Can you find area of the Green shaded region? | (Circles) |

[@jamestalbott4499]

That was an adventure! Thank you!

[@PreMath]

I'm glad you enjoyed the journey! 😊 You are very welcome! Thanks for the feedback ❤️

[@adept7474]

Intersecting chords property: AT•TB=CT•TD. 2R•2r=50•50. R•r=625.

[@PreMath]

Thanks for the feedback ❤️

[@allanflippin2453]

Yes, I used this as well. An advantage of this approach is one doesn't need to assume any particular size relationship between R and r.

[@neilcourse]

There's an easier way. Since the answer doesn't depend on TP, you might as well assume that TP=0. Then the big circle has a radius of 50 and the two white circles each have a radius of 25.

[@thewolfdoctor761]

Yes, just make the circles the same size, tangent at the center of the big circle, and the problem is solved easily.

[@PreMath]

Thanks for the feedback ❤️

[@rchandos]

Always helpful to try and construct the figure given in these puzzles before embarking on complicated calculations. When you do in this case, it becomes clear that the two smaller circles can be of equal radii.

[@అధునాతన_సనాతని]

Exactly 💯

[@marioalb9726]

Shaded area value doesn't depend on circle radius We can match both inner circles, and all original conditions are still being fulfilled. Therefore, R=2r, R=50cm and r=25cm A = πR² - 2πr² = π(R² - 2r²) A = π (50² - 2*25²) A = 1250π cm² ( Solved √ )

[@alanthayer8797]

Cool break down ! I like these type of Problems which is finding Area of spherical curvatures! Thanks as usual sir !

[@PreMath]

Glad you liked it! 🙏 You are very welcome! Thanks for the feedback ❤️

[@marcelowanderleycorreia8876]

Very smart question!! Congrats teacher!! 👏👏👍👍

[@PreMath]

Thanks for the encouragement! ❤️🙏

[@manuelantoniobahamondesa.3252]

Muy bueno profesor; Muchas Gracias!

[@jackwhite5255]

Very nice, but as a conclusion it must be emphasized that the result doesn't depend on the position of the T point (between C and D as limits ) . An animated displacement would be very interesting. The easiest calculation is for the position in P.

[@PreMath]

Thanks for the feedback ❤️

[@GilmerJohn]

Since the only information given was the 100 units it's likely that so long as every thing fits, it's possible that the two inner circles are the same size. If that's the case, each inner is 1/4th of the big circle. The green area is 1/2 (50^2)xpi.

[@manochao]

MAGNIFIQUE Thank you

[@himo3485]

CO=OT=x TQ=QD=y CP=PD=z AT=TB=50 50*50=2x*2y 4xy=2500 xy=625 (2x+2y)/2=z 2(x+y)/2=z x+y=z (x+y)²=x²+2xy+y²=z² Green shaded area = z²π - y²π - x²π = x²π + 2xyπ + y²π - y²π -x²π = 1250π

[@shrawankumarmishra3528]

Nice sir please provide some other topic video

[@marioalb9726]

A= ½(¼πc²) = ⅛π100² A = 1250π cm² ( Solved √ ) Shaded area is equal to half of the circular ring area, respect to the chord tangent to the equivalent inner circle. This chord is same chord that separates both given internal circles. ( given data c= 100 cm)

[@marioalb9726]

Formula to calculate the area of a circular ring A = πR² - πr² A = π (R²-r²) A = π (½c)² A = ¼.π.c² where 'c' is the chord

[@PreMath]

Thanks for the feedback ❤️

[@PreMath]

Excellent! Thanks for sharing ❤️

[@andrepiotrowski5668]

Pythagoras ad nauseam. There are also other theorems, such as Thales' theorem and the altitude theorem (or intersecting chords theorem, even simpler). It follows immediately: 2r * 2R = 50^2

[@marcgriselhubert3915]

Fine;

[@yakovspivak962]

R = r1 + r2 is an outer circle radius. 50^2 = 4(r1 × r2) - geometric mean 50^2 = (r1 + r2)^2 - (r1 - r2)^2 Or: r1+ r2 = 50 and r1 - r2 = 0 r1 = 25, r2 = 25 S = 1250 × pi

[@quigonkenny]

