初中必刷题 数学九年级下册 RJ

方法点拨

由“A”型或者“X”型得到最基本的相似三角形。

刷素养

7.【解】(1) 设 $A(x,\; y)$ , 则由题意可得 $\backslash frac\{y\}\{x\}\; =\; \backslash frac\{3\}\{2\}$ . ①

当 $AO\; =\; AM$ 时, 则 $AO^2\; =\; AM^2$ ,
即 $x^2\; +\; y^2\; =\; (13\; -\; x)^2\; +\; y^2$ . ②

由①②得 $\backslash begin\{cases\}\; y\; =\; \backslash frac\{3\}\{2\}x,\; \backslash \backslash \; x^2\; +\; y^2\; =\; (13\; -\; x)^2\; +\; y^2,\; \backslash end\{cases\}$

解得 $\backslash begin\{cases\}\; x\; =\; \backslash frac\{13\}\{2\},\; \backslash \backslash \; y\; =\; \backslash frac\{39\}\{4\}.\; \backslash end\{cases\}$ 即 $A(\backslash frac\{13\}\{2\},\; \backslash frac\{39\}\{4\})$ .

当 $OA\; =\; OM$ 时, 则 $OA^2\; =\; OM^2$ , 即 $x^2\; +\; y^2\; =\; 169$ . ③

由①③得 $\backslash begin\{cases\}\; y\; =\; \backslash frac\{3\}\{2\}x,\; \backslash \backslash \; x^2\; +\; y^2\; =\; 169,\; \backslash end\{cases\}$

解得 $\backslash begin\{cases\}\; x\; =\; 2\backslash sqrt\{13\},\; \backslash \backslash \; y\; =\; 3\backslash sqrt\{13\}\; \backslash end\{cases\}$ 或 $\backslash begin\{cases\}\; x\; =\; -2\backslash sqrt\{13\},\; \backslash \backslash \; y\; =\; -3\backslash sqrt\{13\}.\; \backslash end\{cases\}$

舍去不合题意的解, 则 $A(2\backslash sqrt\{13\},\; 3\backslash sqrt\{13\})$ .

当 $MA\; =\; OM$ 时, 则 $MA^2\; =\; OM^2$ , 即 $(13\; -\; x)^2\; +\; y^2\; =\; 169$ . ④

由①④得 $\backslash begin\{cases\}\; y\; =\; \backslash frac\{3\}\{2\}x,\; \backslash \backslash \; (13\; -\; x)^2\; +\; y^2\; =\; 169,\; \backslash end\{cases\}$ 解得 $\backslash begin\{cases\}\; x\; =\; 8,\; \backslash \backslash \; y\; =\; 12\; \backslash end\{cases\}$ 或

$\backslash begin\{cases\}\; x\; =\; 0,\; \backslash \backslash \; y\; =\; 0,\; \backslash end\{cases\}$ 舍去不合题意的解, 则 $A(8,\; 12)$ .

综上所述, 如果 $\backslash triangle\; AOM$ 是等腰三角形, 点 $A$ 的坐标是 $(\backslash frac\{13\}\{2\},\; \backslash frac\{39\}\{4\})$ 或 $(2\backslash sqrt\{13\},\; 3\backslash sqrt\{13\})$ 或 $(8,\; 12)$ .

(2) 存在点 $A$ 使 $\backslash triangle\; OMN$ 与 $\backslash triangle\; AOB$ 相似. 点 $A$ 的坐标为 $(4,\; 6)$ 或 $(\backslash frac\{13\}\{2\},\; \backslash frac\{39\}\{4\})$ .

当 $\backslash triangle\; OBA\; \backslash sim\; \backslash triangle\; MON$ 时, $\backslash frac\{AB\}\{NO\}\; =\; \backslash frac\{OB\}\{MO\}$ , 故 $\backslash frac\{ON\}\{OM\}\; =\; \backslash frac\{AB\}\{OB\}\; =\; \backslash frac\{3\}\{2\}$ , 则 $ON\; =\; \backslash frac\{3\}\{2\}OM\; =\; \backslash frac\{39\}\{2\}$ , 所以 $N(0,\; \backslash frac\{39\}\{2\})$ .

直线 $MN$ 的解析式为 $y\; =\; -\backslash frac\{3\}\{2\}x\; +\; \backslash frac\{39\}\{2\}$ . ⑤

由①⑤得 $\backslash begin\{cases\}\; y\; =\; \backslash frac\{3\}\{2\}x,\; \backslash \backslash \; y\; =\; -\backslash frac\{3\}\{2\}x\; +\; \backslash frac\{39\}\{2\},\; \backslash end\{cases\}$ 解得 $\backslash begin\{cases\}\; x\; =\; \backslash frac\{13\}\{2\},\; \backslash \backslash \; y\; =\; \backslash frac\{39\}\{4\}.\; \backslash end\{cases\}$

$\backslash therefore\; A(\backslash frac\{13\}\{2\},\; \backslash frac\{39\}\{4\})$ .

