One-and-two comparison method
For some problems of finding the area, the area of Figure B is often needed when the area of Figure A is known. At this time, it can be solved by finding the relationship between Figure A and Figure B ... This relationship is the magnification (several times) or fraction (several fractions) relationship between the two graphic areas. This idea is often solved by adding appropriate auxiliary lines to form a triangle with equal base and equal height (or other figures with multiple areas).
For example, 1, the area of triangle ABC is 100 square centimeter, and d, e and f are four, five and six equal points of three sides respectively. Find the area of a triangle.
( 1)
Analysis and solution: According to the known conditions in the question, it can be inferred that there is a double relationship between the required area and the known area, because "if two or three angles are the same height, the ratio of their areas is equal to the ratio of the corresponding base length". So let's "create" such a triangle to help solve the problem. Connecting BD, since AF = 5/6ab, the area of triangle AFD accounts for 5/6 of triangle ABD, and the area of triangle ABD is exactly 1/4 of triangle ABC (because AD = 1/4ac), so the ratio of triangle AFD to triangle ABC is 1/4× 5.
Similarly, the ratio of triangle FBE to triangle ECD is 4/5× 1/6 = 2/ 15 and 3/4× 1/5 = 3/20, respectively. Therefore, the fraction of the triangle DEF area is1-5/24-2/15-3/20 = 61120, and its area is100× 61.
Letter substitution method
It is inconvenient to answer some questions directly with arithmetic, so letters can be used instead. These letters can be required quantities or intermediate quantities, and sometimes they just play the role of media. In the process of solving, they participate in the operation as a whole or a number, and cancel each other or be replaced in the calculation. Sometimes it is necessary to find out the numerical values they represent through simple algebraic operations such as comparison and substitution, and then seek the answers to the questions.
Example 2. Enclose a parallelogram frame with a 75-decimeter-long iron wire, and its two heights are required to be 14 decimeter and 16 decimeter respectively. What is the area of this parallelogram?
(2) Analysis and solution: two high lengths are informed in the condition. Because in the same parallelogram, because the area is unchanged, it can be seen from "the area of parallelogram = the height on the bottom × the height on the bottom" that the height is inversely proportional to the corresponding bottom, so we can use the method of letter equivalent substitution to solve the problem. Let two heights correspond to a decimeter and b decimeter, and the area is s square decimeter, so that a=S/14 = B× 16 = S, a=S/14, and B = S/ 16, "a. Therefore, the area of the parallelogram is:
Second, extreme disposal methods
Generally speaking, everything follows certain laws and has its particularity, and its particularity often reflects its universal law. When we answer some questions, we can use a changing point of view to imagine the graphics under special circumstances, so that we can often find a solution to the problem.