Answer:
The volume of the cone is 3π≈9.42.
The surface area of the cone is 9π≈28.27.
Step-by-step explanation:
If we make a section of the sphere and the cone, we have a equilateral triangle inscribed in a circle (see picture attached).
We only know the numerical value of the radius R, that is 2 m.
From the picture, we have
[tex]\bar{KM}=2\cdot R\cdot cos(30^{\circ})=2R\dfrac{\sqrt{3}}{2}=\sqrt{3}R= 2\sqrt{3}[/tex]
The radius of the base of the cone is
[tex]r=\bar{KO}=\bar {KM}/2=(2\sqrt{3})/2=\sqrt{3}[/tex]
The height of the cone can be calculated as:
[tex]h=\bar{LO}=\bar{KL}\cdot cos(30^{\circ})=(2\sqrt{3})*(\sqrt{3}/2})=3[/tex]
The volume of the cone can be calculated as:
[tex]V=\dfrac{1}{3}\pi r^2h=\dfrac{1}{3}\pi (\sqrt{3})^2*3=3\pi\approx9.42[/tex]
The surface area of the cone is:
[tex]S=S_{base}+S_{face}=\pi r^2+\pi r l=\pi r^2+\pi r (\bar{KL})\\\\S=\pi(\sqrt{3})^2+\pi \sqrt{3}*2\sqrt{3}=3\pi+2\pi*3=(3+6)\pi=9\pi\approx 28.27[/tex]
