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

Image of problem

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!

  • 1
    $\begingroup$ As $IJ$ changes, the area changes as well, you sure your conditions are correct? $\endgroup$
    – StAKmod
    Apr 6, 2019 at 2:08
  • $\begingroup$ The diagram is not to scale. The conditions are correct, I made sure the double-check them $\endgroup$ Apr 6, 2019 at 2:28
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    $\begingroup$ 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. $\endgroup$
    – John Doe
    Apr 6, 2019 at 2:31
  • $\begingroup$ @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. $\endgroup$ Apr 6, 2019 at 3:45
  • $\begingroup$ @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 $IG=CH$. $\endgroup$ Apr 6, 2019 at 3:53

1 Answer 1


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.

enter image description here

  • $\begingroup$ +1.... I think an extra condition in the Q went missing somehow. $\endgroup$ Apr 6, 2019 at 3:47

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