The seventh summer vacation mathematics homework answer Jilin education publishing house

Page 1. ( 1) 70，(2)∠FGB∠HGB∠BGH∠FHC(3)AB EC BC 60(4)C(5)C C C(5)C

(6)D (7)D

(8)∠∠AOD:∠BOE = 4: 1

∴∠BOE=？ ∠AOD

∫OE segmentation ∠BOD

∴∠DOE=∠BOE

∴∠boe= 180÷(4+ 1+ 1)= 30

∴∠DOE=∠EOB=30

* Equal share ∠COE

∴∠COF=∠EOF

∫∠EOF =( 180-∠DOE)÷2

=( 180 -30 )÷2

=75

Page 2, (9) Maintain a complementary relationship, because ∠AOB is 180, ∠ COD = 90, 180-90 = 90, so ∠ AOD+∠ BOD.

( 10)∵∠AOD=∠DOB

∴AB⊥CD

∠∠FOC =∠ Doe, ∠COF=? ∠BOE

∴∠doe= 1/3∠dob= 1/3×90 = 30

∴∠AOE=90 +30 = 120

Remarks: (1/3= one third)

( 1 1)(3x+ 10 )×2﹣5 ﹢(3x+ 10 )+(2x﹣ 10)= 180

6x+20 ﹣5 ﹢3x﹢ 10 ﹢2x﹣ 10 = 180

1 1x = 180 ﹢5 ﹣ 10 ﹢ 10 ﹣20

x= 15

(3×15+10 )× 2-5 =105 (degrees)

∴∠AOE= 105

Page 3, (1)2 1, (2) 15, (3)∠AOD=∠DOB (4)3, CD (5) C, (6) B, (7) b (.

Page 4, (9) ? FGM = ∠ GDN, ∠ 1=∠2

∴∠FGD=∠GDC

∵FG⊥AB

∴∠GDC=90

∴CD⊥AB

Page 5, 1. (1)∨, ⊥, ∨, (∨ is parallel) (2) No, the same plane.

2. Omission, 3. C 4 explosive B 5。 C 6。 b，7。 (Figure, omitted

Page 6, (8) (Figure omitted)

9,( 1)∵AD∥BC,AD∥PQ,∴PQ∥BC，

(2) Equation, ∫p is the midpoint of AB, AP=PB, the heights in trapezoidal ABCD are equal, ∴DQ=CQ.

Page 7, (1)c⊥b, (2)ab∨CD, (3)ea∨db (same angle, two straight lines parallel), ed∨BC (same internal dislocation angle, two straight lines parallel), ed∨ab (.

(4) The sentence that judges a thing is set as a conclusion, and the topic is set as (5) If two angles are equal, then their complementary angles are equal.

(6)30cm (7)D, (8) Because A∨C and B∨C, A∨B, if two straight lines are parallel to the third straight line at the same time, then the two straight lines are parallel.

(9) wrong. Every angle smaller than the right angle has a complementary angle. Complementarity means that the sum of two angles with a common vertex and a common edge is equal to 90. Complementarity means that the sum of two angles with a common vertex and a common edge is equal to 180.

P Page 8, (10) ∵∠1∠ 3 =135,

∴∠2= 180 - 135 =45

∵∠4= 135

∴∠2﹢∠4= 180

∴DE∥BC

∵∠2=60

∴∠3+∠ 1= 120

∵∠3-∠ 1=30 ,

∴∠3=75 ,∠ 1=45

∠∠A = 45

∴∠A﹢∠ADF=45 +60 +75 = 180

∴DF∥AC

(1 1) Make a straight line CG parallel to AB.

∫CG∨AB

∴∠ACG= 180 - 130 =50

∴AC⊥CD

∴∠ACD=90

∴∠GCD=90 -50 =40

∠∠CDE = 40

∴∠GCD=∠CDE

∴CG∥ED

∴AB∥ED

( 12)∵∠ 1+∠2= 180 ,∠ 1+∠4= 180

∴∠2=∠4

∠∠3 =∠B

∴∠EDC=∠DCB

∠∠b+∠DCB =∠ADC，∠DCB=∠EDC

∴∠ADE=∠B

∴DE∥BC

∴∠AED=∠ACB

Page 9, (1)0 0 (2) four, two (3) Null (4)(-4, 3) (5) C, (6) d (7) c (8) b.

