2007—2014年陇南市中考题（含答案）(12)

222（2）∵ 参加足球运动项目的学生占所有运动项目学生的比例为

10150?5， ·

·········· 6分 ∴ 扇形统计图中表示“足球”项目扇形圆心角的度数为15?360?72． ··············· 8分 25． 本小题满分10分

解法1：设第一天捐款x人，则第二天捐款（x+50）人， ········································ 1分

由题意列方程 4800x=6000x?50 ． ······························································· 5分

解得 x =200． ·························································································· 7分

检验：当x =200时，x（x+50）≠0， ∴ x =200是原方程的解． ··········································································· 8分 两天捐款人数x+（x+50）=450， 人均捐款

4800x=24（元）． 答：两天共参加捐款的有450人，人均捐款24元． ······································· 10分 说明：只要求对两天捐款人数为450， 人均捐款为24元，不答不扣分． 解法2：设人均捐款x元， ················································································ 1分

由题意列方程

6000x－4800x＝50 ． ························································· 5分 解得 x =24． ··························································································· 7分 以下略．

26． 本小题满分10分

解：（1）如图，过A作AO⊥AC，过B作BO⊥BD，AO与BO相

交于O，O即圆心． ··················································· 3分

说明：若不写作法，必须保留作图痕迹．其它作法略． （2）∵ AO、BO都是圆弧AmB的半径，O是其圆心， ∴ ∠OBA=∠OAB=150°-90°=60°． ······························· 5分 O ∴ △AOB为等边三角形．∴ AO=BO=AB=180． ·············· 7分 ∴ AB?π?60?180180?60π (m)．

∴ A到B这段弧形公路的长为60πm． ························································· 10分

27． 本小题满分10分

证明:（1） ∵ ?ACB??ECD，

∴ ?ACD??BCD??ACD??ACE． A 即 ?BCD??ACE． ······································· 2分 ∵ BC?AC,DC?EC， D

∴ △ACE≌△BCD． ·········································· 4分 （2）∵ ?ACB是等腰直角三角形，

E C B

A（-1，0）， ···················································· 2分 B（3，0）． ······················································ 3分 2）如图14（1），抛物线的顶点为M（1，-4），连结OM． ···························································· 4分

则 △AOC的面积=

32，△MOC的面积=32， △MOB的面积=6， ············································· 5分

图14（１） ∴ 四边形 ABMC的面积

=△AOC的面积+△MOC的面积+△MOB的面积=9． ··································· 6分 说明：也可过点M作抛物线的对称轴，将四边形ABMC的面

积转化为求1个梯形与2个直角三角形面积的和．

3）如图14（2），设D（m，m2?2m?3），连结OD． 则 0＜m＜3，m2?2m?3 ＜0． 且 △AOC的面积=

32，△DOC的面积=32m， 图14（2） △DOB的面积=-32（m2?2m?3）， ···················································· 8分

∴ 四边形 ABDC的面积=△AOC的面积+△DOC的面积+△DOB的面积

=?32m2?92m?6 =?32(m?32)2?758． ········································································ 9分

∴ 存在点D(32，?154)，使四边形ABDC的面积最大为758． ·

·························10分

（4）有两种情况：

图14（3） 图14（4）

（（

如图14（3），过点B作BQ1⊥BC，交抛物线于点Q1、交y轴于点E，连接Q1C． ∵ ∠CBO=45°，∴∠EBO=45°，BO=OE=3． ∴ 点E的坐标为（0，3）．

∴ 直线BE的解析式为y??x?3． ·························································· 12分

由??y??x?3，ì??x1=-2，ì??x2=3， ?y?x2?2x?3 解得í?=5； ??yí1???y2=0.∴ 点Q1的坐标为（-2，5）． ···································································· 13分

如图14（4），过点C作CF⊥CB，交抛物线于点Q2、交x轴于点F，连接BQ2． ∵ ∠CBO=45°，∴∠CFB=45°，OF=OC=3． ∴ 点F的坐标为（-3，0）．

∴ 直线CF的解析式为y??x?3． ·························································· 14分

由??y??x?3，ì?x1=0，ìx2=1， ?y?x2?2x?3 解得?í?????y1=-3； í???y2=-4．∴点Q2的坐标为（1，-4）． ····································································· 15分

综上，在抛物线上存在点Q1（-2，5）、Q2（1，-4），使△BCQ1、△BCQ2是以BC为直角边的直角三角形． ··········································································································· 16分 说明：如图14（4），点Q2即抛物线顶点M，直接证明△BCM为直角三角形同样得2分．

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