If you square rooted the bit inside the square brackets then you would get 2 lengths, which is perfect.

S.A. = π*r*√[r² + (2025/π)r²]

That can be simplified more by combining the two terms inside the bracket to give [(2025/π+1)r²]

The r² can then be taken out of the square root:

S.A. = π*r²√(2025/π+1)

]]>as the formula is S.A= Pi*R*S ...........and s= square of [R2+H2]......so i substituted the height in this formula. Which came to the above term. can it be simplified more???

]]>Every term in a surface area formula has to include exactly 2 lengths.

πr only has 1 length and (r² + 2025/πr²)² is either a contradiction in itself, or it contains 4 lengths, which would be impossible anyway. Are you sure the formula is right?

]]>Does this relate to your question I got part way through here: http://www.mathsisfun.com/forum/viewtopic.php?id=2002 ?

I am sorry, I just haven't had time to finish it.

]]>S.A.= Pi*R* Square of [R2 + 2025/Pi*R2]

---The term in bracket is Square-----

So is there anyway to simplify this more??? Please i really need quick help.

]]>V = 1/3πr²h - 1/27πr²h

These two terms are now identical apart from the constant at the front, so they can be combined:

V = 8/27πr²h

To get r² on its own, divide the right-hand side by everything except it:

27V/(8πh) = r²

To get r, just square root both sides:

√[27V/(8πh)] = r

]]>V=1/3*Pi*r2*h - 1/3*Pi*r2/3*h/3

Can anyone re-arrange to get r or r square please???

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