Imagine a cable that is suspended between two poles at a distance of \(50\) kilometers
in a straight line without sagging. Note this is in practice not feasible but go along
with it for the sake of this theoretical puzzle. On one pole there is \(1\) meter of cable
that is left over that we now want to include in the overall suspension. The total
length thus becomes \(50\) km and \(1\) meter between the two poles.

Question : How much will the middle point of the cable drop down when a very
large weight at that middle point pulls the cable into two straight lines ?
Can you make an educated guess first before going to the answer ?

The answer might surprise you
because the middlepoint will drop a stunning \(~\approx158\) meters
all that just by prolonging the cable with one meter between the fixed poles !
So make sure the two poles are high enough to avoid a dragging cable.

Here is the mathematics behind it :
One of the two sloping cable halves towards the middlepoint is in fact a hypotenuse
of a rectangular triangle. So we can apply here the theorem of Pythagoras.
The hypotenuse has length \(25\) km and half a meter or \(25000.5\) meters.
The long leg is still half the distance of \(50\) km or \(25\) km or \(25000\) meters.
So the short leg is \(\sqrt{25000.5^2-25000^2}~\) which is \(158.11467\ldots\) meters.