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Math Help - 2 parallel chords: finding radius

  1. #1
    Gui
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    2 parallel chords: finding radius

    Hello,

    I have over and over again tried to figure this problem out. I swear I simply am not able to figure it out :/

    Two parallel chords are 16 centimeters and 30 centimeters long are 23 centimeters apart. Find the radius of the cricle.

    Thanks for any help!
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  2. #2
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    Quote Originally Posted by Gui View Post
    Hello,

    I have over and over again tried to figure this problem out. I swear I simply am not able to figure it out :/

    Two parallel chords are 16 centimeters and 30 centimeters long are 23 centimeters apart. Find the radius of the cricle.

    Thanks for any help!
    Draw the 30 cm line.
    Bisect it, the circle centre lies on this bisector.

    draw a parallel line 23 cm away.
    Locate where the bisector cuts that line.
    Draw the 16 cm line by going 8 cm away from the intersection
    of this line and the bisector of the 30 cm line.

    Now take the two extreme points near each other at the end of both lines.
    Join these points and bisect the line.

    The intersection of the 2 bisectors is the circle centre
    because you have found a point equidistant from all 4 endpoints.
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  3. #3
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    Hello, Gui!

    Two parallel chords are 16 cm and 30 cm long are 23 cm apart.
    Find the radius of the cricle.
    Code:
                  * * *
              *        8  *
          C * - - - + - - - * D
           *        |     *  *
               23-x |   * r
          *         | *       *
          *        O*         *
          *         |  * r    *
                   x|     *
         A *- - - - + - - - -* B
            *           15  *
              *           *
                  * * *


    The longer chord is: . AB = 30
    . . It is x cm from the center O.

    The shorter chord is: . CD = 16
    . . It is 23-x cm from the center.

    Let r = the radius: . r = OA = OB = OC = OD.


    In the two right triangles we have:

    . . \begin{array}{ccccc}x^2 + 15^2 &=& r^2 & [1] \\<br />
(23-x)^2 + 8^2 &=& r^2 & [2] \end{array}


    Equate [1] and [2]: . x^2 + 15^2 \:=\:(23-x)^2 + 8^2 \quad\Rightarrow\quad x = 8

    Substitute into [1]: . 8^2 + 15^2 \:=\:r^2 \quad\Rightarrow\quad \boxed{r \:=\:17}

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  4. #4
    Gui
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    Thanks, I understand it now. That was a great answer!
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  5. #5
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    As 15+8=23,

    it's possible to construct 2 identical right-angled triangles
    with the same hypotenuse.

    Or...

    Join the endpoints of the 2 chords (pink slanted line in the attachment).

    Find it's midpoint and notice it's slope is -\frac{23}{7}

    The horizontal pink line's length is 7, the difference between 15 and 8,
    therefore half of that is 3.5, while half of 23 is 11.5.

    Draw the blue perpendicular line going through the midpoint of the slanted pink line.

    It's slope is \frac{7}{23} and it creates an identical triangle of sides 3.5, R and 8.

    Hence R=\sqrt{8^2+15^2}
    Attached Thumbnails Attached Thumbnails 2 parallel chords: finding radius-radius.jpg  
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