Ronald,
Below is part of the plan you sent. I added one line and labeled the corners of the lot.
Since the angle at C is a right angle the length of CD is the height of triangle DBC and hence you can find its area using the standard expression
area of DBC = ^{1}/_{2} base height = ^{1}/_{2} 27.43 32.61 = 447.25 square metres
Again since the angle at C is a right angle I can find the length of DB using Pythagoras theorem.
DB^{2} = BC^{2} + CD^{2} = 27.43^{2} + 32.61^{2} = 1815.81
and hence
DB = √1815.81 = 42.61 metres.
Finally since I now know the lengths of the sides of triangle ABD I can find its area using Heron's Formula.
a = 42.61 metres, b = 9.14 metres and c = 36.9 metres
and hence
s = (42.61 + 9.14 + 36.9)/2 metres
and
area of ABD = Sqrt[s (sa) (sb) (sc)] = 140.92 square metres.
Thus the area of your lot is
447.25 + 140.92 = 588.17 square metres.
Penny
