For this case we can find the volume and area of the cone:
[tex]V = \frac {\pi * r ^ 2 * h} {3}[/tex]
Where:
V: It's the volume
A: It's the radius of the base
h: It's the height
We have to:
[tex]r = \frac {8} {2} = 4 \ units\\h = 6 \ units[/tex]
Substituting:
[tex]V = \frac {\pi * 4 ^ 2 * 6} {3}\\V = \frac {96 * \pi} {3}\\V = 32 \pi \ units ^ 3[/tex]
On the other hand, the area of the cone is given by:
[tex]A = \pi * r * g + \pi * r ^ 2[/tex]
Where:
A: It's the radio
g: It is the generator of the cone.
[tex]g = \sqrt {h ^ 2 + r ^ 2} = \sqrt {6 ^ 2 + 4 ^ 2} = \sqrt {36 + 16} = \sqrt {52} = 7.2[/tex]
SW:
[tex]A = \pi * 4 * 7.2 + \pi * 4 ^ 2\\A = 28.8 \pi + 16 \pi\\A = 44.8 \pi \ units ^ 2[/tex]
Answer:
[tex]V = 32 \pi \ units ^ 3\\A = 44.8 \pi \ units ^ 2[/tex]
