# One question's stumping me

#### helpmeddddd

Yeah I just thought my calculations are correct but all four sections are triangles. How do I fix this?

#### DarnItJimImAnEngineer

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.

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#### helpmeddddd

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?

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#### DarnItJimImAnEngineer

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.

#### helpmeddddd

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.

Does this work?

#### DarnItJimImAnEngineer

Ah, I see. Yes, that works for the top half.

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.

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