If the box's dimensions are x,y,z, then we have
xy = 316
xz = 168
yz = 120
Hmmm. Really? I suspect a typo, since the dimensions are not integers.
316 = 2^2 79
168 = 2^3 3 7
120 = 2^3 3 5
That 79 is a problem.
The area of the bottom of a rectangular box is 316cm squared the area of one side is 168cm squared and the area of the other is 120cm squared. What are the dimensions of the box?
2 answers
Let h=height of the box.
Then
width=120/h
length=168/h
We know the area of the base is
Ab=316
so Ab=width*length, or
316=(120/h)*(168/h)=(120*168)/h².
Solve for h
h=√(316/(120*168)
=12√35/√79.
Use the above equations above to find length and width.
Then
width=120/h
length=168/h
We know the area of the base is
Ab=316
so Ab=width*length, or
316=(120/h)*(168/h)=(120*168)/h².
Solve for h
h=√(316/(120*168)
=12√35/√79.
Use the above equations above to find length and width.