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## Geometry (all content)

### Course: Geometry (all content) > Unit 17

Lesson 1: Worked examples- Challenge problems: perimeter & area
- Challenging perimeter problem
- CA Geometry: Deductive reasoning
- CA Geometry: Proof by contradiction
- CA Geometry: More proofs
- CA Geometry: Similar triangles 1
- CA Geometry: More on congruent and similar triangles
- CA Geometry: Triangles and parallelograms
- CA Geometry: Area, pythagorean theorem
- CA Geometry: Area, circumference, volume
- CA Geometry: Pythagorean theorem, area
- CA Geometry: Exterior angles
- CA Geometry: Pythagorean theorem, compass constructions
- CA Geometry: Compass construction
- CA Geometry: Basic trigonometry
- CA Geometry: More trig
- CA Geometry: Circle area chords tangent
- Speed translation

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# CA Geometry: Circle area chords tangent

71-75, area, chords, tangents of circles. Created by Sal Khan.

## Want to join the conversation?

- What is the proof of "8:14"(9 votes)
- The explanation by Jack Li is correct. Here is a video on youtube with a diagram that might make it clearer. http://www.youtube.com/watch?v=TfRiX7nnM44(1 vote)

- I can't seem to find the video with problems 75-80. Could someone give me the link, please?

Thank you!(5 votes)- I think when he said 80 problems, he was also including the example problems in the first videos of this series.(1 vote)

- Do you have a video explain how to measure the length of a chord?(3 votes)
- 7:47Is there a video where Sal proves this? Thanks(2 votes)
- How do you remember which side is 2x and root 3x?(2 votes)
- Think of the 60 60 60 triangle made by two identical 60 30 90 triangles together back to back. The small side opposite the 30degree angle is obviously half the size of the hypotenuse of the 60 30 90 triangle. Sal draws a diagram to explain this better than I can in words here, https://www.khanacademy.org/math/geometry/right_triangles_topic/special_right_triangles/v/30-60-90-triangle-side-ratios-proof(3 votes)

- how to tell if whether each polygon is inscribed in or circumsribed about the circle(2 votes)
- If the polygon is inscribed in the circle, the vertices of the polygons are on the circle's circumference. If the polygon is circumscribed, the shape's lines are on the outside of the triangle (forming tangents). if you are still confused, visit these websites:

http://www.mathopenref.com/polygonincircle.html (inscribed)

http://www.mathopenref.com/polygoncircumcircle.html (circumscribed)(2 votes)

- the first problem answer was (A)5 because we just have to find the X value and the answer would be just 5 not 5 square root of 2.(2 votes)
- yes, eight years later, I am here, I believe he meant to have one of the variables as y, so they differ(1 vote)

- what is an arc of a circle?(2 votes)
- An arc is a segment of the circle's curve. More info can be found at: Language and notation of the circle at around9:00.(1 vote)

- how do i find tangent ratios?(2 votes)
- well, thats a complicated question.. for me.

and dont be mislead im in the 11th grade haha(1 vote)

- What really is a chord? Not the type when you play guitar, but like in math.(1 vote)
- A chord is a line segment connecting two points along a circle. The diameter is the longest possible chord.(3 votes)

## 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.