Please teach a primary school geometry Olympiad problem, ask for the process, thank you.

Let ab=x and ad=y

xy=48 triangle ade area = 2*y 2=y

Similarly, triangle abf area = x

The area of the triangle CFE = (x-2)*(y-2) 2=xy 2-x-y+2, so the sum of the areas of the three triangles in the blank part is y+x+xy 2-x-y+2=xy 2+2

Because xy=48

So the area of the blank space is 24 + 2 = 26

So the shaded area is 48-26=22

Let the length of the rectangle be x and the width be y, xy=48

s=48-x-y-(x-2)(y-2)/2

48-x-y-xy/2+x+y-2

Based on the numbers on the graph, the calculation is as follows.

Rectangle area - triangle ADE - triangle ABF - triangle ECF = shaded part area.

ad*ab—1/2ad*de—1/2ab*bf—1/2（bc—bf)*(dc—de)

48—1/2ad*2—1/2ab*2—1/2（ad—2）*（ab—2）

48—ad—ab—1 2(ad*ab—2ad—2ab+4)48—ad—ab—1 2ad*ab+ad+ab—2, and the shaded area is 22

Block + circle = 3

Circle-Block=1 Gets Block=1, Circle=2 Block+Triangle=6

Triangle - Block = 4 Get Block = 1, Triangle = 5 Triangle + Circle = 5 + 2 = 7

2001/2002/2001 + 1/2003/12003 and 1/2003

1*4 and 21/97 + 9/19

3. Set up X hectares of land in A.

x/3+2(

4. There are 2 classes of x people, 1 class of 4x people who do not participate, and the number of people in each class is m.

m-4x=1/3(m-x)

m=11x/2

The realization that class 1 did not participate is a fraction of the number of people who did not participate in class 2:

4x/(m-x)=4x/(11x/2-x)=8/9

Solution: s afc=fc*12 2=6*fc=s aeh+quad ehfg+s fgc;

s bfd = bf * 12 2 = 6 * bf = s bef + quadrilateral ehfg + s hgd

Add the two formulas: left = 6 * (fc + bf) = 6 * 12 = 72 right = s aeh + quadrilateral ehfg + s fgc + s bef + quadrilateral ehfg + s hgd = 4 with color + 2 times quadrilateral = 4 with color + 2 * 6, so 4 with color = 72-12 = 60

Because SδDFB = SδABF, the blank part can be understood as overlapping, so it should be subtracted twice, so the colored area is 2 times the S blank part of SδABC = 12 12 2 6 2 = 60cm, supplement: 60 is 60, it will not be wrong, the blank part should be subtracted twice.

Because, (dbf + afc) -2 efgh = colored part area.

And because, area abf=dbf

So, (dbf + afc)= abc = abcd2 i.e., the area is (12 12) 2 - 2 6 = 60

Just a hint: subtract the area of the adjacent area.