Hello, stones44!
We will need Trignometry for this one . . . the Law of Cosines.
One side of the parallelogram is 7, the other is 9.
The long diagonal is 14.
What is the short diagonal? Code:
A 9 B
* - - - - - - - - *
/ * /
/ * /
7/ * / 7
/ * 14 /
/ * /
* - - - - - - - - *
D 9 C
We have parallelogram $\displaystyle ABCD$ with: $\displaystyle AB = DC = 9,\;AD = BC = 7,\;BD = 14$
. . and we want the length of diagonal $\displaystyle AC.$
First, find $\displaystyle \angle A.$
. . In $\displaystyle \Delta ABD\!:\;\;\cos A \;=\;\frac{7^2+9^2-14^2}{2(7)(9)}\;=\;-\frac{11}{21} \quad\Rightarrow\quad A \:=\:121.5881355^o$
Then $\displaystyle \angle D \:=\:\angle ADC \:=\:180^o - 121.5881355^o \:=\:58.41186449^o$
In $\displaystyle \Delta ADC\!:\;\;AC^2\;=\;7^2+9^2-2(7)(9)\cos58.41186449^o \;=\;63.99999999$
Therefore: .$\displaystyle AC \:=\:\sqrt{63.99999999} \:\approx\:8$