Can you please help with this geometry question.

An open box is formed from a square piece of material by cutting 2-in squares from each corner of the material and folding up the sides. If the volume of the box that is formed is to be 72 in cubed, what was the original piece of material?

Thank you for your continued help. Always excellent.

let original piece of material be x by x inches

after cut-out, the base will be x-4 by x-4 and the height will be 2

so 2(x-4)^2 = 72
(x-4)^2 = 36
take √ of both sides
x - 4 = 6
x = 10

the original piece was 10 by 10 inches

Thank you for the kind words! I'd be happy to help you with this geometry question.

To solve this problem, we need to determine the original dimensions of the square piece of material before the corners were cut. Let's call the side length of this square piece of material x inches.

When we cut 2-inch squares from each corner, the dimensions of the folded-up sides will be (x-4) inches by (x-4) inches. The height of the box will be 2 inches.

To find the volume of the box, we need to multiply the length, width, and height. In this case, the volume is given as 72 cubic inches.

Therefore, we have the equation: (x-4) * (x-4) * 2 = 72

To solve this equation, we can divide both sides by 2 and simplify:

(x-4) * (x-4) = 36

Next, we can expand the equation using FOIL (First, Outer, Inner, Last):

x^2 - 8x + 16 = 36

Now, subtract 36 from both sides:

x^2 - 8x = 20

To solve this quadratic equation, we can rearrange it to follow the form: ax^2 + bx + c = 0, where a = 1, b = -8, and c = -20.

Using factoring, completing the square, or the quadratic formula, we find that the possible solutions for x are x = 10 or x = -2 (but since we cannot have negative side lengths, we discard x = -2).

Therefore, the side length of the original piece of material is 10 inches.

In conclusion, the original piece of material was a square with side length 10 inches.