(9) Y axis of the second quadrant, the fourth quadrant and the first quadrant is (10) empty.

Page 10 (1 1), (0,0) (0,4) (4,0) (4,4) (12) are omitted.

Page 1 1, (1) (16,3) (32,0) (N2,3) is empty.

(2)C, (3)A(4) omitted.

Page 12, drawings (omitted)

Page 13 (1)5, (2)No. (3) ∠BAC parallel. (4)D (5)C (6)C (7) Sketch, (8) Sketch

Drawings and descriptions on pages14 (9) ~ (10) are omitted.

Page 15, (1)100,20 (20 (2) 6Δ ace, Δ ADC, Δ ABC AAAD.

(3)A (4)B (5)A

(6) 36×[3÷﹙3+4+5﹚]=9﹙㎝﹚ 36×[3÷﹙3+4+5﹚]= 12(㎝) 36×[3÷﹙3+4+5﹚]= 15(㎝)

They are 9㎝, 12㎝ and 15㎝.

(7)a+b+c+a-b-c+c+a-b

=3a-b+c

Page 16 (8)

(9)∵∠aed = 48 ,∴∠dec= 180-48 = 132

∠∠ACB = 74，∠B=67

∴∠BDF=360 -∠B-∠ACB-∠DEC

=87

P 17 1, 3 2 1 12,10 3, 68 4 ~ 8, AACAD9, (omitted)

P 18 10, solution: Let the intersection of AB and CD be point O∠DOA be ∠ 3 ∫ EB ∨ CD ∠1= 95 ∴ ∠/kloc- The intersection of CE is point O∵CE⊥AB, AD ⊥ BC ∴∠ AEC = 90 ∠ ADB = 90 and ∠ AOE = 58 ∴∠ Bad = 180-90-58 = intersection o ∴∠ bad. When the number of sides of a polygon is 1, the number of sides of the original polygon is 2520 ÷180-1=133.

P209, not good, (because regular pentagons can't be densely laid) (the market prospect is slightly) 10, slightly.

P2 1 1, 0 -22, 0.53, 24-6, BBC7, (1) x = 6 (2) x = 0 (3) x = y =-3y = y = 8, solution: ∨. X=25 Y= 15 when solving equations. A: The speed of object A is 25m/s, and that of object B is15m/s..

P249, (1) solution: ① suppose that X sets of type A and Y sets of type B are purchased, x+y = 501500x+2100y = 90000, and the solution equations x=25 Y=25 satisfy the meaning of the question; (2) suppose you have purchased the Y set of type B and the Z set of type C.. y+z = 502 100y+2500 z = 9000； Y=87.5 Z=-37.5 does not meet the meaning of the question; ③ Suppose that X sets of Class A and Z sets of Class C are purchased; x+z = 50 1500 x+2500 z = 90000； Solve the equation X = 35Z. 25 units of type B, 35 units of type A and 5 units of type C/KLOC-0 (2) Scheme 1: profit: 25× 150+25×200 = 8750 (yuan) Scheme 2: profit: 35×150+/KLOC-. 10, (1), A: There are two ways: ① Play 15 second advertisement twice and 30 second advertisement four times. ② Play 15 second advertisement 4 times and 30 second advertisement 2 times. (2), ① Model income 2×0.6+ 1×4=5.2 (ten thousand yuan) ② Model income 4×0.6+ 1×2=4.4 (ten thousand yuan) 52,000 yuan > 44,000 yuan A: ① Model income is relatively large.

P25

1、-2

2、0.5

3、4000

4~9、DCCCBD

10, solution: suppose this person has saved X yuan in Form A and Y yuan in Form B. ..

x+y=5000

2﹪+2. 1﹪= 103

Solve equation X=2000.

y=3000

Answer: This person deposits 2000 yuan in Form A and 3000 yuan in Form B. ..

P26