 In this last video for section 10.3 we are going to add one more arcs and chords theorem. So you should have a little bit of space before question number three on your note sheet. It should say arcs and chords theorem number three. I need you to go ahead and write this down. Two chords are congruent if and only if they are equidistant from the center. Okay so that means if two chords are the same distance from the center then those chords are congruent. I might have you look down at this picture real quick and you have this picture under number three on your note sheet. What this theorem says is that for example in this picture EF would have to be congruent to GH as long as the distance from the chord to the center so PQ and PR are congruent. So if this is the same distance as this then these chords have to be congruent. Okay so that's arcs and chords theorem number three. We're going to now use that to help us answer the question in number three. So first statement it does say the chords are equidistant from the center. The very first thing that I would do on my picture is mark that those are congruent. And then in the back of my mind I want to make sure that I understand because of that EF has to be congruent to GH so you might make a note of that. Alright it tells us that the radius of this circle is 15. Now when you look at this picture right now there is nothing drawn on here to show the radius. PQ and PR those are not the radii I guess I would say. If you wanted to draw in the radius you could do that in several places. You could draw PF that's the radius, you could draw PH that's the radius, you could draw PE or PG or you could go straight out here and make a new point. But I want you to understand that right now there was not a radius drawn on my picture so I had to draw it in. The other thing they tell us is that EF is 24. Well remember that the theorem we talked about, I think it's theorem number two says that if the chord is perpendicular to the diameter, which that's what we have right here, we do have a perpendicular angle here, then EQ and EF have to be congruent. Well if that's the case and if EF is 24, well then I know that QF is 12, I also know that EQ is 12 and because EF and GH were congruent that means that GR and HR are also 12. Well that answers one of our questions. One of the things we wanted to find was RH so we just figured that out, RH is 12. Alright the other thing that we want to do here is find PR. Okay so PR is this segment right here. I'm going to go ahead and label that with an A and the reason I'm going to label that with an A is that this is a right triangle where I know the leg is 12 and I know PH is the radius which they told us was 15. I can now use Pythagorean Theorem, I'm going to move my page up here a little bit and solve for PR so A squared plus 12 squared equals 15 squared. Go ahead and solve this for A, A squared plus 144 equals to 25 so A squared equals, what is that, 81 and if we solve we find out that A is 9 and A is representing PR so PR is 9. So a lot going on in this picture, you have to use a couple of the different theorems we talked about so EF is congruent to GH only because they are equidistant from the center also EF and GH or because EF and GH are congruent and because we have 90 degree angles here and here when they tell us that EF is 24 we can cut that in half to get the two sides here that are 12 that means that GR and HR also have to be 12 and then from there we have everything we need to figure out PR. Okay so for this last example we have a blank circle. What we're going to do is we're going to use all the information they give us and we're going to draw our own picture. So we're going to start with chords MO and PR and it says that those are both seven units away from the center so I'm going to put where I think the center should go and just estimating here and we know that this is an S and the reason that we know that this is S is if I continue reading the question it says the radius of circle S so if this is circle S then the center has to be S. Alright so MO and PR are both seven units away from the center so what I'm going to do is I'm going to go like this I'm going to say that that's seven and that's chord MO. I'm going to do the same thing this way. I'm going to say that this is also seven and this is chord PR. So we know that these are equidistant which now remember means that MO has to be congruent to PR because if the chords are equidistant from the center then they are congruent. Alright the circle has a radius of 15 it tells us that so I'm going to go ahead and draw in a radius. Now remember I could do that from S to O or from S to M or to SP or SR those are all radius of the circle. I mean I could even go straight out here and put a new point but if you'll notice what we're trying to eventually find is MO and PQ. Well right now I see MO is this but I don't see a Q anywhere in my picture. Well because this is PR and we're trying to find PQ chances are pretty good this is what they meant for us to put as point Q and because we're trying to find PQ and we know that this is a 90 degree angle when I draw my radius I'm going to put it here and label that as 15 because now I know this is seven this is 15 this is a right triangle so to find PQ I'll just label that A I can use Pythagorean theorem. A squared plus seven squared equals 15 squared. Let's go ahead and simplify this or solve I should say before we simplify. A squared plus 49 equals 225 subtract the 49 we get A squared equals 176 and 176 is not a perfect square so we're going to have to go ahead and simplify root 176. We can simplify into root 16 times root 11 which is 4 root 11 so that means we have found PQ which is one of the things we wanted and that's 4 root 11. Okay so the other thing we wanted to find was MO. Let's think back to the theorems we've talked about. If PQ is 4 root 11 that means that QR is also 4 root 11 and remember that PR is congruent to MO. So what we can do is we can say that PR equals 4 root 11 plus 4 root 11 or you can do 4 root 11 times 2 either way we get 8 root 11 and we know that PR back up here remember MO is congruent to PR so since we found PR is that 8 root 11 that means that MO is also 8 root 11. So what we're going to do now is we're going to