According to the relationship between area and area, the mathematical model of quadratic equation with one variable is established, which solves this kind of problem.
Teaching objectives
Master the area method to establish the mathematical model of quadratic equation in one variable and use it to solve practical problems.
Introduce a new lesson by asking questions, solve the problems in the new lesson, and review the area formulas of several special figures.
Difficulties and the key to difficulties
1. Emphasis: According to the equivalent relationship between area and area, the mathematical model of one-dimensional binary equation is established and used to solve practical problems.
2. Difficulties and emphases: According to the equivalence relation between areas, the mathematical model of quadratic equation with one variable is established.
Prepare teaching AIDS and learning tools
Small blackboard
teaching process
First, review the introduction.
(1) What knowledge and methods did you learn last class?
(2) In the last section, we learned to solve the problem of "average growth (decline) rate". Now we should learn to solve the problem of "area and volume".
1. What is the area formula of a right triangle? What is the area formula of a general triangle?
2. What is the formula of square area? What is the area formula of a rectangle?
3. What is the area formula of trapezoid?
4. What is the formula of diamond area?
5. What is the area formula of parallelogram?
6. What is the area formula of a circle?
(Students answer orally, teachers comment)
Second, explore new knowledge.
Now, according to the area formula we reviewed just now, we set up some mathematical models to solve some practical problems.
Example 1. A forest farm plans to build a channel with a length of 750 meters and an isosceles trapezoid cross-section, with a cross-sectional area of 1.6 square meters, a width of the upper opening of which is 2m greater than the depth of the channel, and a bottom of which is 0.4m greater than the depth of the channel.
(1) What is the width of the top and bottom of the channel?
(2) If it is planned to dig 48m3 every day, how many days will it take to finish this passage?
Analysis: Because the canal depth is the smallest, for the convenience of calculation, we assume that the canal depth is xm, the width of the upper mouth is X+2, and the bottom of the canal is x+0.4. Then, the model can be established according to the trapezoidal area formula.
Solution: (1) Let the canal depth be xm.
The bottom of the canal is (x+0.4) meters, and the width of the upper mouth is (x+2) meters.
According to the meaning of the question, (x+2+x+0.4)x= 1.6.
Finishing, get: 5x2+6x-8=0.
Solution: x 1= =0.8m, x2=-2 (s).
∴ The width of the upper mouth is 2.8m, and the bottom of the canal is1.2m..
(2) =25 days
Answer: The width of the upper mouth of the channel is 2.8m, and the depth of the bottom of the channel is1.2m; It takes 25 days to dig the tunnel.
Student activities: Example 2. As shown in the figure, to design the cover of a book, the cover is 27cm long and 2 1 cm wide, and the center is a rectangle with the same aspect ratio as the whole cover. How to design the width of the surrounding side linings (accurate to 0.65433) if the area occupied by the surrounding colored side linings is a quarter of the coverage area, the upper and lower side linings are equally wide, and the left and right side linings are equally wide?
Thinking: (1) What are the quantitative relations in ontology?
(2) The center is rectangular, and the aspect ratio is the same as the whole cover.
(3) How to use the known quantitative relationship to select the unknown and list the equations?
Teacher's comment: According to the meaning of the question, the length-width ratio of the central rectangle is equal to the length-width ratio of the cover = 9:7. From this, it can be judged that the ratio of the width of the upper and lower liners to the width of the left and right liners is 9: 7. If the width of the upper and lower linings is 9X cm, the width of the left and right linings is 7xcm. According to the meaning, the length of the central rectangle is (27- 10.
Because the area around the colored side lining is the coverage area, the area in the middle rectangle is the coverage area.
So (27-18x) (21-14x) = × 27× 21.
Finishing: 16x2-48x+9=0.
Solve the equation and get: x=,
x 1≈2.8cm，x2≈0.2
So: 9x 1=25.2cm (excluding), 9x2 = 1.8cm, 7x2 =1.4cm.
Therefore, the width of the upper and lower linings is 1.8cm, and the width of the left and right linings is1.4cm..
Analysis: The length-width ratio of this book is 9:7. According to the title, the ratio of the two sides of the middle rectangle is also 9:7.
Solution 2: Let the two sides of the middle rectangle be 9xcm and 7xcm respectively. According to the meaning of the question
Solve the equation and get:
Therefore, the widths of the upper and lower linings are:
The widths of the left and right linings are:
Thinking: What are the characteristics of comparing several methods?
Fourth, application expansion.
In order to beautify the campus, a school plans to build several roads on a rectangular field with a length of 32 meters and a width of 20 meters, and the rest will be used as lawns, and all students are invited to participate in the design. Now, two students have each designed a plan (as shown in the picture). According to the two design schemes, the equations are listed to find out the width of the road in the figure. Let the lawn area in Figures (1) and (2) be 540m2.
The exercise is shown in the picture. On the rectangular ground with a width of 20m and a length of 32m, two roads with the same width and perpendicular to the other road will be built, and the remaining six same parts will be used as cultivated land. What is the road width to make the cultivated land area reach 500m2?
Solution 1: Let the width of the road be x, and we use the principle of "the area of the figure will not change after moving" to move the vertical and horizontal roads to make the equation simpler (the purpose is to find out the width of the road surface, and as for the actual construction, the road can still be built according to the original position), and then we can make the equation: (20-x)(32-2x)=500.
