HALPPPPP

area of shaded region  =  area of sector ACD - area of blue segment - area of △ACD

----------

AB  and  DB  are both radiuses of circle B so they are the same length.

AD  =  DB  =  AB  =  20

So △ABD  is an equilateral triangle.

m∠ABD  =  60°

m∠ACD  =  2( m∠ABD )   =   2( 60° )  =  120°

----------

m∠GCD  =  (1/2)(m∠ACD)  =  (1/2)(120°)  =  60°

△CGD is a 30-60-90 triangle.

GD  =  (1/2)(AD)  =  (1/2)(20)  =  10

CG  =  10 / sqrt(3)

CD =   20 / sqrt(3)

----------

area of sector ABD  =  \(\pi\cdot(\text{radius})^2\cdot\big(\frac{\text{angle}}{360^\circ}\big)\ =\ \pi\cdot(20)^2\cdot\big(\frac{60^\circ}{360^\circ}\big)\ =\ \frac{400\pi}{6}\)

area of △ABD  =  \(\frac12\cdot(\text{base})\cdot(\text{height})\ =\ \frac12\cdot(20)\cdot(10\sqrt3)\ =\ 100\sqrt3\)

----------

area of blue segment  =  area of sector ABD - area of △ABD =  \(\frac{400\pi}{6}-100\sqrt3\)

area of sector ACD  =  \(\pi\cdot(\text{radius})^2\cdot\big(\frac{\text{angle}}{360^\circ}\big)\ =\ \pi\cdot\Big(\frac{20}{\sqrt3}\Big)^2\cdot\Big(\frac{120^\circ}{360^\circ}\Big)\ =\ \frac{400\pi}{9}\)

area of △ACD  =  \(\frac12\cdot(\text{base})\cdot(\text{height})\ =\ \frac12\cdot(20)\cdot(\frac{10}{\sqrt3})\ =\ \frac{100}{\sqrt3}\)

----------

area of shaded region  =  \(\frac{400\pi}{9}-(\frac{400\pi}{6}-100\sqrt3)-\frac{100}{\sqrt3}\ \approx\ 46\) square units

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!!!  ]

Hectictar and I agree.....that might be dangerous...... the Sign of the Apocalypse.!!!