Anyone can answer this? Sir cphil?

Hi CierDreaj

You were on for hours after I posted this. I am wondering why you did not comment? 

Was it enough? Did you need more help? Or what?

We like feedback around here you know, interactive teaching and learning is the best kind there is!  :))

Let X be the the intersection point between S and B

I assume you need surface area and volume

To get both you will need length SX

and to get surface area you will also need RS

 

 

\(tan30 = \frac{SX}{RX}\\\\ \frac{1}{\sqrt{3}}= \frac{SX}{12}\\\\ \frac{12}{\sqrt{3}}= SX\\\\ SX=4\sqrt{3}\;\; dm\\\\\)

 

\(cos30=\frac{RX}{RS}\\\\ \frac{\sqrt{3}}{2}=\frac{12}{RS}\\\\ \frac{2}{\sqrt{3}}=\frac{RS}{12}\\\\ \frac{24}{\sqrt{3}}=RS\\\\ RS=8\sqrt{3}\;\;dm\)

 

 

Can you take it from there?

If not just ask for more help :)

 

Oh you need to check my working too.  I think it is right but I don't put too much effort into proof reading because I think that is mainly your job.  If something doesn't make sense or you need more explanation, just let me know.  :)