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Math Help - parallelogram

  1. #1
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    parallelogram

    Prove that parallelograms on the same base and betweem the same parallels are equal in area .
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  2. #2
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    Prove that parallelograms on the same base and betweem the same parallels are equal in area .
    Divide the parallelogram into two equal triangles, using a diagonal. Then the area of the parallelogram = twice the area of one of the triangles. Area of triangle = half base times height. So?

    Grandad
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  3. #3
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    Quote Originally Posted by thereddevils View Post
    Prove that parallelograms on the same base and betweem the same parallels are equal in area .
    Hi thereddevils,

    Draw parallelogram ABCD with AB being the top side. Extend AB to E such that BE = DC.

    Connect D to B and C to E.

    You have just constructed another parallelogram BECD with same base as ABCD and between the same parallels, as required.

    In parallelogram ABCD with diagonal BD, we can prove that the two triangles ABC and CDB are congruent.

    We also have sufficient evidence to prove that triangles CDB and BEC are congruent.

    Now all three triangles are congruent to each other. They have equal areas.
    Attached Thumbnails Attached Thumbnails parallelogram-p.png  
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  4. #4
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    oh thansk a lot , Grandad and masters . i was just a little confused with the meaning of between the parallels .
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  5. #5
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    Hello everyone
    Quote Originally Posted by masters View Post
    Hi thereddevils,

    Draw parallelogram ABCD with AB being the top side. Extend AB to E such that BE = DC.

    Connect D to B and C to E.

    You have just constructed another parallelogram BECD with same base as ABCD and between the same parallels, as required.

    In parallelogram ABCD with diagonal BD, we can prove that the two triangles ABC and CDB are congruent.

    We also have sufficient evidence to prove that triangles CDB and BEC are congruent.

    Now all three triangles are congruent to each other. They have equal areas.
    While this is true, it's a very special case. You should really consider two general parallelograms on the same base and between the same parallels, as I have indicated in the attached drawing. I outlined the proof in my first posting. Here are a few more details.

    In the diagram, ABCD and CDE F are any two parallelograms on the base CD, with the points A,B, E,F in a straight line.

    Construct the diagonals BD and CE. Triangles ABD and CDB can be easily proven congruent; as can triangles DEC and FCE.

    Drop perpendiculars from B, E to meet CD at points X, Y. BEYX is then a rectangle. Therefore BX = EY.

    Area ABCD = 2\times area \triangle CDB = CD \times BX = CD \times EY = area CDE F.

    Grandad
    Attached Thumbnails Attached Thumbnails parallelogram-untitled.jpg  
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