A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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.

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    are you allowed to use calculus?

  4. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  6. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Emeyluv99 did you follow ganeshie's solution?

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Haha It is so simple yeah :)

  11. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.