原創(chuàng)2023學(xué)年中考數(shù)學(xué)預(yù)測適應(yīng)性考試卷(附答案)

原創(chuàng)2023學(xué)年胡文1．截止2023學(xué)年年6月7日12時(shí)，全國各地支援四川地震災(zāi)區(qū)的臨時(shí)安置房已經(jīng)安裝了40600套．這個(gè)數(shù)用科學(xué)記數(shù)法表示為（）A．C．套B．套套0.4061054.06104406102套D．40.61032．如圖1，直線，l分別與）相交，如果，l∥l1l，l2120l1212l1l2那么1的度數(shù)是（2A．B．C．D．75304560圖13．下列事件中是必然事件的是（A．陰天一定下雨）B．隨機(jī)擲一枚質(zhì)地均勻的硬幣，正面朝上C．男生的身高一定比女生高D．將油滴在水中，油會(huì)浮在水面上4．圖2是由幾個(gè)相同的小正方體搭成的一個(gè)幾何體，它的俯視圖是（）A．B．C．D．圖25．下列命題中正確的是（）A．兩條對角線互相平分的四邊形是平行四邊形B．兩條對角線相等的四邊形是矩形C．兩條對角線互相垂直的四邊形是菱形D．兩條對角線互相垂直且平分的四邊形是正方形6．若反比例函數(shù)yk(k0)x的圖象經(jīng)過點(diǎn)(2，1)，則這個(gè)函數(shù)的圖象一定經(jīng)過點(diǎn)（）1/21原創(chuàng)2023學(xué)年胡文A．211，2(1，2)C．B．D．1，(1，2)2≤的解集在數(shù)軸上表示正確的是（7．不等式組213x）x3-301A．-301B．-301C．-301D．8．圖3是對稱中心為點(diǎn)O的正八邊形．如果用一個(gè)含45角的直角三角板的角，借助點(diǎn)O（使角的頂點(diǎn)落在點(diǎn)O處）把這個(gè)正八邊形的面積n等分．那么n的所有可能的值有（A．2個(gè)B．3個(gè)）C．4個(gè)D．5個(gè)圖3二、填空題（每小題3分，共24分）9．分解因式：．x3y4xy10．體育老師對甲、乙兩名同學(xué)分別進(jìn)行了8次跳高測試，經(jīng)計(jì)算這兩名同學(xué)成績的平均數(shù)相同，甲同學(xué)的方差是，乙同學(xué)的方差是28.2，那么這S26.4甲S乙兩名同學(xué)跳高成績比較穩(wěn)定的是同學(xué)．11．一元二次方程x22x10的解是．12．如圖4，分別是△ABC的邊D，E上的點(diǎn)，DE∥BC，AB，ACAD2，則DBS:S．△ADE△ABCADEBC圖5圖413．如圖5，假設(shè)可以在圖中每個(gè)小正方形內(nèi)任意取點(diǎn)（每個(gè)小正方形除顏色外完全相同），那么這個(gè)點(diǎn)取在陰影部分的概率是．14．一個(gè)圓錐底面周長為4cm，母長線為5cm，則這個(gè)圓錐的側(cè)面積是．2/21原創(chuàng)2023學(xué)年胡文15．如圖6，觀察下列圖案，它們都是由邊長為1cm的小正方形按一定規(guī)律拼接而成的，依此規(guī)律，則第16個(gè)圖案中的小正方形有個(gè)．……圖案1圖案2圖案3圖案4圖616.如圖7，直線y3x3x與軸、y軸分別相交于A，By3B兩點(diǎn)，圓心P的坐標(biāo)為(10，)，P與y軸相切于點(diǎn)O．若將PxAOP沿x軸向左移動(dòng)，當(dāng)P與該直線相交時(shí)，橫坐標(biāo)為整數(shù)的點(diǎn)P圖7有個(gè)．三、（每小題8分，共16分）17．先化簡，再求值：3a，其中a2．2aa1a1a1a18．如圖8所示，在網(wǎng)格中建立了平面直角坐標(biāo)系，每個(gè)小正方形的邊長均為1個(gè)單位長度，將四邊形ABCD繞坐標(biāo)原點(diǎn)O按順時(shí)針方向旋轉(zhuǎn)180后得到四邊形．ABCD1111（1）直接寫出D點(diǎn)的坐標(biāo)；1（2）將四邊形平移，得到四邊形，若，畫出平移后的圖D(45，)ABCDABCD1111222223/21原創(chuàng)2023學(xué)年胡文形．（友情提示：畫圖時(shí)請不要涂錯(cuò)陰影的位置哦?。┧?、（每小題10分，共20分）19．如圖9，有四張背面相同的紙牌A，B，C，D圖形，小明將這四張紙牌背面朝上洗勻后隨機(jī)摸出一張，放回后洗勻再隨機(jī)摸出一張．，其正面分別畫有四個(gè)不同的（1）用樹狀圖（或列表法）表示兩次摸牌所有可能出現(xiàn)的結(jié)果（紙牌用A，B，C，D表示）；（2）求兩次摸牌的牌面圖形既是中心對稱圖形又是軸對稱圖形的概率．圖94/21原創(chuàng)2023學(xué)年胡文20.如圖10，AB為O的直徑，D為弦BE的中點(diǎn)，連接OD并延長交O于點(diǎn)F，與過B點(diǎn)的切線相交于點(diǎn)C．若點(diǎn)E為AF的中點(diǎn)，連接AE．CFE求證：．△ABE≌△OCBDABO圖10五、（每小題10分，共20分）21．某中學(xué)開展以“我最喜歡的職業(yè)”為主題的調(diào)查活動(dòng)．通過對學(xué)生的隨機(jī)抽樣調(diào)查得到一組數(shù)據(jù)，下面兩圖（如圖11、圖12）是根據(jù)這組數(shù)據(jù)繪制的兩幅不完整的統(tǒng)計(jì)圖．請你根據(jù)圖中所提供的信息解答下列問題：5/21原創(chuàng)2023學(xué)年胡文（1）求在這次活動(dòng)中一共調(diào)查了多少名學(xué)生？（2）在扇形統(tǒng)計(jì)圖中，求“教師”所在扇形的圓心角的度數(shù)．（3）補(bǔ)全兩幅統(tǒng)計(jì)圖．人數(shù)80其它教師604020醫(yī)生15%軍人公務(wù)員10%職業(yè)0教師醫(yī)生公務(wù)員軍人其它20%圖11圖1222．在“汶川地震”捐款活動(dòng)中，某同學(xué)對甲、乙兩班捐款情況進(jìn)行了統(tǒng)計(jì)：甲班捐款人數(shù)比乙班捐款人數(shù)多3人，甲班共捐款2400元，乙班共捐款1800元，乙班平均每人捐款的錢數(shù)是甲班平均每人捐款錢數(shù)的4倍．求甲、乙兩班各5有多少人捐款？6/21原創(chuàng)2023學(xué)年胡文六、（每小題10分，共20分）23．如圖13，某數(shù)學(xué)興趣小組在活動(dòng)課上測量學(xué)校旗桿高度．