Answer :
Let's solve the problem step by step.
Given:
- The surface area of the cone is [tex]\(216 \pi\)[/tex] square units.
- The height of the cone, [tex]\(h\)[/tex], is [tex]\(\frac{5}{3}\)[/tex] times the radius, [tex]\(r\)[/tex].
The formula for the surface area of a cone is:
[tex]\[ \text{Surface area} = \pi r (r + l) \][/tex]
where [tex]\(l\)[/tex] is the slant height of the cone. The slant height [tex]\(l\)[/tex] can be calculated using the Pythagorean theorem:
[tex]\[ l = \sqrt{r^2 + h^2} \][/tex]
Since [tex]\(h = \frac{5}{3}r\)[/tex], substituting in we get:
[tex]\[ l = \sqrt{r^2 + \left(\frac{5}{3}r\right)^2} = \sqrt{r^2 + \frac{25}{9}r^2} = \sqrt{\left(1 + \frac{25}{9}\right)r^2} = \sqrt{\frac{34}{9}r^2} = \frac{\sqrt{34}}{3}r \][/tex]
Now, substituting [tex]\(l\)[/tex] back into the surface area equation:
[tex]\[ \pi r \left(r + \frac{\sqrt{34}}{3}r\right) = 216 \pi \][/tex]
Removing [tex]\(\pi\)[/tex] from both sides:
[tex]\[ r \left(r + \frac{\sqrt{34}}{3}r\right) = 216 \][/tex]
Simplifying further:
[tex]\[ r \left(r + \frac{\sqrt{34}}{3}r\right) = 216 \][/tex]
[tex]\[ r \left(1 + \frac{\sqrt{34}}{3}\right)r = 216 \][/tex]
[tex]\[ r^2 \left(1 + \frac{\sqrt{34}}{3}\right) = 216 \][/tex]
Let [tex]\(A = 1 + \frac{\sqrt{34}}{3}\)[/tex]:
[tex]\[ r^2 A = 216 \][/tex]
Solving for [tex]\(r^2\)[/tex]:
[tex]\[ r^2 = \frac{216}{A} \][/tex]
Now substituting the numerical values:
[tex]\[ A = 2.9436506316151005 \][/tex]
[tex]\[ r^2 = \frac{216}{2.9436506316151005} = 73.37827311439018 \][/tex]
Thus, the radius [tex]\(r\)[/tex] is:
[tex]\[ r = \sqrt{73.37827311439018} \approx 8.566111901813459 \, \text{feet} \][/tex]
Therefore, the radius of the cone is approximately [tex]\(9\)[/tex] feet to the nearest foot.
Given:
- The surface area of the cone is [tex]\(216 \pi\)[/tex] square units.
- The height of the cone, [tex]\(h\)[/tex], is [tex]\(\frac{5}{3}\)[/tex] times the radius, [tex]\(r\)[/tex].
The formula for the surface area of a cone is:
[tex]\[ \text{Surface area} = \pi r (r + l) \][/tex]
where [tex]\(l\)[/tex] is the slant height of the cone. The slant height [tex]\(l\)[/tex] can be calculated using the Pythagorean theorem:
[tex]\[ l = \sqrt{r^2 + h^2} \][/tex]
Since [tex]\(h = \frac{5}{3}r\)[/tex], substituting in we get:
[tex]\[ l = \sqrt{r^2 + \left(\frac{5}{3}r\right)^2} = \sqrt{r^2 + \frac{25}{9}r^2} = \sqrt{\left(1 + \frac{25}{9}\right)r^2} = \sqrt{\frac{34}{9}r^2} = \frac{\sqrt{34}}{3}r \][/tex]
Now, substituting [tex]\(l\)[/tex] back into the surface area equation:
[tex]\[ \pi r \left(r + \frac{\sqrt{34}}{3}r\right) = 216 \pi \][/tex]
Removing [tex]\(\pi\)[/tex] from both sides:
[tex]\[ r \left(r + \frac{\sqrt{34}}{3}r\right) = 216 \][/tex]
Simplifying further:
[tex]\[ r \left(r + \frac{\sqrt{34}}{3}r\right) = 216 \][/tex]
[tex]\[ r \left(1 + \frac{\sqrt{34}}{3}\right)r = 216 \][/tex]
[tex]\[ r^2 \left(1 + \frac{\sqrt{34}}{3}\right) = 216 \][/tex]
Let [tex]\(A = 1 + \frac{\sqrt{34}}{3}\)[/tex]:
[tex]\[ r^2 A = 216 \][/tex]
Solving for [tex]\(r^2\)[/tex]:
[tex]\[ r^2 = \frac{216}{A} \][/tex]
Now substituting the numerical values:
[tex]\[ A = 2.9436506316151005 \][/tex]
[tex]\[ r^2 = \frac{216}{2.9436506316151005} = 73.37827311439018 \][/tex]
Thus, the radius [tex]\(r\)[/tex] is:
[tex]\[ r = \sqrt{73.37827311439018} \approx 8.566111901813459 \, \text{feet} \][/tex]
Therefore, the radius of the cone is approximately [tex]\(9\)[/tex] feet to the nearest foot.
