Can one deduce whether a given quantity is possible as the area of a triangle when supplied with the length of two of its sides?

Solution 1:

Not an axiom, but yes: The area of a triangle is $\frac{1}{2}ab \sin \theta$, where $a$ and $b$ are two known side lengths and $\theta$ is the angle between the two given side lengths. Since $\sin \theta$ reaches its maximum value when $\theta=90°$ (in which case we have $\sin 90° = 1$), the largest possible area is indeed attained for the case of a right angle.

Solution 2:

Hint: The area of $\Delta ABC$ is $\dfrac{1}{2} \cdot AB \cdot AC \cdot \sin A$. You know that $AB = AC = 2$ and that $0 < \sin A \le 1$, since $0 < A < 180^{\circ}$. So, what is the range of possible areas of $\Delta ABC$?

Solution 3:

The area of the parallelogram proportional to it's height. It is maximal for a rectangle. The triangle is the half parallelogram

Solution 4:

The peak area will be the area when these two legs are perpendicular to each other. But smaller areas are possible, all the way down to $0$, as the two legs become closer and closer to being parallel. If you know that the area is $\frac12(\text{base})(\text{height})$ then you can use one of these legs as the base, and see that height is maximized when the other leg is perpendicular to the base.

Solution 5:

Given two sides have the same length = 2, the minimum area of the triangle will be 0. And maximum area will be 2 (when the triangle is a right triangle).

Thus possible area are A, B, C and D (<=2). E and F are not possible (>2)