# 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?

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 $\frac{1}{6}$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

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)

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