anonymous
  • anonymous
Let $ABCD$ be a square of side length 4. Let $M$ be on side $\overline{BC}$ such that $CM = 1$, and let $N$ be on side $\overline{AD}$ such that $DN = 1$. We draw the quarter-circle centered at $A$. Find x-y.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
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anonymous
  • anonymous
So i know how to find the cuarter circle, and i know the samll rectangle is 4, but i cant figure out the white part of the quarter circle we remove...
anonymous
  • anonymous
are you allowed to use calculus?

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ganeshie8
  • ganeshie8
is the answer \(4\pi-4\) ?
anonymous
  • anonymous
no i cant use calculus @jayzdd and im supposed to write a proof so i dont have the answer @ganeshie8
ganeshie8
  • ganeshie8
let \(a\) be the area of unshaded region in quarter circle. then we have \(x+a = \frac{\pi 4^2}{4}\) \(y+a = 4\times 1\) subtract
anonymous
  • anonymous
just for the fun of it , using integration $$ \large x= \large \int_{0}^{3}\sqrt{ 4^2 - t^2}~dt \\~\\ \large y = \large \int_{3}^{4} \left( 4- \sqrt{ 4^2 - t^2 } \right) ~dt $$ we see that if we subtract x - y, we get a nice expression $$ \large x - y = \large \int_{0}^{3}\sqrt{ 4^2 - t^2}~dt - \int_{3}^{4}\left( 4- \sqrt{ 4^2 - t^2 } \right) ~dt \\ \large = \large \int_{0}^{3}\sqrt{ 4^2 - t^2}~dt - \int_{3}^{4} 4 ~ dt + \int_{3}^{4}\sqrt{ 4^2 - t^2 } ~dt \\ ~\\ = \large \int_{0}^{4}\sqrt{ 4^2 - t^2}~dt - 4\cdot 1 \\ = \large \pi ~\frac{4^2}{4} - 4 = 4\pi - 4 $$
anonymous
  • anonymous
@Emeyluv99 did you follow ganeshie's solution?
anonymous
  • anonymous
@ganeshie8 thanks so much!, i feel so stupid now, it was staring me in the face! @jayzdd Yes, thanks for the calculus solution!
ganeshie8
  • ganeshie8
Haha It is so simple yeah :)

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