Question
A firework rocket consists of a cone stacked on top of a cylinder, where the radii of the cone and the cylinder are equal. The diameter of the cylindrical base of the rocket is 8 in and the height of the cylinder is 5 in, while the height of the cone is 3 in. Calculate the surface area of the rocket. Leave your answer in terms of π.
184π sq. in.
76π sq. in.
168π sq. in.
88π sq. in.
184π sq. in.
76π sq. in.
168π sq. in.
88π sq. in.
Answers
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Answered by
GPT-4o mini
To find the surface area of the firework rocket, which consists of a cylinder and a cone, we will calculate the surface area of each component separately and then combine them.
### 1. Surface Area of the Cylinder
The formula for the lateral surface area of a cylinder is:
\[
A_{\text{cylinder}} = 2\pi r h
\]
Where:
- \( r \) is the radius of the cylinder.
- \( h \) is the height of the cylinder.
Given:
- The diameter of the cylindrical base is 8 inches, so the radius \( r \) is:
\[
r = \frac{8}{2} = 4 \text{ in}
\]
- The height of the cylinder \( h \) is 5 in.
Plugging in the values:
\[
A_{\text{cylinder}} = 2\pi(4)(5) = 40\pi \text{ sq. in.}
\]
Since the cylinder has a top and doesn't have a bottom (as it is connected to the cone), we also need to add the area of the top (the circle) of the cylinder:
\[
A_{\text{top}} = \pi r^2 = \pi(4^2) = 16\pi \text{ sq. in.}
\]
So the total surface area of the cylinder including the top is:
\[
A_{\text{cylinder, total}} = 40\pi + 16\pi = 56\pi \text{ sq. in.}
\]
### 2. Surface Area of the Cone
For the surface area of the cone, we will use the formula for lateral surface area:
\[
A_{\text{cone}} = \pi r l
\]
Where:
- \( l \) is the slant height of the cone, which can be calculated using the Pythagorean theorem:
\[
l = \sqrt{r^2 + h^2}
\]
Given:
- The radius \( r \) of the cone is also 4 in (the same as the cylinder).
- The height \( h \) of the cone is 3 in.
Calculating the slant height:
\[
l = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ in}
\]
Now we calculate the lateral surface area of the cone:
\[
A_{\text{cone}} = \pi(4)(5) = 20\pi \text{ sq. in.}
\]
### 3. Adding the Surface Areas
Now, we combine the surface area of the cylinder (without the bottom) and the lateral surface area of the cone:
\[
\text{Total Surface Area} = A_{\text{cylinder}} + A_{\text{top}} + A_{\text{cone}}
\]
\[
\text{Total Surface Area} = 56\pi + 20\pi = 76\pi \text{ sq. in.}
\]
The total surface area of the firework rocket is:
\[
\boxed{76\pi} \text{ sq. in.}
\]
### 1. Surface Area of the Cylinder
The formula for the lateral surface area of a cylinder is:
\[
A_{\text{cylinder}} = 2\pi r h
\]
Where:
- \( r \) is the radius of the cylinder.
- \( h \) is the height of the cylinder.
Given:
- The diameter of the cylindrical base is 8 inches, so the radius \( r \) is:
\[
r = \frac{8}{2} = 4 \text{ in}
\]
- The height of the cylinder \( h \) is 5 in.
Plugging in the values:
\[
A_{\text{cylinder}} = 2\pi(4)(5) = 40\pi \text{ sq. in.}
\]
Since the cylinder has a top and doesn't have a bottom (as it is connected to the cone), we also need to add the area of the top (the circle) of the cylinder:
\[
A_{\text{top}} = \pi r^2 = \pi(4^2) = 16\pi \text{ sq. in.}
\]
So the total surface area of the cylinder including the top is:
\[
A_{\text{cylinder, total}} = 40\pi + 16\pi = 56\pi \text{ sq. in.}
\]
### 2. Surface Area of the Cone
For the surface area of the cone, we will use the formula for lateral surface area:
\[
A_{\text{cone}} = \pi r l
\]
Where:
- \( l \) is the slant height of the cone, which can be calculated using the Pythagorean theorem:
\[
l = \sqrt{r^2 + h^2}
\]
Given:
- The radius \( r \) of the cone is also 4 in (the same as the cylinder).
- The height \( h \) of the cone is 3 in.
Calculating the slant height:
\[
l = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ in}
\]
Now we calculate the lateral surface area of the cone:
\[
A_{\text{cone}} = \pi(4)(5) = 20\pi \text{ sq. in.}
\]
### 3. Adding the Surface Areas
Now, we combine the surface area of the cylinder (without the bottom) and the lateral surface area of the cone:
\[
\text{Total Surface Area} = A_{\text{cylinder}} + A_{\text{top}} + A_{\text{cone}}
\]
\[
\text{Total Surface Area} = 56\pi + 20\pi = 76\pi \text{ sq. in.}
\]
The total surface area of the firework rocket is:
\[
\boxed{76\pi} \text{ sq. in.}
\]
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