Question 274404
Let's start with the cube.
Let x = the side (length, width and height) of the cube. The faces of the cube are all squares. The area of one of these squares is {{{x^2}}}. There are 6 faces on a cube so the total surface area is {{{6x^2}}}<br>
The box has the same height as the cube so the height is x. The box's length is 4 more than the length of the cube: x+4. The width of the box is 3 less than the cube's width: x-3. The surface area of a rectangular box is 2lw + 2lh + 2wh. With our length, width and height the surface area of our box is 2(x+4)(x-3) + 2(x+4)x + 2(x-3)x.<br>
We are told that the surface areas of the cube and box are the same so:
{{{6x^2 = 2(x+4)(x-3) + 2(x+4)x + 2(x-3)x}}}
To solve this we start by simplifying:
{{{6x^2 = 2(x^2+x-12) + 2(x^2+4x) + 2(x^2-3x)}}}
{{{6x^2 = 2x^2+x-24 + 2x^2+8x + 2x^2-6x}}}
{{{6x^2 = 6x^2+3x-24}}}
Now we want the variable on just one side so we'll subtract {{{6x^2}}} from each side:
{{{0 = 3x-24}}}
Add 24 to each side:
{{{24 = 3x}}}
Divide both sides by 3:
{{{8 = x}}}
Since x = side of the cube (and the height of the box), this is not our answer. We are asked to find the width, x-3, and length, x+4. So the width is 5 and the length is 12.