Can you find area of the Yellow shaded region? | (Quarter Circle) |
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- Опубликовано: 14 окт 2024
- Learn how to find the area of the Yellow shaded region. Important Geometry skills are also explained: Law of Cosines; Law of Sines; Sector area; triangle area; Quadratic formula. Step-by-step tutorial by PreMath.com
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1$t view 😂
Excellent!
Thanks ❤️
Method using sine law:
1. In triangle ODC, by sine law, sin(ODC)/4 = sin(120)/8.
sin(ODC) = (sqrt3)/4
Hence angle ODC = 25,6589
2. In triangle ODC, angle COD = 60 - 25.6589 = 34.3411
3. Area of triangle ODC = (1/2)(8)(4)(sin34.3411) = 9.0259
4. Areaof sector OAD = (34.3411/360)(8)(8)(pi) = 19.1797
5. Area of yellow region = 19.1797 - 9.0259 = 10.1538
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8 /sin 120 = 4 / sin ODC.
Sin ODC = 4 x (sin 120) / 8 = 0.433.
ODC = 25.658 degrees.
COD = 180 - 120 - 25.658 = 34.342 degrees.
Area OCD = 1/2 x 8 x 4 x sin 34.342. = 9.026.
Area sector = Pi x 64 x 34.342 / 360 = 19.180.
Yellow area = 19.180 - 9.026 = 10.154.
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Let's use an orthonormal center O, first axis (OA).
The equation of (CD) is y = tan(60°).(x - 4)
or y = sqrt(3).(x - 4)
The equation of the circle si x^2 + y^2 = 64
At the intersection D we have: x^2 + 3.(x - 4)^2 = 64
or 4.(x^4) - 24.x - 16 = 0, or x^4 - 6.x -4 = 0.
Deltaprime = 13, so x = 3 - sqrt(13) (rejected as negative)
or x = 3 + sqrt(13),
So D(3 + sqrt(13); sqrt(3).(-1 + sqrt(13)),
or D(3 +sqrt(3); sqrt(39) - sqrt(3))
Be t = angleAOD. tan(t) = (sqrt(39) -sqrt(3))/(3 +sqrt(13))
tan(t) = sqrt(3).(4 -sqrt(13)) (about 0.683)
Then t = Arctan(4.sqrt(3) -sqrt(39)). (about 0.6 rad or 34.3°)
The area of sectorAOD is (64.Pi).(t/(2.Pi)) =32.t
The area of triangle OCD is (1/2).OC.(ordinate of D)
so it is (1/2).(4).(sqrt(39) -sqrt(3)) = 2.(sqrt(39) - sqrt(3))
Finally the yellow area is:
32.Arctan(4.sqrt(3) -sqrt(39)) - 2.(sqrt(39) - sqrt(3))
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Sine rule, triangle COD:
sinβ / 4 = sin120°/ 8
β = 25,6589°
Triangle COD:
α = 180° -120° - β
α = 34,3411°
Angular sector AOD:
A = ½αR² = ½ α 8²
A = 19,1797 cm²
Triangle COD:
A = ½ a b sinα = ½ 4 . 8. sinα
A = 9,0259 cm²
Yellow shaded area:
A = A₁ - A₂
A = 10,1538 cm² ( Solved √ )
Seeing vídeo, is absolutely unnecessary to calculate 'x', as we can see in my solution.
( And much less unnecessary to introduce a quadratic equation )
Consider the triangle DOC. Length DO is 8, length OC is 4, and angle OCD is 120 degrees. By law of sines,
DO/sin(120) = 8/sin(120) = 4/sin(ODC) --> sin(ODC) = sin(120)/2
Now find angle DOC:
DOC = 180 - 120 - ODC
DOC = 60 - ODC
DOC = 60 - arcsin(sin(120)/2)
DOC = 34.3411 degrees
Now the y coordinate of D is 8*sin(34.3411) = 4.5129, and then length CD is found from 4.5129/CD = sin(60), CD = 4.5129/sin(60) = 5.2111. Now we can use Heron's formula to calculate the area of the triangle DOC:
p = (8 + 4 + 5.2111)/2 = 8.6056
area(DOC) = sqrt(p*(p-8)*(p-4)*(p-5.2111)
area(DOC) = 9.026
The area of the circle sector DOA is pi*8^2*34.3411/360 = 19.1797. Now the sought after area is just the sector area minus the triangle area:
Yellow_Area = 19.1797 - 9.026
Yellow_Area = 10.1532
Q.E.D.
The way this proved is excellent
Glad to hear that!
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Seems quite complicated.
