2. The circumference of a square = side length ×4 C=4a.
3. Area of rectangle = length× width S=ab
4. Square area = side length x side length s = a.a = a.
5. Area of triangle = base × height ÷2 S=ah÷2.
6. parallelogram area = bottom x height S=ah
7. trapezoidal area = (upper bottom+lower bottom) × height ÷ 2s = (a+b) h ÷ 2.
8. Diameter = Radius× 2D = 2r Radius = Diameter ÷2 r= d÷2
9. The circumference of a circle = π× diameter = π× radius× 2c = π d = 2π r.
10, circular area = pi × radius× radius? =πr
1 1, the surface area of a cuboid = (length× width+length× height+width× height) × 2.
12, cuboid volume = length× width× height V =abh.
13, the surface area of the cube = side length × side length× ×6 S =6a.
14, volume of cube = side length x side length x side length v = a.a.a = a.
15, lateral area of cylinder = circumference of bottom circle × height S=ch.
16, surface area of cylinder = upper and lower bottom area+side area.
s = 2πr+2πRH = 2π(d÷2)+2π(d÷2)h = 2π(c÷2÷π)+Ch
17, cylinder volume = bottom area × height V=Sh
V=πr h=π(d÷2) h=π(C÷2÷π) h
18, volume of cone = bottom area × height ÷3.
v = sh÷3 =πr h÷3 =π(d÷2)h÷3 =π(c÷2÷π)h÷3
19, cuboid (cube, cylinder)
1, number of copies × number of copies = total number of copies/number of copies = total number of copies/number of copies = number of copies.
2. 1 multiple× multiple = multiple1multiple = multiple/multiple = 1 multiple
3. Speed × time = distance/speed = time/distance/time = speed.
4. Unit price × quantity = total price ÷ unit price = total quantity ÷ quantity = unit price
5. Work efficiency × working hours = total workload ÷ work efficiency = working hours ÷ total workload ÷ working hours = work efficiency.
6. Appendix+Appendix = sum, and-one addend = another addend.
7. Minus-Minus = Minus-Minus = Minus+Minus = Minus
8. Factor × factor = product ÷ one factor = another factor.
9. Dividend = quotient dividend = divisor quotient × divisor = dividend
Calculation formula of mathematical graphics in primary schools
1, square c perimeter s area a side length perimeter = side length× 4c = 4a area = side length× side length s = a× a.
2. Cube V: volume A: side surface area = side length × side length× 6s table =a×a×6 volume = side length× side length× side length V = a× a× a.
3. rectangular
Perimeter area side length
Circumference = (length+width) ×2
C=2(a+b)
Area = length × width
S=ab
4. Cuboid
V: volume s: area a: length b: width h: height.
(1) Surface area (L× W+L× H+W× H) ×2
S=2(ab+ah+bh)
(2) Volume = length × width × height
V=abh
5 triangle
S area a bottom h height
Area = bottom × height ÷2
s=ah÷2
Height of triangle = area ×2÷ base.
Triangle base = area ×2÷ height
6 parallelogram
S area a bottom h height
Area = bottom × height
S = ah
7 trapezoid
Height of upper bottom b and lower bottom h in s area a
Area = (upper bottom+lower bottom) × height ÷2
s=(a+b)× h÷2
8 laps
Area c perimeter d= diameter r= radius
(1) circumference = diameter ×∏=2×∏× radius
c =∏d = 2r
(2) area = radius × radius×∈
Cylinder 9
V: volume h: height s; Bottom area r: bottom radius c: bottom perimeter
(1) Transverse area = bottom circumference × height.
(2) Surface area = lateral area+bottom area ×2
(3) Volume = bottom area × height
(4) Volume = lateral area ÷2× radius.
10 cone
V: volume h: height s; Bottom area r: bottom radius
Volume = bottom area × height ÷3
Total number ÷ Total number of copies = average value
Sum-difference problem
(sum+difference) ÷ 2 = large number
(sum and difference) ÷ 2 = decimal
And folding problems.
