Can you find area of the yellow portion? | (Fun Geometry Problem) |

[@phungpham1725]

1/ R= 7
2/ Label ED= x and CD= a
3/ Draw the bisector (the diameter) of CD and intersecting the chord and arc AF at HI.
Note that HI=DE= x
-->a+x= 2R = 14 (1)
And sq (a/2) = DE.DF=14x (2)
From (1) and (2) we have:
X= 14/5 and a= 4x= 4.14/5 =11.2 m
3/ Area of triangle OBE
Height from O to BE= 7-14/5= 4.2m
Area=23.52
4 /Area of sector OBE:
By using cosine law:
Cos angle BOE= -0.28-> angle BOE=106.26 degress
--> Area of sector= pi. 49. 106.26/360 =45.412
Area of segment= 45.412-23.52=21.89 sq m
Area of yellow region = area of rectangle BCDE-area of segment= 9.47 sq m😅😅😅

[@MathandEngineering]

Thank you sir for this amazing method, I hope everyone who comes here benefit like I did from it, thanks

[@santiagoarosam430]

M es punto medio de AB ; N lo es de BE y P lo es de CD. ACDF=2a*2a. ---> AM²+MO²=OA²---> (2a-r)²+a²=r²---> a=4r/5=MO=BN=NE---> BE=8r/5 ; AM=MB=3r/5=ON ; NP=flecha del arco BPE=2r/5. ---> tg(BON)=BN/NO=4/3---> Ángulo BON=53,13º--->Ángulo BOE=2*BON=106,26º --->. Área amarilla =(Rectángulo BCDE) - (Segmento circular BPE) =[(2r/5)(8r/5)]-[(πr²*106,26º/360º)-(8r*3r/5*5*2)]= r²[(28/25)-(π*106,26º/360º)]=(154/π)*[(28/25)-(π106,26°/360°)]= 9,4464..m².
Gracias y un saludo cordial.
-

[@marioalb9726]

Radius of circle :
A = 154m² = πR² --> R=7,00141m
Side of square : (Inters.chords theorem)
s. (2R-s) = (½s)² = ¼s²
2R - s = ¼ s --> s=8/5 .R
s = 11,20225m
Angle of circular sector :
sin ½α = ½s / R = 0,8 --> α=106,26°
Yellow shaded area :
A = A₁ - A₂ - A₃
A = s² - s.2(s-R) - ½R²(α-sinα)
A = 125,4904 - 94,1177 - 21,9262
A = 9,4465 m² ( Solved √) ±0.0001m

[@MathandEngineering]

Wow, thank you so much for this method your shared with us.
It seems you used R to represent 2 different values, one is the radius of he circle and the other, it line 6.
And please how did you get the external segment of the secant to be 2R-s.

[@marioalb9726]

@@MathandEngineering
@MathandEngineering
Hello. Nice video !!!
R is radius of circle, only that (not 2 different values)
Sometimes the text letter size give you an incorrect display,
I modified lightly line 6, was Ok and is Ok now.

[@marioalb9726]

​@@MathandEngineering
I just have applied " Intersecting chords theorem"
One chord is AF, side of square
and this chord was splited in half "½s" and then squared
The other chord is the vertical diameter segment, splited in 2 parts, "s" and "2R-s" that added together is "2R", and multiplied is "s(2R-s)"

[@matthieudutriaux]

Yellow=(56/5)^2-(6/5*7*8/5*7+7^2*arctan(4/3)-7*3/5*7*4/5)
Yellow = 9,44 m² (+/- 0,01 m²)
(with approximation : Pi=22/7)

[@MathandEngineering]

Thank so much for sharing this solution you did, but honestly I am having a hard time understanding how you did all of the solution, please can you make it a little bit more clear than it is right now, thanks

[@matthieudutriaux]

@@MathandEngineering Okay.
Pi*r^2=154
22/7*r^2=154 on average
r^2=7*154/22=49 then r=7
x=AC=CD=DF=AF
Pythagorean Formula : right angled triangle with hypothenuse OA :
r^2=(x/2)^2+(x-r)^2
4*r^2=x^2+(2*x-2*r)^2
196=x^2+(2*x-14)^2
x=56/5=11,2
Area(Yellow)=Area(square ACDF)-Area(rectangle ABEF)-Area(segment)
Area(square ACDF)=x^2=(56/5)^2
Area(rectangle ABEF)=AB*AF=AB*x=2*(x-r)*x=2*(56/5-7)*56/5=2352/25
Area(segment)=Area(sector)-Area(triangle OAF) (same method than you, my dear)
Same angle C than you at 2:55
But i say that C=2*arcsin((x/2)/r) (like you at 7:06) since OAF is an isoscele triangle with sides OA=OF=r=7 and AF=x=56/5
C=2*arcsin(4/5)=2*arctan(4/3)
sin(C)=sin(2*arcsin(4/5))=24/25=0,96
Area(sector)=C*r^2/2=49*arctan(4/3)
Area(triangle OAF)=1/2*OA*OF*sin(C)=1/2*7*7*24/25=588/25 ( =1/2*AF*(1/2*AB) = 1/4*Area(rectangle ABEF) )
Area(segment)=49*arctan(4/3)-588/25
Area(Yellow)=(56/5)^2-2352/25-(49*arctan(4/3)-588/25)
Area(Yellow)=1372/25-49*arctan(4/3)
Area(Yellow)= 9,44 m² (+/- 0,01 m²)