Can you find area of the Pink shaded region?

Can you find area of the Pink shaded region? | (Quarter Circle) |

Рет қаралды 15,006

PreMath

Күн бұрын

Пікірлер: 58

@adogonasidecar1262 Ай бұрын

Pretty cool that we can compute the area with so little information. At first it feels like we would need more ...

@trumpetbob15 Ай бұрын

Not only that, but I realized the diagram itself is drawn in a way that makes the problem more difficult. Since we don't have any measurements for the size of the circles, there is nothing stopping us from shrinking the smaller circle to an area of 0 units by moving the line AB down to the diameter, passing through point D. Thus, you have a fully pink quarter circle with radius 18, a white circle with area of 0, and the same final result of 81*PI (18*18*PI/4).

@PreMath Ай бұрын

Excellent! Thanks for the feedback ❤️

@BKNeifert Ай бұрын

Well, just look at the shapes, and see it has to be. The circle is limited as a shape, so to have a full circle within a semicircle on a tangent of a chord of 18, it would necessarily have to be that way, as it's the only possible way they can fit together, if the circle also has a tangent on the diameter.

@adogonasidecar1262 Ай бұрын

@@BKNeifert Of course I can understand the degrees of freedom and why no more information is required. My point is that it's not the first thing that comes to mind, it takes a couple seconds to figure it out. And that's pretty cool

@adogonasidecar1262 Ай бұрын

@@trumpetbob15 Exactly. There is no unicity of solution in r and R, but unicity in the pink area

@zeroone7500 Ай бұрын

1:21 C, P are collinear and P, D are collinear that doesn't mean all C, P, D are collinear. You can see if angle OAB is not = 90 degree then C, P, D are not collinear.

@ManojGupta-bp5ws Ай бұрын

Very true. Statement that point of contacts and centre are always collinear is wrong. But he is not replying. 😅

@timmcguire2869 Ай бұрын

Just shrink the small circle to radius equals zero. The pink area is then pi18sqr/4

@quigonkenny Ай бұрын

(On assumption that AB and OA are perpendicular) Let r be the radius of circle P and R be the radius of quarter circle O. As AB is tangent to circle P at C and OD is tangent to circle P at D, ∠PCA = ∠ODP = 90°. As ∠CAO and ∠AOD are also 90°, AODC is a rectangle, and CA = OD and DC = OA = 2r. As the pink shaded area is the area of quarter circle O minus circle P, the formula is πR²/4-πr², where as previously noted, R is the radius of O and r is the radius of P. Triangle ∆BAO: BA² + OA² = OB² 18² + (2r)² = R² 324 + 4r² = R² R² - 4r² = 324 R²/4 - r² = 324/4 = 81 πR²/4 - πr² = 81π Aᴘ = 81π sq units

@marioalb9726 Ай бұрын

A = ¼. π c² = ¼ π 18² A = 81π cm² ( Solved √ ) Shaded area is equal to a circle area of diameter 18 cm !!!!

@Micboss1000 Ай бұрын

cool problem with a clever solution, love it

@olesgaiduk5159 4 күн бұрын

Ray OA intersects the great circle at point E. Let triangle EAB be Pythagorean. AB = 18, EA = 13.5; EB = 22.5 Mirroring line ОАЕ, we add a quarter circle. We construct an isosceles triangle, where ВВ’ = 2АВ = 18*2 =36; ЕВ’ = ЕВ . The area of this triangle S_ ЕВВ’ = 0.5*ЕА * ВВ’ = 0.5*13,5* 36 = 243. The radius of the great circle R = (ЕВ’ * ЕВ *В’В) /4S_ ЕВВ’ = 22,5*22,5*36 / 4*243 = 18,75 ; The area of the quarter of the great circle Sb = 0.25 π R^2 =0.25 π (18,75)^2 = 87,89 π; The radius of the small circle r = 0.5(R - ЕА) = 0.5 (18,75 - 13,5) = 2,625; The area of the small circle Sr = π r^2 = π 2,625^2 = 6,89 π. The area of the pink circle Sp = Sb - Sr = 87,89 π- 6,89 π = 81 π.

@Waldlaeufer70 Ай бұрын

When I come across one of the problems where it seems that the result does not depend on a specific size of a part, I tend to move things, so the problem gets easier to solve. In this case, I let r = 0, which makes the white circle disappear. So, R = 18 A(pink) = R²/4 * π A(pink) = 18²/4 * π A(pink) = 18²/2² * π A(pink) = (18/2)² * π A(pink) = 9² * π A(pink) = 81 π In general, we can solve it the way you did it: Pythagoras: R² = 18² + (2r)² R² = 18² + 4r² 4r² = R² - 18² r² = (R² - 18²) / 4 Area: A(pink) = A(quarter circle) - A(white circle) A(pink) = (R²/4 - r²) π A(pink) = ((R² - R² - 18²) / 4) * π A(pink) = 18²/4 * π (for the rest see simplified solution)

@alokranjan4149 Ай бұрын

Beautiful problem ❤

@manuelantoniobahamondesa.3252 27 күн бұрын

Muy simpático su problema, gracias profe!

