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网印经开印字版152千字
2023年2月第1版
2023年12月第2次印刷
书定 号价ISBN 978-7-5762-3604-0
28.00元
+ +赣版权登字-07-2023-116版权所有侵权必究 + +# 初中同步测试卷 + +CHUZHONG TONGBU CESHIJUAN + +金太阳教育研究院编 + +# 梳理重点知识 拓展学习思维 + +本套试卷围绕新一轮基础教育课程改革的新思想、新理念，充分考虑学生的学习体验，以提高学生素质、促进学生学习方式的转变、促进学生积极主动全面的发展为目标，对学生的学习进行同步指导、检测及评价，帮助学生发现优势和不足，促使学生制订改进计划，调整学习策略，从而提高学习效率。 + +主编：刘美平副主编：刘黎明熊林华编委:(按姓氏笔画排序)刘宗政　许　清　宋丽革　杨建余燕婷 　梁东强 　谭 　薇 　熊三勇 + +# 数学北师大族 + +七年级下册 + +# 亮点一 结构合理 + +采用“单元卷+阶段卷+专项卷”的模式编写，思路清晰，结构完整 + +# 亮点 + +# 亮点二 内容新颖 + +紧紧围绕教材编创新题、好题，难易适中，内容丰富，特色鲜明 + +亮点三 卷中悟法 + +参考答案中含重难知识总结、方法归纳、解题技巧等，重点突出 + +亮点四 视频讲解优秀教师详细讲解答题技巧，快速提升学生解题、得分能力 + +本卷特别注重科学性和实用性，在编写过程中，得到了很多一线教师的热心帮助。我们精心设计，细致核对，力求做到尽善尽美。虽然我们竭尽心智，但疏漏之处在所难免，恳请广大师生在使用中提出宝贵的意见，以便我们修改完善。 + +第一章 整式的乘除 + +# 第五章 生活中的轴对称 + +基础过关测试卷能力提优测试卷 + +# $\textcircled{4}$ 阅读理解题 + +第二章 相交线与平行线基础过关测试卷 L能力提优测试卷· 13月考测试卷（一） 17第三章 变量之间的关系基础过关测试卷· 25能力提优测试卷· 29期中测试卷 33 + +第四章 三角形 + +基础过关测试卷 41 + +基础过关测试卷 49 +能力提优测试卷 53 + +月考测试卷（二） 57 + +六章 概率初步 + +基础过关测试卷 65 +能力提优测试卷 69 +专项训练卷（一） 整式的乘除运算 ： 73 +专项训练卷（二） 几何题的计算 77 +专项训练卷（三） 几何题的说理 81 +期末测试卷 85 +答案和解析 93 +新定义/3/18/34 +解题方法/40/52/84 + +# 代数推理 + +等式规律探索/6/19 + +# $\circledcirc$ 真实问题情境 + +汽车探照灯/14 + +山地自行车/78 + +# $\textcircled{4}$ 数学文化 + +《详解九章算法》/35 + +# $\circledcirc$ 综合与实践 + +拓展探究型/48/76 + +$\textcircled{6}$ 填空双空题/59 + +# 第一章 整式的乘除 + +# 基础过关测试卷 + +时间：60分钟 满分：100分 + +
题序二三评卷人总分
得分
+ +# 一、选择题(每小题4分，共32分) + +1.计算 $a ^ { 6 } \bullet a ^ { 2 }$ 的结果是 + +A $a ^ { 3 }$ B. $a ^ { 4 }$ C.a8 D. $a ^ { 1 2 }$ + +2.计算 $2 0 2 4 ^ { \circ } \times 2 ^ { - 1 }$ 的结果是 + +A.-2024 B.-2 C.0 D $\frac 1 2$ + +3.计算 $( 2 a ) \ \bullet \ ( a b )$ 的结果是 + +A. 2ab B. $2 a ^ { 2 } b$ （20 C.3ab D. $3 a ^ { 2 } b$ + +4.若 $( x - 2 ) ( x + a ) = x ^ { 2 } + b x - 6$ ，则 + +A. $a = 3 , b = 1$ $\mathrm { B } _ { \cdot } a { = } 3 , b { = } - 5$ （20 $\stackrel { \cdot } { \scriptscriptstyle \mathscr { O } } . a = - 3 , b = - 1$ I $) . a { = } { - 3 , b { = } { - 5 } }$ + +5.下列运算中，正确的是 + +A. $x ^ { 2 } + 5 x ^ { 2 } = 6 x ^ { 4 } \quad \mathrm { ~ B . ~ } x ^ { 3 } \bullet x ^ { 2 } = x ^ { 6 } \qquad \mathrm { ~ C . ~ } ( x ^ { 2 } ) ^ { 3 } = x ^ { 6 }$ D. $( \boldsymbol { \mathscr { x y } } ) ^ { 3 } = \boldsymbol { \mathscr { x y } } ^ { 3 }$ （204号 + +6.若 $M \bullet ( 3 x - y ^ { 2 } ) = y ^ { 4 } - 9 x ^ { 2 }$ ，则多项式 $M$ 为 + +$ \mathrm { A } . - 3 x + y ^ { 2 } \qquad \mathrm { B } . - y ^ { 2 } - 3 x \qquad \mathrm { C } . ~ 3 x + y ^ { 2 }$ $\mathrm { D } . 3 x - y ^ { 2 }$ + +7.当 $a { = } \frac { 1 } { 3 }$ 时，代数式 $( a - 4 ) ( a - 3 ) - a ( a + 2 )$ 的值为 + +A.9 B.-9 C.3 D $\frac 1 3$ + +8.已知 $\displaystyle a + b = - 4 , a b = - 1 2$ ，则 $a ^ { 2 } + b ^ { 2 }$ 的值为 + +A.46 B.44 C.42 D.40 + +# 二、填空题(每小题4分，共16分) + +9.一种桑蚕丝的直径约为0.0000189米，数据0.0000189用科学记数法表示为 + +10.计算： $1 0 x ^ { 8 } \div 2 x ^ { 2 } =$ + +11.计算： $2 m ^ { 2 } n \bullet ( m ^ { 2 } + n - 1 ) =$ + +12. $\circledcirc$ 视频讲解在化简求 $( a + 3 b ) ^ { 2 } + ( 2 a + 3 b ) ( 2 a - 3 b ) + a ( 5 a - 6 b )$ 的值时，小明把 $a$ 的值看错后代入得结果为10,而小红代人正确的 $\boldsymbol { a }$ 的值得正确的结果也是10,经探究后，发现所求代数式的值与 $b$ 的值无关，则他们俩代入的 $a$ 的值的商为 + +# 三、解答题(本大题6小题，共52分) + +13.(6分)计算： $( 1 ) ( - b ^ { 5 } ) \bullet ( - b ^ { 3 } ) ^ { 4 } ; ( 2 ) ( - x ) ^ { 3 } \bullet ( - x ^ { 2 } ) ^ { 2 } - ( - 2 x ^ { 3 } ) ^ { 2 } \bullet x .$ + +14.（8分)计算： $( 1 ) x ^ { 2 } - { \frac { 1 } { 2 } } x ( 2 - 4 x ) , ( 2 ) ( x + 1 ) ( 7 - x ) - x ( x - 3 ) .$ + +15.(8分)先化简，再求值： $4 ( x - 1 ) ^ { 2 } - ( 2 x + 3 ) ( 2 x - 3 )$ ,其中 $x { = } - 2$ + +16.新考法阅读理解题（8分）规定 $a \ast b = 2 ^ { a } \times 2 ^ { b }$ + +(1)求 $1 \ast 3$ 的值； +(2)若 $2 * ( x + 1 ) = 2 5 6$ ，求 $_ { \mathcal { X } }$ 的值. + +17.(10 分)如果代数式 $\left( a x ^ { - } 1 \right) \left( x ^ { + } 7 \right) - x ^ { 2 } + 7$ 化简后,不含有 $x ^ { 2 }$ 的项. + +(1)求 $a$ 的值； + +(2）求 $( a - 5 ) ( a + 5 ) + ( a - 5 ) ^ { 2 } - a ( 2 a + 1 )$ 的值. + +18. $\circledcirc$ 视频讲解（12分）已知 $P { = } ( x { - } y ) ^ { 5 } \div ( y { - } x ) ^ { 5 } { \bullet } ( x { - } y ) ^ { 2 } , Q { = } [ ( 2 x { - } y ) ^ { 2 } { + }$ $( 2 x + y ) ( 2 x - y ) + 8 x y ] \div 2 x .$ （20 + +(1)化简 $P$ 和 $Q$ + +(2)当 $x = ( 3 - \pi ) ^ { 0 }$ $y = ( - \frac { 1 } { 2 } ) ^ { - 1 }$ 时，分别求 $P$ 和 $Q$ 的值. + +# 能力提优测试卷 + +时间：60分钟 满分：100分 + +
