WEBVTT mathematics/geometry/pyo
00:00:00.000 --> 00:00:02.200
Welcome back to Educator.com.
00:00:02.200 --> 00:00:08.100
For the next lesson, we are going to go over special segments within a circle.
00:00:08.100 --> 00:00:15.300
The first type of segments that we are going to talk about is the chord.
00:00:15.300 --> 00:00:26.600
If you have two chords in a circle (this is AC, and this is chord DE), if you are trying to find the measure of the chords--
00:00:26.600 --> 00:00:33.700
now, we already went over how to find angle measures from the chords; if these two chords intersect,
00:00:33.700 --> 00:00:45.800
then we have interior angles; so then, we found that, to find the interior angle, we would have to take the intercepted arc
00:00:45.800 --> 00:00:50.900
on one side, add it to the intercepted arc on the other side, and then divide it by 2;
00:00:50.900 --> 00:00:59.100
this is a little bit different; we are not looking for angle measures here--we are looking for side measures, segment measures.
00:00:59.100 --> 00:01:17.700
If we have two chords, and we want to find the measures of any segments, it is going to be AB times BC, equals DB times BE.
00:01:17.700 --> 00:01:27.600
You are just taking the product of each of the parts of the chord and making it equal to the product of the parts of the other chord.
00:01:27.600 --> 00:01:37.400
Again, AB times BC equals DB times BE.
00:01:37.400 --> 00:01:47.300
The next part: for secant segments, we know that secants are lines that intersect a circle at two points.
00:01:47.300 --> 00:01:59.600
If we talk about secant segments, then they are just segments that end at the one point, and then it has two endpoints.
00:01:59.600 --> 00:02:06.000
So, it is still intersecting the circle at two points; but it has endpoints.
00:02:06.000 --> 00:02:13.900
If we have two secants that are drawn to a circle from an exterior point, and that point here is A,
00:02:13.900 --> 00:02:21.700
then the product of the measures of one secant segment and its external secant segment is equal
00:02:21.700 --> 00:02:29.600
to the product of the measures of the other secant segment and its external secant segment.
00:02:29.600 --> 00:02:38.700
This is a secant segment; this is another secant segment; you are going to take the whole secant segment, AC,
00:02:38.700 --> 00:02:46.900
times the part of that secant segment that is on the outside of the circle; that is AB.
00:02:46.900 --> 00:03:08.200
So again, starting over: AC, the whole secant segment, times just the outside part, AB, is equal to this whole secant segment, AE, times the external part, AD.
00:03:08.200 --> 00:03:17.500
So, it is AC, the whole thing, times the outside part, AB, equals the whole thing, AE, times the outside part, AD.
00:03:17.500 --> 00:03:23.900
So, it is different than the chord segments, where it is just the part times the part, equaling the part times the other part.
00:03:23.900 --> 00:03:34.400
The secant segment is the whole thing times the outside part, equal to the whole thing here times this outside part.
00:03:34.400 --> 00:03:58.500
Let's say that AC is 6, so the whole thing is 6; then, if the whole thing is 6, times AB is 4, and let's say AE is 8,
00:03:58.500 --> 00:04:11.800
and AD is 3; then when we multiply them together, we know that they have to equal each other: 24 = 24.
00:04:11.800 --> 00:04:19.700
OK, and then, the next part: this is almost the same as the two secant segments.
00:04:19.700 --> 00:04:23.600
In this case, we have a tangent and a secant segment.
00:04:23.600 --> 00:04:30.500
Now, when we have a tangent and a secant segment, it is going to be the tangent, squared,
00:04:30.500 --> 00:04:35.900
equaling the whole secant segment, AD, times the outside part, AC.
00:04:35.900 --> 00:04:41.900
Now, from before, the secant segment was the whole times the outside part;
00:04:41.900 --> 00:04:51.700
well, in the same way, for the tangent, the whole segment and the outside part are the exact same thing.
00:04:51.700 --> 00:05:01.600
It is like saying the whole AB, times the outside part, AB, equals AD times AC.
00:05:01.600 --> 00:05:16.400
The whole segment here is AB, times the outside part (is still AB), equal to AD times AC.
00:05:16.400 --> 00:05:29.600
That is why it became AB²; this is not a new formula--this is the exact same thing as the two secant segments.
00:05:29.600 --> 00:05:41.800
But it just became AB², because the whole segment and the outside segment part are exactly the same, so it is just that segment squared.
