∴△DB′C′=B′O·B′C/2=1/2······························10分 11、∵∠A+∠C+∠E=∠AOE································2分 又∵∠B+∠D+∠F=∠BOD··································4分 又∵∠BOD=∠AOE········································6分 又∵∠AOE=60°·········································8分 ∴∠A+∠B+∠C+∠D+∠E+∠F=60°×2=120°···············10分
12、∵∠EBC=80°，∠CDF=60°···························2分 ∴∠FBE=100°，∠FDE=120°·····························4分 ∴∠FGE=（∠FBE+∠FDE)·······························7分 =×（100°+120°）=110°····························10分
13、（1）小车在送前4人的同时，剩下的人也同时步行不停的往前走， 小车送到火车站后立即再返回接剩下的人。·················2分 设步行的4个人步行了x小时，
x15?15-x
56030解得x=·············································3分
133030355所以共用时间：÷5+（15-）÷60=小时=40（分钟）＜42
131352131212则有：?（分钟），故此方案可行；································4分 （2）先用小汽车把第一批人送到离火车站较近的某一处，让第一批人步行，与此同时第二批人也在步行中；接着小汽车再返回接第二批人，使第二批人与第一批同时到火车站，在这一方案中，每个人不是
七年级数学拔高测试答案 第 4 页 共 5 页
乘车就是在步行，没有人浪费时间原地不动，所以两组先后步行相同的路程。···············································7分 设这个路程为y千米， 则=
y15-y?15-2y·······································8分
360解得：y=2（千米），·····································9分 所用时间为：+
2515-237=小时=37（分钟）＜42（分钟），故此方案6060也可行。···············································10分
七年级数学拔高测试答案 第 5 页 共 5 页