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Bunuel
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It's a pity that this task has no real value which could have been ballparked. Since we have this little shaded region which we need to find, it must be a quarter of something. Because within the square there are 4 of the same regions like the one which is shaded. So we need 1/4(the square - the circle). For me only answer choice C fits. That's my approach on how I would have done it ... no sure thing.

Answer Choice C

According to Option C, \(\pi\) is factor of both area of Square and Area of Circle and that's where this option proves to be wrong as Area of Square will not be a multiple of \(\pi\) for sure.

But Option C might have misled you because there is a closing bracket after \(\pi\) whereas that has no open side :wink:
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Point a is the center of both a circle and a square. The circle, which is fully shown above, is inscribed in the square and the circle is tangent on all sides with the square, which is only partially shown and has both the x-axis and the y-axis as sides. The origin (0,0) is the bottom-left corner of the square and the line DE is a diagonal of the square. If the x-coordinate of point a is \(x_1\), what is the area of the gray shaded region between the circle and the origin (0,0)?

A) \(0.25(x_1)^2(4 - \pi)\)
B) \((x_1)^2 - (x_1)^2\pi\)
C) \(0.25(2(x_1)^2 - (x_1)^2)\pi)\)
D) \(4(x_1)^2 - (x_1)^2\pi\)
E) \((x_1)^2 - (x_1)^2*0.5\pi\)

Source: Platinum GMAT
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Attachment:
00023-1.gif

From the description :
Radius of Circle = x1
Side of square = 2 x1

Area of circle = \pi \((x1)^ 2\)
Area of Square = \((2X1) ^ 2\)

Area of Square- Area of circle = 4 * area of shaded region = \((x_1)^2(4 - \pi)\)
Hence Area of shaded region = \(0.25(x_1)^2(4 - \pi)\)
Option A
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Required value = 1/4( area of square-area of circle)
Point a =x1; therefore diagonal of square = 2x1
Side of square =root2 x1 ; therefore area of square =2x1^2
Radius of circle = side of square/2=x1/root2
Area of circle =pir *x1^2/2
Therefore putting area value in first equation we get =1/8 * (4-pie) *x1^2

Hence answer is A
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Im going with A.

x=x1

(A(Square)-A(Cirlce))/4
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Bunuel


Point a is the center of both a circle and a square. The circle, which is fully shown above, is inscribed in the square and the circle is tangent on all sides with the square, which is only partially shown and has both the x-axis and the y-axis as sides. The origin (0,0) is the bottom-left corner of the square and the line DE is a diagonal of the square. If the x-coordinate of point a is \(x_1\), what is the area of the gray shaded region between the circle and the origin (0,0)?

A) \(0.25(x_1)^2(4 - \pi)\)
B) \((x_1)^2 - (x_1)^2\pi\)
C) \(0.25(2(x_1)^2 - (x_1)^2)\pi)\)
D) \(4(x_1)^2 - (x_1)^2\pi\)
E) \((x_1)^2 - (x_1)^2*0.5\pi\)

Source: Platinum GMAT
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Attachment:
00023-1.gif

Platinum GMAT Official Solution:

The general approach to solving this question is that we want to find:
(Area of Square – Area of Circle)/4
=.25(Area of Square – Area of Circle)
Note: We divide by 4 since we are only interested in the bottom left gray region. This will be one fourth of the total region between the circle and the square since the circle is perfectly inscribed into the square due to the circle being tangent with each side of the square.

Since the x-coordinate of point a is x1, point a is x1 units away from the y-axis. This distance from the y-axis to point a is the exact same distance as the length of the radius AF. Consequently, AF = x1 = length of radius. As a result:
Diameter circle = 2(Radius)
Diameter circle = 2(x1)
Note: The diameter is not important for the next step, but it will be important later.

We can now calculate the area of the circle:
Area circle = πr^2
Area circle = π(x1)^2

The area of the square is the length of a side of the square multiplied by itself. Although we are not told directly the length of a side, since the circle is tangent with the square on all sides, we know that the circle will just fit within the square. Consequently, the length of the side of the square is the same as the length of the diameter of the circle:
Diameter circle = Length square
2(x1) = Length of Side of Square

The area of the square:
Area square = side^2
Area square = (2*(x1))^2

Calculate the area of the gray region:
=.25(Area of Square – Area of Circle)
.25[(2x1)^2 - (x1)2π]
.25[4(x1)^2 - (x1)2π]
.25[(x1)^2*4 - (x1)2π]
.25*(x1)^2(4 – π)

Answer: A.
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We have a circle inside a square
1. Point "a" is in the center of the circle, with x-cooridinate = x1
this means the radius of the circle is of length x1

2. We know that the circle is tangent to the square on all sides, so point a is in the center of the square
this means the length of the square is twice the radius, which is x1+x1 = 2x1

3. The shaded area represents 1/4 or 0.25 of the area of the square minus the circle
Area square = (2x1)^2 = 4x^2
Area circle = pi*r^2 = pi*x^2

0.25(4x^2-pix^2)
0.25(x^2)(4-pi)
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