Question: [![enter image description here](https://i.stack.imgur.com/Ivlaz.png)](https://i.stack.imgur.com/Ivlaz.png)

A GRE question asks whether the area of parallelogram can be found only using the information provided above. 

Area of parallelogram = base x height 

If we cut off one end to form a triangle and place it on the end, we would get a rectangle that looks like: 

[![enter image description here](https://i.stack.imgur.com/GZ37I.png)](https://i.stack.imgur.com/GZ37I.png)

But we still do not know the width of rectangle in order to calculate area. Can we assume that not enough information is provided?

Answer: 
I already discussed this point several times (see [this answer](https://math.stackexchange.com/questions/2379549/proof-of-prop-i-27-in-euclids-elements/2380863#2380863), for instance): [Euclid's exterior angle theorem](https://en.wikipedia.org/wiki/Exterior_angle_theorem) is perfectly valid in [absolute geometry](https://en.wikipedia.org/wiki/Absolute_geometry) (i.e. euclidean geometry without parallel postulate), as its proof does not rely on the parallel postulate (and if you do not believe that, you should point out where this postulate is used in the proof).

Elliptic geometry IS NOT absolute geometry, [because it also does away with Euclid's second postulate](https://en.wikipedia.org/wiki/Non-Euclidean_geometry#Axiomatic_basis_of_non-Euclidean_geometry).
I understand what you are after and I am regretfully not conversant enough with the foundations of geometry to provide an rigorous answer.

Hoping not to waste anybody's time, I wonder though if the following elementary line of reasoning could be re-phrased to meet your requirements: I apologise in advance if it contains gross inaccuracies.

I believe that using the first four postulates one could prove that the time taken travelling at constant unitary speed between two distinct points is strictly positive: let us call this time $t$

So let us take a triangle $ABC$: without loss of generality let the length of the side $AB$ equal $1$.
Now, if the triangular inequality is violated $$\vert BC \vert + \vert AC \vert \leq \vert AB \vert $$
Let us consider first the strict inequality
$$\vert BC \vert + \vert AC \vert < \vert AB \vert $$

Let us define $$\gamma = \frac{\vert BC \vert + \vert AC \vert }{\vert AB \vert } $$ 
and from what is said, $\gamma < 1$.

If an ant had to travel at constant speed and wishes to minimize the travel time between $A$ and $B$, it will certainly choose the path $ACB$, and will travel it in time $\gamma$.

Now, if one builds similar triangles on the sides $BC$ and $AC$ it turns out the optimal path will take time $\gamma^2$ (here I worry I miss essential facts on non-euclidean geometries). Iterating one could prove that there exists a path, along which the travelling time is $\gamma^n $, so one could travel from $A$ to $B$ in a small as desired time, which cannot be right in any geometry. 

We are left with the case whereby $$\vert BC \vert + \vert AC \vert = \vert AB \vert $$ which can be maybe be proven geometrically as follows (again, the "proof" might well be circular, as I might be giving for granted facts about geometry which actually stem from the triangular inequality itself).

Draw a circumference $C\_1$ having $B$ as center and $AB$ as radius. 

Draw a circumference $C\_2$ having $C$ as center and $AC$ as radius.

The line, extension of the segment $AC$, will intersect the circumference in a point $D$. Now $$\vert DB \vert = \vert BC\vert + \vert CD\vert = \vert BC\vert + \vert CA\vert =\vert AB\vert $$ the last equality following from the (assumed) equality in the triangular inequality. This means $D$ should lie on $C\_1$, which is not possible.