91Ó°ÊÓ

The given$V−XYZ$pyramid:

Each edge of pyramid is equal to 6.

The formula for the volume of pyramid is the one third of the product of base area and height of pyramid.

$\text{V}=\frac{1}{3}\text{Bh}$

The base of the pyramid is equilateral triangle with edge of triangle is equal to 6 cm.

The formula for the area of the equilateral triangle is,

$\text{B}=\frac{\sqrt{3}}{4}{a}^2$

Substitute 6 for a in the formula for area.

$\begin{array}{c}\text{B}=\frac{\sqrt{3}}{4}{a}^2\\ =\frac{\sqrt{3}}{4}{\left(6\right)}^2\\ =\frac{\sqrt{3}}{4}\left(36\right)\\ =9\sqrt{3}\end{array}$

Use the Pythagorean theorem in $\mathrm{Δ}VMZ$for the slant height l of the pyramid.

$\begin{array}{c}{\left(VZ\right)}^2={\left(VM\right)}^2+{\left(MZ\right)}^2\\ 6^2=l^2+3^2\\ l=\sqrt{36−9}\\ l=3\sqrt{3}\end{array}$

Use the Pythagorean theorem in $\mathrm{Δ}VOM$for the height h of the pyramid.

$\begin{array}{c}{\left(VM\right)}^2={\left(VO\right)}^2+{\left(OM\right)}^2\\ {\left(3\sqrt{3}\right)}^2={\text{h}}^2+{\left(\sqrt{3}\right)}^2\\ \text{h}=\sqrt{27−3}\\ \text{h}=2\sqrt{6}\end{array}$

Substitute$2\sqrt{6}$ for h and$9\sqrt{3}$ for B in the formula for volume of pyramid.

$\begin{array}{c}\text{V}=\frac{1}{3}\text{Bh}\\ =\frac{1}{3}\left(9\sqrt{3}\right)\left(2\sqrt{6}\right)\\ =3\sqrt{3}×2\sqrt{6}\\ =18\sqrt{2}\end{array}$

Therefore, the volume of the pyramid $V−XYZ$is equal to$18\sqrt{2}$ cm³.