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If PQRO is a square inside a Circle with centre at "O" and

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If PQRO is a square inside a Circle with centre at "O" and [#permalink]

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If PQRO is a square inside a Circle with centre at "O" and radius "a", what is the area of the shaded portion ?


A. a^2((3pi-8)/12)
B. a^2((pi-2)/4)
C. a^2((9pi-16)/12)
D. a((3pi-1)/12)
E. a^2/11
[Reveal] Spoiler: OA

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Re: If PQRO is a square inside a Circle with centre at "O" and [#permalink]

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New post 16 Sep 2012, 03:26
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Pansi wrote:
If PQRO is a square inside a Circle with centre at "O" and radius "a", what is the area of the shaded portion ?


A. a^2((3pi-8)/12)
B. a^2((pi-2)/4)
C.a^2((9pi-16)/12)
D.a((3pi-1)/12)
E.a^2/11


So the square should have a diagonal equal to length of radius of circle. Let x be the side of square.
Hence diagonal of a square with side x= x root2
=> x root2 = a (radius of circle)
=>x= a/root 2
Hence area of square = (a/root 2)^2 = a^2/2.
Now the area of circular quadrant is (pi * a^2)/4
So shaded area = (pi * a^2)/4 - a^2/2, by simplifying
=> a^2((pi-2)/4)
Hence Answer B.
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Re: If PQRO is a square inside a Circle with centre at "O" and [#permalink]

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New post 16 Sep 2012, 06:26
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SOURH7WK wrote:
Pansi wrote:
If PQRO is a square inside a Circle with centre at "O" and radius "a", what is the area of the shaded portion ?


A. a^2((3pi-8)/12)
B. a^2((pi-2)/4)
C.a^2((9pi-16)/12)
D.a((3pi-1)/12)
E.a^2/11


So the square should have a diagonal equal to length of radius of circle. Let x be the side of square.
Hence diagonal of a square with side x= x root2
=> x root2 = a (radius of circle)
=>x= a/root 2
Hence area of square = (a/root 2)^2 = a^2/2.
Now the area of circular quadrant is (pi * a^2)/4
So shaded area = (pi * a^2)/4 - a^2/2, by simplifying
=> a^2((pi-2)/4)
Hence Answer B.


Just a remark: For any quadrilateral with perpendicular diagonals (so obviously also for a square), the area is given by half the product of the diagonals.
(You can easily deduce it by expressing the areas of the triangles formed by the diagonals.)

So, when you know the diagonal of a square, you don't have to compute the side in order to find the area. You just have to square the diagonal and half it.
In the given question, the diagonal of the square is \(a\) (the radius of the circle), so the area of the square is \(a^2/2.\)
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Re: If PQRO is a square inside a Circle with centre at "O" and [#permalink]

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New post 13 Nov 2013, 14:57
SOURH7WK wrote:
Pansi wrote:
If PQRO is a square inside a Circle with centre at "O" and radius "a", what is the area of the shaded portion ?


A. a^2((3pi-8)/12)
B. a^2((pi-2)/4)
C.a^2((9pi-16)/12)
D.a((3pi-1)/12)
E.a^2/11


So the square should have a diagonal equal to length of radius of circle. Let x be the side of square.
Hence diagonal of a square with side x= x root2
=> x root2 = a (radius of circle)
=>x= a/root 2
Hence area of square = (a/root 2)^2 = a^2/2.
Now the area of circular quadrant is (pi * a^2)/4
So shaded area = (pi * a^2)/4 - a^2/2, by simplifying
=> a^2((pi-2)/4)
Hence Answer B.


???? my answer is a square (a square (pi -1)/4) or asquare pi - 4 a square.!!!
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Re: If PQRO is a square inside a Circle with centre at "O" and [#permalink]

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New post 14 Nov 2013, 07:45
Area of one quarter of the circle: (pi*a^2)/4

Area of the square:

Diagonal is equal to radius a. Therefore s(sqrt2) = a --> s = a/(sqrt2)
s^2 = (a^2)/2

Area of the shaded region is area of one quarter of the circle minus area of the square:

(pi*a^2)/4 - (a^2)/2 = [(pi*a^2) - 2(a^2)]/4

Factor out a^2:

a^2[(pi-2)/4]

Answer is B
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Re: If PQRO is a square inside a Circle with centre at "O" and [#permalink]

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New post 09 May 2015, 02:36
Devon wrote:
Area of one quarter of the circle: (pi*a^2)/4

Area of the square:

Diagonal is equal to radius a. Therefore s(sqrt2) = a --> s = a/(sqrt2)
s^2 = (a^2)/2

Area of the shaded region is area of one quarter of the circle minus area of the square:

(pi*a^2)/4 - (a^2)/2 = [(pi*a^2) - 2(a^2)]/4

Factor out a^2:

a^2[(pi-2)/4]

Answer is B


You have reduced qone quadrant with area of Square, I am fine with it. But I have worked out complete Area of circle minus Area of Square which gives= a^2(Pi-1/2). What is wrong in this?
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Re: If PQRO is a square inside a Circle with centre at "O" and [#permalink]

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Re: If PQRO is a square inside a Circle with centre at "O" and [#permalink]

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New post 08 Jun 2016, 10:20
Pansi wrote:
Attachment:
Image.png
If PQRO is a square inside a Circle with centre at "O" and radius "a", what is the area of the shaded portion ?


A. a^2((3pi-8)/12)
B. a^2((pi-2)/4)
C. a^2((9pi-16)/12)
D. a((3pi-1)/12)
E. a^2/11



OQ=a,So OR=\(\frac{a}{\sqrt{2}}\),So area of the Square=(\(\frac{a}{\sqrt{2}}\))^2=\(\frac{a^2}{2}\)

OQ=a,So area of the \(\frac{1}{4}\)th of the Circle=\(\pi\)\(a^2\)/4

So the area of the shaded portion=(\(\pi\)\(a^2\)/4)-\(\frac{a^2}{2}\)=\(a^2\)(\(\pi\)-2/4)

Correct Answer B
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Re: If PQRO is a square inside a Circle with centre at "O" and   [#permalink] 08 Jun 2016, 10:20
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