 # 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. pominjaneh6 2022-08-08 Answered
A circular cylinder with a volume of 8$\pi$ ${m}^{3}$ 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.
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it wietselau
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 $\frac{2}{3}$ +$\frac{1}{3}$ so that the $\frac{2}{3}$ part of one bisector will be equal with r from this circle
so we know that one side equal
$2m\mathrm{cos}30=\frac{h}{2}\sqrt{\frac{3}{2}}=\frac{h}{2}h=\sqrt{3}$ from this result that r= $2\sqrt{\frac{3}{3}}$
so the height of cylinder being H and we know that the volum of cylinder is 8$\pi$ 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 $\pi$r2 = $\pi$($\frac{4}{3}$)
V=$8\pi$
V=$\pi$($\frac{4}{3}$)H
8$\pi$=$\pi$($\frac{4}{3}$)H
8=($\frac{4}{3}$)H
H= 8($\frac{3}{4}$)
H= $\frac{24}{4}$
H=6 m