In a quadrilateral ABCD, AB = BC = x; Angle ABC = 90; Angle BCD = 75;

Hard to explain clearly without drawing a diagram (which is something that would usually be provided in a question like this, if it were a real GMAT question), but: the 75 degree angle is difficult to work with, so we want to divide it up into angles that are familiar. If you draw the line AC, which is a diagonal within the quadrilateral, then because the angle at B is 90 degrees, and because AB = BC, the triangle ABC we've just created is a 45-45-90 triangle. So using the 1 to 1 to √2 ratio, since the short side in this triangle is x, the hypotenuse, which is AC, is √2 x.

Now we have the length of AC in the other triangle ACD. We're told the angle at D is 60 degrees, and the angle ACD is 30 degrees (since it's 75 - 45). So triangle ACD is a 30-60-90 triangle. We found above that the length of the side AC, opposite the 60 degree angle, is √2 x. Using the 30-60-90 triangle ratio, we know the sides are in a 1 to √3 to 2 ratio, so if the middle-length side is √2 x, the short side, AD, is (√2 / √3) x = (√6 / 3) x, and the long side, CD, is (2√2 / √3)x = (2 √6 / 3)x.

We now know the lengths of the four sides of the quadrilateral: they are x, x, (√6/3)x and (2√6/3)x. Adding these we get the perimeter:

2x + (√6/3)x + (2√6/3)x = 2x + (3√6/3)x = 2x + √6 x = x (2 + √6)