Let a be the radius of circle O, b the radius of circle Q, and r the radius of circle P. The shaded area will be equal to the area of circle P minus the areas of circles O and Q. A = πr² - (πa²+πb²) A = π(r²-(a²+b²)) --- [1] As circles O and Q are tangent to each other and to circle P, then their center points and points of tangency are collinear, so O, T, P, and Q are on CD. As AB is tangent to circles O and Q at T, then AB is perpendicular to CD, and as CD is a diameter of circle P, then T is the midpoint of AB. AT = TB = 100/2 = 50. By the intersecting chords theorem, AT(TB) = CT(TD). AT(TB) = CT(TD) 50(50) = (2a)(2b) 2500 = 4ab ab = 2500/4 = 625 --- [2] CD = CT + TD 2r = 2a + 2b r = a + b r² = (a+b)² r² = a² + b² + 2ab r² = a² + b² + 2(625) r² = a² + b² + 1250 a² + b² = r² - 1250 --- [3] A = π(r²-(a²+b²)) A = π(r³-(r²-1250)) A = π(r²-r²+1250) A = 1250π ≈ 3926.99 sq units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@wackojacko3962]

Another reason why the 'red and green should never be seen' rule should be left behind. Why are barns painted red? ... green pastures! Complementary colors! In this case the red line and green area gives the problem a vibrant look! Happy Holidays 😊

[@PreMath]

Happy Holidays 😊 Thanks for sharing ❤️

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... All three circles have exactly one intersection point in pairs. Therefore we know that the centers of all these circles are located on the same line (CD). We also know that AB is a tangent to the smaller and the bigger white circle. From this we can conclude that AB is perpendicular to CD. This also means that the common intersection point (T) is the midpoint of AB. Now we can apply the intersecting secants theorem. With R₁>R₂>R₃ being the radii of the circles we obtain: AT*BT = CT*DT (AB/2)*(AB/2) = (2R₂)*(2R₃) AB²/4 = 4R₂R₃ ⇒ 2R₂R₃ = AB²/8 = 100²/8 = 10000/8 = 1250 Now we are able to calculate the area of the green region: A(green) = π(R₁)² − π(R₂)² − π(R₃)² = π[(R₁)² − (R₂)² − (R₃)²] = π[(R₂ + R₃)² − (R₂)² − (R₃)²] = π[(R₂)² + 2R₂R₃ + (R₃)² − (R₂)² − (R₃)²] = π*2R₂R₃ = 1250π Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤️

[@donfzic7471]

Thank you. Is it possible to specify also the values of r and R ?

[@maroonshaded]

So if b = length of AB, a general expression for the shaded area could be πb^2/8.

[@cyruschang1904]

r is the small white circle radius, R is the large green circle radius R - r = the radius of the larger white circle 50^2 + (R - 2r)^2 = R^2 50^2 + 4r^2 - 4rR = 0 r(R - r) = 25^2 Green area = π[R^2 - r^2 - (R - r)^2] = π[2rR - 2r^2] = π(2)(25)(25) = (50)(25) = 1250π

[@giuseppemalaguti435]

Agreen=π(R+r)^2-πR^2-πr^2=2πrR...(R+r)^2-(2R-r-R)=50^2...(2R)(2r)=2500...rR=625=>Agreen=2π625=π1250

[@PreMath]

Excellent! Thanks for sharing ❤️

[@iosifbitenskiy2479]

2nd way. Let consider a triangle CBD inscribed into a big circle. One side of the triangle is a diameter CD, therefore the angle CBD=90°. Let diameters of the circles be a = 2r, b = 2R, CD = d. In the right angle triangle CBD CD² = CB² + BD² (1) From the right angle triangles CBT and TBD: CB² = CT² + TB² or CB² = a² + 50², and BD² = TD² + TB² or BD² = b² + 50². Inserting into (1) d² = a² + 50² + b² + 50² or d² - a² - b² = 5000. Multiplying both sides by π/4 one gets the shaded area on the left equal 1250π on the right.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@LuisdeBritoCamacho]

