∫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.