00:01
Nautical flags are used to represent letters of alphabets.
00:05
The flag for the letter o consists of a yellow right triangle and a red right triangle join together along their hypotenuse to form a square.
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For our particular flag, the joint hypotenuse of the two triangles is three inches longer than a side of the square.
00:24
We need to find the length of a side of this flag, and we are going to round to the nearest 10th.
00:31
So i have up on the screen an example of what this nautical flag for the letter o may look like.
00:37
You see a yellow right triangle and it is joined together with a red right triangle.
00:44
So according to this problem, what we're looking for is we're looking for the length of a side of this flag.
00:51
So we're looking for this.
00:53
X.
00:54
Now since this flag is in a shape of a square, we know that this side is also an x, this side is also an x, and this side is also an x.
01:06
Right, since squares have all four sides being equal to each other.
01:12
Okay.
01:13
Now let's make use of the information given to us about the hypotenuse of the two triangles.
01:19
We have that the hypotenuse of the two triangles is three inches longer than a side of the square.
01:25
So we know that this hypotenuse is going to be x plus three.
01:30
So three inches longer than a side of the square.
01:36
Okay, how do we solve for x then? well, there is a very famous theorem that we can use.
01:44
The pythagorean theorem.
01:46
Remember, it tells us that if we have a right triangle, then we can take the squares of both of the legs of the right triangle, so a squared and then b squared, add them together, and we should have the square of the hypotenuse, or c squared in this picture.
02:04
For us, we're not going to have a, b, and c.
02:07
Instead, we're going to have x squared.
02:10
That's one of the legs plus x squared equals x plus 3 squared.
02:19
And i highly encourage you to write x plus three in parentheses, because if you write it like this, then you get something entirely different.
02:28
So don't do that.
02:31
Okay.
02:31
And so really we just have to solve this equation right here for x.
02:37
So let's go ahead and get started.
02:39
Well, we have two of the same like terms on the left hand side of the equal sign.
02:45
So i'll add them together.
02:46
We'll have 2x squared.
02:50
And then on the right hand side, again, you want to be a little careful.
02:54
This really tells us that we have to do x plus 3 times x plus 3.
02:59
So we're going to have to do a little bit of foiling.
03:02
So first, outer inner last, we're going to have x squared plus 3x plus 3x plus 3x plus 3x, that would be 9.
03:18
Okay, i do like to try to simplify both sides before i start moving things around.
03:25
So i am going to combine these two like terms together.
03:30
3x plus 3x is 6x.
03:33
And now i feel comfortable with moving things around.
03:37
I have zero on one side of my equation.
03:41
I'm going to actually subtract x squared, subtract 6x, and subtract 9 from both sides of the equation because i do want to try to get a positive coefficient in front of my x squared.
03:57
So doing this would help us get to that.
04:00
2x squared minus 1 x squared is just x squared...