Perimeter of an obtuse-angled triangle with area $10cm^2$ and longest side $10cm$?

Question

What is the maximum and minimum perimeter of an obtuse-angled triangle with area $10cm^2$ and longest side $10cm$?

I have tried using $20= ab \sin C $ and $100=a^2+b^2+2ab \cos C $ but I am not getting the answer.

Please help.

Answer 1

As pointed out by @MichaelHoppe we need to determine one intersection C$_{sup}$ of the line $y=2$ with the Thales circle of $r=5$ around (0|0) the centre of [AB] wich we place onto the $x$-axis. The circumference $u_{sup} $of ABC$_{sup}$, then is the supremum of the circumferences of all possible triangles ABC, which lie all inside the Thales circle. The right intersection point forms a rectangular triangle with hypothenuses of 5 and one side length of 2, consequentially the $y$-coordinate of C$_{sup}$ is $\sqrt{5^2 - 2^2}=\sqrt{21}$. Then we need the distances of this point C$_{sup} = (\sqrt{21}|2)$ to A$=(5|0)$ and B=$(-5|0)$. That is $\sqrt{\sqrt{(5-\sqrt{21})^2 + 4}}$ and $\sqrt{\sqrt{(5+\sqrt{21})^2 + 4}}$, respectively, the sum of which is equal $\sqrt{140}$. Yielding $$u_{sup}=\sqrt{140}+10=10+2\sqrt{35}\approx 21.8$$ The infimum is yielded for C$_{inf}=(0|2)$, which is $$u_{inf}=10+2\sqrt{29}\approx20.8$$