Brahmagupta's formula - Wikipedia, the free encyclopedia
I tink that the above link could shed a little light on your problem. Also, BD can be found by using the law of cosines.
Here is a link to the quadrilateral:
http://img200.imageshack.us/img200/3690/quad.png
I am having a tough time solving for the area. I know you have to split it up into simpler shapes, but it is giving me a hard time.
Please go through the steps for me.
Thanks!
Brahmagupta's formula - Wikipedia, the free encyclopedia
I tink that the above link could shed a little light on your problem. Also, BD can be found by using the law of cosines.
Earboth did it faster than I could, and there's no easier way than the way he describes. I will say this though, in case there was any confusion, BC(sin52.2) is another is a fancy way to describe the height of triangle BCD. So the last line he gives is simply Area of triangle BCD=(1/2)base*height.
use the Cosine Law to determine distance DB
Distance DB being the c (small cap) in the above equation.
Then use pythagoras to deteremine distance DA.
perpendicular dist D is from line BC
The area of a triangle is = base x height / 2
You have two triangles: you know the base of each, and you know the height of each.
There are other methods, but that's the way I would approach it.