资源简介 INCLUDEPICTURE "节.tif" INCLUDEPICTURE "../../../../广东慧源同步上册/广东慧源七年级上册数学/节.tif" \* MERGEFORMAT INCLUDEPICTURE "../../节.tif" \* MERGEFORMAT INCLUDEPICTURE "../../节.tif" \* MERGEFORMAT INCLUDEPICTURE "../节.tif" \* MERGEFORMAT INCLUDEPICTURE "节.tif" \* MERGEFORMAT第六章 几何图形初步 章末作业一、选择题1.下列各角中是钝角的是( B ).A.周角 B.平角C.周角 D.2直角2.如图,下列说法错误的是( A ).INCLUDEPICTURE "../../../../广东慧源同步上册/广东慧源七年级上册数学/SX228.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX228.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX228.tif" \* MERGEFORMAT INCLUDEPICTURE "../SX228.tif" \* MERGEFORMAT INCLUDEPICTURE "SX228.tif" \* MERGEFORMATA.∠AOB也可用∠O来表示B.∠1与∠AOB是同一个角C.图中共有三个角:∠AOB,∠AOC,∠BOCD.∠β与∠BOC是同一个角3.如图,OC是∠AOB的平分线,∠BOD=∠COD,∠BOD=15°,则∠AOC=( C ).INCLUDEPICTURE "../../../../广东慧源同步上册/广东慧源七年级上册数学/SX230.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX230.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX230.tif" \* MERGEFORMAT INCLUDEPICTURE "../SX230.tif" \* MERGEFORMAT INCLUDEPICTURE "SX230.tif" \* MERGEFORMATA.60° B.45°C.30° D.15°【解析】因为∠BOD=∠COD,∠BOD=15°,所以∠COD=3∠BOD=45°,所以∠BOC=∠COD-∠BOD=30°.因为OC是∠AOB的平分线,所以∠AOC=∠BOC=30°.故选C.4.已知∠α是锐角,∠α与∠β互补,∠α与∠γ互余,则∠β-∠γ的度数为( B ).A.45° B.90°C.180° D.无法确定5.如图,直线AB与直线CD相交于点O,若OE平分∠AOC,OF平分∠BOC,∠BOF=40°,则∠COE=( C ).INCLUDEPICTURE "../../../../广东慧源同步上册/广东慧源七年级上册数学/SX232.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX232.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX232.tif" \* MERGEFORMAT INCLUDEPICTURE "../SX232.tif" \* MERGEFORMAT INCLUDEPICTURE "SX232.tif" \* MERGEFORMATA.30° B.40°C.50° D.60°6.若一条铁路线上有10个站,则共需要制作( B )种火车票.A.110 B.90C.65 D.457.如图,已知B,M,C依次为线段AD上的三点,M为AD的中点,MC=CD=AB,若BC=8,则线段AD的长为( D ).INCLUDEPICTURE "../../../../广东慧源同步上册/广东慧源七年级上册数学/SX233.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX233.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX233.tif" \* MERGEFORMAT INCLUDEPICTURE "../SX233.tif" \* MERGEFORMAT INCLUDEPICTURE "SX233.tif" \* MERGEFORMATA.24 B.22C.20 D.18【解析】设MC=x,因为MC=CD=AB,所以CD=2MC=2x,AB=MC=x,所以MD=MC+CD=x+2x=3x.因为M为AD的中点,所以AD=2MD=2×3x=6x,AM=MD=3x,所以BM=AM-AB=3x-x=x.