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CA Geometry: Circle area chords tangent

Video transcript
We're on problem 71. It says, what is the value of x in the triangle below? OK, so we can pull out the Pythagorean theorem here. Although, you might recognize that if these two sides are the same, then these two base angles are going to be the same. And if those base angles are the same, then you have 90 degrees to go between those two angles, so they're going to have to be 45. Because they're the same. So this is a 45, 45, 90 triangle. And if you haven't already, you'll eventually memorize how the sides of a 45, 45, 90 relate to its hypotenuse. But you don't have to memorize. You can prove it here. Sometimes it's just faster on standardized tests and things like that. So what does the Pythagorean theorem tell us? It tells us that this side squared, so let's say x squared, plus this side squared, plus x squared is equal to the hypotenuse squared. Is equal to 10 squared, which is 100. So we get 2x squared is equal to 100. x squared is equal to 50. Dividing both sides by 2. And then what does this turn into? So we can say x is equal to the square root of 50. Is there anything that we can do here to simplify it at all. Let me think. Oh sure, 50 is 25 times 2. So that's equal to the square root of 25 times the square root of 2. Which is equal to 5 times the square root of 2. Choice B. Problem 72. What is the value of x in inches? OK, a couple of problems ago, we saw a 30, 60, 90 triangle, this is another one. 30 degrees, 90 degrees, they have to add up to 180, this one is equal to 60 degrees. And I did that big convoluted drawing where I flipped it and all of that. I think this is a good time to just to memorize the sides of a 30, 60, 90 triangle. Because that's something that one needs to know in life. It's surprisingly useful. Especially once you start taking standardized tests or do trigonometry. So I'll just give you the general rule. So let me just draw another one right here. Let's say this is my other 30, 60, 90 triangle. This is clearly the hypotenuse up here. This is, I call it the 30 degree side, it's opposite the 30 degree angle, or it's the shortest side. So the general rule is, if this side right here is x. Then the hypotenuse is going to be 2x. And we saw that in that previous video. And then you can actually use the Pythagorean theorem here to solve for this last side. You really just need to memorize that the hypotenuse is twice the shortest side. So this case, what's the shortest side? It's opposite the 30 degree side. So it's 7. So the hypotenuse would be twice that, which is 14. And you could use the Pythagorean theorem to figure out x now. Or you could just memorize that the middle side, I guess you could say, or the long non-hypotenuse side, or the 60 degree side, the side opposite the 60 degree angle, that's equal to the square root of 3 times the short side. So in this case, x is the square root of 3 times 7. So x is equal to 7 times the square root of 3. And don't take my word for it. You could take my word for that this is double that. And we proved that in a couple of videos ago. But you could do the Pythagorean theorem here. You could say that 7 squared, which is 49 , plus x squared is going to be equal to the hypotenuse squared. 14 squared is 196. Subtract 49 from both sides. You get x squared is equal to 196 minus 50 would be 157, is that right? Let me make sure I got it. 14 times 14. 4 times 4 is 16. 56. 140, right. 196. And if you were to subtract 49 from that, this is an 8, this is 16, we have a 7. Sorry, 147. It's a good thing I checked that. All right. So x is equal to the square root of 147. 147 is 49 times 3. It's equal to the square root of 49 times 3. Well that's just equal to the square root of 49 times the square root of 3. Which is equal to 7 root 3. Which is what we got. But it might be easier to just memorize that the side opposite the 60 degree side, is going to be the square root 3 times the short side. And the short side is going to be half the hypotenuse. Anyway, the more practice you do, the more it will make sense. OK, a square is circumscribed about a circle. What is the ratio of the circle to the area of the square? Let me draw the circle and the square. Well, I think that's close enough. We know the square is on the outside because it's about the circle. What is the ratio of the area of the circle to the area of the square? So let's say this is the center of the circle right there. This is its radius. Let's call that r. Well what's the area of the square going to be? If that's the radius, this is also the radius. So one side of this square up here, is going to be 2r. So this side is also going to be 2r. It's a square, all the sides are the same. So they want to know the ratio of the area of the circle to the area of the square. The area of the square is just 2r times 2r. Which is 4r squared. Area of the circle is just pi r squared. You hopefully learned the formula for area of a circle. Divide the numerator and the denominator by r squared. You're left with pi/4. That's choice D. Problem 75. In the circle below, AB and CD are chords intersecting at E. Fair enough. If AE is equal to 5, BE is equal to 12, what is the value of DE? CE is equal to 6. What is the value of DE. Let's call that x. Now, I'm not going to prove it here, just for saving time. But there's a neat property of chords within a circle. That if I have two chords intersecting a circle, it turns out that the two segments when you multiply them times each other, are always going to be equal to the same thing. So in this case, 5 times 12. So the two segments of chord AB, so 5 times 12. That's going to be equal to these two segments multiplied by each other. It's going to be equal to x times 6. So you get 60 is equal to 6x. Divide both sides by 6, you get x is equal to 10. And that is choice C. That might be a fun thing for you to think about after this video of why that is. And maybe you want to play around with chords and prove to yourself that that's always the case. At least that it makes intuition for you, makes sense. RB is tangent to a circle. Tangent means that it just touches the outside of the circle right there at only one point. And it's actually perpendicular to the radius at that point. So this is the radius of that point. The center is at A. This is a radius. And it's tangent at point B, so it's perpendicular to the radius at that point. BD is a diameter, OK, fair enough. Well A is the center, so that's kind of obvious. So they want to know what is the measure of angle CBR. So they want to know what this angle is equal to. Well, I kind of did it inadvertently. We know that when a line is tangent to a circle, it's perpendicular to the radius at that point. So this whole angle is 90 degrees. So the angle that we're trying to figure out, let's call that x. That's the complement to 25. x plus 25 is equal to 90. Subtract 25 from both sides. x is equal to 65 degrees. And that is choice B. OK, I'll see you in the next video.