Hi Jack. There's no simple formula based only on the outside lengths.
That's because the outside lengths alone don't tell enough about the shape. See this earlier question for more information. From the diagram you sent us however it may be possible to calculate the area. Are the angles at A and B right angles? I am going to assume they are and show you how to find the area in this situation.
First join the points A and C and C and E to subdivide your property into three triangles.
Since the angle at B is a right angle the triangle ABC is a right triangle and hence Pythagoras theorem
AC^{2} = AB^{2} + BC^{2}
gives AC = 149.56 feet.
The tangent of the angle CAB is ^{BC}/_{AB} and hence the measure of the angle CAB is tan^{1}( ^{BC}/_{AB}) = 23.16 degrees. Since the angle EAB is a right angle
angle EAC = 90  23.16 = 66.84 degrees.
Now you can use the law of cosines with the triangle ACE. This is
CE^{2} = AE^{2} + AC^{2}  2 AE AC cos(EAC)
This gives CE = 148.16 feet.
The area of your property is the sum of the areas of the three triangles. Triangle ABC is a right triangle so AB is the base then BC = 58.833 feet is the height and you can find its area using
triangle area = ^{1}/_{2} base height.
For the other two triangles you can find the areas using Heron's formula.
Stephen and Penny
