Find the volume of a right circular cone that has a height of 12.1 m and a base with a circumference of 17.7 m. Round your answer to the nearest tenth of a cubic meter.

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The volume of a right circular cone is

[tex] V=\dfrac{1}{3}\pi r^2\cdot h [/tex].

If a circumference of a base circle is 17.7 m, then using formula [tex] l=2\pi r [/tex] for circumference of a circle, you can find the radius:

[tex] 17.7=2\pi r,\\ \\ r=\dfrac{17.7}{2\pi} =\dfrac{8.85}{\pi} [/tex].

Then the volume is:

[tex] V=\dfrac{1}{3}\pi r^2\cdot h =\dfrac{1}{3}\pi \left(\dfrac{8.85}{\pi}\right)^2\cdot 12.1=\dfrac{315.90075}{\pi}[/tex].

Answer:

[tex]V_{cone}=100.6 m^{3}[/tex]

Step-by-step explanation:

The volume of a cone is defined as:

[tex]V_{cone}=\frac{1}{3}A_{base}h[/tex]

So, a cone has a circular base, and its area is:

[tex]A_{base}=\pi r^{2}[/tex]

Replacing this relation in the volume equation:

[tex]V_{cone}=\frac{1}{3}\pi r^{2}h[/tex]

Now, the problem gives us the height but no the radius, instead it gives the circumference length, which we can use to find the radius:

[tex]L=2\pi r\\r=\frac{L}{2 \pi}\\ r=\frac{17.7m}{2 \pi}\\r=\frac{8.85}{\pi}m[/tex]

Then, we use this radius to find the volume:

[tex]V_{cone}=\frac{1}{3}\pi (\frac{8.85}{\pi}m)^{2}12.1m[/tex]

[tex]V_{cone}=\frac{315.9}{\pi} m^{3}[/tex]

Using [tex]\pi \approx 3.14[/tex]:

[tex]V_{cone}=\frac{315.9}{3.14} m^{3}=100.6 m^{3}[/tex]

Therefore, the volume is [tex]V_{cone}=100.6 m^{3}[/tex]