WEBVTT mathematics/geometry/pyo
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Welcome back to Educator.com.
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For the next lesson, we are going to go over geometric probability.
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The first thing that we are going to go over is the Length Probability Postulate.
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It is when we are using segments for probability.
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If a point on segment AB is chosen at random, and point C is between A and B, then the probability that the point is on AC
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is going to be (and this is a ratio) segment AC, over AB.
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Now, if you remember probability, probability measures the part over the whole.
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You can also think of the top number as the desired outcome, over the total outcome, the total possible number of all of the different types of possible outcomes.
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So then, it is a desired outcome, what you are looking for, over the total--over the whole thing.
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So, it is just part over whole; this is the most basic way you can remember probability, part over whole.
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Here, the same thing applies to the Length Probability Postulate; you are looking at the part.
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You are looking at what the desired outcome is, which is the point being on AC, over the whole thing; that is AB--that is the whole thing.
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It is AC to AB--always part over whole.
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So, let's say that this right here is 5; CB is 5; the probability of a point landing on AC...what is AC?
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That is the desired outcome; that is the top number, which is 5, over the whole thing (it is not 5; it is not the other half);
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it is the whole thing, which is AB, and that is 10; so when I simplify this, this becomes 1/2.
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The probability of landing on AC is 1/2.
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And for the Area Probability Postulate, when you are talking about the probability of something to do with area, you are looking at space.
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So, you are looking to see, for example, maybe a dart hitting the dartboard; that is area, because it is space that you are looking at.
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If you look to see, maybe, a spinner (we are going to have both of those for our examples) landing on a certain space, that is area.
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So, that has to do with this probability; and this postulate says that if a point in region A
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(this rectangle is region A) is chosen at random, then the probability that the point is in region B
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(which is inside region A) is going to be the area of region B, over...remember: the whole thing is the area of region A.
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Region A is the whole thing; the area of the whole thing is the total,
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and the top one will be the area of the desired outcome, or the part that we were just looking at; and that is region B.
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The area of a sector of a circle: now, a sector is this little piece right here.
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This is the center; the sector is the area of this piece, so it is bounded by the central angle
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(this is a central angle right here; this angle is a central angle, so if I need the θ, that is the central angle) and its intercepted arc.
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This is the intercepted arc; so those are the boundaries, this angle and that.
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This whole thing is called a **sector**; now, I like to refer to the sector as a pizza slice.
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Think of this whole thing as a pizza; this is a slice of pizza; so a sector is a slice of pizza.
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We are finding the area of that slice; to find the area of this, it would be this formula:
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the central angle (which is this angle right here, this central angle) over 360...
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now, why is it over 360?--because going all the way around a full circle, including that, is 360; so it is like the part over the whole,
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the central angle over the whole thing, which is 360...times the area of the circle.
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Now, another way (an easier way, I think) would be (to figure out how to find the area of this):
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instead of looking at this formula, I like to use proportions.
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So, what we can do to find the area of this pizza slice right here: remember: a proportion is a ratio equaling another ratio;
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so, we are going to look at the probability (probabilities are ratios, something to something, which is part to whole)
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of the measures to the areas, because we are looking both: we are looking at measure, and we are looking at area.
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So, for the measures, the part over the whole for the angle measures is going to be the central angle, over the whole thing, which is 360.
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And the probability of the area...isn't that part over whole, also?...so the part will be the area of the sector.
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That is the area of the sector; so let's just call that A for area of the sector...over the whole thing, which is
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(the area of this whole circle is going to be) πr².
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Again, the ratio (or probability) of the part to the whole is the angle measure to the whole thing.
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And the ratio of the areas is going to be the area of the sector, over the area of the circle, because this is part to whole.
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We are going to make them equal to each other; that is our proportion; and you are just basically going to solve.
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Let's say that this angle measure is 40, and the radius, r, is 6.
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If this is 40, that is the part; that is 40 degrees, over...what is the whole thing?...360 degrees, is equal to the area of the sector;
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that is what we are looking for; that is the area of the sector, over the area of the whole thing; that is the circle, so it is πr².
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So, that is π(6)²; and if you are going to solve this out, remember how you solve proportions.
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You do cross-multiplying; so then, the area of the sector, times 360, times a, equals 40 degrees times π(6)².
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And when you solve this out, you divide this by 360, because we are solving for the a.
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Now, if you look at this, this is exactly the same thing as this right here: the central angle,
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the angle of that right there, divided by the 360, the whole circle, times the area of the circle--that is πr².
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It is the same exact thing; if you want to just use this, that is fine--it is the same exact thing.