Sum \ (multiple-1) = decimal
Decimal × multiple = large number
(or sum-decimal = large number)
Difference problem
Difference ÷ (multiple-1) = decimal
Decimal × multiple = large number
(or decimal+difference = large number)
Tree planting problem
1 The problem of planting trees on unclosed lines can be divided into the following three situations:
(1) If trees are planted at both ends of the non-closed line, then:
Number of plants = number of nodes+1 = total length-1.
Total length = plant spacing × (number of plants-1)
Plant spacing = total length ÷ (number of plants-1)
2 If you want to plant trees at one end of the unclosed line and not at the other end, then:
Number of plants = number of segments = total length ÷ plant spacing
Total length = plant spacing × number of plants
Plant spacing = total length/number of plants
(3) If no trees are planted at both ends of the non-closed line, then:
Number of plants = number of nodes-1 = total length-1.
Total length = plant spacing × (number of plants+1)
Plant spacing = total length ÷ (number of plants+1)
The quantitative relationship of planting trees on the closed line is as follows
Number of plants = number of segments = total length ÷ plant spacing
Total length = plant spacing × number of plants
Plant spacing = total length/number of plants
The question of profit and loss
(Profit+Loss) ÷ Difference between two distributions = number of shares participating in distribution.
(Big profit-small profit) ÷ Difference between two distributions = number of shares participating in distribution.
(big loss-small loss) ÷ The difference between two distributions = the number of shares participating in the distribution.
encounter a problem
Meeting distance = speed × meeting time
Meeting time = meeting distance/speed and
Speed Sum = Meeting Distance/Meeting Time
Catch up with the problem
Catch-up distance = speed difference× catch-up time
Catch-up time = catch-up distance ÷ speed difference
Speed difference = catching distance ÷ catching time
Tap water problem
Downstream velocity = still water velocity+current velocity
Countercurrent velocity = still water velocity-current velocity
Still water velocity = (downstream velocity+countercurrent velocity) ÷2
Water velocity = (downstream velocity-countercurrent velocity) ÷2
Concentration problem
Solute weight+solvent weight = solution weight.
The weight of solute/solution × 100% = concentration.
Solution weight × concentration = solute weight
Solute weight-concentration = solution weight.
Profit and discount problem
Profit = selling price-cost
Profit rate = profit/cost × 100% = (selling price/cost-1) × 100%.
Up and down amount = principal × up and down percentage
Discount = actual selling price ÷ original selling price× 1 00% (discount <1)
Interest = principal × interest rate× time
After-tax interest = principal × interest rate × time × (1-20%)
Time unit conversion
1 century = 100 1 year =65438+ February.
The big month (3 1 day) includes:1\ 3 \ 5 \ 7 \ 8 \10 \ 65438+February.
Abortion (30 days) includes: April \ June \ September \165438+1October.
February 28th in a normal year and February 29th in a leap year.
There are 365 days in a normal year and 366 days in a leap year.
1 day =24 hours 1 hour =60 minutes.
1 point = 60s 1 hour = 3600s product = bottom area × height V=Sh.
Sum-difference problem
Knowing the sum and difference of two numbers, the application problem of finding these two numbers is called the sum and difference problem. The general relationship is:
(sum and difference) ÷ 2 = smaller number
(sum+difference) ÷ 2 = larger number.
Example: The sum of two numbers, A and B, is 24, and A is smaller than B by 4. What are the numbers a and b?
(24+4)÷2
=28÷2
= 14 → B number
(24-4)÷2
=20÷2
= 10 → Numbers
A number is 10, and B number is 14.
Difference problem
Given the difference between two numbers and their multiples, the application problem of finding these two numbers is called the difference multiple problem. The basic relationship is:
Difference between two numbers ÷ multiple difference = smaller number
There are two piles of coal, and the second pile is 40 tons more than the first. If you take 5 tons of coal from the second pile to the first pile, the weight of the second pile is exactly three times that of the first pile. How many tons are there in each of the two piles of coal?
