Through observation, we can see that the area of quadrilateral is equal to the area of isosceles right triangle minus S△ADF, so we only need to find out the areas of these two triangles, which requires us to comprehensively use the properties of right triangle and right triangle and the flexible use of trigonometric functions to solve them.

Solution: in △EDB,

∠∠EDB = 90°，∠E = 30°，DE=4√3，

∴DB=DE？ tan30 =4√3*√3/3=4。

∴AD=AB-DB=6-4=2.

∠∠A = 45°,∠AFD = 45 °, FD = ad。

∴s△adf= 1/2*ad^2= 1/2*4=2.

In the isosceles right triangle ABC, the hypotenuse AB=6,

∴AC=BC=3√2.

∴s△abc= 1/2*ac^2= 1/2* 18=9.

∴S quadrilateral DBCF=S△ABC-S△ADF=9-2=7.

Typing is very hard, I hope to adopt it ~ ~ ~