Help on Rectangular Solid Problem

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Help on Rectangular Solid Problem

by EMAN » Thu Oct 01, 2009 4:04 pm
OG 122

What is the volume of a certain rectangular solid?

(1) Two adjacent faces of the solid have areas 15 and 24, respectively.

(2) Each of two opposite faces of the solid has area 40.


Okay, so if we look at statement one, we know that there can be a number of combinations for dimensions that fit the criteria such as 3 x 5 and 3 x 8 or 1 x 15 and 1 x 24. Also, we do not necessarily know if the two opposite sides (top and bottom of the rectangle if you will) has which surface area depending on how it is positioned.

So this part makes sense, but now if you go to statement 2, each of two opposite faces has area of 40. Doesn't a rectangle only have two identical opposite sides and four similar identical sides? If we take both statements together, it seems like there are three sides (area of 40, 24 and 15).

Someone please let me know where my logic fails here.
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EMAN wrote:OG 122

What is the volume of a certain rectangular solid?

(1) Two adjacent faces of the solid have areas 15 and 24, respectively.

(2) Each of two opposite faces of the solid has area 40.


Okay, so if we look at statement one, we know that there can be a number of combinations for dimensions that fit the criteria such as 3 x 5 and 3 x 8 or 1 x 15 and 1 x 24. Also, we do not necessarily know if the two opposite sides (top and bottom of the rectangle if you will) has which surface area depending on how it is positioned.

So this part makes sense, but now if you go to statement 2, each of two opposite faces has area of 40. Doesn't a rectangle only have two identical opposite sides and four similar identical sides? If we take both statements together, it seems like there are three sides (area of 40, 24 and 15).

Someone please let me know where my logic fails here.
No, a rectangular solid has 3 pairs of sides with equal areas. For example, a rectangular solid with sides 1, 2 and 3 has:
- two sides with area 2
- two sides with area 3
- two sides with area 6

Also, in your explanation, you seem to be suggesting that the sides must have integer values, but this isn't stated.

Solution: We can see how statements (1) and (2) alone are not sufficient. Together, they are sufficient.
Let the length of 1 side = x
From here, another side must have length 15/x, since one side has area 15 (so the product of two sides [x and 15/x] must be 15)
Another side must have length x/24 (for a reason similar to above)
Since the last side must have area 40, we know that (x/24)(x/15) must equal 40 . . . (solve for x)




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