In a park, three channels with the same width are built on a rectangular ground ABCD with a length of 40 meters and a width of 26 meters (as shown in the figure), two of which are parallel to AB and the other is vertical to ab. these three channels divide the ABCD into small rectangular blocks with an area of 144 square meters, and plant six colors of flowers and plants respectively to calculate the width of the channel

In a park, three channels with the same width are built on a rectangular ground ABCD with a length of 40 meters and a width of 26 meters (as shown in the figure), two of which are parallel to AB and the other is vertical to ab. these three channels divide the ABCD into small rectangular blocks with an area of 144 square meters, and plant six colors of flowers and plants respectively to calculate the width of the channel

Let the width of the road be x meters. From the meaning of the title: 40 × 26-2 × 26x-40x + 2x2 = 144 × 6, the solution is: x2-46x + 88 = 0, and the solution is: X1 = 2, X2 = 44. When x = 44, the width of the road exceeds the length and width of the rectangular field, so it is not appropriate to give up. Answer: the width of the road is 2 meters

As shown in the figure, in a rectangular land with length a and width b, build two roads with width x, then the area of the remaining land is

Idea 1: total area - horizontal road - vertical road + one intersection
(this overlaps an intersection, so it needs to + the area of an intersection)
a×b－a× X－b×X+X×X
Idea 2: roads do not intersect / / such problems will not appear
a×b－a× X－b×X

A city wants to build a quadrilateral garden on the open space of a parallelogram ABCD. The area of the garden is half of the area of the parallelogram ABCD
And the four vertices of the quadrilateral garden should be on the four sides of the parallelogram ABCD
How to draw a picture? Just say the method. I'd better draw it. I'm very grateful

On the four sides AB BC CD Da of parallelogram ABCD, take the midpoint m n p q to connect mnpq in turn, then the quadrilateral formed is the required garden shape

In a residential area, a rectangular garden ABCD is to be built on an open space with one side against the wall (the wall is 15m long). One side of the garden is against the wall, and the other three sides are for general use
The garden is enclosed by a fence with a length of 40m, the width x area y. find out the relationship between Y and X, and write out the value range X of the independent variable x, when y is the largest

Let the width x be perpendicular to the wall, then the length parallel to the wall = 40-x-x = 40-2x,
y=（40-2x）x
The value range of the independent variable x,
40-2x

There is a zigzag road in the park
There is a "Z" shaped road ABCD in the park, as shown in the figure (the figure can't be opened). AB / / CD has a small stone bench e, F, m beside AB, CD, BC, and be = CF, M is at the midpoint of BC. Try to explain that the three stone benches e, F, m are just in a straight line
Answer neatly and logically

∵ ab ∥ CD, (known)
The angle between two parallel lines is equal
∵ m is the midpoint of BC
∴BM=CM,
In △ BEM and △ CFM,
Be = CF (known)
∠ B = ∠ C (proved)
BM = cm (midpoint definition)
∴△BEM≌△CFM（SAS）．
∴∠BME=∠CMF,
And ∠ BMF + ∠ CMF = 180 °,
∴∠BMF+∠BME=180°,
E, m, f are in a straight line

As shown in the figure, a plot plans to build a small road with the same width on a rectangular field ABCD with a length of 32m and a width of 20m, and plant grass on the rest. If the lawn area is 540m2, the width of the road should be calculated?