Choose origin at O = (0, 0)
Equation of Circle is
x^2 + y^2 = 8^2
C = (4, 0)
Equation of line CD is
(y - 0) = tan(60) (x - 4)
y = sqrt(3) (x - 4)
Assuming D = (a, b)
and substituting on line
b = sqrt(3) (a - 4)
and substituting in circle
a^2 + b^2 = 8^2
a^2 + (sqrt(3) (a - 4))^2 = 64
a^2 + 3 (a-4)^2 - 64 = 0
a^2+ 3 (a^2 - 8a + 16) - 64 = 0
a^2 + 3 a^2 - 24 a + 48 - 64 = 0
4 a^2 - 24 a - 16 = 0
a^2 - 6 a - 4 = 0
a = (6 +/- sqrt(36 + 16)) / (2 (1))
a = 3 +/- sqrt(13)
a = 3 + sqrt(13), after discarding negative value
b = sqrt(3)(3 + sqrt(13) - 4)
b = sqrt(39) - sqrt(3)
theta = angle(AOD)
cos(theta) = a / r = (3 + sqrt(13)) / 8
theta = arccos((3 + sqrt(13)) / 8)
Yellow Area = Area of sector AOD - Area of triangle COD
= pi r^2 (theta / 2 pi) - 4 b / 2
= pi 8^2 (theta / 2 pi) - 2 b
= 32 theta - 2 b
= 32 * arccos((3 + sqrt(13)) / 8) - 2 (sqrt(39) - sqrt(3))
= 32 * 0.599365154 - 2 * 4.51294719
= 10.15379 sq units
Connect O to D
Angle OCD=180°-60°=120°
OC=AC=4
In ∆ OCD
Let CD=x
OD^2=OC^2+x^2-2(OC)(x)cos(120°)
8^2=4^2+x^2-8x(-1/2)
So x=2√13-2
(2√13-2)^2=4^2+8^2-2(4)(8)cos(COD)
So COD=34.34°
Yellow area=(π)(8^2)(34.34°/360°)-1/2(4)(8)sin(120°)=10.15 square units.❤❤❤
Excellent!
Thanks for sharing ❤️
That was pretty much the method I used, but I like to round at the end, so I got
Area = 32 arctan(√3(4−√13)) − 2√3(√13−1)) ≈ 10.15379
NOTE: Calculator has to be set to radians, and I used the following radian formula for sectors
A = πr² * α/(2π) = r²α/2
Great service thanks 🙏🏻
You are very welcome!
Thanks for the feedback ❤️
Let's find the area:
.
..
...
....
.....
Let's assume that O is the center of the coordinate system and that OA and OB are located on the x-axis and y-axis, respectively. Then we obtain the following coordinates from the sketch:
O: ( 0 ; 0 )
A: ( 8 ; 0 )
B: ( 0 ; 8 )
C: ( 4 ; 0 )
D: ( xD ; yD )
The line CD is represented by the following function:
y = [(yD − yC)/(xD − xC)]*(x − xC) + yC = tan(60°)*(x − 4) + 0 = √3*(x − 4)
Since D is located on the quarter circle, the coordinates of D can be calculated in the following way:
xD² + yD² = R² = 8² = 64
xD² + 3*(xD − 4)² = 64
xD² + 3*xD² − 24*xD + 48 = 64
4*xD² − 24*xD − 16 = 0
xD² − 6*xD − 4 = 0
xD = 3 ± √(3² + 4) = 3 ± √(9 + 4) = 3 ± √13
Since xD>0, the only useful solution is:
xD = 3 + √13
yD = √3*(xD − 4) = √3*(3 + √13 − 4) = √3*(√13 − 1)
tan(∠AOD) = tan(∠COD)
= yD/xD
= √3*(√13 − 1)/(√13 + 3)
= √3*(√13 − 1)(√13 − 3)/[(√13 + 3)(√13 − 3)]
= √3*(13 − 4√13 + 3)/(13 − 9)
= √3*(16 − 4√13)/4
= √3*(4 − √13)
sin(∠AOD) = sin(∠COD) = yD/R = √3*(√13 − 1)/8
Now we are able to calculate the area of the yellow region:
A(yellow)
= A(circle sector OAD) − A(triangle OCD)
= πR²*[∠AOD/(2π)] − (1/2)*OC*OD*sin(∠COD)
= R²*arctan(yD/xD)/2 − (1/2)*(R/2)*R*sin(∠COD)
= 8²*arctan[√3*(4 − √13)]/2 − (1/2)*(8/2)*8*[√3*(√13 − 1)/8]
= 32*arctan[√3*(4 − √13)] − 2√3*(√13 − 1)
≈ 10.15
Best regards from Germany
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Thanks for sharing ❤️
As OA = r, OC = CA = r/2. As ∠DCA = 60°, ∠OCD = 180°-60° = 120°. By the law of cosines:
Triangle ∆OCD:
OD² = OC² + CD² - 2OC(CD)cos(120°)
r² = (r/2)² + CD² - 2(r/2)CD(-1/2)
r² = r²/4 + CD² + rCD/2
4r² = r² + 4CD² + 2rCD
4CD² + 2rCD - 3r² = 0
CD = [-(2r)±√((2r)²-4(4)(-3r²))]/2(4)
CD = -r/4 ± √(4r²+48r²)/8
CD = -r/4 ± √(52r²)/8
CD = -r/4 + √13r/4 --- CD > 0
CD = (√13-1)r/4
Drop a perpendicular from D to E on CA. Let ∠DOA = θ.
