设另一腰长为AB=xcm,过D作DE⊥BC垂线,垂足为E在直角梯形ABCD中有,AD∥BC,AB⊥BC,∴DE∥AB∴四边形ABED为平行四边形∴AB=ED=xcm在RT△CED中,有CD=10cm,∠C=45°∴DE=CE=xcm则DE^2+CE^2=CD^,即x^2+x^2=10^2解得：x=5√...