SOLUTION: Please help me with this--Find the area of a parallelogram ABCD, if m Ang A = 60 degrees, AB=10, and AD = 4sqrt(3). I am lost!

Question 84019: Please help me with this--Find the area of a parallelogram ABCD, if m Ang A = 60 degrees, AB=10, and AD = 4sqrt(3). I am lost!
Answer by vertciel(183) (Show Source): You can put this solution on YOUR website!
*I would assume that this problem involves the use of trigonometry. I was not sure so I am offering a solution without trigonometry. If it actually involves trig, I can offer another solution.
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� Draw a tidy diagram of this problem. Label what you know.
� Denote the intersection of angles (perpendicular bisector in parallelograms) any letter you want. I'll call it O.
� A parallelogram's area = Base x Height. However, height is not given here.
� We can also find area by another way. Twice the area of the triangles formed by the diagonals = the area of the parallelogram.
So, the area of the parallelogram = Area of triangles ACD + ABD or Area of triangles BDC + BAC (either set works).

I'll stick with ACD + ABD:, so all you need to do is calculate the area of either ACD or ABD. I'll pick ACD.
You know AD is the base, but you don't know the height. However, O is perpendicular so triangle COD is a right angle triangle. You have CD = 10 and AD = 4sqrt(3). Use the Pythagoras Theorem to find out OC, which is rounded to 9.38.
So now you know triangle ACD's area = 4sqrt(3) x 9.38
= 64.987
So double that, and you get 129.974 sq. units, which is area of the entire parallelogram.

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