Can you calculate area of the Red Rectangle? | (Quarter Circle) |

[@Atharva-jf7cq]

nice question thanks from India

[@PreMath]

Glad to hear that! You are very welcome! Thanks for the feedback ❤️🙏

[@MathTrick-t7v]

I Like Your solution

[@marcelowanderleycorreia8876]

Very nice!!!👍👍

[@PreMath]

Glad you liked it! ❤️ Thanks for the feedback ❤️

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! ❤️ Thanks for the feedback ❤️

[@MathTrick-t7v]

I like you solution

[@jimlocke9320]

A more straightforward but less creative way to solve: let the intersection of arc AD and line segment BC be denoted E. Construct OE. Let the area of rectangle ABCO be denoted R. OE divides the quarter circle ADO into two sectors (⌔), ⌔AEO and ⌔EDO. The leftmost green area is R - ΔECO - ⌔AEO. The rightmost green area is ⌔EDO - ΔECO. We equate the two. R - ΔECO - ⌔AEO = ⌔EDO - ΔECO We add ⌔AEO + ΔECO to both sides: R = ⌔AEO + ⌔EDO. However, ⌔AEO + ⌔EDO is the area of the quarter circle. Skip ahead to 5:15.

[@alanthayer8797]

THANKS for doing Semi circle Area problems ! Do u ever do NON EUCLIDEAN geometry ?

[@PreMath]

You are very welcome! Any specific question/example on that? Thanks for the feedback ❤️

[@alanthayer8797]

@ nothing special except doin some Non Euclidean problems ! I like your simplicity sir ! Keep up the great work sir ! I watch u everyday !

[@marcgriselhubert3915]

Fine.

[@PreMath]

Glad to hear that! Thanks for the feedback ❤️

[@giuseppemalaguti435]

8r=(1/4)πr^2...r=32/π...Arett=256/π

[@PreMath]

Excellent! Thanks for sharing ❤️

[@cyruschang1904]

Length of the rectangle = radius = r 8r - A + A = (πr^2)/4 8 = (πr)/4 r = 32/π Rectangle area = 8r = 256/π

[@PreMath]

Excellent! Thanks for sharing ❤️

[@cyruschang1904]

@ Thank YOU

[@zawatsky]

Сперва быстренько доказываем, что два зелёных сегмента складываются в один прямоугольник. Это несложно - части одной окружности проходят по обоим координатам одно и то же расстояние (разницу высоты и основания по одной, только высоту по второй), что бывает только с дугами одной длины. Они отсекают от прямоугольника квадрат со стороной 8. Итого общая площадь будет 64+2А. Диагональ квадрата со стороной 8 будет 8√2, она же радиус четверти круга и она же основание прямоугольника. Т. е. искомая площадь будет 64√2. Мы можем даже найти теперь A: 64+2A=64√2⇔2A=64√2-64=64(√2-1)⇔A=32(√2-1).

[@santiagoarosam430]

El rectángulo y el cuadrante tienen la misma superficie ---> 8r=πr²/4---> r=32/π---> Área ABCO =π32²/4π² =256/π. Gracias y saludos

[@PreMath]

Excellent! Thanks for sharing ❤️

[@alexundre8745]

Bom dia Mestre Vou tentar fazer sozinho o exercicio

[@PreMath]

Bom dia, querido! Ótima ideia. Thanks for sharing ❤️

[@himo3485]

AO=CO=r ABCD=AOD AOD=r²π/4 ABCD=8r 8r=r²π/4 32r=r²π 　　　　　　32=rπ　 r=32/π Red Rectangle area (ABCD) = 8*32/π = 256/π

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Alishbafamilyvlogs-bm4ip]

Idont think we need to some areas for us to friend ❤

[@jadenlee3379]

By the way how can I calculate area A??

[@peterhui7023]

The external green area equal to the internal green area is lack of proof .

[@quigonkenny]

There's no proof needed. That the two green areas are the same is a starting condition of the diagram. The point is to find the area of the rectangle (for which you must find the radius of the quarter circle) such that the two green areas are the same.

[@JLvatron]

It was part of the question.

[@peterhui7023]

@@quigonkenny Thanks !

[@peterhui7023]

@@JLvatron Thanks !

