Example 1: Xiaoming needs to walk 36 steps from 1 building to the third floor, and how many steps does Xiaoming need to walk from 1 building to the sixth floor?

36(3- 1)= 18 (horizontal)

18x(6- 1)=90 (level)

A: Xiaoming needs to walk 90 steps from the 1 building to the 6th floor.

Example 2: Plant trees around the oval fish pond. The circumference of the fish pond is 1000 meters. If every 50m species 1 tree, how many trees will be planted in a * * *?

1000÷50=20 (tree)

A * * * can plant 20 trees.

Example 3: There is a square flower bed in the school with a side length of 50m. Now we should plant trees around the flower bed, in four corners, with a distance of 5m between every two adjacent trees. How many trees will be planted?

50x4÷5=40 (tree)

A * * * can plant 40 trees.

Example 4: The construction team needs to pile on the foundation with a length of 150m and a width of 60m. Piling should be done at all four corners, and one pile should be driven every 2.5m How many piles should be driven around the foundation of this building?

(150+60) x2 = 420m.

420÷2.5= 168 (root)

A: 168 piles need to be driven around the foundation of this building.

Mathematical wide angle-planting trees mainly discusses the problem of planting trees at both ends. Through some common practical problems in real life, students are guided to explore the one-to-one relationship between the number of trees planted and the number of intervals by means of line graphs. Inspire students to discover laws through phenomena.

Understand the strategy to solve the problem of planting trees, extract the mathematical model, and then solve the practical problems in life with the discovered laws.

Mathematics wide angle-knowledge points of tree planting;

1. method: make it smaller or simpler, draw a picture, list it, and then summarize the application.

2. Tree planting problem: (1), planting at both ends: interval number = total length ÷ spacing; Total length = spacing x number of intervals; Tree = interval+1; Interval = Tree-1 (similar problems include: vertical telephone poles, flags inserted at both ends ...). (2) No planting at both ends: number of intervals = total length ÷ spacing; Total length = spacing x interval number: tree = interval number-1; Interval number = tree+1 (similar problems are: sawing wood, cutting wires ...).

(3) Plant one head and not plant the other: interval number = total length ÷ spacing; Total length = spacing x number of intervals; Tree = number of intervals: number of intervals = tree (similar problems are: ringing the bell to listen to the sound, time to go upstairs ...).

3. Sawing problem: number of segments = times+1: times = number of segments-1; Total time = x times each time.

4. Square matrix problem: the outermost layer number is: side length x4-4 or (side length-1)x4: single side length = (outermost layer number +4)÷4 The total number of the whole square matrix is: side length x side length.

5. Closed figures (such as circles and ellipses): total length ÷ spacing = number of intervals; Tree = interval number.

6. Total length of bridge crossing problem = body length+vehicle distance x vehicle distance+bridge (road length) speed = total length ÷ time.

7, taxi billing (letter postage, photo printing) and other issues. The calculation is divided into two parts. (1) standard parts. If you already know the total price, it doesn't count. If you don't know the total price, it counts. (2) Excess. Excess quantity x exceeds the unit price. Finally add it up.