Part (a);
[tex]\begin{gathered} \text{The area of the semi-circle with a radius of 5 yds;} \\ A=\pir^2^{} \\ \text{Foe a semi-circle (}\frac{1}{2}\text{ circle)} \\ \text{Area}=\frac{\pir^2}{2} \\ \text{Area}=\frac{3.14\times5^2}{2} \\ \text{Area}=\frac{78.5}{2} \\ \text{Area}=39.25yd^2 \\ \text{Area of the triangle=}\frac{1}{2}bh \\ b=\text{base (}15) \\ h=vertical\text{ height (10) that is radius x 2} \\ \text{Area}=\frac{1}{2}\times15\times10 \\ \text{Area}=75 \\ \text{Area of the entire figure=39.25 + 75} \\ \text{Area of the entire figure=114.25 yds}^2 \end{gathered}[/tex]Part (b);
[tex]\begin{gathered} \text{Area of the circle=}\pir^2 \\ \text{radius}=5\text{yds} \\ \text{Area}=3.14\times5^2 \\ \text{Area}=3.14\times25 \\ \text{Area}=78.5yd^2 \\ \text{Area of the triangle=}\frac{1}{2}\times bh \\ b=\text{base (16)} \\ h=\text{vertical height (16)} \\ \text{Area}=\frac{1}{2}\times16\times16 \\ \text{Area}=128yd^2 \\ Area\text{ of shaded region=Area of triangle-Area of circle} \\ \text{Area of shaded region=128-78.5} \\ \text{Area of shaded region=}49.5yd^2 \end{gathered}[/tex]