1. ## Triangle 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?

2. ## Maximum area

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3. Hello everyone
Originally Posted by Henryt999
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 $\displaystyle a=1, b=2, c=3$, the 'triangle' is a straight line. (Note: $\displaystyle 1+2=3$.) So it has zero area.

4. ## Thanks

Oh lord, You are correct Im sorry that was very very wrong. Sorry Hope he didnīt wright that down.

5. ## Acctually I might be wrong again but....

Hello everyoneSorry, but this is certainly not correct. With $\displaystyle a=1, b=2, c=3$, the 'triangle' is a straight line. (Note: $\displaystyle 1+2=3$.) So it has zero area.
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....

6. Hello everyone
Originally Posted by Henryt999

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....
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 $\displaystyle b=2$ 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.