Answer:
The volume of pyramid B is 81 cubic units
Step-by-step explanation:
Given
Pyramid A
[tex]s = 4[/tex] -- base sides
[tex]V = 36[/tex] -- Volume
Pyramid B
[tex]s = 6[/tex] --- base sides
Required
Determine the volume of pyramid B [Missing from the question]
From the question, we understand that both pyramids are equilateral triangular pyramids.
The volume is calculated as:
[tex]V = \frac{1}{3} * B * h[/tex]
Where B represents the area of the base equilateral triangle, and it is calculated as:
[tex]B = \frac{1}{2} * s^2 * sin(60)[/tex]
Where s represents the side lengths
First, we calculate the height of pyramid A
For Pyramid A, the base area is:
[tex]B = \frac{1}{2} * s^2 * sin(60)[/tex]
[tex]B = \frac{1}{2} * 4^2 * \frac{\sqrt 3}{2}[/tex]
[tex]B = \frac{1}{2} * 16 * \frac{\sqrt 3}{2}[/tex]
[tex]B = 4\sqrt 3[/tex]
The height is calculated from:
[tex]V = \frac{1}{3} * B * h[/tex]
This gives:
[tex]36 = \frac{1}{3} * 4\sqrt 3 * h[/tex]
Make h the subject
[tex]h = \frac{3 * 36}{4\sqrt 3}[/tex]
[tex]h = \frac{3 * 9}{\sqrt 3}[/tex]
[tex]h = \frac{27}{\sqrt 3}[/tex]
To calculate the volume of pyramid B, we make use of:
[tex]V = \frac{1}{3} * B * h[/tex]
Since the heights of both pyramids are the same, we can make use of:
[tex]h = \frac{27}{\sqrt 3}[/tex]
The base area B, is then calculated as:
[tex]B = \frac{1}{2} * s^2 * sin(60)[/tex]
Where
[tex]s = 6[/tex]
So:
[tex]B = \frac{1}{2} * 6^2 * sin(60)[/tex]
[tex]B = \frac{1}{2} * 36 * \frac{\sqrt 3}{2}[/tex]
[tex]B = 9\sqrt 3[/tex]
So:
[tex]V = \frac{1}{3} * B * h[/tex]
Where
[tex]B = 9\sqrt 3[/tex] and [tex]h = \frac{27}{\sqrt 3}[/tex]
[tex]V = \frac{1}{3} * 9\sqrt 3 * \frac{27}{\sqrt 3}[/tex]
[tex]V = \frac{1}{3} * 9 * 27[/tex]
[tex]V = 81[/tex]
