2009年山西省中考數(shù)學試卷及答案-(word整理版)

這是一份2009年山西省中考數(shù)學試卷及答案-(word整理版)，共7頁。試卷主要包含了填空題，選擇題，解答題等內(nèi)容，歡迎下載使用。
2009年山西省中考數(shù)學試卷-(word整理版)一、填空題（每小題2分，共20分）1．比較大?。?/span>             （填“＞”、“=”或“＜“）．2．山西有著豐富的旅游資源，如五臺山、平遙古城、喬家大院等著名景點，吸引了眾多的海內(nèi)外游客，2008年全省旅游總收入739.3億元，這個數(shù)據(jù)用科學記數(shù)法可表示為             ．3．請你寫出一個有一根為1的一元二次方程：            ．4．計算：=          ．5．如圖所示，、、、是圓上的點， 則          度．6．李師傅隨機抽查了本單位今年四月份里6天的日用水量（單位：噸）結(jié)果如下：7，8，8，7，6，6，根據(jù)這些數(shù)據(jù)，估計四月份本單位用水總量為             噸． 7．如圖，與是位似圖形，且頂點都在格點上，則位似中心的坐標是             ．          8．如圖，的對角線、相交于點，點是的中點，的周長為16cm，則的周長是          cm．9．若反比例函數(shù)的表達式為，則當時，的取值范圍是          ．10．下列圖案是晉商大院窗格的一部分，其中“○”代表窗紙上所貼的剪紙，則第個圖中所貼剪紙“○”的個數(shù)為          ．      二、選擇題（每小題3分，共24分）11．下列計算正確的是（    ）A．         B．        C．      D． 12．反比例函數(shù)的圖象經(jīng)過點，那么的值是（    ） A．          B．         C．          D．6 13．不等式組的解集在數(shù)軸上可表示為（    ）14．解分式方程，可知方程（    ）A．解為      B．解為     C．解為       D．無解 15．如圖是由幾個相同的小正方體搭成的幾何體的三視圖，則搭成這個幾何體的小正方體的個數(shù)是（    ）      A．5 B．6              C．7  D．816．如圖，是的直徑，是的切線，點在上，，則的長為（    ）  A．    B． C．       D．     17．如圖（1），把一個長為、寬為的長方形（）沿虛線剪開，拼接成圖（2），成為在一角去掉一個小正方形后的一個大正方形，則去掉的小正方形的邊長為（    ）A． B．         C．  D．18．如圖，在中，的垂直平分線交的延長線于點，則的長為（    ）A．            B．          C．  D．2三、解答題（本題共76分）19．（每小題4分，共12分）（1）計算：                  （2）化簡：    （3）解方程：                20．（本題6分）已知每個網(wǎng)格中小正方形的邊長都是1，圖1中的陰影圖案是由三段以格點為圓心，半徑分別為1和2的圓弧圍成．（1）填空：圖1中陰影部分的面積是             （結(jié)果保留）；（2）請你在圖2中以圖1為基本圖案，借助軸對稱、平移或旋轉(zhuǎn)設計一個完整的花邊圖案（要求至少含有兩種圖形變換）．                       21．（本題8分）根據(jù)山西省統(tǒng)計信息網(wǎng)公布的數(shù)據(jù)，繪制了山西省2004~2008固定電話和移動電話年末用戶條形統(tǒng)計圖如下：（1）填空：2004~2008移動電話年末用戶的極差是        萬戶，固定電話年末用戶的中位數(shù)是             萬戶；（2）你還能從圖中獲取哪些信息？請寫出兩條．                  22．（本題8分）某商場為了吸引顧客，設計了一種促銷活動：在一個不透明的箱子里放有4個相同的小球，球上分別標有“0元”、“10元”、“20元”和“30元”的字樣．規(guī)定：顧客在本商場同一日內(nèi)，每消費滿200元，就可以在箱子里先后摸出兩個球（第一次摸出后不放回）．商場根據(jù)兩小球所標金額的和返還相應價格的購物券，可以重新在本商場消費．某顧客剛好消費200元．（1）該顧客至少可得到       元購物券，至多可得到        元購物券；（2）請你用畫樹狀圖或列表的方法，求出該顧客所獲得購物券的金額不低于30元的概率．                 23．（本題8分）有一水庫大壩的橫截面是梯形，為水庫的水面，點在上，某課題小組在老師的帶領下想測量水的深度，他們測得背水坡的長為12米，迎水坡上的長為2米，求水深．（精確到0.1米，）                                          24．（本題8分）某批發(fā)市場批發(fā)甲、乙兩種水果，根據(jù)以往經(jīng)驗和市場行情，預計夏季某一段時間內(nèi)，甲種水果的銷售利潤（萬元）與進貨量（噸）近似滿足函數(shù)關系；乙種水果的銷售利潤（萬元）與進貨量（噸）近似滿足函數(shù)關系（其中為常數(shù)），且進貨量為1噸時，銷售利潤為1.4萬元；進貨量為2噸時，銷售利潤為2.6萬元．（1）求（萬元）與（噸）之間的函數(shù)關系式．（2）如果市場準備進甲、乙兩種水果共10噸，設乙種水果的進貨量為噸，請你寫出這兩種水果所獲得的銷售利潤之和（萬元）與（噸）之間的函數(shù)關系式．并求出這兩種水果各進多少噸時獲得的銷售利潤之和最大，最大利潤是多少？                                  25．（本題12分）在中，將繞點順時針旋轉(zhuǎn)角得交于點，分別交于兩點．（1）如圖1，觀察并猜想，在旋轉(zhuǎn)過程中，線段與有怎樣的數(shù)量關系？并證明你的結(jié)論；（2）如圖2，當時，試判斷四邊形的形狀，并說明理由；（3）在（2）的情況下，求的長．                                    26．（本題14分）如圖，已知直線與直線相交于點分別交軸于兩點．矩形的頂點分別在直線上，頂點都在軸上，且點與點重合．    （1）求的面積；（2）求矩形的邊與的長；（3）若矩形從原點出發(fā)，沿軸的反方向以每秒1個單位長度的速度平移，設移動時間為秒，矩形與重疊部分的面積為，求關于的函數(shù)關系式，并寫出相應的的取值范圍．                         2009年山西省中考數(shù)學試卷答案1．＞       2．      3．答案不唯一，如      4．      5．306．210      7．（9，0）      8．8      9．      10．題  號1112131415161718答  案DCDDBAAB19．（1）解：原式=·····················································（2分）               =·····························································（3分）               =．