x and y are similar shapes. the total surface area of x is 900cm squared. the total surface area of y is 1600cm squared. the volume of x is

Yes, you are correct.  I answered a similar question a couple of days ago

 http://web2.0calc.com/questions/the-volumes-of-two-mathematically-similar-solids-are-in-the-ratio-27-125-the-surface-area-of-the-smaller-solid-is-36cm-2-work-out-the-surfa#r102051

x and y are similar shapes. the total surface area of x is 900cm squared. the total surface area of y is 1600cm squared. the volume of x is 540cm squared. calculate the volume of y

---------------------------------------------------------------------------------------------------------------------------

(I'm going to assume that you meant the volume of x to be 540cm.³ )

Let's assume that we could take all the "surfaces" of x and lay them flat on a table.

Note that, if x's surface area is 900cm², and it volume is 540cm³, the other dimension must be .6cm.

And the "scaling" factor  between the larger object y and the smaller similar object x is given by SQRT(y's surface area/ x's surface area) = SQRT(1600/900) = 40/30 = 4/3

To see that this is true, let's suppose that the surface area of x is just a square with a side = 30cm - it may not be a square, but let's suppose that it is !!  Again, let's suppose that the surface area of y is a square, too, with a side of 40cm. Then, the scaling factor is 4/3....for each dimension of x, the same dimension in y is 4/3 as long......since they're "similar" objects.

Well, if that's true, the other dimension of our hypothetical y must be = (4/3) * .6cm = .8cm.

So , if we could take y's "surfaces" and lay them flat, too, it's total volume would be !600cm * .8cm. = 1280cm³

Finally, no matter what the dimensions of x, each dimension of y is (4/3) as much. So, if the width (w) * length (l) * height (h) of x = volume = 540cm³, then the dimensions of y = (w*4/3) (l*4/3) (h*4/3) = (w* l * h)  * (4/3)^{3 ₌}

(volume of x) * (scaling factor)³ = (540cm³) * (4/3)³ = 1280cm³.......which is just what we thought!!!

Sorry 

I have to rethink

If you keep reading, I'm not claiming I know WHAT it is......I was just giving some examples....whatver it is, since we don't have (pi) involved, it's probably not a curved surface......

Yes I am sorry Chris.  I did try to delete almost immediately but the system wouldn't let me.

Thanks Chris,

I was struggling with this question.  

I guessed that  

$$V=540\times\left(\sqrt{\frac{1600}{900}}\right)^3$$

but you helped me understand why this is correct.

If anyone else would like to put another take on it i would be very interested. 

To be honest, I kinda' struggled with it, too. I don't "do" 3-D as well as "2-D," so if I can lay something out in terms of area, it usually helps me see its volume....basically, my answer was just me "thinking aloud!!!" I believe it's correct, but  Alan, Rom or Bertie  - or someone else - might verify it (or not)!!!

Thanks Alan,

Thanks for the address too - it is a really good idea to reference posts like this.  

I used the link and examined your previous post and I would have arrived at the same answer but I have always maintained that there are differerent levels of understanding.  

I think Chris and I both  accept this answer but I think I only understand it at a superficial level.

Is there anything you can say that will help?  Maybe I just need to think about it and work through a few different shapes to cement an understanding.  Yes, that should work.

Don't worry, I am just thinking with my fingers.