I have a school assignment, and it requires me to list a few of the real world applications of Pythagoras Theorem. However, most of the ones I found are rather generic, and not special at all.
What are some of the real world applications of Pythagoras' Theorem?
There was an older person nearby who used to use it in his job making window frames. He needed them square or rectangular with 90 degree angles. So he'd measure the two sides and a diagonal and if Pythagoras worked he'd got the 90 degree right.
"How far does the second baseman have to throw the ball in order to get the runner out before he slides into home plate?" ...and other neat examples found at: http://www.suite101.com/content/the-pythagorean-theorem-a21010
Here is a true life application of the Pythagorean theorem (the 3-dimensional version, which is a corollary of the 2-dimensional version).
My wife and I needed to have a long iron rod manufactured for us, to use as a curtain rod.
I measured the length $L$ of the rod we wanted.
But we forgot to take into account that we live on the 24th floor of an apartment building and therefore the only way the rod could get into our apartment was by coming up the elevator.
Would the rod fit in the elevator?
My wife measured the height $H$, the width $W$, and the depth $D$ of the elevator box. She then calculated the diagonal of the elevator box by applying the Pythagorean theorem: $\sqrt{H^2 + W^2 + D^2}$. She compared it to $L$, and thankfully, it was greater than $L$. The rod would fit!
I would like to say that we realized this problem BEFORE we asked them to manufacture the rod, but that would be a lie. However, at least my wife realized it before the manufacturers arrived at our apartment building with the completed curtain rod, and she quickly did the measurements, and the Pythagorean Theorem calculation, and the comparison. So PHEW, we were saved.
Say you're playing frisbee with a few of your friends, and the frisbee gets stuck in a tree. You want to get a ladder to reach it, but you don't know how long the ladder needs to be. You can make a point where you want the ladder to touch the ground, and then measure from there, whereas $ds^2=dx^2+dy^2$ is true, so we can say, as an example, that $dx=3 \text{m}$ long and $dy=4\text{m}$ long. Now, we must solve for $ds$. Here's my calculations:
$ds^2=3^2+4^2=9+16=25$
$\sqrt{ds^2}=ds$
$\sqrt{25}=5$
$ds=5\text{m}$
You'd need a ladder that is $5\text{m}$ long to reach the frisbee.
