# A copper wire when bent in the form of a square encloses an area of 121cm^2. If the same wire is bent into the form of a circle, what is the area of the circle?

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Nov 30, 2015

$\frac{484}{\pi} c {m}^{2} \approx 154.061984913 c {m}^{2}$

#### Explanation:

First we need to find the length of the wire.
To do that, we should try to figure out the length of 1 side of the square

Area of a square $= {\left(s i \mathrm{de}\right)}^{2}$
that means that a $\left(s i \mathrm{de}\right) = \sqrt{121}$ if the Area of the square was 121.
therefore one side of the square would be $\sqrt{121} = 11$

The Perimeter of the square would be equal to the full length of the copper wire.
Perimeter of a square $= 4 \left(s i \mathrm{de}\right)$

since we just found that $\left(s i \mathrm{de}\right) = 11$
we can determine that the Perimeter would be $11 \cdot 4 = 44$

Therefore the length of the copper wire would be $44 c m$ long

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If we make a circle from this wire, the length of the wire would then be the Circumference of the Circle.
Circumference of a Circle $= 2 \pi \left(r a \mathrm{di} u s\right)$

from this we can get that $\left(r a \mathrm{di} u s\right) = \frac{44}{2 \pi} = \frac{22}{\pi}$

Then, the Area of the Circle would be
Area of a Circle $= \pi {\left(r a \mathrm{di} u s\right)}^{2}$

Since we found that $\left(r a \mathrm{di} u s\right) = \frac{22}{\pi}$
we can figure out that the Area of the Circle would be
$\pi {\left(\frac{22}{\pi}\right)}^{2} = \frac{484}{\pi} \approx 154.061984913$ $\leftarrow$ answer

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