Can you find the Area of the Green shaded Region? | Quick & Easy Tutorial

[@mahalakshmiganapathy6455]

Super explanation i like it very much

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome Mahalakshmi😀 Love and prayers from the USA!

[@prbmax]

What makes your content so great is that I have no clue where to start but by the end, you can't help but learn something.

[@PreMath]

Thank you for your feedback! Cheers! You are awesome Paul 😀 Love and prayers from Arizona, USA!

[@lraoux]

These are addictive! Figured it out as soon as I saw that BD=8. Keep 'em coming, please.

[@EddieDraaisma]

Rotate the top-right green sector anti-clockwise around B by 60 deg (C -> D), and after that fold half of the moved sector below AB anti-clockwise by 180 deg around the middle of AB (D' to C'). All the green then forms a 30 deg pie slice with radius 8...

[@PreMath]

Nice tip Thank you for your feedback! Cheers! You are awesome Eddie😀

[@jefffinkbonner9551]

You’re going to have to make your own video and post it to demonstrate that…

[@jaiprashanth2005]

Any one can explain briefly about this pls ?

[@stephenrosenfeld833]

If only this medium and this quality of explanation were available to me 55 years ago!! I always enjoyed high school math, but this is so much more fun!

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome Stephen😀

[@AnonimityAssured]

My own method was a bit of a shortcut. I could see that if the green segment on the right were halved, and the two halves attached to the green shape on the left to form a 30-degree sector, then that sector would have the same area as the two green areas, which would be a twelfth of the circle's area.

[@PreMath]

Thank you for your feedback! Cheers! You are awesome Julian😀

[@hstrinzel]

Wow, MUCH BETTER approach.!

[@murdock5537]

Awesome question, excellent way to solve it, many thanks, Sir! φ = 30° → BC = 8 → AC = 16 → AB = 8√3 DBC = 2φ → AD' = BD' = AB/2 = 4√3 → area ∆BCD = 16√3 r = 8 → πr^2 = 64π → 64π(60/360) = 32π/3 → 32π/3 - 16√3 = (16/3)(2π - 3√3) DAB = ABD = φ → area ∆ABD = 16√3 64π(30/360) = 16π/3 → green area = (16/3)(2π - 3√3) + (16/3)(3√3 - π) =16π/3 or: because ED'D = (1/2)(1/3)(32π - 48√3) → green area = (1/3)(32π - 48√3) + 8√3 - (1/6)(32π - 48√3) = (1/6)(32π - 48√3 + 48√3) = 16π/3

[@242math]

very well explained, your step-by-step teaching is so easy to follow and understand, thanks for sharing this area problem

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome my friend😀

[@Gargaroolala]

Area of ABC = (0.5)(8)(8/tan 30) = 55.42cm2. Area of quadrant = (0.25)(pi)(8)(8) = 50.27 cm2. Area of triangle DBC = (0.5)(8)(8)(sin 60) = 27.71 cm2. Area of sector BDE = (30/360)(pi)(8)(8) = 16.76 cm2. Area of green shaded area = 55.42 + 50.27 - 2(27.71 + 16.76) = 16.75 cm2.

[@PreMath]

Excellent Thank you for your feedback! Cheers! You are awesome Garrick😀

[@michalswiderski507]

I love your videos, keep me on my toes and like very much from time to time refresh my math basics. so many cool stuff you presented. great job. THANKS and Merry Christmas.

[@PreMath]

Glad you like them! Thank you for your feedback! Cheers! You are awesome Michal😀 Merry Christmas. Love and prayers from the USA!

[@seagulld100]

Great problem! Very interesting. I’ve learned how to solve from watching your previous videos.

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome David.😀

[@Mathematician6124]

It was very easy to me. My friend pointed it out without cosine rule and I did point it out with cosine rule. I love to figure out two methods of solution of a single problem. Because on one hand it increases skill on the other hand, it enhances our creativity. It also makes it possible to teach a junior student. A boy of class 7 must not know cosine rule, for him I would like to think another solution which would make him understand it.

