is this algebra correct for geometry problem?

[dwill]

What is the volume of a cube with a surface area of 54T^2?
Surface area of cube = 6S^2
Vol of cube = S^3

54T^2 = 6S^2
9T^2 = S^2
9T = S

Volume = 9T^3

Is the algebra correct, I am not sure what to do with the exponents.

 

[galactus]

You are correct that the area of one side is \(\displaystyle 9t^{2}\)

The length of a side is then \(\displaystyle 3t\)

Therefore, the volume is \(\displaystyle (3t)(3t)(3t)=27t^{3}\)

 

[jonboy]

Hi dwill!

Since we know \(\displaystyle \L V\,=\,{s}^3\) we need to find \(\displaystyle \L s\)

We know:\(\displaystyle \L \;{6s}^2\,=\,{54t}^2\)

Square root both sides:\(\displaystyle \L \;\sqrt{6}s\,=\,3\sqrt{6}t\)

Divide by \(\displaystyle \sqrt{6}\):\(\displaystyle \L \;s\,=\,3t\)

.....So \(\displaystyle \L \;{V\,=\,{(3t)}^3\,\Rightarrow\,27{t}^3}\)

 

[dwill]

Hi dwill!

Since we know \(\displaystyle \L V\,=\,{s}^3\) we need to find \(\displaystyle \L s\)

We know:\(\displaystyle \L \;{6s}^2\,=\,{54t}^2\)

Square root both sides:\(\displaystyle \L \;\sqrt{6}s\,=\,3\sqrt{6}t\)

Divide by \(\displaystyle \sqrt{6}\):\(\displaystyle \L \;s\,=\,3t\)

.....So \(\displaystyle \L \;{V\,=\,{(3t)}^3\,\Rightarrow\,27{t}^3}\)

Thanks for the explanation.

 

[jonboy]

Hi dwill....

Thanks for the explanation.

You're welcome, though I like galactus' solution better because it's easier, so thank you galactus!