Excellent solution! Congratulations.

@PreMath

5 ай бұрын

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The location of point G is not given, so the problem statement implies that the solution is the same for all placements which fit the diagram. So, we assume point G is on line segment BC, point O is above line segment EF and look for a special case which is straightforward to solve. I choose the limiting case as the distance between EF and O becomes infinitesimal. Then, the circle has diameter 12, making the rectangle's height 12, and the rectangle's base is 22. Rectangle area = (12)(22) = 264 sq. units. I note that yboboN found the same solution, but assumed that point O is allowed to be on line segment EG.

@jimleahy3858

Ай бұрын

For this to be a complete solution you need to prove the area of the rectangle is invariant under these conditions which the given solution effectively does.

Wonderful, teacher! Enjoyed the solution, thank you.☀

@PreMath

5 ай бұрын

You're very welcome! Thanks❤️

This channel will turn u in a genius overnight

@PreMath

5 ай бұрын

Thanks ❤️🌹

@michaelgarrow3239

5 ай бұрын

Which night?

@apexpredatorbilliardstraining

5 ай бұрын

@@michaelgarrow3239 for you maybe a decade cause u are slow

@michaelgarrow3239

5 ай бұрын

@@apexpredatorbilliardstraining - You must feel insecure. Lashing out at random people is just a coping mechanism. Didn’t your mom ever hug you?

@SherlockHomeless1

5 ай бұрын

@@michaelgarrow3239 dont ask dumb questions then

Exquisite 😍🥰

Very clever solution! Much appreciated!

@PreMath

5 ай бұрын

Glad it helped! Thanks❤️

@PreMath

5 ай бұрын

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Nice work hard🎉❤

@PreMath

5 ай бұрын

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I fiddled about with various dead-end routes for ages before I stumbled on your method.

Very clever

Good presentation.

@PreMath

5 ай бұрын

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Yeah, I did this problem in the exact same way. But only up until the connection between the length and width. I knew which proportion to use, but I didn’t realize EP and ET were the length and width for a moment.

@PreMath

5 ай бұрын

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Simply clever, many thanks, Sir, this is amazing.

@PreMath

5 ай бұрын

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Very nice!

@PreMath

5 ай бұрын

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Very nice

@PreMath

5 ай бұрын

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There is a confusion at 7:57 time. How can say that EP/EG=EF/ET. You should say that EP/EG=ET/EF. Suppose,there is 2 similar triangles which have ab, ac and ap,ag length. The ratio will be ab/ac=ap/ag.

@PreMath

5 ай бұрын

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So, it was not explicitly stated that the blue line of length 12 had endpoint at point E. Even though it LOOKS like the blue line ends at point E, in the general case you cannot assume this.

Amazing that we didn't even need to solve the radius!

@PreMath

5 ай бұрын

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Fantastic solution 👍, thank you teacher 🙏.

@PreMath

5 ай бұрын

Glad you liked it! Thanks❤️

I love that you didn't even need to bother with calculating the diameter. Well played, sir! :)

@PreMath

5 ай бұрын

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Beautiful solution

@PreMath

5 ай бұрын

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b = 2r (AD) cos(alpha) = 6/r (equilateral triangle EFO) cos(alpha) = a/12 (AB/12) => a/(12+10) = 6/(b/2) => ab = A = 22*12 = 264

I believe many people who tried to solve the problem, setup themselves to find first the size of width and the size of length, and them find the area of the rectangular. The video just begins with that. I believe not many people were expected to find the area without calculating sizes of the rectangular. That is the highlight of this problem particular solution

@PreMath

5 ай бұрын

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This was especially clever, using only the two lengths provided to determine the rectangle's area.

@PreMath

5 ай бұрын

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Perfecto 👍

@PreMath

5 ай бұрын

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Thank you

@PreMath

5 ай бұрын

Thank you too❤️

First I tried it the cheap way - if this is true at all, but hasn't specified location of G, then it's true when G is midpoint of BC, so AB = 22, diameter = 12, area 12*22. Then redid it properly and turned out to be ok. Neat problem.

@PreMath

5 ай бұрын

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Гениально!

So the area can be determined but not the length or width - or diameter of circle.

