What is the greatest area difference between two "nice triangles"? We call a triangle "nice if all angles are between $45$ and $90$ degrees (including $90$ and $45$ itself) and all sides are between $1$ and $2$ (including $1$ and $2$ itself). What is the greatest area difference between two "nice triangles"?
My attempt: Because we have side and angle limits the best way to finding area is using the formula $S=bc\cos{A}$. We should find the greatest and lowest area. But here I got stuck and I don't know how to have both limits with each other. First I thought that the maximum area is $A=90^\circ $ and $b=c=2$. But then I saw that then we have$a=2\sqrt{2}>2$. Could you please give a way?

By the isoperimetric inequality the largest area of a nice triangle is $\sqrt{3}$, i.e. the area of a equilateral triangle with side length $2$. By the formula $2\Delta = ab\sin\gamma$ the smallest area of a nice triangle is $\frac{1}{2}\cdot 1\cdot 1\cdot\sin 60^\circ = \frac{1}{4}\sqrt{3}$.
It follows that the largest area difference between two nice triangles is $\color{red}{\frac{3}{4}\sqrt{3}}$.
