PAGE專題02函數(shù)與幾何綜合題（填空壓軸題）通用的解題思路：函數(shù)與幾何綜合題通常涉及將函數(shù)的性質(zhì)與幾何圖形相結(jié)合1.理解題意：首先，仔細(xì)閱讀題目，確保理解題目中的每一個(gè)條件和要求。確定題目中涉及的是哪種類型的函數(shù)（反比例、二次、三角函數(shù)等）和哪種幾何圖形（三角形、圓、四邊形等）。2.建立函數(shù)關(guān)系：根據(jù)題目條件，嘗試建立幾何圖形與函數(shù)之間的關(guān)系。例如，如果題目涉及到一個(gè)拋物線與直線或另一個(gè)拋物線的交點(diǎn)，那么可能需要設(shè)置等式來(lái)找到這些交點(diǎn)。3.利用函數(shù)性質(zhì)：利用函數(shù)的性質(zhì)來(lái)簡(jiǎn)化問(wèn)題。4.幾何分析：使用幾何知識(shí)來(lái)分析函數(shù)圖像的性質(zhì)。例如，分析角度、長(zhǎng)度、面積等幾何量，并嘗試將它們與函數(shù)的性質(zhì)聯(lián)系起來(lái)。5.代數(shù)運(yùn)算：進(jìn)行必要的代數(shù)運(yùn)算，如解方程、不等式等。代數(shù)運(yùn)算可以幫助找到函數(shù)與幾何圖形之間的具體關(guān)系。6.驗(yàn)證答案：檢查答案是否符合題目中的所有條件。如果可能，使用圖形計(jì)算器或繪圖軟件來(lái)驗(yàn)證答案的正確性。7.總結(jié)與反思：總結(jié)解題過(guò)程中的關(guān)鍵步驟和使用的技巧。反思是否有其他方法可以解決這個(gè)問(wèn)題，以及這些方法之間的優(yōu)劣。1．（2023·江蘇鹽城·中考真題）如圖，在平面直角坐標(biāo)系SKIPIF1<0中，點(diǎn)SKIPIF1<0，SKIPIF1<0都在反比例函數(shù)SKIPIF1<0的圖象上，延長(zhǎng)SKIPIF1<0交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0軸于點(diǎn)SKIPIF1<0，連接SKIPIF1<0并延長(zhǎng)，交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，連接SKIPIF1<0.若SKIPIF1<0，SKIPIF1<0的面積是SKIPIF1<0，則SKIPIF1<0的值為.

【答案】6【分析】過(guò)點(diǎn)B作SKIPIF1<0于點(diǎn)F，連接SKIPIF1<0，設(shè)點(diǎn)A的坐標(biāo)為SKIPIF1<0，點(diǎn)B的坐標(biāo)為SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，證明SKIPIF1<0，則SKIPIF1<0，得到SKIPIF1<0，根據(jù)SKIPIF1<0，進(jìn)一步列式即可求出k的值.【詳解】解：過(guò)點(diǎn)B作SKIPIF1<0于點(diǎn)F，連接SKIPIF1<0，設(shè)點(diǎn)A的坐標(biāo)為SKIPIF1<0，點(diǎn)B的坐標(biāo)為SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，

∵SKIPIF1<0軸于點(diǎn)SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0的面積是SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，故答案為：6【點(diǎn)睛】此題考查反比例函數(shù)的圖象和性質(zhì)、相似三角形的判定和性質(zhì)等知識(shí)，求出SKIPIF1<0是解題的關(guān)鍵.2．（2023·江蘇無(wú)錫·中考真題）二次函數(shù)SKIPIF1<0的圖像與x軸交于點(diǎn)SKIPIF1<0、SKIPIF1<0，與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0的直線將SKIPIF1<0分成兩部分，這兩部分是三角形或梯形，且面積相等，則SKIPIF1<0的值為．