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?
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....
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