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ABCD is a square inscribed in a circle with circumference

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ABCD is a square inscribed in a circle with circumference  [#permalink]

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New post 08 Jun 2015, 09:04
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Question Stats:

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ABCD is a square inscribed in a circle with circumference \(2\pi{\sqrt{x}}\). What is the area of the shaded region in the diagram above?

A. 2x
B. \(\pi{x}-2x\)
C. \(\pi{x}-x\sqrt{2}\)
D. \(1-\frac{2}{\pi}\)
E. \(1-\frac{2}{x}\)

Attachment:
2015-06-08_2004.png
2015-06-08_2004.png [ 13.43 KiB | Viewed 10069 times ]


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Re: ABCD is a square inscribed in a circle with circumference  [#permalink]

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New post 08 Jun 2015, 09:31
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Radius of the circle is equal to √x (2πr = 2π√x).
Diagonal of the square is twice the radius of the circle = 2√x. Therefore side of the square is equal to √2x. (a√2 = 2√x)

Area of the circle is equal to πx (πr^2). Area of the square is equal to 2x. (a^2)

Therefore, area of the shaded part = πx - 2x.
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Re: ABCD is a square inscribed in a circle with circumference  [#permalink]

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New post 08 Jun 2015, 09:34
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Bunuel wrote:
Image
ABCD is a square inscribed in a circle with circumference \(2\pi{\sqrt{x}}\). What is the area of the shaded region in the diagram above?

A. 2x
B. \(\pi{x}-2x\)
C. \(\pi{x}-x\sqrt{2}\)
D. \(1-\frac{2}{\pi}\)
E. \(1-\frac{2}{x}\)

Attachment:
2015-06-08_2004.png


Kudos for a correct solution.


circumference = \(2\pi{R} = 2\pi{\sqrt{x}}\)
So radius = \(\sqrt{x}\) and diagonal of square = \(2\sqrt{x}\)

Let's find side of square y:
\(y^2+y^2 = (2\sqrt{x})^2\) --> \(y = \sqrt{2x}\)
Area of square = 2x and area of circle = \(\pi{R^2}\) --> \(\pi{x}\)

Difference between areas of circle and square = \(\pi{x} - 2x\)

Answer is B
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Re: ABCD is a square inscribed in a circle with circumference  [#permalink]

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New post 08 Jun 2015, 10:39
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Ok I tried my best here;

What do we need to find is A(shaded region) = A(circle) - A(square)

A(circle) = r^2*pi
A(square) = s^2

We are given the circumference of the circle, which also operates as the diagonal of the square.
The circumference is 2*pi*sqrt(x)

The circumference normally is diameter * pi
Since the circumference given already has pi in it, the diameter must be 2*sqrt(x)
Since the radius is diameter divided by 2 then the radius would be the sqrt(x)
So the area of the circle would be sqrt(x)^2 * pi which would be x*pi

Okay now the square;
the diameter of the circle is the diagonal of the square, splitting it in two 45-45-90 triangles.
We know that 45-45-90 triangles have a ratio of x : x : x*sqrt(2)
Since the diagonal is already given which is 2*sqrt(x), would make the side the same as 2*x since, sqrt(2) * 2 = 2 and sqrt(x) * x = x

Therefore the answer should be B.

Hope i'm right :)
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Re: ABCD is a square inscribed in a circle with circumference  [#permalink]

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New post 08 Jun 2015, 10:50
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Bit tricky question to understand...

Lets take x=4, so area of circumference 2πr=2π. Then r=1 and Diameter=2r=2.

Diameter of circle=Diagonal of square. 2=√(L^2+L^2). Therefore Length of square L=√2.

Area of circle - Area of Square=πr^2 - L^2=π-2.

In the answer choices, only B will fit.

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Re: ABCD is a square inscribed in a circle with circumference  [#permalink]

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New post 09 Jun 2015, 01:54
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Bunuel wrote:
Image
ABCD is a square inscribed in a circle with circumference \(2\pi{\sqrt{x}}\). What is the area of the shaded region in the diagram above?

A. 2x
B. \(\pi{x}-2x\)
C. \(\pi{x}-x\sqrt{2}\)
D. \(1-\frac{2}{\pi}\)
E. \(1-\frac{2}{x}\)


Ans: B

Solution: given that 2pr=rp * x^1/2
this means r=x^1/2 then diagonal of square= 2* x^1/2
so the side of square is (2x)^1/2
and area is therefor 2x

area of the shaded region = Area of circle- Area of square
px-2x
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Re: ABCD is a square inscribed in a circle with circumference  [#permalink]

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New post 10 Jun 2015, 02:20
If the circumference is \(2\pi\sqrt{x}\) then the diameter is \(2\sqrt{x}\). This is also the diagonal of the square. Therefore the area of the square is (using area using the diagonal formula) \(\frac{(2\sqrt{x})^2}{2}=2x\). The area of the circle is \(x\pi\) so the shaded area is \(x\pi-2x\) or B
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Re: ABCD is a square inscribed in a circle with circumference  [#permalink]

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New post 14 Jun 2015, 04:45
Radius of the circle is equal to rootx (2πrootx).
Diagonal of the sq= 2rootx. Side= root2x.

Area of the circle = πx.
Area of the sq= 2x

Area of the shaded part = πx - 2x.
Ans B
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Re: ABCD is a square inscribed in a circle with circumference  [#permalink]

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New post 15 Jun 2015, 03:41
Bunuel wrote:
Image
ABCD is a square inscribed in a circle with circumference \(2\pi{\sqrt{x}}\). What is the area of the shaded region in the diagram above?

A. 2x
B. \(\pi{x}-2x\)
C. \(\pi{x}-x\sqrt{2}\)
D. \(1-\frac{2}{\pi}\)
E. \(1-\frac{2}{x}\)

Attachment:
The attachment 2015-06-08_2004.png is no longer available


Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

The circumference of the circle = \(2\pi{r}=2\pi{\sqrt{x}}\), so \(r=\sqrt{x}\).

The area of the circle = \(\pi{r^2}=\pi{({\sqrt{x}})^2}=\pi{x}\).

The area of a square is side^2. The diagonal of this square is the diameter of the circle = \(2r=2\sqrt{x}\). The diagonal of a square is always \(\sqrt{2}(side)\), so side = \(\frac{diagonal}{\sqrt{2}}\). Therefore, side = \(\frac{2\sqrt{x}}{\sqrt{2}}=\sqrt{2x}\)
and the area of square ABCD is: \(side^2=(\sqrt{2x})^2=2x\)
Attachment:
2015-06-15_1439.png
2015-06-15_1439.png [ 35.38 KiB | Viewed 7580 times ]


The shaded area is the area of the circle minus the area of the square = πx – 2x.

The correct answer is B.
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Re: ABCD is a square inscribed in a circle with circumference  [#permalink]

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New post 14 Jan 2019, 18:46
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Re: ABCD is a square inscribed in a circle with circumference   [#permalink] 14 Jan 2019, 18:46
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