One more interesting and challenging problem. Thank you, Prof for the solution.

@PreMath

Жыл бұрын

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

@sergeyvinns931

11 ай бұрын

Какой на хрен экс и уай, если это одно и то же, то есть радиус вписанной окружности?

THANK YOU FOR PITOT THEOREM PROOF!

Dear Professor, you are preventing me from acquiring Alzheimer’s dementia. Love from Australia

@PreMath

Жыл бұрын

I call these math puzzles "gymnastics for the mind!" They make us think and improve mental agility! Thanks for your feedback! Cheers! You are awesome, Jack. Keep smiling👍 Love and prayers from the USA! 😀

Awesome👍 Thanks for sharing😊😊

Marvelous example, I had no idea such a theorem as Pitot existed, thanks for broadening my view, geometry is just so much more interesting than I ever imagined. Thank you very much You are indeed the Alchemist of geometry, its as if you say "let it be so" And it is so !

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

Generally the radius is ab/(a+b), where a, b is the length of upper and lower sides, it can be proved without any theorem, even pythagoras' theorem, just considering two pairs of similar tiangles. In the particular case, r=9x21/30=6.3.

Taking the right triangle, we have: (2r)²+(21-9)²=(21-r+9-r)² 4r²+144=(30-2r)² 4r²+144=30²-120r+4r² 120r=756 r=756/120 r=6,3 cm Area = π r² Area = 124,7cm² ( Solved √ )

(21-9)²+(2r)²=(21-r+9-r)²　　　144+4r²=900-120r+4r²　 120r=756 r=63/10 63/10*63/10*π=3969π/100 (Area of Blue shaded Circle)

@PreMath

Жыл бұрын

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

Wow. I don't ever see this theorem. Thank u!

@PreMath

Жыл бұрын

Keep watching... You are very welcome! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

For those who didn't know the Pitot theorem: I am assuming that viewers will find that lengths DF and AE are r and, in ΔBCE, find the side lengths of 2r and 12. After designating length FC as z and BE as w, two tangent theorem produces the hypotenuse length as z + w. We derive 2 equations: r + z = 9 for the trapezoid's top side and r + w = 21 for the trapezoid's bottom side, and add them: 2r + z + w = 30. Subtract 2r from both sides and z + w = 30 - 2r. So, our hypotenuse length z + w = 30 - 2r and we can proceed to solve for r using the Pythagorean theorem, as PreMath did. Then, we compute the area of the circle from r.

By Pythagorean theorem, 2r•2r + 12•12 = (30-2r)^2 Solve for r Then area of blue circle = pi•r^2

Love your work, Professor!🥂❤👍

@PreMath

Жыл бұрын

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

@bigm383

Жыл бұрын

@@PreMath 😂😄👌

if we take the internal right triangle whose legs are '9-r' and 'r', and the other internal right triangle whose legs are '21-r' and 'r', we can apply similarity of triangles (9-r)/r = r/(21-r) r² = (9-r).(21-r) r² =189-30r+r² 30r=189 r=189/30 r=6,3 cm Area = π r² Area = 124,7cm² ( Solved √ ) and was not necessary to apply the pitot theorem

@Benbou969

Жыл бұрын

There needs a proof those triangles are similar.

I like very much your lessons. Thanks

@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! 😀

EBNO and CFON are similar deltoids, because the sum of the angles at O is 180, so (9-r)/r=r/(21-r) => r=63/10

Excellent working

@PreMath

Жыл бұрын

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

I found the BC by the tangent theorem 30-2r, not by the Pitot theorem, I haven't heard of this before. I also dropped a perpendicular line onto AB Then you will find the solution using the Pythagorean theorem

V nice explanation.

@PreMath

Жыл бұрын

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

@nirupamasingh2948

Жыл бұрын

@@PreMath 🙏🙏

Let R be the point of tangency between the circle and the side BC. Join O with C and B to form the angles COP=COR = α and BOR = BOQ = β. The sum of these 4 angles is 180° and consequently α and β are complementary. Triangles CPO and OQB are similar (both are right triangles, in P and Q respectively, and have α and β as one of the other angles). So we can write: PC:PO = OQ:QB Observe that PC = DC-DP=9-r QB=AP-AQ=21-r So we have: (9-r):r=r:(21-r) --> (9-r)·(21-r) = r² --> 9·21-9r-21r+r²=r² --> 9·21=r(9+21) --> r=3²·7·3/3·2·5 --> r=63/10. [...]

