CA Geometry: Circle Area Chords Tangent 71-75, area, chords, tangents of circles
CA Geometry: Circle Area Chords Tangent
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- We're on problem 71.
- It says, what is the value of x in the triangle below?
- OK, so we can just 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 this is a... well, if this is 90,
- so 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 kind of how
- the sides of a 45, 45, 90 relate to its hypotenuse.
- But you don't have to memorize it.
- 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.
- OK, 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.
- 140, right.
- 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?
- So the square is circumscribed about the circle.
- Let me draw the circle and the square. Well, if that's my circle, then if I want to draw a square, See if I can, nope that's not what I wanted to do . I wanted to draw a solid square. That's... let me see if I can draw it without pressing the shift key.
- Oh, I think that's close enough.
- We know the square is on the outside because
- it's about the circle. It's circumscribed 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.
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