The volume of cone whose area of base is 8π[tex]mm^{2}[/tex] is 32π/3 [tex]mm^{3}[/tex].
Given area of base of the cone 8π[tex]mm^{2}[/tex].
We are required to find the volume of the cone.
We know that base of a cone used to be in circle so the area of base of cone is equal to π[tex]r^{2}[/tex].
Area=8π (given)
π[tex]r^{2}[/tex]=8π
[tex]r^{2}[/tex]=8
r=[tex]\sqrt{8}[/tex]
r=2[tex]\sqrt{2}[/tex]
Volume of cone=1/3*π[tex]r^{2} h[/tex]
=1/3*π[tex](2\sqrt{2} )^{2}[/tex]*4
=1/3*8*4π
=32π/3
(We are not required to put value of π so our answer will be 32π/3.)
Hence the volume of cone whose area of base is 8π is 32π/3.
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Question is incomplete as it should also include height of cone be 4 mm.
