for those math people?

Question

A rectangular box has demensions 1.5ft x 2ft X 3ft, what is the length of the longest object that can be put in the box, if the object can be placed in any position?

the answer is not 3, its either 3.6 ft, 3.9 ft., 6.5 ft., or 15.25ft

Answer 1

Use the Pythagorean Theorem. First, figure out the length of the diagonal along the bottom of the box.

Assuming that 1.5 is the height, then you need to figure out the length of the diagonal across the rectangle that measures 2 by 3 feet. The Pythagorean Theorem shows us that:
c^2=a^2+b^2
Let a = 2 and b = 3. Then c = SQRT(13)

Now that you know that, you can use c to define the leg of the right triangle formed by c and the height of the box (1.5).

Applying the Pythagorean Theorem once more, we see that the square of the diagonal across the box is equal to (SQRT(13))^2 + (1.5)^2 = 13 + 2.25 = 15.25.

The square root of 15.25 is about 3.9.

It'd be easier to show with a picture. For now, just imagine the rectangle divided by the diagonal. That diagonal forms another triangle so that you have a new diagonal that goes from the top left corner to the bottom right corner.

Answer 2

It's the length of the hypotenuse of the triangle having sides 2ft & 3ft.
H = sqrt(2^2 + 3^2)

Answer 3

it's 3.9 ft

to describe how the object would be situated it would start at the bottom left front corner and end at the top back right corner if the box were 2ft wide 3ft tall and 1.5 ft deep

Answer 4

everyone is trying to figure it out using the wrong idea. I could put a very long strand of string in the box that would far out do 15 feet. It would depend on the object. Remember she did say ANY position. I just don't feel like doing the math.

Answer 5

3.9 = sqrt (1.5^2 + 2^2 + 3^3)

Distance from lower left hand front corner to upper right hand back corner.