Question
Question
Use the image to answer the question.
A cone shows a radius of 6 and hypotenuse or side as 11.
What is the surface area of the cone? Use 3.14 for pi and round to the nearest tenth, if necessary.
(1 point)
square units
Use the image to answer the question.
A cone shows a radius of 6 and hypotenuse or side as 11.
What is the surface area of the cone? Use 3.14 for pi and round to the nearest tenth, if necessary.
(1 point)
square units
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To calculate the surface area of a cone, we use the formula:
\[
\text{Surface Area} = \pi r^2 + \pi r l
\]
where:
- \( r \) is the radius of the base,
- \( l \) is the slant height (hypotenuse), and
- \( \pi \) is approximately 3.14.
Given:
- Radius, \( r = 6 \)
- Slant height, \( l = 11 \)
Now we can calculate each component:
1. Calculate the area of the base:
\[
\pi r^2 = 3.14 \times (6^2) = 3.14 \times 36 = 113.04
\]
2. Calculate the lateral surface area:
\[
\pi r l = 3.14 \times 6 \times 11 = 3.14 \times 66 = 207.24
\]
3. Add both areas together to get the total surface area:
\[
\text{Surface Area} = 113.04 + 207.24 = 320.28
\]
4. Rounding to the nearest tenth:
\[
\text{Surface Area} \approx 320.3 \text{ square units}
\]
Thus, the surface area of the cone is **320.3 square units**.
\[
\text{Surface Area} = \pi r^2 + \pi r l
\]
where:
- \( r \) is the radius of the base,
- \( l \) is the slant height (hypotenuse), and
- \( \pi \) is approximately 3.14.
Given:
- Radius, \( r = 6 \)
- Slant height, \( l = 11 \)
Now we can calculate each component:
1. Calculate the area of the base:
\[
\pi r^2 = 3.14 \times (6^2) = 3.14 \times 36 = 113.04
\]
2. Calculate the lateral surface area:
\[
\pi r l = 3.14 \times 6 \times 11 = 3.14 \times 66 = 207.24
\]
3. Add both areas together to get the total surface area:
\[
\text{Surface Area} = 113.04 + 207.24 = 320.28
\]
4. Rounding to the nearest tenth:
\[
\text{Surface Area} \approx 320.3 \text{ square units}
\]
Thus, the surface area of the cone is **320.3 square units**.
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