Question

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.

Answer 1

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...

Answer 2

are you allowed to use calculus?

Answer 3

is the answer \(4\pi-4\) ?

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

@Emeyluv99 did you follow ganeshie's solution?

Answer 8

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

Answer 9

Haha It is so simple yeah :)