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d, e, and f are the midpoints, and the following conclusions can be drawn:

The area of the triangle ABC is equal to 8 times the area of the triangle FDE and also equal to the area of the triangle ADE by 4 times (the reason is that the middle line of the triangle divides into two triangles with equal base and height and equal area).

According to the graph and the known conditions, it can also be concluded that SFDE=SGD+SEFG, SADE=SGD+SEADG, so that the area of the triangle ADE minus the area of the triangle EDF is also equal to 4

With the above two conditions, it can be obtained that one-fourth of the triangle ABC area minus one-eighth of the triangle ABC area is equal to 4, so the triangle ABC area can be calculated as 32 square centimeters.

The triangular ADG FGE is similar.

and FE is the median of the triangular ADC.

fe=1/2ad

The triangular ADG FGE has an area ratio of 4:1

The triangular ADG is 4 square centimeters more than the FGE.

Then ADG is 16 3 fge and 4 3 square centimeters.

The area of ADG+FGE is trapezoidal area, and the area of ADFE is 1 2, that is, 40 3 square centimeters.

According to the nature of the median line, the area ratio of triangular FEC to trapezoidal ADFE is 1 3, that is, 40 9

Then the triangular FEC plus the trapezoidal ADFE 160 9 is the triangular ADC area.

So the ABC area is 320 9 cm2

Easy to verify: EFG ADG, and their similarity ratio is 1:2 and their area ratio is 1:4

Let the area of the EFG be S

Then the area of ADG is 4s

4s-s=4

s=4 3 area of adg = 4 3*4 = 16 3eg ag = fe ad=1 2

Area of deg = 8 3

Area of ADE = 16 3 + 8 3 = 8

Area of the ADC = 16

Area of ABC = 32

32 square centimeters.

ADE and EFD are the same as low and equal height, and the ratio of height is 2:1 Let the DEG area be X, the ADE area is N, and the FDE area is M

then n+x-(m+x)=n-m=4

m+x=4

There is also a contour of the same height, so the area is 4*2*2*2=32

Let the abc area be S

The ADG area is A

The AFG area is B

The DGE area is C

The EFG area is D

The EFC area is E

a+c=b+d+e=s/4

a+b=c+b+e=s/4

c+d=e=s/8

a-d=4a-d=e=s/8

s=4*8=32

The ABC area is 32 cm2

Because the ratio of selling in the afternoon and in the morning is 3 2

Since the morning sold 1 3

Then what is sold in the afternoon is.

1/3)x(3/2)=1/2

Bought 1 2 in the afternoon

A total of 1 3 + 1 2 = 5 6 was bought

1 6 is 9 left.

That's 6x9=54 in total.

1.The length of the cuboid is 6cm, the ratio of length to width is 2:1, and the height is two-thirds of the width, find the surface area and volume of this cuboid.

The width is 3cm and the height is 2cm

The surface area of the cuboid is 2 (6*3+3*2+6*2)=72cm, and the volume of the cuboid is 6*2*3=36cm3

2.Xiao Ming's mother brought 150 yuan, went to the mall to buy things, shared 130 yuan, and when she passed the fruit stall, she bought 2 kilograms of bananas with 3 yuan per kilogram, as well as kilograms of cherries, and there was still a balance after paying; If you bought a kilogram of bananas and 2 kilograms of cherries at that time, the money you brought with you would not be enough. Liquidate out the ** range of cherries.

Set the **x2x+ of the cherry

Solution: 1) The length of the cuboid is 6cm, and the ratio of length to width is 2:1, so the width is 3 cm, and the height is two-thirds of the width, so the height is 2 cm, so the surface area of this cuboid is 2*(6*3+3*2+6*2)=72 square centimeters Volume: 6*3*2=36 cubic centimeters.

2) Set the ** of cherries to be x yuan kg, 3 * 2 + < 150-130 <

31/4

Find the degree, push it slowly, four equations and four unknowns, all at once. Guess.