We can divide the figure in two regular polygons:
then, the area and perimeter of the rectangle are the following:
[tex]\begin{gathered} A_r=22(10)=220 \\ P_r=2(22+10)=2(32)=64 \end{gathered}[/tex]
then, the area and perimeter of the semicircle are the following:
[tex]\begin{gathered} A_c=\frac{(3.1416)(5)^2}{2}=\frac{78.54}{2}=39.27 \\ P_c=\frac{(3.1416)(10)}{2}=15.71 \end{gathered}[/tex]
then, if we add the areas and the perimeters, we get:
[tex]\begin{gathered} A_t=A_r+A_c=220+39.27=259.27 \\ P_t=P_r+P_c=64+15.71=79.71 \end{gathered}[/tex]
therefore, the area is 259.27 squared units and the perimeter is 79.71 units
