Solution
Step 1
Determine the number of shapes in figure 1
Figure 1 is made up of a cone and a hemisphere
Step 2
Write out the expression for the area of a cone and a hemisphere
![\begin{gathered} \text{The area of a cone = }\pi* r* l \\ \text{The area of a hemisphere = 2}*\pi* r^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/63kvmr773lz7ovp1eyaaiditu10njjnjkw.png)
Where
pi =3.14
r = 7 inches
l = 15 inches
Step 3
Substitute in the values and find the area of the shape
![\begin{gathered} \text{Total area of the shape is given as} \\ \text{area of cone + area }of\text{ hemisphere} \\ =\text{ 2 x 3.14 }*7^2+(3.14\text{ }*7*15) \\ =\text{ 307.72 + 329.7} \\ =637.42inch^2\text{ to 2 decimal places} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zpvqm1bw2psfyiky7mjv3qu7mh1p1uv151.png)
Area = 637.42 square inches
Step 4
Determine the number of shapes in figure 2
Figure 2 is made up of a cone and a cylinder
Step 5
Write an expression for the area of a cylinder
![\text{The area of a cylinder = 2}*\pi* r* h\text{ + }\pi* r^2](https://img.qammunity.org/2023/formulas/mathematics/college/nbj9485i02w3wv23361pdvi8krfykivpsd.png)
where h = 13yards
radius(r) = 11/2 = 5.5inches
l = ?
To find l, the slant height we use the Pythagoras theorem
so that
![\begin{gathered} l^2=11^2+9^2 \\ l^2=202 \\ l\text{ =}\sqrt[]{202} \\ l\text{ = 14.2126704 in} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jy0rgqa0w0ka0fgwa6rgclnw9ey1jypxcd.png)
Step 6
Substitute in the values and find the area of the shape
![\begin{gathered} \text{Area of figure 2 is given as} \\ \pi* r* l\text{ + (2}*\pi* r* h\text{ + }\pi* r^2) \\ (3.14\text{ }*\text{ 5.5}*14.2126704)+((2*3.14*5.5*13)+\text{ (3.14 }*5.5^2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2n4d5er7j0jup1sipb0cvzaebjnah0kq6x.png)
![\begin{gathered} 245.4528179+94.985+449.02=789.4578179Inches^2 \\ To\text{ 2 decimal places }\approx789.46inches^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/t98hwltcqrglgc0q8wymzpqj1mtuxckp7y.png)
Area = 789.46 square inches