Can you find area of the Blue shaded region? | (Semicircle) |
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- Опубліковано 10 чер 2024
- Learn how to find the area of the Blue shaded region. Important Geometry and Algebra skills are also explained: Congruent triangles; area of the sector formula. Step-by-step tutorial by PreMath.com
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Good explanation
Glad to hear that, Ravi ji 🌹
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Beautiful question and nice solution
I found a way to solve this but your solution was so elegant!
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Elegant solution. Thank you.
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Very interesting problem.
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Nice!
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α = 60° ; β = 30°
A = A2 - A1
A = ½R² (2α- sin2α) - ½R² (2β - sin 2β)
A = ½3² (⅔π -√3/2) - ½3² (⅓π - √3/2)
A = 5,527 - 0,815
A = 4,71 cm² ( Solved √ )
Beautiful solution. Thank you, sir.
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Divided by six equal parts 6×30°, the area of the first segment is 1/6×3^2 pi+3^2sqrt(3)×(1/2)=(3/2)pi+(9/2)sqrt(3), the area of the third segment is (3/2)pi-(9/2)sqrt(3), therefore the answer is (9/2) pi-3pi=(3/2)pi.😮 the interesting answer, that is one sixth xircle.
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Liked this a lot
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Brilliant solution!
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Love and prayers from the USA! 😀
Very clever indeed
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Area of Blue region=2(30/360)(π)(3^2)=3π/2 square units.=4.71 square units.❤❤❤
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Los cuartos de circulo están trisecados en ángulos de 30°.
Área BNMC =(Sector circular de 120° - Triángulo OBN) - (Sector circular de60° - Triángulo OCM).
Los triángulos OBN Y OCM tienen igual superficie ---> Área BNMC =Sector 120° - Sector 60° =Sector 60° =π3^2/6 =9π/6 =3π/2.
Gracias y un saludo cordial.
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As the lengths of all 6 arcs are the same, we can determine the angle between rhem, as the full arc of a semicircle is 180°:
180°/6 = 30°
Thus CM covers an arc of 2(30°) = 60° and BN covers an arc of 4(30°) = 120°. We can determine the area between them by finding the difference between the circular segment defined by CM and the circular segment defined by BN. The formula for a circular segment is θ/360°πr²-r²sin(θ)/2, or the area of a sector of angle θ minus the corresponding isosceles triangle of side length r. In this case our radius is 6/2 = 3, so we already have all the information we need:
Blue area:
A = (θ₁/360°πr²-r²sin(θ₁)/2) - (θ₂/360°πr²-r²sin(θ₂)/2)
A = (120/360°πr²-r²sin(120)/2) - (60/360°πr²-r²sin(60)/2)
A = πr²/3 - πr²/6
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Brilhante solução!!!! Dayvis Jr (Brazil)
Fico feliz em ouvir isso! 🌹
Obrigado pelo feedback ❤️🙏
Amor e orações dos EUA! 😀
S=3π/2≈4,71
I did it a clunky way area of sectors then minus the triangle up to the chord then subtract the very top area. Worked out the same way
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there is a calculation using the formula, see line 50, and there is a numeric one:
10 rem print ln(pi^4+pi^5); "aus youtube":vdu5:for a=0 to 15:gcola:print a:next a
20 print "premath-can you find area of the blue shaded region"
30 ne=6:nu=15:ne=ne-1:nuk=100:rem anzahl der unterteilungen des halbkreises und nu fstr$vposr die grafik
40 dia=6:r=dia/2:wt=pi/(ne+1):wo=wt*2:wu=wt:ho=r*sin(wo):hu=r*sin(wu):h=ho-hu
50 xl=-r*cos(wo):xr=-xl:da=r*r*(wt/2-sin(wt/2)*cos(wt/2)):wud=pi-2*wt:wod=pi-4*wt: print wod*180/pi
60 lu=r*sqr(2)*sqr(1-cos(wud)):lo=r*sqr(2)*sqr(1-cos(wod))
70 masx=1200/2/r:masy=850/r:if masx
The area can also be calculated from the original diagram, without dividing into 2 quarter circles, as follows: Blue region area = (sector BON - ΔBON) - (sector COM - ΔCOM) = (sector BON - sector COM) - (ΔCOM - ΔBON). The difference in areas of the sectors equals the area of a 60° sector, 1/6 the area of the full circle, or (1/6)(π(3)² = 3π/2. ΔCOM is an equilateral triangle, sides equal to the radius of the circle. ΔBON can be divided into 2 30°-60°-90° special triangles with hypotenuse equal to circle's radius, and assembled into an equilateral triangle with sides equal to the radius, so ΔCOM and ΔBON have equal area. Blue region = 3π/2, as PreMath also found.
