The dimensions of a triangular lot are 184 feet by 161 feet

The dimensions of a triangular lot are 184 feet by 161 feet by 173 feet. If the price of such land is $3 per square foot, how much does the lot cost?

Solution

Let the lengths of the three sides of the triangle be denoted a, b, and c (in feet for this problem). Define s as s = (a + b + c)/2 (the \"semi-perimeter\"). Then the area A of the triangle (in square feet for this problem) is given by

A = (s(s - a)(s - b)(s - c))

and the value (in dollars) of the lot is 3A = 3 (s(s - a)(s - b)(s - c)).

S = 259 feet

A = (s(s - a)(s - b)(s - c)) = (259(259 - 184)(259 - 161)(259 - 173)) = 12795.073 square feet

And Value = 3A = $38385.22