Beautiful question and more beautiful explanation 👍. Thank you teacher 🌹🙏.

@PreMath

Жыл бұрын

You are very welcome! Thanks for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

تمرين جميل جيد. رسم واضح مرتب. شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا . تحياتنا لكم من غزة فلسطين

This was difficult, I never could have done this without the video.. I nearly always find it difficult to get started. Ah well, onward and downward 🤓 But thoroughly enjoyed your solution, thanks again 👍🏻

@PreMath

Жыл бұрын

Thanks for your feedback! Cheers! You are amazing, Ian. Keep smiling👍 Love and prayers from the USA! 😀

تمرين جميل جيد .رسم واصح مرتب. شرح واصح مرتب .شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا . تحياتنا لكم من غزة غلسطين

@PreMath

Жыл бұрын

شكرًا عزيزي

Great 👍 Thanks for sharing😊

Very good!

Brilliant! Loved it, thanks

@PreMath

Жыл бұрын

Glad you think so! Thanks for your feedback! Cheers! You are awesome, Danny. Keep smiling👍 Love and prayers from the USA! 😀

I was not able to get this on my own. I like a good challenge. Great video PreMath!

well very useful

Beautiful question. Thank you Sir.

@PreMath

Жыл бұрын

Glad you think so! Thanks for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

What a fantastic solution, you have taught me so much from this. I give all my thanks to you.

@PreMath

11 ай бұрын

You are most welcome😀

Another technique in solving 96r²-500r-625= 0 Is by using the _Edge Test_ factoring method 24 25 4 -25 Then by cross multiplying and adding, we get the value of b (24)(-25)+(4)(25) -600+100 -500 So our factors are (24r+25)(4r-25) r= -25/24, 25/4 Since this is a geometric figure, we take the positive value, we have 6.25 units or 25/4

@PreMath

Жыл бұрын

Excellent! Thanks for your feedback! Cheers! You are awesome, Kevin. Keep it up 👍 Love and prayers from the USA! 😀

@ghhdcdvv5069

11 ай бұрын

شكرا جزيلا لكم والله يحفظكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين ​@@PreMath

@PreMath

11 ай бұрын

@@ghhdcdvv5069 Thanks dear

Thanks for video.Good luck sir!!!!!!!!

@PreMath

Жыл бұрын

You are very welcome! Thanks for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

Nice. I did not see an obviously solution that could be guessed at, or found by inspection. At always, the building blocks like the quadratic and Pythagoras formulas were there, and the solution lies in putting things together the right way which makes it interesting.

Thanks Professor, another gnarly little problem!🥂❤

@PreMath

Жыл бұрын

You are very welcome! Thanks for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

@bigm383

Жыл бұрын

@@PreMath 🍺😃👌

Very nice!

@PreMath

Жыл бұрын

Glad you think so! You are awesome, Alexander. Keep smiling👍

Many thanks, Sir, this is a real challenge - and a masterpiece solving it. btw: sin⁡(φ) = 2/(r - 2) = 8/17 → sin⁡(δ) = 3/(r - 3) = 12/13 → φ + δ > 90°

I absolutely love "Famous Identities" !🙂

@PreMath

Жыл бұрын

Glad to hear that! Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

we draw Auxiliary lines ON , OT for center and point of contact According Phythagorean thereom Radius rof semicircle r (r-2)²=4+x² x=√(r²-6) (5-x)²+2²=(r-2)² 25-10x+x²+4=r²-4r+4 r=6.25

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Made a sketch in CAD and eyeballed the semi circle. Measured it and is was about 6,3 so it must be 2pi ;)

Sir can we possibly work out the length of the vertical line M?

@PreMath

Жыл бұрын

Good question! Never thought about that...

@ybodoN

Жыл бұрын

When the green circles are in the ratio 3:2, the vertical line is equal to the diameter of the larger of the two circles 😉

@timeonly1401

10 ай бұрын

I poked around a found the length of the vertical line segment MC (If we call the point C where that vertical intersects the semicircle): From the video, we have r=25/4, and that x=√(r²-6r) = √{(25/4)[(25/4)-6]} = √(25/16) = 5/4. In right triangle AOP, the long leg AO = 5 - x = 5 - (5/4) = 15/4. Since AM = 2, and AO = 15/4, we have MO = AO - AM = 15/4 - 2 = 7/4. Draw radius OC. Then the 3 segments MO, MC & OC form a right triangle. By Pythag's Thm: MC² = OC² - MO² = r² - MO² = (25/4)² - (7/4)² = 576/16, and MC = √(576/16) = 24/4 = 6. Done!! [I'm STUNNED that that vertical turned out to be a "nice number" 6. 😅 ]

it took longer for me than expected because once the root must be substracted and once it must be added, which confused me: 10 r1=2:r2=3:sw=.01:ru=sw:ym1=r1:ym2=r2:ng=r1^2+r2^2:goto 60 20 dis1=(ru-r1)^2-ym1^2:dis2=(ru-r2)^2-ym2^2 30 if dis1

@PreMath

Жыл бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@zdrastvutye

Жыл бұрын

@@PreMath thanks for your attention, there was a problem with the result inside the sqr( ). now i have worked this out and have edited and updated the comment

Finally, a problem that someone didn't claim to solve in 35 seconds in their head.

