Consider the attached figure. The height CD cuts the triangle exactly in half. This means that
![\overline{AD}=\overline{BD}=(1)/(2)\overline{AB}=10](https://img.qammunity.org/2020/formulas/mathematics/college/xpxf0203cfbkk2eyxtug14tcxzhzsvqywd.png)
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
![\overline{CD}=\sqrt{\overline{AC}^2-\overline{AD}^2}=√(400-100)=√(300)](https://img.qammunity.org/2020/formulas/mathematics/college/6jwibi3znzxlp6yiwvetq92by4ms62h5qq.png)
So, the area is
![A=(bh)/(2)=\frac{\overline{AB}\cdot\overline{CD}}{2} = (20\cdot√(300))/(2)=10√(300)\approx 173](https://img.qammunity.org/2020/formulas/mathematics/college/a8idrprah0czp4778nek4wolocu21chnxb.png)