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I am trying to find the area of the parallelogram with vertices (4,1), (6, 6), (7, 7), and (9, 12).

So I believe the way to solve this problem is through the cross product and then taking the magnitude. However, I got the wrong answer and I am now rethinking my methods.

My original work:

Point A:(4,1) Point B: (6,6) Point C:(7,7) Point D: (9,12)

I took the cross product of AB and AC and took the magnitude of that and ended up getting sqrt(369).

I would appreciate the help. I am kind of stumped.

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  • $\begingroup$ Probably AB or AC is a diagonal rather than a side. Draw a picture and see, otherwise your approach is good and the error is just an arithmetic mistake. $\endgroup$ – user142299 May 14 '14 at 0:40
  • $\begingroup$ I drew it multiple times on a large paper and it was still very ambiguous. I am not sure if I am taking the right points...? $\endgroup$ – Rohit Tigga May 14 '14 at 0:42
  • $\begingroup$ The cross product is much shorter than you have calculated. Did you subtract all the components of A from B to get the vector AB? Please show how you got $\sqrt {369}$ It is true that neither AB nor AC is a diagonal. $\endgroup$ – Ross Millikan May 14 '14 at 0:50
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See this illustration on how to compute the area of a parallelogram in an easy manner.

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  • $\begingroup$ That's actually how I got this far I looked at the page; however, I am not sure why my setup is wrong $\endgroup$ – Rohit Tigga May 14 '14 at 0:49
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area should be 3 square units.

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  • $\begingroup$ Why?${}{}{}{}{}$ $\endgroup$ – vonbrand May 14 '14 at 1:21
  • $\begingroup$ AB=2i+5j, AC=3i+6j. So, AB x AC=-3k. Area- magnitude= 3 sq. units. $\endgroup$ – abstract May 14 '14 at 1:30

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