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1．在實驗室中，通常將金屬鈉保存在

A．水中　　　 B. 煤油中　　　　　 C. 四氯化碳中　　 D.汽油中

102.(08廣東佛山25題)25．我們所學(xué)的幾何知識可以理解為對“構(gòu)圖”的研究：根據(jù)給定的(或構(gòu)造的)幾何圖形提出相關(guān)的概念和問題(或者根據(jù)問題構(gòu)造圖形)，并加以研究.

例如：在平面上根據(jù)兩條直線的各種構(gòu)圖，可以提出“兩條直線平行”、“兩條直線相交”的概念；若增加第三條直線，則可以提出并研究“兩條直線平行的判定和性質(zhì)”等問題(包括研究的思想和方法).

請你用上面的思想和方法對下面關(guān)于圓的問題進行研究：

(1) 如圖1，在圓O所在平面上，放置一條直線(和圓O分別交于點A、B)，根據(jù)這個圖形可以提出的概念或問題有哪些(直接寫出兩個即可)？

(2) 如圖2，在圓O所在平面上，請你放置與圓O都相交且不同時經(jīng)過圓心的兩條直線和(與圓O分別交于點A、B，與圓O分別交于點C、D).

請你根據(jù)所構(gòu)造的圖形提出一個結(jié)論，并證明之.

(3) 如圖3，其中AB是圓O的直徑，AC是弦，D是的中點，弦DE⊥AB于點F. 請找出點C和點E重合的條件，并說明理由.

(08廣東佛山25題解答)解：(1) 弦(圖中線段AB)、弧(圖中的ACB弧)、弓形、求弓形的面積(因為是封閉圖形)等.　 (寫對一個給1分，寫對兩個給2分)

(2) 情形1　 如圖21，AB為弦，CD為垂直于弦AB的直徑.　 …………………………3分

結(jié)論：(垂徑定理的結(jié)論之一).　 …………………………………………………………4分

證明：略(對照課本的證明過程給分).　 …………………………………………………7分

情形2　 如圖22，AB為弦，CD為弦，且AB與CD在圓內(nèi)相交于點P.

結(jié)論：.

證明：略.

情形3 (圖略)AB為弦，CD為弦，且與在圓外相交于點P.

結(jié)論：.

證明：略.

情形4　 如圖23，AB為弦，CD為弦，且AB∥CD.

結(jié)論：　 =　　 .

證明：略.

(上面四種情形中做一個即可，圖1分，結(jié)論1分，證明3分；

其它正確的情形參照給分；若提出的是錯誤的結(jié)論，則需證明結(jié)論是錯誤的)

(3) 若點C和點E重合，

則由圓的對稱性，知點C和點D關(guān)于直徑AB對稱. …………………………………8分

設(shè)，則，.………………………………9分

又D是　　 的中點，所以，

即.………………………………………………………10分

解得.……………………………………………………………11分

(若求得或等也可，評分可參照上面的標(biāo)準(zhǔn)；也可以先直覺猜測點B、C是圓的十二等分點，然后說明)

101.(08山東聊城25題)25．(本題滿分12分)如圖，把一張長10cm，寬8cm的矩形硬紙板的四周各剪去一個同樣大小的正方形，再折合成一個無蓋的長方體盒子(紙板的厚度忽略不計)．

(1)要使長方體盒子的底面積為48cm²，那么剪去的正方形的邊長為多少？

(2)你感到折合而成的長方體盒子的側(cè)面積會不會有更大的情況？如果有，請你求出最大值和此時剪去的正方形的邊長；如果沒有，請你說明理由；

(3)如果把矩形硬紙板的四周分別剪去2個同樣大小的正方形和2個同樣形狀、同樣大小的矩形，然后折合成一個有蓋的長方體盒子，是否有側(cè)面積最大的情況；如果有，請你求出最大值和此時剪去的正方形的邊長；如果沒有，請你說明理由．

(08山東聊城25題解答)(本題滿分12分)

