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Math Help - Quadrilateral with area and 2 sides only

  1. #1
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    Quadrilateral with area and 2 sides only

    Hi

    I have 2 triangles when joined by a common side form a quadrilateral:

    1. right angled triangle with the opposite side about 17 metres
    2. non-right angled triangle with one side of about 21 metres.

    Joined together their area is about 200m squared.

    The adjacent side of the right angled triangle joins the non-right angled triangle so they share a side.

    I'm trying to find the length of the shared side and the unknown side of the non-right angled triangle.

    I'm thinking something like 200=x+y
    Where x is area of right triangle and y is area of non right triangle

    x=0.5(17)(shared side)
    y=0.5(shared side)h(somehow?).

    I was thinking of Herons rule but I dont have enough information and it would be a lot of working.

    Any ideas?

    Regards,
    Neverquit
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  2. #2
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    Quote Originally Posted by Neverquit View Post
    Any ideas?
    Not unless you provide a diagram.
    Plus the EXACT wording of the original problem.

    The way you got it worded, with approximates, means we can theoretically
    make the 2nd triangle also a right triangle, work it out, then change a side
    of the 2nd triangle by .000000001 or so, so it is no longer "quite right!";
    get my drift?
    Last edited by Wilmer; February 5th 2011 at 11:25 PM.
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  3. #3
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    ---------------------------------------
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  4. #4
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    picture . this is the best I can do

    Hi I have done a rough drawing attached. This is the best I can do.

    I was thinking that the triangle bcd could be solved with sine rule for area. I am trying to work out length of bd and cd..

    Thanks,
    Neverquit
    Attached Thumbnails Attached Thumbnails Quadrilateral with area and 2 sides only-maths-fun.png  
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  5. #5
    MHF Contributor Unknown008's Avatar
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    I don't think that you have enough information.

    You could very well have different values for the shared side and different values for the height of triangle BCD, while keeping the area as 200 m^2, meaning that there is no exact solution to your problem.
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  6. #6
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    Hi,

    how would you go about finding the diffferent values for the lengths bd and cd? Use excel trial and error?

    Regards,
    Neverquit
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  7. #7
    MHF Contributor Unknown008's Avatar
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    Ok, let BD = 4 m, x = 34 m^2, y = 166 m^2, h = 83 m

    Let BD = 8 m, x = 68 m^2, y = 132 m^2, h = 33 m

    Need I say more? [I used even numbers for BD so that I have whole numbers]

    The area of BCD does not depend on the side BC which is 21 m because no angle is specified and the vertex can be adjusted to get BC = 21 m in all cases.
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  8. #8
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    WHY are you considering it solved?

    Seems like an "interesting" problem IF given sufficient information.

    Why don't you "experiment"; say take an actual integer-sided right triangle,
    say 12-16-20, then construct a non-right triangle using side=12 ...
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  9. #9
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    Hi okay,

    I tried the integers 12-16-20 which gives triangle ABC area of 96m^2
    200-96=104.
    104=0.512h
    h=17.3333

    (BA-x)^2=21^2-17.33^2=140.6711^2. sqrt(140.6711)=11.8557
    x=12-11.8557=0.1443

    other triangle right triangle with side CD:
    CD^2=17.3333^2+0.1443^2=300.4641. sqrt(300.4641)=17.3339.

    Its the same as h to 3 decimal places? So that would indicate that triangle BCD is a right triangle too!

    That is interesting...........
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  10. #10
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    I kinda meant set up something ENTIRELY with integers; like:
    Code:
                       B
    
                       E          D
    
    
    A                  C
    Right triangle ABC: AB=50, AC=48, BC=14 ; area = 336

    Now tack on non-right triangle BCD: BD=13, CD=15 ; area 84
    Height DE=12; BE=5, CE=9

    Now your original problem would be worded:
    given right triangle ABC, AC=48;
    non-right triangle BCD added, CD=15;
    total area of resulting quadrilateral ABDC = 420

    It will now be easier for you to "see" that more info is required.
    Suggestion: pretend height DE=12 is given instead of CD=15;
    see how far you get with that...
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  11. #11
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    Hi
    it becomes 2 right angled triangles. share side, corresponding angles are the same. then triangles are congruent.
    So they have equal area. 11.77 works best for shared side.
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