已知小明的眼睛與地面的距離(AB)是1.7m，看旗桿頂部M的仰角為45；小紅的眼睛與地面的距離(CD)是1.5m，看旗桿頂部M的仰角為30．兩人相距28米且位于旗桿兩側(cè)（點(diǎn)B，N，D在同一條直線上）．請求出旗桿MN的高度．（參考數(shù)據(jù)：，，結(jié)果保留整數(shù)）2≈1.43≈1.7M45°A30°CBND圖1324．2023學(xué)年胡文年6月1日起，我國實(shí)施“限塑令”，開始有償使用環(huán)保購物袋．為了滿足市場需求，某廠家生產(chǎn)A，B兩種款式的布質(zhì)環(huán)保購物袋，每天共生產(chǎn)4500個(gè)，兩種購物袋的成本和售價(jià)如下表，設(shè)每天生產(chǎn)A種購物袋x個(gè)，每天共獲利y元．成本（元/售價(jià)（元/個(gè)）個(gè)）7/21原創(chuàng)2023學(xué)年胡文232.33.5AB（1）求出y與x的函數(shù)關(guān)系式；（2）如果該廠每天最多投入成本10000元，那么每天最多獲利多少元？七、（本題12分）25.如圖14，在Rt△ABC中，，，，另有一等腰梯形DEFGA90ABACBC42（GF∥DE）的底邊DE與BC重合，兩腰分別落在AB，AC上，且G，F(xiàn)分別是AAB，AC的中點(diǎn)．（1）求等腰梯形DEFG的面積；GF(D)BC(E)圖148/21原創(chuàng)2023學(xué)年胡文（2）操作：固定△ABC，將等腰梯形DEFG以每秒1個(gè)單位的速度沿BC方向向右運(yùn)動(dòng)，直到點(diǎn)D與點(diǎn)C重合時(shí)停止．設(shè)運(yùn)動(dòng)時(shí)間為x秒，運(yùn)動(dòng)后的等腰梯形為DEFG（如圖15）．探究1：在運(yùn)動(dòng)過程中，四邊形BDGG能否是菱形？若能，請求出此時(shí)x的值；若不能，請說明理由．AGFGFEBDC圖15探究2：設(shè)在運(yùn)動(dòng)過程中△ABC與等腰梯形DEFG重疊部分的面積為y，求與yx的函數(shù)關(guān)系式．9/21原創(chuàng)2023學(xué)年胡文八、（本題14分）26．如圖16，在平面直角坐標(biāo)系中，直線y3x3x與軸交于點(diǎn)A，與y軸交于點(diǎn)C，拋物線23yax2xc(a0)經(jīng)過A，B，C三點(diǎn)．3（1）求過三點(diǎn)拋物線的解析式并求出頂點(diǎn)F的坐標(biāo)；A，B，C（2）在拋物線上是否存在點(diǎn)P，使△ABP為直角三角形，若存在，直接寫出P點(diǎn)坐標(biāo)；若不存在，請說明理由；（3）試探究在直線AC上是否存在一點(diǎn)M，使得△MBF的周長最小，若存在，求出M點(diǎn)的坐標(biāo)；若不存在，請說明理由．yxOBACF圖1610/21原創(chuàng)2023學(xué)年胡文11/21原創(chuàng)2023學(xué)年胡文答案一、選擇題（每小題3分，共24分）題號(hào)答案1B2345678BCDDADA二、填空題（每小題3分，共24分）10．甲9．11．712．13．xy(x2)(x2)xx114:925214．10cm2（丟單位扣1分）三、（每小題8分，共16分）15．13616．317．解法一：原式3a(a1)a(a1)a1·········································2分2a1a22a4········································································································6分當(dāng)a2時(shí)，原式2248···································································8分解法二：原式3a(a1)(a1)a(a1)(a1)······························2分a1aa1a2a4········································································································6分當(dāng)a2時(shí)，原式2248···································································8分18．解：（1）（2）····························································································2分描對一個(gè)點(diǎn)給1分．·········································6分B，C，D222D(3，1)1，A2畫出正確圖形（見圖1）········································································8分12/21原創(chuàng)2023學(xué)年胡文圖1四、（每小題10分，共20分）19．（1）解法一：第二次第一次ABCDAB（A，A）（A，B）（A，C）（A，D）（B，A）（B，B）（B，C）（B，D）（C，A）（C，B）（C，C）（C，D）（D，A）（D，B）（D，C）（D，D）···················6分CD（2）從表中可以得到，兩次摸牌所有可能出現(xiàn)的結(jié)果共有16種，其中既是中心對稱圖形又是軸對稱圖形的有9種．·············································8分故所求概率是9．················································································10分1619．（1）解法二：開始第一次牌面的字母第二次牌面的字母ABCDABCDABCDABCDABCD所以可能出現(xiàn)的結(jié)果：（A，A），（A，B），（A，C），（A，D），（B，A），（B，B），（B，C），（B，D），（C，A），（C，B），（C，C），（C，D），（D，A），（D，B），13/21原創(chuàng)2023學(xué)年胡文（D，C），（D，D）．························································6分（2）以下同解法1．20.