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Math Help - geometry finding length

  1. #1
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    geometry finding length



    how do you find segment AD and segment DH?

    i was thinking of the trapezoid and separating it into 3 different shapes; a rectangle, a triangle and a triangle.
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  2. #2
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    Hello, algebra2!

    Code:
              A 2 H
              *---*
           2/ |     \
        B *---*P      * G
         2|   |       |
          |   |       |
        C *---*Q      * F
           2\ |     /
              *---*
              D 2 E
    We have a regular octagon with side 2.
    Draw BP \perp AD \text{ and }CQ \perp AD
    Draw DH.

    We have isosceles right triangle APB with hypotenuse 2.
    . . Hence: . AP = \sqrt{2}
    Similarly: . QD = \sqrt{2}

    Hence: . AD \;=\;AP + PQ + QD \;=\;\sqrt{2} + 2 + \sqrt{2}

    . . Therefore: . \boxed{AD \;=\;2 + 2\sqrt{2}}


    In right triangle H\!AD\!:\;\;DH^2 \:=\:AH^2 + AD^2 \;=\;2^2 + (2+2\sqrt{2})^2 \;=\;16 + 8\sqrt{2}

    . . Therefore: . \boxed{DH \:=\:2\sqrt{4 + 2\sqrt{2}}}

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