题序二三评卷人总分
得分
+ +# 一、选择题(每小题4分，共32分) + +1.下列运算正确的是 + +A. $a ^ { 2 } \cdot a ^ { 3 } = a ^ { 6 }$ （20 B. $a ^ { 8 } \div a ^ { 4 } = a ^ { 2 }$ （204号$\mathrm { C } , 5 a - 3 a = 2 a$ $\operatorname { D } . ( - 2 a b ^ { 2 } ) ^ { 2 } = - 4 a ^ { 2 } b ^ { 4 }$ + +2.熔喷布，俗称口罩的“心脏”,是口罩中间的过滤层，能过滤细菌，防止病菌传播.经测量，医用外科口罩的熔喷布厚度约为0.000156米，数据0.000156用科学记数法表示为 （ ） + +A. $0 . 1 5 6 \times 1 0 ^ { - 3 }$ （20 $\mathrm { B } . 1 . 5 6 { \times } 1 0 ^ { - 4 }$ （204号 $: 1 5 . 6 \times 1 0 ^ { - 5 }$ （20 I $) . 1 . 5 6 \times 1 0 ^ { - 3 }$ （20 + +3.一个多项式除以 $4 . x ^ { 2 }$ ,商是一个三次三项式,那么这个多项式是 （） + +A.五次三项式 B.四次三项式C.五次四项式 D.四次四项式 + +4.已知 $( a + b ) ^ { 2 } = 4 9 , a ^ { 2 } + b ^ { 2 } = 2 5$ ，则 $a b$ 的值为 + +A.12 B.20 C.24 D.48 + +5.若 $x { + } m$ 与 $x - 5$ 的乘积中不含 $\mathcal { X }$ 的一次项,则 $m$ 的值为 + +A.-5 B.5 C.0 D.3 + +6 $\begin{array} { r l r } { . ( - 5 a ^ { 2 } + 4 b ^ { 2 } ) ( } & { { } } & { ) = 2 5 a ^ { 4 } - 1 6 b ^ { 4 } } \end{array}$ ,则括号内应填 + +A. $5 a ^ { 2 } + 4 b ^ { 2 }$ $3 , 5 a ^ { 2 } - 4 b ^ { 2 }$ $\mathrm { C } , - 5 a ^ { 2 } + 4 b ^ { 2 }$ $\mathrm { D } . - 5 a ^ { 2 } - 4 b ^ { 2 }$ + +7.长方体的长、宽、高分别是 $4 x - 3 , x$ 和 $2 x$ ，则它的体积等于 + +A. $4 x ^ { 3 } - 3 x ^ { 2 }$ （204号 $\mathrm { B } , 4 x ^ { 2 } - 3 x$ C. $8 x ^ { 3 } - 6 x ^ { 2 }$ D. $2 \mathcal { x } ^ { 2 }$ （204号 + +8.新考法代数推理 $\circledcirc$ 视频讲解观察下列各式及其展开式： + +$$ +\begin{array} { l } { { ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 } ; } } \\ { { \ } } \\ { { ( a + b ) ^ { 3 } = a ^ { 3 } + 3 a ^ { 2 } b + 3 a b ^ { 2 } + b ^ { 3 } ; } } \\ { { \ } } \\ { { ( a + b ) ^ { 4 } = a ^ { 4 } + 4 a ^ { 3 } b + 6 a ^ { 2 } b ^ { 2 } + 4 a b ^ { 3 } + b ^ { 4 } ; } } \\ { { \ } } \\ { { ( a + b ) ^ { 5 } = a ^ { 5 } + 5 a ^ { 4 } b + 1 0 a ^ { 3 } b ^ { 2 } + 1 0 a ^ { 2 } b ^ { 3 } + 5 a b ^ { 4 } + b ^ { 5 } ; } } \end{array} +$$ + +请你猜想 $( a + b ) ^ { 1 4 }$ 的展开式第三项的系数是 + +A.66 B.78 C.91 D.105 + +# 二、填空题(每小题4分，共16分) + +9.已知 $a \cdot ( - b ^ { 2 } ) ^ { 2 } { = } 3$ ，则 $a ^ { 2 } b ^ { 8 }$ 的值为 + +10.在一次“学数学，爱数学"的主题班会上，主持人小明同学亮出了 $A , B , C$ 三张卡片，上面分别写有 $\begin{array} { r } { \boxed { - 2 a b } \boxed { a ^ { 2 } + 2 a b + b ^ { 2 } } \boxed { ( a - b ) ^ { 2 } } } \end{array}$ ,其中有两张卡片上的式子相乘,所得的积为一 $2 a ^ { 3 } b + 4 a ^ { 2 } b ^ { 2 } - 2 a b ^ { 3 } .$ 这两张卡片是 和 + +11.如图，将完全相同的四个长方形纸片拼成一个大的正方形，用两种不同的方法表示这个大正方形的面积,则可以得出的一个等式为 + +![](images/ecdef52c2a8dc6203035fd1b01da09e2ebb0ab20a8580483e15c2230bdc3388e.jpg) + +12.若规定 $\left| { \begin{array} { l l } { a } & { b } \\ { c } & { d } \end{array} } \right| = a d - b c$ ,当 $\left| { \begin{array} { l l } { x - 2 } & { x - 2 } \\ { x + 2 } & { x - 2 } \end{array} } \right| = 0$ 时， $_ { \mathcal { X } }$ 的值为 + +# 三、解答题(本大题6小题，共52分) + +13.（6分）计算： $( 1 ) a ^ { 4 } { \bf \lambda } { \bf \dot { \alpha } } a ^ { 2 } - ( 3 a ^ { 3 } ) ^ { 2 } + ( - 4 a ^ { 2 } ) ^ { 3 } { \bf \dot { \beta } } ( 2 ) \left| - 5 \right| - ( 1 - \pi ) ^ { 0 } +$ （20 $( - { \frac { 1 } { 3 } } ) ^ { - 1 } .$ + +14.(8分)先化简，再求值： $( x - 3 ) ^ { 2 } + ( x + 3 ) ( x - 3 ) + 2 x ( 2 - x )$ ,其中 $x { = } - \frac { 1 } { 2 }$ + +15.(8分)小红在计算 $\alpha ( 1 + a ) - ( a - 1 ) ^ { 2 }$ 时,解答过程如下： + +a(1+a)-(a-1)² +=a+a²-(a²-1)·第一步$| = a + a ^ { 2 } - a ^ { 2 } - 1 \cdots$ ·第二步=a-1……第三步 + +小红的解答从第 步开始出错，请写出正确的解答过程. + +16.（8 分)小明在进行两个多项式的乘法运算时,不小心把乘 $\frac { x + y } { 2 }$ 错抄成乘 $\cdot \frac { x } { 2 }$ 结果得到 $3 x ^ { 2 } - 5 x y$ ,则第一个多项式是多少？正确的结果应该是多少？ + +17.(10分)已知等式 $\left( x + a \right) \left( x + b \right) = x ^ { 2 } + m x + 2 8$ 其中 $a , b , m$ 均为正整数，你认为 $m$ 可取哪些值？它与 $a , b$ 的取值有关吗？请你写出所有满足题意的 $m$ （20的值. + +18. $\circledcirc$ 视讲（12分)一张长方形铁皮如图1所示，四个角都剪去边长为30（cm）的正方形，再四周折起，做成一个有底无盖的铁盒如图2所示,铁盒底面长方形的长是 $4 a ( \mathrm { c m } )$ ，宽是 $3 a ( \mathrm { c m } )$ ，这个无盖铁盒各个面的面积之和称为铁盒的表面积. + +(1)请用含 $a$ 的代数式表示图中原长方形铁皮的面积. + +(2)若要在铁盒的各个外表面漆上某种油漆，每元钱可漆的面积为 ${ \frac { a } { 5 0 } } ( \mathrm { c m } ^ { 2 } )$ ，则油漆这个铁盒需要多少钱？（用含 $a$ 的代数式表示） + +![](images/4ac5e58a9e1d159b053e76e8c0b1b24106f6a1fcab84cea3d4665b5abfe54cb9.jpg) +图1 + +![](images/471d45e26a1d2b0f9e1d3455956d33f1779c774a984bca221b90d10b7147b7e2.jpg) +图2 + +# 第二章 相交线与平行线 + +# 基础过关测试卷 + +时间：60分钟 满分：100分 + +
题序二三评卷人总分
得分
