00:01
We have a flagpole with a height which i've called h, and it's held up by two wires, each of which has a length 5 feet longer than the flagpole, which i've called h plus 5.
00:09
Additionally, the wires touch the ground h plus 5 units away from each other, with the flagpole right in the middle.
00:15
Now let's solve for h.
00:16
You may notice here that we can break this up into two different right triangles.
00:20
So if you look at the left one, you'll notice that it has sides equal to h, h plus 5, and this third side, which is half of h plus 5, or h plus 5.
00:30
Plus 5 over 2.
00:32
So since this is a right triangle we can use the pythagorean theorem.
00:37
So h squared plus h plus 5 over 2 squared is equal to h plus 5 squared.
00:49
That is a squared plus b squared equals c squared.
00:53
Now let's rearrange this equation because it's a bit foreboding.
00:56
So we have h squared plus now we can square the top and the bottom separately.
01:01
The bottom, the squared, that's easy, that's just four.
01:04
The top we need to use the distributive property, which you may have learned is the foil method.
01:08
This comes out to h squared plus 10h plus 25.
01:13
And this is equal to, once again, distributing h squared plus 10h plus 25.
01:22
Okay, now there's an h squared on both sides...