SOLUTION: If someone can PLEASE help me with the poblem that would be fantastic! i cannot understand for the life of me this question. thank you. the surface areas of two simular cylinders

Algebra ->  Surface-area -> SOLUTION: If someone can PLEASE help me with the poblem that would be fantastic! i cannot understand for the life of me this question. thank you. the surface areas of two simular cylinders       Log On


   

Question 565452: If someone can PLEASE help me with the poblem that would be fantastic! i cannot understand for the life of me this question. thank you.
the surface areas of two simular cylinders are 8pi in^2 and 18piin^2. Find the following ratios.
Area1/Area2= a^2/b^2=
r1/r2=h1/h2=a/b=
Volume1/Volume2=a^3/b^3=
again thank you for helping!!!!
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
A figure (or solid) similar to another is a scaled-up or scaled-down (or maybe even identical) version of the other one. All dimensions (length, width, radius, height, etc) are changed by the same factor. The area is changed by that factor squared, and the volume is changed by that factor cubed.
That's written in the equalities you show.
So if the ratio (scale factor) for height and radius of those cylinders is the fraction/ratio a%2Fb
r%5B1%5D%2Fr%5B2%5D=h%5B1%5D%2Fh%5B2%5D=a%2Fb,
the surface areas are in the ratio Area%5B1%5D%2FArea%5B2%5D=%28a%2Fb%29%5E2=a%5E2%2Fb%5E2
You know that
Area%5B1%5D%2FArea%5B2%5D=%28a%2Fb%29%5E2=a%5E2%2Fb%5E2=8%2Api%2F%2818%2Api%29=4%2F9
%28a%2Fb%29%5E2=a%5E2%2Fb%5E2=4%2F9 --> a%2Fb=sqrt%284%2F9%29=sqrt%284%29%2Fsqrt%289%29=2%2F3 -->a%5E3%2Fb%5E3=%28a%2Fb%29%5E3=2%5E3%2F3%5E3=8%2F27
So the surface area of the samaller cylinder is 4/9 of the surface area of the larger cylinder. The radius and height are 2/3 of those for the larger cylinder, and the volume is 8/27 of the volume of the larger cylinder.