∫l 1∑L2，

∴∠DHE=90，

∴∠DHE=∠BAE=90，

∫ The quadrilateral BFDE happens to be a diamond,

∴BE=DE，

In △DEH and △ BEI, ∠ DHE = ∠ BEI = 90 ∠ DEH = ∠ BEI = DE,

∴△DEH≌△BEA(AAS)，

∴AB=DH，

∫DH = 2×6 = 12cm，

∴AB= 12cm，

BG = 6cm，

∴∠bak=30；

Solution 2:∫ straight line l 1∪L2∪L3∪l4∪l5, and the distance between two adjacent parallel lines is 6cm.

According to the bisection theorem of parallel lines, DE=2AE,

In addition, the quadrilateral BFDE happens to be a diamond,

∴BE=DE=2AE，

And BAE = 90,

∴∠ABE=30，

∫l 1∑L2，

∴∠bak=∠abe=30；

(2)BAK = 30 degrees, BAE=90 degrees,

∴∠DAM=90 -30 =60，

∵BG=6cm，DM= 18cm，

∴AB=6sin30 = 12cm，

AD= 12sin60 = 123cm，

The area of rectangular ABCD is12×123 =1443cm2.