【答案】SKIPIF1<0或SKIPIF1<0或SKIPIF1<0【分析】先求得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，直線SKIPIF1<0解析式為SKIPIF1<0，直線SKIPIF1<0的解析式為SKIPIF1<0，1）、當(dāng)分成兩個(gè)三角形時(shí)，直線必過(guò)三角形一個(gè)頂點(diǎn)，平分面積，必為中線，則①如圖1，直線SKIPIF1<0過(guò)SKIPIF1<0中點(diǎn)，②如圖2，直線SKIPIF1<0過(guò)SKIPIF1<0中點(diǎn)，直線SKIPIF1<0解析式為SKIPIF1<0，SKIPIF1<0中點(diǎn)坐標(biāo)為SKIPIF1<0，待入直線求得SKIPIF1<0；③如圖3，直線SKIPIF1<0過(guò)SKIPIF1<0中點(diǎn)，SKIPIF1<0中點(diǎn)坐標(biāo)為SKIPIF1<0，直線SKIPIF1<0與SKIPIF1<0軸平行，必不成立；2）當(dāng)分成三角形和梯形時(shí)，過(guò)點(diǎn)SKIPIF1<0的直線必與SKIPIF1<0一邊平行，所以必有SKIPIF1<0型相似，因?yàn)槠椒置娣e，所以相似比為SKIPIF1<0．④如圖4，直線SKIPIF1<0SKIPIF1<0SKIPIF1<0，根據(jù)相似三角形的性質(zhì)，即可求解；⑤如圖5，直線SKIPIF1<0SKIPIF1<0SKIPIF1<0，⑥如圖6，直線SKIPIF1<0SKIPIF1<0SKIPIF1<0，同理可得SKIPIF1<0，進(jìn)而根據(jù)SKIPIF1<0，即可求解．【詳解】解：由SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，設(shè)直線SKIPIF1<0解析式為SKIPIF1<0，∴SKIPIF1<0解得：SKIPIF1<0∴直線SKIPIF1<0解析式為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則直線SKIPIF1<0與y軸交于SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴點(diǎn)SKIPIF1<0必在SKIPIF1<0內(nèi)部．1）、當(dāng)分成兩個(gè)三角形時(shí)，直線必過(guò)三角形一個(gè)頂點(diǎn)，平分面積，必為中線設(shè)直線SKIPIF1<0的解析式為SKIPIF1<0∴SKIPIF1<0解得：SKIPIF1<0則直線SKIPIF1<0的解析式為SKIPIF1<0①如圖1，直線SKIPIF1<0過(guò)SKIPIF1<0中點(diǎn)，，SKIPIF1<0中點(diǎn)坐標(biāo)為SKIPIF1<0，代入直線求得SKIPIF1<0，不成立；

②如圖2，直線SKIPIF1<0過(guò)SKIPIF1<0中點(diǎn)，直線SKIPIF1<0解析式為SKIPIF1<0，SKIPIF1<0中點(diǎn)坐標(biāo)為SKIPIF1<0，待入直線求得SKIPIF1<0；③如圖3，直線SKIPIF1<0過(guò)SKIPIF1<0中點(diǎn)，SKIPIF1<0中點(diǎn)坐標(biāo)為SKIPIF1<0，SKIPIF1<0直線SKIPIF1<0與SKIPIF1<0軸平行，必不成立；2）、當(dāng)分成三角形和梯形時(shí)，過(guò)點(diǎn)SKIPIF1<0的直線必與SKIPIF1<0一邊平行，所以必有SKIPIF1<0型相似，因?yàn)槠椒置娣e，所以相似比為SKIPIF1<0．④如圖4，直線SKIPIF1<0SKIPIF1<0SKIPIF1<0，∴SKIPIF1<0∴SKIPIF1<0，∴SKIPIF1<0，解得SKIPIF1<0；

⑤如圖5，直線SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0∴SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴不成立；⑥如圖6，直線SKIPIF1<0SKIPIF1<0SKIPIF1<0，同理可得SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，解得SKIPIF1<0；綜上所述，SKIPIF1<0或SKIPIF1<0或SKIPIF1<0．【點(diǎn)睛】本題考查了二次函數(shù)的綜合問(wèn)題，解直角三角形，相似三角形的性質(zhì)與判定，熟練掌握以上知識(shí)，并分類討論是解題的關(guān)鍵．3．（2023·江蘇徐州·中考真題）如圖，點(diǎn)SKIPIF1<0在反比例函數(shù)SKIPIF1<0的圖象上，SKIPIF1<0軸于點(diǎn)SKIPIF1<0軸于點(diǎn)SKIPIF1<0．一次函數(shù)SKIPIF1<0與SKIPIF1<0交于點(diǎn)SKIPIF1<0，若SKIPIF1<0為SKIPIF1<0的中點(diǎn)，則SKIPIF1<0的值為．