+ +# 一、选择题(每小题4分，共32分） + +1.下面的四个图形中， $\angle 1$ 与 $\angle 2$ 是对顶角的是 + +![](images/09bfb80391d7cada3c9ec9b3a50bcc972fe2be22482e9edb69b73cbcc4dd3341.jpg) + +2.如果一个角的度数为 $6 0 ^ { \circ }$ ,那么这个角的补角的度数是 + +A.30° $\mathrm { B } , 6 0 ^ { \circ }$ （204号 C.90° D.120° + +3.如图，在立定跳远中，体育老师是这样测量运动员的成绩的，将一块直角三角板的一边附在起跳线上，另一边与拉直的皮尺重合，这样做的理由是（ ） + +A.垂线段最短 B.过两点有且只有一条直线C.两点之间，线段最短 D.过一点可以作无数条直线 + +![](images/fc6396eb4338900f4250315d3b56eefad5a7e9d27f45773d6194e94ab89b1e1f.jpg) +第3题图 + +![](images/b6551db5d54d147f026bdbe8e65a5ae41b948e73a18b43a33013d6bd04c360ec.jpg) +第4题图 + +![](images/172f6ccad68b806503983e360dcb7e6280b3d054388784683e1d2d20527bb09d.jpg) +第5题图 + +4.如图，直线 $a$ 和直线 $b$ 被直线 $c$ 所截，则 $\angle 1$ 与 $\angle 2$ 是 + +A.同位角 B.内错角 C.同旁内角 D.对顶角 + +5.如图，直线 $l _ { 1 } / / l _ { 2 }$ $\angle 1 = 5 5 ^ { \circ }$ ，则 $\angle 2$ 的度数为 + +A. $3 5 ^ { \circ }$ （204号 B.45° C. $5 5 ^ { \circ }$ （20 D.125° + +6.如图，直线 $a$ 和直线 $b$ 被直线 $c$ 所截，下列条件不能判定直线 $a$ 与 $b$ 平行的是（ + +A. $\angle 1 = \angle 3$ （204号 ${ \mathrm { B } } _ { \cdot } \angle 2 + \angle 4 = 1 8 0 ^ { \circ }$ C. $\angle 1 = \angle 4$ （204号 D. $\angle 3 = \angle 4$ （20 + +![](images/91164ed58178049018dbc2f1d9b78c460d93442e77e103f840bb4930a4d29bd4.jpg) +第6题图 + +![](images/5fbf38418b9e922932108778435b4529a39ba2195adc5abe2cc04689eb1a0165.jpg) +第7题图 + +![](images/35309b3e10db9ba87ed1b981dce53a1e4a08e26526cf858e4d0d225d5866ba5d.jpg) +第8题图 + +7.如图，直线 $l _ { 1 } , l _ { 2 } , l _ { 3 }$ 相交于点 $O ,$ 直线 $l _ { 4 } / / l _ { 1 }$ ，若 $\angle 1 = 1 2 4 ^ { \circ }$ $\angle 2 = 8 8 ^ { \circ }$ ，则 $\angle 3$ 的 度数为 + +A $2 6 ^ { \circ }$ （204号 B. $3 6 ^ { \circ }$ C.46° D.56° + +8. $\circledcirc$ 视频讲解如图所示，与 $\angle \alpha$ 构成同位角的角的个数为 + +A.1 B.2 C.3 D.4 + +# 二、填空题(每小题4分，共16分) + +9.如图所示，直线AB 和 $C D$ 相交于点 $O$ ，若 $\angle A O D$ 与 $\angle B O C$ 的度数之和为$2 1 6 ^ { \circ }$ ，则 $\angle A O C$ 的度数为 + +![](images/a7c545ac432eacbb1f095732f2ac012aa58a6fcf0a4b3a545bc7c4cf5571f578.jpg) +第9题图 + +![](images/89a47e96a23454043b98d7de694711411966dd8e19f7ce38df84c205e94e76c9.jpg) +第10题图 + +![](images/0703d9e526f150120fafab88f4c90b2d747dd0cec779a9a0e6bc6f09554eb308.jpg) +第11题图 + +![](images/7f5b9c17c03d868a3191d665c09d5f4945b02ea8b2dc82041decb5ad56fdcdc9.jpg) +第12题图 + +10.如图，直线 $a / / b$ ，直线 $c$ 与直线 $a , b$ 分别相交于点 $A , B ,$ 若 $\angle 1 = 6 0 ^ { \circ }$ ，则 $\angle 2$ 的度数为 + +11.如图, $\angle A = 7 0 ^ { \circ }$ ，点 $O$ 是 $A B$ 上一点，直线 $O D$ 与AB的夹角 $\angle B O D { = } 8 2 ^ { \circ }$ ，要 使 $O D / / A C$ ，则直线 $O D$ 绕点 $O$ 按逆时针方向至少旋转__.度. + +12.如图,AB//CD, $\angle D = 7 5 ^ { \circ }$ $\angle C A D : \angle B A C = 2 : 1 ,$ 则 $\angle C A D$ 的度数为 + +# 三、解答题(本大题6小题，共52分) + +13.(6 分)(1)如果把下图看成是直线 $A B , E F$ 被直线 $C D$ 所截,那么 $\angle 1$ 与 $\angle 2$ 是一对什么角？ $\angle 2$ 与 $\angle 3$ 呢？ + +(2)如果把下图看成是直线 $A B , C D$ 被直线 $E F$ 所截,那么 $\angle 2$ 与 $\angle 5$ 是一对什么角？ $\angle 4$ 与 $\angle 5$ 呢？ $\angle 5$ 与 $\angle 6$ 呢？ + +![](images/8db2e574dc67bba88b40bc4bd6c72772a8ca950c03ebfb22f8650375bee08dc3.jpg) + +14.(8分)如图,利用尺规作图,在 $A C$ 的上方作 $\angle C A D = \angle A C B$ ,并说明AD// CB.(尺规作图要求保留作图痕迹,不写作法) + +![](images/8375a430c0dc71ec1d71a2d66fefd88eb325471d8b9eeeb14b16c40146bc402a.jpg) + +15.(8分)如图所示，点 $O$ 为直线 $B D$ 上一点， $O C \bot O A$ ，垂足为点 $\quad O , \angle C O D =$ $2 \angle B O C$ ，求 $\angle A O B$ 的度数. + +![](images/47510441fb6eba02ad4fe86fd36e42f373e75f6162192f5d537aa58ad91540f7.jpg) + +16.(8分)如图 $, A D / / B C , \angle E A D = \angle C , \angle F E C = \angle B A E ,$ $\angle E F C = 5 0 ^ { \circ }$ + +(1)试说明线段 $A E$ 与线段 $C D$ 平行； + +(2)求 $\angle B$ 的度数. + +![](images/c717d5c11b67ec92f2ac5f0ef8ce816efa0d52e5e4beb39e003fd4e179e8c887.jpg) + +17.(10分)如图， $B$ 处在 $A$ 处的南偏西 $4 5 ^ { \circ }$ 方向， $C$ 处在 $B$ 处的北偏东 $8 0 ^ { \circ }$ 方向. + +(1)求 $\angle A B C$ 的度数. + +(2)要使 $C D / / A B , D$ 处应在 $C$ 处的什么方向？ + +![](images/d68deb47892e9e184f3fd8876399a1a5286deb2b82416cab43b0208f02b3969d.jpg) + +18. $\circledcirc$ 视频讲解（12分）（1） $\textcircled{1}$ 如图1,已知 $A B / / C D$ $\angle A B C = 6 0 ^ { \circ }$ ，根据，可得 $B C D =$ 。； + +$\textcircled{2}$ 如图2,在 $\textcircled{1}$ 的条件下,若 $C M$ 平分 $\angle B C D$ ，则 $\ _ { - } B C M =$ 。$\textcircled{3}$ 如图3，在 $\textcircled{1}$ 和 $\textcircled{2}$ 的条件下,若 $C N \bot C M$ 则 $\angle B C N =$ + +(2)尝试解决下面的问题：如图 $4 , A B / / C D$ $\angle B = 4 0 ^ { \circ }$ ,CN是 $\angle B C E$ 的平分线， $C N \bot C M ,$ 求 $\angle B C M$ 的度数. + +![](images/59f4f05a35ec3e2742ee566a14c2141eb76caf3765da9206ceb645aaabeecc9e.jpg) +图1 图2 + +![](images/3b4311782887b174853f5596e1952c5e6430b135cfd859f9661ed93b7b5dfdcc.jpg) +图3 + +![](images/3b3b7271fae7ce7bab8dd69f5296b2ba9a3764cba158e476b000b79930d11b4d.jpg) +图4 + +
