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 = 27.432 + 32.612 = 1815.81
|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
s = (42.61 + 9.14 + 36.9)/2 metres
area of ABD = Sqrt[s (s-a) (s-b) (s-c)] = 140.92 square metres.
Thus the area of your lot is
447.25 + 140.92 = 588.17 square metres.