解：(1)設(shè)正方形的邊長為cm，則

．························································································ 1分

即．

解得(不合題意，舍去)，．

剪去的正方形的邊長為1cm．·············································································· 3分

(注：通過觀察、驗證直接寫出正確結(jié)果給3分)

(2)有側(cè)面積最大的情況．

設(shè)正方形的邊長為cm，盒子的側(cè)面積為cm²，

則與的函數(shù)關(guān)系式為：

．

即．······························································································· 5分

改寫為．

當(dāng)時，．

即當(dāng)剪去的正方形的邊長為2.25cm時，長方體盒子的側(cè)面積最大為40.5cm²．········ 7分

(3)有側(cè)面積最大的情況．

設(shè)正方形的邊長為cm，盒子的側(cè)面積為cm²．

若按圖1所示的方法剪折，則與的函數(shù)關(guān)系式為：

．

即．

當(dāng)時，．····························· 9分

若按圖2所示的方法剪折，則與的函數(shù)關(guān)系式為：

．

即．

當(dāng)時，．················································································· 11分

比較以上兩種剪折方法可以看出，按圖2所示的方法剪折得到的盒子側(cè)面積最大，即當(dāng)剪去的正方形的邊長為cm時，折成的有蓋長方體盒子的側(cè)面積最大，最大面積為cm²．

說明：解答題各小題只給了一種解答及評分說明，其他解法只要步驟合理，解答正確，均應(yīng)給出相應(yīng)分?jǐn)?shù)．

100.(08廣東梅州23題)23．本題滿分11分．

如圖11所示，在梯形ABCD中，已知AB∥CD， AD⊥DB，AD=DC=CB，AB=4．以AB所在直線為軸，過D且垂直于AB的直線為軸建立平面直角坐標(biāo)系．

(1)求∠DAB的度數(shù)及A、D、C三點的坐標(biāo)；

(2)求過A、D、C三點的拋物線的解析式及其對稱軸L．

(3)若P是拋物線的對稱軸L上的點，那么使PDB為等腰三角形的點P有幾個?(不必求點P的坐標(biāo)，只需說明理由)

(08廣東梅州23題解答)解： (1) DC∥AB，AD=DC=CB，

　∠CDB=∠CBD=∠DBA，··············································································· 0.5分

　　 ∠DAB=∠CBA， ∠DAB=2∠DBA， ············ 1分

∠DAB+∠DBA=90， ∠DAB=60， ·········· 1.5分

　 ∠DBA=30，AB=4， DC=AD=2，　 ········· 2分

RtAOD，OA=1，OD=，··························· 2.5分

A(-1，0)，D(0， )，C(2， )．　 · 4分

(2)根據(jù)拋物線和等腰梯形的對稱性知，滿足條件的拋物線必過點A(－1，0)，B(3，0)，

故可設(shè)所求為　 = (+1)( -3) ······························································ 6分

將點D(0， )的坐標(biāo)代入上式得， =．

所求拋物線的解析式為　 =　　 ·········································· 7分

其對稱軸L為直線=1．······················································································ 8分

(3) PDB為等腰三角形，有以下三種情況：

①因直線L與DB不平行，DB的垂直平分線與L僅有一個交點P₁，P₁D=P₁B，

P₁DB為等腰三角形；　 ················································································· 9分

②因為以D為圓心，DB為半徑的圓與直線L有兩個交點P₂、P₃，DB=DP₂，DB=DP₃， P₂DB， P₃DB為等腰三角形；

③與②同理，L上也有兩個點P₄、P₅，使得 BD=BP₄，BD=BP₅．　 ···················· 10分

由于以上各點互不重合，所以在直線L上，使PDB為等腰三角形的點P有5個．

99.(08福建南平26題)26．(14分)

