olivia60

[olivia60]

Find the surface area of a box with a length of 8 in., width of 6.5 in. and height of 4 in., by evaluating the polynomial?

 

[Mrspi]

Find the surface area of a box with a length of 8 in., width of 6.5 in. and height of 4 in., by evaluating the polynomial?

I'm GUESSING (and gee, I HATE to do that!) that perhaps this question is an extension of a discussion of a rectangular solid with length x, width y, and height z.

If you draw and label such a "box," you should see that it has six "faces," each of which are rectangles. The top and bottom are identical, the front and back are identical, and the left and right sides are identical.

The area of the top of the box is x * y units, or xy units squared. There are TWO surfaces with that area (top and bottom), so the total area of the top and bottom is 2xy units squared.

The area of the FRONT of the box is x units * z units, or xz units squared. The front and the back BOTH have that same area, so total area of front and back is 2xz units squared.

The area of the left end is y units * z units, or yz units squared. The left and right ends have the same area, so the total for the twp ends is 2yz units squared.

The total surface area of the box, then, is (2xy + 2xz + 2yz) units squared.

I THINK that the polynomial you're expected to use for the surface area is (2xy + 2xz + 2yz) square units. Remember that x, y and z represent the length, width and height of the box.

 

[mmm4444bot]

Ha, ha -- very funny, Denis. You git hit in the head with a puck, again?