Consider the attached figure. The height CD cuts the triangle exactly in half. This means that
[tex]\overline{AD}=\overline{BD}=\dfrac{1}{2}\overline{AB}=10[/tex]
Moreover, since CD is the height of the triangle, we know that ACD is a right triangle. We know the hypothenuse AC to be 20 feet because it is a side of the triangle, and we just found out that AD is 10. We can use the pythagorean theorem to deduce
[tex]\overline{CD}=\sqrt{\overline{AC}^2-\overline{AD}^2}=\sqrt{400-100}=\sqrt{300}[/tex]
So, the area is
[tex]A=\dfrac{bh}{2}=\dfrac{\overline{AB}\cdot\overline{CD}}{2} = \dfrac{20\cdot\sqrt{300}}{2}=10\sqrt{300}\approx 173[/tex]
