Area of the circle =\(\frac{x^2}{4\pi } \; square \; inches\)
Area of square = \(\frac{x^2}{16}\) square inches
Given :
A piece of wire that is x inches long . It is bent into shape of a circle
Perimeter of circle = x inches
we know that , perimeter of circle =\(2\pi r\)
\(2\pi r=x\\solve \; for \; r\\divide \; both \; sides \; by \; 2\pi \\r=\frac{x}{2\pi } \\r=\frac{x}{2\pi }\)
Now we find area of the circle using the radius
\(Area =\pi r^2\\Area=\pi (\frac{x}{2\pi } )^2\\Area=\pi (\frac{x^2}{4\pi ^2} )\\\\Area=\frac{x^2}{4\pi }\)
Area of the circle =\(\frac{x^2}{4\pi } \; square \; inches\)
Perimeter of square = 4(side)= 4s
4s=x
Divide both sides by 4
\(s=\frac{x}{4}\)
Area of square =\(s^2=\frac{x}{4} ^2=\frac{x^2}{16}\)
Area of square = \(\frac{x^2}{16}\)
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