What is the greatest area difference between two "nice triangles"?

Question

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?

Answer 1

The Maximal triangle has two sides of length $2.$

for any triangle that had only 2 sides of lenght 2, we could extend the second longest side until it was length 2, and create a larger triangle.

$Area = \frac 12 BC\sin a\\ B,C = 2$

maximize $2\sin a$

constrained by: $4 \sin \frac {a}{2} \le 2$

$\sin \frac {a}{2}\le \frac 12\\ \frac {a}{2}\le 30\\ a\le 60$

Since $\sin a$ is strictly increasing between 0 and 90.

$a = 60\\ Area = 2\sin 60 = \sqrt 3$

then minimize $\frac 12 \sin a$

constrained by: $4 \sin \frac {a}{2} \ge 1$

and again $a = 60$

Answer 2

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 $\color{red}{\frac{3}{4}\sqrt{3}}$.