学校
+ +# 能力提优测试卷 + +时间：60分钟 满分：100分 + +
题序二三评卷人总分
得分
+ +# 一、选择题(每小题4分，共32分) + +1.下列说法不正确的是 + +
姓名
+ +
学号
+ +A.对顶角相等 +B.若 $\angle \alpha = 1 4 ^ { \circ } 2 6 ^ { \prime }$ ，则 $\angle \alpha$ 的余角为 $7 5 ^ { \circ } 3 4 ^ { \prime }$ +C.如果 $a / / b , a / / c$ ,那么 $b / / c$ +D.如果 $a \perp b , a \perp c$ ,那么 $b \perp c$ + +2.如图，与 $\angle 1$ 是同旁内角的是 + +A. $\angle 2$ B. $\angle 3$ （20 C.∠4 D.∠5 + +![](images/332e60154b6f3e43a3bbb5388b06a36d38ee5df1e5fbbf090a393e27f3028460.jpg) +第2题图 + +![](images/07bb78fa0f38d2df5b32bafe41f2587c5e9f9f317e7ffdb7b6c4ef32956216b5.jpg) +第3题图 + +![](images/18a8baa75a52fd0254186604a494678fca22efafb70779e7bdea77dfcf8d7f1f.jpg) +第4题图 + +![](images/fc050aec774b96e9a21717f13e3e7dde74ac3eae765a30d8319d135c05ff9238.jpg) +第5题图 + +3.如图，直线 $A B , C D$ 相交于点 $O , O E \bot A B$ 于点 $O , \angle C O E { = 5 5 ^ { \circ } }$ ，则 $\angle B O D$ 的 度数是 + +A. $3 5 ^ { \circ }$ （20 B.45° C.30° D.40° + +4.如图， $A B / / C D$ $\angle F E B = 7 0 ^ { \circ }$ $\angle E F D$ 的平分线 $F G$ 交 $A B$ 于点 $G$ ，则 $\angle E F G$ 的度数为 （ + +A. $6 3 ^ { \circ }$ （20 B.53° C.65° D.55° + +5.如图，直线 $A C / / B D , A O , B O$ 分别是∠BAC, $\angle A B D$ 的平分线，则 $\angle B A O$ 与$\angle A B O$ 的关系一定为 （ ） + +A.互余 B.相等 C.互补 D.无法判断 + +6.如图， $A B / / C D$ ，直线 $E F$ 与 $A B , C D$ 分别交于点 $G , H , I G \bot E F$ 于点 $G$ ，$\angle A G I = 4 3 ^ { \circ }$ ，则 $\angle E H D$ 的度数为 （ ） + +A. $5 7 ^ { \circ }$ （204号 B.53° C.47° D. $4 3 ^ { \circ }$ + +![](images/bc1fe50ae877da38d6391aa2cc376cc1c465d26dafe2ceadb65d11fe414732b7.jpg) +第6题图 + +![](images/60eee65a70c665e81791f2c056c7087cf32c545520c3eb0c1de6e3be16a18662.jpg) +第7题图 + +![](images/14f1642520409e0e620e9b695a564541f244d266cada4fbc8fb64c1ab22b8a0a.jpg) +第8题图 + +7.如图， $A B / / C D$ ,则根据图中标注的角，下列关系中,成立的是 + +A $\angle 1 = \angle 3$ $3 . \angle 2 + \angle 3 = 1 8 0 ^ { \circ }$ C. $\angle 2 + \angle 4 < 1 8 0 ^ { \circ }$ $\mathrm { D } . \angle 3 + \angle 5 = 1 8 0 ^ { \circ }$ + +8. 新考法真实问题情境视频讲解如图，这是一汽车探照灯纵剖面，从位于点$O$ 的灯泡发出的两束光线 $O B , O C$ 经过反射以后平行射出，若 $\angle A B O = \alpha$ ，$\angle D C O { = } \beta$ 则 $\angle B O C$ 的度数是 （） + +$$ +\mathrm { A } . \alpha + \beta \qquad \mathrm { B } . 1 8 0 ^ { \circ } - \alpha \qquad \mathrm { C } . \frac { 1 } { 2 } ( \alpha + \beta ) \qquad \mathrm { D } . 9 0 ^ { \circ } + \alpha + \beta +$$ + +# 二、填空题(每小题4分，共16分) + +9.如图，已知 $\angle \alpha = 6 0 ^ { \circ }$ $\angle \beta = 7 5 ^ { \circ }$ ,根据尺规作图的痕迹，可得 $\angle A O C$ 的度数为 + +![](images/88c1fda811c8bf79a9007645cc017af0f7f52a851cc3b3eaab1df757afbbac25.jpg) +第9题图 + +![](images/bcb1a86fa5f96dceded4306f00855951fa8cf1dbd48a1ee4e1e525590985fd6c.jpg) +第10题图 + +10.如图,要使得 $A D / / B F$ ，则需要添加的条件是 (写出一个即可). + +11.如图,已知AE//BC, $\angle B A C = 1 0 0 ^ { \circ }$ $\angle D A E = 5 0 ^ { \circ }$ ，则 $\angle C =$ + +![](images/a216bdbc67b8b1a80bce392e2beee3cb6f3539d673a1ac866a2607398b044152.jpg) +第11题图 + +![](images/56dd743c1c148a11b2c243aea3431640a12631ecdd9e67b15f43ac32de026cc7.jpg) +第12题图 + +12如图，点 $A , C , F , B$ 在同一直线上, $C E$ 平分 $\angle A C D , F G / / C D$ ，若 $\angle E C A = \alpha ^ { \circ }$ ， 则 $\angle G F B$ 的度数为 （用含 $\alpha$ 的代数式表示). + +# 三、解答题(本大题6小题，共52分) + +13.(6分)若一个角的余角比它的补角的 $\frac 1 2$ 小 $2 0 ^ { \circ }$ ,求这个角的度数. + +14.(8分)如图,在四边形 $A B C D$ 中， $B D$ 是连接两个顶点的线段,若 $\angle 1 = \angle 2$ ，$\angle A = 5 5 ^ { \circ } 1 6 ^ { \prime }$ ，求 $\angle A D C$ 的度数. + +![](images/c5c0f0993a8fa7f653f8d2a4adc1640ee61714313fd0ca568f607bfc32f36aad.jpg) + +15.(8分)如图 $, A B / / C D , B G$ 平分∠ABD, $\angle E D F { = } 7 0 ^ { \circ }$ ，求 $\angle F B G$ 的度数. + +A FBGC D E + +16.(8分)如图， $C D \perp A B , E F \perp A B , \angle E = \angle E M C , C D$ 是 $\angle A C B$ 的平分线吗? 请说明理由. + +![](images/755f858ea92f95d87afe6a0237eb29ed31baa404119df315fbc46caa85ad399a.jpg) + +17.