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But this way, you just know that you are looking at the part, the angle measure, over the whole, the circle's angle measure.
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That is equal to the area of the sector (that is the part), over the area of the whole circle.
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It is part over whole, for angle measures, equals part over whole, for the areas.
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And that is just a way for you to be able to solve this out without having to memorize this formula.
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And then, we solve this out; and you can just do that on your calculator; I have a calculator here on my screen.
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I get that my area is 12.57; and again, that is the area of this sector, the pizza slice; and that is units of area, squared.
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That is the area of a sector.
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Let's go ahead and do some more examples: What is the probability that a point is on XY?
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Again, for probability, we are looking at part to whole; so the desired outcome, the part that we are looking for, is XY;
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that is going to be my numerator--that is the top part of my ratio--so XY is from 0 to 2; that is 2 units.
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It is XY; again, we are looking at XY over the whole thing, which is XZ, so that is 2 over...the whole thing, from X to Z, is 10.
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If I simplify this, this becomes 1/5, because 2 goes into both; it is a factor of both 2 and 10.
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I can divide this by 2 and divide that by 2, and I get 1/5; that is the probability that a point will land on XY.
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Find the probability of the spinner landing on orange, this space right here; here is that spinner.
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This angle measure is 72, and the radius is 4; so, if this is 4, then we know that any segment from the center to the circle is going to be 4.
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OK, so then, I want to use that proportion: the angle measure, over the measure of the circle, the total angle measure,
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is equal to the area of the sector, over the whole thing (is going to be the area of the circle; and that just means "circle").
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This is my proportion: the angle measure...any time I am dealing with the part (since it is always part over whole),
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it is always going to be about the sector, this piece right here, the orange; and then, any time I am talking about the whole,
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it is going to be the whole circle...(that is that) is 72 degrees, over the whole thing (is 360), is equal to
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the area of the sector (and again, that is what I am looking for, so I can just say A for area of the sector),
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over the area of the circle (that is the whole thing); and that is πr².
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My radius is 4, squared; so then, I can go ahead and cross-multiply.
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360A = 72(π)(4²); then, to solve for A, divide the 360; divide this whole thing by 360.
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And then, from there, you can just use your calculator: 72 times π times the 4²...and then divide 360; you get 10.05.
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And we have inches here for units, so it is inches squared; that is the area of this orange.
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Now, to find the probability...we found the area of this orange; and be careful, because,
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if they ask you for the area of this base right here, then that would be our answer; but they are asking for a probability
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of landing on orange; and any time you are looking at probability, you are always looking at part over whole.
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And again, since we are talking about area, it is the area of the orange, over the area of the whole thing.
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I found the area of a sector; now, to find the area of the whole thing, the area of the circle is πr².
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And all you have to do is...we know that r is 4, so 16 times π is 50.27 inches squared; that is the area of the circle.
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And then, the probability is going to just be (I'll write it on this side) 10.05/50.27.
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You can change this to a decimal, so you can go ahead and divide this; or maybe you can just leave it like that,
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depending on how your teacher wants you to write the answer.
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You can definitely have probability as a decimal; you can just go ahead and take this and divide it by this number; and that would be your answer.
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This is the probability, part over whole, the area of the orange over the area of the circle.
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The circle is circumscribed about a square; if a dart is thrown at the circle, what is the probability that it lands in the circle, but outside the square?
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We want to know what the probability is of landing in the gray area: it said "in the circle, but outside the square"; that is all the gray area.
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That is the probability: they are not asking for the area of that part; they are asking for the probability of landing on that part.
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So then, we have to make sure that we are going to do the part over the whole.
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First, I have to find the area of that gray area, because that is my desired outcome; that is my part.
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The desired outcome is the area of the gray, over the whole thing, which would be the area of the circle, because that is the whole thing.
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So then, my part is going to be, again, area of gray over the area of the circle.
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To find the area of the gray region, we have to first find the area of the circle and subtract the area of the square.
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The area of the gray is going to be the circle, minus the square.
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The area of the circle is πr², minus...the area of this is going to be side squared.
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We know that the radius is 10, because, from the center of the circle to the point on the circle, it is π(10)²;
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minus...do we know what the side is?...we actually don't, because this is from the center to the vertex of this square.
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So, let me make a right triangle: I know that this angle right here (let me just draw the triangle out again--that doesn't look good;
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this is more accurate)...this is that triangle here: this is 10, and I want to know either this or this.
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Let's say that we are going to call that x.
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Now, this is a right angle; we know that this is a 45-degree angle, because it is half the square; in squares, everything is regular.
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So, to find the other sides of a 45-45-90 degree triangle, since we know that it is a special right triangle, we are going to use that shortcut.
