Question

What is the ratio of these volumes if the dimensions were length = 4, width = 3, and height = 2? Fill out the table below.

Answer 1

Step-by-step explanation:

Linear escale factor 1.

In this case the original volume of the parallelepiped is:

`Volume=4*2*3=24\text{ u}^3`

24 cubic units and the new volume must be:

`NewVolume=\left(4*1\right)\left(2*1\right)\left(3*1\right)=24u^3`

So in this case the ratio of volumes is 1:1

Linear escale factor 2.

In this case the original volume of the parallelepiped is the same: 24 cubic units.

The new voume must be:

`NewVolume=\left(4*2\right)\left(2*2\right)\left(3*2\right)=8*4*6=192u^3`

So in this case the ratio of volumes is 24:192, that is 1:8

Linear escale factor 3.

In this case the original volume of the parallelepiped is the same: 24 cubic units.

The new voume must be:

`NewVolume=(4*3)(2*3)(3*3)=12*6*9=648u^3`

So in this case the ratio of volumes is 24:648, that is 1:27

Linear escale factor 4.

In this case the original volume of the parallelepiped is the same: 24 cubic units.

The new voume must be:

`NewVolume=(4*4)(2*4)(3*4)=16*8*12=1536u^3`

So in this case the ratio of volumes is 24:1536, that is 1:64

Linear escale factor r.

In this case the original volume of the parallelepiped is the same: 24 cubic units.

The new voume must be:

`NewVolume=(4*r)(2*r)(3*r)=4r*2r*3r=24r^3\text{ }u^3`

So in this case the ratio of volumes is 24:24r^3, that is:

`1:r^3`