Before watching: construct triangle OCD, which is a 30-60-90 triangle. Since CD is 1 (area of the circle is r^2pi), OC is 2. Thus the radius of the large circle is 3. The yellow area is 1/6 of 3^2*pi (or 3/2pi) minus pi. So interestingly the yellow area is pi/2 or half of the green area!
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Área azul =πr² =π → r²=1→r=1 → ∆OCD es la mitad de un triángulo equilátero: Si CD=r=1→ OC=2CD=2 → Radio del sector circular =R =OC+CF =2+r=2+1=3 → Área amarilla =(60πR²/360) - π =(9π/6)-π =π/2 Gracias y un saludo.
@@anibalarostegui5574 OF es bisectriz de ∠BOA → ∠COD=30º y ∆OCD es rectángulo en D → ∠OCD=60º → ∆OCD es la mitad de un triángulo equilátero de altura OD=√3, lado OC=2 y medio lado CD=r=1. Por otra parte, usted parece decir que ∆OCA es isósceles; afirmación errónea, puesto que sus tres lados tienen longitudes distintas: OC=2 , OA=Radio R=3=OD+DA=√3+(3-√3) y CA=√(CD²+DA²)=1.6148 Gracias por su observación. Un saludo cordial.
No peeking, no calculator, not even a pencil, all mental. Letters designating line segments are filled in from your diagram later. Since the blue circle has area πr^2 = π. The radius must be 1. Drop a perpendicular from the center of the blue circle to the bottom horizontal line (CD) and also connect the center of the blue circle to the left-hand vertex of the yellow arc (CO). This forms a 30-60-90 triangle OCD. CD, the side opposite the 30-degree angle is 1, so the hypotenuse OC is 2. Draw a line from the left-hand vertex (O) through the center of the blue circle (C) to the outer arc (OF). This has length 2 + 1 = 3; so the radius of the arc (OF) is 3. The area of the arc is 60/360 of a complete circle of radius 3: that is, (1/6)(π)(3^2) = (9/6)π = 3π/2. The yellow region is this area minus the area of the blue circle, or 3 π/2 - π = π/2. Cheers. 🤠
If R = 3 then the length of the arc AB = ⅙ 2R π = π. So the area of the circular sector is ½ R AB = 3/2 π. Subtract 2/2 π from 3/2 π. The yellow region is π/2.
So 1 is the radius of the inside circle, then OF is the radius of the outside circle, it is OC +CF=2+1=3, hence the answer is 3^2 pi/6-pi=pi/2=1.57 approximately. 😃
∆OCD = ∆OCE = 30-60-90 triangle a = x b = x√3 c = 2x a = CD = CE = 1 cm b = OD = OE = √3 cm c = OC = 2 cm Radius of O = 2 cm + 1 cm = 3 cm Area of O = π r² (angle / 360°) O = π (3 cm)² (60°/360°) = π(9 cm²)(1/6) = (3/2)π cm² O - C = 3/2π cm² - π cm² = 3/2π cm² - 2/2π cm² = ½π cm² ≈ 1.57 cm²
Let r be the radius of the blue circle and R be the radius of the sector: A = πr² π = πr² r² = π/π = 1 r = 1 Draw EC, DC, and OC. As OA and OB are tangent to blue circle, ∠ODC and ∠OEC are each 90°. By Two Tangent property OE = OD, and as DC = EC = r, ∆ODC and ∆OEC are congruent. α = ∠EOC = ∠DOC = (1/2)60° = 30° CE/OC = sin(α) 1/OC = sin(30°) = 1/2 OC = 2 R = OP = OC + r = 2 + 1 = 3 Yellow area = sector - blue circle A = (θ/360)πR² - π A = (60/360)π(3²) - π = (1/6)π(9) - π A = (3/2)π - π = π/2 cm²
So elegant!
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Thanks to you I managed to do the whole thing in my head in about 30 seconds. I learned the 30-60-90 trick from you.
Before watching: construct triangle OCD, which is a 30-60-90 triangle. Since CD is 1 (area of the circle is r^2pi), OC is 2. Thus the radius of the large circle is 3. The yellow area is 1/6 of 3^2*pi (or 3/2pi) minus pi. So interestingly the yellow area is pi/2 or half of the green area!
Excellent!
Thanks for sharing! Cheers!
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Love and prayers from the USA! 😀
CD = CE = CF = 1. OD = 1/tan30 = 1.732 (sqrt3). OC = 2, OF = 3.
Area of sector - Area of circle = pi*3*3/6 - pi = 1/2pi = 1.57
Great explanation👍
Thanks for sharing😊
Wow! Exquisite Geometry problem
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Very good Explanation
Sir, you have shown a simple method to solve the problem. Please include problems on Calculus too.
Sure! Keep watching...
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Delightful!