Let the width of the road be XM, and the path that can be built by the graph can be equivalent to: a horizontal path and a vertical path. The lengths of the two paths are 32m and 20m respectively, but the overlapping intersection of the paths is calculated twice, so the total area of the path is: (32 + 20) x-x2. From the meaning of the question, we can get: (32 + 20) x-x2 = 32 × 20-540（ The width of the road should be 2m

It is known that the area of quadrilateral ABCD is 1, and points e, F, G and H are on edges AB, BC, CD and Da of quadrilateral ABCD respectively
(1) If the points e, F, G and H are the middle points of each side, as shown in Figure 1, please write the area of the quadrilateral efgh directly
(2) If AE / EB = BF / FC = CG / Gd = DH / ha = 2, as shown in Figure 2, the area of quadrilateral efgh is requested;
(3) If AE / EB = BF / FC = CG / Gd = DH / ha = K (k is a positive number), as shown in Figure 3, please use the algebraic expression of K to express the area of efgh and explain the reason

（1）1/2
(2)S△ AEH:S Δ abd = AE / AB * ah / ad = 2 / 3 * 1 / 3 = 2 / 9 CFG:S △CBD=2/9
S △ AEH + s △ CFG = 2 / 9s quadrilateral ABCD
Similarly, s △ bef + s △ DGH = 2 / 9s quadrilateral ABCD
So efgh = (1-2 / 9) ABCD = 7 / 9
(3)S△ AEH:S Δ abd = AE / AB * ah / ad = 1 / (K + 1) * k / (K + 1) = K / (K + 1) & # 178; similarly, s △ CFG:S △CBD=k/(k+1)²
S △ AEH + s △ CFG = K / (K + 1) &# 178; s quadrilateral ABCD
Similarly, s △ bef + s △ DGH = K / (K + 1) & 178; s quadrilateral ABCD
So s quadrilateral efgh = [1-k / (K + 1) & # 178;] s quadrilateral ABCD = (K & # 178; + K + 1) / (K + 1) & # 178;

As shown in the figure, in the known circle O inscribed quadrilateral ABCD, ab = 1, BC = 2, CD = 3, Da = 4, find (1) the area of quadrilateral ABCD; (2) the radius r of circle o

(1) By the cosine theorem in △ ABC, we get ac2 = AB2 + bc2-2ab · bccos ∠ ABC = 12 + 22-2 × 1 × 2cos ∠ ABC = 5-4cos ∠ ABC (3 points). By the cosine theorem in △ ACD, we get ac2 = ad2 + dc2-2ad · DCCOs ∠ ADC = 42 + 32-2 × 4 × 3cos ∠ ADC = 25-24cos ∠ ADC (6 points)

In rectangular paper ABCD, ab = 3, ad = 5. As shown in the figure, fold the paper so that point a falls at a 'on the edge of BC, and the crease is PQ. When point a' moves on the edge of BC, the end point P.Q of the crease also moves. If the limiting points P and Q move on the edge of AB and ad respectively, the maximum distance that point a 'can move on the edge of BC is ()
A. 1B. 2C. 3D. 4

As shown in Figure 1, when point d coincides with point Q, according to the folding symmetry, we can get a ′ d = ad = 5. In RT △ a ′ CD, a ′ D2 = a ′ C2 + CD2, that is 52 = (5-a ′ b) 2 + 32, we can get a ′ B = 1. As shown in Figure 2, when point P coincides with point B, according to the folding symmetry, we can get a ′ B = AB = 3, ∵ 3-1 = 2, and the maximum distance that a ′ can move on the edge of BC is 2

As shown in the figure, there is a quadrilateral piece of paper ABCD, ab ‖ CD, ad ‖ BC, ∠ a = 60 degrees. Fold the paper along the crease Mn and PQ respectively, so that point a coincides with point E on the edge of AB, point C coincides with point F on the edge of CD, eg bisection ∠ MEB intersects CD at g, FH bisection ∠ PFD intersects AB at h. try to explain: (1) eg ‖ FH; (2) me ‖ PF

(1) ∵ point a is folded along Mn to coincide with point E, point C is folded along PQ to coincide with point F, ∵ mea = ∵ a, ∵ PFC = ∵ C, (1 point) ∵ ab ∥ CD (known), ∵ a = 60 °, ∵ D + ∥ a = 180 ° (two lines are parallel, and the inner angles of the same side are complementary), ∵ d = 120 °, ∵ ad ∥ BC (known), ∵ C + ∥ d = 180