Question

3D Pythagorean Theorem: If you want the diagonal, d, to be least 41.5 inches, what would be a possible length, width, and height for the box? (20 inches, 21 inches, 26 inches; 20 inches, 20 inches, 27 inches; 21, inches, 24 inches, 26 inches; 23 inches, 24 inches, 25 inches)

Answer 1

To determine whether the diagonal \(d\) of a rectangular box meets or exceeds 41.5 inches using the 3D Pythagorean theorem, we can use the formula:

\[
d = \sqrt{l^2 + w^2 + h^2}
\]

where \(l\), \(w\), and \(h\) are the length, width, and height of the box.

Now, let's calculate the diagonal for each of the provided sets of dimensions.

1. **20 inches, 21 inches, 26 inches:**
\[
d = \sqrt{20^2 + 21^2 + 26^2} = \sqrt{400 + 441 + 676} = \sqrt{1517} \approx 38.93 \text{ inches}
\]

2. **20 inches, 20 inches, 27 inches:**
\[
d = \sqrt{20^2 + 20^2 + 27^2} = \sqrt{400 + 400 + 729} = \sqrt{1529} \approx 39.11 \text{ inches}
\]

3. **21 inches, 24 inches, 26 inches:**
\[
d = \sqrt{21^2 + 24^2 + 26^2} = \sqrt{441 + 576 + 676} = \sqrt{1693} \approx 41.15 \text{ inches}
\]

4. **23 inches, 24 inches, 25 inches:**
\[
d = \sqrt{23^2 + 24^2 + 25^2} = \sqrt{529 + 576 + 625} = \sqrt{1730} \approx 41.60 \text{ inches}
\]

Now, comparing these calculated values of \(d\) with the desired minimum diagonal length of 41.5 inches:

- Setup 1 (20, 21, 26) gives \(d \approx 38.93\) inches (does not meet)
- Setup 2 (20, 20, 27) gives \(d \approx 39.11\) inches (does not meet)
- Setup 3 (21, 24, 26) gives \(d \approx 41.15\) inches (meets the requirement)
- Setup 4 (23, 24, 25) gives \(d \approx 41.60\) inches (meets the requirement)

Thus, the possible dimensions for which the diagonal is at least \(41.5\) inches are:

- **21 inches, 24 inches, 26 inches**
- **23 inches, 24 inches, 25 inches**