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If this is n, then this is n, and this is n√2; and in this case, I should label this n, because that is the side opposite the 45, which is n.
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The side opposite this 45 is n, and then the side opposite the 90 is 10.
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Here is the shortcut; I am given the 90-degree side, which is this right here, so I am going to make those equal to each other,
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because this is n, and this is n√2, which is 10; so n√2 = 10.
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Divide the √2 to both sides: I am going to solve for n; n = 10/√2...what do I do here?
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Well, this square root is in the denominator, so I have to rationalize it; when I do that, this becomes 10√2/2; simplify this out; this becomes 5√2.
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So again, what did I do? I took this...because I have this right here, the hypotenuse of this right triangle, I want to find this side right here.
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I am going to use special right triangles, since this is a 45-45-90 degree triangle: n, n...the side opposite the 90 is n√2.
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That is the side that I am given, so I am going to make that equal to n√2: n√2 = 10.
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Solve for n by dividing the √2; let me rationalize the denominator, because I can't have a radical in the denominator.
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So then, this becomes 10√2, over...√2 times √2 is just 2; simplify that out, and I get 5√2.
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That means that n is 5√2; this side is 5√2; this side is 5√2.
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Well, if this is 5√2, then what is this whole thing? We labeled that as s.
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So, if this is 5√2, then this is 5√2; so you basically have to just multiply it by 2, because this is half of this whole side.
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My side is 5√2 times 2, which is 10√2.
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So, to find the area of the square, I am going to do 10√2 times 10√2, base times height (or side squared): 10√2, squared.
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And then, I am going to use my calculator: this is 3.14 times 100, which is 314, minus 10√2 times 10√2;
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that is 100 times 2; that is 200 (if you want, you can just double-check on your calculator).
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This right here is (I'll just show you really quickly) 10√2 times 10√2.
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10 times 10 is 100; √2 times √2 is times 2, so it is 100 times 2, which is 200.
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Then, this is going to be 114; so the area of the gray is 114, because I took the area of the circle,
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which was πr², 314, and then I found the area of the square, which is 200, 10√2 times 10√2.
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And then, I got 114; now, that is just the area of the gray; we are looking for the probability that it lands in the gray.
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That is the area of the gray, over the area of the whole thing, which is the circle.
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We take 114 over the area of the circle (where is the area of the circle?), which is 314; and that is the probability.
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Now, we know that both of these numbers are even, so I can simplify it.
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So then, if you were to cut this in half, this is going to be 57; if you cut this in half, this is going to be 157.
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And that would be the probability, 57/157.
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So again, the probability is going to be the area of the gray over the area of the whole thing, which is the circle.
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The fourth example: we have a hexagon, and I am just going to go ahead and write that this is a regular hexagon with side length of 4 centimeters.
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It is inscribed in a circle; what is the probability of a random point being in the hexagon?
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"Inscribed": now, I don't have a diagram to show you, so I am going to have to draw it out.
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"Inscribed" means that it is inside, so the hexagon is inside the circle; but all of the vertices of the hexagon are going to be on the circle.
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They have to be intersecting; so let me first draw a circle, and the regular hexagon ("hexagon" means 6 sides).
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I am going to try to draw this as regular as I possibly can, something like that, so it will look like it is inscribed...something like that.
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What else do we have? Side length is 4 centimeters; what is the probability of a random point being in the hexagon?
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OK, so then, again, we are giving a probability, so it is part over whole.
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What is the part? The part will be the hexagon; inside the hexagon is the desired outcome--that is what we are looking for.
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So, it is going to be the area of the hexagon, over...the whole thing is going to be the area of the circle.
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Let's see, now: to find the area of this hexagon...remember: to find the area of a regular polygon,
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if we were to take this hexagon and then break this up into triangles, we have 1, 2, 3, 4, 5, 6 triangles.
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Each triangle is going to be 1/2 base times height; and then, we have 6 of them...times 6.
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Now, if we take the base (the base is right here), and we multiply this base with this 6, isn't that the same thing as the perimeter?
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The base, with the 6, is going to be the perimeter; the height, this right here, we call the apothem.
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I am going to draw arrows to show that the base and the 6 together became the perimeter, and the height became the apothem.
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1/2 just stayed as 1/2; this is the formula for the area of a regular polygon: it is 1/2 times the perimeter of the polygon, times the apothem.
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The apothem is from the center, the segment going, not to a vertex, but to the center of the side; so it is perpendicular.
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Now, we don't know what the apothem is, so I am going to have to look for it.
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Now, remember: you always want to use right triangles, if you possibly can; we can, because the apothem is perpendicular to the side.