[@steverobbins4872]

You got the right answer, but there is an easier way. 1) Draw a line perpendicular to CD, at the midpoint of CD. This line bisects the upper green area into two equal pieces. 2) Join these two pieces together along their long sides, so they are back-to-back. 3) Join this to the lower green section, and you get an area shaped like a pizza slice. The 30 degree angle makes this 1/12 of the pizza. 4) The answer is now obvious: Compute the area of a circle with radius=8, then divide by 12.

[@MrPaulc222]

I got there in the end. I got the right-hand green bit with (64/6)pi - 16*sqrt(3), then the left hand part with 16*sqrt(3) - (64/12)pi (what a palaver). Then add up. The 16*sqrt(3) parts cancel each other so I got to (64/6)pi - (64/12)pi which reduces a bit to (32/6)pi then (16/3)pi for the16.7...sq units

[@seroujghazarian6343]

Area of top green sector=area of the section tended by the arc CD-Area of BCD=1/2×64×pi/3-1/2×64×sqrt(3)/2=32pi/3-16sqrt(3) Area of bottom green section=Area of ADB-Area of section subtended by the arc DE=16sqrt(3)-1/2×64×pi/6=16sqrt(3)-16pi/3 Area of all green parts=32pi/3-16sqrt(3)+16sqrt(3)-16pi/3=16pi/3

[@tieutucvinh]

Thank guy,the way easiest for the normal lesson.

[@zplusacademy5718]

Extremely beautiful and amazing solution dear sir keep.... showing... always us Neo ideas and methods....❤️🙏❤️🙏❤️❤️❤️🙏🙏🙏🙏

[@PreMath]

So nice of you dear Thank you for your feedback! Cheers! You are awesome 😀

[@zawatsky]

Более прямолинейное решение: ▫ΔBCD - равнобедренный, BC=BD=8; ▫∠BCD=180-30-90=60°. ΔBCD равнобедренный => ∠BDC=∠BCD=60°. Т. о. ∠CBD=180-60-60=60° => ΔBCD - равносторонний (правильный); ▫находим площадь сектора BCD, находим и отнимаем площадь ΔBCD. Получаем площадь верхнего сегмента DC; ▫находим площадь ΔABC; ▫находим площадь четверти круга BCE; ▫площадь нижнего сегмента ADE=площадь ΔABC + площадь сегмента DC - площадь четверти круга BCE; ▫поскольку нам нужна общая площадь обоих сегментов, она выражается как площадь ΔABC + 2*(площадь сегмента DC) - площадь четверти круга BCE; Можно ещё развернуть выражение: четверть круга плюс сектор 60° это будет сектор 150°, от него уже надо отнимать ΔBCD.

[@johnbrennan3372]

Great flow of interesting problems.

[@PreMath]

Keep watching! Thank you for your feedback! Cheers! You are awesome John.😀

[@JLvatron]

“Excellent!”

[@PreMath]

Super Thank you for your feedback! Cheers! You are awesome JL😀

[@ikeetkroketjes8431]

there is a 30* and a 90* so its a 1:sqrt3:2 triangle •8, so AB=8sqrt3, AC=16 the area of AED=ABC-EBCD. EBCD is the 90* part of the circle-the green shaded aread of the circle. since angle ACB=60*, if you construct triangle BCD, it has 2 equal sides of 8, since those are both the radius. those stand on the opposite side of angle BCD and BDC, so those angles are also equal. angle C in ABC= angle C in BCD, so angle BCD=60* and that is equal to BDC so that is also 60*, this means that angle CBD=60* as well by substraction. since we know the angle and the radius of the green shaded part of the circle, we can calculate its area: its equal to 60*/360*•area of the circle=64π/6=10.2π. now we substact the triangle BCD:. since its a special 1:1:1 triangle, the area is 2•side•sqrt3= 1/2•8•4sqrt3=16sqrt3. so the green shaded part of the circle is 10.2π-16sqrt3. recall ADE=ABC-(circle/4-green shaded pt. circle)=ABC-(16π-(10.2π-16sqrt3))=ABC-5.8π+16sqrt3. now all thats left is calculate the area of ABC. the area of a triangle is (b•h)/2, we will use b=AB and h=BC=8 , since ABC is a 1:sqrt3:2 triangle, AB=sqrt3•8=8sqrt3. so the area of ABC=(8sqrt3•8)/2=32sqrt3 so ADE=ABC-5.8π+16sqrt3=48sqrt3-5.8π. now add the green shaded area of the circle to it and we’re done: (48sqrt3-5.8π)+(10.2π-16sqrt3)=32sqrt3+4,4π

