《2025年通城学典非常课课通九年级数学下册苏科版江苏专版》

8. (2024·宜宾)如图，正五边形ABCDE的边长为4，则这个正五边形的对角线AC的长是______.

答案： $2\sqrt{5}+2$

9. (2024·连云港)如图，将一张矩形纸片ABCD上下对折，使之完全重合，打开后，得到折痕EF，连接BF.再将矩形纸片折叠，使点B落在BF上的点H处，折痕为AG.若G恰好为线段BC最靠近点B的一个五等分点，AB = 4，则BC的长为______.

答案： $2\sqrt{10}$ 解析：设$AG$与$BF$交于点$M$，$BG = a$，则$BC = 5a$.$\because$四边形$ABCD$是矩形，$\therefore\angle ABC=\angle C = 90^{\circ}$，$AB = CD = 4$.$\because$上下对折后得到折痕$EF$，$\therefore CF=\frac{1}{2}CD = 2$.$\because AG$是折痕，点$B$的对应点为$H$，$\therefore AG\perp BH$.$\therefore\angle BMG = 90^{\circ}$.$\therefore\angle CBF+\angle BGA = 90^{\circ}$.$\because\angle BAG+\angle BGA = 90^{\circ}$，$\therefore\angle BAG=\angle CBF$.$\because\angle ABG=\angle C = 90^{\circ}$，$\therefore\triangle ABG\backsim\triangle BCF$.$\therefore\frac{AB}{BC}=\frac{BG}{CF}$.$\therefore\frac{4}{5a}=\frac{a}{2}$，解得$a=\frac{2}{5}\sqrt{10}$（负值舍去）.$\therefore BC = 5a = 2\sqrt{10}$.

10. (2023·南京改编)如图，在菱形ABCD中，E是AB上一点，将△BCE沿CE翻折得到△FCE，CF与AD交于点G.若CF⊥AD，FG = 1，DG = 3，则BE = ______.

答案：
$\frac{25}{7}$ 解析：如图，过点$E$作$EH\perp BC$于点$H$，则$\angle BHE=\angle EHC = 90^{\circ}$.设$CG = x$，则$CF = x + 1$.$\because CE$是折痕，$\therefore\angle BCE=\angle FCE$，$BC = CF = x + 1$.$\because$四边形$ABCD$是菱形，$\therefore\angle B=\angle D$，$AD// BC$，$CD = BC = x + 1$.$\because CF\perp AD$，$\therefore\angle CGD = 90^{\circ}$.$\therefore$在$Rt\triangle CGD$中，$CG^{2}+DG^{2}=CD^{2}$，即$x^{2}+3^{2}=(x + 1)^{2}$，解得$x = 4$.$\therefore CG = 4$，$BC = 5$.$\because AD// BC$，$\therefore\angle BCG=\angle CGD = 90^{\circ}$.$\therefore\angle BCE=\angle FCE=\frac{1}{2}\angle BCG = 45^{\circ}$.$\therefore\angle HEC = 90^{\circ}-\angle BCE = 45^{\circ}=\angle BCE$.$\therefore EH = CH$.$\because\angle B=\angle D$，$\angle BHE=\angle DGC = 90^{\circ}$，$\therefore\triangle BHE\backsim\triangle DGC$.$\therefore\frac{BH}{DG}=\frac{EH}{CG}$，即$\frac{BH}{EH}=\frac{DG}{CG}=\frac{3}{4}$.设$BH = 3k$，则$EH = CH = 4k$.$\because\angle BHE = 90^{\circ}$，$\therefore BE=\sqrt{BH^{2}+EH^{2}}=\sqrt{(3k)^{2}+(4k)^{2}} = 5k$.$\because BC = 5$，$\therefore 3k + 4k = 5$，解得$k=\frac{5}{7}$.$\therefore BE = 5\times\frac{5}{7}=\frac{25}{7}$.

11. 如图，在Rt△ABC中，∠ACB = 90°，AC = 9，BC = 12. 在Rt△DEF中，∠DFE = 90°，DF = 3，EF = 4.用一根始终绷直的弹性染色线连接CF，Rt△DEF从起始位置(点D与点B重合)平移至终止位置(点E与点A重合)，且斜边DE始终在线段AB上，则Rt△ABC的外部被染色的区域的面积是______.

