TITLE: How might one actually prove, in a rigorous way, that the area of a parallelogram is given by the magnitude of the cross product?
QUESTION [6 upvotes]: In $3$-dimensions, we can compute areas of a parallelograms using the cross product, like so: $$H_2(\{ax+by : a \in [0,1],b \in [0,1]\}) = |x \times y|,$$ where $H_2$ is the $2$-dimensional Hausdorff measure. Of course, it's rarely formalized like this. So much of elementary geometry is never formalized. In any event, the above formula seems like the most reasonable interpretation of this well-known fact.

Question. How would you actually prove such a formula rigorously?

By "prove rigorously" I do not mean draw a persuasive geometric diagram and make a reasonable-sounding argument. Rather, I mean something that could, in principle at least, be formalized in a sufficiently powerful interactive theorem prover one day. A diagram is fine, but please explain how you would actually implement the proof implicitly contained in that diagram.

REPLY [4 votes]: Here is a possible approach. First, check that for any two vectors $u$, $v$, in $3$-space we have 
$$|u|^2 \cdot |v|^2 = (u\cdot v)^2 + |u\times v|^2$$
That can be shown using the definitions involving the coordinates of $u$ and $v$.
Now, if you also know that $u\cdot v = |u|\cdot |v| \cdot \cos \theta$, and also using $\cos^2 \theta + \sin^2 \theta = 1$ we conclude
$$|u\times v | = |u|\cdot |v| \cdot \sin \theta$$
( $\theta$, the angle between $u$ and $v$, lies in the interval $[0, \pi]$). 
From here, it may be almost done, assuming that we know what the area of a parallelogram is.