解：（1）證明：如圖2．CEFAB是O的直徑．DE90·······························································1分ABO又BC是O的切線，OBC90圖2EOBC·························································3分OD過圓心，BDDE，EFFBBOCA．··························································································6分E為AF中點(diǎn)，EFBFAEABE30·······························································································8分E90AE1ABOB························································································9分2△ABE≌△OCB．·················································································10分五、（每小題10分，共20分）21．（1）被調(diào)查的學(xué)生數(shù)為40200（人）···········································2分20％（2）“教師”所在扇形的圓心角的度數(shù)為115％20％10％70100％36072··············································5分20014/21原創(chuàng)2023學(xué)年胡文（3）如圖3，補(bǔ)全圖·············································································8分如圖4，補(bǔ)全圖·····················································································10分人數(shù)80教師其它20%60402035%醫(yī)生15%公務(wù)員軍人10%職業(yè)0教師醫(yī)生公務(wù)員軍人其它20%圖3圖422．解法一：設(shè)乙班有人捐款，則甲班有(x3)人捐款．···········1分x根據(jù)題意得：····························································································5分240041800x35x解這個(gè)方程得x45．·············································································8分經(jīng)檢驗(yàn)x45是所列方程的根．·····························································9分x348（人）答：甲班有48人捐款，乙班有45人捐款．·································10分解法二：設(shè)甲班有人捐款，則乙班有(x3)人捐款．····················1分x根據(jù)題意得：····························································································5分240041800x5x3解這個(gè)方程得x48．·············································································8分經(jīng)檢驗(yàn)x48是所列方程的根．·····························································9分x345（人）答：甲班有48人捐款，乙班有45人捐款．·································10分六、（每小題10分，共20分）23．解法一：解：過點(diǎn)作于，過點(diǎn)作于，·····················1分AAEMNECCFMNF15/21原創(chuàng)2023學(xué)年胡文則在EFABCD1.71.50.2···································································2分Rt△AEM中，AEM90，MAE45AEME···································································································3分設(shè)AEMEx（不設(shè)參數(shù)也可）MMFx0.2，5412-0·································5分45°AE30°C在Rt△MFC中，，F(xiàn)MFC90MCF30BND圖5MFCFtanMCF3(28x)·········································7分x0.23x≈10.0···································································································9分MN≈12答：旗桿高約為12米．·····································································10分解法二：解：過點(diǎn)作于，過點(diǎn)作于，····1分AAEMNECCFMNF則在EFABCD1.71.50.2···································································2分Rt△AEM中，，AEM90MAE45AEME設(shè)在AEx，則MFx0.2·········································································3分Rt△MFC中，，MFC90MCF30CFMFtan603(x0.2)·······································································5分BNNDBDx3(x0.2)28·····················································································7分解得x≈10.2···································································································9分MN≈12答：旗桿高約為12米．