The overlapping part of two rectangles is equal to 1 / 6 of the area of large rectangle and 1 / 4 of the area of small rectangle

The overlapping part of two rectangles is equal to 1 / 6 of the area of large rectangle and 1 / 4 of the area of small rectangle

[solution 1] first, find out how many overlapping areas the large rectangle contains, then find out how many overlapping areas the small rectangle contains, and finally find out the ratio of the two rectangular areas
1. How many overlapping areas does the large rectangular area contain?
1÷1/6=6
2. How many overlapping areas does the small rectangle area contain?
1÷1/4=4
3. What is the area ratio of two rectangles?
6：4=3：2
[solution 2] the area of a large rectangle is regarded as "L", the area of a small rectangle is equivalent to a fraction of the area of a large rectangle, and then the ratio of the area of two rectangles is calculated
1. How much of the area of a small rectangle is that of a large rectangle?
2. What is the ratio of the area of two rectangles?
[solution 3] the area of a small rectangle is regarded as "L". First, the area of a large rectangle is equal to several times of that of a small rectangle, and then the area ratio of two rectangles is calculated
How many times the area of a large rectangle is that of a small one?
2. What is the area ratio of two rectangles?
[solution 4] the inverse ratio of the ratio of the overlapping parts to the area of the large and small rectangles is the ratio of the area of the large and small rectangles
[solution 5] we can draw several auxiliary lines as shown in the figure below. It is easy to see that the area of large rectangle is 6 times that of overlapping part, and that of small rectangle is 4 times that of overlapping part. Then we can find two rectangular surfaces
Product ratio
6：4=3：2
In the above two simple operations, solution 4 is the most direct, and solution 4 is a. in the above five solutions, the idea of solution 1 is easy to understand, and the simplification is relatively simple; solution 5 uses the method of auxiliary line, which is clear at a glance, and is also a better solution; in the other three solutions, it is easy to understand and simplify, The way of thinking is a little tortuous, and the simplification is not as simple as the above two methods