To find the surface area of a right cone, we need to consider both the base area and the lateral (side) surface area. We'll proceed step by step to calculate both areas.
Step 1: Calculate the base area
The base of the cone is a circle with radius \( r = 2 \). The area \( A_{\text{base}} \) of a circle is given by the formula:
[tex]\[ A_{\text{base}} = \pi r^2 \][/tex]
Substituting the given radius:
[tex]\[ A_{\text{base}} = \pi (2)^2 \][/tex]
[tex]\[ A_{\text{base}} = \pi \cdot 4 \][/tex]
[tex]\[ A_{\text{base}} \approx 12.566370614359172 \][/tex]
Step 2: Calculate the lateral surface area
The lateral surface area \( A_{\text{lateral}} \) of a cone can be found using the formula:
[tex]\[ A_{\text{lateral}} = \pi r l \][/tex]
where \( l \) is the slant height. Given that the slant height \( l = 6 \) and \( r = 2 \), we substitute these values into the formula:
[tex]\[ A_{\text{lateral}} = \pi \cdot 2 \cdot 6 \][/tex]
[tex]\[ A_{\text{lateral}} = \pi \cdot 12 \][/tex]
[tex]\[ A_{\text{lateral}} \approx 37.69911184307752 \][/tex]
Step 3: Calculate the total surface area
The total surface area \( A_{\text{total}} \) of the cone is the sum of the base area and the lateral surface area:
[tex]\[ A_{\text{total}} = A_{\text{base}} + A_{\text{lateral}} \][/tex]
Using the areas we calculated:
[tex]\[ A_{\text{total}} \approx 12.566370614359172 + 37.69911184307752 \][/tex]
[tex]\[ A_{\text{total}} \approx 50.26548245743669 \][/tex]
Step 4: Compare the calculated total surface area with the provided answer options
Given the following answer options:
A. \( 48 \pi \) units\(^2\)
B. \( 24 \pi \) units\(^2\)
C. \( 12 \pi \) units\(^2\)
D. \( 16 \pi \) units\(^2\)
We can see that none of these options exactly match the calculated total surface area (approximately 50.26548245743669). However, when expressed in terms of \(\pi\), our calculated surface area is close to one of the numerical results:
[tex]\[ A_{\text{total}} \approx 50.26548245743669 \approx 16 \pi \text{ units}^2 \][/tex]
Hence, the correct answer choice is:
D. [tex]\( 16 \pi \)[/tex] units[tex]\(^2\)[/tex]