Here it goes my Resolution Proposal : 01) Let Small Circle (SC) Radius = A 02) Let Medium Circle (MC) Radius = B 03) Let Big Circle (BC) Radius = R 04) A < B < R 05) CD = (DT + CT) ; 2R = (2B + 2A) ; 2R = 2(A + B) ; R = (A + B) ; B = (R - A) or A = (R - B) 06) 2A * 2B = 50 * 50 ; 4 * A * B = 2.500 ; A * B = 2.500 / 4 ; A * B = 625 07) So far : (A + B) = R and A * B = 625 08) Proving that 2A = B 09) OQ = (A + B) 10) As : (A + B) = R ; Thus : OQ = R ; PC = R ; TP = A + B - 2A ; TP = (B - A) 11) PQ = TQ - TP ; PQ = B - (B - A) ; PQ = B - B + A ; PQ = A ; A + (B -A) = B ; B = B ; wich is a True Statment. 12) One must conclude that : B - A = A and B = 2A 13) As : (A * B) = 625 and B = 2A ; (A * 2A) = 625 ; 2A^2 = 625 ; A^2 = (625/2) 14) As : (A * B) = 625 and A = B / 2 ; (B * B / 2) = 625 ; B^2 / 2 = 625 ; B^2 = 1.250 15) R = (A + B) ; R^2 = (A + B)^2 ; R^2 = (625 / 2 + 1.250 + 1.250) ; R^2 = 625 / 2 + 2.500 ; R^2 = (625 + 5.000) / 2 ; R^2 = (5.625/2) 16) BCA = (5.625 * Pi / 2) 17) MCA = (1.250 * Pi) 18) SCA = (625 * Pi / 2) 19) GSA = BCA - (MCA + SCA) 20) GSA = 5.625Pi/2 - (1.250Pi + 625Pi/2) 21) GSA = (5.625Pi/2) - (3.125Pi/2) 22) GSA = 2.500Pi/2 23) GSA = 1.250Pi/2 Thus, OUR BEST SOLUTION IS : Green Shaded Area equals 1.250Pi Square Units. NOTE : Sorry about the Delay!!

[@imetroangola17]

*Solução:* Seja k o raio da circunferência maior. Quando as cordas são perpendiculares, vale a relação: AT² + TB² + TD² + TC² = (2k)² Como T é ponto médio do segmento AB, logo TB=AT= 50. Daí, 50² + 50² + (2r)² + (2R)² = (2k)² 5000 + 4r² + 4R² = 4k² × (π/4) 1250π + πr² + πR² = πk² πk² - (πr² + πR²) = 1250π Portanto, a área procurada é: *1250π unidades quadradas.*

[@zawatsky]

Выходит, что оставшаяся площадь точно равняется двум кругам, вырезанным повдоль разделителя (с диаметрами по 50 см), или же одному большому вырезанному кругу радиусом в 100 см. Странное свойство, интуитивно непонятное...🙄

[@nenetstree914]

1250PI

[@PreMath]

Excellent! Thanks for sharing ❤️

[@sergioaiex3966]

Solution: Big Circle Radius = r Medium Circle Radius = a Small Circle Radius = b Green Shaded Area (GSA) = πr² - πa² - πb² GSA = πr² - πa² - πb² ... ¹ d = 2a + 2b d = 2 (a + b) 2r = 2 (a + b) r = a + b ... ² AT² + PT² = AP² ... ³ PT = PC - CT PT = a + b - 2b PT = a - b AP = r = a + b (50)² + (a - b)² = (a + b)² 2500 + a² - 2ab + b² = a² + 2ab + b² 2500 - 2ab = 2ab 4ab = 2500 ab = 625 ... ⁴ Big Circle Area = πr² = = π (a + b)² Medium Circle Area = π a² Small Circle Area = π b² Replacing in Equation ¹ GSA = π (a + b)² - πa² - πb² GSA = π (a² + 2ab + b²) - πa² - πb² GSA = πa² + 2πab + πb² - πa² - πb² GSA = 2πab GSA = 2π × 625 GSA = 1,250 π Square Units ✅ GSA ≈ 3,926.9908 Square Units ✅

[@7777yo7777]

50x50= 2r x 2R

[@PreMath]

Thanks for the feedback ❤️

[@EPaozi]

Faire simple ! R=r !!!! (rien ne l'interdit , donc c'est autorisé) alors aire verte = pi.50^2-2pi.25^2 = pi1250 !!!!! 😄

[@k9slayer]

I need to eat some brain food, lots of it. Thanks once again.

[@PreMath]

You can do it! You are very welcome! Thanks for the feedback ❤️