Q12. - Geometry

Question

Each edge of pyramid V- XYZ is 6 cm. Find the following numerical answers.

$\text{Volume}=\_\_\_$

Step-by-Step Solution

Answer

The volume of the pyramid $V-XYZ$is equal to$18\sqrt{2}{\text{\hspace{0.17em}\hspace{0.17em}cm}}^3$ .

Step 1. Given information.

The given$V-XYZ$ pyramid:

Each edge of pyramid is equal to 6.

Step 2. Write the concept.

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,

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

Step 3. Determine the volume.

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 $\Delta VMZ$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 $\Delta VOM$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.

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

 

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