Solution: r = Radius of the circle. One can connect O with the touch points of the circle, down E, on the right side F and up G. And one can connect O with B and C. Then, because O lies on the bisecting line, you will see, that triangle EBO is congruent to OBF and triangle OFC is congruent to OCG. ⟹ EB = 21-r = FB and GC = 9-r = CF ⟹ CB = FB+CF = 21-r+9-r = 30-2r ⟹ Pythagoras: (2r)²+(21-9)² = CB² = (30-2r)² ⟹ 4r²+144 = 900-120r+4r² |-4r²-144+120r ⟹ 120r = 756 |/120 ⟹ r = 756/120 = 6,3 ⟹ Blue Area = π*r² = π*6,3² = 39,69π ≈ 124,6898

63/10 = 6.3 , If you are using 3.14 and working without a calculator the answer would be 124.6266 sq. un. 124.69 sq. un. is found when you use the Pi button on the calculator and round off to two decimal places. Great presentation made possible by knowledge of the Tangential Quadrilateral Theorem.

Circlactular sir🎉🎉🎉kiddos to your noble work. I adore your presentation. Iam mathurious student ,could you please make a sum of all important geometrical theorems and formulations together in one whole video series.dhanyavaad from india

@PreMath

Жыл бұрын

You are very welcome! Nice suggestion. So nice of you, Ramesh dear Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

S=39,69π≈124,69

Prof. is it possible to solve such problem without this pitot theorem? Is there another way to sove it? Thank you so much!!

Let r be the radius, so CB is 9-r+21-r=30-2r, by pythagorean theorem, (2r)^2+(21-9)^2=(30-2r)^2, 4r^2+144=900-120r+4r^2, 120r=900-144=756, r=756/120=6.3, area is 6.3^2 pi=124.7 approximately.🙂

@PreMath

Жыл бұрын

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

make a point P for point of tangent on the line CB According to Two-Tangent theorem line CP = 9-R line PB = 21-R therefore, line CB = 30-2R According to Pythagorean theorem line OC² = (9-R)² + R² line OB² = (21-R)² + R² line CB² = {(9-R)² + R²} + {(21-R)² + R²} = (30-2R)² 4R² - 60R + 522 = 4R² -120R + 900 60R = 378 R = 63/10 πR² = 3969π/100 ≈ 124.69

👏👏👏👏👏👏👏👏👏

Tricky at first but became easy midway

(2R)^2+(12)^2=(30-2R)^2 R=6.3 --> Lblue=pi*R^2=pi*(6.3)^2=124.69

Yay!, I solved it with 39.69*(pi).

@PreMath

Жыл бұрын

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

How is (2r)squared = 4rsquared at 8:42??

When you call AB a and CD b the solution is r = ab/(a + b)😀😀😀

This one was really hard for me

@ 2:48 the answer was blurted out by the dog barking...it's a pretty smart dog. 🙂

@PreMath

Жыл бұрын

My neighbor's dogs😀

Don't think its possible to come to this channel and not find something new and worthwhile!

@PreMath

Жыл бұрын

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

∆OPC is Similar to ∆OQB. (9-r)/r = r/(21-r) r = 63/10

@ashieshsharmah1326

Жыл бұрын

How did you know these are similar

@harikatragadda

Жыл бұрын

@@ashieshsharmah1326 If ∠QOB = θ, then, ∠POC = ½(π-2θ) = ½π - θ Hence ∠OCP = θ By *AAA* , ΔOPC is Similar to ΔOQB.

@ashieshsharmah1326

Жыл бұрын

@@harikatragadda thanks 🤩 , but could you please explain why poc = 1/2(π-theta)

@harikatragadda

Жыл бұрын

@@ashieshsharmah1326 If K is the point of tangency of CB, then ∆POC is Congruent to ∆CKO, and hence ∠POC = ∠KOC. Similarly, ∠QOB = ∠KOB =θ Hence, ∠POC = ½∠POK = ½(π-∠QOK) = ½(π-2θ)

@ashieshsharmah1326

Жыл бұрын

@@harikatragadda thanks 🤩 . Very clever .

((9-r)+(21-r))^2=(2r)^2+(21-r)^2...r=sqrt1980-39... Boh?? Ho sbagliato é 21-9...

This guy made it WAY too complicated. In less than a minute I had the answer. You subtract CD from AB (=12). That's the diameter of the circle, making the radius = 6. Square it = 36. Multiply 36 by pi (3.1415). Answer is 113.094