2013年中考二模数学试卷及答案(白下区)(11)

（2）解法一：设小轿车离甲地的路程y2（km）与行驶时间x（h）的函数关系式是

y2＝mx．

代入点（1.5，150），得150＝1.5m．解得m＝100．

所以，小轿车离甲地的路程y2（km）与行驶时间x（h）的函数关系式是y2＝100x．

由（1）知，货车离甲地的路程y1（km）与行驶时间x（h）的函数关系式是y1＝240－60x．

当y1＝0时，代入y1＝－60x＋240，得x1＝4． 当y2＝300时，代入y2＝100x，得x2＝3．

答：小轿车先到达目的地． ······························································· 8分

解法二：根据图象，可得小轿车的速度为150÷1.5＝100． 货车到达甲地用时240÷60＝4（h）． 小轿车到达乙地用时300÷100＝3（h）．

答：小轿车先到达目的地． ······························································· 8分

25．（本题8分）

解：（1）画图正确． ······························································································ 3分

（2）BC与⊙O相切．理由如下：

连接CO．∵∠A＝∠B＝30°，

∴∠COB＝2∠A＝60°．

∴∠COB＋∠B＝30°＋60°＝90°．

∴∠OCB＝90°，即OC⊥BC． 又BC经过半径OC的外端点C，

∴BC与⊙O相切． ·················································································· 8分

26．（本题8分）

解：（1）可画出下面的反例： 图中，AB＝CD，DA∥BC．

此时，虽有∠A＝∠C，但△AOD与△COB不全等． ································ 4分

（2）答案不唯一，如OA＝OC． 理由如下：

∵AB＝CD，OA＝OC，

∴AB－OA＝CD－OC，即OB＝OD． ∵∠AOD＝∠COB，

∴△AOD≌△COB． ················································································ 8分

B

C