@devondevon4366 Ай бұрын

81 pi Let the radius of the quarter circle = N then its area = pi N^2/4 Let the radius of a small circle = r the its area = pi r^2 Hence, area of shaded region = pi N^2/4 - pi r^2 = piN^2/4 - 4pi r^2/4 give both a common denominator of 4 pi/4 (N^2 - 4 r^2) This is the area of the shaded region in terms of both radii Construct a right triangle by drawing a line from the end of line 18 to the center of the quarter circle. The sides are N (the hypotenuse), 18, and 2r (since the small circle distance on that line = diameter, which is twice the radiusZ0 Employing Pythagorean N^2= 18^2 + (2r)^2 N^2 = 324 + 4 r^2 N^2 - 4r^2 = 324 (subtract 4r^2 from both sides of the equation). Equation B The reason to put it in this form is that it has something in common with pi/4 (N^2- 4r^2), which is an area of the shaded region, which is what we are trying to find Hence, pi/4 ( N^2 -4 r^2) = pi/4 ( 324) = pi (324/4 = pi (81) answer

@coyotejc1 Ай бұрын

It is a beautiful solution. I had to watch it 4 times for it to make sense.

@patriciaceli1536 Ай бұрын

Wow! Love this kind of problems! Gracias! ❤

@jacquespictet5363 27 күн бұрын

(1) It could be worth mentioning that the result is independant of both radius and why; (2) Points C and D play no role here.

@als2cents679 Ай бұрын

Since the answer is a constant, just keep everything the same but choose R such that the circle is exactly inscribed into the quarter circle. Then calculate the pink area.

@murdock5537 Ай бұрын

Nice, many thanks, Sir! :-)

@FriendSanjay Ай бұрын

Fantastic

@Birol731 Ай бұрын

My way of solution ▶ If we draw a line through point P and OB, it would be equal to the radius (R) of the quarter circle, so: [OB]= R [BA]= 18 [AO]= 2r if we apply the Pythagorean theorem for the ΔOBA, we get: [OB]²= [BA]²+[AO]² R²= 18²+(2r)² R²= 324 + 4r² R²-4r² = 324........(1) the pinkt area, Apink Apink= πR²/4 - πr² Apink= π(R²/4 - r²) if we take the equation (1) and divide by 4 we get: R²-4r² = 324 R²/4 - r²= 81 multiplied by π : π(R²/4 - r²)= 81π ⇒ Apink= 81π Apink ≈ 254,469 square units

@murdock5537 Ай бұрын

Nice, many thanks, Sir! 🙂

@suntipongvisetchaisri9084 Ай бұрын

Picture of complete. This method must be done

@BBMathTutorials Ай бұрын

🎉love to watch your videos sir❤

@user-ll5wl9gq4x Ай бұрын

Respect! An interesting task! The only strange thing is that you take so long to describe the solution! Perhaps this is because you do not need to prepare students for the Russian Unified State Exam! I have to teach not just to solve, but to solve as quickly as possible!🙂👍

@wackojacko3962 Ай бұрын

Of 247 shades of pink you had too pick the one I'm color blinded of. And it's not on the visible light spectrum . So no worries, I have my handy spectrophotometer by my side at all times. Proceeded to readily solve. 🙂

@jamestalbott4499 Ай бұрын

Thank you!

@marioalb9726 Ай бұрын

Red circle: A = ¼πR² --> Quarter circle 4A = πR² --> Whole circle White circle tangent to R: A = πr² New x4 bigger white circle, concentric to red circle: 4A = 4πr² = π(2r)² Area of Red circular ring: A = A₁ - A₂ = πR² - π(2r)² A = ¼πc² = ¼π36² A = 324π cm² Area of Red Quarter Circular ring: A = ¼A = 81π cm² ( Solved √ ) Red Shaded area is equal to area of quarter circular ring (with white circle turned into a white concentric quarter circle with same area)

@ManojGupta-bp5ws Ай бұрын

At 1:21, your statement that centre and point of contacts are collinear is totally wrong. So your solution is correct only and only if angle OAB is 90 degree. So don't make fool . first answer my point.dont ignore it

@BKNeifert Ай бұрын

That's clever. I was racking my brain to figure out why you were going about it that way.

