The name of the shape you’re describing is “pentagon”. One thing about pentagons - their angles have to sum to 540°. So those angles you gave can’t be right. As it turns out, you don’t need all the angles - two will do, as long as you have the lengths of all five sides. So here’s what you do. Measure the five sides. In order, call them A, B, C, D, and E. Measure the angle between A and B and call the cosine of it X. Measure the angle between D and E and call the cosine of it Y. Now we can go:
Divide the pentagon into three triangles, ABF, DEG, and CGF. You can get the lengths of sides F and G with the law of cosines:
F = sqrt(A[Sup]2[/Sup] + B[Sup]2[/Sup] - 2ABX)
G = sqrt(D[Sup]2[/Sup] + E[Sup]2[/Sup] - 2DEY)
Now, in my opinion, the best way to get the area of a general triangle is by the semiperimeter method. So define:
H = (A + B + F) / 2
J = (D + E + G) / 2
K = (C + G + F) / 2
So you can get the areas of each of the triangles:
sqrt(H(H - A)(H - B)(H - F))
sqrt(J(J - D)(J - E)(J - G))
sqrt(K(K - C)(K - G)(K - F))
Sum these up and you’re done! As an example, let’s use your numbers for side lengths, A=12, B=14, C=16, D=18, E=20. For the angles, let’s say the angle between A and B is 66° and the angle between D and E is 99°. Then X=cos(66°)=0.4067 and Y=cos(99°)=-0.1564. This is a really irregular shape, but I think it’ll work. I get the following numbers:
F = 14.260
G = 28.924
H = 20.130
J = 33.462
K = 29.592
And the three areas are 76.739, 177.785, and 63.436. The sum of these three numbers is 317.96, which is your total square footage. I’m pretty bad at arithmetic, so I may have gotten some of the numbers wrong, but I think you’ll get the gist. Good luck!