Question

The areas of the faces of a rectangular box are 48 m2, 96 m2, and 288 m2.
A second box is cubical and each of its faces has area 16 m2.
Find the ratio of the volume of the first box to the volume of the second box.

Answer 1

let the sides of the first box be
a, b , and c m
so
ab = 48 #1
ac = 96 #2
bc = 288 #3

divide #1 by #2 --> b/c = 1/2 or c = 2b
divide #2 by #3 --> a/b = 1/3 or b = 3a
divide #1 by #3 --> a/c = 1/6 or c = 6a

we know ab = 48
a(3a) = 48
3a^2 = 48
a^2 = 16
a = 4
then b = 12
and c = 24

the volume of the first box must be
4x12x24 or 1152 m^3

the second box must have each side as 4 m
so its volume is 4x4x4 = 64 m^3

so the ratio of theri volumes = 1152:64
= 18:1