Grade 7 math and geometry application problems

Solution: (1)**2 Conclusion: bbc=

7. The reasons are as follows:

BH and CH are the angular bisectors of ABC and ACD, respectively, 1=PaBC, 2=

acd, and acd is an outer angle of arc, acd= a+ abc, 2=x

a+∠abc）=u

a+ 1, 2 is an outer angle of boc, boc= 8- 1=va+ 1- 1=

a；2）**3：∠obc=

a+∠acb），∠ocb=

a+∠abc），boc=1u0°-∠0bc-∠ocb，180°-∠a+∠acb）-

a+∠abc），180°-

a+ abq+ aqb), conclusion boc=90°-a 1) According to the information provided, according to the sum of two inner angles of a triangle that are not adjacent to it, 2 is represented by a and 1, and 2 is represented by o and 1, and then the relationship between boc and o can be obtained.

2) OBC and OCB are expressed according to the definition of one outer angle of a triangle equal to the sum of two inner angles not adjacent to it and the bisector of the angle, and then the solution can be obtained according to the inner angle of the triangle and the theorem column

1.There is an iron wire, the first time he used half of his less than 1 meter, the second time he used half of the remaining iron wire and 1 meter more, but this iron wire still has meters left, ask how many meters of this wire is originally long?

2.Pour the full bucket of water in the cylindrical bucket with an inner diameter of 200mm into a rectangular iron box with internal length, width and height respectively, just pour it full, find the water height in the cylindrical bucket?

3.The train was blocked in the middle of the way, delayed for 6 minutes, and then increased the speed from the original 40 kilometers per hour to 50 kilometers per hour.

4.The seventh grade (1) class of a school organizes extracurricular activities and prepares to hold a badminton competition, so I go to the store to buy badminton rackets and shuttlecocks, 25 yuan per racket and 2 yuan per finger chain"Badminton and rackets are 9% off", said B store"Buy a pair of rackets and get 2 shuttlecocks as a gift, (1) The school is going to spend 90 yuan to buy 2 pairs of badminton rackets and a number of badminton balls, which store is more cost-effective to buy?

2) If I have to buy 2 pairs of badminton rackets, how many shuttlecocks should I buy when I go to both stores?

5.Students A, B, and C donated books to children in poor areas, and it is known that the ratio of the number of books donated by these three students is 5:6

9. If the sum of the number of books donated by students A and C is 12 more than twice the number of books donated by B, how many books do they donate?

x-[1 2(1 Yunchang2x-1)+1]=

x=4 2.Solution: The height is xmm

100·100·л·x＝300·300·80

x＝720л

3.Solution: Walk x km.

x/50=[x-（40·6/60）]/40

x=4 4.A: After 9% discount, the racket is: yuan ball is yuan.

Racket Yuan Ball: (90-45) pcs).

B: 25·2 50 (yuan).

Balls to buy: (90-50) 2=20 (only).

Total ball: 20+2=22 (only).

A can buy 25 from there, while B can only buy 22.

Therefore, A is more cost-effective.

5.Dissolution: x per serving

A: 5x B: 6x C: 9x

5x+9x=6x·2+12

x=6 So: A: 5·6=30 (Ben).

B: 6·6=36 (Ben).

C: 9·6 = 54 (Ben).

Can you give me your email address, or I won't be able to pass it on.

This question is very basic, even AC and AD. Knowable ABC AED

So ac=ad and am are the midlines of ACD.

According to the three-line junction certain reason, so am cd hoped.

∵∠acb=∠adb=90°

ab=ab∴△acb≌△adb

cd=df (the height on the same side of the congruent triangle is equal).

Solution: Let the side length of the small square be x

360-2x)(265-2x)=720

95400-530x-720x+4x^2=7204x^2-1250x+94680=0

x^x^2-2*

x = (about) 27+

x1=x2=rounded).

A: The height of the box is.

Are you sure the question is correct?

Because the speed of a passenger ship is faster than that of a cargo ship, if the cargo ship moves from point D to point A, it will definitely not be able to catch up if you plan to chase the passenger ship from behind. Then the freighter can only drive to point C first, and then go to point B, that is, to the passenger ship head-on, and then meet the passenger ship.

Then the meeting point E of the two ships should be on BC, so that no matter how long the two ships have been driving, the distance they have traveled together is AB+BC+CD=200+200+200*(root number 2) = 400 + 200* (root number 2), which is about equal to nautical miles.

:(1)B(2) Let the freighter sail X nautical miles from departure to the meeting of the two ships Df CB, the vertical foot is F, and the connection DE is DE=X, AB+BE=2X In the isosceles right triangle ABC, AB=BC=200, D is the midpoint of AC, DF=100, EF=300-2X

In RT def, de2=df2+ef2, x2 1002+(300-2x)2

Solution, get x = 200 100 6 3

200+100 6 3>200, de=200-100 6 3 Answer: The freighter sailed a total of 200-100 6 3 nautical miles from the departure to the meeting of the two ships

Answer your question in the answer to the netizen "zxcvbnm13c", according to the given answer, it means that the direction of the freighter is the direction of DE, and E is the intersection of the passenger ship and the freighter, so the length of DE is twice the length of AB+BE, you will know it by drawing, the question is given that the two set off at the same time, driving at a constant speed, the speed of the passenger ship is twice that of the freighter, and the distance traveled is also twice the relationship.

Solution: EF BC, EOB= OBC, OBC= OBE, EOB= EB, EO=EB, similarly FO=FC, AEF Perimeter=AE+EO+of+AF=AE+EB+FC+AF=AB+AC=28

ABC circumference = AB + AC + BC = 28 + 7 = 35 (cm).

The lack of a condition in this question is that EF is parallel to BC

After adding it, it is easy to prove that df=fc and de=eb (two isosceles triangles).

Then the circumference of ABC is 28+7=35cm

Because folded past.

So abc= cbo, and because abc+ cbo+ dba=180°, abc=(180°- dba) 2

Because dba 70°

So abc is equal to 55°

55° to the right extends db at point o, because the folded one is abc = cbo, and because abc + cbo + dba = 180°

Because dba is 70°, abc is equal to 55°