Thank you for that. For some reason I was thinking that the cube needed to have a whole number on each side which is obviously untrue.
WholeLottaLove wrote:
The base of a rectangular block has an area of 60 square centimeters. Is the block a cube?
(1) The area of the front face of the block is 60 square centimeters.
(2) The area of a side of the block is 60 square centimeters.
I figured the answer was C and I was correct. If you are given the base area of the rectangle and the front face of it then you have the depth of the rectangle, the width and the height, however, just having the base and one side isn't enough to "lock in" the actual shape of the figure. You need to know more. For example, the base could be 4x15 and the height could be 4x15 meaning this figure is a rectangular block, not a cube. On the other hand, the base could be √60x√60 and the height could be the same meaning that this figure is a cube. I am having a bit of difficultly with the algebra though and would like help figuring out how to solve it. For #1 I had:
w*h = 60. Also, we are given that w*d = 60 --> I ended up with hd=d or h = 1 which would lead me to believe that the figure has a height of 1 and thus, is a rectangle but that doesn't make sense. Can someone elaborate?
From the stem we have:
width*depth = 60;
From (1) we have that:
width*height = 60.
Thus:
width*depth = width*height, which gives that
depth = height. Now, if
depth = height = 1 and
width = 60, then the block is NOT a cube but if
depth = height = width = \(\sqrt{60}\), then the block is a cube.
The same logic applies to (2): we get that
width = height.
For (1)+(2) we have that
depth = height and
width = height, therefore
depth = height = width, which means that the block is a cube.
Answer: C.
Hope it helps.