Because of the relationship, S=(BD? /2)*sin∠CDE=2sin30 = 1

Let's prove the general situation:

As shown in figure 1, AB=BC, CD=DE, let ∠CDE=α, ∠ABC=β, α = 180-β, BD = A.

In figure 1, rotate δ△△DC' b counterclockwise around point d to get Δ DC' b'.

CD = DE

∴CD coincides with DE, point E is point C', and BD=DB'. Therefore, the area in figure 1 is equal to the ABDB area in figure 2.

The area of ∴∴abdb'e is S△DEB'+S△DEB+S△ABE.

∫∠ABC =β

∴ The included angle between B'E and AB obtained by rotating BC counterclockwise is α+β = 180.

∴B'E∥AB

Connect BE, AB'

BC = B 'E = AB

A quadrilateral is a parallelogram.

The area of two triangles divided by the diagonal of the parallelogram is equal.

∴S△ABE=S△BEB'

∴ area+0 in figure 65438 = s △ DBB' ∠BDB'=α in figure 4, BD = DB' = A.

∴S Pentagon ABCDE = s△ DBB' = (BD× DB'/2 )× sin α = (a? /2)×sinα (sine theorem)

If you haven't studied sine theorem, you can also use the auxiliary line method to find the area of isosceles triangle S△DBB'.

(For the height of waist, h=asinα, then S=ah/2=a? sinα/2)

So this S=(a? /2)×sinα

Applicable to general situations, provided that:

1. convex pentagon

2. Diagonal complementarity

3. Two complementary corners have equal sides.

Haha, I finally finished it. This is actually a junior high school topic, which makes me feel ashamed.

Hope to adopt! The application is full!