91Ó°ÊÓ

The length of the side BC is equal to 6 in the given pyramid.

The sideAM is the altitude of the equilateral triangle with side length equal to6 .

Calculate the length of the altitude AM.

$\begin{array}{c}AM=\frac{\sqrt{3}}{2}\left(BC\right)\\ =\frac{\sqrt{3}}{2}\left(6\right)\\ =3\sqrt{3}\end{array}$

The length of the sideOM is always the one third of the length of altitude of the triangle i.e.AM .

$\begin{array}{c}OM=\frac{1}{3}\left(AM\right)\\ =\frac{1}{3}\left(3\sqrt{3}\right)\\ =\sqrt{3}\end{array}$

The length of the sideAO is the subtraction of the length of sideOM from the length of sideAM .

$\begin{array}{c}AO=AM−OM\\ =3\sqrt{3}−\sqrt{3}\\ =2\sqrt{3}\end{array}$

Therefore, the length of the sides AM and AO is equal to$3\sqrt{3}$ and $2\sqrt{3}$, respectively.

The length of the side BC is equal to 6 in the given pyramid.

The length of the side MC is the half of the length of the side BC as M is the midpoint of the side BC.

So, the length of the side MC is equal to 3.

The length of the lateral edges of the pyramid is equal. So, the length of side VC is equal to the length of the side VA.

So, the length of the side VC is equal to 4.

Use the Pythagoras theorem in$\mathrm{Δ}VMC$ for the length of side VM .

$\begin{array}{c}{\left(VC\right)}^2={\left(VM\right)}^2+{\left(MC\right)}^2\\ {\left(VM\right)}^2={\left(4\right)}^2−{\left(3\right)}^2\\ VM=\sqrt{16−9}\\ =\sqrt{7}\end{array}$

So, the length of l is equal to$\sqrt{7}$ .

As calculated in part (a), the value of the length of the sideOM is equal to $\sqrt{3}$.

Use the Pythagoras theorem in$\mathrm{Δ}VOM$ for the length of sideVO .

So, the length of h is equal to 2.

$\begin{array}{c}{\left(VM\right)}^2={\left(VO\right)}^2+{\left(OM\right)}^2\\ {\left(VO\right)}^2={\left(\sqrt{7}\right)}^2−{\left(\sqrt{3}\right)}^2\\ VO=\sqrt{7−3}\\ VO=2\end{array}$

Therefore, the value of h and l is equal to 2 and$\sqrt{7}$ , respectively.