又因为BC=8,所以BC=BM+MC=x+x=8,所以x=3,所以AD=6x=6×3=18.故选D.8.如图,将一张长方形纸片ABCD沿对角线BD折叠后,点C落在点E处,BE交AD于点F,再将△DEF沿DF折叠后,点E落在点G处,若DG刚好平分∠ADB,则∠BDC的度数为( D ).INCLUDEPICTURE "../../../../广东慧源同步上册/广东慧源七年级上册数学/SX234.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX234.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX234.tif" \* MERGEFORMAT INCLUDEPICTURE "../SX234.tif" \* MERGEFORMAT INCLUDEPICTURE "SX234.tif" \* MERGEFORMATA.57° B.56°C.55° D.54°【解析】由折叠可知,∠BDC=∠BDE,∠EDF=∠GDF.因为DG平分∠ADB,所以∠BDG=∠GDF,所以∠EDF=∠BDG,所以∠BDE=∠EDF+∠GDF+∠BDG=3∠GDF,所以∠BDC=∠BDE=3∠GDF,∠BDA=∠GDF+∠BDG=2∠GDF.因为∠BDC+∠BDA=90°=3∠GDF+2∠GDF=5∠GDF,所以∠GDF=18°,所以∠BDC=3∠GDF=3×18°=54°.故选D.二、填空题9.将25.3°用度、分表示为________.【答案】25°18′10.一个角的补角的比它的余角小15°,则这个角的大小是________°.【答案】40【解析】设这个角的大小为x°.由题意,得(90-x)-(180-x)=15,解得x=40.即这个角的大小为40°.11.一副三角尺按如图方式摆放,∠1的度数是∠2度数的2倍,则∠2的度数为________.INCLUDEPICTURE "../../../../广东慧源同步上册/广东慧源七年级上册数学/SX235.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX235.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX235.tif" \* MERGEFORMAT INCLUDEPICTURE "../SX235.tif" \* MERGEFORMAT INCLUDEPICTURE "SX235.tif" \* MERGEFORMAT【答案】30°12.线段AB=1,C1是AB的中点,C2是C1B的中点,C3是C2B的中点,C4是C3B的中点,…,依此类推,线段AC2 025的长为________.【答案】1-()2 025【解析】由题意可知:C1B=AB,C2B=C1B=×AB,…,由此可知其规律为CnB=()nAB ,因此可知C2 025B=()2 025AB,因此可求得AC2 025=1-C2 025B=1-()2 025.三、解答题13.如图,C是线段AB上一点,M是线段AC的中点,N是线段BC的中点.INCLUDEPICTURE "../../../../广东慧源同步上册/广东慧源七年级上册数学/SX237.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX237.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX237.tif" \* MERGEFORMAT INCLUDEPICTURE "../SX237.tif" \* MERGEFORMAT INCLUDEPICTURE "SX237.tif" \* MERGEFORMAT(1)如果AB=30 cm,AM=9 cm,求NC的长;(2)如果MN=9 cm,求AB的长.【解】(1)因为M是线段AC的中点,所以AC=2AM.因为AM=9 cm,所以AC=18 cm.因为AB=30 cm,所以BC=AB-AC=30-18=12(cm).因为N是线段BC的中点,所以NC=BC=×12=6(cm).(2)因为M是线段AC的中点,N是线段BC的中点,所以BC=2NC,AC=2MC.因为MN=NC+MC=9 cm,所以AB=BC+AC=2×9=18(cm).14.如图所示,已知直角三角形纸板ABC,直角边AB=4 cm,BC=6 cm.INCLUDEPICTURE "../../../../广东慧源同步上册/广东慧源七年级上册数学/SX281.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX281.