Pyramid A has a triangular base where each side measures 4 units and a volume of 36 cubic units. Pyramid B has the same height, but each side of its base is 6 units long.

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]

Answer:

81

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