✔ 最佳答案
Let ABC be the fixed card and DEF be the moving card.
To begin with, B and E touches each other and ABED formed a straight line with length (2 + 2) = 4 cm. When EF moves along BC, D also moves until ACFD formed a straight line also with length 4 cm. Therefore, area of region covered by the moving figure
= Area of sector AEDFCA - Area of sector ABC + 2 x (area of sector DEF - area of triangle DEF).
Area of sector AEDFCA = (pi)(4)^2 x 60/360 = 8(pi)/3.
Area of sector ABC = (pi)(2)^2 x 60/360 = 2(pi)/3.
Area of sector DEF = area of sector ABC = 2(pi)/3.
Area of triangle DEF = (1/2)(2)(2)sin 60 = sqrt 3.
Therefore, Area = 8(pi)/3 - 2(pi)/3 + 4(pi)/3 - 2sqrt 3.
= 10(pi)/3 - 2sqrt 3 = 31.4/3 - 2 x 1.732 = 7 cm^2.