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May 12th, 2014, 04:45 AM   #1
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Finding the similar area of a square and circle

If a piece of 1m wire is cut into 2 pieces to make a make a square and a circle with the same area how would you find this out?
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May 12th, 2014, 05:40 AM   #2
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Area of a circle (lets all it Ac) is [pi]r^2
Circumference of a circle, c, is 2[pi]r
Therefore the area of a circle, with respect to circumference is:
Ac = 4[pi]^3 / c^2
We know that the area of a square (lets call it As) is x^2, where x is the length of a side
Ac = As as per the question, so:
4[pi]^3 / c^2 = x^2
we also know that c + 4x = 1 meter
Which can be arranged as
c = 1 - 4x
We now have 2 equations with 2 unknowns, just substitute the 'c'
4[pi]^3 / (1 - 4x)^2 = x^2
The numbers in this are horrible....
I'll leave it at that stage (tired!)
But that's how you go about solving these types of question.
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May 12th, 2014, 07:09 AM   #3
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Math Focus: Abstract algebra
For the circle with circumference $C$ and radius $r$: $C^2=4\pi^2r^2=4\pi A$ $\implies$ $A=\dfrac{C^2}{4\pi}$

For the square with perimeter $P$ and side $x$: $P^2=16x^2=16A$ $\implies$ $A=\dfrac{P^2}{16}$.

And $P+C=1$ $\implies$ $P=1-C$.

$\displaystyle \begin{align*}
\therefore\ \frac{C^2}{4\pi}\ &=\ \frac{(1-C)^2}{16}
\\\\ 4C^2\ &=\ \pi(1-2C+C^2)
\\\\ (4-\pi)C^2+2\pi C-\pi\ &=\ 0
\\\\ C\ &=\ \frac{-2\pi+\sqrt{4\pi^2-4(4-\pi)(-\pi)}}{2(4-\pi)}
\\\\ &=\ \frac{2\sqrt{\pi}-\pi}{4-\pi}
\\\\ &\approx\ 0.47

So the wire should be cut to pieces of lengths approximately $\mathrm{47\ cm}$ and $\mathrm{53\ cm}$, the former to make the circle and the latter to make the square.
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May 12th, 2014, 07:32 AM   #4
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Haha, got to laugh at my mistake there...
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