[@HassanLakiss]

Thank you for a very interesting question and a very good explanation. With respect, if the radii of the sectors are the same and one angle of the sector is twice the other, then the required area is simple half the area of the larger sector or just the area of the smaller sector. Furthermore, it may be done by rotating the 60° sector about B anticlockwise and with a bit of folding, it would become clear that the area of the green region is 16π/3. Unfortunately I cannot upload a photo of the arrangement I am talking about. Thank you. God Bless

[@PreMath]

Great tip! Thank you for your nice feedback! Cheers! You are awesome Hassan😀 Love and prayers from the USA!

[@HassanLakiss]

@@PreMath Thank you Sir.

[@HassanLakiss]

@@PreMath Thank you Sir. All the very best for 2022.

[@philipkudrna5643]

It took me some calculations and I hope I haven‘t messed it up, but leveraging on the properties of 30/60/90 triangles, the properties of an equilateral triangle and the area of a circle, I arrived at 16/3*pi or 16,75 square units. After watching: realizing that the two triangles have the same area would have made my live easier and explains, why in the end everything neatly cancelled away…!

[@PreMath]

Thank you for your nice feedback! Cheers! You are awesome Philip.😀

[@calspace]

I did it slightly differently, but got the same answer, although it took me three times to make sure my math was correct. The green area is triangle ABC - quarter circle B +2 times the green arc DC. ABD = 32sqrt3. Quarter circle B = 16π Curve BC =1/6 circle B - triangle BCD = (64/6)π - 16sqrt3 So… 32sqrt3 - 16π + 2[ (64/6)π - 16sqrt3] 32sqrt3 - 16π + (64/3)π - 32sqrt3 (64/3)π - 16π (64/3)π - (48/3)π (16/3)π

[@PreMath]

Excellent Thank you for your feedback! Cheers! You are awesome 😀

[@ROCCOANDROXY]

In general, let r be the radius of the quarter circle with m(angle(A)) = 30(degrees) implies Area (Green) = pi/12 * r^2 = 1/3 * (pi/4 * r^2) = 1/3 * Area (quarter circle) implies Area (Green)/A (quarter circle) = 1/3.

[@marioalb9726]

Area of quadrant : Aq= π R²/4 Aq= π 8²/4 Aq= 50.265 cm² Base of right triangle: tan β = h / b b = h / tan β b = 8 / tan 30° b = 13,856 cm Area of right triangle: At = ½ b. h At = ½ 13,856 . 8 At = 55,426 cm² Area of green circular segment: Acs = ½ R² (α - sin α) Acs = ½ 8² (60° - sin 60°) Acs = 5,797 cm² Green Shaded Area: A = At - Aq + 2.Acs A = 55,426 -50,265 + 2 x 5,797 A = 16.755 cm² ( Solved √ )

[@luigipirandello5919]

Amazing solution. Thank you.

[@PreMath]

You're welcome Luis! Glad to hear that! Thank you for your feedback! Cheers! You are awesome 😀

[@shango6164]

That was beautifully explained. A very interesting problem. Thanks.....

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome Shango😀

[@bobeyay]

Its amazing we can solve this kind of problems with just basic geometry knowledge and with a little bit of logic. I mean sure you can use trigonometry but that’s higher lv math than this

[@mkpet13]

I probably would have solved this using the segment of a circle formula A = (½) × r 2 × [(π/180) θ - sin θ], then use half of the green circle segment to create a right triangle with new point F. ADF having side AD=8, DF=4, a=30*. Solve area of ADF, then add half the area of circle segment.

[@PreMath]

Thank you for your feedback! Cheers! You are awesome 😀

[@susennath6035]

Excellent. Specially sector analysis

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome Susen.😀

[@sharadaprasaddhakal9213]

I liked your video on premath . Thank you sir !

[@PreMath]

So nice of you Sharada Thank you for your feedback! Cheers! You are awesome 😀 Love and prayers from the USA!

