Help me with this math problem! I choose best answer :D

Harry needs to put his 4' long broom in a box. The first box he found was only 36 inches long, but he was able to fit his broom flat on the bottom of it. What's the smallest the width of the box could be?

4' = 48 inches. Assuming this broom fit the box inside the boxes diagonal (which would give the box the smallest width), this would give us two measures. The hypotenuse of a triangle would be 48, the length 36, and the width is the unknown. According to the Pythagorean theorem:

w^2 + 36^2 = 48^2 where w is the width. Solving for w:

w^2 + 1296 = 2304

w^2 = 1008

w = 31.75 inches or 2' 7 and 3/4"

If he places the broom diagonally, it will fit with the least width. We have to assume that the box is rectangular with right angle (90 deg) corners. Diagonal creates two identical right triangles.
Using 4 ft. as the diagonal or hypoteneuse of a right triangle with one of the sides at 36 in. or 3 ft., calculate the other side using c^2 = a^2 + b^2 (Pythagorean theory)
c = 4, a = 3
4^3 = 3^2 + b^2
b^2 = 16 - 9
b^2 = 7
b = +/-sqrt 7
Since width has to be a positive number, only positive sqrt is valid.
b = + sqrt 7 or approximately 2.65 ft or 31.8 in.

31.75 inches. The box is a rectangle, the broom is put in the box by being a diagonal from one corner to the other. Being the diagonal it is the hypotenuse of a triangle. the other side is 36inches so 48^2-36^2=1008, take the square root to get the wifth and it would be 31.75 inches or 32 inches if you are asked to round up.

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There is a diagram showing what it would look like. The broom is the diagonal in the rectangle.

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