Analysis: Short distance is the length of (AD+CD+BC-AB). The crossing point D is DH⊥AB in H and DG∑CB in G. The trapezoidal problem is transformed into a triangle to solve.

Solution: As shown in the figure, the intersection point D is DH⊥AB in H, and dg∨CB intersects AB in G. 。

∫DC∨AB，

Quadrilateral DCBG is a parallelogram.

∴DC=GB,GD=BC= 1 1.

The difference between the two routes is ad+DG-ag.

At Rt△DGH,

DH=DG？ sin37 ≈ 1 1×0.60=6.60，

GH=DG？ cos37 ≈ 1 1×0.80≈8.80。

In Rt△ADH,

AD=？ 2DH≈ 1.4 1×6.60≈9.3 1。

AH=DH≈6.60。

∴ad+dg-ag=(9.3 1+ 1 1)-(6.60+8.80)≈4.9(km).

That is to say, the distance from A to B is 4.9 kilometers less than before.