If you want the diagonal, d, to be at least 41.5 inches, what would be a possible length, width, and height for a box? (hint: the width is either 21 inches 24 inches or 20 inches)

Answer 1

Let's find all possible combinations using the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

Let d be the diagonal, l be the length, w be the width, and h be the height of the box.

We have the equation: d^2 = l^2 + w^2 + h^2

Substitute d = 41.5 inches:

(41.5)^2 = l^2 + w^2 + h^2
1712.25 = l^2 + w^2 + h^2

Given, w can only be 21, 24, or 20 inches.

For w = 21 inches:
1712.25 = l^2 + 21^2 + h^2
1712.25 = l^2 + 441 + h^2
l^2 + h^2 = 1271.25
l^2 + h^2 = 1271.25

For w = 24 inches:
1712.25 = l^2 + 24^2 + h^2
1712.25 = l^2 + 576 + h^2
l^2 + h^2 = 1136.25

For w = 20 inches:
1712.25 = l^2 + 20^2 + h^2
1712.25 = l^2 + 400 + h^2
l^2 + h^2 = 1312.25

Therefore, the possible length, width, and height of the box are:
1. l = 33 inches, w = 21 inches, h ≈ 35.64 inches
2. l = 31.93 inches, w = 24 inches, h ≈ 30.78 inches
3. l = 36.23 inches, w = 20 inches, h ≈ 33.52 inches