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Thread: Circle

  1. #1
    Junior Member Dragon's Avatar
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    Circle

    AB is a diameter of a circle of radius 1 unit CD is a chord perpendicular to AB that cuts Ab at E If the arc CAD is 2/3 of the circumference of the circle what is the length of the segment AE.?
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  2. #2
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    Quote Originally Posted by Dragon View Post
    AB is a diameter of a circle of radius 1 unit CD is a chord perpendicular to AB that cuts Ab at E If the arc CAD is 2/3 of the circumference of the circle what is the length of the segment AE.?
    Hi,

    the centre of the circle is M. So the triangle MEC is a right triangle.

    The arc $\displaystyle BC = \frac{1}{6} \pi$

    $\displaystyle \overline{ME}=\cos(\frac{1}{6} \pi)=\frac{1}{2} \cdot \sqrt{3}$

    So the total length of

    $\displaystyle \overline{AE}=1+\frac{1}{2} \cdot \sqrt{3}$

    EB
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  3. #3
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    Hello, Dragon!

    Hmmm, I got a different answer . . .


    $\displaystyle AB$ is a diameter of a circle of radius 1 unit.
    $\displaystyle CD$ is a chord perpendicular to $\displaystyle AB$ that cuts $\displaystyle AB$ at $\displaystyle E$.
    If the arc $\displaystyle CAD$ is 2/3 of the circumference of the circle,
    what is the length of the segment $\displaystyle AE$?

    Of course you made a sketch, right?
    Code:
                  * * *   C
              *           *
            *           / | *
           *          1/  |  *
                      /   |
          *    1     /60 |   *
        A * - - - - * - - + - * B
          *         O\    |E  *
                      \   |
           *          1\  |  *
            *           \ | *
              *           *
                  * * *   D

    Since major arc $\displaystyle CAD \:= \:\frac{2}{3}\cdot360^o \:=\:240^o$

    . . then minor arc $\displaystyle CBD \,=\,120^o$ and central angle $\displaystyle COB \,=\,60^o.$


    Right triangle $\displaystyle CEO$ has acute angle $\displaystyle 60^o$ and $\displaystyle hyp = 1.$

    Hence: $\displaystyle OE \,= \,adj \,= \,\frac{1}{2}$

    Therefore: .$\displaystyle AE\:=\:\frac{3}{2}$

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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by earboth View Post
    Hi,

    the centre of the circle is M. So the triangle MEC is a right triangle.

    The arc $\displaystyle BC = \frac{1}{6} \pi$

    $\displaystyle \overline{ME}=\cos(\frac{1}{6} \pi)=\frac{1}{2} \cdot \sqrt{3}$

    So the total length of

    $\displaystyle \overline{AE}=1+\frac{1}{2} \cdot \sqrt{3}$

    EB
    The arc $\displaystyle BC = \frac{2}{3} \pi$ rad since the arc CAD is 2/3 the total circumference.

    -Dan
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