B
分析:首先作等腰梯形的两条高,易得四边形AEFD是矩形,Rt△ABE≌Rt△DCF;根据题意,可得BE=CF=4cm,由勾股定理,即可求得等腰梯形的腰长.
解答:
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解:过点A作AE⊥BC于E,过点D作DF⊥BC于F,
∴AE∥DF,∠AEB=∠DFC=90°,
∵AD∥BC,
∴四边形AEFD是矩形,
∴AD=EF,AE=DF,
∵AB=CD,
∴Rt△ABE≌Rt△DCF(HL),
∴BE=CF,
∴BE=CF=
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(BC-AD)=
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×(10-2)=4(cm),
∵AE=3cm,
∴AB=5cm.
故选B.
点评:本题考查了等腰梯形的性质,矩形的判定与性质、全等三角形的判定与性质等知识.注意梯形中常做的辅助线:①作高;②平移一腰;③平移对角线;④反向延长两腰.