1, connecting the center of the square with the other two vertices, it can be proved that two triangles with 45 degrees are congruent, and then the area of the shadow part is equal to the area of the square, that is? .

N squares, the number of shaded parts is n- 1, and the sum of ∴ areas is (n- 1)/4.

2、

Assume that the army will reinforce the riverbank by X meters every day after receiving the instruction, and then reinforce it by (x- 15) meters every day before receiving the instruction.

According to the meaning, you get 40/ (? x- 15)？ +( 150-40)/? x？ =3

Solution, x 1=55, x2= 10.

It is verified that x 1=55 and x2= 10 are the roots of the original equation, but when x= 10 and x-15 =10-15 < 0,

∴x= 10 doesn't matter, just take x=55.

3、

The quadrilateral ACDE is an isosceles trapezoid, ∫ diagonal AE=CD, DE∨AC (to be proved according to the angle).

Perimeter only needs to calculate DE, that is, the length of FH.

AB = 4，AD=3，

∴AC=5，

∵DF？ AC=DA？ DC (right triangle area method)

∴DF=EH=2.4？ ,

∴AF=CH AF？ =AD？ -DF？

∴FH=DE=AC-2AF

Then the area should be available.