[tex]\bf \textit{volume of a pyramid}\\\\
V=\cfrac{1}{3}Bh\qquad
\begin{cases}
B=base\\
h=height\\
----------\\
B=width\cdot length\\
length=3\qquad width=3\\
B=3\cdot 3=9\\
h=5
\end{cases}\\\\
-----------------------------\\\\
\textit{what if we increase the slant height by 4 and height by 2?}
\\\\\\
V=\cfrac{1}{3}Bh\qquad
\begin{cases}
B=width\cdot length\\
length=3\qquad width=3\\
B=3\cdot 3=9\\
h=7
\end{cases}[/tex]
the slant-height plays no role on that equation to get the volume, only the height does, and the Base, so the slant-height going from 7 to 11, has no bearing on the volume, since we know the height
[tex]\bf V=\cfrac{1}{3}\cdot 9\cdot 5\implies V=3\cdot 5\implies V=15
\\\\\\
V=\cfrac{1}{3}\cdot 9\cdot 7\implies V=3\cdot 7\implies V=21[/tex]
so hmmm, it was 15, then it went up to 21, 21-15 = 6, went up by 6 units
