Oh, you said one piece was non-triangular, so I assumed that upper line was drawn just a little bit away from point B. If you divide triangle ABC any other way besides through one of the vertices (such as how Skeeter drew it), you get a non-triangular piece.
That's what I was thinking, but for the ABC how do I divide it so that I know three things (i.e. a side/ angles) so I can find the area of the one triangle in ABC and, therefore, work out the non-triangle area? If this makes sense!
Edit: do I have to make one a sector or something? If so, how would that work?
There are actually an infinite number of possibilities. You just have to make some decisions.
If you take the line that is passing through point B and move its endpoint just 1 m closer to point A, you technically have a triangle and a quadrilateral, and their areas will still satisfy the 700 mÂ² requirement. If you want to make it more aesthetically pleasing, you can move that endpoint a little closer towards A and move the other endpoint a little closer towards C. It will take a little algebra (or guess-and-check) to figure out how much to move it.
I'm going to try with a sector for ABC in the top right side. so if I want the area to be half of 1462=731 then I do r=âˆš731Ã·38.648/360Ã·pi r=46.55555 so the area of the sector is 731m^2 meaning the other area is about 731 also.
In the bottom half, you now have a triangle which doesn't use the pipe as a side. If you go back to your old solution (bisect the 60Â° angle) for the bottom half, and use your new solution for the top half, it works beautifully.