Hello, ViresArcanum!

Is there more information?

As stated, there is no unique solution.

i've been trying to calculate the perimeter of a parallelogram

by only knowing the length of the diagonals (which are 7 and 9 units). Code:

S R
* - - - - - - - - *
/ * * /
/ * O * /
/ * / b
/ 4½ * θ * 3½ /
/ * * /
* - - - - - - - - *
P a Q

We have parallelogram $\displaystyle PQRS$ with diagonals $\displaystyle PR = 9,\:QS = 7$

. . intersecting at $\displaystyle O\!:\;\;OP = 4.5,\:OQ = 3.5,\;\angle POQ = \theta$

Law of Cosines: .$\displaystyle a^2 \;=\;4.5^2 + 3.5^2 - 2(4.5)(3.5)\cos\theta $

. . Hence: .$\displaystyle a \;=\;\sqrt{32.5 - 31.5\cos\theta} $

It can be shown that: .$\displaystyle b \;=\;\sqrt{32.5 + 31.5\cos\theta}$

Hence, the perimeter, $\displaystyle 2a + 2b$, is a function of $\displaystyle \theta$, not a constant.