Question

In the given fig., ABCD is a trapezium with AB||CD and , ∠BCD = 60°, If BFEC is a sector of a circle with centre C and AB = BC = 7 cm and DE = 4 cm,then find the area of the shaded region.(Use π = 22/7, √3 =1.732).

Answer 1

GIVEN:
∠BCD = 60°, AB = BC = 7 cm &  DE = 4 cm

CE = CB = 7 cm [ radius of a sector of a circle with Centre C]
CD = CE +DE = 7+4= 11 cm
CD = 11 cm

In ∆CLB ,
sin 60° = BL/BC = Perpendicular/ Hypotenuse
sin 60° = BL /7
√3/2 = BL/7
BL = 7√3/2 cm

Area of trapezium = ½(sum of parallel sides) × distance between parallel sides(BL)
AREA OF TRAPEZIUM (ABCD) =
½ (7+11)×7√3/2 = ½(18)×7√3/2 = (63√3)/2 cm²

AREA OF SECTOR =  (θ /360) ×πr²
Area of sector BFEC = (60°/360°) ×( 22/7) × 7²
= ⅙ (22×7) = (11×7)/3 = 77/3 cm²

Area of shaded region = Area of trapezium -  AREA OF SECTOR BFEC

Area of shaded region = (63√3)/2 - 77/3
= (63 ×1.732)/2 - 25.666
= (108.116 /2) - 25.666
= 54.558 - 25.666
= 28.89cm²
Area of shaded region  = 28.89 cm²

Hence, the Area of shaded region is 28.89 cm².

HOPE THIS WILL HELP YOU...

Answer 2

Hi here is the answer