00:05:41.800 --> 00:05:55.900
So again, the whole segment, AB, times the outside segment, AB, equals AD times AC.
00:05:55.900 --> 00:06:03.300
Let's actually try some problems: Find the value of x.
00:06:03.300 --> 00:06:13.700
These are chords, where they form interior angles; in this case, it would just be the part times the part, equals the part times the part.
00:06:13.700 --> 00:06:40.800
It is 5 times x, equal to 3 times 7; so this will be 5x = 21; if you divide the 5, then x is going to be 21/5.
00:06:40.800 --> 00:06:48.100
The same thing happens here: you have chord segments, so it is going to be this part, times this part;
00:06:48.100 --> 00:07:00.200
just look up here; so it is x + 5, times 6, is equal to x times 12.
00:07:00.200 --> 00:07:09.700
Don't forget to distribute the 6; so it would be 6x + 30 = x times 12, is 12x.
00:07:09.700 --> 00:07:25.600
If I subtract the 6x here, 30 = 6x; divide the 6; so then, x becomes 5.
00:07:25.600 --> 00:07:30.000
Don't forget: if they are chords, then it is just the part times the part, equaling the part times the part.
00:07:30.000 --> 00:07:39.700
Now, keep in mind: you have seen this in a few different lessons, but if they are asking for the segments, this is the theorem that you use.
00:07:39.700 --> 00:07:44.800
If they are asking for interior angles, then you are going to have to use the intercepted arc,
00:07:44.800 --> 00:07:51.500
so it is different formulas, different theorems, for whatever they are looking for.
00:07:51.500 --> 00:08:00.600
Then, angles...this is another theorem; and then, here we are talking about segments.
00:08:00.600 --> 00:08:10.600
The next example: Find the value of x; here we have the secant segments at an external point.
00:08:10.600 --> 00:08:24.000
Then, remember: you are going to take the whole thing, and multiply it by the outside part; set it equal to this whole thing, times the outside part.
00:08:24.000 --> 00:08:31.200
Now, I am only given these parts; be careful that you don't do the part times the part, like the chords.
00:08:31.200 --> 00:08:36.100
With this one, you are going to have to do the whole thing, times the outside part.
00:08:36.100 --> 00:08:44.700
And also, be careful when you look for the whole thing; be careful not to multiply these two numbers together.
00:08:44.700 --> 00:08:57.400
This is 4x, and this is 5; if I give you a segment, and I tell you that this is 4 and this is 3, what is the measure of the whole thing?
00:08:57.400 --> 00:09:06.000
It is the 7, because you add them together; you don't multiply them; if this is 4 inches long, and this is 3 inches long, together they are 7 inches long.
00:09:06.000 --> 00:09:17.200
So, when you are finding the whole thing, don't multiply these numbers; this is 4x + 5; the whole thing is 4x + 5,
00:09:17.200 --> 00:09:33.400
times the outside part, 5, is equal to this whole secant segment, which is 6x + 4, times the outside part, 4.
00:09:33.400 --> 00:09:46.800
Then, we are going to distribute this; this is 20x + 25 = 24x + 16.
00:09:46.800 --> 00:10:10.300
If I subtract this 20x, we get 25 = 4x + 16; subtract the 16; I get 9 = 4x, so then x is going to give me 9/4.
00:10:10.300 --> 00:10:21.600
Again, do not multiply these numbers together; that is the most common mistake: 4x + 5 is the measure of the whole segment.
00:10:21.600 --> 00:10:28.900
The next one: here, we have a tangent instead of a secant, but it is the same thing--don't get it confused.
00:10:28.900 --> 00:10:38.300
I know that there is a separate theorem for it, but just think of it as the exact same thing: the whole segment times the outside segment.
00:10:38.300 --> 00:10:49.600
So then, here, this is the whole segment; it is not 80; remember: it is the 8, plus the 10, so then the whole segment here is going to be 18;
00:10:49.600 --> 00:11:04.000
multiply it by the outside segment; it is 8; then, the whole segment, x...multiply it by the outside segment, x.
00:11:04.000 --> 00:11:27.000
This is going to be 144; that is equal to x²; then, I am going to take the square root of that, so x is going to be 12.
00:11:27.000 --> 00:11:39.400
Find the value of x: For this one right here, we didn't go over two tangents for this section.
00:11:39.400 --> 00:11:49.200
But I had this problem here, because I wanted to show you that it is the same concept as the two tangents that we went over in the previous lesson.
