# Practice GMAT Problem Solving Question 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 x1, what is the area of the gray shaded region between the circle and the origin (0,0)?
 A) .25(x1)2[4 - π] B) x12 - x12π C) .25[2(x1)2 - x12π] D) 4(x1)2 - x12π E) x12 - x12.5π
1. 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.
2. 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:
Diametercircle = 2x1
Note: The diameter is not important for the next step, but it will be important later.
3. We can now calculate the area of the circle:
Areacircle = πr2
Areacircle = π(x1)2
4. 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:
Diametercircle = Lengthsquare
2x1 = Length of Side of Square
5. The area of the square:
Areasquare = side2
Areasquare = (2x1)2
6. 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)24 - (x1)2π]
.25*x12(4 – π)