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Turn the whole drawing Pi/2 rad clockwise. The blue area is twice the area of FECB with tha abscissa of F beeing 3.cos(Pi/3) = 3/2, the abscissa of E beeing 3.cos(Pi/6) =(3/2).sqrt(3), and the equation of the circle x^2 + y^2 = 9 or y = sqrt(9 -x^2)
So the blue area is 2. the integral from 3/2 to (3/2).sqrt(3) of sqrt(9 -x^2) dx, or 6.the integral from 3/2 to (3/2).sqrt(3) of sqrt(1 - (x/3)^2) dx.
We note x/3 = sin(t), dx = 3.cos(t).dt
The blue area is then 18. the integral from Pi/6 to Pi/3 of sqrt(1 -sin(t)^2).cos(t) dt, or 18.the integral from Pi/6 to Pi/3 of cos(t)^2 dt, or 9.the integral from Pi/6 to Pi/3 of
(1 + cos(2.t)).dt.
Finally 9.(F(Pi/3) - F(pi/6)) with F(t) = t - (1/2).sin(2.t)
or 9.(Pi/3) - sin((2/3).Pi) - Pi/6) + sin(Pi/3)) = 9.(Pi/3 - Pi/6) = 9.(Pi/6) = (3/2).Pi.
It's very simple but not easy to write with the computer without special signs.
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I went straight ahead to integral calculus to reach the same result.
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Die blaue Fläche ist 1/3 der gesamten Fläche. 1,5 π
Yes! What is the difference of congruent triangles OCE and OBF? ...similar question could be asked, what's the difference between a frog? 🤔
😀
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Good morning sir
Same to you, dear 🌹❤️
😉
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Ab=2(9π/4-3π/2)=3π/2
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1/ Area of the blue region = Area of segment BCMN- Area of the small upper white segment.
2/ Notice that the triangle COM is an equilateral triangle of which the side = radius= 3. Moever, the triangle BON is formed by two special 30-90-60 of which the side is also= radius= 3, which means that the areas of the 2 triangles mention above are equal.
So, Area of the segment BCMN = Area of sector 120 degree - Area of triangle BON (1)
Area of small white segment= Area of 60 degree sector- Area of triangle COM (2)
(1) -(2) ->Area of shaded region= Area of 120 degree sector- Area of 60 degree sector = pi sqr/3- pi sqr/6= 3pi-3 pi/2= 3pi/2 sq units😅
blue Area=Semi circle Area - (above white Area+ below white Area)
above white Area=Area(Sector OCM)-Area(Triangle OCM)=(60/360)*9*Pi - (3*3*sin(60))/2=
(9*Pi)/6 - ( (9*sqrt(3))/4 )
below white area=Area(Sector OAB)+Area(Triangle OBN)+Area(sector ONP)
=(9*Pi/12)+(9*sqrt(3)/4 )+(9*Pi/12)=(9*Pi/6)+(9*sqrt(3)/4 )
blue Area=Semi circle Area - (above white Area+ below white Area)
blue Area=(9*Pi/2) - [ (9*Pi)/6 - ( 9*sqrt(3)/4 ) +(9*Pi/6)+(9*sqrt(3)/4 ) ]
=(9*Pi/2)-[ 18*Pi/6 ]
=( 27*Pi- 18* Pi)/6 =(9*Pi)/6
=(3*Pi)/2
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First comment and first like
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Is the blue area a semicircle 😂