00:11:49.200 --> 00:11:58.600
It applies the same as this section here, because, if you have two tangents that meet at an external point,
00:11:58.600 --> 00:12:05.900
we know that these two tangents are congruent; so then, for this, x is just equal to 5.
00:12:05.900 --> 00:12:14.000
That is the answer; but I also wanted to apply the same theorem from this section to this problem.
00:12:14.000 --> 00:12:22.500
So then, the outside, times the whole thing, is x² (the whole thing is x; the outside is x);
00:12:22.500 --> 00:12:39.200
that is equal to...the whole thing is 5, times the outside part is 5; so then, x² is equal to 25; x is going to equal 5.
00:12:39.200 --> 00:12:49.200
Although we know that two tangents are congruent if they meet at an external point (from the same circle),
00:12:49.200 --> 00:13:01.000
all of those secants and tangent...secant, secant, tangent, secant, tangent, tangent--it is the same concept.
00:13:01.000 --> 00:13:14.300
And then, for this one here, again, the same thing happens: it is going to be the whole thing, which is x + 5,
00:13:14.300 --> 00:13:24.200
times...now, again, don't multiply these together...it is x + 5, times...the outside part is x;
00:13:24.200 --> 00:13:32.600
it is equal to...the whole thing is 6; the outside part is 6; so 6 times 6, which is 36.
00:13:32.600 --> 00:13:48.600
Distribute this: x² + 5x = 36; OK, now, here, you have a quadratic equation, because you have this x².
00:13:48.600 --> 00:13:56.400
So, to solve for x, we are going to have to factor this out.
00:13:56.400 --> 00:14:05.800
I am going to subtract 36, so I can make it equal to 0; now, hopefully, you remember how to factor these.
00:14:05.800 --> 00:14:18.900
If not, then you can use what I call the x method: you are going to put this number up here,
00:14:18.900 --> 00:14:24.500
and then this number times this number in front--it is the constant times the leading coefficient;
00:14:24.500 --> 00:14:37.000
this is 1, so it is just -36; and then, you are going to find two numbers that add to get this and multiply to get this.
00:14:37.000 --> 00:14:43.600
They add to get that, and multiply to get that.
00:14:43.600 --> 00:14:51.700
The easiest way to do this is to find a factor pair; we know that, since this is a negative--they are two numbers
00:14:51.700 --> 00:14:58.200
that multiply to get a negative--that means that one of them is going to be a negative number.
00:14:58.200 --> 00:15:17.400
Some factor pairs of 36 are going to be 6 times 6, 9 times 4, 12 times 3; which pair is going to give us a positive 5?
00:15:17.400 --> 00:15:23.600
Now, again: one is going to be a negative number, because they have to multiply to get -36.
00:15:23.600 --> 00:15:31.100
That means that one of them has to be a negative number; and I know that, if I have a 9 and 4,
00:15:31.100 --> 00:15:41.400
+9 and -4 are going to add to get 5, and they are going to multiply to get -36.
00:15:41.400 --> 00:15:50.300
So then, I am going to use these two numbers; I don't have a leading coefficient (or it is just 1),
00:15:50.300 --> 00:15:55.000
so I don't have to worry about anything else; these are just my numbers.
00:15:55.000 --> 00:16:09.600
So, it is going to be (x + 9)(x - 4) = 0; and if you want to just double-check, you can FOIL it out.
00:16:09.600 --> 00:16:21.800
x²...and then this is -4x + 9x; that is +5x; +9 times -4 is -36.
00:16:21.800 --> 00:16:24.600
Then, from here, now that we have factored, we are going to have to solve for x.
00:16:24.600 --> 00:16:37.900
We make each of these equal to 0; x + 9 = 0; x - 4 = 0; so here, x is -9; x = +4.
00:16:37.900 --> 00:16:47.900
Now, we have two answers, but here is the thing: sometimes we can just say that both of those are going to be our answer;
00:16:47.900 --> 00:16:56.700
but this x is a measure--that is the length of this side right here, this segment.
00:16:56.700 --> 00:17:04.200
We can't have a negative number; I can't have a pen that is -5 inches long.
00:17:04.200 --> 00:17:10.700
If you have a negative measure, you would have to cancel that out; you have to cross that out; you can't have a negative measure.
00:17:10.700 --> 00:17:16.000
If it is finding the distance, finding the length of something, then it can't be a negative number.
00:17:16.000 --> 00:17:24.100
So then, my answer is going to be just 4.
