Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:
First, draw the coordinate plane and add in point A, which has coordinates (d,d). Since d is greater than 0, this point lies in the first quadrant, as shown:
Notice that the origin, the point (0, d), the pint (d, 0), and point A (d,d) form the corners of a square:
Now, we draw the line segment between (0,0) and (d,d). This is the diameter of a circle, which we also draw:
Notice that the answer doe not depend on d. All that matters is that the circle contains a square. The fraction of the circle’s area in the first quadrant will be the same, no matter what. Thus, we can drop d as a variable and create a new variable r for the radius of the circle. It will be easier to compute areas in terms of r (which will also cancel out in the end).
The area of the circle is just ?r2.
The fraction we are looking for is this:
So we need the shaded area of the circle, which consists of the square (
) and 2 of the four “leaves of the table” ( 
).
The square’s area can be found this way:
Now the area of the 4 leaves ( 
) is the area of the circle ( 
) minus the area of the square (
). So the area of the 4 leaves is ?r2 – 2r2 = (? – 2)r2.
This means that the area of 2 leaves is 
.
The whole shaded area is thus 
.
Finally, the desired fraction is 
.
The correct answer is D.