Analysis: It turns out that the second pile of coal is 40 tons more than the first pile. After giving the first pile 5 tons, the second pile of coal is only 40-5× 2 tons more than the first pile. The basic relationship is as follows:
(40-5×2)÷(3- 1)-5
=(40- 10)÷2-5
=30÷2-5
= 15-5
= 10 (ton) → weight of the first pile of coal
10+40 = 50 (ton) → the weight of the second coal pile
A: The first coal pile has 10 tons, and the second coal pile has 50 tons.
Reduction problem
The problem of finding the original unknown after knowing the result of some changes in a number is generally called reduction problem.
The reduction problem is the inverse solution of the application problem. Generally speaking, according to the relationship between reciprocal operations of addition, subtraction, multiplication and division. Start with the order of topic description, think in reverse order, start with the last known condition, and get the result in reverse.
Example: There is some rice in the warehouse, and the weight sold on the first day is 12 ton, which is less than half of the total. The weight sold the next day was less than the remaining half 12 tons, leaving 19 tons. How many tons of rice are there in this warehouse?
Analysis: If the remaining half is just sold the next day, it should be 19+ 12 tons. After being sold on the first day, the remaining tonnage is (19+ 12) × 2 tons. The following analogy.
Formula: [(19+12) × 2-12 ]× 2.
=[3 1×2- 12]×2
=[62- 12]×2
=50×2
= 100 (ton)
This warehouse used to have 65,438,000 tons of rice.
Replacement problem
There are two unknowns in the problem, and one of them is often regarded as the other for the time being, and then the hypothetical operation is carried out according to the known conditions. The results are often inconsistent with the conditions, and then appropriate adjustments are made to get the results.
Example: A stamp collector bought stamps of 10 and 20 cents *** 100, with a total value of180 cents in 8 yuan. How many stamps did this stamp collector buy?
Analysis: Assuming that every 100 stamp bought is 20 cents, the total value should be 20× 100 = 2000 (minutes), which is 2000- 1880 = 120 (minutes) more than the original total value. And this extra 120 points means that every point of 10 is regarded as 20 points, and every point is 20- 10 = 10 (points), so how many points can you get from 10 points?
Formula: (2000-1880) ÷ (20-10)
= 120÷ 10
= 12 (sheets) → 10 Number of sheets per sheet.
100- 12 = 88 (sheets) → 20 minutes for each sheet.
Or first find the number of sheets with 20 points, and then find the number of sheets with 10 points. The method is as above. Please note that the total value is less than the original total value.
Profit and loss problem (insufficient profit)
There are often two distribution schemes in the topic, and the result of each distribution scheme will be more (surplus) or less (deficit). This kind of problem is usually called profit and loss problem (also called the problem of insufficient surplus).
To solve this kind of problem, we should first compare the two distribution schemes, find out the change of remainder caused by the change of each share, find out the total number of shares participating in the distribution, and then find out the number of items to be distributed according to the meaning of the question. Its calculation method is:
When one time is left and the other time is insufficient:
Per share = (remainder+deficiency) ÷ twice the difference per share.
When there are two remainders:
Total number of copies = (larger remainder-smaller number) ÷ twice the difference per copy.
When neither time is enough:
Total number of copies = (large shortage-small shortage) ÷ twice the difference per copy.
Example 1. A certain class of a PLA unit participated in afforestation activities. If each person plants 5 seedlings, there are still 14 seedlings left; If everyone plants seven trees, there will be a shortage of four seedlings. How many people are there in this class? A * *, how many seedlings?
Analysis: according to the conditions, this question belongs to the first case.
Formula: (14+4) ÷ (7-5)
= 18÷2
= 9 (person)
5×9+ 14
=45+ 14
= 59 (tree)
Or: 7× 9-4
=63-4
= 59 (tree)
A: There are 9 people in this class, and each class has 59 seedlings.
Age problem
The main feature of the age problem is that the age difference between two people remains the same, but the multiple difference has changed.
The commonly used calculation formula is:
Age times = age difference ÷ (multiple-1)
Age a few years ago = small gift-multiplied by small age
Age in a few years = times its age-its age when it was very young now.
