well, the pyramid is a square pyramid, so, the "rectangle" at the bottom, is really a square, well, a square is a rectangle as much as it's a quadrilateral
like the one in the picture below
now [tex]\bf \textit{volume of a pyramid}\\\\
V=\cfrac{1}{3}Bh\qquad
\begin{cases}
B=\textit{base area}\\
h=height\\
-------\\
B=70\\
V=140
\end{cases}\implies 140=\cfrac{1}{3}\cdot 70h[/tex]
solve for "h"