(10 分)如图,已知 $A D / / B C , \angle A = \angle B .$ （20 + +(1)试判断 $A F$ 与 $B E$ 的位置关系,并说明理由； + +(2)若 $\angle B O D { = } 3 \angle B$ ，求 $\angle A$ 的度数. + +![](images/b2a7e854eb7ea0fed18167c8c1bc2d43e6da8b69dbb6be2fc8e4d613dff394b2.jpg) + +18. $\circledcirc$ 视频讲解（12分）如图，点 $C$ 在 $\angle A O B$ 的一边 $O A$ 上，过点 $C$ 的直线 $D E$ /$O B , C F$ 平分 $\_ A C D , C G \bot C F$ ，垂足为点 $C .$ + +(1)若 $\angle A O B = 4 0 ^ { \circ }$ ，求 $\angle E C F$ 的度数. + +(2)试说明： $C G$ 平分 $\angle O C D$ + +(3）当 $\angle A O B$ 为多少度时， $C D$ 平分 $\angle O C F ?$ 请说明理由. + +![](images/a93e0a7d161995dd4ca25e09ea8a0f9156650a6357b1377fec9adc11b2da533c.jpg) + +# 月考测试卷（一） + +时间：90分钟 满分：120分考试范围：整式的乘除 + +
题序二三评卷人总分
得分
+ +# 一、选择题(本大题共10小题，每小题3分，共30分) + +1.下列运算结果为 $a ^ { 6 }$ 的是 + +A. $a ^ { 2 } \cdot a ^ { 3 }$ B. $a ^ { 1 2 } \div a ^ { 2 }$ （20 C. $( a ^ { 3 } ) ^ { 2 }$ （204号 D. $( \frac { 1 } { 2 } a ^ { 3 } ) ^ { 2 }$ + +2.某种病毒的平均直径大约在 $8 0 \sim 1 4 0$ 纳米之间，已知1纳米 $= 1 0 ^ { - 9 }$ 米，则140纳米用科学记数法可表示为 （ + +A. $1 4 0 \times 1 0 ^ { - 9 }$ 米 B. 1. $4 \times 1 0 ^ { - 7 }$ 米C. $1 4 \times 1 0 ^ { - 8 }$ 米 D. 1. $4 \times 1 0 ^ { - 8 }$ 米 + +3.如图,阴影部分的面积是${ \frac { 7 } { 2 } } x y$ B ${ \frac { 9 } { 2 } } x y$ C.4xy D.2xy + +![](images/55c7078cd429e2c982ad8e76aa7baa8cf6a53417ce7bb7dc538d5c1195669578.jpg) + +4.计算 $2 8 x ^ { 4 } y ^ { 2 } \div ( - 7 x ^ { 3 } y )$ 的正确结果是 + +A.4xy B.-4xy C.4x²y D.4xy² + +5.化简 $x ( 2 x - 1 ) - x ^ { 2 } ( 2 - x )$ 的结果是 $\mathrm { A } , - x ^ { 3 } - x \qquad \mathrm { B } , x ^ { 3 } - x \qquad \mathrm { C } , - x ^ { 2 } - 1$ D. $x ^ { 3 } - 1$ （20 + +6.已知 $m + n { = } 2 , m n { = } - 2$ ，则 $( 1 + m ) ( 1 + n )$ 的值为 + +A.6 B.-2 C.0 D.1 + +7.如图，从边长为 $( a + 4 ) { \mathrm { ~ c m } }$ 的正方形纸片中剪去一个边长为 $( a + 1 ) \ \mathrm { c m }$ 的小正方形（ $_ { ( a > 0 ) }$ ，剩余部分沿虚线又剪拼成一个长方形(不重叠、无缝隙)，则长方形的面积为 （ ） + +A. $( 2 a ^ { 2 } + 5 a$ ）cm² B.( $3 a + 1 5$ ）cm² C. $\left( 6 a + 1 5 \right) ~ \mathrm { c m } ^ { 2 }$ D $\ : ( 8 a + 1 5 ) \ : \mathrm { ~ c m ^ { 2 } }$ + +![](images/4e0e9cd28098091d41436ac2d2af9075c51c928ee09515918460739a3df88a2f.jpg) + +8.豆豆认为下列括号内都可以填 $a ^ { 4 }$ ,你认为使等式成立的只能是 + +A $\mathrm { ~ ~ \cdot ~ } a ^ { 1 2 } = ( \mathrm { ~ ~ \rho ~ } ) ^ { 2 }$ $\mathrm { B } , a ^ { 1 2 } = ( \phantom { a } ) ^ { 3 }$ $\mathrm { ~ \cal ~ C ~ } _ { \bullet } a ^ { 1 2 } = ( \mathrm { ~ ~ { ~ \partial ~ } ~ } ) ^ { 4 }$ （204号 $\mathrm { D } . { a } ^ { 1 2 } = ( \phantom { a } ) ^ { 6 }$ + +9.若 $9 ^ { x + 2 } 3 ^ { x + 5 } = 2 7 ^ { 2 x - 1 }$ ，则 $\mathcal { X }$ 等于 + +A.3 B.-3 C.4 D.-4 + +10. $\circledcirc$ 视频讲解对于等式 $( a + b ) ^ { 2 } = a ^ { 2 } + b ^ { 2 }$ ,甲、乙、丙三人有不同看法，则下列说法正确的是 （ ） + +
甲：无论a和b取何值， 等式均不能成立.乙：只有当a=0时，丙：当a=0或b=0 等式才成立.时,等式成立.
+ +A.只有甲正确 B.只有乙正确C.只有丙正确 D.三人说法均不正确 + +# 二、填空题(本大题共5小题，每小题3分，共15分) + +11.一颗人造地球卫星的速度为 $2 . 8 8 \times 1 0 ^ { 7 }$ 米/时，一架喷气式飞机的速度为$1 . 8 \times 1 0 ^ { 6 }$ 米/时，则这颗人造地球卫星的速度是这架喷气式飞机的速度的倍. + +12.如果一个三角形的底边长为 $2 x ^ { 2 } y + x y - y ^ { 2 }$ ，这条边上的高为 $6 x y$ ,那么这个三角形的面积为 + +13.新考法阅读理解题规定一种新运算“ $\divideontimes$ ”,对于任意实数 $a$ 和 $b$ ，有 $a \times b$ $= a \div b + 1$ ，则 $( 6 x ^ { 3 } y - 3 x y ^ { 2 } ) * 3 x y =$ + +14.随着数学学习的深人,数系不断扩充,引入新数 $i$ ，规定 $i ^ { 2 } = - 1$ ，并且新数 $i$ 的加、减、乘、除运算和整式的加、减、乘、除运算类似，则 $\left( 1 + i \right) \left( 2 - i \right)$ 的运算结果是 + +15.新考法代数推理 $\circledcirc$ 视频讲解请你计算： $( 1 - x ) ( 1 + x ) , ( 1 - x ) ( 1 + x + x ^ { 2 } ) , \cdots ,$ 猜想 $( 1 - x ) ( 1 + x + x ^ { 2 } + \cdots + x ^ { 2 0 2 4 } )$ 的结果是 + +# 三、解答题(本大题共8小题，满分75分) + +16.(10分)（1)计算： $a ^ { 3 } \bullet a + ( - a ^ { 2 } ) ^ { 3 } \div a ^ { 2 }$ (2）计算： $( \pi - 3 . 1 4 ) ^ { \circ } - \Bigl ( - \frac { 1 } { 2 } \Bigr ) ^ { - 2 } + ( - 2 ) ^ { 2 } \times ( - 1 ) ^ { 2 0 2 4 } .$ + +17.(9分)已知 $a ^ { m } = 4 , a ^ { n } = \frac { 1 } { 2 }$ ，求 $a ^ { 2 m - 3 n }$ 的值. + +18.(9分)先化简,再求值： $( 2 a + b ) ( 2 a - b ) - ( 4 a b ^ { 3 } - 8 a ^ { 3 } b ) \div 2 a b \cdot$ 其中 $a = - 1$ $b { = } - 2 .$ （204号 + +![](images/c754a00b53adf7d98dd422ef604f3c9d1e2c30444d0237522348b5ba7a369518.jpg) + +19.(9分)已知 $( x { + } 7 )$ 和 $( x { - } 3 )$ 的积为 $x ^ { 2 } + m x ^ { - } n$ ，求 $( 5 m - n ) ^ { 2 0 2 4 }$ 的值. + +20.（9分)已知 $\ x ^ { 2 } + 2 x { = } 4$ ，求代数式 $x ( x - 2 ) ^ { 2 } - x ^ { 2 } ( x - 6 ) - 3$ 的值. + +21.