Find area of blue shaded region in a circle | Chord lengths are 4 and 8 | Important skills explained

[@bigm383]

Thanks, Professor. Great fun.❤🥂

[@PreMath]

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

[@اممدنحمظ]

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

[@JLvatron]

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

[@theoyanto]

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

[@murdock5537]

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

[@nandisaand5287]

Wow, much of this went right over my head.

[@jaimeyomayuza6140]

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

[@sohail2556]

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

[@sujatasahoo2949]

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

[@melihpuskulcu8335]

Excellent solution!

[@himanshusharma2387]

Fantastic problem😊

[@PreMath]

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

[@jakkima1067]

Very Nice!!! Thank You!

[@KAvi_YA666]

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

[@PreMath]

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

[@jeanmarcbonici9525]

Nice Thank you

[@mehulpunia6174]

Thanks sir it was very new concept

[@Teamstudy4595]

Sir Interesting Question indeed !

[@stephenbrand5779]

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.

[@Okkk517]

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.

[@eulerthegreatestofall147]

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

[@amitsinghbhadoriya6318]

Thanks

[@PreMath]

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

[@rjpattnaik4781]

Well good evening from India master.

[@santiagoarosam430]

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]

∆EDB is a 30-60-90 triangle

[@santiagoarosam430]

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

[@TDSONLINEMATHS]

thanks professor

[@tariqmumtaz9362]

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 .

[@atiqaked838]

شكرأ جزيلا من الجزائر 🇩🇿

[@michaelkouzmin281]

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...

[@Ankitsingh-y4j8m]

Good question sir

[@PreMath]

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

[@cmmp5510]

Really dificult to start. Beautiful problem

[@alexundre8745]

Good morning MASTER Thanks Sir

[@PreMath]

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! 😀

[@alexundre8745]

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

[@srinivasansundararajan456]

Super sir

[@ashumishra5618]

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]

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

[@del66404]

👏👏👏👏👏👏👏👏👏👏

[@shadmanhasan4205]

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.

[@bramont6225]

Buenos días profesor bonito problema Gracias por esta clase

[@PreMath]

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