Part A.
In this case, we need to find the perimeter of the figure, which is equal to the lenght of the three side and the semicircle.
Since the arc of the semicircle has a lenght of one half the circunference,
![\text{ arc lenght=}\frac{2\pi r\text{ }}{2}=\pi r](https://img.qammunity.org/2023/formulas/mathematics/college/7yljpqogha0vygyewa0x7fvjtyp83qzhsn.png)
where, in our case, the radius is
![r=(11)/(2)=5.5](https://img.qammunity.org/2023/formulas/mathematics/college/sq46o295jt0tm3qt14ig2mm854zbj4orr2.png)
then, the arc lenght of the semicircle is3
![\text{arc lenght=3.1416}*5.5=17.28\text{ ft}](https://img.qammunity.org/2023/formulas/mathematics/college/spsbrvuz46aqizoxj1z0q2mtv2l5t255sb.png)
Therefore, the Perimeter of the figure is
![\begin{gathered} \text{Perimeter}=14+11+14+17.28 \\ \text{Perimeter}=56.28\text{ ft} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/u8a6zqgb7f9dps7m2cuqlfe7iqf6sra879.png)
Therefore, the answer for part A is 56.28 feet.
Part B.
In this case, we need the area of the semicircle and the area of the rectangle below it, that is,
The area of the semicircle is half the area of the circle, that is,
![\begin{gathered} \text{area semicircle=}(\pi r^2)/(2)=((3.1416)(5.5^2))/(2) \\ \text{area semicircle=}47.5167ft^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/op3o5h820mdxjb6qos95n6b4io5zthc1x1.png)
and the area of the rectangle is
![\text{area rectangle=11}*14=154ft^2](https://img.qammunity.org/2023/formulas/mathematics/college/bbdmbej2th4vrci52jlah3220q0y975yxv.png)
Therefore, the total area is
![47.5167+154=204.5167ft^2](https://img.qammunity.org/2023/formulas/mathematics/college/dw2tnrotnnl2s1d2339ynlrcu2vegrze4l.png)
Therefore, the answer for part B is: 204.52 square feet