# Thread: circumference of a parallelogram

1. ## circumference of a parallelogram

hey guys,

i've been trying to calculate the circumference of a parallelogram by only knowing the length of the diagonals (which are 7 and 9 units).

i drew many triangles, tried to use pythagoras and some trig stuff, but failed until now...
thanks a lot for your help!

2. Hello, ViresArcanum!

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.

3. Originally Posted by ViresArcanum
hey guys,

i've been trying to calculate the circumference of a parallelogram by only knowing the length of the diagonals (which are 7 and 9 units).

i drew many triangles, tried to use pythagoras and some trig stuff, but failed until now...
thanks a lot for your help!
Unfortunately that isn't possible:

The parallelogram consists of 2 congruent triangles or 2 pairs of congruent triangles. To determine one triangle unambiguously you need 3 pieces. Therefore you need the value of one more piece: an angle or the height or the length of one side.

To demonstrate how the circumference can change without changing the initial conditions (length of diagonals are constant) I've made a sketch.

4. Those were both really great posts.