00:17:24.100 --> 00:17:39.900
To review for this problem (because we did spend a little bit of time on this): you are going to do the whole thing, x + 5, times x, equals 36.
00:17:39.900 --> 00:17:50.700
Then, you distribute the x, and then make it equal to 0, because this is now a quadratic equation that you have to factor.
00:17:50.700 --> 00:17:57.600
Find two numbers that it is going to factor out into; we get (x + 9)(x - 4) = 0.
00:17:57.600 --> 00:18:10.400
When you solve out for x, then you get -9, and you get +4; this can't be a negative number; therefore, the answer is that x is 4.
00:18:10.400 --> 00:18:17.200
OK, and then, the final example: we are going to find the values of x and y.
00:18:17.200 --> 00:18:23.800
I have two tangent circles, two circles that are tangent; they meet at one point.
00:18:23.800 --> 00:18:30.500
That means that this line is a tangent, also; here is a secant segment and a secant segment.
00:18:30.500 --> 00:18:42.300
Now, if I look at this circle, this secant, and this tangent, I know that I can use that theorem that says
00:18:42.300 --> 00:18:46.700
that the whole thing times the outside equals the whole thing times the outside.
00:18:46.700 --> 00:18:56.200
But I have two variables: x and y are both from the segments of this circle.
00:18:56.200 --> 00:19:02.600
So, let's look at the other circle: here, I have a secant, and I have a tangent.
00:19:02.600 --> 00:19:08.100
Now, the whole thing, times the outside, equals the whole thing, times the outside.
00:19:08.100 --> 00:19:13.600
So, since I only have one variable, I can go ahead and solve using these segments.
00:19:13.600 --> 00:19:37.200
I am going to solve for y first; that means the whole thing is 18 (10 + 8 is 18), times the outside (is 10), is equal to y²,
00:19:37.200 --> 00:19:42.600
because that is the whole segment; it is y, and then the outside segment is y; so that is y².
00:19:42.600 --> 00:19:56.700
This is 180; that is equal to y², so y, then, is going to be √180.
00:19:56.700 --> 00:20:06.000
Now, I can simplify that out; but I want to leave it for now, because I am just thinking of...
00:20:06.000 --> 00:20:13.800
I know that I am going to have to use this value to find x, and I have a feeling that I am going to have to square this back anyway.
00:20:13.800 --> 00:20:19.000
So, just leave it like that for now, and then you can simplify it in a second.
00:20:19.000 --> 00:20:37.700
Here, to find x, it is going to be the whole thing times the outside; that is y²; we know that y is √180, times √180.
00:20:37.700 --> 00:20:46.800
That is equal to 9 + x (that is the whole thing), times 9.
00:20:46.800 --> 00:21:13.100
So, here we are going to get...this is 180, equals 81 + 9x; subtract the 81; I am going to get 99 = 9x.
00:21:13.100 --> 00:21:25.900
Then, divide the 9; x is going to be 11.
00:21:25.900 --> 00:21:38.000
Now, back to y: we can either simplify that, or we can just use our calculator and find the decimal for it,
00:21:38.000 --> 00:21:42.400
because we know that 180 is not a perfect square.
00:21:42.400 --> 00:21:52.500
How do we simplify √180? The easiest way to factor out a radical is to do a factor tree.
00:21:52.500 --> 00:22:04.200
Let's see, we know that 18 times 10 is 180; this is 9 times 2; this is 5 times 2.
00:22:04.200 --> 00:22:13.200
When you have a prime number, circle it: here we have 3 times 3, so then all of our numbers are circled.
00:22:13.200 --> 00:22:20.600
Then, whenever you have a pair, whenever you have two of the same number, you are going to write that on the outside.
00:22:20.600 --> 00:22:28.900
That is going to come outside; so that is going to be 3...and then we have, also, a 2;
00:22:28.900 --> 00:22:34.100
so, you are going to multiply those two numbers together; and then, what is not crossed out? It is the 5.
00:22:34.100 --> 00:22:44.300
That means that that has to stay inside, because the only time it can come out of this radical is when it has two of the same numbers.
00:22:44.300 --> 00:22:57.700
If it is 3², then that 3 will come out; this is 2², so a 2 came out; so this will be 6√5 is y.
00:22:57.700 --> 00:23:03.300
Or again, you can just use your calculator and just find the decimal of that.
00:23:03.300 --> 00:23:06.700
There is y, and there is x; and that is it for this lesson.
00:23:06.700 --> 00:23:08.000
Thank you for watching Educator.com.