Scheme 2: 20×32-2×20x-32x+2x2=500.
Example 4. As shown in figures (a) and (b), in △ABC, ∠ b = 90, AB = 6 cm, BC = 8 cm, point P moves from point A to point B at a speed of 1cm/s, and point Q moves from point B to point C at a speed of 2 cm/s. 。
(1) If P and Q start from A and B at the same time, after a few seconds, S△PBQ=8cm2.
(2) If P and Q start from A and B at the same time, P will continue to advance on BC side after arriving at B, and Q will continue to advance on CA side after arriving at C.. After a few seconds, the area of △PCQ is equal to 12.6cm2 (friendly reminder: Q is DQ⊥CB, and the vertical foot is D, then:)
Analysis: (1) Suppose that after x seconds, S△PBQ=8cm2, then AP=x, PB=6-x, QB=2x, and the mathematical model of a quadratic equation with one variable can be obtained from the area formula.
(2) let y seconds pass, where y >；; Let the area of △PCQ be equal to 12.6cm2, because AB=6 and BC=8, which is obtained by Pythagorean theorem: AC= 10, and because PA=y, CP=( 14-y) and CQ=(2y-8), which is also true.
Solution: (1) Let x seconds, point P be on AB and point Q be on BC, and let the area of △PBQ be 8cm2.
Then: (6-x) 2x = 8
Solution: x 1=2, x2=4.
After 2 seconds, the distance from point P to point A 1×2=2cm, from point Q to point B 2×2=4cm, from point P to point A 1×4=4cm and from point Q to point B 2×4=8cm all meet the requirements.
(2) Let point P move to BC after y seconds, with CP=( 14-y)cm, point Q move to CA, CQ=(2y-8)cm, intersection point Q is DQ⊥CB, and vertical foot is D, then there is
AB = 6，BC=8
According to Pythagorean theorem, AC= = 10.
∴DQ=
Then: (14-y) = 12.6.
Finishing, we get: y2- 18y+77=0.
Solution: y 1=7, y2= 1 1.
That is, after 7 seconds, point P is apart from point C7 cm on BC (CP =14-y = 7) and point Q is apart from point C6 cm on CA (CQ = 2y-8 = 6), so that the area of △PCD is12.6cm2.
1 1 sec later, point p on BC is 3cm away from point c, and point q on CA > is 3 cm away from point c 14cm: 10,
∴ Point Q is beyond the scope of CA, that is, this solution does not exist. There is only one solution to this small problem, y 1=7.
Verb (abbreviation of verb) abstract
This lesson should master: using the area formula of special graphics we have learned, establish the mathematical model of quadratic equation with one variable and use it to solve practical problems.
Distribution of intransitive verbs
1. The textbook P53 comprehensively uses 5 and 6 to explore all.
2. Select the job design: 1. Multiple choice problem
1. If the sum of two right angles of a right triangle is 7 and the area is 6, the hypotenuse is ().
A.B.5 C. D.7
2. There are two boards, the length of the first board is twice as wide as that of the first board, and the length of the second board is 2m less than that of the first board, and the width is three times as wide as that of the first board. It is known that the area of the second board is larger than that of the first board 108m2, and the length and width of these two boards are () respectively.
A. The first board is 18m long and 9m wide, and the second board is 16m long and 27m wide.
B the first board is 12m in length and 6m in width, and the second board is 10m in length and18m in width;
C The first board is 9m long and 4.5m wide, and the second board is 7m long and13.5m wide;
D. none of the above is true
3. Cut a rectangle with a width of 2cm from the square iron sheet, and the remaining area is 48cm2, so the area of the original square iron sheet is ().
A.8cm B.64cm C.8cm2 D.64cm2
Second, fill in the blanks
1. If the perimeter of the rectangle is 8 and the area is 1, then the length and width of the rectangle are _ _ _ _ _ _ _.
2. If the length of a rectangle is 4 cm more than the width and the area is 60 cm 2, then its circumference is _ _ _ _ _ _.
3. As shown in the figure, it is a schematic plan view of a rectangular chicken farm, with one side against the wall and the other three sides surrounded by bamboo fences. If the total length of the bamboo fence is 35m and the enclosed area is 150m2, the length and width of the rectangular chicken farm are _ _ _ _ _ _ _ _ _.
Figure 22- 10
Third, comprehensively improve the problem.
1. As shown in the figure, the section of the waterproof dam (trapezoidal) has a crest width of 3m, backwater gradient 1:2 and inlet gradient 1: 1. If the dam is 30m long and the earthwork used to complete the dam is 4500m2, what should be the height of the dam? (Description: reverse slope =, incoming slope) (accurate to 0. 1m)
2. In the center of a rectangular plane with a length of 12m and a width of 8m, leave a place to build an area of 8 m2 \ If the width of the flower bed is the same, what is the width?
3. Who can say the width of the way out?
As shown in Figure 22- 10, there is a rectangle of ABCD, and a rectangular flower auxiliary EFGH should be built in the middle, so that its area is half of this land, and the width of roads around the flower garden is equal. Today, there are no measuring tools, only a rope long enough without scale. How to measure the width of the road?
Please use your own mathematical knowledge to solve this practical problem. I believe you can do it.
Please adopt it.