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Is there some data missing, like an angle? I'm not sure if there is sufficient information, unless, there is a property of a parallelogram that I missed.

How do we solve this or whats the OE? _________________

I'm not sure if that formula applies to a parallelogram. Looks more like what would apply to a kite (adjacent sides equal and diagonals perpendicular).

Couldn't find this anywhere through google. Does anyone have a link? _________________

If anyone doesn't believe the answer is 28,
you can work it out by drawing all the right angles in and applying Pythagoras' theorem several times and substituting.

So draw a rectangle which is 18+x wide, y high. In the top left take out a triangle x wide, with hypotenuse of 14. Then 14^2 = y^2 + x^2

How do the diagonals come in the picture?
Richard.. can you elaborate on that? I tried like you said but somehow the diagonals dont get included on any of the pythagorean equations... What am I missing?

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

I'm not sure if that formula applies to a parallelogram. Looks more like what would apply to a kite (adjacent sides equal and diagonals perpendicular).

Couldn't find this anywhere through google. Does anyone have a link?

I too was wondering where from this new formula came. A little trignometry helped derive formula though. Anyone interested in derivation can look into attached doc.