(9分)小颖在计算某一个多项式乘一 $3 \mathcal { x } ^ { 2 }$ 时，因抄错运算符号，写成了加上$- 3 x ^ { 2 }$ ，结果得到 $x ^ { 2 } - 4 x + 1$ ,那么原题正确的计算结果是什么？请计算出正确的结果. + +22.（10分)如图，长方形ABCD的周长为16,以长方形的四条边为边长向外作四个正方形，若这四个正方形的面积之和为68,求长方形ABCD的面积. + +![](images/fd0b649c896175f81415e08d645f3e92a3e847d55cfb5c7c73e3166f1e87da47.jpg) + +23. $\circledcirc$ 视讲(10分)我们通常用作差法比较代数式的大小.例如：已知 $M { = } 2 x +$ $3 , N { = } 2 { x } { + } 1$ ，比较 $M$ 和 $N$ 的大小.先求 $M - N$ ，若 $M { - } N { > } 0$ ，则 $M { > } N$ ；若$M { - } N { < } 0$ ，则 $M { < } N$ ；若 $M { - } N { = } 0$ ，则 $M { = } N$ ,反之亦成立.本题中,因为 $M$ $- N { = } 2 x { + } 3 { - } ( 2 x { + } 1 ) { = } 2 { > } 0$ ，所以 $M { \ > } N .$ + +(1)如图1,这是一个边长为 $a$ 的正方形,将正方形一边不变,另一边增加4,得到如图2所示的新长方形，此长方形的面积为 $S _ { 1 }$ ;将图1中正方形边长增加2得到如图3所示的新正方形,此正方形的面积为 $S _ { 2 }$ $\textcircled{1}$ 用含 $a$ 的代数式分别表示 $S _ { 1 }$ 和 $S _ { 2 }$ （结果需要化简）；$\textcircled{2}$ 请用作差法比较 $S _ { 1 }$ 与 $S _ { 2 }$ 的大小. + +(2）若 $M { = } a ^ { 2 }$ ， $N { = } 4 { - } ( a { + } 1 ) ^ { 2 }$ ，且 $M { = } N$ 求 $\alpha ( a { + } 1 )$ 的值. + +![](images/c6829e9bc8ea0bef99a2be237868c01d2f9fc2926798f18bb8ece99391e16b6f.jpg) + +# 第三章 变量之间的关系 + +# 基础过关测试卷 + +时间：60分钟 满分：100分 + +
题序二三评卷人总分
得分
+ +# 一、选择题(每小题4分，共32分) + +1.某超市某种商品的单价为70元/件，若买 $_ { \mathcal { X } }$ 件该商品的总价为 $y$ 元,则其中的常量是 （ ） + +A.70 B. $_ { \mathcal { X } }$ C.y D.不确定 + +2.在利用太阳能热水器来加热水的过程中，热水器里的水温随所晒时间的长短而变化，这个问题中因变量是 （ ） + +A.太阳光的强弱B.水的温度 C.所晒时间 D.热水器的容积 + +3.在用图象表示变量之间的关系时，下列说法最恰当的是 + +A.用水平方向的数轴上的点表示因变量B.用竖直方向的数轴上的点表示自变量C.用横轴上的点表示自变量·D.用横轴或纵轴上的点表示自变量 + +4.变量 $y$ 与 $\mathcal { X }$ 之间的关系式是 $\scriptstyle { y = 2 x - 3 }$ ,当因变量 $y = 6$ 时，自变量 $\mathcal { X }$ 的值是（ + +A.15 B.9 C.4.5 D.1. 5 + +5.为了加强爱国主义教育，每周一学校都要举行庄严的升旗仪式，同学们凝视着冉冉上升的国旗，下列哪个函数图象能近似地刻画上升的国旗离旗杆顶端的距离与时间的关系 （ ） + +![](images/2501e7cb5a037909c2831198f8f79bec0e5f06e9987b438dc4e641928241582f.jpg) + +6.根据如图所示的程序计算变量 $y$ 的对应值，若输入变量 $\mathcal { X }$ 的值为一1,则输出 （20 $y$ 的值为 （ + +A.-2 B.2 C.-1 D.0 + +![](images/7b3b272397e842106fd0d1d45d2d9552309372f280068667b34554d54ca2ed4c.jpg) +第6题图 + +![](images/f882ee50ed053feaf7c27ff63e62c33b654001f1acef23d3e15f23d47359bdcc.jpg) +第8题图 + +7.某大剧场地面的一部分为扇形，观众席的座位数按下表所示的方式设置： + +
排数x1234
座位数y50535659
+ +有下列结论： $\textcircled{1}$ 排数 $\mathcal { X }$ 是自变量，座位数 $y$ 是因变量； $\textcircled{2}$ 排数 $\mathcal { X }$ 是因变量，座位数 $y$ 是自变量； $\textcircled { 3 } y = 5 0 + 3 x$ $\textcircled { 4 } y = 4 7 + 3 x .$ 其中正确的结论有 ） + +A.1个 B.2个 C.3个 D.4个 + +8. $\circledcirc$ 视讲如图，该图反映的是小明从家先去食堂吃早餐，接着去图书馆读报，然后回家的过程，其中 $\mathcal { X }$ 表示时间， $y$ 表示小明离家的距离，小明家、食堂、图书馆在同一直线上，根据图中提供的信息，下列说法正确的有 （）$\textcircled{1}$ 食堂离小明家 $0 . 4 \ \mathrm { k m }$ $\textcircled{2}$ 小明从食堂到图书馆用了 $3 ~ \mathrm { m i n }$ $\textcircled{3}$ 图书馆在小明家和食堂之间； $\textcircled{4}$ 小明从图书馆回家的平均速度是 $0 . 0 4 ~ \mathrm { k m / m i n }$ + +A.4个 B.3个 C.2个 D.1个 + +# 二、填空题(每小题5分，共20分) + +9.“早穿皮祅午穿纱，围着火炉吃西瓜"这句谚语反映了我国新疆地区一天中，温度随时间的变化而变化，其中自变量是 ，因变量是 + +10.李老师带领 $\mathcal { X }$ 名学生到某动物园参观，已知成人票每张20元，学生票每张10元.设门票总费用为 $y$ 元，则 $y$ 与 $\mathcal { X }$ 之间的关系式为 + +11.某人购进一批苹果到市场上零售，已知卖出苹果数量 $\mathcal { X }$ 与售价 $y$ 的关系如下表. + +
数量x/kg12345
售价y/元8.51725.53442.5
+ +写出用 $\mathcal { X }$ 表示 $y$ 的关系式： + +12.如图,这是小明从学校到家行进的路程s(米)与时间t(分)的函数图象.观察图象，从中得到如下信息：$\textcircled{1}$ 学校离小明家1000米； $\textcircled{2}$ 小明用了20分钟到家；$\textcircled{3}$ 小明前10分钟走了路程的一半； $\textcircled{4}$ 小明后10分钟比前 10 分钟走得快.其中正确的是 . (填序号) + +![](images/54572e5cbe3e37f31b117863c30a7a3aedb56a5454dfe05a88cf39770bc4a75a.jpg) + +# 三、解答题(本大题4小题，共48分) + +13.(10分)若用c表示摄氏温度，f表示华氏温度，则c与f之间的关系式为c$= \frac { 5 } { 9 } ( f - 3 2 )$ .试分别求： + +(1)当 $f { = } 6 8$ 和 $f { = } { - } 4$ 时， $c$ 的值; +(2)当 $c { = } 1 0$ 时， $f$ 的值. + +14.(10分)温度的变化是人们经常谈论的话题,请根据下面的图象与同伴讨论这一天温度变化的情况. + +（1)这一天的最高温度是多少？是在几时到达的？最低温度呢？（2)这一天的温差是多少？从最低温度到最高温度经过多长时间? + +![](images/a39cd68baf6c32d935fb6ec24d8ee727c56e70cfafa71e05ab881ada43bb741f.jpg) + +15.(13分)如图所示,将一个边长为 $1 2 \ \mathrm { c m }$ 的正方形的四个角都剪去一个大小相等的小正方形，当小正方形的边长由小到大变化时，图中阴影部分的面积也随之发生变化. + +(1)在这个变化过程中，自变量、因变量各是什么？ +(2)如果小正方形的边长为xcm,图中阴影部分的面积为ycm²,请写出y与 $\mathcal { X }$ 之间的关系式. +(3)当小正方形的边长由 $1 ~ \mathrm { c m }$ 变化到 $5 ~ \mathrm { c m }$ 时,图中阴影部分的面积是怎样变化的？ + +![](images/9c785ebffdea8a32ab4f53393056256e9576d2053466b11a9560c14e910e66eb.jpg) + +16.视濒讲（15分)某机动车出发前油箱内有油42L,行驶若干小时后，途中在加油站加油若干升,油箱中剩余油量 $Q ( \mathrm { L } )$ 与行驶时间 $t ( \mathrm { h } )$ 之间的关系如图所示.根据图象回答下列问题： + +（1)机动车行驶多少小时后加油？ +(2)途中加油多少升？ +(3)已知加油站距目的地还有 $2 3 0 ~ \mathrm { k m }$ ，车速为 $4 0 ~ \mathrm { k m / h }$ ,若要到达目的地,油箱中的油是否够用？请说明理由. + +Q/L 4236302418126 01234567891011t/h + +# 能力提优测试卷 + +时间：60分钟 满分：100分 + +班级 + +