【答案】4【分析】根據(jù)題意可設(shè)點(diǎn)P的坐標(biāo)為SKIPIF1<0，則SKIPIF1<0，把SKIPIF1<0代入一次函數(shù)解析式中求出m的值進(jìn)而求出點(diǎn)P的坐標(biāo)，再求出k的值即可．【詳解】解：∵SKIPIF1<0軸于點(diǎn)SKIPIF1<0軸于點(diǎn)SKIPIF1<0，∴點(diǎn)P的橫縱坐標(biāo)相同，∴可設(shè)點(diǎn)P的坐標(biāo)為SKIPIF1<0，∵SKIPIF1<0為SKIPIF1<0的中點(diǎn)，∴SKIPIF1<0，∵SKIPIF1<0在直線SKIPIF1<0上，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∵點(diǎn)SKIPIF1<0在反比例函數(shù)SKIPIF1<0的圖象上，∴SKIPIF1<0，故答案為：4．【點(diǎn)睛】本題主要考查了一次函數(shù)與反比例函數(shù)綜合，正確求出點(diǎn)P的坐標(biāo)是解題的關(guān)鍵．4．（2023·江蘇連云港·中考真題）如圖，矩形SKIPIF1<0的頂點(diǎn)SKIPIF1<0在反比例函數(shù)SKIPIF1<0的圖像上，頂點(diǎn)SKIPIF1<0在第一象限，對(duì)角線SKIPIF1<0軸，交SKIPIF1<0軸于點(diǎn)SKIPIF1<0．若矩形SKIPIF1<0的面積是6，SKIPIF1<0，則SKIPIF1<0．

【答案】SKIPIF1<0【分析】方法一：根據(jù)SKIPIF1<0的面積為SKIPIF1<0，得出SKIPIF1<0，SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，得出SKIPIF1<0，根據(jù)勾股定理求得SKIPIF1<0，根據(jù)SKIPIF1<0的幾何意義，即可求解．方法二：根據(jù)已知得出SKIPIF1<0則SKIPIF1<0，即可求解．【詳解】解：方法一：∵SKIPIF1<0，∴SKIPIF1<0設(shè)SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0∵矩形SKIPIF1<0的面積是6，SKIPIF1<0是對(duì)角線，∴SKIPIF1<0的面積為SKIPIF1<0，即SKIPIF1<0∴SKIPIF1<0在SKIPIF1<0中，SKIPIF1<0即SKIPIF1<0即SKIPIF1<0解得：SKIPIF1<0在SKIPIF1<0中，SKIPIF1<0∵對(duì)角線SKIPIF1<0軸，則SKIPIF1<0，∴SKIPIF1<0,∵反比例函數(shù)圖象在第二象限，∴SKIPIF1<0，方法二：∵SKIPIF1<0，∴SKIPIF1<0設(shè)SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，故答案為：SKIPIF1<0．【點(diǎn)睛】本題考查了矩形的性質(zhì)，反比例函數(shù)SKIPIF1<0的幾何意義，余弦的定義，熟練掌握反比例函數(shù)的性質(zhì)是解題的關(guān)鍵．5．（2022·江蘇鹽城·中考真題）《莊子?天下篇》記載“一尺之錘，日取其半，萬(wàn)世不竭．”如圖，直線SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0軸的平行線交直線SKIPIF1<0于點(diǎn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0軸的平行線交直線SKIPIF1<0于點(diǎn)SKIPIF1<0，以此類推，令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0對(duì)任意大于1的整數(shù)SKIPIF1<0恒成立，則SKIPIF1<0的最小值為．【答案】2【分析】先由直線SKIPIF1<0與SKIPIF1<0軸的夾角是45°，得出SKIPIF1<0，SKIPIF1<0，…都是等腰直角三角形，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，…，得出點(diǎn)SKIPIF1<0的橫坐標(biāo)為1，得到當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，SKIPIF1<0，點(diǎn)SKIPIF1<0的橫坐標(biāo)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，得出點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，以此類推，最后得出結(jié)果．