[@montynorth3009]

Nice one. Well explained, thanks.

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome Monty.😀

[@msafasharhan]

DR.How the two rectangles equal

[@asilaalkhmisi7922]

thank you Mr💪🏻

[@raya.pawley3563]

Thank you

[@lavoiedereussite922]

شكرا.

[@isaacquevedoalmarza8736]

Es un buen ejercicio mental, pero ya existen soluciones a estos problemas con teoremas.

[@davidfromstow]

Excellent

[@PreMath]

Thank you for your feedback! Cheers! You are awesome David😀

[@IAMAHEROHAHAHA]

There’s another easier method, no need to calculate so much, just compare the relationship to know: the shaded area is equal to 1/12 of the circle area.

[@loveislam8189]

Neat solution;I love it😚

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome 😀 Love and prayers from the USA!

[@loveislam8189]

@@PreMath ... I got stuck because I could not figure out how to sum the areas up... But I really appreciate the solution you gave...

[@pranavamali05]

Brilliant question thnx a lot

[@PreMath]

Most welcome Pranav dear Thank you for your feedback! Cheers! You are awesome 😀

[@zatho5421]

it is weird. Triangle ABD looks a lot larger than BCD, but math says they are identical in size.

[@DB-lg5sq]

من المغرب شكرا على المجهودات للتبسيط نأخذ الوحدة هي u=8

[@sameerqureshi-kh7cc]

Premathers rocks 😊👍

[@PreMath]

Thank you for your feedback! Cheers! You are awesome Sameer.😀 Love and prayers from the USA!

[@jas2819]

Wow........⭐️⭐️⭐️

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome James😀

[@theoyanto]

Great problem, I didn't quite get your methodology but I got very close to the right answer, I can't just see at the moment where I went wrong, my final answer was 0.66 short 🙄 ah well tomorrow is another day 👍🏻 Ah...just found my error, I didn't add my areas properly... Doh!

[@AmirgabYT2185]

S=16π/3≈16,76)

[@josecarlosmyersbenavides8755]

PRESENTO MIS EXCUSAS POR QUE NO HABIA VISTO EL VIDEO COMPLETO SI ESTA CORRECTAMENTE RESUELTO

[@giuseppemalaguti435]

I miei calcoli sono Averde=(16/3)pi..... Ho visto che tu hai utilizzato un altro metodo, non il mio(intersezione e integrale)

[@PreMath]

Super

[@stickmanbattle997]

My answer 16.40 , i think my wrong was the construction

[@dllord]

My ans is 17,53... Im sтuрid

[@giuseppemalaguti435]

Devo lavorare di geometria analitica e calcolare lintersezione tra la retta e la circonferenza e usare gli integrali.. Non ci sono altri sistemi più semplici

[@PreMath]

Grazie per il tuo feedback! Saluti! Sei fantastico.😀

[@josecarlosmyersbenavides8755]

CON MUCHA EXTRAÑEZA VEO UNA ANALISIS Y RESPUESTA INCORRECTA AL RESOLVER ESTE PROBLEMA. TODO EL TRABAJO MATEMATICO DESARROLLARDO EN ESTE EJERCICIO NO CORRESPONDE A ORIENTAR LA RESPUESTA DE LA PREGUNTA DEL AREA VERDE QUE SE MUESTRA LA FIGURA. A MI MANERA DE VER EXINTES DOS FROMAS SENCILLAS DE RESOLVERLO, Y EN AMBAS QUE INVOLUCRAR EL CALCULO DEL AREA DEL TRIANGULO RECTANGULO, SI NO SE HACE NO ESPOSIBLE LLEGAR A LA RESPUESTA QUE SOLICITAN, YA QUE SE PREGUNTA POR PARTE DEL AREA DE ESTE TRIANGULO Y POR ESO NO LO PODEMOS DESCARTAR

[@25rupendrakumar31]

International mathematics olempyad. 👍

[@КоляКонор-н7я]

А=16,746🎉

[@srividhyamoorthy761]

1st comment

[@PreMath]

Super Srividhya Thank you for your feedback! Cheers! You are awesome 😀

[@Teamstudy4595]

1st comment

[@createyourownfuture3840]

You are not the first.

[@PreMath]

Super