Check the picture below.
[tex]\bf \textit{volume of a pyramid}\\\\ V=\cfrac{1}{3}Bh~~ \begin{cases} B=area~of\\ \qquad its~base\\ h=height\\[-0.5em] \hrulefill\\ B=\stackrel{\textit{triangular base}}{\frac{1}{2}(14)(18)}\\[1em] h=30 \end{cases}\implies V=\cfrac{1}{3}\left( \cfrac{1}{2}(14)(18) \right)(30) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill V=1260~\hfill[/tex]