Very surprising puzzle😅 for unexpected simple solution. 😮 Let the rectangle be 2r×s, 2r cos x=12, s/cos x=22, 264=12×22=2r cos x× s/cos x=2r×s, that is the answer.😊

@PreMath

5 ай бұрын

Well done! Thanks❤️

I used a completely different way to solve this. So at the right top corner u can form a square with radius which is DMOE, in that square all corners are 90°, and then u draw another line with same length as EF from M to form a (idk what it was called), in order to make

Yay, I solved the problem. I defined the terms in relations to r, the radius of the circle. The width of the rectangle becomes 2r. I let a = length of rectangle minus 2r. Proportional triangles are drawn like the video, but my ratios were 12/2r = (2r + a)/(10 + 12). Cross multiplying and solving for "a" gives gives the additional length of the rectangle beyond 2r. a = (132/r) -2r. so the length of the rectangle becomes 2r + [(132/r) - 2r]. Multiplying length and width gives (2r + 132/r - 2r) * (2r) = (132/r) * (2r) = 264 square units.

@PreMath

5 ай бұрын

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form angle 0EG: width/(10+22)=(12/2)/radius => area =2*width*radius =(10+22)*12=264

@PreMath

5 ай бұрын

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@sergiykanilo9848

5 ай бұрын

it should be 10+12 instead 10+22 I wrote 22 originally, and then decided to change it to (10+12)

I like you so much❤

@PreMath

5 ай бұрын

Thanks dear❤️

Easy and fast to find the area using proportional triangles, but what are the dimensions L and H of the rectangle ? The solution is not unique: The largest rectangle is 22x12 (trivial solution), the highest is 21.10 x 12.51. Any solution between these limits (respecting LxH = 264) is possible.

A = B x H = (22 x cos(α)) x (2 x r) = 44 x r x cos(α) = 44 x (6 / cos(α)) x cos(α) = 44 x 6 = 264

At minimum width and maximum length, ABCD is a 12 × 22 rectangle. At maximum width and minimum length, ABCD is a square of 264 u².

@PreMath

5 ай бұрын

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Through out the mathematics formula is correct....got it

@PreMath

5 ай бұрын

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Interesting that the circle diameter, the length and width of the rectangle are not fixed. For example, if G is moved up to coincide with P and also F with T, then length = 12 + 10 = 22, and width = dia = 12.

@firstname4337

5 ай бұрын

"if G is moved up to coincide with P" -- then it wouldn't be a triangle anymore ... so who the hell cares

@georgebliss964

5 ай бұрын

Good point, it needs to be at least an atom's width short.@@firstname4337

@flash24g

5 ай бұрын

Indeed, possibly the first thing I noticed is that there is this degree of freedom. As such, a shortcut solution is to realise that, for the problem to be valid, the answer must be constant wrt the degree of freedom, and therefore solve it for this limit case.

@thewolfdoctor761

5 ай бұрын

Yes, G can be placed anywhere on CB and the problem hasn't changed.

@PreMath

5 ай бұрын

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What is the proof that angle P will be 90 degrees? The point P could be anywhere on CB... The solution is right but we need to prove that angle P is 90 degree...

12 : AB = 2r : (12+10) AB*2r = 12*22

@PreMath

5 ай бұрын

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where are you from i am from India 😅

@PreMath

5 ай бұрын

Thanks dear❤️ Love and prayers from the USA! 😀

264 HK bus

@PreMath

5 ай бұрын

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Oikein laaditun tehtävän periaate.(BG=CG) 5 s. päässälaskua.

@PreMath

5 ай бұрын

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Second to None !!!!!!!! kkkkkkkkkkkkkkkkkkkk

@PreMath

5 ай бұрын

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Ah ! oui ! Mais il fallait dire que E était le MILIEU de AD !!!!!!!!

@PreMath

5 ай бұрын

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Ist

@PreMath

5 ай бұрын

Thanks ❤️🌹

@davewright2561

5 ай бұрын

this is wrong as the line given as 12 is not the diameter of the circle so the diameter must be larger therefore the width of the rectangle must be larger than 12@@PreMath

Hi, there's a simple way to solve it. Because it's only a question of keeping ratio : just consider the Line passing thru the centre of the circle. So the diameter is 12 and lenght 12+10 So area is 12*22 = 264 That's all folks Keep in mind to Always take a step backward