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

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:

$NewVolume=(4\times3)(2\times3)(3\times3)=12*6*9=648u^3$

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:

$NewVolume=(4\times4)(2\times4)(3\times4)=16\times8\times12=1536u^3$

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:

$NewVolume=(4*r)(2*r)(3*r)=4r\times2r\times3r=24r^3\text{ }u^3$

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

$1:r^3$