过梯形ABCD的顶点D作DF⊥BC于F.
∵∠B＝45° ∴∠BAE=45° ∴BE＝AE＝6 再有勾股定理可得AB=6√2
∵∠C＝60° ∴∠FDC＝30° ∴FC＝½DC 再有勾股定理得DC＝4√3
即求斜坡AB,CD的的长为6√2,4√3.