Farmer's Triangular Plot and Pond

Let the isosceles triangle $ABC$ have base $BC = 300$ feet and altitude $AD = 200$ feet from $A$ to $BC$. Thus $D$ is the midpoint of $BC$, so $BD = DC = 150$ feet. Side $AC = \sqrt{200^2 + 150^2} = 250$ feet. The center of the circle lies on $BC$ at $D$, so radius $r$ is perpendicular to $AC$ at point $E$. In right triangle $ADC$ and triangle $DEC$, $\angle C$ is common and both are right triangles, so $\Delta ADC \sim \Delta DEC$. Therefore, $\frac{DE}{AD} = \frac{DC}{AC} \implies \frac{r}{200} = \frac{150}{250} \implies r = 120$ feet. Area of the semi-circle outside the plot $= \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 120^2 = 22,628.57$ sq ft. Cost $= 22,628.57 \times 1400 \approx \text{Rs. } 3,16,80,000$.