Answer:
The surface area \( A \) of the rocket is the sum of the surface areas of the cone and the cylinder. Let's denote \( r \) as the radius of the base.
1. **Surface area of the cylinder \( A_{\text{cylinder}} \):**
\[ A_{\text{cylinder}} = 2\pi r^2 + 2\pi rh \]
where \( r \) is the radius, and \( h \) is the height of the cylinder.
Given \( r = \frac{8}{2} = 4 \) inches and \( h = 5 \) inches:
\[ A_{\text{cylinder}} = 2\pi(4)^2 + 2\pi(4)(5) \]
2. **Surface area of the cone \( A_{\text{cone}} \):**
\[ A_{\text{cone}} = \pi r^2 + \pi r\sqrt{r^2 + h^2} \]
where \( r \) is the radius, and \( h \) is the height of the cone.
Given \( r = 4 \) inches and \( h = 3 \) inches:
\[ A_{\text{cone}} = \pi(4)^2 + \pi(4)\sqrt{(4)^2 + (3)^2} \]
Now, add the two surface areas to get the total surface area \( A \).
\[ A = A_{\text{cylinder}} + A_{\text{cone}} \]
After evaluating this expression, we can compare the result with the given options to find the correct answer.
Let's calculate the surface area:
1. **Surface area of the cylinder:**
\[ A_{\text{cylinder}} = 2\pi(4)^2 + 2\pi(4)(5) = 32\pi + 40\pi = 72\pi \]
2. **Surface area of the cone:**
\[ A_{\text{cone}} = \pi(4)^2 + \pi(4)\sqrt{(4)^2 + (3)^2} = 16\pi + 4\pi\sqrt{16 + 9} = 16\pi + 4\pi\sqrt{25} = 16\pi + 20\pi = 36\pi \]
Now, add the two surface areas:
\[ A = A_{\text{cylinder}} + A_{\text{cone}} = 72\pi + 36\pi = 108\pi \]
The correct option closest to this result is:
**c) 168π square inches**
It seems there might be a discrepancy in the provided options, or there could be a mistake in the question. Please double-check the options or provide additional information if needed.