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:
-
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).
- This point would be at (12, 0, 35), where:
-
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).
- This point would be at (0, 16, 0), where:
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:
- \(x_2 - x_1 = 0 - 12 = -12\)
- \(y_2 - y_1 = 16 - 0 = 16\)
- \(z_2 - z_1 = 0 - 35 = -35\)
Now substituting these values back into the distance formula:
\[ d = \sqrt{(-12)^2 + (16)^2 + (-35)^2} \]
Calculating each term:
- \((-12)^2 = 144\)
- \(16^2 = 256\)
- \((-35)^2 = 1225\)
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.