Example 1. Father is 54 years old and son 12 years old. A few years later, the father was four times as old as his son.
(54- 12)÷(4- 1)
=42÷3
= 14 (years old) → the age of the son in a few years.
14- 12 = 2 (year) →2 years later
Two years later, the father was four times older than his son.
Example 2: My father is 54 years old and my son is 12 years old this year. A few years ago, the father was seven times as old as his son.
(54- 12)÷(7- 1)
=42÷6
= 7 (years old) → the age of my son a few years ago
12-7 = 5 (years) →5 years ago
A: Five years ago, my father was seven times older than my son.
Ex. 3: The total age of Wang Gang's parents this year is 148 years old, and the difference between his father's age and his mother's age is more than three times that of his age and four years old. How old are Wang Gang's parents this year?
( 148×2+4)÷(3+ 1)
=300÷4
= 75 (years old) → father's age
148-75 = 73 (years old) → mother's age
A: Wang Gang's father is 75 years old and his mother is 73 years old.
Or: (148+2) ÷ 2
= 150÷2
= 75 years old
75-2 = 73 years old
Chicken and rabbit problem
Knowing the total number of chickens and rabbits and finding the number of chickens and rabbits, there is a kind of application problem called chicken and rabbit problem, which is also called "turtle crane problem" and "replacement problem"
It is generally assumed that they are all chickens (or rabbits), and then rabbits (or chickens) are used instead of chickens (or rabbits). Commonly used basic formulas are:
(Total number of feet-number of chicken feet × total number of chickens) The difference between the number of feet of each chicken and rabbit = number of rabbits.
(The number of rabbits × the total number of rabbits-the total number of rabbits) ÷ The difference between the number of feet of each chicken and rabbit = the number of chickens.
Example: There are 24 chickens and rabbits in the same cage. There are 64 legs. How many chickens and rabbits are there in the cage?
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(64-2×24)÷(4-2)
=(64-48)÷(4-2)
= 16 ÷2
= 8 (only) → Number of rabbits
24-8 = 16 (only) → number of chickens
Answer: There are 8 rabbits and 16 chickens in the cage.
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The problem of cattle eating grass (the problem of ship leaking)
Several cows are grazing on a limited meadow. Cattle eat grass and grass grows on grass. When increasing (or decreasing) the number of cattle, how long will it take before the grass on this grassland has just been eaten?
Example 1. A piece of grassland can feed 15 cows 10 days and feed 25 cows for 5 days. If the grass grows at the same speed every day, how many days can this grass feed 10 cows?
Analysis: Generally, the daily grazing amount of 1 cow is regarded as one share, so 15 cows ate 10 days, including the original grass on the grassland, and the grass grew on this grassland for 10 days, and so on ... Among them, it can be found that the grazing amount of 25 cows for 5 days is/kloc-. The reason is because, first, it takes less time; Second, the corresponding grass grows less. The difference is that grass grows on this grassland for 5 days. The grass that grows every day can feed five Niu Yi days. In this way, when feeding 10 cows, 5 cows are taken out every day to eat the long grass, and the rest cows eat the original grass on the grassland.
( 15× 10-25×5)÷( 10-5)
=( 150- 125)÷( 10-5)
=25÷5
= 5 (head) → You can feed 5 heads in Niu Yi for one day.
150- 10×5
= 150-50
= 100 (head) → The original grass on the grassland can feed 100 cows a day.
100÷( 10-5)
= 100÷5
= 20 days
Answer: If you feed 10 cows, you can eat them for 20 days.
Example 2: A well gushes water upward at a uniform speed, and four pumps can discharge water for 100 minutes; If you use six identical pumps, it can be emptied in 50 minutes. Now, with seven identical pumps, how many minutes can you pump the water out of this well?
( 100×4-50×6)÷( 100-50)
=(400-300)÷( 100-50)
= 100÷50
=2
400- 100×2
=400-200
=200
200÷(7-2)
=200÷5
= 40 minutes
With seven identical pumps, the water in this well can be drained in 40 minutes.