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Thread: Geometry: Area of Trapezoid Circumscribed About A Circle

  1. #1
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    Geometry: Area of Trapezoid Circumscribed About A Circle

    Hiii, this is a problem that I have encountered and I need help ASAP.
    This is the figure:

    Thanks a lot!

    So far what I have done are:
    draw 2 lines down. from point D and C. So it will form a rectangle in inside the trapeoid. Then according to that, I can figure out the bottom base line (the 18 one) can be divided into 5, 8, 5. =/ I need some help to move on =/ I can't seem to be able to figure out it. Thanks alot

  2. #2
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    You have not yet used the circle. Two possibilities come to mind:

    1) Construct a line through the center, parallel to the top and bottom. It will be half way between the top and bottom. That might lead somewhere.

    2) Construct a line segment from the center of the circle to all four points where the circle is tangent to the trapezoid. These form right angles with the sides they meet. That might lead to something.

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    Hmm here.. this one is the one drawn with the line segment. Looks to me that it is an isoscele trapezoid. IF it is, that would make D8 (8 as the middle point forget to label sorry ) congruent to 8C, and A18 (18 as the middle point forget to label also .. ) congruent to18B. So would means D8 is 4, and A18 is 9. Which would means DA is 13.


    Then in here, DC = EF, meaning AE is 5 (So is FB). I draw a rectangle down to form the right triangle (Triangle DAE), since DA is 13, AE is 5, I can use the pythegrean Theorem. So, 13^2 = 5^2 + x^2 which gets the heigh of 12.
    Area = 1/2*12*(8+18)
    Area = 1/2*12*26
    Area = 156

    However that is only if the trapezoid is isoscele.. Well =/ I am not sure how to prove the trapezoid is isoscele, could you please provide information on it? =D
    Thanks!

  4. #4
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    Sorry, I don't follow. Your first picture gave the trapazoid as isos by the two equal base angles but
    So would means D8 is 4, and A18 is 9. Which would means DA is 13
    I agree d8 is 4 and a18 is 9 but I don't see how that gives DA = 13. I took the long way around.
    I drew a perp-bisector thru 18 & 8 up to H then extended DA to H. The line from the center (O) of the circle of radius R meets DA at X (the tangent of the circle)
    I now have three similar triangles
    D8H, HOX and A18H.
    H8/4=(H8+2R)/9 gives
    H8=8R/5
    and
    HD=sqrt(H8+4)
    HD/4=(H8+R)/R
    Substituting
    sqrt((8R/5)+4)/4=(8R/5+R)/5R=13/5
    Solving that gave me R=6 and area = 156 as you said. How didja do it???
    I hope this helps. If you need more, come back with a post-reply.
    You are better off putting problems in the general areas. More people are avauilable to answer them there.
    Gene

  5. #5
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    Hello, yumi!
    Code:
                    D  4  :  4  C
                    * * o o o * *
                    o           o  4
                  o             | o
                 *              |  * E
                                |
                o               |   o
              * o         o     |   o * 9
                o               |   o
                                |
           *     o              |  o     *
                  o             | o
                    o           o
        * * * * * * * * o o o * + * * * * * *
        A        9        :  4  F     5     B
    Let [tex]E[/tex] be the point of tangency of side [tex]CB[/tex] to the circle.
    [tex]\;\;[/tex]Then [tex]EC\,=\,4,\;BE\,=\,9\;\;\Rightarrow\;\;CB\,=\,13[/tex]

    Draw perpendicular [tex]CF[/tex] from vertex [tex]C[/tex] to base [tex]AB.[/tex]
    [tex]\;\;[/tex]We find that [tex]FB\,=\,5[/tex]

    In right triangle [tex]CFB:\;CF\,=\,12\;\;[/tex] . . . and we have the height!
    I'm the other of the two guys who "do" homework.

  6. #6
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    OK, that's a shortcut I didn't think of
    --------------
    G
    I hope this helps. If you need more, come back with a post-reply.
    You are better off putting problems in the general areas. More people are avauilable to answer them there.
    Gene

  7. #7
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    Also, remember that what we have here is a circle inscribed
    in an isosceles TRIANGLE; extend AD and BC to meet.
    I'm just an imagination of your figment !

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    Thanks a lot all for the replies helped me a lot
    I have a question about the isosceles TRIANGLE, I do see a triangle if extend AD and BC to meet as you said. But umm, I have a question, what does it prove? What I can see is sorta like provide they are congruent? Not sure
    Once again thanks a lot everyone for your helps

  9. #9
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    Quote Originally Posted by Yumi
    I have a question about the isosceles TRIANGLE, I do see a triangle if extend AD and BC to meet as you said. But umm, I have a question, what does it prove?
    Not intended to "prove" anything...
    I was trying to let you know that if the problem was presented differently, as an
    isosceles triangle with base 18 and the length of the equal sides was a given,
    then Soroban's short cut could be used to find the length of CD, if such was the
    question being asked.
    Of course, CD would be described as tangent to the incircle, and parallel
    to the triangle's base.
    I'm just an imagination of your figment !

  10. #10
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    oooooh! Thanks for explaination

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