·····································································10分16/21原創(chuàng)2023學(xué)年胡文（注：其他方法參照給分）24．解：（1）根據(jù)題意得：y(2.32)x(3.53)(4500x)0.2x2250·········2分（2）根據(jù)題意得：2x3(4500x)≤10000············································5分解得x≥3500元··························································································6分，隨增大而減小·························································8分k0.20yx當(dāng)x3500時(shí)y0.2350022501550·········································································9分答：該廠每天至多獲利1550元．····················································10分七、（本題12分）25．解：如圖6，（1）過點(diǎn)作于．GGMBCM，，，為中點(diǎn)ABACBAC90BC42GABAGM2．······································1分G又分別為G，F(xiàn)的中點(diǎn)AB，ACFGF1BC22·······························2分(D)BC(E)M2圖6S1(2242)262梯形DEFG等腰梯形DEFG的面積為6．·····························································3分（2）能為菱形·························································································4分如圖7，由BG∥DG，GG∥BC四邊形BDGG是平行四邊形········6分A當(dāng)BDBG1AB2時(shí)，四邊形BDGG為菱形，GFGF2此時(shí)可求得x2EBDMC當(dāng)秒時(shí)，四邊形BDGG為菱形．8分x2圖717/21原創(chuàng)2023學(xué)年胡文（3）分兩種情況：0≤x22時(shí)，方法一：①當(dāng)，GM2S2xBDGG重疊部分的面積為：y62x當(dāng)0≤x22時(shí)，與的函數(shù)關(guān)系式為y62x······················10分yx方法二：當(dāng)0≤x22時(shí)，A，，F(xiàn)G22xDC42xGM2FGPF重疊部分的面積為：Gy(22x)(42x)262xBEDQC2圖8當(dāng)0≤x22時(shí)，與的函數(shù)關(guān)系式為y62x···················10分=yx②當(dāng)22≤x≤42時(shí)，設(shè)FC與DG交于點(diǎn)P，則PDCPCD45，CPD90PCPD作PQDC于Q，則PQDQQC1(42x)2重疊部分的面積為：y1(42x)1(42x)1(42x)21x222x8························12分2244八、（本題14分）26．解：（1）直線y3x3與x軸交于點(diǎn)A，與y軸交于點(diǎn)C．，A(10，)C(0，3)··················································································1分點(diǎn)A，C都在拋物線上，23330aca3c33c18/21原創(chuàng)2023學(xué)年胡文323x3·············································3分3拋物線的解析式為yx23······················································································4分43頂點(diǎn)F1，3（2）存在··································································································5分····································································································7分P(0，3)1····································································································9分P(2，3)2（3）存在·······························································································10分理由：解法一：延長BC到點(diǎn)B，使BCBC，連接交直線AC于點(diǎn)M，則點(diǎn)M就是所求的點(diǎn)．BF··································································11分過點(diǎn)作于點(diǎn)H．BBHABy323x23B點(diǎn)在拋物線x3上，