Question

A circular cylinder with a volume of 8pi m^3 is circumscribed about a right prism whose base is an equilateral triangle with one side that measures 2m. What is the altitude of the cylinder?

Can anyone solve this? Its a problem on solid mensuration

Answer 1

so you need to know that in one equilateral triangle one degree mesure is 60 and the bisectores are heights and again you need to know that those heights where will intersected so this point will be the central point of this circule what is the base of this cylinder

we know again that those heights (bisectores) part in report 2/3 +1/3 so that the 2/3 part of one bisector will be equal with r from this circle

so we know that one side equal 2m
cos30 = h/2
sqrt3/2 = h/2
h=sqrt3

from this result that r= 2sqrt3/3

so the height of cylinder being H and we know that the volum of cylinder is 8pi m3
and we know that the volum of one cylinder is equal with area of base multiply the height of cylinder

area of base is pir2 = pi(4/3)

V=8pi
V=pi(4/3)H
8pi=pi(4/3)H
8=(4/3)H
H= 8(3/4)
H= 24/4
H=6 m