当 $\backslash triangle\; OAB\; \backslash sim\; \backslash triangle\; NMO$ 时, $\backslash frac\{AB\}\{MO\}\; =\; \backslash frac\{OB\}\{NO\}$ , 故 $\backslash frac\{OM\}\{ON\}\; =\; \backslash frac\{AB\}\{OB\}$ , 则 $ON\; =\; \backslash frac\{OB\}\{AB\}\; \backslash cdot\; OM\; =\; \backslash frac\{2\}\{3\}\; \backslash times\; 13\; =\; \backslash frac\{26\}\{3\}$ , 所以 $N(0,\; \backslash frac\{26\}\{3\})$ .

直线 $MN$ 的解析式为 $y\; =\; -\backslash frac\{2\}\{3\}x\; +\; \backslash frac\{26\}\{3\}$ . ⑥

由①⑥得 $\backslash begin\{cases\}\; y\; =\; \backslash frac\{3\}\{2\}x,\; \backslash \backslash \; y\; =\; -\backslash frac\{2\}\{3\}x\; +\; \backslash frac\{26\}\{3\},\; \backslash end\{cases\}$ 解得 $\backslash begin\{cases\}\; x\; =\; 4,\; \backslash \backslash \; y\; =\; 6,\; \backslash end\{cases\}$

$\backslash therefore\; A(4,\; 6)$ .

综上所述, 当点 $A$ 的坐标为 $(4,\; 6)$ 或 $(\backslash frac\{13\}\{2\},\; \backslash frac\{39\}\{4\})$ 时, $\backslash triangle\; OMN$ 与 $\backslash triangle\; AOB$ 相似.

3. 3.2 或 5 【解析】 $\backslash because\; AB$ 是 $\backslash odot\; O$ 的直径, $\backslash therefore\; \backslash angle\; ACB\; =\; 90^\backslash circ$ . 在 $\backslash text\{Rt\}\backslash triangle\; ABC$ 中, $AC\; =\; 4$ , $BC\; =\; 3$ , $\backslash therefore\; AB\; =\; \backslash sqrt\{4^2\; +\; 3^2\}\; =\; 5$ . $\backslash therefore\; l\; \backslash parallel\; AB$ , $\backslash therefore\; \backslash angle\; ACP\; =\; \backslash angle\; CAB$ . $\backslash therefore$ 以点 $P,\; A,\; C$ 为顶点的三角形与 $\backslash triangle\; ABC$ 相似, $\backslash therefore\; \backslash frac\{PC\}\{AB\}\; =\; \backslash frac\{AC\}\{AC\}$ 或 $\backslash frac\{AC\}\{AB\}\; =\; \backslash frac\{PC\}\{AC\}$ , $\backslash therefore\; \backslash frac\{PC\}\{5\}\; =\; 1$ 或 $\backslash frac\{4\}\{5\}\; =\; \backslash frac\{PC\}\{4\}$ , 解得 $PC\; =\; 5$ 或

3. 2. 综上可知, 若 $\backslash triangle\; ABC$ 与 $\backslash triangle\; PAC$ 相似, 则 $PC\; =\; 3.2$ 或 5.

4. 1 或 3 或 8 【解析】设 $AP\; =\; x$ , 则 $PB\; =\; 9\; -\; x$ . $\backslash therefore$ 以 $A,\; C,\; P$ 为顶点的三角形与以 $B,\; D,\; P$ 为顶点的三角形相似, $\backslash angle\; A\; =\; \backslash angle\; B\; =\; 90^\backslash circ$ , $\backslash therefore$ 分两种情况讨论: ① 当 $\backslash frac\{AC\}\{BD\}\; =\; \backslash frac\{AP\}\{PB\}$ 时, $\backslash frac\{2\}\{4\}\; =\; \backslash frac\{x\}\{9-x\}$ , 解得 $x\; =\; 3$ . ② 当 $\backslash frac\{AC\}\{BP\}\; =\; \backslash frac\{AP\}\{BD\}$ 时, $\backslash frac\{2\}\{9-x\}\; =\; \backslash frac\{x\}\{4\}$ , 解得 $x\; =\; 1$ 或 8. $\backslash therefore$ 当以 $A,\; C,\; P$ 为顶点的三角形与以 $B,\; D,\; P$ 为顶点的三角形相似时, $AP$ 的长为 1 或 3 或 8. 故答案为 1 或 3 或 8.

