第十一章三角形 达标测试卷
一、选择题(每题3分,共30分)
1.如图,∠1的大小等于( )
A.40° B.50° C.60° D.70°
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(第1题) (第4题)
2.下列长度的三条线段,能组成三角形的是( )
A.2 cm,3 cm,4 cm B.2 cm,3 cm,5 cm
C.2 cm,5 cm,10 cm D.8 cm,4 cm,4 cm
3.在△ABC中,能说明△ABC是直角三角形的是( )
A.∠A∶∠B∶∠C=1∶2∶2 B.∠A∶∠B∶∠C=3∶4∶5
C.∠A∶∠B∶∠C=1∶2∶3 D.∠A∶∠B∶∠C=2∶3∶4
4.如图,在△ABC中,∠A=80°,∠B=40°,D,E分别是AB,AC上的点,且DE∥BC,则∠AED的度数是( )
A.40° B.60° C.80° D.120°
5.在下列各图形中,分别画出了△ABC中BC边上的高,其中正确的是( )
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6.如图,△ABC的角平分线BE,CF相交于点O,且∠FOE=121°,则∠A的度数是( )
A.52° B.62° C.64° D.72°
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(第6题) (第7题) (第9题)
7.如图,在△ABC中,∠C=90°,D,E是AC上两点,且AE=DE,BD平分∠EBC.下列说法不正确的是( )
A.BE是△ABD的中线 B.BD是△BCE的角平分线
C.∠1=∠2=∠3 D.BC是△ABE的高
8.一个多边形的内角和比它的外角和的3倍少180°,这个多边形的边数是( )
A.8 B.7 C.6 D.5
9.如图,在△ABC中,∠C=75°,若沿图中虚线截去∠C,则∠1+∠2=( )
A.360° B.180° C.255° D.145°
10.如图,∠A,∠B,∠C,∠D,∠E五个角的和等于( )
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(第10题)
A.90° B.180° C.360° D.540°
二、填空题(每题3分,共24分)
11.人站在晃动的公交车上,若分开两腿站立,还需伸出一只手抓住栏杆才能站稳,这是利用了_________________________________________________.
12.正十边形每个外角的度数是________.
13.已知三角形三边长分别为1,x,5,则整数x=________.
14.将一副三角尺按如图所示放置,则∠1=________.
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(第14题) (第16题) (第18题)
15.一个多边形从一个顶点出发可以画9条对角线,则这个多边形的内角和为________.
16.如图,在△ABC中,AD是BC边上的中线,BE是△ABD中AD边上的中线,若△ABC的面积是24,则△ABE的面积是________.
17.当三角形中一个内角α是另一个内角β的一半时,我们称此三角形为“半角三角形”,其中α称为“半角”.若一个“半角三角形”的“半角”为20°,则这个“半角三角形”最大内角的度数为________.
18.已知△ABC,有下列说法:
(1)如图①,若P是∠ABC和∠ACB的平分线的交点,则∠P=90°+∠A;
(2)如图②,若P是∠ABC和外角∠ACE的平分线的交点,则∠P=90°-∠A;
(3)如图③,若P是外角∠CBF和∠BCE的平分线的交点,则∠P=90°-∠A.
其中正确的有______个.
三、解答题(23题12分,24题14分,其余每题10分,共66分)
19.如图,一艘轮船在A处看见巡逻艇C在其北偏东62°的方向上,此时一艘客船在B处看见巡逻艇C在其北偏东13°的方向上.试求此时在巡逻艇上看这两艘船的视角∠ACB的度数.
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(第19题)
20.如图,BD,CE是△ABC的两条高,它们交于O点.
(1)∠1和∠2的大小关系如何?并说明理由.
(2)若∠A=50°,∠ABC=70°,求∠3和∠4的度数.
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(第20题)
21.如图,已知AD是△ABC的角平分线,CE是△ABC的高,AD,CE相交于点P,∠BAC=66°,∠BCE=40°.求∠ADC和∠APC的度数.
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(第21题)
22.如图,六边形ABCDEF的内角都相等,CF∥AB.
(1)求∠FCD的度数;
(2)求证AF∥CD.
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(第22题)
23.如图,在△ABC中,∠A=30°,一块直角三角尺XYZ放置在△ABC上,恰好三角尺XYZ的两条直角边XY,XZ分别经过点B,C.
