As sin 30=1/2, the ratio of radii of two circles is 1:2, if let r be the radius of the small circle, hence 3r^2pi=23 pi, r=sqrt(23/3), therefore the answer is 6rpi=6(sqrt(23/3))pi=52.2.😊
OC = r, AC = R. 30/60/90 so AC = 2*OC. green area = pi*(ACsq - OCsq) = 3*pi*rsq = 23*pi. r = sqrt(23/3). Perimeter = 2*pi*(AC+OC) = 6*pi*r = 6pi*(sqrt(23/3)) = 52.19
Si el ángulo en "A" es 30º → R=2r → ⊓(R²-r²)=⊓(4r²-r²)=3⊓r²=23⊓ → r²=23/3 → r=√69 → Perímetro corona circular verde = 2⊓(2r+r) =6⊓r=6⊓√69 =52.1920 Gracias y saludos cordiales.
sin 30° = r / R = 1/2 R = 2 r Area = π (R²-r²) = 23π R² - r² = 23 4r² - r² = 23 r² = 23/3 r = 2,769 cm R = 5,538 cm Perimeter: P = 2πR + 2πr P = 2π (R + r) P = 52,19 cm ( Solved √ )
Before running the video for hints: the 30 degree angle alludes to a 30-60-90 triangle with sides of r, r*sqrt(3), and 2r, and that R=2r. As the area of the annulus is 23pi units^2 and the area of a circle is pi*r^2, I think that means that R-r=sqrt(23) or alternatively that 2r-r=sqrt(23). I think that means that r=sqrt(23). Therefore, R=2*sqrt(23) and D=4*sqrt(23). Therefore, the larger circumference (pi*D) = 4pi*sqrt(23)=60.27(2dp). I wish I had the guts to post this before checking the video :) . Nope. I messed it up but will post it anyway as it's only right to be open about both successes and failures.
Awfully exciting for the astrophysicists too use this math too too determine the mass of the annular rings of Saturn and determine the mass of Saturn's core with flybys of Cassini-Huygens mission satellite. 🙂
Please give me solution this problem The lengths of three sides and two included angles of a quadrilateral are given. AB= 4cm BC= 5cm CD= 5.5cm Angle : ABC= 120° BCD=75° Find value of angle DAB and CDA
Solution: area = perimeter ⇒ π (R² − r²) = 2π (R + r) ⇒ (R − r) = 2 which is remarkable! So here we have π (R² − r²) = 23π ⇒ (R + r) (R − r) = 23 ⇒ (R + r) 2 = 23 ⇒ R + r = 23/2. Subtracting R − r = 2 from R + r = 23/2 will give 2r = 23/2 − 4/2 ⇒ r = 19/4 ⇒ R = 27/4.
![Can you find area of the red Triangle ABC? | Square and Equilateral triangle | [Math Olympiad]](http://i.ytimg.com/vi/OpiFYrNo8to/mqdefault.jpg)
![Can you find area of the red Triangle ABC? | Square and Equilateral triangle | [Math Olympiad]](/img/tr.png)
You are awasome 😊
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As sin 30=1/2, the ratio of radii of two circles is 1:2, if let r be the radius of the small circle, hence 3r^2pi=23 pi, r=sqrt(23/3), therefore the answer is 6rpi=6(sqrt(23/3))pi=52.2.😊
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Excellent, Pre-Math suggestion, if I may now, teach & lecture on architecture & engineering equations, please.
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Great explanation👍
Thanks for sharing😊
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Nice! I like coming up with strange ways of doing these puzzles...
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What does perimeter of ring mean?
Thanks for video.Good luck sir!!!!!!!!!!!
So nice of you, dear
Area = ¼ π C²
C² = 4 A / π
C² = 4 . 23π / π = 92
C = √92
sin 30° = r / R = 1/2
R = 2 r
Intersecting tangent-secant theorem:
(C/2)² = (R+r).(R-r) = (3r)r = 3r²
¼C² = 3 r²
r² = C² /12 = 92/12
r = 2,769 cm
R = 5,538 cm
Perimeter:
P = 2πR + 2πr
P = 2π (R + r)
P = 52,19 cm ( Solved √ )
Thanks for sharing! Cheers!
