Find the RED area! - Geometry problem

[@darkblade9986]

Because the inner circle doesn't have any conditions for its size, we can think of it as an area of 0. If we do that, the 20m will go from one edge to another through the center, and that will give us that the ring has a diameter of 20m. A = pi × r² A = pi × (20m ÷ 2)² A = pi × 100m²

[@Dank_Lulu]

Love seeing a specialist doing their craft! The problem got me a headache when I saw it but as soon as you drew x touching with the tangent, it was an "oh" moment. I think I might brush-up my math skills as a grown adult now that it's no longer something others are forcing on me. Thank you!

[@Doc_Fartens]

I really like seeing problems like this. Thanks for the video, Susanne!

[@jeansylvain2596]

Great and usefull video. Love the way you explain the approach used. Thank you Suzanne.

[@thisnamewastakentoo_]

let the inner circle have an area of 0. then the R of the outer circle is 10.

[@lusoverse8710]

Exactly!

[@ElvisSaturn]

The tangent with 90°: Triangle in the circle: R² = r² + 10² AND A= π.(R² - r²) -> A = π.10² = 314,159

[@andysymes8406]

There's a very simple way of solving problems of this type, although you do have to assume that a unique solution exists. Since the inner radius y is not specified, its value cannot have a bearing on the red area (or else there would not be a unique solution). So let's set y to zero. y=0. Then the 20 m stick becomes a diameter of the red circle, whose radius is 10, and area 100 pi. You can easily do this sort of thing in your head.

[@f5673-t1h]

Yes, you can imagine the inner radius shrinking, and the tangent being pulled with it, but in order for the tangent to touch the outer circle, the outer circle must also shrink. So you can imagine that there are infinitely-many such setups with different inner and outer radii, and because the question implicitly tells us there's a definite answer, that means all these setups have the same answer. So take the limit for the inner radius as it goes to 0, where it will be like you said, and the limit for the area will be the same area as we started with (because it's constant).

[@alejomdp]

I did the same, it's much faster.

[@michab4083]

Love to see your new international channel - I wish you every possible success!

[@lusoverse8710]

I came to the same result by reasoning in the following manner: The only value given in the diagram is the length of the straight line. Therefore, the value of the area should always be constant regardless of where the line is placed, or the size of the circle(s), provided that its ends touch the outer circle, and it makes a tangent with the inner white circle, _regardless of the size of the white circle_. Therefore, this should also be the case when the area of the white circle is zero, so that the line in fact forms a diameter of the outer circle. The area of such a circle with a diameter of 20m is 100*pi metres.

[@neilbeacher17]

Great solution. It seems simpler when you get to <a href="#" class="seekto" data-time="274">4:34</a> to just substitute in what you know x squared equals (y squared + 100) and then you just have an equation with y as the only unknown. They cancel out and you’re left with A= 100pi

[@gilbertbarnett8225]

Thank you for your video. I have one suggestion to make the solution a bit clearer for students. You mentioned that the tangent cord is perpendicular to the radius of the circle, but didn't specifically mention that the radius is a perpendicular bisector of the tangent cord. When you diagrammed the right triangle you mentioned that you only needed half of cord's length for the side of the triangle, but didn't provide much of an explanation as to why it was exactly half.

[@allanwrobel6607]

I do so Like the fact that every step is laid out. I've often been stuck in maths at one step, that I just can't see, therefor cannot follow to the conclusion. Thanks

[@markleeks1116]

I can't do any of these problems but it's fascinating to see how it's figured out. Make's no sense to be but enjoyable to watch

[@-I-l-I-l-I-l-I-l-I-l-I-l-]

As math admirer I really love your videos my perfect comfort entertainment thank you so much keep shinning with that wholesome smile of yours ! :)

[@mensaswede4028]

You might observe that it’s actually impossible to compute either circle’s radius. There are an infinite number of radii that satisfy the initial given problem, but the area is always the same for all these radii.

[@UberHummus]

Dooooope tusind tak!

[@alwaysprepared]

It's magic!! LOL Very well explained and at a relaxed speed! Love your presentations!!

[@hellsong23]

Don't forget your units! You'll lose marks (heheh)

[@fouadsharif2327]

Well done Tiger 🐅....👍👍💯💯

[@brendanward2991]

<a href="#" class="seekto" data-time="200">3:20</a> - How do you know that the point of contact is the midpoint of the chord?

[@JessWLStuart]

My Grandad used to joke: "Pie are not square! Cornbread are square! Pie are round!"

[@johnkrogman4708]

Coffee, 'Hello my lovelies' and a math problem are a great way to start my day!

[@MathQueenSusanne]

Aaaaw, this sounds like a wonderful morning! I’m proud of you starting your day like this!

