 In this video, Practice Problem 1, we're asked to find X given the information that the long leg of the triangle is tangent to the circle. Remember, we can't ever assume that a segment is tangent by appearance only. We have to be either told as we are in this problem that it's tangent or given the right angle symbol. We're told that it's tangent, so I know that I do have a right triangle here. And anytime we have a right triangle, we can solve missing pieces using information we learned in chapter 8, either trig function, special right triangles, or Pythagorean theorem. In this right triangle, we don't know any information about the angle measures besides the 90 degree angle. And so when we only have information about sides of a right triangle, we're going to be using the Pythagorean theorem. One missing piece of the triangle, though, is right here. We know that this measure is 8, but we don't know anything about this missing piece here. And that's where you have to realize that if this measure is X, this is a radius of the circle, and any radius in that circle is also going to be X, which includes this missing piece. So if I'm told the radius here is X, this measure is going to be the same, and I'm going to mark that as X. So now we have all the pieces of our right triangle. Short leg X, long leg is 12, and then the hypotenuse, if we add those together, is X plus 8. And we can set up the information for our equation. We know that a leg squared plus another leg squared is going to equal the hypotenuse squared. And remember when we do that, we have to put it in parentheses, and when we square that, we're going to have to foil. We're going to expand that out if that helps remind us to foil. So now we're simplifying the equation. The left side, X squared, 12 squared is 144. And then when I foil this, we're going to get X squared plus 8X plus 8X, and then 8 times 8, 64. When I combine like terms, the left side stays the same. And the right side, 8X plus 8X is 16X plus 64. And now that I've combined like terms on each side, I can go ahead and simplify it. You'll notice that I have X squared on each side, so those are going to cancel out. It looked like we might have had a quadratic equation, but the X squareds cancel out. And so now I'm just going to solve for X by getting the X term on one side and the non-X term on the other side. 144 minus 64 is 80. And then my last step here when I divide is going to be 80 divided by 16. And I get X equals 5. And I'm done because I was asked to find X. I found X. Not a bad idea to just kind of plug things in. If X equals 5, I'll plug these values in here, and I can just double check real quick. This whole hypotenuse side then, 5 plus 8, would be 13. And I'm just going to double check that leg squared plus leg squared, 5 squared plus 12 squared, does that equal the hypotenuse squared? And when I simplify that, 25 plus 144 does equal 169. And I'm done.