Question

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Answer 1

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here is your answer

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.

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