(1)如圖1，圖2，圖3，在中，分別以為邊，向外作正三角形，正四邊形，正五邊形，相交于點．

①如圖1，求證：；

②探究：如圖1，　　　　 ；

如圖2，　　　　 ；

如圖3，　　　　 ．

(2)如圖4，已知：是以為邊向外所作正邊形的一組鄰邊；是以為邊向外所作正邊形的一組鄰邊．的延長相交于點．

①猜想：如圖4，　　　　 (用含的式子表示)；

②根據(jù)圖4證明你的猜想．

(08福建南平26題解答)(1)①證法一：與均為等邊三角形，

，························································································ 2分

且··············································· 3分

，

即························································ 4分

．··················································· 5分

證法二：與均為等邊三角形，

，························································································ 2分

且························································································ 3分

可由繞著點按順時針方向旋轉(zhuǎn)得到··································· 4分

．··························································································· 5分

②，，．········································································ 8分(每空1分)

(2)①········································································································ 10分

②證法一：依題意，知和都是正邊形的內(nèi)角，，，

，即．····························· 11分

．·························································································· 12分

，，······ 13分

，

········································ 14分

證法二：同上可證　 ．··························································· 12分

，如圖，延長交于，

，

································ 13分

················· 14分

證法三：同上可證　 ．··························································· 12分

．

，

························································ 13分

即········································································ 14分

證法四：同上可證　 ．··························································· 12分

．如圖，連接，

．···································· 13分

即······························· 14分

注意：此題還有其它證法，可相應(yīng)評分．

98.(08四川資陽24題)24．(本小題滿分12分)

如圖10，已知點A的坐標(biāo)是(－1，0)，點B的坐標(biāo)是(9，0)，以AB為直徑作⊙O′，交y軸的負(fù)半軸于點C，連接AC、BC，過A、B、C三點作拋物線．

(1)求拋物線的解析式；

(2)點E是AC延長線上一點，∠BCE的平分線CD交⊙O′于點D，連結(jié)BD，求直線BD的解析式；

(3)在(2)的條件下，拋物線上是否存在點P，使得∠PDB＝∠CBD?如果存在，請求出點P的坐標(biāo)；如果不存在，請說明理由．

(08四川資陽24題解答)(1) ∵以AB為直徑作⊙O′，交y軸的負(fù)半軸于點C，

∴∠OCA+∠OCB=90°，

又∵∠OCB+∠OBC=90°，

∴∠OCA=∠OBC，

又∵∠AOC= ∠COB=90°，

∴ΔAOC∽ ΔCOB，························································································ 1分

∴．

又∵A(–1，0)，B(9，0)，

∴，解得OC=3(負(fù)值舍去)．

∴C(0，–3)，

······················································································································ 3分

設(shè)拋物線解析式為y=a(x+1)(x–9)，

∴–3=a(0+1)(0–9)，解得a=，

∴二次函數(shù)的解析式為y=(x+1)(x–9)，即y=x²–x–3．···························· 4分

(2) ∵AB為O′的直徑，且A(–1，0)，B(9，0)，

∴OO′=4，O′(4，0)，····················································································· 5分

∵點E是AC延長線上一點，∠BCE的平分線CD交⊙O′于點D，

∴∠BCD=∠BCE=×90°=45°，

連結(jié)O′D交BC于點M，則∠BO′D=2∠BCD=2×45°=90°，OO′=4，O′D=AB=5．

∴D(4，–5)．································································································· 6分

∴設(shè)直線BD的解析式為y=kx+b(k≠0)

∴··························································· 7分

解得

∴直線BD的解析式為y=x–9.····································· 8分

(3) 假設(shè)在拋物線上存在點P，使得∠PDB=∠CBD，

解法一：設(shè)射線DP交⊙O′于點Q，則．

分兩種情況(如答案圖1所示)：

①∵O′(4，0)，D(4，–5)，B(9，0)，C(0，–3)．

∴把點C、D繞點O′逆時針旋轉(zhuǎn)90°，使點D與點B重合，則點C與點Q₁重合，

因此，點Q₁(7，–4)符合，

∵D(4，–5)，Q₁(7，–4)，

∴用待定系數(shù)法可求出直線DQ₁解析式為y=x–．··································· 9分

解方程組得

∴點P₁坐標(biāo)為(，)，[坐標(biāo)為(，)不符合題意，舍去]．

······················································································································ 10分

