How to divide a right angle trapezoid into four congruent figures

How to divide a right angle trapezoid into four congruent figures

First divide it into 12 triangles, and then see the figure below

How to divide an equilateral triangle into four isosceles triangles? Three congruent isosceles trapezoids? Four figures with equal area, including three congruent trapezoids?

To divide into four isosceles is to take any two points on either side, but the distance from these two points to any vertex is equal. Then connecting these two points must be parallel to the next edge. Then take the key points of the parallel edges and connect the two points just now to form four isosceles triangle. If the four areas are equal, take any side and divide them into four segments, equal ah, four segments, and connect the vertices, the four areas will be equal, They are all triangles. Because they have the same slant height, equal base and equal height, and the same area. Elder sister, you are too greedy to ask so many questions at one breath. I can't seem to remember the other two puzzles. However, I'll try it at school. It's fun

How to divide an isosceles trapezoid into four figures with the same area? As the title

The upper side (short side) and lower side (long side) are equally divided into four equal parts, and the corresponding points are connected. Because the trapezoid height is equal and the length of upper and lower sides is equal, the area is equal
Like this:
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As shown in the figure, the side length of each small square in the square grid is 1. The vertex of each small grid is called a grid point. With the grid point as the vertex, draw triangles (with shadow) according to the following requirements (1) In Figure 1, draw a triangle so that its three sides are rational numbers; (2) In Figure 2 and figure 3, draw a right triangle so that its three sides are irrational

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As shown in the figure, △ ABC in the square grid, if the side length of the small grid is 1, then △ ABC is:______ Triangle

∵AC2=22+32=13，AB2=62+42=52，BC2=82+12=65，
 ac2 + AB2 = BC2,  ABC is a right triangle

As shown in the figure, △ ABC in the square grid, if the side length of the small grid is 1, then △ ABC is:______ Triangle

∵AC2=22+32=13，AB2=62+42=52，BC2=82+12=65，
 ac2 + AB2 = BC2,  ABC is a right triangle

As shown in the figure, △ ABC in the square grid, if the side length of the small grid is 1, then △ ABC is:______ Triangle

∵AC2=22+32=13，AB2=62+42=52，BC2=82+12=65，
 ac2 + AB2 = BC2,  ABC is a right triangle

A frog jumps on the grid points (the vertex of the small square) of the square (the side length of each small square is 1) as shown in Figure 8 × 8, and the maximum distance the frog jumps each time is 5. If the frog jumps from point a six times in a row, and just jumps back to point a, the maximum area of the enclosed figure is______ .

As shown in the figure, the frog jumps from point a six times in a row, and just jumps back to point a. the graph formed by the line segments it skips is a hexagon, and the side length is
The area of hexagon is 12
So the answer is: 12

As shown in the figure, the square is divided into 16 small squares, each of which is a square with a side length of 1. You can make a line segment of length 5 from the grid point in the figure

The problem is very simple, as shown in the figure below
(according to the Pythagorean theorem, a hook is 3, a strand is 4, and a chord must be 5)

The square pattern of 4 times 4 has? Squares, and that of 5 times 5 has? Squares Work out the number

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