【詳解】解：SKIPIF1<0直線SKIPIF1<0與SKIPIF1<0軸的夾角是45°，SKIPIF1<0，SKIPIF1<0，…都是等腰直角三角形，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，…SKIPIF1<0點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，SKIPIF1<0點(diǎn)SKIPIF1<0的橫坐標(biāo)為1，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0點(diǎn)SKIPIF1<0的橫坐標(biāo)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，SKIPIF1<0，……以此類推，得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，……，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的最小值為2．【點(diǎn)睛】本題考查了此題考查一次函數(shù)圖象上的點(diǎn)的坐標(biāo)特征，探究以幾何圖形為背景的問(wèn)題時(shí)，一是要破解幾何圖形之間的關(guān)系，二是實(shí)現(xiàn)線段長(zhǎng)度和點(diǎn)的坐標(biāo)的正確轉(zhuǎn)換，三是觀察分析所得數(shù)據(jù)并找出數(shù)據(jù)之間的規(guī)律．1．如圖，點(diǎn)SKIPIF1<0在反比例函數(shù)SKIPIF1<0的圖像上，點(diǎn)SKIPIF1<0在反比例函數(shù)SKIPIF1<0的圖像上，SKIPIF1<0，連結(jié)SKIPIF1<0交SKIPIF1<0的圖像于點(diǎn)SKIPIF1<0，若SKIPIF1<0是SKIPIF1<0的中點(diǎn)，則SKIPIF1<0的面積是．【答案】SKIPIF1<0【分析】如圖所示，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0軸于點(diǎn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0軸于點(diǎn)SKIPIF1<0，可證SKIPIF1<0，根據(jù)相似三角形的性質(zhì)，反比例系數(shù)與幾何圖形面積的計(jì)算方法可得SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，根據(jù)點(diǎn)SKIPIF1<0是中點(diǎn)，且在反比例函數(shù)SKIPIF1<0的圖象上，可得SKIPIF1<0，由此即可求解．【詳解】解：如圖所示，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0軸于點(diǎn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0軸于點(diǎn)SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，∵點(diǎn)SKIPIF1<0是中點(diǎn)，且在反比例函數(shù)SKIPIF1<0的圖象上，∴SKIPIF1<0，∴SKIPIF1<0，整理得，SKIPIF1<0，∴SKIPIF1<0，故答案為：SKIPIF1<0．【點(diǎn)睛】本題主要考查反比例函數(shù)與幾何圖形的綜合，掌握反比例函數(shù)圖象的性質(zhì)，相似三角形的判定和性質(zhì)，乘法公式的運(yùn)用，中點(diǎn)坐標(biāo)的計(jì)算方法，掌握反比例函數(shù)的性質(zhì)，相似三角形的判定和性質(zhì)是解題的關(guān)鍵．2．如圖，在平面直角坐標(biāo)系xOy中，SKIPIF1<0的頂點(diǎn)C在x軸負(fù)半軸上，SKIPIF1<0軸，點(diǎn)B在反比例函數(shù)SKIPIF1<0的圖象上，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的值為，k的值為【答案】3SKIPIF1<0【分析】本題考查了反比例函數(shù)圖象上點(diǎn)的坐標(biāo)特征：反比例函數(shù)SKIPIF1<0為常數(shù)，SKIPIF1<0的圖象是雙曲線，圖象上的點(diǎn)SKIPIF1<0的橫縱坐標(biāo)的積是定值SKIPIF1<0，即SKIPIF1<0．