Question

can the 90 - cubic - foot box fit through the locker door if the door is 6 feet wide and 6 feet tall? why or why not?

Explanation:

Step1: Find the area of the door

The area of a rectangle (door) is $A = w\times h$, where $w = 6$ feet and $h = 6$ feet. So $A=6\times6 = 36$ square - feet. But we need to consider the diagonal of the door to see the maximum length an object can pass through. Using the Pythagorean theorem $d=\sqrt{w^{2}+h^{2}}$, where $w = 6$ and $h = 6$. So $d=\sqrt{6^{2}+6^{2}}=\sqrt{36 + 36}=\sqrt{72}\approx8.49$ feet.

Step2: Assume the box is a cube (for worst - case scenario)

If the box has a volume $V = 90$ cubic - feet and assume it is a cube, then the side length $s$ of the cube is $s=\sqrt[3]{V}$. So $s=\sqrt[3]{90}\approx4.48$ feet.
Since the diagonal of the door ($\approx8.49$ feet) is greater than the side - length of the box (assuming cube, $\approx4.48$ feet), the box can fit through the door.

Answer:

Yes, because the diagonal of the 6 - feet by 6 - feet door is approximately 8.49 feet, and assuming the box is a cube with a volume of 90 cubic feet, its side - length is approximately 4.48 feet, which is less than the diagonal of the door.