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Math Help - Test on 4/14. Geometry. Picture included Please help ASAP please

  1. #1
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    Test on 4/14. Geometry. Picture included Please help ASAP please



    The ones I know are know are correct but no well understood:

    a) 1:1 because both triangles have the same base? So 16:16 = 1:1 right?
    b) same method as ^
    d) 9:16 because using the similar triangles area ratio formula. The method of similar trianges i used was AA because of the parallel lines right?

    The ones I don't understand at all but I know the answers. I'm all confused:

    c) The answer is 1:1. I think you can assume that the height of both triangles are the same? How can it be 1:1 if there are no other measurements of the triangles?
    e) The answer is 3:4. I figured maybe you do simplfy 12/16 to get 3:4 but I have no clue why because the the triangles that the question is asking for doesn't have 12 or 16?

    Im so confused. please help ASAP. Thanks... :cry:
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  2. #2
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    Quote Originally Posted by AirForceOne
    ...
    The ones I don't understand at all but I know the answers. I'm all confused:
    c) The answer is 1:1. I think you can assume that the height of both triangles are the same? How can it be 1:1 if there are no other measurements of the triangles?
    e) The answer is 3:4. I figured maybe you do simplfy 12/16 to get 3:4 but I have no clue why because the the triangles that the question is asking for doesn't have 12 or 16?
    Im so confused. please help ASAP. Thanks... :cry:
    Hello,

    let H be the height of the trapezoid and h the height of triangle(ZYP) (I#ve attached a diagram to demonstrate, what I'll calculate).
    Then you get the proportion:
    \frac{h}{16}=\frac{H-h}{12}. Solve for h and you'llget h = 4/7*H.

    to e.: You get the area of triangle(XPY) by:
    A_{\Delta XPY}=A_{\Delta WXY}-A_{\Delta WXP}

    \frac{1}{2} \cdot 12 \cdot H-\frac{1}{2} \cdot 12 \cdot \frac{3}{7} \cdot H=\frac{1}{2} \cdot 12 \cdot H \cdot \frac{4}{7}
    That means:
    \frac{A_{\Delta XPY}}{A_{\Delta WXP}}=\frac{\frac{1}{2} \cdot 12 \cdot H \cdot \frac{4}{7}}{\frac{1}{2} \cdot 12 \cdot \frac{3}{7} \cdot H}=\frac{4}{3}

    to d.: As I've shown above you can calculate the areas of the triangles in question by calculating the differences of two triangles.
    Unfortunately I'm a little bit in a hurry to complete the problem, but I'm certain that you now know how to handle the problem.

    Greetings and Happy Easter to you.

    EB
    Attached Thumbnails Attached Thumbnails Test on 4/14. Geometry. Picture included Please help ASAP please-trap_teilfl.gif  
    Last edited by earboth; April 13th 2006 at 11:37 PM.
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  3. #3
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    Quote Originally Posted by AirForceOne
    ...
    The ones I don't understand at all but I know the answers. I'm all confused:
    c) The answer is 1:1. I think you can assume that the height of both triangles are the same? How can it be 1:1 if there are no other measurements of the triangles?
    ...
    Hello,

    to c.: as you've demonstrated:
    A_{\Delta ZYW}=A_{\Delta ZYX} (same base, same height). Thus

    A_{\Delta ZYW}-A_{\Delta ZYP}=A_{\Delta ZYX}-A_{\Delta ZYP}

    A_{\Delta WZP}=A_{\Delta XYP}. Thus the ratio is 1:1.

    Greetings

    EB
    Last edited by earboth; April 13th 2006 at 11:41 PM.
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