(1)∠ABC+∠ACB=________,∠XBC+∠XCB=________,∠ABX+∠ACX=________.
(2)若改变直角三角尺XYZ的位置,但三角尺XYZ的两条直角边XY,XZ仍然分别经过点B,C,则∠ABX+∠ACX的大小是否变化?请说明理由.
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(第23题)
24.已知∠MON=40°,OE平分∠MON,点A,B,C分别是射线OM,OE,ON上的动点(点A,B,C均不与点O重合),连接AC交射线OE于点D,设∠OAC=x°.
(1)如图①,若AB∥ON,则
①∠ABO的度数是________.
②当∠BAD=∠ABD时,x=________;当∠BAD=∠BDA时,x=________.
(2)如图②,若AB⊥OM,是否存在这样的x的值,使得△ADB中有两个相等的角?若存在,求出x的值;若不存在,说明理由.
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(第24题)
答案
一、1.D 2.A 3.C 4.B 5.B 6.B
7.C 8.B 9.C 10.B
二、11.三角形具有稳定性 12.36°
13.5 14.105° 15.1 800° 16.6
17.120° 18.2
三、19.解:由题意可得AD∥BF,
∴∠BEA=∠DAC=62°.
∵∠BEA是△CBE的一个外角,
∴∠BEA=∠ACB+∠CBE.
∴∠ACB=∠BEA-∠CBE=62°-13°=49°.
答:此时在巡逻艇上看这两艘船的视角∠ACB的度数为49°.
20.解:(1)∠1=∠2.理由如下:
∵BD,CE是△ABC的两条高,
∴∠AEC=∠ADB=90°.
∵∠A+∠1+∠ADB=180°,
∠2+∠A+∠AEC=180°,
∴∠1=∠2.
(2)∵∠A=50°,∠ABC=70°,∠A+∠ABC+∠ACB=180°,
∴∠ACB=60°.
∵在△AEC中,∠A+∠AEC+∠2=180°,
∴∠2=40°.
∴∠3=∠ACB-∠2=20°.
∵在四边形AEOD中,∠A+∠AEO+∠4+∠ADO=360°,∠A=50°,∠AEO=∠ADO=90°,
∴∠4=130°.
21.解:∵CE是△ABC的高,
∴∠AEC=90°.
∴∠ACE=180°-∠BAC-∠AEC=24°.
∵AD是△ABC的角平分线,
∴∠DAC=∠BAC=33°.
∵∠BCE=40°,
∴∠ACB=40°+24°=64°.
∴∠ADC=180°-∠DAC-∠ACB=83°.
∴∠APC=∠ADC+∠BCE=83°+40°=123°.
22.(1)解:∵六边形ABCDEF的内角都相等,内角和为(6-2)×180°=720°,
∴∠B=∠A=∠BCD=720°÷6=120°.
∵CF∥AB,
∴∠B+∠BCF=180°.
∴∠BCF=60°.
∴∠FCD=∠BCD-∠BCF=60°.
(2)证明:∵CF∥AB,
∴∠A+∠AFC=180°.
∴∠AFC=180°-120°=60°.
∴∠AFC=∠FCD.
∴AF∥CD.
23.解:(1)150°;90°;60°
(2)∠ABX+∠ACX的大小不变.
理由:在△ABC中,∠A+∠ABC+∠ACB=180°,∠A=30°,
∴∠ABC+∠ACB=180°-30°=150°.
∵∠YXZ=90°,
∴∠XBC+∠XCB=90°.
∴∠ABX+∠ACX=(∠ABC-∠XBC)+(∠ACB-∠XCB)=
(∠ABC+∠ACB)-(∠XBC+∠XCB)=150°-90°=60°.
∴∠ABX+∠ACX的大小不变.
24.解:(1)①20° ②120;60
(2)存在.
①当点D在线段OB上时,
若∠BAD=∠ABD,则x=20;
若∠BAD=∠BDA,则x=35;
若∠ADB=∠ABD,则x=50.
②当点D在射线BE上时,
易知∠ABE=110°,又三角形的内角和为180°,
∴只有∠BAD=∠BDA,此时x=125.
综上可知,存在这样的x的值,使得△ADB中有两个相等的角,且x=20,35,50或125.