Question

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 1

Solution :-

In trapezium ABCD,

AB II DC

∠ BCD = 60°

AB = BC = 7 cm and DE = 4 cm

C is the centre of the sector. So, BC and EC are radii of this section.

BC = EC = 7 cm

∴ DC = DE + EC = 4 + 7 = 11 cm

Area of the sector BEFC = 60/360*πr²

⇒ 1/6*22/7*7*7

= 25.67 cm²

Area of trapezium = [(a + b)/2*h]

Here, a = AB = 7 cm ; b = DC = 11 cm and height = BG

For BG in Δ BCG,

∠ BCG = 60°

Sin 60° = BG/BC

⇒ √3/2 = BG/BC

⇒ √3/2 = BG/7

⇒ BG = (√3*7)/2

⇒ BG = 12.124/2

⇒ BG = 6.062 cm or height = 6.062 cm

Area of trapezium = [(7 + 11)/2]*6.062

⇒ 18*6.062/2 

⇒ 109.116/2

⇒  54.558 cm²

Area of shaded region = Area of trapezium - Area of sector BEFC

⇒ 54.558 - 25.67 

= 28.888 cm²

Answer.

Answer 2

In trapezium ABCD,

AB II DC

∠ BCD = 60°

AB = BC = 7 cm and DE = 4 cm

C is the centre of the sector. So, BC and EC are radii of this section.

BC = EC = 7 cm

∴ DC = DE + EC = 4 + 7 = 11 cm

Area of the sector BEFC = 60/360*πr²

⇒ 1/6*22/7*7*7

= 25.67 cm²

Area of trapezium = [(a + b)/2*h]

Here, a = AB = 7 cm ; b = DC = 11 cm and height = BG

For BG in Δ BCG,

∠ BCG = 60°

Sin 60° = BG/BC

⇒ √3/2 = BG/BC

⇒ √3/2 = BG/7

⇒ BG = (√3*7)/2

⇒ BG = 12.124/2

⇒ BG = 6.062 cm or height = 6.062 cm

Area of trapezium = [(7 + 11)/2]*6.062

⇒ 18*6.062/2 

⇒ 109.116/2

⇒  54.558 cm²

Area of shaded region = Area of trapezium - Area of sector BEFC

⇒ 54.558 - 25.67 

= 28.888 cm²

I hope it,s clear to u

Thanx