[@kelumo7981]

You comment is totally inscrutable dude

[@Z-eng0]

I was able to get the rectangle's length on my own, but my mind totally wandered after that and I went on to get the are A and totally forgot that the question was asking for the rectangle area and not A. (Anyway for anyone wondering that area turned out to be ≈ 20u, I also turned it into the area of a full circle segment with angle 2*arccos(8/(32/π))

[@PreMath]

Thanks for sharing your thought process! ❤️

[@jeffreyhudale5633]

256/pi = rectangle area

[@PreMath]

Excellent! Thanks for sharing ❤️

[@wasimahmad-t6c]

8×11.3137=90.5

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : 01) Red Rectangle Area (RRA) = (8*R) sq un 02) Quarter of Circle Area = (Pi * R^2) / 4 03) A = A ; thus : 8R = (Pi*R^2) / 4 04) 32R = Pi*R^2 ; PiR^2 - 32R = 0 ; R*(Pi*R - 32) = 0 ; R = 0 or R = 32 / Pi 05) R = 32 / Pi ; R ~ 10,186 lin un 06) RRA = 8 * (32 / Pi) ; RRA = (256 / Pi) sq un ; RRA ~ 81,5 sq un Therefore : OUR BEST ANSWER : Red Rectangle Area equal 81,5 Square Units. Exact Form = (256/Pi) Square Units

[@PreMath]

Excellent! Keep rocking😀 Thanks for sharing ❤️

[@marioalb9726]

¼πR² = 8R R = 4*8/π = 32/π cm A = 8*32/π = 256/π cm² ( Solved √ )

[@peterhui7023]

no cm specified, the size is in units not in cm or cm².

[@marioalb9726]

@peterhui7023 Does not exist in all the universe, adimensional figures !!! You are wrong !!! Only in your dreams !!

[@PreMath]

Excellent! Thanks for sharing ❤️

[@peterhui7023]

@@marioalb9726 The question doesn't specified length dimension in cm unit.

[@marioalb9726]

@@peterhui7023 I corrected the question Now with my correction, adding the length unit, all is correct !!! Try to buy 10 meters of hose, for example, and don't say the length unit to the seller, and then pay attention to how he looks at you !! Or, If you insist, Refrain from purchasing items where it is essential to manufacture them using an international system unit. Only then, would you be consistent with what you write.

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... The two green regions have the same area. Since the rectangle and the quarter circle share the white area within the rectangle, they also have the same area: A(rectangle) = A(quarter circle) AB*AO = π*R²/4 AB*AO = π*AO²/4 8*AO = π*AO²/4 ⇒ AO = 32/π ⇒ A(rectangle) = AB*AO = 8*32/π = 256/π ≈ 81.49 Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤️

[@sergioaiex3966]

Solution: Since the two Green Area are equal and worth "A" and we have three regions in the figure, let's label White Area as "B" 1ª Green Region: A 2ª White Region: B 3ª Green Region: A Therefore, the figure is compound by a Quarter Circle and a Rectangle Area Quarter Circle = Green Area + White Area (π r²)/4 = A + B ... ¹ Area Rectangle = Green Area + White Area 8r = A + B ... ² Equaling ¹ and ² (π r²)/4 = 8r π r = 32 r = 32/π Red Rectangle Area (RRA) = 8r RRA = 8 . 32/π RRA = 256/π Square Units ✅ RRA ≈ 81.4873 Square Units ✅

[@PreMath]

Excellent! Thanks for sharing ❤️

[@angshukNag]

i dont think we need two same areas for us to find solution

[@PreMath]

Thanks for the feedback ❤️

[@kelumo7981]

Come on dude your are dumbing down everybody by painstakingly performing the basic operations as if you are teaching primary school kids

[@wackojacko3962]

Beginning @ 7:14 , I like too simplify things and get this over with so this degenerate can go back too bed. Because the answer my friend according to Bob Dylan is blowing in the wind, and I gotta get some sleep! 0⁰=1 0!=1 Pi and 256 are weird 2⁸=256 and in comp sci needs 2 bytes to represent. Pi is irrational but approximated too 31 Trillion Digits 😂 I'm tired and going with r=0. ...Next question please. 😊

[@PreMath]

😀 Thanks for the feedback ❤️