Suppose AB is base. Suppose h is the perpendicular distance between AB and CD. Then the area of ABCD is (1/2)(a+b)h. EF || AB. EF lies half way between AB and CD. Therefore the perpendicular distances between AB and EF and between DF and CD are each (1/2)h. The length of EF lies half way between that of AB and that of CD, so EF = (1/2)(a+b). So the area of ABFE is (1/2)(a+(1/2)(a+b))(1/2)h = (1/8)(3a+b)h and the area of EFCD is (1/2)((1/2)(a+b)+b)(1/2)h = (1/8)(a+3b)h. Therefore the ratio of areas of ABFE and EFCD is (3a+b)/(a+3b).
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3a+b:2a+2b is the answer.... please mark as brainliest