Edward,
We had a similar problem from Susan, but her pentagon had a side length of 16 feet where your has a side length of 6 feet. Look at the diagram I drew for her problem. In your situation the length of AP is ^{6}/_{2} = 3 feet. Thus
tan( 36^{o}) = ^{3}/_{h} so 0.7265 h = 3 and h = 4.13 feet.
Thus the area of the triangle ACB is 3 4.13 square feet and hence the area of the pentagon is
5 3 4.13 = 61.9 square feet.
Penny
