The volume [tex]V[/tex] is given by
[tex]V=\frac{L^2 H}{3}[/tex]
https://en.wikipedia.org/wiki/Square_py ... e_pyramidsTo solve for [tex]H[/tex], we need to do identical operations on both sides of the equal sign that eventually leaves [tex]H[/tex] only on one side.
[tex]V=\frac{L^2 H}{3}\quad[/tex] (Starting formula. First we'll try and get rid of 3 on the right hand side.)
[tex]3V = \frac{3 L^2 H}{3}\quad[/tex] (Multiply both sides by 3)
[tex]3V = L^2 H\quad[/tex] (The two 3s on the right hand side cancel each other out. Next we'll try and get rid of the [tex]L^2[/tex].)
[tex]\frac{3V}{L^2} = \frac{L^2 H}{L^2}\quad[/tex] (Divide both sides by [tex]L^2[/tex])
[tex]\frac{3V}{L^2} = H\quad[/tex] (The two [tex]L^2[/tex] terms on the right hand side cancel each other out)
So [tex]H = \frac{3V}{L^2}[/tex].
For part (b) substitute [tex]V=144[/tex] and [tex]L=12[/tex] into the above formula to get [tex]H=3[/tex] meters.
Hope this helped,
R. Baber.