5. 40 或 5 【解析】设 $BF\; =\; x$ , $\backslash therefore\; BF\; =\; B\text{'}F\; =\; x$ , $\backslash therefore\; FC\; =\; BC\; -\; BF\; =\; 10\; -\; x$ . $\backslash therefore\; \backslash angle\; FCB\text{'}\; =\; \backslash angle\; BCA$ , $\backslash therefore$ 当 $\backslash frac\{CF\}\{CB\}\; =\; \backslash frac\{CB\text{'}\}\{CA\}$ 时, $\backslash triangle\; CFB\text{'}\; \backslash sim\; \backslash triangle\; CBA$ , 即 $\backslash frac\{10-x\}\{8\}\; =\; \backslash frac\{x\}\{8\}$ , 解得 $x\; =\; \backslash frac\{40\}\{9\}$ ; 当 $\backslash frac\{CF\}\{CA\}\; =\; \backslash frac\{CB\text{'}\}\{CB\}\; =\; \backslash frac\{FB\text{'}\}\{AB\}$ 时, $\backslash triangle\; CFB\text{'}\; \backslash sim\; \backslash triangle\; CAB$ , 即 $\backslash frac\{10-x\}\{8\}\; =\; \backslash frac\{x\}\{8\}$ , 解得 $x\; =\; 5$ . 综上所述, 当 $BF\; =\; \backslash frac\{40\}\{9\}$ 或 5 时, 以点 $B\text{'},\; F,\; C$ 为顶点的三角形与 $\backslash triangle\; ABC$ 相似.

6. ①④ 【解析】 $\backslash because$ 四边形 $ABCD$ 为正方形, $\backslash therefore\; \backslash angle\; ADC\; =\; \backslash angle\; BCD\; =\; 90^\backslash circ$ , $AD\; =\; CD$ . $\backslash therefore\; E,\; F$ 分别为 $BC,\; CD$ 的中点, $\backslash therefore\; DF\; =\; EC\; =\; 2$ . $\backslash therefore\; \backslash triangle\; ADF\; \backslash cong\; \backslash triangle\; DCE\; (\backslash text\{SAS\})$ , $\backslash therefore\; \backslash angle\; AFD\; =\; \backslash angle\; DEC$ , $\backslash angle\; FAD\; =\; \backslash angle\; EDC$ . $\backslash therefore\; \backslash angle\; EDC\; +\; \backslash angle\; DEC\; =\; 90^\backslash circ$ , $\backslash therefore\; \backslash angle\; EDC\; +\; \backslash angle\; AFD\; =\; 90^\backslash circ$ , $\backslash therefore\; \backslash angle\; DGF\; =\; 90^\backslash circ$ , 即 $DE\; \backslash perp\; AF$ , 故①正确. $\backslash therefore\; AD\; =\; 4$ , $DF\; =\; \backslash frac\{1\}\{2\}CD\; =\; 2$ , $\backslash therefore\; AF\; =\; \backslash sqrt\{4^2\; +\; 2^2\}\; =\; 2\backslash sqrt\{5\}$ . $\backslash therefore\; \backslash frac\{1\}\{2\}AD\; \backslash cdot\; DF\; =\; \backslash frac\{1\}\{2\}DG\; \backslash cdot\; AF$ , $\backslash therefore\; DG\; =\; \backslash frac\{AD\; \backslash cdot\; DF\}\{AF\}\; =\; \backslash frac\{4\backslash sqrt\{5\}\}\{5\}$ , 故②错误. $\backslash because\; H$ 为 $AF$ 的中点, $\backslash therefore\; HD\; =\; HF\; =\; \backslash frac\{1\}\{2\}AF\; =\; \backslash sqrt\{5\}$ , $\backslash therefore\; \backslash angle\; HDF\; =\; \backslash angle\; HFD$ . $\backslash because\; AB\; \backslash parallel\; DC$ , $\backslash therefore\; \backslash angle\; HDF\; =\; \backslash angle\; HFD\; =\; \backslash angle\; BAG$ . $\backslash therefore\; AG\; =\; \backslash sqrt\{AD^2\; -\; DG^2\}\; =\; \backslash frac\{8\backslash sqrt\{5\}\}\{5\}$ , $AB\; =\; 4$ , $\backslash therefore\; \backslash frac\{AB\}\{DH\}\; =\; \backslash frac\{4\backslash sqrt\{5\}\}\{5\}\; =\; \backslash frac\{AG\}\{DF\}$ , $\backslash therefore\; \backslash triangle\; ABG\; \backslash sim\; \backslash triangle\; DHF$ , 故④正确. $\backslash therefore\; \backslash angle\; ABG\; =\; \backslash angle\; DHF$ , 而 $AB\; \backslash neq\; AG$ , 则 $\backslash angle\; ABG$ 和 $\backslash angle\; AGB$ 不相等, $\backslash therefore\; \backslash angle\; AGB\; \backslash neq\; \backslash angle\; DHF$ , $\backslash therefore\; HD$ 与 $BG$ 不平行, 故③错误. 故答案为①④.

易错警示

本题未明确相似三角形的对应关系, 注意分类讨论, 避免漏解.

D12

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