学校
+ +
题序二三评卷人总分
得分
+ +# 一、选择题(每小题5分，共30分) + +
姓名
+ +1.2023 年10月19日，广东省茂名市遭受台风"三巴"和弱冷空气共同影响，大部分地区发生强降雨，某条河受暴雨袭击，一天的水位记录如下： + +
学号
+ +
时间/时04812162024
水位/米22.534.5568
+ +观察表中的数据，水位上升最快的时段是 + +A. $8 \sim 1 2$ 时 B. $1 2 { \sim } 1 6$ 时 $C . 1 6 { \sim } 2 0$ 时 D. $2 0 { \sim } 2 4$ 时 + +2.自动测温仪记录的图象如图所示，它反映了齐齐哈尔市的春季某天气温 $T$ 如何随时间 $t$ 的变化而变化，下列从图象中得到的信息正确的是 （ ） + +A.0点时气温达到最低 B.最低气温是零下 $4 ~ ^ { \circ } C$ C.0点到14点之间气温持续上升 D.最高气温是 $8 ~ ^ { \circ } C$ + +![](images/81b3700ab667276ebded8d3e2de13cc85cbb1dd67f5cc4cde677cd9d66480076.jpg) +第2题图 + +![](images/2910ef549cd54d24ceafe5cdf94a7a6027caeb5867d867509e637bbf8556c3b5.jpg) +第6题图 + +3.下面的表格列出了一个试验的统计数据，表示皮球从高处落下时，弹跳高度 $b$ （ $\mathrm { c m } )$ 与下降高度 $d ( \mathrm { c m } )$ 的关系，下列式子中，能表示这种关系的是 （ ） + +
考生注意
+ +
d/cm5080100150
b/cm25405075
+ +A. $b { = } d ^ { 2 }$ （204号 B. $b { = } 2 d$ $\mathrm { C } . \mathit { b } \mathrm { = } \frac { \mathit { d } } { 2 }$ $\mathrm { D } , b { = } d { + } 2 5$ + +4.某工厂有甲、乙两个大小相同的蓄水池，如图所示，且中间有管道连通.现要向甲池中注水,若单位时间内的注水量不变，则从注水开始，乙池水面上升的高度 $h$ 与注水时间 $t$ 之间的关系的图象可能是 （ ） + +![](images/cfc7cf376608c06fbf1ee069161c5e40282fc2ca2816867a2f193058976fe29e.jpg) + +5.向一个容器内均匀地注入水，液面升高的高度 $y$ 与注水时间 $x$ 满足如图所示的图象，则符合图象条件的容器为 （ ） + +![](images/a309d4823d11785bf0b72388263392fdf0cefc512bebd43e52c5ed3246627b5c.jpg) + +6. $\circledcirc$ 视频讲解如图1，在矩形 $M N P Q$ 中，动点 $R$ 从点 $N$ 出发，沿 $N { } P { } Q { } M$ 方向运动至点 $M$ 处停止,设点 $R$ 运动的路程为 $x , \triangle M N R$ 的面积为 $y$ ,如果 $y$ 关于 $\mathcal { X }$ 的函数图象如图2所示，那么当点 $R$ 运动到 $M Q$ 的中点时， $\triangle M N R$ 的面积为 （） + +A.5 B.9 C.10 D.不可确定 + +# 二、填空题(每小题5分，共20分) + +7.气象观测小组进行活动，一号探测气球从海拔 $5 ~ \mathrm { m }$ 处出发，以 $1 . 2 ~ \mathrm { m / m i n }$ 的速度上升，气球所在位置的海拔 $y$ （单位： $\mathrm { m }$ 与上升时间 $_ { \mathcal { X } }$ （单位： $\operatorname* { m i n }$ 之间的关系式为 + +8.小明从家跑步到学校，接着马上原路步行回家.如图，这是小明离家的路程 $y$ （米)与时间 $t$ （分)的函数图象，则小明回家的速度是每分钟步行 米. + +→y/米 800 0 5 15分 + +9.共享单车为市民的出行带来了方便.某共享单车公司规定，首次骑行需交99元押金，第一次骑行收费标准如下表所示：(不足半小时的按半小时计算) + +
骑行时间t/小时0.511.52
骑行费用y/元99+199+299+399+4
+ +则第一次骑行费用 $y$ (元)与骑行时间 $t$ （小时)之间的关系式为 + +10.小明步行，小颖骑车，他们同时从小颖家出发，以各自的速度 y/米匀速到公园游玩，小颖先到并停留了8分钟，发现相机忘在了家里，于是沿原路以同样的速度回家去取.若小明的步行可i0 20 1020/分钟速度为180米/分，他们各自距离出发点的路程 $y$ 与出发时间 $x$ 之间的关系图象如图所示，则当小明到达公园时小颖离家 米. + +# 三、解答题(本大题5小题，共50分) + +11.(8分)地壳的厚度约为 $8 ~ \mathrm { k m }$ 到 $4 0 ~ \mathrm { k m }$ ，在地表以下不太深的地方，温度可按关系式 $\scriptstyle { y = 3 5 x + t }$ 计算,其中 $_ { \mathcal { X } }$ 是深度 $( \mathrm { k m } ) , t$ 是地球表面温度， $y$ 是所达深度的温度.当地表温度为2℃时，分别计算当x为1km,5km,10 km,$2 0 ~ \mathrm { k m } , 3 0 ~ \mathrm { k m }$ 时地壳的温度. + +12.(10 分)在空中，自地面算起,每升高1千米,气温下降若干 $\mathrm { { } ^ { \circ } C }$ .某地空中气温 $t ( ^ { \circ } \mathrm { C } )$ 与高度 $h$ (千米)之间的图象如图所示. + +(1)该地地面气温为 $\mathrm { { } ^ { \circ } C }$ (2)当高度为 千米时,气温为 $0 ~ ^ { \circ } C$ (3)求出空中气温 $t ( ^ { \circ } \mathrm { C } )$ 与高度 $h$ (千米)之间的关系式. + +t/C241680 1234h千米 + +13.（10分)某公司计划购买A,B两种型号的电脑，已知购买一台A型电脑需0.6万元，购买一台 $B$ 型电脑需0.4万元，该公司准备投人资金 $y$ 万元，全部用于购进35台这两种型号的电脑,设购进 $A$ 型电脑 $\mathcal { X }$ 台. + +(1)求 $y$ 与 $\mathcal { X }$ 之间的关系式. +(2)若购进B型电脑的数量恰好为A型电脑数量的2.5倍，则该公司需要投人资金多少万元？ + +14.(10 分)甲、乙两家体育用品商店出售同样的乒乓球拍和乒乓球,乒乓球拍每副定价30元,乒乓球每盒定价5元.现两家商店搞促销活动，甲店：每买一副球拍赠一盒乒乓球.乙店：按定价的9折优惠.某班级需购买球拍4副，乒乓球若干盒(不少于4盒). + +(1)设购买乒乓球为 $\mathcal { X }$ 盒，在甲店购买的付款金额为 $y _ { \mathbb { H } }$ 元,在乙店购买的付款金额为 $y _ { Z }$ 元,分别写出在两家商店购买的付款金额与乒乓球盒数 $\mathcal { X }$ 之间的关系式. + +（2)购买多少盒乒乓球时，在两家商店购买的付款金额一样多? + +15.视频讲（12分)新能源纯电动汽车的不断普及让很多人感受到了它的好处，其中最重要的一点就是对环境的保护.如图，这是某型号新能源纯电动汽车充满电后,蓄电池剩余电量y(千瓦·时)与已行驶路程 $\mathcal { X }$ (千米)之间的关系图象. + +(1)图中点 $A$ 表示的实际意义是什么？(2)当汽车行驶了120千米时，求蓄电池的剩余电量.(3)当汽车行驶了多少千米时，剩余电量降至20千瓦·时？ + +↑y千瓦·时 +60 A +35 +10 +0 150200x/干米 + +# 期中测试卷 + +时间：90分钟 满分：120分考试范围：第一章～第三章 + +
题序二三评卷人总分
得分
