Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Pythagorean theorem and a right triangle. The sum of the squares of a side of a right triangle equal the sum of the hypotenuse. So if we thought about the sides being A&B, the right angle is what attaches the two sides, and the hypotenuse is always opposite the right angle. So a ^2 + b ^2 = C ^2. If we wanted to think about solving for C, we could actually think of the distance of C being sqrt a ^2 + b ^2. So we want to figure out what the unknown side is. In this first example, we have 3 ^2. We don't know our b ^2 and we know our C is 5 S a ^2 + b ^2 = C ^2. So B is going to equal sqrt 25 - 9 and that's 16 and we know sqrt 16 is just four so side B is going to be 4. This next one, the two sides are actually the same length and this is called isosceles, and this is an isosceles right triangle because the two sides are the same and there is a right angle. So a ^2 + b ^2 = C ^2. We don't know our C so we get 9 + 9 or 18. Sqrt 18 equals our C18 we could think of as 9 * 2 and sqrt 9 is 3. So we get C equalling 3 square roots of two. The next example, we're going to have root 3 and we're going to square it plus 2 ^2 equaling our C ^2 because we know our A and our B square root and the square are going to cancel. So we get 3 + 4 equaling C ^2 or C is just sqrt 7. We can't go any further with that one. This next one, a ^2 plus we don't know our B this time, so we're going to have b ^2 equaling sqrt 5 ^2. Sqrt 5 ^2 is just going to give us 5, so we're going to get b ^2. If we subtract the one, we get b = sqrt 4 or just two. So now there's a couple for you to try. When we look at the distance formula, the distance formula is based on two points. Let's call it X1 and Y1 and X2 and Y2. And the distance formula is actually going to be based off of the Pythagorean theorem. We're going to draw a right triangle. When I draw this right triangle, we can now figure out the distance here. What the distance here is just going to be the change of the heights. I don't know why I wrote a two that should have been a 1X1Y1. The little subscript numbers just tell us that that pair goes together. X2Y2 tells us the X and the Y go together. So this height was Y 2 -, y one, and this length here the furthest to the right minus the furthest to the left in a number line we always would think about subtracting. So 9 -, 5 would give us a distance of four. So when we look at this, we really want to find the distance here. So the distance squared is just going to equal the sides squared a ^2 + b ^2 = C ^2. Well, if we want to actually solve for the distance, we just say D equal the square root of X2 minus X 1 ^2 + y two minus Y 1 ^2. Truly we get it straight off of the Pythagorean theorem. So when we look at these next two examples, and it doesn't matter which point we put in first, I'm going to show that also. So the distance equals 4 - 8 ^2 + 7 - 10 ^2. So 4 - 8 would give me -4 ^2 7 - 10, negative 3 ^2. So we'd get 16 + 9 ^2 of 25, or a distance of five. Now, if we decided we wanted to put the other point first, we should see that we get the same answer. We have to remember to pair the XS and pair the YS, but other than that, the order doesn't matter. 8 - 4 ^2 + 10 - 7 ^2. Well, 8 - 4 is 4 and 10 - 7 is 3. So 16 + 9 ^2 of 25, which is five. Again, the reason it doesn't matter is whatever number is on the inside, when I subtract, we're going to square it. So if it was a negative and we square it, we get a positive. But if it was a positive and squared, we get a positive. Our next example, we're going to have 6 - 4 ^2. Our X is paired together. Plus -4 - -2 ^2 6 - 4 is 2 ^2 -4 -, a negative two. That would give me a -2 ^2. So we get 4 + 4 sqrt 8. But remember sqrt 8? We could think of a sqrt 4 * 2 or two square roots of two. And then there are a couple for you to try. Thank you and have a wonderful day.