Question

talia is packing a moving box. she has a square - framed poster with an area of 16 square feet. the cube - shaped box has a volume of 68 cubic feet. will the poster lie flat in the box? explain? an area of 16 square feet means the square poster has dimensions 4 ft×4 ft. if the poster were the side of a cube, that cube would have a volume of 64 ft³. the box has a larger 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

The area of the square - framed poster is $A = 16$ square feet. For a square, $A=s^{2}$, where $s$ is the side - length. Solving $s^{2}=16$ for $s$, we get $s = 4$ feet (since $s>0$).

Step2: Analyze the box's side - length

The volume of the cube - shaped box is $V = 68$ cubic feet. For a cube, $V = a^{3}$, where $a$ is the side - length of the cube. We need to find $a$. Taking the cube - root of both sides, $a=\sqrt[3]{68}$. Since $4^{3}=64$ and $5^{3}=125$, and $64<68<125$, we have $4 <\sqrt[3]{68}<5$.

Step3: Determine if the poster fits

Since the side - length of the box $\sqrt[3]{68}$ is greater than the side - length of the poster ($4$ feet), the poster will lie flat in the bottom of the box.

Answer:

Yes, the poster will lie flat in the box because the side - length of the cube - shaped box ($\sqrt[3]{68}\approx4.08$ feet) is greater than the side - length of the square poster ($4$ feet).