設(shè)SKIPIF1<0，SKIPIF1<0，利用含30度的直角三角形三邊的關(guān)系得到SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，根據(jù)反比例函數(shù)圖象上點(diǎn)的坐標(biāo)特征得SKIPIF1<0，再利用SKIPIF1<0得到SKIPIF1<0，所以SKIPIF1<0．【詳解】解：SKIPIF1<0SKIPIF1<0軸，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0和SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0點(diǎn)SKIPIF1<0在反比例函數(shù)SKIPIF1<0的圖象上，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0．故答案為：3，SKIPIF1<0．3．如圖，在平面直角坐標(biāo)系中，等腰直角三角形SKIPIF1<0的斜邊SKIPIF1<0軸于點(diǎn)B，直角頂點(diǎn)A在y軸上，雙曲線SKIPIF1<0SKIPIF1<0經(jīng)過(guò)SKIPIF1<0邊的中點(diǎn)D，若SKIPIF1<0，則SKIPIF1<0．【答案】SKIPIF1<0【分析】本題考查了反比例函數(shù)的性質(zhì)，解題的關(guān)鍵是熟練掌握反比例函數(shù)圖象上點(diǎn)的特征，過(guò)點(diǎn)A作SKIPIF1<0于E，根據(jù)直角三角形斜邊的中線等于斜邊的一半可求得SKIPIF1<0，即可得出點(diǎn)A和點(diǎn)C的坐標(biāo)，再根據(jù)中點(diǎn)坐標(biāo)公式即可求出點(diǎn)是D的坐標(biāo)，從而可得結(jié)論，【詳解】解：如圖，過(guò)點(diǎn)A作SKIPIF1<0于E，∵等腰直角三角形SKIPIF1<0的斜邊SKIPIF1<0軸于點(diǎn)B，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∵D是SKIPIF1<0的中點(diǎn)，∴SKIPIF1<0，∴SKIPIF1<0．故答案為：SKIPIF1<0．4．如圖，點(diǎn)A，C在雙曲線SKIPIF1<0上，點(diǎn)B，D在雙曲線SKIPIF1<0上，SKIPIF1<0軸，且四邊形SKIPIF1<0是平行四邊形，則SKIPIF1<0的面積為．【答案】8【分析】本題考查了已知比例系數(shù)求特殊圖形的面積，由平行于y軸的直線上的點(diǎn)橫坐標(biāo)相等，設(shè)出點(diǎn)的坐標(biāo)，再根據(jù)平行四邊形面積公式求解即可．【詳解】解：設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0軸，四邊形SKIPIF1<0是平行四邊形，SKIPIF1<0軸，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0邊上的高SKIPIF1<0，SKIPIF1<0SKIPIF1<0的面積SKIPIF1<0，故答案為：8．5．如圖，矩形SKIPIF1<0的頂點(diǎn)SKIPIF1<0在反比例函數(shù)SKIPIF1<0的圖象上，頂點(diǎn)SKIPIF1<0、SKIPIF1<0在第一象限，對(duì)角線SKIPIF1<0軸，交SKIPIF1<0軸于點(diǎn)SKIPIF1<0．若矩形SKIPIF1<0的面積是16，SKIPIF1<0，則SKIPIF1<0．【答案】SKIPIF1<0【分析】本題考查了反比例函數(shù)SKIPIF1<0值的幾何意義，熟練掌握反比例函數(shù)SKIPIF1<0值的幾何意義是關(guān)鍵．根據(jù)反比例函數(shù)SKIPIF1<0值的幾何意義計(jì)算出三角形SKIPIF1<0的面積即可．【詳解】解：SKIPIF1<0矩形SKIPIF1<0的面積是16，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0軸，∴SKIPIF1<0，∵矩形SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，又∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，反比例函數(shù)圖象在第二象限，SKIPIF1<0．