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

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- Jan 9th 2010, 08:09 AMjames_bondTriangle with 0<a<1<b<2<c<3
In a triangle with sides $\displaystyle a$, $\displaystyle b$, $\displaystyle c$ the following holds: $\displaystyle 0<a\le 1\le b\le 2\le c\le 3$. What can be the maximum area?

- Jan 9th 2010, 10:17 AMHenryt999Maximum area
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- Jan 9th 2010, 11:19 AMGrandad
- Jan 9th 2010, 11:23 AMHenryt999Thanks
Oh lord, You are correct Im sorry that was very very wrong. Sorry Hope he didnīt wright that down.

- Jan 9th 2010, 01:09 PMHenryt999Acctually I might be wrong again but....
You are correct the third side can not be 3 But has to be less.

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

And Area is given by: $\displaystyle 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: $\displaystyle 1;2$ the last side would be $\displaystyle sqrt:5$??

If this is the correct solution its not pretty, hope it was somewhat helpful.... - Jan 9th 2010, 10:19 PMGrandad
Hello everyoneYes, 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 $\displaystyle b=2$ as base, the maximum height possible is 1, where $\displaystyle a = 1$ and the angle between these two sides is $\displaystyle 90^o$. So $\displaystyle c = \sqrt5$, and the area is $\displaystyle 1$ sq unit.

Grandad