# Sum of Areas Geometry Problem. Cannot understand how provided answer is derived.

Given that $$CH=8$$, $$HI=3$$, $$DC=10$$, and $$FG=9$$, I have to find the area of parallelogram $$DEFJ$$.

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 $$122$$, 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!

• As $IJ$ changes, the area changes as well, you sure your conditions are correct? – StAKmod Apr 6 '19 at 2:08
• The diagram is not to scale. The conditions are correct, I made sure the double-check them – SacredCobalt Apr 6 '19 at 2:28
• IG is also equal to $8$ by symmetry. The issue is finding the vertical height of the triangle - it seems that IJ and HE have freedom to vary, which would greatly change the area. If you could even fix a single angle anywhere in the diagram, I think that would fix it. – John Doe Apr 6 '19 at 2:31
• @JohnDoe . Given CH=8, HI=3, DC=10, take $any$ point G on the line CI to the right of I, and take F such that FG=9. Then take J such that IJ$\le$ 9, and take E such that DE is parallel to FJ. Then DEFJ is a parallelogram. So IG is undetermined.... And the answer by Strichcoder shows that its area is also undetermined. – DanielWainfleet Apr 6 '19 at 3:45
• @DanielWainfleet The length of $IG$ is not undetermined. We know that $DE=JF$ and that they are parallel. Thus we conclude by similar triangles that $IG=CH$. – Strichcoder Apr 6 '19 at 3:53

Here is a version that is to scale. Moving the point $$J$$ will result in different areas of the parallelogram (as pointed out by StAKmod). I think your problem is not well defined.