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×ft. if the poster were the side of a cube, that cube would have a volume of ft³. the box has a volume, so its sides must be than the sides of the poster, meaning that the poster lie flat in the bottom of the box.

Explanation:

Step1: Find side - length of the square poster

The area formula for a square is $A = s^{2}$, where $A$ is the area and $s$ is the side - length. Given $A = 9$ square feet, we solve for $s$: $s=\sqrt{A}=\sqrt{9}=3$ ft. So the dimensions of the square poster are $3$ ft $\times$ $3$ ft.

Step2: Consider the cube case

If the side of a cube is the same as the side of the poster ($s = 3$ ft), the volume of the cube $V=s^{3}=3^{3}=27$ $ft^{3}$.

Step3: Compare with box volume

The box has a volume of $33$ cubic feet. Since the side - length of the poster is $3$ ft, and the box's volume is $33$ $ft^{3}$, the side - length of the box $a$ (assuming it is a cube - like situation for comparison, although not necessarily a cube) is such that $a^{3}=33$, so $a=\sqrt[3]{33}\approx3.21$ ft. Since $3.21>3$, the sides of the box are larger than the sides of the poster.

Answer:

Yes, the poster will lie flat in the box because the side - length of the box (implied by its volume) is larger than the side - length of the poster.