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hình học 7
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I- TRC NGHIM:
N TP I 7 CHNG I
I- TRC NGHIM: 1- Cho ng thng d v mt im A nm ngoi ng thng d. Qua A ta v c:
A. V s ng thng song song vi d. B. t nht mt ng thng song song vi d.
C. Nhiu nht mt ng thng song song vi d. D. Khng v c ng thng no song song vi d.
2- Hai ng thng ct nhau to thnh bao nhiu cp gc i nh:
A. 1
B. 2
C. 3
D. 4
3- Qua 1 im ngoi ng thng cho trc, ta v c bao nhiu ng thng song song vi ng thng cho trc:
a. 1
b. 2
c. 3
d. 4
4- Cho 2 ng thng m v n song song. ng thng d to vi ng thng m gc bng 30o. Gc to bi ng thng d v ng thng n l:
A. 60o
B. 160o
C. 30o
D. 120o5- Cho a // b. ng thng c vung gc vi ng thng a th:
A. c // b
B. c ( b
C. c khng ct bD. c trng vi b
6- ng thng xy l trung trc ca on thng MN khi:
A. xy i qua im I ca MN ; B. xy ( MN ; C. xy ( MN ti I v IM = IN; D. xy // MN v IM = IN
II- T LUN: Bi 1: Cho hnh v, bit BAC = 50o, ACD = 110o, CDE = 60o. Gii thch v sao AB // DE.Bi 2: Cho (ABC c = . V tia CD l tia i ca tia CA. Trn na mt phng b AC cha nh B v tia Cx // AB. Chng minh Cx l tia phn gic ca Bi 3: Cho hnh v v cc s liu v ln cc gc. Chng minh: Ax//By//Cz
Bi 4: Cho hnh v, bit Ax//Cz,
= 20o, = 42o, Ct l tia phn gic ca . Tnh s o ca
Bi 5:Cho hnh v, bit My//Ax, = 30o, = 48o, Az l tia phn gic ca Tnh s o ca
I.Trc nghm 1/ Ba ng thng ct nhau ti O. Tng s cc cp gc i nh (khng k gc bt ) lA.3 cp B. 6 cp C. 9 cp D.. 12 cp
2/ Ba ng thng xx, yy, zz cng i qua O. Gc no sau y l gc k b vi xy:
A. yx B. xy C. yz v zy D. yx v xy 3/ ) Cho 2 gc ph v x0y. V 0z, 0t ln lt l tia phn gic ca cc gc x0y, x0y. Bit gc x0y = 30 th s o gc z0x bng:
A. 60
B. 70
C. 75
D. 90
4/ Cho hai gc xy = zt nh hnh v. Pht biu no sau y l sai: t y
450 z O x A. xy v zt i nh B. xy v yz l hai gc k b C. xy v zt khng i inh D. yt = 900
5/ Gc to bi hai tia phn gic ca hai gc k b bng :
A. 1800
B. 600
C. 900
D. 450 II. T lun Bi 1 : Cho ng thng a v M ( a , N( a
a/ v ng thng b vung gc vi a ti M
b/ V ng thng c i qua N v c//a. Bi 2 :Hnh v sau y cho bit a // b , = 300 , = 400 . Tnh s o gc AOB . Ni r v sao tnh c nh vy .
B b
400
O
300 a
A
Bi 3 : Trn hnh bn, cho AD ( Ax, AD ( Dy, gc ABC = 600. Tnh s o gc
Bi 4 : Cho hnh v bit xx // yy , = 420 a) Tnh gc xAB
b) V tia phn gic ca ABy ct xx ti M, tnh gc AMB
c) V Bt l tia phn gic ca gc mBy
Chng minh BM b Bt l 2 tia i nhau.
Bi 5 : Cho hnh v, bit m // n m A
Tnh gc mAB? B
80
n 55 C
I-Trc nghim:
1) Hai ng thng xy v zt ct nhau ti im M. Ta cA) i nh vi .
C) i nh vi
B) i nh vi
D) i nh vi
2) Hai ng thng song song l hai ng thng:
A) Khng c im chung
C) C mt im chung
B) C v s im chung
D) C hai im chung3) Nu c hai ng thng:
A) Vung gc th ct nhau.
C) Ct nhau th to thnh bn gc bng nhau
B) Ct nhau th vung gc.
D) Ct nhau th to thnh bn cp gc i nh4) Gc to bi hai ng thng vung gc l:
A) Gc vung
B) Gc nhn
C) Gc t D) Gc btII/ T lun: Bi 1: Hnh v bn cho bit: a // b v = 420. Tnh ;;;
Bi 2: Hnh v sau cho bit: = 300; = 600; = 1500. Chng t: Ax // By x A
B y
Bi 3: Chng minh rng: Nu hai gc nhn xOy v xOy c Ox // Ox, Oy // O y th =
Bi 4: Cho hai ng thng xx v yy ct nhau ti A. Bit x,tnh x
EMBED Equation.3 ,x
EMBED Equation.3 Bi 5: Cho hnh v bit Ax // Cy v = 1100, = 500. Tnh ?
x A
1100
B
500
yC
Bi 6: Cho hnh v, bit = 110o .
Chng minh: Ax // Cy
Bi 7 : Cho hnh v, bit ;
a- Chng minh xy // zt
b- V tia phn gic ca ct xy ti C. Tnh .
c- V Bq l tia phn gic ca
Chng minh BC v Bq l 2 tia i nhau.Bi 8: Cho hnh v bit : Ax // Cy ; = 1300 ; = 1100 Tnh ? B
y CBi 9: Cho hnh v bit = 1500 ; = 1100; = 800 z MChng minh: Ma // Nb. B N t
Bi 10: Chng minh rng: Nu hai gc nhn x Ay v zB t c Ax // Bz, Ay // Bt th =
Bi 11: Cho hnh v, bit = 50o, = 110o, = 60o. Gii thch v sao AB // DE.
Bi 12: Hnh v bn cho bit: a // b v = 1420. Tnh ;;;
Bi 13:Hnh v sau cho bit: = 400; = 700; = 1500. Chng t: Ax // By
x A
B y
Bi 14: Cho hnh v bit : Ax // Cy ; = 1500; = 1300 Tnh ? B
y CBi 15: Cho hnh v bit = 1400; = 1100; = 700 z MChng minh: Ma // Nb. B N t
D
A
E
(
C
(
B
110o
50o
60o
120o
x
y
z
A
B
C
130o
110o
130o
20o
z
y
t
x
A
B
C
42o
30o
t
x
y
z
A
M
C
48o
A
D
C
B
x
y
x
y
y'
x'
n
m
B
A
O
B
(
C
(
E
A
y
t
120o
60o
m
n
A
B
xc
zc
O
40o
50o
90o
D
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