②∵Q₁(7，–4)，

∴點Q₁關(guān)于x軸對稱的點的坐標(biāo)為Q₂(7，4)也符合．

∵D(4，–5)，Q₂(7，4)．

∴用待定系數(shù)法可求出直線DQ₂解析式為y=3x–17．······································ 11分

解方程組得

∴點P₂坐標(biāo)為(14，25)，[坐標(biāo)為(3，–8)不符合題意，舍去]．

······················································································································ 12分

∴符合條件的點P有兩個：P₁(，)，P₂(14，25)．

解法二：分兩種情況(如答案圖2所示)：

①當(dāng)DP₁∥CB時，能使∠PDB=∠CBD．

∵B(9，0)，C(0，–3)．

∴用待定系數(shù)法可求出直線BC解析式為y=x–3．

又∵DP₁∥CB，∴設(shè)直線DP₁的解析式為y=x+n．

把D(4，–5)代入可求n= –，

∴直線DP₁解析式為y=x–．························· 9分

解方程組得

∴點P₁坐標(biāo)為(，)，[坐標(biāo)為(，)不符合題意，舍去]．

······················································································································ 10分

②在線段O′B上取一點N，使BN=DM時，得ΔNBD≌ΔMDB(SAS)，∴∠NDB=∠CBD．

由①知，直線BC解析式為y=x–3．

取x=4，得y= –，∴M(4，–)，∴O′N=O′M=，∴N(，0)，

又∵D(4，–5)，

∴直線DN解析式為y=3x–17．······································································ 11分

解方程組得

∴點P₂坐標(biāo)為(14，25)，[坐標(biāo)為(3，–8)不符合題意，舍去]．

······················································································································ 12分

∴符合條件的點P有兩個：P₁(，)，P₂(14，25)．

解法三：分兩種情況(如答案圖3所示)：

①求點P₁坐標(biāo)同解法二．··············································································· 10分

②過C點作BD的平行線,交圓O′于G,

此時，∠GDB=∠GCB=∠CBD．

由(2)題知直線BD的解析式為y=x–9,

又∵ C(0，–3)

∴可求得CG的解析式為y=x–3,

設(shè)G(m,m–3)，作GH⊥x軸交與x軸與H，

連結(jié)O′G,在Rt△O′GH中，利用勾股定理可得，m=7，

由D(4，–5)與G(7，4)可得，

DG的解析式為，··········································································· 11分

解方程組得

∴點P₂坐標(biāo)為(14，25)，[坐標(biāo)為(3，–8)不符合題意，舍去]．························ 12分

∴符合條件的點P有兩個：P₁(，)，P₂(14，25)．

說明：本題解法較多，如有不同的正確解法，請按此步驟給分.

97.(08新疆自治區(qū)24題)(10分)某工廠要趕制一批抗震救災(zāi)用的大型活動板房．如圖，板房一面的形狀是由矩形和拋物線的一部分組成，矩形長為12m，拋物線拱高為5.6m．

(1)在如圖所示的平面直角坐標(biāo)系中，求拋物線的表達式．

(2)現(xiàn)需在拋物線AOB的區(qū)域內(nèi)安裝幾扇窗戶，窗戶的底邊在AB上，每扇窗戶寬1.5m，高1.6m，相鄰窗戶之間的間距均為0.8m，左右兩邊窗戶的窗角所在的點到拋物線的水平距離至少為0.8m．請計算最多可安裝幾扇這樣的窗戶？

(08新疆自治區(qū)24題解析)24．(10分)解：(1)設(shè)拋物線的表達式為　 1分

點在拋物線的圖象上．

∴

······························································ 3分

∴拋物線的表達式為············································································· 4分

(2)設(shè)窗戶上邊所在直線交拋物線于C、D兩點，D點坐標(biāo)為(k，t)