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Nice! OC = r; OA = r + k; sin(φ) = OC/OA = 1/2 = r/(r + k) → r = k →
3r^2 = 23 → r = √69/3 → 2π(2r) + 2πr = 6πr = 2π√69
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I won't go into the rigorous details, but (assuming we count both the outside and inside perimeter), I get P = 2π(sqrt(92/3)+sqrt(23/3)) = 52.2
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Area = ¼ π C²
C² = 4 A / π
C² = 4 . 23π / π
C² = 92
C = √92
R = ½C / cos30°
R = 5,538 cm
R² = (C/2)² + r²
r² = R² - (½C)²
r² = 5,538² -92/4
r² = 7,666
r = 2,769 cm = ½ R
Perimeter:
P = 2πR + 2πr
P = 2π (R + r)
P = 52,19 cm ( Solved √ )
Thanks for sharing! Cheers!
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😍❤
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OC = r, AC = R. 30/60/90 so AC = 2*OC. green area = pi*(ACsq - OCsq) = 3*pi*rsq = 23*pi. r = sqrt(23/3).
Perimeter = 2*pi*(AC+OC) = 6*pi*r = 6pi*(sqrt(23/3)) = 52.19
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Si el ángulo en "A" es 30º → R=2r → ⊓(R²-r²)=⊓(4r²-r²)=3⊓r²=23⊓ → r²=23/3 → r=√69 → Perímetro corona circular verde = 2⊓(2r+r) =6⊓r=6⊓√69 =52.1920
Gracias y saludos cordiales.
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Is the inner circle be included in the perimeter?
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Basta mettere a sistema le 2 equazioni.. 1)R^2-r^2=23..2)r/R=sin30
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sin 30° = r / R = 1/2
R = 2 r
Area = π (R²-r²) = 23π
R² - r² = 23
4r² - r² = 23
r² = 23/3
r = 2,769 cm
R = 5,538 cm
Perimeter:
P = 2πR + 2πr
P = 2π (R + r)
P = 52,19 cm ( Solved √ )
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Before running the video for hints: the 30 degree angle alludes to a 30-60-90 triangle with sides of r, r*sqrt(3), and 2r, and that R=2r. As the area of the annulus is 23pi units^2 and the area of a circle is pi*r^2, I think that means that R-r=sqrt(23) or alternatively that 2r-r=sqrt(23). I think that means that r=sqrt(23). Therefore, R=2*sqrt(23) and D=4*sqrt(23). Therefore, the larger circumference (pi*D) = 4pi*sqrt(23)=60.27(2dp). I wish I had the guts to post this before checking the video :) . Nope. I messed it up but will post it anyway as it's only right to be open about both successes and failures.
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Awfully exciting for the astrophysicists too use this math too too determine the mass of the annular rings of Saturn and determine the mass of Saturn's core with flybys of Cassini-Huygens mission satellite. 🙂
😀
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P=2√69π≈52,19 cm²
Why trigonometry applications is vast?
Please give me solution this problem The lengths of three sides and two included angles of a quadrilateral are given. AB= 4cm BC= 5cm CD= 5.5cm Angle : ABC= 120° BCD=75° Find value of angle DAB and CDA
Bonus question: what should be the length of R and r so that the perimeter of a 23π cm² annulus is 23π cm?
Great!
Solution: area = perimeter ⇒ π (R² − r²) = 2π (R + r) ⇒ (R − r) = 2 which is remarkable!
So here we have π (R² − r²) = 23π ⇒ (R + r) (R − r) = 23 ⇒ (R + r) 2 = 23 ⇒ R + r = 23/2.
Subtracting R − r = 2 from R + r = 23/2 will give 2r = 23/2 − 4/2 ⇒ r = 19/4 ⇒ R = 27/4.
π ( R² -- r² ) = 23πcm²
R² -- r² 23
( 2r )² -- r² =23
Ll
Circumstances =2π ( R + r )
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By taking cos 30° instead of sin 30° we can do it fast 😊 6:07