···························································（4分）（2）解：原式=·······················································（2分）             =·····························································（3分）             =1．··························································（4分）（3）解：移項，得配方，得··············································（2分）         ∴∴····························································（4分）          （注：此題還可用公式法，分解因式法求解，請參照給分）20．解：（1）；························································（2分）（2）答案不唯一，以下提供三種圖案．    （注：如果花邊圖案中四個圖案均與基本圖案相同，則本小題只給2分；未畫滿四個“田”字格的，每缺1個扣1分．）21．（1）935.7，859.0；··················································（4分）  （2）解：2004~2008移動電話年末用戶逐年遞增．           2008年末固定電話用戶達803.0萬戶．····································（8分）           （注：答案不唯一，只要符合數(shù)據(jù)特征即可得分）22．解：（1）10，50；···················································（2分）       （2）解：解法一（樹狀圖）：     ··································································（6分）從上圖可以看出，共有12種可能結(jié)果，其中大于或等于30元共有8種可能結(jié)果，因此（不低于30元）= （8分）解法二（列表法）：第一次第二次01020300 1020301010 3040202030 5030304050 ······························································（6分）（以下過程同“解法一”）·········································（8分）23．解：分別過作于于過作于則四邊形為矩形．∴在中，∴································································（3分）在中，····························································（6分）∴································································（7分）答：水深約為6.7米．··················································（8分）（其它解法可參照給分）24．解：（1）由題意，得：解得············································（2分）           ∴······························································（3分）（2）    ∴ （5分）      ∴時，有最大值為6.6. ···········································（7分）∴（噸）.答：甲、乙兩種水果的進貨量分別為4噸和6噸時，獲得的銷售利潤之和最大，最大利潤是6.6萬元.  （8分）25．解：（1）··························································（1分）證明：（證法一）由旋轉(zhuǎn)可知，∴·············································（3分）∴又∴即···········································（4分）（證法二）由旋轉(zhuǎn)可知，而∴·············································（3分）∴∴即·············································（4分）       （2）四邊形是菱形. ··················································（5分）證明：同理∴四邊形是平行四邊形. ···································（7分）又∴四邊形是菱形. ·······································（8分）       （3）（解法一）過點作于點，則在中，……（10分）由（2）知四邊形是菱形，∴∴·················································（12分）（解法二）∴在中， ··················································（10分）∴·················································（12分）（其它解法可參照給分）26．（1）解：由得點坐標為由得點坐標為∴·························································（2分）由解得∴點的坐標為···········································（3分）∴·························································（4分）   （2）解：∵點在上且            ∴點坐標為·······················································（5分）又∵點在上且∴點坐標為··················································（6分）∴·························································（7分）   （3）解法一：當時，如圖1，矩形與重疊部分為五邊形（時，為四邊形）．過作于，則        ∴即∴∴即···················································（10分）