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Math Help - 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 BC = \frac{1}{6} \pi

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

    So the total length of

    \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 . . .


    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?

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

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

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


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

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

    Therefore: . 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 BC = \frac{1}{6} \pi

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

    So the total length of

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

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

    -Dan
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