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Homework Help: Inscribed Circle Geometry

  1. Jun 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the radius of a circle inscribed in a quadrant of a circle with radius 5

    2. Relevant equations

    3. The attempt at a solution
    I worked this but I'm not sure if its correct. I looked at the first quadrant so a quarter of a circle with radius 5. I drew the radius that would bisect the angle at the origin into 45° angles. I found the midpoint of the radius and dropped a line to the x-axis from that midpoint. This forms a 45-45-90 triangle which would make the radius of the inscribed circle 2.5/√2. Is this correct??
  2. jcsd
  3. Jun 9, 2012 #2
    Your method seems correct, but I got [itex]\frac{5}{1+\sqrt{2}}[/itex].

    Is it possible you can show a picture of where you are making your triangle?
  4. Jun 9, 2012 #3
    I am looking at the first quadrant. Center of the circle with radius 5 is at the origin, so I draw my radius from the origin and bisecting the 90° angle. Then from the midpoint of that radius (2.5)(which I think will be the center of the inscribed circle) I drop the segment to the x-axis which will be a leg of the 45-45-90 and also the radius of the inscribed circle.?? so the triangle has a hypotenuse of 2.5 and dividing by √2 would be the leg/radius ??
  5. Jun 9, 2012 #4
    Is the center circle here the one you are trying to find the radius of?

  6. Jun 9, 2012 #5
    Yes! I love the drawing!
  7. Jun 9, 2012 #6
    The problem with your work I think was assuming that the line dropped down from the center of that circle bisected the bottom line, which isn't true.
  8. Jun 9, 2012 #7
    No I wasn't assuming that. The first part of the bottom line would be 2.5/√2 as well since the triangle formed is a 45-45-90.
  9. Jun 9, 2012 #8
    Oh oh oh, I see now. But if that were true, then wouldnt the radius also have to be 2.5? Since the segment from the center to the outside of the larger circle is 2.5 and that is another radius of the circle.
  10. Jun 9, 2012 #9
    I don't think the midpoint of the 5 radius is the center of the inscribed circle anymore because you are correct about the 2.5 but the perpendicular line would also have to be a radius and it is 2.5√2 so that must not be the center
  11. Jun 9, 2012 #10
    If it helps, I used the diagonal line in order to solve for the radius.

    I don't want to say too much however.
  12. Jun 9, 2012 #11
    Are you talking about the radius drawn in that bisects the 90° at the origin??
  13. Jun 9, 2012 #12
    Yeah, the one that is 45 degrees above the horizontal.
  14. Jun 9, 2012 #13
    I'm assuming you didn't use 2.5,2.5 as your center??
  15. Jun 9, 2012 #14

    What is the length of that segment? How can you label its parts in terms of r?
  16. Jun 9, 2012 #15
    I'm not sure I need to look at it some more :(
  17. Jun 9, 2012 #16
    Ok now I got (10-√2)/4 its close to your answer but still not the same. I need to go back and see if I made a calculation error.
  18. Jun 9, 2012 #17
    Explain how you got to that.
  19. Jun 9, 2012 #18
    Its wrong I was assuming that perpendicular segment was the radius of the inscribed circle. I need to add 2.5 to that piece that I keep thinking is a radius (but not) to get the diameter but I don't know how to get it
  20. Jun 9, 2012 #19
    The perpindicular segment is a radius though.

    And you never use the number 2.5 here.
  21. Jun 9, 2012 #20
    Ok so 2 of those segments make the diameter of the inscribed circle. and that little piece outside the circle would need to be subtracted from 5. Should I be able to find that segment length first??
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