Question

talia is packing a moving box. she has a square - framed poster with an area of 16 square feet. an area of 16 square feet means the square poster has dimensions (square) ft×(square) ft. if the poster were the side of a cube, that cube would have a volume of (square) ft³. the box has a (square) volume, so its sides must be (square) than the sides of the poster, meaning that the poster (square) lie flat in the bottom of the box. will the poster lie flat in the box? explain.

Explanation:

Step1: Find side - length of square poster

For a square with area $A = s^{2}$, where $A = 16$ square feet. Solving for $s$, we have $s=\sqrt{A}$.
$s=\sqrt{16}=4$ feet. So the dimensions of the square poster are $4$ ft $\times$ 4 ft.

Step2: Find side - length of cube if poster is its side

If the side of the poster is the side of a cube, the volume of a cube $V = s^{3}$, with $s = 4$ feet. Then $V=4^{3}=64$ cubic feet.

Step3: Compare volumes

The box has a volume of 70 cubic feet. Since $70>64$, the poster will lie flat in the box because the side - length of the base of the box (assuming it can fit the side - length of the poster) and the overall volume of the box are large enough.

Answer:

The dimensions of the square poster are 4 ft $\times$ 4 ft. The volume of the cube with side - length equal to the side of the poster is 64 ft³. The box has a volume of 70 ft³, so the poster will lie flat in the box.