Pitot Theorem Proof | Find area of the blue shaded circle | Tangential Quadrilateral | Trapezoid

[@vara1499]

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

[@PreMath]

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

[@sergeyvinns931]

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

[@Abby-hi4sf]

THANK YOU FOR PITOT THEOREM PROOF!

[@jackrubin6303]

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

[@PreMath]

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! 😀

[@tecevik6196]

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

[@PreMath]

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

[@theoyanto]

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 !

[@mariangorski]

I like very much your lessons. Thanks

[@PreMath]

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

[@sorourhashemi3249]

Perfect. Thanks

[@spiderjump]

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

[@KAvi_YA666]

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

[@HappyFamilyOnline]

Awesome👍 Thanks for sharing😊😊

[@marioalb9726]

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 √ )

[@ramanivenkata3161]

Excellent working

[@PreMath]

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

[@nirupamasingh2948]

V nice explanation.

[@PreMath]

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]

@@PreMath 🙏🙏

[@himo3485]

(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]

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

[@bigm383]

Love your work, Professor!🥂❤👍

[@PreMath]

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

[@bigm383]

@@PreMath 😂😄👌

[@jimlocke9320]

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.

[@zsoltszigeti758]

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

[@batavuskoga]

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

[@marioalb9726]

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]

There needs a proof those triangles are similar.

[@neelaramramesh3829]

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]

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! 😀

[@kennethstevenson976]

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.

[@AmirgabYT2185]

S=39,69π≈124,69

[@marcelowanderleycorreia8876]

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

[@EnnioPiovesan]

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. [...]

[@gelbkehlchen]

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

[@Copernicusfreud]

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

[@PreMath]

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

[@Law_Abiding_Citizen_ok]

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

[@dolphintoss]

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

[@mohamadtaufik5770]

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

[@josealmariferreiraferrreir3374]

👏👏👏👏👏👏👏👏👏

[@triplem1812]

This one was really hard for me

[@alster724]

Tricky at first but became easy midway

[@johnryder1713]

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

[@PreMath]

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

[@wackojacko3962]

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

[@PreMath]

My neighbor's dogs😀

[@mayeradmin5342]

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

[@harikatragadda]

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

[@ashieshsharmah1326]

How did you know these are similar

[@harikatragadda]

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

[@ashieshsharmah1326]

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

[@harikatragadda]

@@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]

@@harikatragadda thanks 🤩 . Very clever .

[@giuseppemalaguti435]

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

[@TTSchedule]

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