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Question 233495: Alright so I looked at this problem and thought it was quite simple from the way it was worded on how to solve it. But then I got a little confused by what it said:
"The area of the top of a rectangular box is 324 in.^2, the area of the front of the box is is 135 in.^2, and the area of the end is 60 in.^2. What is the volume of the box?"
Well for me it's been quite a while since I've worked with area and volume so if anyone has a suggestion on how to go about solving this problem or who would know how to do so, please let me know! I have a couple of other problems quite like this one and I'm stumped.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! "The area of the top of a rectangular box is 324 in.^2, the area of the front of the box is is 135 in.^2, and the area of the end is 60 in.^2. What is the volume of the box?"
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Volume = x*y*z
xy = 324
xz = 135
yz = 60
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x = 324/y
x = 135/z
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yz = 60
324/y = 135/z
z = 60/y
324/y = 135y/60
135y^2 = 19440
y^2 = 144
y = 12
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z = 5
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x = 27
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Vol = 12*5*27 = 1620 cu inches
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