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Math Help - Finding area problem

  1. #1
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    Finding area problem



    Find area of CED and BDF.

    ABDC is a quadrangle. Sorry for my drawing skills, but CBEF is also quadrangle.
    CB = EF
    CE=BF
    AC = 2

    AB = 4

    We get that CB is 2\sqrt{5}. Now I am a bit confused, as I can't move from this on. What should I do next?
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  2. #2
    Member HappyJoe's Avatar
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    Are all angles in the quadrilateral BCEF right angles? (And how about the angles in ABDC for that matter?)

    One idea is to find the area of the quadrilateral BCEF (which I hope is a rectangle), and then subtract the area of the triangle BCD. This gives you the sum of the areas of CED and BDF.

    Or did you want the individual areas of the two grey triangles?
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  3. #3
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    Quote Originally Posted by HappyJoe View Post
    Are all angles in the quadrilateral BCEF right angles? (And how about the angles in ABDC for that matter?)

    One idea is to find the area of the quadrilateral BCEF (which I hope is a rectangle), and then subtract the area of the triangle BCD. This gives you the sum of the areas of CED and BDF.

    Or did you want the individual areas of the two grey triangles?
    Yes. But how do I find area of BCEF, do I need to do anything with angles? thanks.
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  4. #4
    Member HappyJoe's Avatar
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    You probably know already that the area of a rectangle is base times height. You can take BC as being the base (of BCEF). Then if you draw the line segment from the point D to the diagonal BC (and call the point of intersection for G), you have that DG is the height of BCEF.

    See if you can find this height. Notice that you have two right angled triangles BDG and CDG, where the sought-for height is a leg of both.
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  5. #5
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    E is right-angle, I assume CD is C's bisector = 45 degrees. If I am right, we get that CE = \sqrt{5} , and area of CED = about 4 ...I doubt it...
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  6. #6
    Member HappyJoe's Avatar
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    Quote Originally Posted by Ellla View Post
    E is right-angle, I assume CD is C's bisector = 45 degrees. If I am right, we get that CE = \sqrt{5} , and area of CED = about 4 ...I doubt it...
    No, because CD is the bisector of C only if ABDC is a square, which we cannot assume here.
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  7. #7
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    Well, I have no idea what to do next...as I have prove why sum of areas CED and BDF is 4.
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  8. #8
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    Actually quite easy...
    let u = DF and v = BF ; leaves DE = 2sqrt(5) - u

    right triangle BDF: u^2 + v^ 2 = 4 [1]

    right triangle CDE: (2sqrt(5) - u)^2 + v^2 = 16 ; simplify:
    20 - 4usqrt(5) + u^2 + v^2 = 16 [2]

    Substitute [1] in [2]:
    20 - 4usqrt(5) + 4 = 16
    4usqrt(5) = 8 ; u = 2 / sqrt(5)

    Substitute that in [1] to get v

    OK? I'll let you finish...
    Last edited by Wilmer; May 8th 2011 at 04:13 PM.
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  9. #9
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    Triangle ABC is simular to BDF and CED. Segments BF, DF, and DE can be obtained by ratios Areas of the two shaded rt triangles follow



    bjh
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  10. #10
    Member HappyJoe's Avatar
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    Look at post #9. That is the simple way to go.
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  11. #11
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    Nice "catch", BJ!

    So we really don't need point A here; problem is same as:
    rectangle CBFE with D on FE such that angle BDC = 90 degrees.
    In other words, the rectangle is divided into 3 right triangles.

    Smallest case with integer solution: BD = 15, CD = 20; so CB = EF = 25.
    Makes CE = BF = 12, DE = 16 and DF = 9.
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  12. #12
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    Got it.

    I appreciate your help.
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