According to the meaning of the question: CG/AB=CD/(CD+BC)

EH/AB=EF/(EF+CE+BC)

Substituting the known conditions:1.5/ab =1/(1+BC)

1.5/AB=2/(5+BC)

It can be concluded that AB = 6m and BC = 3m. So the height AB of street lamp A is 6 meters.

2. Solution: Schematic diagram Figure 2.

(1) According to the meaning of the question,

AC = BD，CD=DC

∴Rt△ACD≌Rt△BDC

∴∠ADC=∠BCD

∠∠FQD =∠EPC = 90，FQ=EP

∴Rt△FQD≌Rt△EPC

∴QD=CP

∴CD=2CP+PQ=2CP+ 12

And PE/BD=CP/CD=CP/(2CP+ 12).

The solution is CP = 3m.

So CD = 2 * 3+12 = 6+12 =18m. That is, the distance between two street lamps is18m.

(2) The shadow of alternating current on the street lamp is x meters.

According to the meaning of the question: 1.6/9.6=X/(X+ 18)

Solution: x=3.6

So when Wang Hua walked towards the street lamp BD, his shadow under the street lamp AC was 3.6 meters long.

3. Solution: Schematic diagram Figure 3.

(1) Xiaoming's line of sight is blocked by a wall, so the lowest point of his line of sight is the highest point G of the blind area. For example, the blind area in the schematic diagram is between point F and point G. ..

AB/FG=EA/EF=EA/(EA+EF) is known.

CD/FG=EC/EF=EA/(EA+EF)

It can be concluded that FG=7.5 meters, so the blind area is on the Y axis.

The range of is greater than or equal to 0 and less than 7.5.

(2) The area of blind area is the area of trapezoidal DCFG.

S = 0.5 * cf * (CD+FG) = 0.5 * 5 (2+7.5) = 23.75 (square meter)

So the area of the blind area is 23.75 square meters.