Question

Suppose a spider was able to create one thread that would extend from the top-right back corner of its tank to the bottom-left front corner. Approximately how long would that thread need to be for the tank below? The path of the spider's thread is shown with a red dotted line. Round your answer to the nearest tenth of an inch. Show your work. The length is 35 in, 16 in high and 12 in in width

Answer 1

To find the length of the thread that extends from the top-right back corner of the tank to the bottom-left front corner, we can use the 3D distance formula derived from the Pythagorean theorem.

Given the dimensions of the tank:

• Length (width) = 12 inches
• Height = 16 inches
• Depth = 35 inches

The coordinates of the two corners we are considering can be defined as follows:

1. Top-right back corner (let's call this point A):

   • This point would be at (12, 0, 35), where:
     • 12 is the width (x-coordinate),
     • 0 is the height (y-coordinate),
     • 35 is the depth (z-coordinate).

2. Bottom-left front corner (let's call this point B):

   • This point would be at (0, 16, 0), where:
     • 0 is the width (x-coordinate),
     • 16 is the height (y-coordinate),
     • 0 is the depth (z-coordinate).

Now that we have the coordinates, we can use the 3D distance formula:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

Substituting in our values:

• \((x_1, y_1, z_1) = (12, 0, 35)\)
• \((x_2, y_2, z_2) = (0, 16, 0)\)

Calculating each part:

1. \(x_2 - x_1 = 0 - 12 = -12\)
2. \(y_2 - y_1 = 16 - 0 = 16\)
3. \(z_2 - z_1 = 0 - 35 = -35\)

Now substituting these values back into the distance formula:

\[ d = \sqrt{(-12)^2 + (16)^2 + (-35)^2} \]

Calculating each term:

1. \((-12)^2 = 144\)
2. \(16^2 = 256\)
3. \((-35)^2 = 1225\)

Now summing those results:

\[ d = \sqrt{144 + 256 + 1225} = \sqrt{1625} \]

Calculating the square root:

\[ d \approx 40.31 \]

Rounding to the nearest tenth:

\[ d \approx 40.3 \]

Therefore, the length of the spider's thread would need to be approximately 40.3 inches.

Answer 2

you didn't really need most of that work. You had the dimensions of the tank, so you just needed

Length (width) = 12 inches
Height = 16 inches
Depth = 35 inches
diagonal is d = √(144+256+1225) = √1625 = 40.3

Answer 3

You're correct! I appreciate your feedback and understand that the solution can be simplified significantly by directly using the dimensions.

To summarize succinctly:

1. Given the tank dimensions:

   • Length (width) = 12 inches
   • Height = 16 inches
   • Depth = 35 inches

2. We calculate the diagonal length (d) using: \[ d = \sqrt{(width)^2 + (height)^2 + (depth)^2} \]

   Substituting the given values: \[ d = \sqrt{(12)^2 + (16)^2 + (35)^2} \] \[ d = \sqrt{144 + 256 + 1225} = \sqrt{1625} \] \[ d \approx 40.3 \text{ inches} \]

So indeed, the length of the spider's thread should be approximately 40.3 inches. Thank you for pointing out the more straightforward approach!