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 = 360º​
 ∴ sector BFEC is $\frac{1}{6}$^th of the total circle
so area of sector BFEC = $\frac{1}{6}{\mathrm{\pi r}}^2$
                                  = $\left(\frac{1}{6}\right)\left(\frac{22}{7}\right){\left(7\right)}^2$
                                 = (0.1666)(3.1428)(49)
                                = 25.6651 $c{m}^2$
For area of ABCD
we know area of a trapezium = $\left(\frac{a+b}{2}\right)h$
Here a = AB = 7 cm
       b = DC = 11 cm
      h = BG
For BG In $∆BCG$
 we know ∠ BCG = 60º​
we know $\sin 60° = \frac{BG}{BC}$
           $$\begin{gathered} \frac{\sqrt{3}}{2} =\frac{BG}{BC} \\ \frac{\sqrt{3}}{2} = \frac{BG}{7} \\ BG = \frac{\sqrt{3}}{2}\times 7 \end{gathered}$$
          BG = 6.062
        ∴ h  = 6.062 cm
substitute values of a, b and h to find area of trapezium, we get:
              =  $\left(\frac{7+11}{2}\right)\left(6.062\right)$
             =  54.552 $c{m}^2$
Area of shaded section  = area of trapezium - area of section BFEC
                                  = 54.552 - 25.6651
                                  = 28.8929 $c{m}^2$                       (Ans)