Using Clifford algebra $Cl_{0,3}(\mathbb{R})$ one can define a 1-vector as:
$$ \mathbf{v}=X\sigma_x+Y\sigma_y+Z\sigma_z $$
and the Pythagorean can be obtained by the geometric product $\mathbf{v}\mathbf{v}$:
$$ \mathbf{v}\mathbf{v}=(X\sigma_x+Y\sigma_y+Z\sigma_z)^2\\ =X^2+Y^2+Z^2==||\mathbf{v}||^2 $$
Try the same thing but with 2-vectors:
$$ \mathbf{a}=A\sigma_x\sigma_y+B\sigma_x\sigma_z+C\sigma_z\sigma_y $$
As the 2-basis elements are orthogonal, one can define an "area-distance" as follows:
$$ \mathbf{a}\mathbf{a}=(A\sigma_x\sigma_y+B\sigma_x\sigma_z+C\sigma_z\sigma_y)^2\\ =-A^2-B^2-C^2=||\mathbf{a}||^2 $$
$$ \implies ||\mathbf{a}||=i\sqrt{A^2+B^2+C^2} $$
What is the geometric interpretation of the area-equivalent of the Pythagorean theorem?
For a vector $\mathbb v=v_1\sigma_1+v_2\sigma_2+v_3\sigma_3$ (which can be interpreted as a line segment), $v_1$ is $\pm$ the length of its projection onto the $\sigma_1$-axis. The Pythagorean theorem says that the squared length of the segment is the sum of the squared lengths of its projections on all axes.
For a bivector $\mathbb a=a_{12}\sigma_1\sigma_2+a_{13}\sigma_1\sigma_3+a_{23}\sigma_2\sigma_3$ (which can be interpreted as a plane segment), $a_{12}$ is $\pm$ the area of its projection onto the $\sigma_1\sigma_2$ plane (its "shadow"). The equivalent theorem says that the squared area of the segment is the sum of the squared areas of its projections on all coordinate planes.