Question

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 1

Answer:

The volume of pyramid B is 81 cubic units

Step-by-step explanation:

Given

Pyramid A

$s = 4$ -- base sides

$V = 36$ -- Volume

Pyramid B

$s = 6$ --- 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:

$V = \frac{1}{3} * B * h$

Where B represents the area of the base equilateral triangle, and it is calculated as:

$B = \frac{1}{2} * s^2 * sin(60)$

Where s represents the side lengths

First, we calculate the height of pyramid A

For Pyramid A, the base area is:

$B = \frac{1}{2} * s^2 * sin(60)$

$B = \frac{1}{2} * 4^2 * \frac{\sqrt 3}{2}$

$B = \frac{1}{2} * 16 * \frac{\sqrt 3}{2}$

$B = 4\sqrt 3$

The height is calculated from:

$V = \frac{1}{3} * B * h$

This gives:

$36 = \frac{1}{3} * 4\sqrt 3 * h$

Make h the subject

$h = \frac{3 * 36}{4\sqrt 3}$

$h = \frac{3 * 9}{\sqrt 3}$

$h = \frac{27}{\sqrt 3}$

To calculate the volume of pyramid B, we make use of:

$V = \frac{1}{3} * B * h$

Since the heights of both pyramids are the same, we can make use of:

$h = \frac{27}{\sqrt 3}$

The base area B, is then calculated as:

$B = \frac{1}{2} * s^2 * sin(60)$

Where

$s = 6$

So:

$B = \frac{1}{2} * 6^2 * sin(60)$

$B = \frac{1}{2} * 36 * \frac{\sqrt 3}{2}$

$B = 9\sqrt 3$

So:

$V = \frac{1}{3} * B * h$

Where

$B = 9\sqrt 3$ and $h = \frac{27}{\sqrt 3}$

$V = \frac{1}{3} * 9\sqrt 3 * \frac{27}{\sqrt 3}$

$V = \frac{1}{3} * 9 * 27$

$V = 81$

Answer 2

Answer:

81

Step-by-step explanation: