Question 150915
We know the in finding the {{{Volume}}} of the container = {{{pi(r^2)h}}}
The {{{height}}} is given which is {{{(1/3)(12cm)}}}, but we don't have the radius, {{{r}}}.
Since {{{Area}}} is given, we can get it there, with the formula,
{{{A=2pi(r^2)+2pi(r)h}}}---- area on the top and base + area on the sides
{{{100=2pi(r^2+(rh))}}}
{{{(100/2pi)=r^2+rh}}}
Substituting & re-arranging,
{{{r^2+12r-(50/pi)=0}}}
by Quadratic eqn, {{{x=r}}}, {{{a=1}}}, {{{b=12}}},{{{c= -50/pi}}}
{{{x=(-b+-sqrt(b^2-4ac))/2a}}}
{{{x=(-12+-sqrt(12^2-4*1*-50/pi))/(2*1)}}}
{{{x=(-12+-sqrt(144+200/pi))/2}}}
{{{x=(-12+-sqrt(207.66198))/2}}}
{{{x=(-12+-14.41)/2}}}
{{{x=(-12+14.41)/2}}}-----> {{{x=1.205}}}
{{{x=(-12-14.41)/2}}}-----> {{{x=-13.205}}}
USE {{{x=1.205=r}}}
So {{{V=pi(1.205^2)(12/3)}}}
{{{V=5.81(pi)}}}cc
Thank you,
Jojo