DE = CDsin(60°) = ((√13-1)r/4)(√3/2) = (√39-√3)r/8
sin(θ) = DE/OD = ((√39-√3)r/8)/r = (√39-√3)/8
The yellow shaded area is equal to the area of the sector subtended by minor arc DA minus the area of triangle ∆OCD:
Yellow shaded area DCA:
Aʏ = (θ/360°)πr² - r(r/2)sin(θ)/2
Aʏ = 64θπ/360 - 16((√39-√3)/8)
Aʏ = 8θπ/45 - 2(√39-√3)
Aʏ = 8πsin⁻¹((√39-√3)/8)/45 - 2√3(√13-1)
Aʏ ≈ 10.15 sq units
Very nice sharing ❤❤❤❤
Thank you! Cheers!❤️
CD=a..DOC=α R=8 ..legge del coseno...R^2=4^2+a^2-2*4*a*cos120...a=2(√13-1)..R/sin120=a/sinα..α=arcsin((√39-√3)/8)..A(ODC)=4asin120/2=√3a=2(√39-√3)...Asett(ODA)=αR^2/2=32arcsin((√39-√3)/8)... Ayellow=32arcsin((√39-√3)/8)-2(√39-√3)=10,154....
Excellent!
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merci sire , excellent exercice
You are very welcome!
Thanks for the feedback ❤️
Very good!
Yellow Area = [DOA circle wedge] - [DOC]. Let DC=X, then 8^2 = 4^2 + X^2 - 2*4*X*cos (120) -> 64 = 16 + X^2 + 4x -> X^2 - 4X - 48 = 0 -> X^2-4X+4 = (X+2) ^2 = 52 -> X+2 = 2*sqrt (13) -> X = 2sqrt (13) - 2 = 2*[sqrt (13) - 1]. Find [DOC]: X/sin (DOC) = 8/sin (120) -> sin (DOC) = X*sin (120)/8 = 2*[sqrt (13)-1] * (sqrt (3)/2)/8 = [sqrt (39) - sqrt (3)]/8 -> DOC=34.3411. Now Yellow Area = (DOC/360) * Pi*R^2 - 8*4/2 * sin (DOC) = [(34.3411/360) * 64*Pi] - [16*sin (34.3411)] = 19.1797 - 9.0259 = 10.1538 sq. units
Love the vid as usual, but at 7:43, should that not be a double inversion and the root 3 be multiplied by 8, not the 2 multiplied by it?
First we had 2 sides and an angle. Apply Law of Cosines. Next we have 2 sides and an angle and need to find a missing angle. Apply Law of Sines. Find Sector area. Next find area of Triangle OCD. Use calculator! 🙂 ...cue some Mariachi music and tiptoe thru the tulips with a little bit of subtraction and wallah , problem solved! 🙂
Bravo👍
Thanks for the awesome feedback ❤️
Thank you!
You are very welcome!
Thanks for the feedback ❤️
Another way to do it is by finding angle odc .using sine rule and then you have angle doc. Then get the area
"E" es la proyección ortogonal de D sobre OA y CE=a--->ED=a√3 ---> 8²-(4+a)²=(a√3)² ---> a=-1+√13---> tg(DOA)=a√3/(4+a) ---> Área amarilla =(8²π*34,34/360)-(4a√3/2) =10,1532...
Gracias y saludos.
Esta solución es correcta. Incluyo paso intermedio omitido: Potencia de "E" respecto a la circunferencia =[8-(4+a)]*[8+(4+a)] =8²-(4+a)² =(a√3)²
If we draw a perpendicular from D to AC, is it possible to solve with a special triangle?
Once you e have calculated the value of x, isn't it easier to calculate the area of a circle whom radius is X (2(-1+sqare root of 13) and next divide it by six ? Because the angle of CDA is 60° ?
It will not be circle since x which is CD=(2(-1+sqare root of 13) and CA= 4
I used integral calculus to reach the same result. 👍🏼
Nice work!
Thanks for the feedback ❤️
I am Bangladeshi student. We are facing a dangerous situation.
I've seen the footage as well. Our thoughts and prayers are with them 🙏
Pi /6 ×16 ?
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