👏

@PreMath

Жыл бұрын

Thanks dear for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

Tough problem involving two variables and superior insight to get the problem set up and produce the final answer. Again you gave each step required to produce the answer. I could not do this problem!

@PreMath

Жыл бұрын

Thanks for your feedback! Cheers! You are awesome, Kenneth. Keep smiling👍 Love and prayers from the USA! 😀

Кстати, 2-е решение (когда радиус отрицательный) тоже имеет место быть, оно вполне реальное, и его можно даже построить на рисунке. Задание: подумайте, как это изобразить 😜 === By the way, the 2nd solution (when the radius is negative) also takes place, it is quite real, and it can even be built in the figure. Task: think about how to portray it 😜

@PreMath

Жыл бұрын

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@photonomist6345

11 ай бұрын

I was wondering and thinking it must mean something, but I am not clever enough to visualise it!

@alexandermorozov2248

11 ай бұрын

@@photonomist6345 :) I can build a drawing with visualization especially for you.

@alexandermorozov2248

11 ай бұрын

@@photonomist6345 A negative radius can be represented as a radius laid in the opposite direction (hint).

@photonomist6345

11 ай бұрын

@@alexandermorozov2248 Thank you! I have it!

How do we know that ON and OT pass through P and Q

@rchandos

Жыл бұрын

We don't know that in advance,. He assumes that they do, contingent on locating the center of the semicircle such that ON equals OT.

@PreMath

Жыл бұрын

Dear TB, I believe I covered this topic in some of videos. Very simple explanation. I'll cover again pretty soon. No worries. Keep watching😀

Solution: R = radius of semicircle. From the center of the semicircle to the centers of the two small circles it is R-2 and R-3. These are the hypotenuses of two right-angled triangles whose centers are 2+3 = 5 apart. This gives the equation: √[(R-2)²-2²]+√[(R-3)²-3²] = 5 ⟹ √[R²-4R+4-4]+√[(R²-6R+9-9] = 5 ⟹ √[R²-4R]+√[R²-6R] = 5 |-√[R²-6R] ⟹ √[R²-4R] = 5-√[R²-6R] |()² ⟹ R²-4R = 25-10*√[R²-6R]+R²-6R |-R²+4R+10*√[R²-6R] ⟹ 10*√[R²-6R] = 25-2R |/10 ⟹ √[R²-6R] = 2,5-0,2R |()² ⟹ R²-6R = 6,25-R+0,04R² |-0,04R²+R-6,25 ⟹ 0,96R²-5R-6,25 = 0 |/0,96 ⟹ R²-125/24*R-625/96 = 0 |p-q-formula ⟹ R1/2 = 125/48±√[(125/48)²+625/96] = 125/48±√[15625/2304+15000/2304] = 125/48±√[30625/2304] = 125/48±175/48 ⟹ R1 = 125/48+175/48 = 300/48 = 6,25 and R2 = 125/48-175/48 = -50/48 = -25/24 [invalid in geometry]

I would like to get in contact with you premath. How can I do so?

@PreMath

Жыл бұрын

You may email us: premathchannel@gmail.com Thanks for asking. Cheers

Ok, I knew that. Yep. So, what's for dinner?

Why are we sure that OT and ON go through the centers of green circles?

@rchandos

Жыл бұрын

They have to so that both the semicircle and the smaller circles are tangent at T and N. His assumption turns out to be justified because he manages to locate the center of the semicircle such that OT=ON.

@PreMath

Жыл бұрын

Hello dear, I believe I covered this topic in some of videos. Very simple explanation. I'll cover again pretty soon. No worries. Keep watching😀

@4iriska

Жыл бұрын

@@rchandos Thank you

Rather difficult, this puzzle, 🤔 two circles do not touch each other, but the common vertical tangent, do horizontal distance between P and Q is 2+3=5, let r be radius of the semicircle, so square root((r-2)^2-4)+square root((r-3)^2-9)=5, we can find r by solving this equation, a little bit troublesome.🤨

@PreMath

Жыл бұрын

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Where is the proof that these respective points are collinear? It looks obvious, but how can one be sure!

Sir what is your age

@PreMath

Жыл бұрын

Dear Vaibhav, I'd talk about myself pretty soon. I can understand your curiosity. Thanks for your continued love and support! You are awesome. Keep smiling👍

@wackojacko3962

Жыл бұрын

I absolutely love "Famous Identities"!🙂

@SohailKhan-xx8wi

Жыл бұрын

35years old.

@vaibhavgupta5359

Жыл бұрын

@@PreMath sir you have a sharp brain I want to know how do u relate all the concepts of maths so easily

@maturaczynicuda

Жыл бұрын

@@vaibhavgupta5359 true, the possibility of learning from such a great teacher is a great privilege

(R^2-6R)^0.5+(R^2-4R)^0.5=2V6 --> R=6.223

@ybodoN

Жыл бұрын

If the two circles were tangent, this would be the correct answer...

@mohamadtaufik5770

Жыл бұрын

I thought there were tangent, thanks for the correction

Difficult

Ok I need help with this one

First go to a school and lean how to pronounce English Language. Your Math skills are good but pronunciation is to be improved

You can still solve even if the green circles were tangential