Given that $\displaystyle CH=8\ ,\ HI=3,\ DC=10,EH=16,\ and\ FG=9$, I have to find the area of parallelogram $\displaystyle DEFJ$
I'm really stuck here. I have tried assigning coordinates to each vertex, and I have tried to use similar polygon ratios. My teacher says the answer is $\displaystyle 122$, but I do not know how it is possible to get that answer. Any tips or solutions would be greatly appreciated. Thank you so much in advance!
The area of the parallelogram is twice the area of $DEF$. That, in turn is the area of the entire figure, minus the area of the $CDFG$ trapeze. The only trick in this problem is to show that the two triangles that are in the parallelogram are congruent. You have all angles the same, and the vertical lines have the same length. This means that the perpendiculars from $D$ and $F$ onto the vertical lines have the same length. The conclusion is that $IG=CH$. Then $$A_{DEFJ}=2(A_{CDEH}+A_{HEFG}-A_{CDFG})=122$$