Is the convex hull in $\mathbb{R}^2$ of $\Bigg\{\begin{pmatrix}0\\0\end{pmatrix}, \begin{pmatrix}3\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}\Bigg\}$ scissor equivalent/congruent to the convex hull of $\Bigg\{\begin{pmatrix}0\\0\end{pmatrix}, \begin{pmatrix}2\\1\end{pmatrix}, \begin{pmatrix}1\\2\end{pmatrix}\Bigg\}$?
The background for this is Hilbert's Third Problem.
My thoughts on these are:
The convex hull of the first set has a convex hull which is a triangle of area $1.5$, and the second set has a convex hull which is a triangle of area $1.5$. These two triangles have the same area, so must be scissor congruent.
I'm really not sure how wrong or how right this thought process is - I'm struggling to understand my lecture notes regarding this.
The following figure shows that the second triangle is scissor congruent to a $1\cdot{3\over2}$ rectangle:
I leave it to you to produce a scissor congruence of the first triangle with a $1\cdot{3\over2}$ rectangle. The two figures taken together then show that the two given triangles are scissor equivalent.