In rectangular ABCD, AE⊥BD is in E,CF⊥BD is in F, BF= 1, EF=2, then the area of parallelogram ABCD is equal to the required speed.

Triangle ABE is similar to triangle AED, AE/BE=DE/AE, that is, the square of AE is equal to ED, which is easy to get, BE=DF, so DE=2+ 1=3, AE= root number 3, BD= 1+2+ 1=4, so triangle.

Or use this method:

And symmetry, DF=BE= 1,

cf^2=df*(be+ef)= 1*( 1+2)=3；

CF= root number 3

Rectangular area =BD*CF=(BE+EF+DF)*CF=4 times the root number 3.