By how much is the volume of a square-based pyramid increased when all the dimensions are doubled?
To find the increase in volume when all the dimensions of a square-based pyramid are doubled, we need to compare the original volume to the new volume.
Let's consider the original square-based pyramid:
- Base length = x
- Height = h
The original volume of the pyramid is given by: V1 = (1/3) * x^2 * h
Now, when we double all the dimensions, the new pyramid will have:
- Base length = 2x
- Height = 2h
The new volume of the pyramid is given by: V2 = (1/3) * (2x)^2 * (2h)
Simplifying, we get:
V2 = (1/3) * 4x^2 * 2h
V2 = (8/3) * x^2 * h
To calculate the increase in volume, we subtract the original volume from the new volume:
Increase in volume = V2 - V1
Increase in volume = (8/3) * x^2 * h - (1/3) * x^2 * h
Increase in volume = (8/3 - 1/3) * x^2 * h
Increase in volume = (7/3) * x^2 * h
Therefore, the volume of a square-based pyramid is increased by a factor of 7/3 (or approximately 2.33) when all the dimensions are doubled.