@devondevon4366 Ай бұрын

81 pi

@emilioricou Ай бұрын

Parece fácil despues de explicado

@marcgriselhubert3915 Ай бұрын

In an orthonormal center O and first axis (OD) the equation of the big circle is x^2 + y^2 = R^2, with R its radius. At point B we have x = 18, so y = sqrt(r^2 - 18^2). Sqrt(R^2 - 18^2) is also the ordinate of C and the diameter of the little circle, so the radius of the little circle is r = (1/2).sqrt(R^2 - 18^2) and its area is (Pi/4).(R^2 - 18^2) The area of the big quater circle is (Pi/4).R^2 So, by difference, the pink area is (Pi/4).(18^2) = 81.Pi.

@user-ll5wl9gq4x Ай бұрын

Респект! Интересное задание! Странно лишь, что Вы так долго расписываете решение! Возможно это потому, что Вам нет необходимости готовить учеников к Российскому ЕГЭ! Мне приходится учить не просто решать, но по возможности решать быстрее!👍🙂

@santiagoarosam430 Ай бұрын

π18²/4=81π. La corona circular delimitada por dos círculos concéntricos de radios OB y CD tiene una superficie 4 veces mayor que el área rosa de la figura propuesta, puesto que esa es la razón entre las áreas de los círculos de radios CD y PD. Gracias y saludos.

@prossvay8744 Ай бұрын

Let R is a Radius quarter circle ; and r is a small circle. Pink area=1/4(π)(R^2)-πr^2=π(R^2/4-r^2) OA =CD=2r Connect O to B OA^2+AB^2=OB^2 (2r)^2+(18)^2=R^2 4r^2+324=R^2 So R^2/4-r^2=81 square units So Pink area=81π square units.❤❤❤

@giuseppemalaguti435 Ай бұрын

R^2=(2r)^2+18^2=4r^2+324.. .Apink=πR^2/4-πr^2=π324/4=π81

@misterenter-iz7rz Ай бұрын

Smartest way to solve is assuming AB is radius, then the answer is simply 18^2/4 pi=81pi. Vigorous solution is not too difficult, R^2=18^2+(2r)^2, the area is 1/4 R^2 pi-r^2pi=(18^2/4+r^2-r^2)pi=18^2/4 pi=81pi.😊

@phungpham1725 Ай бұрын

1/ Draw the full circle Let R and r be the radius of the big and small circle. By using chord theorem sq AB= (R+OA)(R-OA) Or sq18=(R+2r)(R-2r)=sqR- 4sqr -> sqR/4 -sqr = sq18/4=81 -> pi. sqR/4-pi.sqr=81.pi Area of the red region= 81.pi sq units😅😅😅

@himo3485 Ай бұрын

CP=PD=r AO=CD=2r (2r)²+18²=R² 4r²+324=R² Pink area = R²π/4 - r²π = (4r²+324)π/4 - r²π = 81π

@unknownidentity2846 Ай бұрын

Let's find the area: . .. ... .... ..... Today we do not think outside the box, but we think outside the quarter circle. If we think of a complete big circle instead of a quarter circle only, we can apply the intersecting chords theorem. With R being the radius of the quarter circle and r being the radius of the circle inside the quarter circle we obtain: 18*18 = (R + 2*r)*(R − 2*r) 18*18 = R² − 4*r² 18*18/4 = R²/4 − r² 81 = R²/4 − r² Now we are able to calculate the area of the pink region: A = A(quarter cirle) − A(circle) = πR²/4 − πr² = π*(R²/4 − r²) = 81π Best regards from Germany

@LuisdeBritoCamacho Ай бұрын

STEP-BY-STEP RESOLUTION PROPOSAL (Integer Solutions) : 01) Let : R = Radius of Quarter of Big Circle 02) Let : r = Radius of Small Circle 03) Let : D = Diameter of Small Circle 04) r = (D / 2) or D = 2r 05) Pink Area (PA) = ((Pi * R^2) / 4) - (Pi * r^2) sq un ; PA = (Pi * (R^2 / 4 - r^2)) ; PA = Pi * ((R^2 / 4) - (4r^2 / 4)) ; PA = Pi * [((R^2 - 4r^2) / 4] sq un 06) As : D = 2r one can conclude that D^2 = 4r^2 07) One can see by the Picture that : (R^2 - D^2) = 18^2 ; (R^2 - D^2) = 324. And, as D = 2r : (R^2 - 4r^2) = 324 08) As : PA = Pi * [((R^2 - 4r^2) / 4] sq un 09) PA = Pi * (324 / 4) sq un ; Pa = (Pi * 81) ; PA = 81Pi sq un Therefore, OUR BEST ANSWEWR : The Pink Area is equal to 81Pi Square Units. Or, if you want, approx. equal to 255 Square Units. Best Wishes from "The Great Islamic Institute of Mathematical Sciences" in Cordoba Caliphate.