Q14. - Geometry

Question

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

$\text{L.A.}=\_\_\_$

Step-by-Step Solution

Answer

The lateral area of the pyramid$V-XYZ$ is equal to$27\sqrt{3}{\text{\hspace{0.17em}\hspace{0.17em}cm}}^2$$27\sqrt{3}{\text{\hspace{0.17em}\hspace{0.17em}cm}}^2$ .

Step 1. Given information.

The given$V-XYZ$ pyramid:

Each edge of pyramid is equal to 6.

Step 2. Write the concept.

The lateral area of the pyramid is the product of number of faces and the area of each face.

$\text{L.A.}=n\times B$

 

The area of each face is equal to the area of the base.

 

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 Lateral Area.

Substitute 6 for a in the formula for area.

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

 

Substitute$9\sqrt{3}$ for b and 3 for in n the formula for lateral area.

$\begin{array}{c}\text{L.A.}=n\left(\text{B}\right)\\ =3\left(9\sqrt{3}\right)\\ =27\sqrt{3}\end{array}$

 $\begin{array}{c}\text{L.A.}=n\left(\text{B}\right)\\ =3\left(9\sqrt{3}\right)\\ =27\sqrt{3}\end{array}$

Therefore, the lateral area of the pyramid$V-XYZ$ is equal to $27\sqrt{3}{\text{\hspace{0.17em}\hspace{0.17em}cm}}^2$.