《2025年名校调研跟踪测试卷九年级数学下册人教版》

23. 如图，在正方形$ABCD$中，$E$，$F$分别是边$AD$，$CD$上的点，$AE = DE$，$DF = \frac{1}{4}DC$，连接$EF$并延长交$BC$的延长线于点$G$.
(1)求证：$\triangle ABE\sim\triangle DEF$；
(2)若正方形的边长为$4$，求$BG$的长.
　　　　

答案：
(1)证明：$\because$四边形$ABCD$为正方形，$\therefore AD = AB = DC = BC$，$\angle A=\angle D = 90^{\circ}$. $\because AE = ED$，$\therefore DE=\frac{1}{2}AD=\frac{1}{2}AB$. $\because DF=\frac{1}{4}DC$，$\therefore DF=\frac{1}{4}AD=\frac{1}{2}AE$，$\therefore\frac{DE}{AB}=\frac{DF}{AE}=\frac{1}{2}$，$\therefore\triangle ABE\sim\triangle DEF$.
(2)解：$\because$四边形$ABCD$为正方形，$\therefore ED// BG$，$\therefore\triangle EDF\sim\triangle GCF$，$\therefore\frac{ED}{CG}=\frac{DF}{CF}$. 又$\because DF=\frac{1}{4}DC$，$AE = DE$，正方形的边长为$4$，$\therefore DF = 1$，$CF = 3$，$ED = 2$，$\therefore\frac{2}{CG}=\frac{1}{3}$，$\therefore CG = 6$，$\therefore BG = BC + CG = 10$.

24. 如图，已知一次函数$y = ax + b$与反比例函数$y = \frac{m}{x}(x＜0)$的图象交于$A(-2,4)$，$B(-4,2)$两点，且与$x$轴和$y$轴分别交于点$C$，点$D$.
(1)根据图象直接写出不等式$\frac{m}{x}＜ax + b$的解集；
(2)求反比例函数与一次函数的解析式；
(3)点$P$在$y$轴上，且$S_{\triangle AOP} = \frac{1}{2}S_{\triangle AOB}$，请求出点$P$的坐标.
　

答案： 解：
(1)$-4\lt x\lt -2$.
(2)将$A(-2,4)$代入$y=\frac{m}{x}$得，$m = -8$，$\therefore$反比例函数的解析式为$y = -\frac{8}{x}$. 将$A(-2,4)$，$B(-4,2)$代入$y = ax + b$得，$\begin{cases}-2a + b = 4\\-4a + b = 2\end{cases}$，解得$\begin{cases}a = 1\\b = 6\end{cases}$，$\therefore$一次函数的解析式为$y = x + 6$.
(3)在$y = x + 6$中，当$y = 0$时，$x = -6$，$\therefore C(-6,0)$. $\therefore S_{\triangle ABO}=S_{\triangle AOC}-S_{\triangle BOC}=\frac{1}{2}×6×(4 - 2)=6$，$\therefore S_{\triangle AOP}=\frac{1}{2}×6 = 3$. 设$P(0,n)$，则$\frac{1}{2}×2×|n| = 3$，$\therefore n = 3$或$-3$，$\therefore P(0,3)$或$(0,-3)$.