故答案為：SKIPIF1<0．6．如圖，在直角SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，將SKIPIF1<0繞點(diǎn)O順時(shí)針旋轉(zhuǎn)SKIPIF1<0至SKIPIF1<0的位置，點(diǎn)E是SKIPIF1<0的中點(diǎn)，且點(diǎn)E在反比例函數(shù)SKIPIF1<0的圖象上，則k的值為．【答案】SKIPIF1<0/SKIPIF1<0【分析】本題考查求反比例函數(shù)的解析式，旋轉(zhuǎn)的性質(zhì)，解直角三角形，特殊角的三角函數(shù)，掌握相關(guān)的知識(shí)是解題的關(guān)鍵．作SKIPIF1<0軸于點(diǎn)SKIPIF1<0，根據(jù)SKIPIF1<0可得SKIPIF1<0，根據(jù)旋轉(zhuǎn)的性質(zhì)可得SKIPIF1<0，進(jìn)一步算出SKIPIF1<0，根據(jù)勾股定理和中點(diǎn)定義可得SKIPIF1<0，再解直角三角形求出點(diǎn)SKIPIF1<0的坐標(biāo)，代入反比例函數(shù)解析式即可求解．【詳解】作SKIPIF1<0軸于點(diǎn)SKIPIF1<0，如圖所示，∵在直角SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∵將SKIPIF1<0繞點(diǎn)O順時(shí)針旋轉(zhuǎn)SKIPIF1<0至SKIPIF1<0的位置，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∵點(diǎn)E是SKIPIF1<0的中點(diǎn)，∴SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，∴點(diǎn)SKIPIF1<0的坐標(biāo)是SKIPIF1<0，∵點(diǎn)SKIPIF1<0在反比例函數(shù)SKIPIF1<0的圖象上，∴SKIPIF1<0故答案為：SKIPIF1<0．7．已知，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，……都是邊長(zhǎng)為2的等邊三角形，按下圖所示擺放．點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，……都在SKIPIF1<0軸正半軸上，且SKIPIF1<0，則點(diǎn)SKIPIF1<0的坐標(biāo)是．【答案】SKIPIF1<0【分析】本題考查正三角形的性質(zhì)以及點(diǎn)的坐標(biāo)的規(guī)律性，掌握正三角形的性質(zhì)和點(diǎn)的坐標(biāo)的變化規(guī)律是解決問(wèn)題的關(guān)鍵．根據(jù)正三角形的性質(zhì)以及三角形的排列規(guī)律可得點(diǎn)SKIPIF1<0橫坐標(biāo)為1，點(diǎn)SKIPIF1<0橫坐標(biāo)為2，點(diǎn)SKIPIF1<0橫坐標(biāo)為3，點(diǎn)SKIPIF1<0橫坐標(biāo)為4，SKIPIF1<0因此點(diǎn)SKIPIF1<0橫坐標(biāo)為2024，再根據(jù)這些正三角形的排列規(guī)律得出點(diǎn)SKIPIF1<0在x軸上，進(jìn)而得出答案．【詳解】解：如圖，過(guò)點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0分別作SKIPIF1<0軸的垂線，SKIPIF1<0SKIPIF1<0是邊長(zhǎng)為2正三角形，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0點(diǎn)SKIPIF1<0橫坐標(biāo)為1，由題意可得，點(diǎn)SKIPIF1<0橫坐標(biāo)為2，點(diǎn)SKIPIF1<0橫坐標(biāo)為3，點(diǎn)SKIPIF1<0橫坐標(biāo)為4，SKIPIF1<0因此點(diǎn)SKIPIF1<0橫坐標(biāo)為2024，∵SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；分布在第一、四象限，其余的分布在x軸上，所以每隔六個(gè)作為一循環(huán)，SKIPIF1<0，SKIPIF1<0點(diǎn)SKIPIF1<0在x軸上，∴點(diǎn)SKIPIF1<0，故答案為：SKIPIF1<0．