答案：
21 解析：如图，设$CF$交$AB$于点$M$，连接$CF'$，交$AB$于点$N$，过点$F$作$FG\perp AB$于点$G$，过点$F'$作$F'H\perp AB$于点$H$，连接$FF'$，则易得四边形$FGHF'$是矩形，$Rt\triangle ABC$的外部被染色的区域是梯形$MFF'N$.在$Rt\triangle DEF$中，$\because\angle DFE = 90^{\circ}$，$\therefore DE=\sqrt{DF^{2}+EF^{2}}=\sqrt{3^{2}+4^{2}} = 5$.在$Rt\triangle ABC$中，$\because\angle ACB = 90^{\circ}$，$\therefore AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{9^{2}+12^{2}} = 15$.$\because S_{\triangle DEF}=\frac{1}{2}DF\cdot EF=\frac{1}{2}DE\cdot FG$，$\therefore FG=\frac{12}{5}$.$\because\angle BGF = 90^{\circ}$，$\therefore BG=\sqrt{BF^{2}-FG^{2}}=\sqrt{3^{2}-(\frac{12}{5})^{2}}=\frac{9}{5}$.$\therefore GE = BE - BG=\frac{16}{5}$.同理，可得$AH = GE=\frac{16}{5}$，$F'H = FG=\frac{12}{5}$.$\therefore FF' = GH = AB - BG - AH = 10$.$\because$易得$BF// AC$，$\therefore$易得$\triangle BMF\backsim\triangle AMC$.$\therefore\frac{BM}{AM}=\frac{BF}{AC}=\frac{1}{3}$.$\therefore BM=\frac{1}{4}AB=\frac{15}{4}$.同理，可得$AN=\frac{1}{4}AB=\frac{15}{4}$.$\therefore MN = AB - BM - AN=\frac{15}{2}$.$\therefore Rt\triangle ABC$的外部被染色的区域的面积$=\frac{1}{2}\times(10+\frac{15}{2})\times\frac{12}{5}=21$.

12. (2023·无锡)二次函数y = a(x - 1)(x - 5)(a ＞ $\frac{1}{2}$)的图像与x轴交于点A、B(点A在点B的左侧)，与y轴交于点C，过点M(3，1)的直线将△ABC分成两部分，这两部分是三角形或梯形，且面积相等，则a的值为______.

