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October 19th, 2014, 09:49 AM   #1
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Calculate the radius of the cylinder if its height is 2.80m?

The total SA of a closed cylindrical container is 20.0cm^2. Calculate the radius of the cylinder if its height is 2.80m

I know the SA formula is 2pir+2pirh=20cm
but from here?
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October 19th, 2014, 10:45 AM   #2
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Originally Posted by mathsheadache View Post
The total SA of a closed cylindrical container is 20.0cm^2. Calculate the radius of the cylinder if its height is 2.80m

I know the SA formula is 2pir+2pirh=20cm
but from here?
correction ...

$\displaystyle A = 2\pi r^2 + 2\pi r h $

sub in 2.8 for h and 20 for A ... solve for r
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October 19th, 2014, 10:52 AM   #3
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Originally Posted by skeeter View Post
correction ...

$\displaystyle A = 2\pi r^2 + 2\pi r h $

sub in 2.8 for h and 20 for A ... solve for r
I am not so great with rearranging for r. There is always r^2
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October 19th, 2014, 10:58 AM   #4
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Originally Posted by mathsheadache View Post
I am not so great with rearranging for r. There is always r^2
for $\displaystyle ar^2 + br + c = 0$

$\displaystyle r = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

look familiar?
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October 20th, 2014, 02:04 PM   #5
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$\displaystyle \textbf{Hello, mathsheadache (However, maths is not a headache.)}$

$\displaystyle \textbf{We know,}$

$\displaystyle \textbf{Total Surface Area of a cylinder}=2 \pi r(r+h)$

$\displaystyle \textbf{Here, TSA is given to be 20} \textbf{cm}^{2}$

$\displaystyle \textbf{Height = 2.80 m but we need to be consistent with units, so let's convert in cm.} $

$\displaystyle \textbf{So, h = 280 cm.}$

$\displaystyle \textbf{According to the question,}$

$\displaystyle 2 \pi r^{2} + 2 \pi r \times 280 = 20 \ \ \textbf{cm}^{2}$

$\displaystyle \textbf{Divide each term by}\ \ 2 \pi$

$\displaystyle r^{2} + 280r = 3.183$

$\displaystyle \textbf{Rearrange this into:}$

$\displaystyle r^{2} + 280r - 3.183 = 0$

$\displaystyle \textbf{A quadratic equation can be solved in three ways :}$

$\displaystyle \textbf{1. By middle-term-splitting method.} \\ \textbf{2. By completing the square.} \\ \textbf{3. By using the quadratic formula.}$

$\displaystyle \textbf{In this case, we use the Quadratic Formula :}$

$\displaystyle \textbf{A Quadratic Equation of the form } ax^{2}+bx+c=0 \ \ \textbf{can be solved by the following formula :}$

$\displaystyle x=-b \pm \frac{\sqrt{b^{2}-4ac}}{2a} \ \ \textbf{where x represents an unknown, and a, b, and c are constants with }a \neq 0$

$\displaystyle \textbf{If you solve that using the usual quadratic formula, the positive value for r is 0.01 cm.}
$

$\displaystyle \textbf{Is the height = 2.80 m, or is it 2.80 cm?}$

$\displaystyle \textbf{If it is the latter, then :}$

$\displaystyle r^{2} + 2.8r - 3.183 = 0$ $\displaystyle \textbf{would give:}$

$\displaystyle r = 0.868 \ \ \textbf{cm}$

Last edited by Prakhar; October 20th, 2014 at 02:22 PM.
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October 20th, 2014, 07:38 PM   #6
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Quote:
Originally Posted by prakhar View Post

$\displaystyle \textbf{A Quadratic Equation of the form } ax^{2}+bx+c=0 \ \ \textbf{can be solved by the following formula :}$

$\displaystyle x=-b \pm \frac{\sqrt{b^{2}-4ac}}{2a} \ \ \textbf{where x represents an unknown, and a, b, and c are constants with }a \neq 0$
That should be $\displaystyle \ \ x \ = \ \dfrac{-b \pm \sqrt{b^{2}-4ac \ }}{2a}. $
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