I would have dropped out of highschool if it weren't for this video THANKS!!!!
@EulersAcademyАй бұрын
Thanks for watching. I'm happy to hear that the video was helpful.
@jakelm4256 Жыл бұрын
When I did this blind, my parallelograms overlapped on the top parallel. The proof was a little different, but it still worked.
@EulersAcademy Жыл бұрын
Thanks for watching.
@mnicole-l2n Жыл бұрын
thanks bubba!! u rlly save lives ❤️
@EulersAcademy Жыл бұрын
Thanks for watching. Glad I could help.
@michaeljagdharry3 жыл бұрын
What if the second parallelogram was just a rectangle going directly up to AF from the base BC? This rectangle and the parallelogram ABCD would be unequal, as the respective interior angles between them would differ. Would anyone care to tell me what I am missing here?
@OzoneTheLynx3 жыл бұрын
Probably too late for you, but for anyone else reading this. We only want to show that they have the same area. The proof is a little different. You get to the equality of the triangles ABE=DCF the same way. Now you don't have to remove and add those other triangles to get to the two paralleograms. Instead you'll find the triangles are just seperated by some trapezoid which you can add to the triangles. One triangle and the trapezoid gives you the parallelorgramm and the other triangle and the trapezoid gives you the rectangle.
@Enlightenchannel2 жыл бұрын
Yea this confused me too. The language here is ambiguous. It should say that the area of the parallelogram is equivalent, not that the parallelogram is equivalent, because that gives you the impression that the interior angles are also equal, when they're obviously not. But I understand it now, thanks to this comment! LOL