Longest Line In a Box

Consider the blue rectangle that creates the top face. Drawing a diagonal (draw it between, say, the vertices to the farthest left and right on the top face), we can find the length through the Pythagorean Theorem. Assume 20 and 12 are the base and length of the top face. The length of this diagonal, a, is given through the Pythagorean Theorem: $\sqrt{20 ^ 2+12 ^ 2}=a$ .

Now, imagine cutting the entire box in half with the diagonal (make a cut from the farthest left/right on the top face to the bottom face). Now, facing you, would be a rectangle with a base and length of 9 and a, the length of the diagonal.

Now, you can create another diagonal in this square. If you have imagined this correctly, it should be fairly obvious and pretty intuitive why this is the longest line in the box. We can once again use the 2-d Pythagorean theorem to get

$\sqrt{9 ^ 2+a ^ 2}$

$=\sqrt{81+(\sqrt{20 ^ 2+12 ^ 2})^2}$

$=\sqrt{81+400+144}$

$=\sqrt{625}$

$=25=x=\text{length of the longest line in the box!}$

Therefore, the length of the longest line that can be drawn in this box is $\boxed{25}$ !

A visual representation of what we just derived

Longest line $=\sqrt{20^2+12^2+9^2}=\boxed{25}$

Diagonal of a cuboid=(l^2+b^2+h^2)^1/2 =(400+144+81)^1/2=625^1/2=25

We can consider the box as a three dimensional co-ordinate system. The longest line will be the diagonal. The last point of the diagonal will have the co-ordinate (20,12,9). We know that in a three dimensional co-ordinate system, the formula for figuring out a line is, l=\sqrt{x^{2}+y^{2}+z^{2}} So, the answer will be, longest line = \sqrt{20^{2}+12^{2}+9^{2}} = \sqrt{625} =\boxed{25}