Question

talita 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 3 ft×3 ft. 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 larger 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 poster

For a square poster with area $A = 9$ square feet, using the formula $A = s^{2}$ (where $s$ is the side - length of the square), we solve for $s$: $s=\sqrt{A}=\sqrt{9}=3$ feet.

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

If the side of the cube was equal to the side of the poster ($s = 3$ feet), the volume of the cube $V_{1}=s^{3}=3^{3}=27$ cubic feet.

Step3: Compare volumes

The actual volume of the box is $V = 33$ cubic feet. Since $33>27$, the side - length of the box $l$ (where $V = l^{3}$ and $l=\sqrt[3]{33}\approx3.21$ feet) is greater than the side - length of the poster ($3$ feet).

Answer:

Yes, the poster will lie flat in the box because the side - length of the box is greater than the side - length of the square poster.