《2026年金考卷中考试题汇编45套数学山东专版》

19. (本小题满分9分)
如图,在平面直角坐标系$xOy$中,已知$\triangle OAB$为等边三角形,$AB = 6$,点$C$为$AB$的中点,反比例函数$y=\frac{k}{x}(x ＞ 0)$的图象经过$A$,$B$两点,且与$OC$交于点$D$,$\angle BOE = 15^{\circ}$,点$B$的横、纵坐标之和为$3\sqrt{6}$.
(1)点$C$的坐标为$(\frac{3\sqrt{6}}{2}, \frac{3\sqrt{6}}{2})$;(请直接写出结果)
(2)求反比例函数的解析式;
(3)求线段$CD$的长度.

答案：

(1)$(\frac{3\sqrt{6}}{2}, \frac{3\sqrt{6}}{2})$
解法提示:如图,分别过点B,A作BM⊥x轴,AN⊥y轴,垂足分别为M,N,则∠BMO = ∠ANO = 90°.

∵△OAB为等边三角形,
∴∠AOB = 60°,OA = OB.
又
∵∠BOE = 15°,
∴∠AON = 15°,
∴∠AON = ∠BOE,
∴△OBM≌△OAN(AAS),
∴BM = AN,OM = ON.
设B(m,n),则A(n,m).
∵点C为AB的中点,
∴C$(\frac{n + m}{2}, \frac{m + n}{2})$.
∵点B的横、纵坐标之和为$3\sqrt{6}$,
∴m + n = $3\sqrt{6}$,
∴C$(\frac{3\sqrt{6}}{2}, \frac{3\sqrt{6}}{2})$.
(2)设B(m,n),则$m^2 + n^2 = OB^2 = AB^2 = 36$. ①
∵点B的横、纵坐标之和为$3\sqrt{6}$,即m + n = $3\sqrt{6}$,②
将②两边同时平方,得$(m + n)^2 = m^2 + 2mn + n^2 = 54$,③
将①代入③,整理得2mn = 18,解得mn = 9.
∴反比例函数的解析式为y = $\frac{9}{x}(x ＞ 0)$. (5分)
(3)
∵C$(\frac{3\sqrt{6}}{2}, \frac{3\sqrt{6}}{2})$,
∴OC = $\sqrt{(\frac{3\sqrt{6}}{2})^2 + (\frac{3\sqrt{6}}{2})^2} = 3\sqrt{3}$.
∵△OAB为等边三角形,点C为AB的中点,
∴∠BOC = $\frac{1}{2}$∠AOB = 30°(依据:等腰三角形“三线合一”),
∴∠COE = ∠BOC + ∠BOE = 45°,
∴OC所在直线的解析式为y = x.
联立,得$\begin{cases} y = x, \\ y = \frac{9}{x} \end{cases}$解得$\begin{cases} x = 3, \\ y = 3, \end{cases}$
∴D(3,3).
∴OD = $\sqrt{3^2 + 3^2} = 3\sqrt{2}$,
∴CD = OC - OD = $3\sqrt{3} - 3\sqrt{2}$. (9分)