Asked by jake really needs help
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?
Answers
Answered by
Steve
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.
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.
Answered by
MathMate
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.
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