A simple math problem, thanks so much, I m about to fall asleep

Let ab=x, bc=y, and the height of the trapezoid is h, x -(y-x) 4=h 3x+y=40

h/[(y-x)/2]=tg45°=1

Solve the above three equations to get:

x=,y=,h=

The area of the trapezoid at this time: (

Therefore, the area of this trapezoid is smaller than the area of the square, and the difference is by square meters.

Solution: After a and d, they are respectively AE BC, the vertical foot is E, DF BC, the vertical foot is F, ab=ad=dc, b=45, so that ab=ad=dc=x, then ae= 2 2, ef=xab+bc+cd+ad=40, 4x+ 2x=40, the solution is x

The area of the trapezoid is s = (2 2) x + (1 2) x 66

The area of this trapezoid is smaller than that of a square. The difference is 34m

Since the angle b=45 degrees, and ab=ad=dc, let the trapezoidal height be x, we can see dc + 3 2x = 40, dc = 3 2x where.

2 = , i.e. 6 2x = 40 calculates x, and then calculates the area s=2 2x*x.

The classmate is 13+x years old, and the teacher is 45+x years old.

then 13+x=(45+x) 3

Solution x=3

13+x=1/3(45+x)

13+x is the age of the student after x years.

Equal to one-third (45+x) is the age of the teacher.

Isosceles right triangle.

This is because the DAE can coincide with the DCF after rotation.

So the angle ade = angle cdf, de=df

Because the quadrilateral ABCD is a square.

So the angle adc = 90

So the angle edf=90

So def is an isosceles right triangle.

Proof triangle congruence:

The quadrilateral ABCD is a square.

CF is an extension of BC.

Angular DAE = Angular DCF

ad=dccf=ae

ade≌△cdf

DE = DF Angle DAE = Angle DCF

Angle ADE + Angle EDC = 90 degrees = Angle EDF

So def is isosceles right triangle.

The length of the rope is xm, then:

1/2 x -9=1/3 x -2

Solution: x=42

So, from the meaning of the title:

42 2-9 = well depth.

Then: 42 2-9 = 12m

The rope is 42m long and the well is 12m deep.

Solution: x>-1 x>=3 x<5 where x>=3 is rounded to -1

Let the weight of A be ma, then the weight of B is na

Let the heat emitted by A be nk, then the heat emitted by B is mk, then the cooling of A is nk ( ma) and the cooling of B is mk ( na) so that the cooling ratio of A and B is:

nk/(ρma):mk/(ρna)=n/m:m/n=n^2:m^2

An object made of the same material cools down equally after emitting the same amount of heat with the same weight, and the cooling is proportional to the heat released and inversely proportional to the weight.

The ratio of A and B cooling = the ratio of A and B cooling The ratio of A and B weight = n: m m: n = n : m

Liu Chang 123456liu, hello:

Cylinder diameter: 40 2 5 4 (Li Zhi Bi cm).

Cylindrical radius: 4 2 2 (centicoma meters).

Cylindrical volume: cubic centimeters).

1. B1D ABC, then B1D AC; And BC AC. Get the AC plane BCC1B1.

two, ab1 bc1, and ad b1d; then we get [adb1 plane bcc1b1].

AC BC, and A1C1 B1C1; then [plane acc1a1 plane bcc1b1] is obtained.

So, there is adb1 planar acc1a1 and therefore, a1c adb1.

Note: There is no need for some tentative guides made in the diagram, please delete them as appropriate!

(1) The projection of point B1 on the bottom surface D falls on BC, so B1D plane ABC, so B1D AC, by the known AC BC, so AC plane BB1CC1

2) Even ab1, cb1

From (1) it is known that AC plane bb1cc1, so ac bc1, and ab1 bc1, so bc1 plane acb1

So BC1 CB1, because the diagonal of the parallelogram B1BCC1 BC1 and CB1 are bisected by each other, so BC1 is the perpendicular line of CB1, so BB1=BC, since the angle B1BC=60°, so the triangle BCB1 is an equilateral triangle. Since B1D BC, D is the midpoint of BC.

Let the intersection of the two diagonals of the parallelogram abb1a1 be e, and if you connect de, you will have de a1c

Since E is the midpoint of ab1, DE is a straight line on the plane ADB1, so A1C is the plane ab1d

1) Because the projection of B1 on the bottom surface is D, B1D plane ABC can obtain B1D AC

Because c = 90°, ACBC

Because bc and b1d are on the same plane bb1cc1 and cross, ac plane bb1cc1

Proof that: (1) b1d bc,d is the projection of b1 on the bottom bb1cc1 a1b1c1

c=90°, i.e., ac bc, a1c1 b1c1 ac plane bb1cc1(2).