17. As shown in the figure, in the pyramid p-abcd, the bottom surface is a right angled trapezoid, ad ‖ BC, ∠ bad = 90 °, PA ⊥ bottom surface ABCD, and PA = ad = AB = 2BC, m, n respectively If the first question is proved, ask the second question. Please don't use vectors

17. As shown in the figure, in the pyramid p-abcd, the bottom surface is a right angled trapezoid, ad ‖ BC, ∠ bad = 90 °, PA ⊥ bottom surface ABCD, and PA = ad = AB = 2BC, m, n respectively If the first question is proved, ask the second question. Please don't use vectors

17. As shown in the figure, in the pyramid p-abcd, the bottom is a right angled trapezoid, ad ∥ BC, ∠ bad = 90 °, PA ⊥ bottom ABCD, and PA = ad = AB = 2BC, m, n respectively

As shown in the figure, m and N are the midpoint of AB and PC respectively in the plane of PA ⊥ rectangle ABCD. (1) prove: Mn ⊥ plane pad; (2) prove: Mn ⊥ CD; (3) if ∠ PDA = π 4, prove: plane PMC ⊥ plane PCD

(1) Take PD midpoint e, connect AE and en, then en ‖. 12CD ‖. 12ab ‖. Am, so the quadrilateral amne is parallelogram ⊂ AE and AE ⊂ planar pad, Mn ⊄ planar pad ⊄ Mn ‖ planar pad (5 points) (2) ⊥ PA ⊥ planar ABCD ⊥ PA ⊥ ab ⊥ ab ⊥ planar pad ⊥ ab ⊥ AE, that is ab ⊥ Mn and CD ⊥ AB, ⊥ Mn ⊥ CD (10 points) (3) ⊥ PA ⊥ planar ABCD ⊥ PA ⊥ ad ⊥ APD = 45 °, e is p The results show that the center point of D is ∩ AE ⊥ PD and AE ∥ Mn ⊥ Mn ⊥ PD and Mn ⊥ CD, and PD ∩ CD = D ∩ Mn ⊥ plane PCD and Mn ⊂ plane PMC, ∩ plane PMC ⊥ plane PCD (14 points)

As shown in the figure, in trapezoidal ABCD, ab ‖ CD, m and N are the midpoint of CD and ab respectively, and Mn ⊥ ab. trapezoidal ABCD must be isosceles. Please explain the reason in two different ways

Proof 1: connecting am and BM, ∵ n is the midpoint of AB, ∵ an = BN, and ∵ Mn ⊥ AB, ∵ am = BM, ∵ amn = ∵ BMN, ∵ m is the midpoint of CD, ∵ cm = DM, and ∵ am = BM, ? mAb = ? MBA = ∵ AMD, ? MBA = ? BMC, ≌ ADM ≌ BCM, ≌ ad =

In order to beautify the environment, a rectangular garden ABCD should be built on a piece of open space with one side against the wall (the wall length is 19m)
The other three sides are surrounded by exhibition fortress with a total length of 36cm, as shown in the figure. The BC side length of ruoshe garden is XM, and the garden area is y m2
Q:
What is the maximum area of the garden? What is the maximum area?

BC = x, then ad = x, ab = CD = (36-x) / 2, the area is:
y=BC×AB=x×（36-x）/2=18x-x^2/2=-1/2(x^2-36x)=-1/2(x^2-36x+324-324)=-1/2(x^2-18)+162
When x = 18, the maximum value of Y is 162
Because I don't see the picture, I think BC is the side against the wall

A surveyor stood at point P and measured the distance between the three vertices of ABCD on rectangular land of other roads. PA = 60, Pb = x, PC = 70, PD = 20
A survey team member stood at point P and measured the distance between the three vertices of ABCD on rectangular land of other roads, PA = 60, Pb = x, PC = 70, PD = 20. In order to determine the distance x from him to the fourth vertex, do you need to measure other data?

There is no need. The specific measures are as follows:
Suppose the coordinates of ABCD are (0,0) (0, a) (B, a) (B, 0), and the coordinates of point P are (x, y)
So your problem is to find x2 + y2 =?
The solution is x2 + y2 = 1700
The results came out
It should be noted that although the distance can be obtained, the shape of the rectangle cannot be determined

This is a rectangular ABCD field, length AB = 62, width ad = 41
The width of the road at the entrance of a and B is 1, the junction of the two paths is 2, and the area of lawn planted in the rest part?

It can be seen from the figure: after removing the path in rectangular ABCD, the lawn can just be put together into a new rectangle, and its length is 62-2 = 60, width is 41-1 = 40
So the area of lawn should be length × width = 60 × 40 = 2400,

This is a rectangular ABCD site with length AB = 102m and width ad = 51m. The width of the middle road from the entrance of a and B is 1m, and the width of the road at the junction of the two paths is 2m
The answer is as follows:
It can be seen from the picture that the remaining part of the lawn can just be put together into a rectangle. The length of the rectangle is 102-2 = 100m, and the width of the rectangle is 51-1 = 50m. Therefore, the area of the lawn is 50 × 100 = 5000 square meters
Why subtract "1" of width? Why?

The entrance is one meter wide, ad minus that entrance, so it's one meter wide

The picture shows a rectangular field ABCD, with length AB = 102m and width ad = 51m. The road width at the junction of two paths at the entrance of a and B is 2m,

38. As shown in the picture, it is a rectangular ABCD site with length AB = 102m and width ad = 51m. The middle road from entrance a and B is 1m wide,
The width of the road at the junction of the two paths is 2m, and the area of lawn is ()
（A）5050m2 （B）4900m2 （C）5000m2（D）4998m2
The answer is C

Draw an inscribed rectangle ABCD on a round field with a radius of 20 meters. How to make the rectangle area the largest?

The square is the largest, that is, draw two vertical diagonals to get the square

The ratio of field AB to width ad of a rectangular field ABCD is 2:1, de ⊥ AC at point E, BF ⊥ AC at point F, connecting be and DF,
It is planned to plant flowers and plants in the quadrilateral debf area, and calculate the area ratio of quadrilateral debf to rectangular ABCD
I don't need it tomorrow! I will be scolded by the teacher very miserably!

Let B, D to ab be H. ∵ ad = BC, ad ∥ BC, de ⊥ AC, BF ⊥ AC, ≌ AED ≌ BCF. Similarly, ≌ AEB ≌ DCF. ≌ AE = CF, s △ AED = s △ BCF = s △ AEB = s △ DCF