8．如圖，已知點(diǎn)SKIPIF1<0，點(diǎn)B為直線SKIPIF1<0上的一動(dòng)點(diǎn)，點(diǎn)SKIPIF1<0，SKIPIF1<0于點(diǎn)C，連接SKIPIF1<0．若直線SKIPIF1<0與x軸正半軸所夾的銳角為α，那么當(dāng)SKIPIF1<0的值最大時(shí)，n的值為．【答案】SKIPIF1<0/SKIPIF1<0【分析】本題考查了二次函數(shù)的性質(zhì)，解直角三角形等，當(dāng)SKIPIF1<0的值最大時(shí)，則SKIPIF1<0值最大，即當(dāng)SKIPIF1<0最大時(shí)，SKIPIF1<0的值最大，設(shè)SKIPIF1<0，由SKIPIF1<0，得到SKIPIF1<0，進(jìn)而求解．【詳解】解：過(guò)點(diǎn)A作SKIPIF1<0軸于點(diǎn)M，作SKIPIF1<0交于點(diǎn)N，∵直線SKIPIF1<0與x軸平行，∴SKIPIF1<0，當(dāng)SKIPIF1<0的值最大時(shí)，則SKIPIF1<0值最大，故SKIPIF1<0最小，即SKIPIF1<0最大時(shí)，SKIPIF1<0最大，即當(dāng)SKIPIF1<0最大時(shí)，SKIPIF1<0的值最大，設(shè)SKIPIF1<0，則SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，m取得最大值，故SKIPIF1<0，故答案為：SKIPIF1<0．9．如圖，在矩形SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0．分別以SKIPIF1<0，SKIPIF1<0所在直線為SKIPIF1<0軸、SKIPIF1<0軸建立如圖所示的平面直角坐標(biāo)系．SKIPIF1<0為SKIPIF1<0邊上的一個(gè)動(dòng)點(diǎn)（不與SKIPIF1<0，SKIPIF1<0重合），過(guò)點(diǎn)SKIPIF1<0的反比例函數(shù)SKIPIF1<0的圖像與邊SKIPIF1<0交于點(diǎn)SKIPIF1<0，連接SKIPIF1<0．（1）SKIPIF1<0；（2）將SKIPIF1<0沿SKIPIF1<0折疊，點(diǎn)SKIPIF1<0恰好落在邊SKIPIF1<0上的點(diǎn)SKIPIF1<0處，此時(shí)SKIPIF1<0的值為．【答案】2SKIPIF1<0/SKIPIF1<0/6.75【分析】（1）首先根據(jù)矩形的性質(zhì)可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，結(jié)合題意確定點(diǎn)SKIPIF1<0的坐標(biāo)，進(jìn)而可得SKIPIF1<0，SKIPIF1<0，然后根據(jù)SKIPIF1<0求解即可；（2）過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0于SKIPIF1<0，根據(jù)折疊的性質(zhì)可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，證明SKIPIF1<0，由相似三角形的性質(zhì)可解得SKIPIF1<0，在SKIPIF1<0中，由勾股定理可得SKIPIF1<0，代入并解得SKIPIF1<0的值即可．【詳解】解：（1）∵四邊形SKIPIF1<0為矩形，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0為SKIPIF1<0邊上的一點(diǎn)，過(guò)點(diǎn)SKIPIF1<0的反比例函數(shù)SKIPIF1<0的圖像與邊SKIPIF1<0交于點(diǎn)SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0；（2）由（1）可知，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，如下圖，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0于SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，由折疊知，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．故答案為：2；SKIPIF1<0