答案：
$\frac{9}{10}$或$\frac{2+\sqrt{2}}{5}$或$\frac{\sqrt{2}+1}{2}$ 解析：在$y = a(x - 1)(x - 5)$中，令$y = 0$，得$x_{1}=1$，$x_{2}=5$.$\therefore$点$A$的坐标为$(1,0)$，点$B$的坐标为$(5,0)$.令$x = 0$，得$y = 5a$.$\therefore$点$C$的坐标为$(0,5a)$.$\because$点$M$的坐标为$(3,1)$，$\therefore$易得直线$BM$对应的函数表达式为$y = -\frac{1}{2}x+\frac{5}{2}$.$\therefore$直线$BM$与$y$轴交于点$(0,\frac{5}{2})$.$\because a＞\frac{1}{2}$，$\therefore 5a＞\frac{5}{2}$.$\therefore$点$M$必在$\triangle ABC$内部.当分成两个三角形时，直线必过$\triangle ABC$的一个顶点，若两个三角形的面积相等，则过点$M$的直线必为$\triangle ABC$的中线所在的直线.① 如图①，若直线$AM$平分$\triangle ABC$的面积，$\because$点$A$的坐标为$(1,0)$，点$M$的坐标为$(3,1)$，$\therefore$易得直线$AM$对应的函数表达式为$y=\frac{1}{2}x-\frac{1}{2}$.$\because BC$的中点的坐标为$(\frac{5}{2},\frac{5}{2}a)$，$\therefore$把$(\frac{5}{2},\frac{5}{2}a)$代入$y=\frac{1}{2}x-\frac{1}{2}$，得$\frac{5}{2}a=\frac{1}{2}\times\frac{5}{2}-\frac{1}{2}$，解得$a=\frac{3}{10}$.$\because\frac{3}{10}＜\frac{1}{2}$，$\therefore$此种情况不存在.② 如图②，若直线$BM$平分$\triangle ABC$的面积，$\because$直线$BM$对应的函数表达式为$y = -\frac{1}{2}x+\frac{5}{2}$，$AC$的中点的坐标为$(\frac{1}{2},\frac{5}{2}a)$，$\therefore$把$(\frac{1}{2},\frac{5}{2}a)$代入$y = -\frac{1}{2}x+\frac{5}{2}$，得$\frac{5}{2}a = -\frac{1}{2}\times\frac{1}{2}+\frac{5}{2}$，解得$a=\frac{9}{10}$.$\because\frac{9}{10}＞\frac{1}{2}$，$\therefore$此种情况存在.③ 如图③，若直线$CM$平分$\triangle ABC$的面积，$\because AB$的中点的坐标为$(3,0)$，点$M$的坐标为$(3,1)$，$\therefore$直线$CM// y$轴.$\therefore$此种情况不存在.当分成三角形和梯形时，过点$M$的直线必与$\triangle ABC$的一边平行.$\therefore$分成的三角形与$\triangle ABC$相似.$\because$分成的三角形和梯形的面积相等，$\therefore$分成的三角形和$\triangle ABC$的面积之比为$1:2$.$\therefore$分成的三角形和$\triangle ABC$的相似比为$\frac{\sqrt{2}}{2}$.① 如图④，直线$ME// AB$，交$y$轴于点$E$，交$CA$于点$N$.$\therefore$易得$\frac{CE}{CO}=\frac{\sqrt{2}}{2}$.$\therefore\frac{5a - 1}{5a}=\frac{\sqrt{2}}{2}$，解得$a=\frac{2+\sqrt{2}}{5}$.$\because\frac{2+\sqrt{2}}{5}＞\frac{1}{2}$，$\therefore$此种情况存在.② 如图⑤，直线$ME// AC$，交$x$轴于点$E$，过点$M$作$MN\perp x$轴于点$N$，则点$N$的坐标为$(3,0)$.$\therefore$易得$\frac{BE}{BA}=\frac{\sqrt{2}}{2}$.$\because BA = 5 - 1 = 4$，$\therefore BE = 2\sqrt{2}$.$\because BN = 5 - 3 = 2$，$2＜2\sqrt{2}$，$\therefore BN＜BE$，即点$N$在点$E$的右侧.$\therefore$此种情况不存在.③ 如图⑥，直线$ME// BC$，交$x$轴于点$E$，过点$M$作$MN\perp x$轴于点$N$，则点$N$的坐标为$(3,0)$.$\therefore$易得$\frac{AE}{AB}=\frac{\sqrt{2}}{2}$，$\angle MEN=\angle CBO$，$AN = 3 - 1 = 2$，$MN = 1$.$\because AB = 4$，$\therefore AE = 2\sqrt{2}$.$\therefore NE = AE - AN = 2\sqrt{2}-2$.$\because\angle MEN=\angle CBO$，$\angle MNE=\angle COB = 90^{\circ}$，$\therefore\triangle MNE\backsim\triangle COB$.$\therefore\frac{MN}{CO}=\frac{NE}{OB}$.$\therefore\frac{1}{5a}=\frac{2\sqrt{2}-2}{5}$，解得$a=\frac{\sqrt{2}+1}{2}$.$\because\frac{\sqrt{2}+1}{2}＞\frac{1}{2}$，$\therefore$此种情况存在.综上所述，$a$的值为$\frac{9}{10}$或$\frac{2+\sqrt{2}}{5}$或$\frac{\sqrt{2}+1}{2}$.

13. (2023·镇江)如图，正比例函数y = - 3x与反比例函数y = $\frac{k}{x}$的图像交于点A、B(1，m)，点C在x轴的负半轴上，且∠ACO = 45°.
(1)m = ______，k = ______，点C的坐标为______；
(2)点P在x轴上，若以B、O、P为顶点的三角形与△AOC相似，求点P的坐标.

