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Math Help - Triangle with 0<a<1<b<2<c<3

  1. #1
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    Triangle with 0<a<1<b<2<c<3

    In a triangle with sides a, b, c the following holds: 0<a\le 1\le b\le 2\le c\le 3. What can be the maximum area?
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  2. #2
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    Maximum area

    .
    Last edited by Henryt999; January 9th 2010 at 11:22 AM. Reason: I was wrong
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  3. #3
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    Hello everyone
    Quote Originally Posted by Henryt999 View Post
    Maximum area is when the sides are maximum so a =1 b = 2 c = 3

    Do you know how to go from here?
    Law of cosines?

    C^2 = a^2 + b^2 -2ab*cos(c) this gives you angle c = arccos(((3^2)-(1^2)-(2^2))/(-2*1*2)) that gives you the angle.
    Now to find the area use: A = (ab*sin(x))/2 where x is the angel between the sides a;b.
    Hope this was not too confusing, I can follow you through if you want.
    Sorry, but this is certainly not correct. With a=1, b=2, c=3, the 'triangle' is a straight line. (Note: 1+2=3.) So it has zero area.

    I'm still thinking about how to answer the question correctly...

    Grandad
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    Thanks

    Oh lord, You are correct Im sorry that was very very wrong. Sorry Hope he didnīt wright that down.
    Last edited by Henryt999; January 9th 2010 at 11:34 AM.
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  5. #5
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    Acctually I might be wrong again but....

    Quote Originally Posted by Grandad View Post
    Hello everyoneSorry, but this is certainly not correct. With a=1, b=2, c=3, the 'triangle' is a straight line. (Note: 1+2=3.) So it has zero area.
    Grandad
    You are correct the third side can not be 3 But has to be less.

    If for example the angle is (pi)/2 then it has area: (1^2+2^2)^(0.5) = sqrt(5)

    And Area is given by: A = (a*b*sin(x))/2 = (1*2*sin(x))/2 = sin(x); That would give me a maximum area of 1??

    If sides are: 1;2 the last side would be sqrt:5??
    If this is the correct solution its not pretty, hope it was somewhat helpful....
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  6. #6
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    Hello everyone
    Quote Originally Posted by Henryt999 View Post

    You are correct the third side can not be 3 But has to be less.

    If for example the angle is (pi)/2 then it has area: (1^2+2^2)^(0.5) = sqrt(5)

    And Area is given by: A = (a*b*sin(x))/2 = (1*2*sin(x))/2 = sin(x); That would give me a maximum area of 1??

    If sides are: 1;2 the last side would be sqrt:5??
    If this is the correct solution its not pretty, hope it was somewhat helpful....
    Yes, I think this is correct. I didn't have any more time to think about this yesterday (it's now 7:15 am UK time), but I had come to the same conclusion overnight. With b=2 as base, the maximum height possible is 1, where a = 1 and the angle between these two sides is 90^o. So c = \sqrt5, and the area is 1 sq unit.

    Grandad
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