[@KrytenKoro]

r2+10^2=R^2 A1=pi*R2-pi*r2=pi(100-r2+r2)=100pi

[@StuartSimon]

You forgot to mention the theorem that allows us to assume that the side length is 10. The theorem: If a radius of a circle is perpendicular to a chord of the same circle, then the radius bisects the chord. I still remember having to be that rigorous in high school. It's entirely possible you don't know how to state the theorem in English, but you still speak very well. After all, I don't know how to state it in German.

[@ClNoBody]

Thanks - came looking for that and appreciate you mentioning the theorem.

[@utedalheimer1742]

Dankeschön ❤❤

[@Jefuslives]

I immediate saw that the area would be the same as a 20m diameter circle no matter what distance the 20m chord was from the center.

[@m.h.6470]

Solution: Assume the inner circle has a radius of 0, then the red ring becomes a red circle itself with a diameter of 20m, so the area is: A = πr² = π(20/2)² = π10² = 100π [m²] Done. But to prove it: The 20m line is twice the length of one of the legs of a right angle triangle. The radius of the inner circle (r) is the other leg, the radius of the entire ring (R) is the hypotenuse. Therefore with pythagoras, we have: r² + (20/2)² = R² |-r² R² - r² = 100 The area of the ring is: A = πR² - πr² = π(R² - r²) We know R² - r² = 100, therefore no matter how large or small the inner circle is, the area is fixed at 100π [m²], as long as the given 20m value is true.

[@merkulus-n5v]

Is it too early in the day for pi?? Never!

[@peetiegonzalez1845]

We have to assume there's a unique solution based on the question, regardless of the ratio of the diameters of the circles, so if the diameter of the inner circle is zero, the area of the red portion is π10². i.e. 314.16m². Took about 20 seconds. Haven't watched the video yet but if it's 6 minutes long surely I'm wrong... let's see... Edit: Spoiler alert. I wasn't wrong.

[@UncleKennysPlace]

I come here to be reminded of what I've forgotten.

[@snaramor]

It hurts my brain that you can't solve for X and Y. X can be anything ≥10 and the area of the outer ring will always be 100π.

[@roland3et]

Just shrink the inner circle to zero, and you'll get A = 10²π = 100π immediately 😉. 🙂👻

[@odalesaylor]

Should we mention that the area of the red part is in square meters?

[@nickwhite2996]

Notice that the answer does not depend on either radius, so any two circles that you can fit a 10 unit stick between differ by 10.pi in area

[@HxTurtle]

100*

[@jonathandavidson2042]

This is pure witchcraft, lol.

[@Lambrequin]

that's a long stick.

[@champfisk5613]

A bit tricky eh

[@pjgrieco]

How do you know that the the longer leg of the triangle is 10? What is the rule or reasoning that tells you that it is?

[@grahamjones9888]

The line is a tangent with a length of 20. The radius forming the other side of the triangle is at 90 degrees - therefore 10 units.

[@Potencyfunction]

@@grahamjones9888 That is Math level B chapter analytical geometry. there u learn about tangent and secant on a circle. The area is learn in early elementary school.

[@pjgrieco]

@@Potencyfunction ok but how do we know that the midpoint of the secant is the same as the the point where the line is tangent to the smaller circle?

[@johncirillo9544]

The 20 unit segment is a chord of the larger circle. This chord is a tangent segment to the smaller circle. The radius of the smaller circle drawn to the point of tangency is perpendicular to this tangent segment. Furthermore, If we imagine this radius to extend to the larger circle creating its radius, we have the perpendicular bisector of the chord by theorem. This is why the 20 unit chord is separated into two 10 unit segments.

[@bitpony]

The stick extends to the outer edge of the ring on both sides. Connect these two points to the origin of the outer circle (edge). You get an isosceles triangle (they are radii of the outer circle, with radius x). For such a triangle, the height (from the point at the angle between the two same length sides) will land on the middle of the opposite side. To prove, just take the small triangles (in the video with sides x, y and 10), and use a and b instead of 10 for the length of the two parts of the stick being divided by the height (so a+b = 20). Since it is a right angled triangle, we'll see that x^2 = y^2 + a^2 (from one side ) and x^2 = y^2 +b^2. Therefore we get: a^2 = x^2 - y^2 = b^2, so a = b (since a and b are distances we remove the negative answers), or 2a = 2b = a + b = 20 or a=b=10

[@FrankAnzalone]

i seem to remember a chord theorem that worked with less math from a puzzle book

[@apismelliferanaturalis]

You give a call to Math Queen and she says yes possible to solve then you answer I don't need more help, I found it ! 😇

[@davidmazzola5547]

God darn I have a crush.