Question

talia is packing a moving box. she has a square - framed poster with an area of 9 square feet. the cube - shaped box has a volume of 33 cubic feet. will the poster lie flat in the box? explain. an area of 9 square feet means the square poster has dimensions ft x . if the poster were the side of a cube, that cube would have a volume of 27 ft³. the box has a smaller volume, so its sides must be smaller than the sides of the poster, meaning that the poster would lie flat in the bottom of the box.

Explanation:

Step1: Find side - length of the square poster

For a square with area $A = s^{2}$, where $A = 9$ square feet. Solving for $s$, we have $s=\sqrt{A}$. So, $s=\sqrt{9}=3$ feet.

Step2: Find side - length of the cube if the poster was its side

If the side of the cube is the same as the side of the square poster ($s = 3$ feet), the volume of the cube $V_{1}=s^{3}=3^{3}=27$ cubic feet.

Step3: Compare volumes

The volume of the moving box is $V_{2} = 33$ cubic feet. Since $V_{2}>V_{1}$, and the side - length of the poster is 3 feet, and we know that the box has a larger volume, the sides of the box must be larger than the sides of the poster.

Answer:

Yes, the poster will lie flat in the box because the volume of the box (33 cubic feet) is larger than the volume of a cube with side - length equal to the side - length of the square poster (27 cubic feet), which implies that the sides of the box are larger than the sides of the poster.