Answer
$15\sqrt{3}\text{ }mi \text{ OR approximately }25.98\text{ }mi
$
Work Step by Step
Using the given lengths of the triangle, then the semi-perimeter, $s,$ given by $
s=\dfrac{1}{2}(a+b+c)
,$ is equal to
\begin{array}{l}\require{cancel}
s=\dfrac{1}{2}(6+10+14)
\\
s=\dfrac{1}{2}(30)
\\
s=15
.\end{array}
Using $
A=\sqrt{s(s-a)(s-b)(s-c)}
$ or the Heron's Formula, then the area of the triangle, $A,$ is
\begin{array}{l}\require{cancel}
A=\sqrt{15(15-6)(15-10)(15-14)}
\\
A=\sqrt{15(9)(5)(1)}
\\
A=\sqrt{675}
\\
A=\sqrt{225\cdot3}
\\
A=\sqrt{(15)^2\cdot3}
\\
A=15\sqrt{3}
\\\text{OR}\\
A\approx25.98
.\end{array}
Hence, the area, $A,$ is $
15\sqrt{3}\text{ }mi \text{ OR approximately }25.98\text{ }mi
.$