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So, if I just maybe draw it bigger to show: this is the apothem.
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If the whole side measures 4 (see how this is 4), then this half is going to be 2.
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And then, I want to look for this angle measure, because I don't have this side.
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If I have this side, then I can use the Pythagorean theorem, because then I have a² + b² = c².
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But I don't, so instead, I want to see: this is a circle; the whole thing, all the way around, is 360 degrees.
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If I break this up into parts, this is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
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I basically just look to see how much this triangle is from the whole 360; if the whole thing measures 360,
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remember how we said that this is actually one of the 6 triangles; so 360 divided by 6 is 60.
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Then, since this whole thing measures 60 degrees, what is this half right here?
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This half is 30 degrees, so this is 30 and this is 60; and again, all I did was just...
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I know that the whole thing, the full circle, measures 360; since I know that this right here is 60,
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because I have 6 triangles, so it is like 6 triangles sharing 360 degrees;
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then this angle right here is 60 degrees, which means that this half right here is 30.
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A 30-60-90 triangle is a special right triangle; if this is n, the side opposite the 60 is n√3; the side opposite the 90 is 2n.
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This is the special right triangle; what do I have--what am I given?
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The side opposite the 30 is 2; so I am going to make those two equal to each other: n = 2.
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That means that the side opposite the 60 is going to be 2√3; the side opposite the 90 is going to be 2 times n, which is 4.
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That means that a, which is the side opposite my 60, is 2√3; isn't that my apothem, my a?
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So, I know that that took a while; but it is just going over the area of a regular polygon.
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It is 1/2 perimeter (which is 6 times 4, because there are 6 sides, and each side is 4...so perimeter is 24)...my apothem is 2√3.
00:34:13.500 --> 00:34:24.700
And then, I can just cut this in half; so 24/2 is 12, times 2 is 24, √3.
00:34:24.700 --> 00:34:54.000
To make that into a decimal, 24 times √3...we get 41.57 units squared...oh, we have centimeters, so this is centimeters squared.
00:34:54.000 --> 00:35:10.100
And again, this is the area of the hexagon; so this is the hexagon, and then I want to find the area of the circle.
00:35:10.100 --> 00:35:13.500
Now, if it were just the area of the hexagon that we were looking for, then this would be the answer.
00:35:13.500 --> 00:35:24.000
But again, we are looking at probability: what is the probability of a random point being in the hexagon?
00:35:24.000 --> 00:35:30.700
It is the area of the hexagon, over the area of the full circle.
00:35:30.700 --> 00:35:44.100
So then, I look at it: here is the hexagon; here is the circle; the area is πr².
00:35:44.100 --> 00:35:58.900
π...do I know r?...r would be this length right here; this is the radius, because this is the center of the circle; this is a point on the circle.
00:35:58.900 --> 00:36:07.800
From here to here...if this triangle is this triangle...what do we find about this side?
00:36:07.800 --> 00:36:39.100
That side is opposite the 90, and that is 4; so this is 4 squared; this is 16π; 16 times π is 50.27 centimeters squared.
00:36:39.100 --> 00:37:19.300
So then, hexagon over circle is 41.57, over 50.27; so let me just do that on a calculator: 41.57/50.27...and I get (so the probability is) 0.83.
00:37:19.300 --> 00:37:26.800
Now, one thing to mention here: when you have a decimal, when you change your probability fraction into a decimal,
00:37:26.800 --> 00:37:37.500
you have to make sure that it is less than 1, because, if you are looking at a part over the whole,
00:37:37.500 --> 00:37:42.600
it is going to be a proper fraction; the part is going to be smaller that the whole.
00:37:42.600 --> 00:37:49.600
If the whole is everything--it is the whole thing--well, then, the part can only be a fraction of it.
00:37:49.600 --> 00:37:56.300
So, the only time you can get anything greater than this...
00:37:56.300 --> 00:38:04.400
I'm sorry: the biggest number you can get for probability, when you change it to a decimal, is 1, because,
00:38:04.400 --> 00:38:09.600
when you look at the fraction, it can just be the whole thing over the whole thing.
00:38:09.600 --> 00:38:16.100
And when you have whole over whole, well, that is just going to equal 1, because it is the same number over itself.
00:38:16.100 --> 00:38:24.000
So, make sure that your probability decimal is not greater than 1, unless you are talking about the whole thing.
00:38:24.000 --> 00:38:37.800
Then, it is going to be 100%, all of it, which is 1; but otherwise, if the part is smaller than the whole, then your decimal has to be less than 1.
00:38:37.800 --> 00:38:40.000
That is it for this lesson; thank you for watching Educator.com.