Thanks, Professor. Great fun.❤🥂

@PreMath

Жыл бұрын

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

@jordanborissov5043

5 ай бұрын

УуууАауауауауау

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

Wow this was challenging, but amazing. Thank you, professor!

This was particularly challenging, I couldn't get started... Your step by step method was fascinating to watch. Fabulous fabulous fabulous 😎👍🏻

Excellent, Sir, many thanks! This is awesome! AP = a; BP = b; CP = c; DP = d; AC = 4; BD = 8; OA = OB = OC = OD = OE = r ab = cd → a = cd/b a(a) + c(c)=16 b(b) + d(d) = 64 → 4c^2((d/b)^2 + 1) → 2c = b → ab = cd = 2ac = cd → 2a = d → ∆ADP → AP = a → DP = d = 2a → AD = a√5 → sin⁡(φ) = AP/AD = √5/5 → cos⁡(φ) = (1 - sin^2(φ)) = 2√5/5 → sin⁡(2φ) = 2sin⁡(φ)cos⁡(φ) = 4/5 → cos⁡(2φ) = (1 - sin^2(2φ)) = 3/5 16 = 2r^2(1 - cos⁡(2φ)) → r = 2√5 → BE = OB + OE = 2r → (BE)^2 - (DE)^2 = 80 - 64 = 16 → DE = 4 → EOD = 2φ → DOB = 2β = 180° - 2φ 64 = 2r^2(1 - cos⁡(2β)) → cos⁡(2β) = -3/5 DE/BE = sin⁡(φ) = √5/5 → area ∆DOE = ∆AOC = ∆DBO = 8 → blue area = 20π(2φ/360°) + 20π(2β/360°) - 16 = 20π(0,1476 + 0,3524) - 16 = 10π - 16 ≈ 15,4159 btw: ADC = ABC = φ BAD = BCD = β → BOD = 2β → DOE = 180° - 2β = 2φ → OED = EDO = β → ABC = DBE = φ → DE = AC = 4

Wow, much of this went right over my head.

Fantastic problem😊

@PreMath

Жыл бұрын

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

Wonderful Solution. Thanks Sir. From Bogota D.C. Colombia

Excellent solution!

wonderful sir , this wasn't an easy problem , what an explanation ! 👍👍👍

Very Nice!!! Thank You!

Keep going with this content literally mind boggling questions 🙂🙂🙂 Amazing work 👍👍👍

thanks professor

Thanks sir it was very new concept

Nice Thank you

Sir Interesting Question indeed !

One way to approach this problem is analytic geometry: let P the intersection point of AB and CD. Then let AD=x, BP=w, CP=y and DP=z. Now if we take point P as the origin we deduce point O((x-w)/2,(y-z)/2). Moreover AO^2=(x^2+w^2)/4 + (y^2+z^2)/4 can be determined using the intersecting cord theorem. However, AO^2=the radius r^2. By observing that x^2+y^2=4^2 and w^2+z^2=8^2 we find r=sqrt(20). Next once r is found we can easily calculate the blue region as 10*pi-16.

Well good evening from India master.

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

@PreMath

Жыл бұрын

Thank you too You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

شكرأ جزيلا من الجزائر 🇩🇿

Super sir

Potencia de P respecto a la circunferencia: CPxPD=APxPB → CP/AP=PB/PD → ∆APC∼∆DPB → ∠A=∠D; ∠C=∠B→∠A+∠B=90º → Hacemos simetría de ∆APC tomando como eje el diámetro vertical → A se superpone a B y C se transforma en C´ → El nuevo triángulo ∆C´BD es rectángulo puesto que ∠A+∠B=90º y por tanto está inscrito en un semicírculo → La hipotenusa C´D es diámetro del círculo → C´D=2r → 8²+4²=(2r)²→ r²=20 → Área azul = (Área del semicírculo) - (∆C´BD) =(πr²/2)-(8x4/2) =10π-16 Gracias y saludos cordiales.

@alster724

Жыл бұрын

∆EDB is a 30-60-90 triangle

@santiagoarosam430

Жыл бұрын

@@alster724 No es así. ¿Qué quieres decir con esa afirmación?. Un saludo.

Like a magic show🥰

I got same results using similar procedure at the beginning, but totally different when estimating areas, i.e, the blue shaded region.

Thanks

@PreMath

Жыл бұрын

You are very welcome! Thank you! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Really dificult to start. Beautiful problem

Area of the semi circle also includes the area between the line ED and arc ED, therefore we also have to calculate this area ("a") as well and only then the blue shaded area can be calculated: AREA OF SEMICIRCLE - (AREA OF TRIANGLE BED + AREA "a"). Therefore, I am not so sure about your solution .

This is the most interesting and challenging problem that I have managed to solve so far. But I used a completely different method, similar triangles and trigonometry to find the radius and angles subtending AC and BD. Then the standard formula to find segment areas. I think your solution is more elegant.

Good question sir

@PreMath

Жыл бұрын

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

I managed to solve it numerically (15.416 sq.units) only findig alpha = D64.435 and beta = D26,565, then passing to respective center angles. r^2 and r via cosine theorem etc...

Good morning MASTER Thanks Sir

@PreMath

Жыл бұрын

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

@alexrocha9191

Жыл бұрын

@@PreMath Continuo firme e forte seguindo o Sr Grato pelos ensinamentos

👏👏👏👏👏👏👏👏👏👏

It seems to be very difficult😮. Clearly there are two similar right angles triangles with bases 4 and 8, ... Maybe out of my ability to solve it.😅

@PreMath

Жыл бұрын

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

Sir, I've a doubt that (1= -1) Proof: (1)²=1 and (-1)²=1 R.H.S are equal of both the equations then L.H.S will have to equal (1)²=(-1)²=> 1=-1 [because mª=nª=> m=n] Sir how is it possible ( 1=-1 ). Please make a video about this topic. Sir I humble request of you please.

@RajeshKumar-ms3br

Жыл бұрын

See so if we simplify -1², we get -1*-1. Same applies to 1,then- 1²=(-1)² =1*1=-1*-1 So you are basically multiplying both the sides by different no. So this is false. Though this is my way of seeing, could be wrong. Sorry if this is wrong

Ah I get it. The triangle with side AC and the triangle with side BD are Congruent triangles. Still don't know wtf ro do 😅. Edit: ok. Let's assume that both triangles are ones that have 30 dgree and 60 degree angles: The triangle with AC has 2Root(3) = 3.464 as longer side-lenghth and 2.0 as shorter side-lenghth. Meanwhile, triangle with BD has the shorter side-lenghth of 4.0 and the longer side-lenghth of 4Root(3) = 6.928.

Buenos días profesor bonito problema Gracias por esta clase

@PreMath

Жыл бұрын

¡De nada! ¡Gracias por tus comentarios! Usted es maravilloso. Sigue así 👍 ¡Amor y oraciones desde los EE. UU.! 😀