答案：

(1) -3 -3 (-4,0) 解析：$\because$点$B(1,m)$在正比例函数$y = -3x$的图像上，$\therefore m = -3\times1 = -3$.$\therefore$点$B$的坐标为$(1,-3)$.$\because$点$B$在反比例函数$y=\frac{k}{x}$的图像上，$\therefore k = 1\times(-3)=-3$.$\because$正比例函数与反比例函数的图像都是关于原点$O$的中心对称图形，$\therefore$点$A$与点$B$关于原点$O$成中心对称.$\therefore$点$A$的坐标为$(-1,3)$.如图，过点$A$作$AH\perp x$轴于点$H$，则$OH = 1$，$AH = 3$，$\angle AHC = 90^{\circ}$.$\therefore\angle CAH = 90^{\circ}-\angle ACO = 90^{\circ}-45^{\circ}=45^{\circ}=\angle ACO$.$\therefore AH = CH = 3$.$\therefore OC = OH + CH = 1 + 3 = 4$.$\because$点$C$在$x$轴的负半轴上，$\therefore$点$C$的坐标为$(-4,0)$.
(2) 由（1），得$\angle AHC=\angle AHO = 90^{\circ}$，$AC=\sqrt{AH^{2}+CH^{2}}=\sqrt{3^{2}+3^{2}} = 3\sqrt{2}$，$BO = AO=\sqrt{AH^{2}+OH^{2}}=\sqrt{3^{2}+1^{2}}=\sqrt{10}$.$\because OH＜AH$，$\therefore\angle OAH＜45^{\circ}$.$\therefore\angle CAO＜90^{\circ}$.易知$\angle AOC＜90^{\circ}$，$\therefore\triangle AOC$为锐角三角形.$\therefore$与$\triangle AOC$相似的$\triangle BOP$也为锐角三角形.$\because\angle BOC＞90^{\circ}$，$\therefore$点$P$在$x$轴的正半轴上.设点$P$的坐标为$(n,0)$（$n＞0$），则$OP = n$.$\because\angle BOP=\angle AOC$，$\therefore$分两种情况：① 当$\triangle BOP\backsim\triangle AOC$时，$\frac{OP}{OC}=\frac{BO}{AO}$.$\therefore\frac{n}{4}=\frac{\sqrt{10}}{\sqrt{10}}$，解得$n = 4$.$\therefore$点$P$的坐标为$(4,0)$.② 当$\triangle BOP\backsim\triangle COA$时，$\frac{OP}{OA}=\frac{BO}{CO}$，即$\frac{n}{\sqrt{10}}=\frac{\sqrt{10}}{4}$，解得$n=\frac{5}{2}$.$\therefore$点$P$的坐标为$(\frac{5}{2},0)$.综上所述，点$P$的坐标为$(4,0)$或$(\frac{5}{2},0)$

14. (2023·南京)如图，玻璃桌面a与地面b平行，桌面上台灯的点光源O与铅笔AB在同一竖直平面上.在灯光照射下，AB在地面上形成的影子为CD(不考虑折射)，AB//CD.
(1)在桌面上沿着AB方向平移铅笔，试判断CD的长是否改变.
(2)桌面上一点P恰在点O的正下方，且OP = 36 cm，PA = 18 cm，AB = 18 cm，桌面的高度为60 cm.在点O与AB所确定的平面内，将AB绕点A旋转，使得CD的长最大.
①画出此时AB所在位置的示意图；
②CD的长最大为______cm.

答案：

(1) 如图①，过点$O$作$OF\perp$地面$b$于点$F$，交桌面$a$于点$E$.$\because a// b$，$\therefore OF\perp$桌面$a$.$\because AB// CD$，$\therefore\triangle OAB\backsim\triangle OCD$.$\because OE$、$OF$分别是$\triangle OAB$、$\triangle OCD$对应边上的高，$\therefore\frac{AB}{CD}=\frac{OE}{OF}$.$\therefore CD=\frac{AB\cdot OF}{OE}$.根据题意，得$AB$、$OE$、$OF$的长为定值，$\therefore CD$的长不变
(2) ① 如图②，$AB'$的位置即为所求 解析：$\because$点$A$固定，$\therefore$点$C$固定.$\therefore$若想$CD$的长最大，则需要点$D$离点$C$最远.$\because$点$B$在以点$A$为圆心、$AB$长为半径的圆上，$\therefore$离点$C$最远的点$D$和点光源$O$的连线与$\odot A$相切.如图②，以点$A$为圆心、$AB$长为半径作$\odot A$，以点$O$为圆心、$OP$长为半径画弧，交$\odot A$于另一点$B'$，连接$OB'$并延长，交地面$b$于点$D$，交桌面$a$于点$G$，则此时$CD$的长最大.
② 80 解析：设$AG = x\ cm$，则$PG=(x + 18)\ cm$.$\because$易得$\angle AB'G = 90^{\circ}$，$\angle OPG = 90^{\circ}$，$\therefore\angle AB'G=\angle OPG$.$\because\angle B'GA=\angle PGO$，$\therefore\triangle AB'G\backsim\triangle OPG$.$\therefore\frac{AG}{OG}=\frac{B'G}{PG}=\frac{AB'}{OP}=\frac{18}{36}=\frac{1}{2}$.$\therefore OG = 2AG = 2x\ cm$.$\therefore B'G=(2x - 36)\ cm$.$\therefore\frac{2x - 36}{x + 18}=\frac{1}{2}$，解得$x = 30$.经检验，$x = 30$是原方程的解，且符合实际.$\therefore AG = 30\ cm$.如图②，延长$OP$交地面$b$于点$Q$，则$PQ = 60\ cm$.同理于
(1)，得$CD=\frac{AG\cdot OQ}{OP}=\frac{30\times(36 + 60)}{36}=80(cm)$.