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX281.tif" \* MERGEFORMAT INCLUDEPICTURE "../SX281.tif" \* MERGEFORMAT INCLUDEPICTURE "SX281.tif" \* MERGEFORMAT(1)将直角三角形纸板绕三角形的边所在的直线旋转一周,能得到几种大小不同的几何体?(2)分别计算绕三角形直角边所在的直线旋转一周,得到的圆锥的体积.(圆锥的体积=πr2h,其中π取3)【解】(1)3种.(2)以AB为轴:×3×62×4=×3×36×4=144(cm3);以BC为轴:×3×42×6=×3×16×6=96(cm3).答:以AB为轴得到的圆锥的体积是144 cm3,以BC为轴得到的圆锥的体积是96 cm3.15.如图,已知点O为直线AB上一点,∠BOC=110°,∠COD=90°,OM平分∠AOC.INCLUDEPICTURE "../../../../广东慧源同步上册/广东慧源七年级上册数学/SX238.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX238.tif" \* MERGEFORMAT INCLUDEPICTURE "../../SX238.tif" \* MERGEFORMAT INCLUDEPICTURE "../SX238.tif" \* MERGEFORMAT INCLUDEPICTURE "SX238.tif" \* MERGEFORMAT(1)求∠MOD的度数;(2)若∠BOP与∠AOM互余,求∠COP的度数.【解】(1)因为∠BOC=110°,∠COD=90°,所以∠BOC+∠COD=110°+90°=200°.因为∠AOB=180°,所以∠AOD=200°-180°=20°,∠AOC=180°-110°=70°.因为OM平分∠AOC,所以∠AOM=∠AOC=×70°=35°.所以∠MOD=∠AOD+∠AOM=20°+35°=55°.(2)因为∠BOP与∠AOM互余,所以∠BOP+∠AOM=90°.因为∠AOB=180°,所以∠MOP=180°-90°=90°.由(1)知∠AOM=35°,所以∠COM=35°,所以∠COP=∠MOP-∠COM=90°-35°=55°.第六章 几何图形初步 章末作业一、选择题1.下列各角中是钝角的是( ).A.周角 B.平角C.周角 D.2直角2.如图,下列说法错误的是( ).A.∠AOB也可用∠O来表示B.∠1与∠AOB是同一个角C.图中共有三个角:∠AOB,∠AOC,∠BOCD.∠β与∠BOC是同一个角3.如图,OC是∠AOB的平分线,∠BOD=∠COD,∠BOD=15°,则∠AOC=( ).A.60° B.45°C.30° D.15°4.已知∠α是锐角,∠α与∠β互补,∠α与∠γ互余,则∠β-∠γ的度数为( ).A.45° B.90°C.180° D.无法确定5.如图,直线AB与直线CD相交于点O,若OE平分∠AOC,OF平分∠BOC,∠BOF=40°,则∠COE=( ).A.30° B.40°C.50° D.60°6.若一条铁路线上有10个站,则共需要制作( )种火车票.A.110 B.90C.65 D.457.如图,已知B,M,C依次为线段AD上的三点,M为AD的中点,MC=CD=AB,若BC=8,则线段AD的长为( ).A.24 B.22C.20 D.188.如图,将一张长方形纸片ABCD沿对角线BD折叠后,点C落在点E处,BE交AD于点F,再将△DEF沿DF折叠后,点E落在点G处,若DG刚好平分∠ADB,则∠BDC的度数为( ).A.57° B.56°C.55° D.54°二、填空题9.将25.3°用度、分表示为________.10.一个角的补角的比它的余角小15°,则这个角的大小是________°.11.一副三角尺按如图方式摆放,∠1的度数是∠2度数的2倍,则∠2的度数为________.12.线段AB=1,C1是AB的中点,C2是C1B的中点,C3是C2B的中点,C4是C3B的中点,…,依此类推,线段AC2 025的长为________.三、解答题13.如图,C是线段AB上一点,M是线段AC的中点,N是线段BC的中点.(1)如果AB=30 cm,AM=9 cm,求NC的长;(2)如果MN=9 cm,求AB的长.14.如图所示,已知直角三角形纸板ABC,直角边AB=4 cm,BC=6 cm.(1)将直角三角形纸板绕三角形的边所在的直线旋转一周,能得到几种大小不同的几何体?(2)分别计算绕三角形直角边所在的直线旋转一周,得到的圆锥的体积.(圆锥的体积=πr2h,其中π取3)15.如图,已知点O为直线AB上一点,∠BOC=110°,∠COD=90°,OM平分∠AOC.(1)求∠MOD的度数;(2)若∠BOP与∠AOM互余,求∠COP的度数. 展开更多...... 收起↑ 资源列表 第六章 几何图形初步 章末作业 - 学生版.doc 第六章 几何图形初步 章末作业.doc