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@@PreMath Thank you and namaskar. 🙏
Thanks for video.Good luck sir!!!!!!!!!
So nice of you.
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Thanks premath
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Thanks professor for this amazing question
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So nice of you, Manikant
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Área azul =πr² =π → r²=1→r=1 → ∆OCD es la mitad de un triángulo equilátero: Si CD=r=1→ OC=2CD=2 → Radio del sector circular =R =OC+CF =2+r=2+1=3 → Área amarilla =(60πR²/360) - π =(9π/6)-π =π/2
Gracias y un saludo.
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Hay un error de interpretación, el triángulo OCD es la mitad de un triángulo ISÓSCELES (OCA). Por lo demás todo está bien. 😑🙃🤩 --- 🇦🇷⭐⭐⭐ ---
@@anibalarostegui5574 OF es bisectriz de ∠BOA → ∠COD=30º y ∆OCD es rectángulo en D → ∠OCD=60º → ∆OCD es la mitad de un triángulo equilátero de altura OD=√3, lado OC=2 y medio lado CD=r=1.
Por otra parte, usted parece decir que ∆OCA es isósceles; afirmación errónea, puesto que sus tres lados tienen longitudes distintas: OC=2 , OA=Radio R=3=OD+DA=√3+(3-√3) y CA=√(CD²+DA²)=1.6148
Gracias por su observación. Un saludo cordial.
No peeking, no calculator, not even a pencil, all mental. Letters designating line segments are filled in from your diagram later.
Since the blue circle has area πr^2 = π. The radius must be 1.
Drop a perpendicular from the center of the blue circle to the bottom horizontal line (CD) and also connect the center of the blue circle to the left-hand vertex of the yellow arc (CO). This forms a 30-60-90 triangle OCD. CD, the side opposite the 30-degree angle is 1, so the hypotenuse OC is 2.
Draw a line from the left-hand vertex (O) through the center of the blue circle (C) to the outer arc (OF). This has length 2 + 1 = 3; so the radius of the arc (OF) is 3.
The area of the arc is 60/360 of a complete circle of radius 3: that is, (1/6)(π)(3^2) = (9/6)π = 3π/2.
The yellow region is this area minus the area of the blue circle, or 3 π/2 - π = π/2.
Cheers. 🤠
Excellent!
Thanks for sharing! Cheers!
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@@PreMath Thanks. We need all the love and prayers we can get.
If R = 3 then the length of the arc AB = ⅙ 2R π = π.
So the area of the circular sector is ½ R AB = 3/2 π.
Subtract 2/2 π from 3/2 π. The yellow region is π/2.
Excellent!
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A good mental exercise if one sees the hidden triangles.
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So 1 is the radius of the inside circle, then OF is the radius of the outside circle, it is OC +CF=2+1=3, hence the answer is 3^2 pi/6-pi=pi/2=1.57 approximately. 😃
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∆OCD = ∆OCE = 30-60-90 triangle
a = x
b = x√3
c = 2x
a = CD = CE = 1 cm
b = OD = OE = √3 cm
c = OC = 2 cm
Radius of O = 2 cm + 1 cm = 3 cm
Area of O = π r² (angle / 360°)
O = π (3 cm)² (60°/360°) = π(9 cm²)(1/6) = (3/2)π cm²
O - C = 3/2π cm² - π cm² = 3/2π cm² - 2/2π cm² = ½π cm² ≈ 1.57 cm²
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Yay! I solved it, (pi)/2.
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Let r be the radius of the blue circle and R be the radius of the sector:
A = πr²
π = πr²
r² = π/π = 1
r = 1
Draw EC, DC, and OC. As OA and OB are tangent to blue circle, ∠ODC and ∠OEC are each 90°. By Two Tangent property OE = OD, and as DC = EC = r, ∆ODC and ∆OEC are congruent.
α = ∠EOC = ∠DOC = (1/2)60° = 30°
CE/OC = sin(α)
1/OC = sin(30°) = 1/2
OC = 2
R = OP = OC + r = 2 + 1 = 3
Yellow area = sector - blue circle
A = (θ/360)πR² - π
A = (60/360)π(3²) - π = (1/6)π(9) - π
A = (3/2)π - π = π/2 cm²
Pi*3^2/6-pi=pi/2
pi/2=1.57 cm^2
R/r=3 because the small circle is 1 of the 7 circles inscribed in the big one.
Wow
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@@PreMath 👍👍👍😁
Can someone explain to me why OC＝2 please😢
sin 30=1/OC
△DCO is half of an equilateral triangle with DC half of its base. DC is 1, so the full length of the side CO is 2.
Got the answer without help on first try (79 year - old on the Pacific Coast)
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@@PreMath innovation is the kt..👍❤️
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@@PreMath 😀🦘
Meanwhile here I was, cheating by remembering you can pack six circles around a seventh, and going from there
Too easy 😞