ABCD is a trapezium AB ||DC and angle BCD =60 degree.if befc is sector with center C and ab=bc=7cm and de=4 cm ,then find the area of the shaded region?

Answer:
           We have a trapezium that have AB||DC
                                               ∠ BCD = 60º
                                       AB = BC = 7 cm
                                                      DE = 4 cm 
                             
Here C is the centre of sector so BC and EC is radius of this section 
                                                 BC = EC = 7cm
                                           
∴ DC = DE + EC = 4 + 7 = 11 cm
For area BFEC,
                    C is the centre of the sector and that forms 
60º angle.
We know total angle of a circle = 3
60º​
 
∴ sector BFEC is 16th of the total circle
so area of sector BFEC = 16πr2
                                  = 1622772
                                 = (0.1666)(3.1428)(49)
                                = 25.6651 cm2

For area of ABCD
we know area of a trapezium = a+b2h
Here a = AB = 7 cm
       b = DC = 11 cm
      h = BG
For BG In BCG
 we know 
∠ BCG = 60º​
we know sin60° = BGBC
           32 =BGBC32 = BG7BG = 32×7
         
 BG = 6.062
        h  = 6.062 cm
substitute values of a, b and h to find area of trapezium, we get:
              =  7+1126.062

             =  54.552 cm2
Area of shaded section  = area of trapezium - area of section BFEC
                                  = 54.552 - 25.6651
                                  = 28.8929 cm2                       (Ans)  

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