area of Shadded region = Area of Square - Area of Circle

Or, Area of Shadded region = \(8^2 - π*4\sqrt{2}*\sqrt{2}\)

Or, Area of Shadded region = \(64- π*32\) => \(32(2- π)\), Answer must be (A)

Shouldn't we minus square from circle as square is inscribed inside the circle?

Posted from my mobile device


Rectified, thanks for pointing out...
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Re: ABCD is a square inscribed in a circle. If the length of BC = 8cm. Fin   [#permalink] 02 Jun 2020, 09:41

ABCD is a square inscribed in a circle. If the length of BC = 8cm. Fin

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