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diamong [38]
3 years ago
6

Give the digits in the tens place and the tenths place. 12.05

Mathematics
1 answer:
Alina [70]3 years ago
3 0
The digit in the tens place is 1.

The digit in the tenths place is 0.
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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,
omeli [17]

Answer:

The volume of pyramid B is 81 cubic units

Step-by-step explanation:

Given

<u>Pyramid A</u>

s = 4 -- base sides

V = 36 -- Volume

<u>Pyramid B</u>

s = 6 --- base sides

Required

Determine the volume of pyramid B <em>[Missing from the question]</em>

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

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