Question 790139
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Let the radius of the inner circle be *[tex \Large x].  Then the radius of the outer circle is *[tex \Large x\ +\ 2]


There are 2 ways to describe the area of the outer circle.  One way is the standard formula for the area of a circle, namely *[tex \Large \pi r^2]


But since the radius of the outer circle is *[tex \Large x\ +\ 2], the area would be *[tex \Large A_o\ =\ \pi(x\ +\ 2)^2].


But the area of the outer circle can also be defined as the area of the inner circle plus the area of the shaded region.  The area of the inner circle is just *[tex \Large \pi x^2] and the area of the shaded region is *[tex \Large 176 cm^2].  Since these two ways of describing the same area must be equal:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pi(x\ +\ 2)^2\ =\ \pi x^2\ +\ 176]


Solve for *[tex \Large x]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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