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CPhill

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Nombre de usuarioCPhill
Puntuación104674
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Preguntas 52
Respuestas 32492

 #1
avatar+104674 
+2
23 jun. 2019
 #4
avatar+104674 
+3

 

It appears that the chord AD is bisected by  BE

So angle BED is right   and ED  =  1/2 AD  =  10

And we can find angle EBD using the Law of Sines

sin BED          sin (EBD)

_______  =  __________

  BD                 ED 

 

sin 90              sinEBD

______  =       ______

  20                     10

 

sinEBD  = (10/20) sin 90

sin EBD  = (1/2) (1)

sin EBD  = 1/2

So....EBD must be  30° 

So.....by symmetry....angle ABD  must be twice this  = 60°  

So....now....we can compute the area of sector AEBD =

pi* BD^2(60/360)  = pi*  (20)^2 (1/6)  = pi * 400/6  = 200/3 pi  units^2

 

 

And since angle ACD is twice angle ABD.....then   (1/2)ACD = ABD  =  60° = angle ECD

 

Then....since DEC is right, then triangle ECD is a 30-60-90 right triangle

Then  

CD / sin DEC  =   ED / sin ECD

CD / 1  =  10/  [sqrt(3)/2 ]

CD  =  20 /sqrt (3)

 

And the area of sector   AFDC =

pi (CD^2) (120/360)  = pi * (400/3)(1/3)  = (400/9) pi units^2      (1)

 

Now....we need to caclculate the area of triangle ABD

(1/2)(AB)(BD) sin (60°)   =

(1/2)(20)(20)sqrt (3)/2)  =  100sqrt(3) units^2

 

So.....the area between chord AD and arc AED  =

 

Area of  AEBD  - area of triangle ABD  =

 

  ( [200/3] pi  - 100sqrt(3)  )    (2)  

 

Lastly....we need to calculate the area of triangle ACD  [note CD = AC ]  =

(1/2)(CD)(AC)sin(120°)  =  (1/2)(20/sqrt3)^2 sqrt(3)/2  =   (1/2)(400/3)sqrt(3)/2 = 100/sqrt(3) = 100sqrt(3)/3   (3)

 

So....the area we are looking for is : 

Area of sector AFDC  - area between chord AD and arc AED  - area of triangle ACD  =

 

(1)   - (2)  - (3)  =

 

[400/9] pi  - [ 200/3 ]pi + 100sqrt(3) - 100sqrt(3)/3=

 

[ 400/9]pi - [600/9]pi  + (2/3)100sqrt(3)  =

 

(2/3)100sqrt(3) - (200/9)pi  ≈  45.66 units^2  =  46 units^2

 

[If I didn't make any errors....LOL!!!  ]

 

cool cool cool

23 jun. 2019
 #7
 #5