Junior high school mathematics: the area of the algebraic value for value, the use of the supplementary method, and the use of the transformation method

Junior high school mathematics: the area of ​​the algebraic value for value, the use of the supplementary method, and the use of the transformation method

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In this section of the algebra (the combination and subtraction method), there is a type of area problem, and many students feel difficult. There are two main reasons to cause this situation: (1) the first contact algebraic formula, accustomed to digital calculation area, not accustomed to referring to letters; Essence

01 Type 1: Direct use of area formula

Example 1: As shown in the figure, Xiao Ming's housing structure plan, (unit: meter), when decorating the house, he planned to spread the floor tiles outside the bedroom.

(1) If the price of floor tiles is 80 yuan/square meter, how much does it cost to buy floor tiles? (Represents generation digital);

(2) The height of the known house is 3 meters. Now if you want to stick to the wallpaper on the walls of the living room and bedroom, how many square meters of wallpaper is needed (the area of ​​doors and windows is ignored)? (Represents generation digital);

(3) If x = 4, y = 5, and the price of floor tiles per square meter is 90 yuan, and the price of per square meter wallpaper is 15 yuan. So, how much does Xiaoming spend in these two decorations? (Various small losses do not count).

Analysis: (1) Find the bathroom area = y (4x-x-2x) = xy, kitchen area = x (4y-2y) = 2xy, living room area = 2x4y = 8xy, and then find total area, you can find land tile The cost of spending; (2) The area of ​​the wall of the living room and the bedroom is the area of ​​the required wallpaper;

Solution: (1) The area of ​​the bathroom = y (4x-x-2x) = xy,

Kitchen area = x (4y-2y) = 2xy,

Living room area = 2x4y = 8xy,

积 The area of ​​the floor tile = xy+2xy+8xy = 11xy,

费 The cost of paving floor tiles is 880xy yuan;

(2) The wallpaper of the bedroom = (2Y+2Y+2X+2X) × 3 = (12x+12y) square meters,

The wallpaper of the living room = 2 (2x+4y) × 3 = (12x+24y) square meters,

12 A total of 12X+12Y+12X+24Y = (24X+36y) square meters of wallpaper are required;

(3) When x = 4, y = 5, the floor tiles need to cost: 90 × 11 × 4 × 5 = 19800 (dollars),

The wallpaper needs to cost: (24 × 4+36 × 5) × 15 = 4140 (dollars),

8 Xiaoming spent 19800+4140 = 23940 (dollars).

Direct use of the area of ​​the plane graphics to be familiar with the area formula of the common plane graphics, such as rectangular (rectangular), square, parallel quadrangle, question type, triangle, and so on. Pay attention to the solution of the circular area. In the junior high school stage, π is π, which cannot be written as 3.14, unless the decimal is required in the topic.

02 Type 2: Cutting Method

Example 2: As shown in the figure, a rectangular shape. (1) According to the size in the figure, the area s in the shadow part with the algebra containing X is used; (2) If x = 2, find the value of s.

Analysis: Because the shadow part is irregular, the cutting method can be considered to reduce the area of ​​the two triangles with the area of ​​the rectangular area to indicate the area of ​​the shadow part, and then the data is substituted for calculation.

Solution: (1) S shadow part = s rectangular-s triangular ABC-S triangle DEF

= 1/2 × 6-12 × 1/2 × 6-1/2 × 6 × (6-x) = 72-36-18+3X = 18+3X;

(2) When x = 2, s = 18+3 × 2 = 24.

When the graphics are required to be irregular graphics, the cutting method needs to be used to convert it into common geometric graphics. For example, Example 2 calculates the shadow part to the rectangle, or it can also be divided into two triangles, connecting EC, and decomposing it to △ AEC and △ CEF. Both triangles are regular triangles. You can directly select the area formula for calculation.

03 Type 3: Transformation Method

Example 3: A park is preparing to build a rectangular lawn, which grows A meter and BM is wide. The cross road shown in the lawn is built on the lawn, and the cross road is known to be 2 meters wide.

(1) Use the algebraic type of A and B to represent the area of ​​the cross road built.

(2) If A = 30, B = 20, find the area of ​​the lawn (shadow part).

Analysis: There are two ideas in this question. The first is to directly find the sum of the area of ​​the two roads, and then subtract the part that appears in the middle; the second is to use the conversion method to move all the two roads to the side. The area of ​​the large square can be reduced to the area of ​​the small square.

Solution: According to the meaning: (2A+2B-4) square meters;

(2) When A = 30, B = 20, AB- (2A+2B-4) = 600-96 = 504 (square meter), the area of ​​the lawn is 504 square meters.

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