One math problem with eight strokes, one math problem with eight strokes

The bottom edge is the number of Geng (20 square + 20 square) = 20 times the number of Geng 2, and the note in the middle is 10 times the number of Geng 2

The lower one is (10 times the number of 2 + 20 times the number of the number 2) divided by two = 15 times the number of the number 2 of the number of the number of 2 and the number of the top is 10 times the number of the number 2 divided by 2 = 5 times the number of the number of the number 2

The whole adds up to 10 times the number of 2 + 15 times the number of 2 + 5 times the number of 2 = 30 times the number of 2

Because it's an isosceles right angle, and it's a quarter of the high cd on the hypotenuse, so each small segment = 5, think about 45, you should know how to solve it, I don't know, ask me again, I'll definitely reply.

It should be 60 times the root number 2, and the process is as follows:

Because ac=ab=20cm

So cd = 10 times the root number 2

Because of the high cd on the hypotenuse.

So the width of all rectangles is equal to 1 4cd = times the root number 2, and the length of the small rectangle is equal to 2 * width = 5 times the root number 2

The length of the medium rectangle is equal to 4 * width = 10 times the root number 2

The length of the large rectangle is equal to 6 * width = 15 times the root number 2

So the circumference of the three rectangles = 5 times the root number 2 + 6 * times the root number 2 + 10 times the root number 2 + 2 * 15 times the root number 2

60 times the root number 2

This is to find the circumference, if you want to find the length...）

Long sum = 5 times the root number 2 + 10 times the root number 2 + 15 times the root number 220 times the root number 2

Landlord, please make the problem clear next time, whether the length refers to the circumference or the sum of the length.

Use whichever of the two answers you want.

Solution: Let collapse bf=x eb=y

Rectangular ABCD area = AB S1 = 1 2A(B-X) S2 = 1 2B(A-Y) S3 = 1 2XY

s1=s2=1 volt disturbance 2 (s3+s4).

s1+s2=s3+s4=1/2ab

s4=1/2ab-s3

s1=s2=1/2（s3+s4)

s4=s1+s2-s3

1/2ab-1/2ax+1/2ab-1/2by-s3ab-(1/2ax+1/2by)-s3

s4=1/2ab-s3

ab-(1/2ax+1/2by)-s3=1/2ab-s3ax+by=ab

Regimental Hall Rock s1=s2

1 2ab-1 2ax=1 2ab-1 2byax=by to obtain a system of binary linear equations.

ax+by=ab

ax=by solution x=1 2b ,y=1 2as1=s2=1 4ab

s3=1/8ab

s4=3/8ab

Solution: Let eg=x, fh=y

According to the median segment chain theorem of the trapezoid.

4+x=2y

y+10=2x

Solve the hole and burn the surplus.

x=8y=6

So eg=6, fh=8

Because dc ab, the angle aed = angle edc, because e is the midpoint and dc = 1 2ab, so ae = dc, ed = ed, so the triangle aed is all equal to the triangle edc, and the same reason proves that the triangle bec is all equal to the triangle edc, so the triangle aed is all equal to the triangle ebc

36-4 root number 6, this is the question on the eight.