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

3 answers

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.

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

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!