A Very Nice Geometry Problem | You should be able to solve this! | 3 Different Methods

[@soli9mana-soli4953]

There would be two other possible methods, the first with a trigonometric approach, the second with a logical approach: With trigonometry: Since the area requested is equal to Area = (R²/4 - r²)*π (1) we can find the two radii like this: R = 8/cos α r = R/2*sin α and substituting R r = 4 sin α/cos α therefore the area (1) is: Area = (8²/cos²α*1/4 - 4²* sin²α/cos²α)*π Area = 16/cos²α*(1 - sin²α)*π Area = 16*π In the logical approach, observing that we have no particular constraints either for the position of the chord CD or for the radius of the smaller circle, then we can imagine that the radius of the smallere circle is equal to zero and so the chord CD coincides with the radius of the quarter circle, therefore the area will be equal to 8²/4 = 16

[@ناصريناصر-س4ب]

Let R be the radius of the larger circle and r be the radius of the smaller circle, and the required area is equal to πR²/4-πr²=(π(R²-(2r)²)/4=π(OD²-OC²)/4=π*64/4=16π

[@ThenmozhiGovindaraj-sj1hi]

Correct

[@ThenmozhiGovindaraj-sj1hi]

OD^2-OC^2=8^2=64

[@mahmoudmohammed1986]

It's a very beautiful Question and the three different solutions are great solutions too

[@marioalb9726]

A = ¼(¼πc²)= 1/16 π(2*8)² A = 16π cm² ( Solved √ ) Too completated vídeo solution Just need to move the small circle at the quadrant center, and then apply the formula of circular ring area, respect to the chord, and then divide by 4.

[@russcooper9628]

If you consider the limiting case where the circle shrinks to nothing, then OB = CD = 8. From there, you just have a quarter circle with radius 8.

[@jimlocke9320]

The problem statement implies that CD may be located anywhere above OB as long as a valid figure is produced. Moving CD closer and closer to OB results in a smaller and smaller circle which would disappear if CB were allowed to align with OB, the limiting case. For a multiple choice test or a test where you are not required to show your work, solve the limiting case and either choose the corresponding multiple choice answer or write down that answer, as appropriate. If you are required to solve the general case and show your work, the limiting case can be used as a check that you've solved the problem correctly.

[@spdas5942]

Very nice sir 👍

[@prossvay8744]

Let R is Radius of the sector and r is the Radius of the circle. Area of the sector=1/4(π)R^2 (1)) Area of the circle=(π)r^2 (2) Connect O to D In ∆OCD OC^2+CD^2=OD^2 (2r)^2+8^2=R^2 4r^2+64=R^2 So R^2-4r^2=64 Blue area=1/4πR^2-πr^2=π/4(R^2-4r^2)=π/4(64)=16π.❤

[@santiagoarosam430]

El círculo pequeño de radio OC/2, tiene una superficie de 1/4 de la de otro círculo de radio =OC → Área sombreada = (1/4)*(Área de la corona circular delimitada por las circunferencias concéntricas de radios OD y OC) = π(OD²-OC²)/4 =π*CD²/4 = π8²/4 =16π. Gracias y saludos.

[@michaeldoerr5810]

The answer is 16pi. Also if I can summarize ALL three methods-which are familiar to me fully-the first method requires direct application of the Pythagorean Theorem and makes the substitution of R^2-4delta^2 for 64 getting us pi/4*64. The second method requires making a circle which shows that because of the symmetry of CD and CF, you can make use of the interecting chords theorem. And lastly the third method make use of the circle theorem that justifies the HL similarity that involves beta-alpha-right angle comparison. And these I almost know by heart and I hope that this means thay I definitely SHOULD be able to know how to do this!!!

[@quigonkenny]

Let R be the radius of the quarter circle and r the radius of the smaller circle. As the shaded area is equal to the difference in the areas between the quarter circle and the smaller circle, the formula is as follows: Aₛ = πR²/4 - πr² Aₛ = (πR²-4πr²)/4 Aₛ = (R²-4r²)π/4 Triangle ∆DCO: DC² + OC² = OD² 8² + (2r)² = R² 64 + 4r² = R² R² - 4r² = 64 (π/4)(R²-4r²) = (π/4)64 (R²-4r²)π/4 = 16π [ Aₛ = 16π ]

[@gelbkehlchen]

Lösung: R = Radius des Viertelkreises, r = Radius des ganzen Kreises. Farbige Fläche = π*R²/4-π*r² = π*[8²+(2r)²]/4-π*r² = π*[64+4r²]/4-π*r² = 16π+π*r²-π*r² = 16π ≈ 50,2655 Solution: R = radius of the quarter circle, r = radius of the whole circle. Colored area = π*R²/4-π*r² = π*[8²+(2r)²]/4-π*r² = π*[64+4r²]/4-π*r² = 16π+π*r²- π*r² = 16π ≈ 50.2655

[@tontonbeber4555]

First idea that come Area quarter (blue + yellow) = pi r² / 4 r² = 64 + x² where x is diameter of yellow circle Area quarter = pi (16 + x²/4) Area yellow circle = pi x²/4 And so blue area = 16 pi

[@yakupbuyankara5903]

16×3,14

[@RealQinnMalloryu4]

(8)^2=64.90°ABCDO/64=1.26 1.2^13 1.2^13^1 1.2^1^1 2^1 (ABCDO ➖ 2ABCDO+1).

[@ItsKaranrana]

Don't use google voice bro

[@yohanesfreedy4488]

betul, merusak konsentrasi

[@AmirgabYT2185]

S=16π

[@nenetstree914]

16 pi

[@joegillian6781]

に、日本語やん！