In the known Cartesian coordinate system, the vertex coordinates of the polygon ABCDEF are A 2,0 B

There are two ways to do this: one is to divide it into small rectangles and small right triangles, and the other is to complete it into a large rectangle and subtract the added parts!

Solution: 5 + 3

Let (2,0) be g (5,0) be h (6,0) be i (6,5) be js abg=1 x 3 x 1 2=3 2s bchg=(3+6) x 3 x 1 2=27 2s ch i j=(5+6) x 1 x 1 2=11 2s jde=3 x 1 x 1 2=3 2s eif=3 x 1 x 1 2=3 2 2 s total = (3 + 27 + 11 + 3 + 3) 2 = 47 2 I have limited ability....I guess I'm not mistaken...There are other methods, but this one is the most conventional, you can try to translate a part of the image to see if you can get a more regular shape, and then do the math

Hee-hee, are you a junior high school student? Be sure to recover!

My consistent method is to check directly, or complete, which is easy to use and convenient, but it is too troublesome to calculate.

This kind of professional knowledge should go to professional**.

After passing the point C as CE AB and intersecting AD at the point E, the ordinate of the point E can be 5, the linear equation where AD is located is Y=KX, and the coordinates of the point D are substituted into the system of equations 2K=7, and the solution is K=7 2

So the equation for the straight line where ad is located is y=7 2x

Let y=5, solve x=10 7, then the point E coordinates (10 7,5)CE divide the quadrilateral ABCD into trapezoidal ABCD and CDE, where the upper and lower edges of trapezoidal ABCE=7-10 7=39 7, the lower bottom edges, AB=9, and the height is 5. The cde side CE=39 7, and the height on the CE side is 7-5=2

So sabcd=sabce+s cde=1 2 5 (39 7+9)+1 2 2 39 7

The area of this quadrilateral.

s△cde+s△cae+s△abc

ce|*|2|/2+|ce|*|4|/2+|bc|*|3-2|/2

3*|ce|+

Mark the ABCD points in the coordinate system, it can be seen that the CD is on the X axis, and the area is required to be over the two points of AB to do the vertical intersection of the X axis, and the X axis is EF two points.

1) Finding the area of the quadrilateral is transformed into finding the sum of the area of the triangle BCE and the triangle ADF and the trapezoidal ABEF.

The detailed process is as follows: AF=8, BE=6, CD=14, FD=2, ED=11 can be seen from ABCD coordinates

So ef=ed-fd=11 2=9

ce=cd=ed=14－11=3

s△afd＝2*8／2＝8

s△bce=3﹡6/2=9

S trapezoidal (6 8) 9 2 63

s quadrilateral 8 9 63 80

2) The ordinate remains the same, and the abscissa increases, which is equivalent to translating the figure without changing the shape of the figure, so the area remains the same.

Set up three datum points E (2,0), F(2,5), G(7,0) and so the quadrilateral ABCD is divided into a combination of triangular ADE, triangular CDF, triangular BCG and square CFEG.

The area is calculated separately as follows:

Triangle ADE : 2*7 2=7

Triangular CDF : 2*5 2=5

Triangular BCG: 2*5 2=5

Square CFEG: 5*5=25

Sum: 7 + 5 + 5 + 25 = 42 unit area.

Solution: After d, c do de, cf is perpendicular to ab, then there is:

s=s△oed+sefcd+s△cfb

Therefore, the area of the quadrilateral ABCD is 42 square units

The area of the quadrilateral ABCD should be 42, and it is possible to cut the figure into two right-angled triangles and a right-angled trapezoid, and add the calculated areas to get 42

If D is made by DE perpendicular to the X-axis to E, and by C as CF perpendicular to the X-axis by F, the area of the quadrilateral ABCD can be found to be 42

In the planar Cartesian coordinate system, the vertex coordinates of the quadrilateral ABCD are A(1,0), B(5,0), C(3,3), D(2,4), respectively.

Solution: Make the CE x-axis at the point E, and the DF x-axis at the point F then the area of the quadrilateral ABCD = S ADF + S BCE + S trapezoidal GDFE = 12 (2-1) 4+ 12 (5-3) 3+ 12 (3+4) (3-2)=

The P point is obviously on a straight line parallel to BC, and there are two such lines, because BC is neither parallel to the X axis nor parallel to the Y axis, then the two lines required will have an intersection point with the X and Y axes, and there are 4 such intersection points.

According to the area can be obtained to obtain the perpendicular distance between the straight line and BC, and find the equations of these two straight lines, so that x=0 and y=0 respectively, and it is natural to find the four coordinates of p. It's better to push the detailed process yourself.

There are countless such points.

bc = root number under = root number 29, the area is 50, then the height on the side of bc is: 100 root number 29.

The point P is parallel to the straight line BC and has a distance of 100 and number 29.

Set it up to find it, and turn the polyline into a straight line.