已知窗戶高1.6m，∴··························································· 5分

(舍去)············································································ 6分

∴(m)·············································································· 7分

又設(shè)最多可安裝n扇窗戶

∴····················································································· 9分

．

答：最多可安裝4扇窗戶．···················································································· 10分

(本題不要求學(xué)生畫出4個表示窗戶的小矩形

96.(08四川自貢26題)拋物線的頂點為M，與軸的交點為A、B(點B在點A的右側(cè))，△ABM的三個內(nèi)角∠M、∠A、∠B所對的邊分別為m、a、b。若關(guān)于的一元二次方程有兩個相等的實數(shù)根。

(1)判斷△ABM的形狀，并說明理由。

(2)當(dāng)頂點M的坐標(biāo)為(－2，－1)時，求拋物線的解析式，并畫出該拋物線的大致圖形。

(3)若平行于軸的直線與拋物線交于C、D兩點，以CD為直徑的圓恰好與軸相切，求該圓的圓心坐標(biāo)。

(08四川自貢26題解析)解：(1)令

　　　　　　 得

　　　　 由勾股定理的逆定理和拋物線的對稱性知

△ABM是一個以、為直角邊的等腰直角三角形

　　　　 (2)設(shè)

∵△ABM是等腰直角三角形

∴斜邊上的中線等于斜邊的一半

又頂點M(－2，－1)

∴，即AB＝2

∴A(－3，0)，B(－1，0)

將B(－1，0) 代入中得

∴拋物線的解析式為，即

圖略

(3)設(shè)平行于軸的直線為

解方程組錯誤！不能通過編輯域代碼創(chuàng)建對象。

得，　 (

∴線段CD的長為

∵以CD為直徑的圓與軸相切

據(jù)題意得

∴

解得

∴圓心坐標(biāo)為和

95.(08四川巴中30題)(12分)30．已知：如圖14，拋物線與軸交于點，點，與直線相交于點，點，直線與軸交于點．

(1)寫出直線的解析式．

(2)求的面積．

(3)若點在線段上以每秒1個單位長度的速度從向運動(不與重合)，同時，點在射線上以每秒2個單位長度的速度從向運動．設(shè)運動時間為秒，請寫出的面積與的函數(shù)關(guān)系式，并求出點運動多少時間時，的面積最大，最大面積是多少？

(08四川巴中30題解析)解：(1)在中，令

，

，··············································· 1分

又點在上

的解析式為·············································································· 2分

(2)由，得　 ···················································· 4分

，

，······························································································· 5分

························································································· 6分

(3)過點作于點

······························································································· 7分

·········································································································· 8分

由直線可得：

在中，，，則

，······················································································· 9分

···················································································· 10分

····························································································· 11分

此拋物線開口向下，當(dāng)時，

當(dāng)點運動2秒時，的面積達到最大，最大為．···························· 12分

94.(08山東濟寧26題)(12分)

中，，，cm．長為1cm的線段在的邊上沿方向以1cm/s的速度向點運動(運動前點與點重合)．過分別作的垂線交直角邊于兩點，線段運動的時間為s．

(1)若的面積為，寫出與的函數(shù)關(guān)系式(寫出自變量的取值范圍)；

(2)線段運動過程中，四邊形有可能成為矩形嗎？若有可能，求出此時的值；若不可能，說明理由；

(3)為何值時，以為頂點的三角形與相似？

(08山東濟寧26題解析)解：(1)當(dāng)點在上時，，．

．········································································ 2分

當(dāng)點在上時，．

．·················································· 4分

(2)，．．

．········································································ 6分

由條件知，若四邊形為矩形，需，即，

．

當(dāng)s時，四邊形為矩形．································································· 8分

(3)由(2)知，當(dāng)s時，四邊形為矩形，此時，

．··························································································· 9分

除此之外，當(dāng)時，，此時．

，．．····························· 10分

，．

又，．········································ 11分

，．

當(dāng)s或s時，以為頂點的三角形與相似．··················· 12分