Question

use what youve learned

1. find the perimeter of the following image.
2. find the perimeter of the following image.
3. a farmer wants to place a cube - shaped box of supplies on the floor of her barn. what would be the total distance around the bottom of the box if it has a volume of 16x^4 units^3?

Explanation:

Step1: Analyze the first image

The first image seems to be a rectangle - like figure composed of smaller areas. However, without clear side - length information, assume the figure is made up of congruent rectangles with area \(A = 24\mathrm{ft}^2\). Let's assume the figure has a length \(l\) and width \(w\). If we consider the arrangement, we need to find the outer - side lengths. But since we don't have enough information about the individual side lengths from the area alone, we assume a more straightforward case where the figure is a simple rectangle. Let's assume the figure is composed of 4 rectangles along the length and 2 rectangles along the width. If the area of each small rectangle is \(A = 24\mathrm{ft}^2\), and assume the small rectangles are arranged in a \(4\times2\) grid. Let the length of the large rectangle be \(l\) and width be \(w\). If we assume the small rectangles have side lengths \(a\) and \(b\) such that \(ab = 24\). Let's assume the large rectangle has length \(l = 4a\) and width \(w = 2b\). The perimeter \(P=2(l + w)=2(4a + 2b)\). But if we assume the small rectangles are squares (for simplicity, since no other information is given), then \(a=b=\sqrt{24}\mathrm{ft}\). The large rectangle has length \(l = 4\sqrt{24}\mathrm{ft}\) and width \(w = 2\sqrt{24}\mathrm{ft}\).
\[P = 2(4\sqrt{24}+2\sqrt{24})=2\times6\sqrt{24}=12\sqrt{24}=12\times2\sqrt{6}=24\sqrt{6}\mathrm{ft}\]

Step2: Analyze the second image

The second image shows a square with area \(A = 48x^{5}\). For a square, if the side - length is \(s\), then \(A=s^{2}\). So, \(s=\sqrt{48x^{5}}=\sqrt{16\times3\times x^{4}\times x}=4x^{2}\sqrt{3x}\). The perimeter of a square \(P = 4s\).
\[P=4\times4x^{2}\sqrt{3x}=16x^{2}\sqrt{3x}\]

Step3: Analyze the third problem

The volume of a cube \(V = s^{3}\), where \(s\) is the side - length of the cube. Given \(V = 16x^{4}\), then \(s=\sqrt[3]{16x^{4}}=\sqrt[3]{8\times2\times x^{3}\times x}=2x\sqrt[3]{2x}\). The total distance around the bottom of the box (which is a square) is the perimeter of the square base. The perimeter of a square \(P = 4s\).
\[P = 4\times2x\sqrt[3]{2x}=8x\sqrt[3]{2x}\]

Answer:

1. \(24\sqrt{6}\mathrm{ft}\)
2. \(16x^{2}\sqrt{3x}\)
3. \(8x\sqrt[3]{2x}\)