摘要：25． 我们将能完全覆盖某平面图形的最小圆称为该平面图形的最小覆盖圆．例如线段的最小覆盖圆就是以线段为直径的圆． (1)请分别作出图1中两个三角形的最小覆盖圆(要求用尺规作图.保留作图痕迹.不写作法), (2)探究三角形的最小覆盖圆有何规律?请写出你所得到的结论, (3)某地有四个村庄.现拟建一个电视信号中转站.为了使这四个村庄的居民都能接收到电视信号.且使中转站所需发射功率最小(距离越小.所需功率越小).此中转站应建在何处?请说明理由． 25．解:(1)如图所示:································ 4分 (注:正确画出1个图得2分.无作图痕迹或痕迹不正确不得分) (2)若三角形为锐角三角形.则其最小覆盖圆为其外接圆,··················· 6分 若三角形为直角或钝角三角形.则其最小覆盖圆是以三角形最长边为直径的圆．···························································································································· 8分 (3)此中转站应建在的外接圆圆心处(线段的垂直平分线与线段的垂直平分线的交点处)．········································································· 10分 理由如下: 由. .. 故是锐角三角形. 所以其最小覆盖圆为的外接圆. 设此外接圆为.直线与交于点. 则． 故点在内.从而也是四边形的最小覆盖圆． 所以中转站建在的外接圆圆心处.能够符合题中要求． ················································································ 12分 16一列快车从甲地驶往乙地.一列慢车从乙地驶往甲地.两车同时出发.设慢车行驶的时间为.两车之间的距离为.图中的折线表示与之间的函数关系． 根据图象进行以下探究: 信息读取 (1)甲.乙两地之间的距离为 km, (2)请解释图中点的实际意义, 图象理解 (3)求慢车和快车的速度, (4)求线段所表示的与之间的函数关系式.并写出自变量的取值范围, 问题解决 (5)若第二列快车也从甲地出发驶往乙地.速度与第一列快车相同．在第一列快车与慢车相遇30分钟后.第二列快车与慢车相遇．求第二列快车比第一列快车晚出发多少小时? 28． 解:(1)900,································································································· 1分 (2)图中点的实际意义是:当慢车行驶4h时.慢车和快车相遇．······ 2分 (3)由图象可知.慢车12h行驶的路程为900km. 所以慢车的速度为,····························································· 3分 当慢车行驶4h时.慢车和快车相遇.两车行驶的路程之和为900km.所以慢车和快车行驶的速度之和为.所以快车的速度为150km/h．··············································· 4分 (4)根据题意.快车行驶900km到达乙地.所以快车行驶到达乙地.此时两车之间的距离为.所以点的坐标为． 设线段所表示的与之间的函数关系式为.把.代入得 解得 所以.线段所表示的与之间的函数关系式为．······· 6分 自变量的取值范围是．································································· 7分 (5)慢车与第一列快车相遇30分钟后与第二列快车相遇.此时.慢车的行驶时间是4.5h． 把代入.得． 此时.慢车与第一列快车之间的距离等于两列快车之间的距离是112.5km.所以两列快车出发的间隔时间是.即第二列快车比第一列快车晚出发0.75h．·········· 10分

网址：http://m.1010jiajiao.com/timu_id_2041027[举报]