Let's solve the given questions step by step:
a. Find the volume of the pyramid.
The formula for the volume of a pyramid with a square base is:
[tex]\[ V = \frac{1}{3} \cdot \text{base area} \cdot \text{height} \][/tex]
Given that the base side is 10 meters and the height is 15 meters, we can calculate the base area as follows:
[tex]\[ \text{base area} = \text{base side}^2 = 10^2 = 100 \ \text{m}^2 \][/tex]
Now, we can calculate the volume of the pyramid:
[tex]\[ V = \frac{1}{3} \cdot 100 \ \text{m}^2 \cdot 15 \ \text{m} = \frac{1}{3} \cdot 1500 \ \text{m}^3 = 500 \ \text{m}^3 \][/tex]
b. If the top 6m of the pyramid are removed, what is the volume of the remaining frustum?
Firstly, we calculate the volume of the top part of the pyramid that is removed, which is also a smaller pyramid with a height of 6 meters.
This part still has the same proportions as the original pyramid. To find the size of the smaller base side, we apply similar triangles to obtain the ratio between the new height and the original height:
[tex]\[ \frac{\text{smaller base side}}{10 \ \text{m}} = \frac{6 \ \text{m}}{15 \ \text{m}} \][/tex]
Solving for the size of the smaller base side:
[tex]\[ \text{smaller base side} = 10 \ \text{m} \cdot \frac{6 \ \text{m}}{15 \ \text{m}} = \frac{10 \times 6}{15} \ \text{m} = \frac{60}{15} \ \text{m} = 4 \ \text{m} \][/tex]
Now we can calculate the base area of the smaller pyramid:
[tex]\[ \text{smaller base area} = \text{smaller base side}^2 = 4^2 = 16 \ \text{m}^2 \][/tex]
Using the volume formula for the smaller pyramid, we get:
[tex]\[ \text{smaller pyramid volume} = \frac{1}{3} \cdot 16 \ \text{m}^2 \cdot 6 \ \text{m} = \frac{1}{3} \cdot 96 \ \text{m}^3 = 32 \ \text{m}^3 \][/tex]
To find the volume of the remaining frustum, we need to subtract the volume of the smaller (top) pyramid from the volume of the original pyramid:
[tex]\[ \text{frustum volume} = \text{pyramid volume} - \text{smaller pyramid volume} \][/tex]
Substitute the values we've calculated:
[tex]\[ \text{frustum volume} = 500 \ \text{m}^3 - 32 \ \text{m}^3 = 468 \ \text{m}^3 \][/tex]
So the volume of the frustum, after removing the top 6 meters of the pyramid, is 468 cubic meters.
