anonymous one year ago 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.

1. anonymous

2. 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...

3. anonymous

are you allowed to use calculus?

4. ganeshie8

is the answer $$4\pi-4$$ ?

5. anonymous

no i cant use calculus @jayzdd and im supposed to write a proof so i dont have the answer @ganeshie8

6. 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

7. 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 $$\large x - y = \large \int_{0}^{3}\sqrt{ 4^2 - t^2}~dt - \int_{3}^{4}\left( 4- \sqrt{ 4^2 - t^2 } \right) ~dt \\ \large = \large \int_{0}^{3}\sqrt{ 4^2 - t^2}~dt - \int_{3}^{4} 4 ~ dt + \int_{3}^{4}\sqrt{ 4^2 - t^2 } ~dt \\ ~\\ = \large \int_{0}^{4}\sqrt{ 4^2 - t^2}~dt - 4\cdot 1 \\ = \large \pi ~\frac{4^2}{4} - 4 = 4\pi - 4$$

8. anonymous

@Emeyluv99 did you follow ganeshie's solution?

9. anonymous

@ganeshie8 thanks so much!, i feel so stupid now, it was staring me in the face! @jayzdd Yes, thanks for the calculus solution!

10. ganeshie8

Haha It is so simple yeah :)