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

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Joined: 02 Sep 2009
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ABCD is a square inscribed in a circle. If the length of BC = 8cm. Fin  [#permalink]

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02 Jun 2020, 03:14
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35% (medium)

Question Stats:

79% (01:23) correct 21% (01:37) wrong based on 34 sessions

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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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02 Jun 2020, 03:25
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Bunuel wrote:

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

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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02 Jun 2020, 03:34
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1
Bunuel wrote:

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 $$= π(4√2)^2 - 8^2 = 32π - 64 = 32(π-2)$$

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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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02 Jun 2020, 03:59
1
Bunuel wrote:

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

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)

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

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02 Jun 2020, 09:13
1
Bunuel wrote:

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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02 Jun 2020, 09:28
Abhishek009 wrote:
Bunuel wrote:

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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02 Jun 2020, 09:41
yashikaaggarwal wrote:
Abhishek009 wrote:
Bunuel wrote:

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...
_________________
Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )
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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