Difficult to describe.
Draw a big rectangle on a sheet of paper.
Its height is y.
Its length is 18+x.
On the top mark x from the left.
On the bottom mark x from the right.
Draw in the parallelogram inscribed within this rectangle.
(By drawing line from bottom left corner to point on top x from left.)
Mark a rectangle on the left, which is x wide, y high.
The diagonal is 14. Side of parallelogram.
So x^2 + y^2 = 14^2 = 196 (E1)
Let z = 18 - x (E2), z is the middle width portion of the rectangle.
Then you can split the bottom line, into three parts,
x, z, x.
Now we know the length of the shorter diagonal 16.
But this is the hypotenuse of a triangle.
The sides are y and z.
So y^2 + z^2 = 16^2 = 256 (E3)
(E3) - (E1)
z^2 - x^2 = 60
From (E2) z = 18 - x
So z^2 = 18^2 - 36x + x^2
(324 - 36x + x^2) - x^2 = 60
(324 - 60) = 36x
x = 264 / 36 = 22 / 3
Hence (E2)
z = 18 - 22/3 = 32/3
And (E1)
y^2 = 196 - x^2 = 196 - (22/3)^2
y^2 = (1764 - 484) / 9 = 1280/9
Width of rectangle,
w = 18 + x = (54+22)/3 = 76/3
Diagonal d
d^2 = w^2 + y^2
d^2 = (76^2)/9 + 1280/9
d^2 = (5776+1280)/9 = 7056/9
d = 84/3 = 28
But there would not be time to do this in the real GMAT.