Question

A third box is designed so that it is also a cube that has sides that are half of solid 1's. find the volume of solid 3. then find the ratio of the volume of solid 1 to the volume of solid 3. explain why the ratio is not 2 to 1

Answer 1

Given :

Solid 1:

The side of the cube is, a = 8 cm.

Solid 3:

The side of the cube is, b = 4 cm.

The volume of the third cube can be calculate as,

$V=b^3=(4cm)^3=64cm^3$

The volume of first cube is,

$V^{\prime}=a^3=(8cm)^3=512cm^3$

Thus, the ratio can be calculated as,

$\frac{V^{\prime}}{V}=\frac{512}{64}=\frac{8}{1}$

Thus the required ratio is 8:1.

Since the volume is proprtional to the cube of the side, the ratio will also be in the range of cube of the fractional difference between the sides.