The area of the hemisphere is
[tex]\begin{gathered} A=\frac{1}{2}(4\times\pi\times r^2)+\pi\times r^2 \\ A=2\times\pi\times r^2+\pi\times r^2 \\ A=3\text{ }\pi r^2 \end{gathered}[/tex]
Since r = 5 in, then
[tex]\begin{gathered} A=3\times\pi\times(5)^2 \\ A=75\pi \end{gathered}[/tex]
Now we need to subtract from it the 6 faces of the removable cube
The area of each face is
[tex]a=s^2[/tex]
Where s is the edge of the cube
Since s = 3 in
Then the area of the 6 faces is
[tex]\begin{gathered} T\mathrm{}a=6(3)^2 \\ T\mathrm{}a=6\times9 \\ T\mathrm{}a=54 \end{gathered}[/tex]
Now we will subtract it from the area of the hemisphere
[tex]\begin{gathered} A=75\pi-54 \\ A=181.619449 \end{gathered}[/tex]
Round it to the nearest tenth
A = 181.6 square inches
