Question 984218
You might want to compare their volumes.


a:
Area is 6.
{{{x^2*3=6}}}
{{{x=sqrt(2)}}}


b:
Area is 8.
{{{3y^2=8}}}
{{{y=sqrt(8/3)}}}
{{{y=(2/3)sqrt(6)}}}


Question now means, compare y^3 to x^3, as the ratio.
{{{((2/3)sqrt(6))^3/(sqrt(2))^3}}}


{{{((4/9)sqrt(2)^3*sqrt(3)^3)/(sqrt(2))^3}}}


{{{(4(sqrt(3))^3)/9}}}


{{{(4*3sqrt(3))/(3*3)}}}


{{{highlight(4sqrt(3)/3)}}}