Mathematical geometry problem, find AB GE CD FG BE EF FG triangle AEG area is 7 find AEFD area

It should be be=ef=fc, not fg

Connect BG, CG

Because ab eg, the distance from points a, b to eg is equal, i.e. beg and aeg are equal in height and have the same base (eg), so the area of beg and aeg is equal and equal to 7.

And because the bases of beg and gef are equal (be=ef) and have the same height (the distance from the point g to bf), the area of gef and beg is equal, equal to 7.

Because of GF DC, the distance from point D, C to GF is equal, i.e. the high of GFD and GFC is equal and has the same base (GF), so the area of GFD and GFC is equal.

And because the bases of GFC and GEF are equal (EF=FC) and have the same height (the distance from the point G to EC), the area of GFC and GEF is equal and equal to 7.

So the areas of AEG, GEF, and GFD are equal, all of which are 7.

So the area of the quadrilateral AEFD is the sum of the above 3 triangles:

s(aefd) = 3×7 = 21

Connecting BG, the area of the triangle BGE is equal to the area of the triangle AEG (equal height at the base) is equal to 7, and the area of the triangle GEF is equal to the area of the triangle BGE (equal height at the base) is also equal to 7, and the area of the triangle FGD is also equal to 7, so the total area is 21.

If it is be=ef=fc, then the area is 21, otherwise there is no solution.

Draw a line parallel to be through G, and the area of AB and P and CD extension line at GDQ is easily equal. So the area is equal to 7*5=35

The area of the square CFF is 169, Fc=13, and in RT ACF, such as Af=12, CF=13, AC=CFAF

5, slag in right-angled triangle ABC, ac=5, ab=4, bc=acab

The length ratio of BF and FC is 10:8, so the area of ABF is 60, E is the midpoint of AF, ABF and DEF are respectively high at point A and point E, the height ratio is 2:1, and the ratio of BF and DF at the bottom edge is 10:

6, so the area ratio of the triangle def to abf is 10:3, so the area of def is 18, and the area of the quadrilateral is 60-18=42

The triangle ABF area is 60, and the triangle EDF area is 18, thus having 42

Find her height according to the triangle AFC.

Hold the game and connect the AD to divide the quadrilateral ACDF into two triangles, triangle ADF + triangle ACD = the area of the quadrilateral segment burning ACDF.

So the area of the quadrilateral ACDF = 2 8 2 + 5 4 2 = 18 (square centimeters of virtual meters).

Proof: The quadrilateral ABCD is a diamond shape, and the rent is a= c, b= d, ab=bc=cd=da

ae=ah=cf=cg,be=bf=dh=dg, aeh cgf, bef dgh,eh=fg,ef=gh,quadrilateral efgh is a parallelogram, a+ d=180°, ahe+ dhg=90°, ehg=90°, quadrilateral efgh is rectangular