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Math Help - Two problems

  1. #1
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    Two problems

    1)http://img.photobucket.com/albums/v2...kid/wrench.gif

    ABCDEFGH is a regular octagon with each side of length 2. Find AF and AE.

    2)http://img.photobucket.com/albums/v2...id/hexagon.png

    Is AF 3 and AE 4 (it's probably not, because the answer seems too simple)?

    Thanks!
    Last edited by icebreaker09; July 24th 2007 at 05:02 PM.
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  2. #2
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    Is there any more information? Like are any of these line segments medians? Maybe bisectors?
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  3. #3
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    Below is a the picture. There are 8 triangles. And each one has 45 degrees. Now use trigonometry (or whatever) to find the length of the "red" line. Double it to get x.
    Attached Thumbnails Attached Thumbnails Two problems-picture25.gif  
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  4. #4
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    Quote Originally Posted by Jonboy View Post
    Is there any more information? Like are any of these line segments medians? Maybe bisectors?
    Sorry, that's all. My teacher has a tendency to provide us with vague problems... that's why I went here--I was wondering if I was missing something.

    ThePerfectHacker: Just to clarify: that equation is for which problem? The octagon? If it is, what is x equal to AE or AF?
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  5. #5
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    Quote Originally Posted by icebreaker09 View Post


    ABCDEFGH is a regular octagon with each side of length 2. Find AF and AE.


    Is AF 3 and AE 4 (it's probably not, because the answer seems too simple)?

    Thanks!
    See regular hexagon & octagon properties
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  6. #6
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    Quote Originally Posted by icebreaker09 View Post
    ...
    2)http://img.photobucket.com/albums/v2...id/hexagon.png

    Is AF 3 and AE 4 (it's probably not, because the answer seems too simple)?

    Thanks!
    Hello,

    to #2.:

    I'll pick up TPH's hint how to do this problem.
    I've modified your drawing of an octogon (which you saved as a hexagon ).

    1. AE = 2r
    Take one of the 8 isosceles triangles and split it into 2 right triangles. Then you get:
    \frac{1~cm}{r}=\sin(22.5^\circ)~~\Longrightarrow ~~r=\frac{1~cm}{\sin(22.5^\circ)}\approx 2.613... ~cm

    Therefore: AE = 2r \approx 5.226...~cm

    2. AF = 2x
    \frac{1~cm}{x}=\tan(22.5^\circ)~~\Longrightarrow~~ x=\frac{1~cm}{\tan(22.5^\circ)}\approx 2.414...~cm

    Therefore: AE = 2r \approx 4.828...~cm
    Attached Thumbnails Attached Thumbnails Two problems-octogon.gif  
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  7. #7
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    Quote Originally Posted by icebreaker09 View Post
    1)http://img.photobucket.com/albums/v2...kid/wrench.gif

    ABCDEFGH is a regular octagon with each side of length 2. Find AF and AE.

    2)http://img.photobucket.com/albums/v2...id/hexagon.png

    Is AF 3 and AE 4 (it's probably not, because the answer seems too simple)?

    Thanks!
    Hello,

    I've attached a modified drawing of the nut.
    You deal with a regular hexagon which consists of 6 equilateral triangles. Split one of these triangles into 2 right triangles and use Pythagorean theorem.

    For confirmation only: x = \sqrt{3}~cm \approx 1.732...~cm
    Attached Thumbnails Attached Thumbnails Two problems-screw6gon.gif  
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  8. #8
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    Hello, icebreaker09!

    2) Find AF and AE.
    Code:
                  F   2   E
                  * * * * *
             2  * | _       *
              *   |√2         *
            *     |             *
        G * - - - +               * D
          *       |               *
        2 *       | 2             *
          *       |               *
        H * - - - +               * C
            *     | _           *
           2  *   |√2         *
                * |         *
                  * * * * *
                  A   2   B

    In the upper-left and lower-left, we have isosceles right triangles.
    . . Their hypotenuse is 2, hence their sides are \sqrt{2}

    Therefore: . AF \;=\;2+2\sqrt{2}


    Draw AF.

    In right triangle AFE, we have: . AE^2\:=\:AF^2 + FE^2

    That is: . AE^2 \;=\;(2 + 2\sqrt{2})^2 + 2^2\;=\;16 + 8\sqrt{2}\;=\;4(4+2\sqrt{2})

    Therefore: . AE \;=\;2\sqrt{4+2\sqrt{2}}

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