+ +# 一、选择题(本大题共10小题，每小题3分，共30分） + +1.下列运算正确的是 + +A. $( a ^ { 3 } ) ^ { 4 } = a ^ { 1 2 }$ B $( - 2 a ) ^ { 2 } = - 4 a ^ { 2 }$ $\mathrm { C } . a ^ { 3 } \cdot a ^ { 3 } = a ^ { 9 }$ $\operatorname { D } . a ^ { 6 } \div a ^ { 2 } = a ^ { 3 }$ + +2.如图， $A C \bot B C$ ，直线 $E F$ 经过点 $C$ ，若 $\angle 1 = 3 5 ^ { \circ }$ ，则 $\angle 2$ 的度数为 （） + +A $3 5 ^ { \circ }$ （20 B.45° C.55° D. 65° + +![](images/e58b9b8f0d5b3b107d85b633a189613f9cdd05e3f3678dd6b6412a0b97abcad5.jpg) +第2题图 + +![](images/30f13a80499e1dd8be7addf79730ba9a13bd52821386add2efe896d309bd4d12.jpg) +第6题图 + +3.据专家介绍，某种病毒的直径大约为0.000000125米，它与飞沫等体液结合后体积会变大，正确佩戴口罩能有效预防.数据0.000000125用科学记数法表示为 （ ） + +A. $0 . 1 2 5 { \times } 1 0 ^ { - 8 }$ $\mathrm { B } . 1 2 . 5 { \times } 1 0 ^ { - 6 }$ （204号 $\mathrm { C } . 1 . 2 5 { \times } 1 0 ^ { - 6 }$ （20 （204号 $\mathrm { D } . 1 . 2 5 { \times } 1 0 ^ { - 7 }$ + +4.一根弹簧长 $8 ~ \mathrm { c m }$ ，它所挂物体的质量不能超过 $5 ~ \mathrm { k g }$ ，并且所挂物体的质量每增加 $1 ~ \mathrm { k g }$ ，弹簧就伸长 $0 . 5 ~ \mathrm { c m }$ ,则挂上物体后弹簧的长度 $y ( \mathrm { c m } )$ 与所挂物体的质量 $x ( \mathrm { k g } ) ( 0 { \leqslant } x { \leqslant } 5 )$ 之间的关系式为 C ） + +A. $y { = } 0 . 5 ( x { + } 8 )$ $\mathrm { B } . \mathrm { { } } y { = } 0 . 5 x { - } 8 $ +C. $y { = } 0 . 5 ( x { - } 8 )$ $\mathrm { D } . { \it y } = 0 . 5 x ^ { + } 8$ + +5.一个长方体的高为 $x \ \mathrm { c m }$ ,长比高的3倍少 $4 \mathrm { \ c m }$ ，宽是高的2倍，那么这个长方体的体积是 （ ） + +A. $( 3 x ^ { 3 } - 4 x ^ { 2 } ) \mathrm { { c m } ^ { 3 } }$ B $\dot { \cdot } ( 6 x ^ { 3 } + 8 x ^ { 2 } ) \mathrm { c m } ^ { 3 }$ C. $( 6 x ^ { 3 } - 8 x ^ { 2 } ) \mathrm { c m } ^ { 3 }$ $\mathrm { D } . ( 6 x ^ { 2 } - 8 x ) \mathrm { c m } ^ { 3 }$ + +6.如图，下列条件中,能判定 $A B / / C D$ 的是 + +$$ +\Delta . \angle 1 = \angle 2 \qquad \mathrm { B } . \angle 3 = \angle 4 \qquad \mathrm { C } . \angle B = \angle D \qquad \mathrm { D } . \angle D = \angle 5 +$$ + +7.当 $a = - 1$ 时,式子 $( 2 8 a ^ { 3 } b - 1 4 a ^ { 2 } b + 7 a b ) \div ( - 7 a b )$ 的值是 + +$$ +\mathrm { A } , - 7 \mathrm { B } , - 6 \mathrm { C } , - 3 \mathrm { D } , - 2 +$$ + +8. $\circledcirc$ 视频讲解如图，直线 $A B / / C D / / E F$ ，点 $O$ 在直线 $E F$ 上，下列结论正确的是（ + +A. $\angle \alpha + \angle \beta - \angle \gamma = 9 0 ^ { \circ }$ B $. \angle \alpha + \angle \beta + \angle \gamma = 1 8 0 ^ { \circ }$ C.∠x+∠β-∠α=180° D.∠a+∠γ-∠β=180° + +![](images/56a0385115498cecfdbd7e93fd3f31282b8bea3e63526f27793646bc5af918cc.jpg) +第8题图 + +![](images/a991c53c7a4836cc578c6967bea3ccde4ae581755849596f0c4aac971f9129f5.jpg) +第10题图 + +9. 新考法阅读理解题定义一种新的运算：若 $a \neq 0$ ，则有 $a \mathbf { A } b = a ^ { - 2 } + a b +$ |-b|.那么 $( - \frac { 1 } { 2 } ) \pmb { \Delta } 2$ 的值是 + +A.-3 B.5 C.13 D $\frac { 3 } { 2 }$ + +10.小明和小华是同班同学，也是邻居.某日早晨,小明7：40 先出发去学校，走了一段路后，在途中停下吃了早餐，后来发现上学时间快到了，就跑步到学校；小华离家后直接乘公共汽车到了学校.他们从家到学校已走的路程s(米)和所用时间t(分钟)的关系图象如图所示.则下列说法中，错误的是 （ ） + +A.小明吃早餐用时5分钟B.小华到学校的平均速度是240米/分C.小明跑步的平均速度是100米/分D.小华到学校的时间是7：55 + +# 二、填空题(本大题共5小题，每小题3分，共15分) + +11.某人购进一批葡萄到市场上零售，已知卖出葡萄质量 $\mathcal { X }$ 与销售额 $y$ 之间的关系如下表： + +
质量x/千克12345
销售额y/元612182430
+ +则当卖出葡萄质量 $_ { \mathcal { X } }$ 为10千克时,销售额 $y$ 为 元 + +12.已知 $x ^ { + } y { = } 2 x y$ ，则 $( 2 x - 1 ) ( 2 y - 1 )$ 的值为 + +13.如图,已知直线 $E F \bot M N$ 垂足为点 $F$ ，且 $\angle 1 = 1 3 8 ^ { \circ }$ ，则当 $\angle 2 =$ 时， $A B / / C D$ + +![](images/caa23f6889a06123a156d3a4016accbafbce5e2197abc3c12cfcfb3d31cb3378.jpg) +第13题图 + +![](images/4a1afd0387d10b5e51a2da8da622db3945bf77e731d5f4cb88bf281dd9c32d4f.jpg) +第14题图 + +图1 + +![](images/25e90cfea0d3fa9ef4299f4d0e3f7d76e149c168355275316bb73dff4ae70512.jpg) +第15题图 + +14.如图,AB//DE, $\angle 1 = 1 3 5 ^ { \circ }$ $A C \bot C D$ ，则 $\angle D$ 的度数为 + +15.新考法数学文化 $\circledcirc$ 靓频讲我国宋朝数学家杨辉在他的著作《详解九章算法》中提出的“杨辉三角"如图1所示，并观察如图2所示的等式.根据前面各式的规律，请你猜想 $( 2 x - 1 ) ^ { 6 }$ 的展开式中含 $x ^ { 5 }$ 项的系数是 + +![](images/ee7768ebb741ee407069dbb2ecdd824b00d93e87e1e3292a714bc3f50cac3bcf.jpg) +图2 + +# 三、解答题(本大题共8小题，满分75分） + +16.(10分)（1)计算： $( x + 2 ) ( 4 x - 1 ) - 2 x ( 2 x - 1 ) .$ + +(2)如图， $A B / / C D$ ，点 $E$ 在 $A B$ 上，点 $F$ 在 $C D$ 上，如果 $\angle C F E : \angle E F B =$ （204号 $3 : 4$ $\angle A B F { = } 4 0 ^ { \circ }$ ，求 $\angle B E F$ 的度数. + +17.(9分)已知 $5 x ^ { 2 } - x - 2 = 0$ ，求代数式 $( 3 x + 2 ) ( 3 x - 2 ) + x ( x - 2 )$ 的值. + +18.(9分)如图，把一些相同规格的碗整齐地叠放在水平桌面上，这擦碗的高度随着碗的数量变化而变化的情况如下表所示： + +
碗的数量/只12345：
高度/cm45.26.47.68.8：
+ +![](images/b77c674c3552af08200c0107c317ddee910577b2c1ae9325d7d7639e74b3bfd0.jpg) + +(1)若用 $h ( \mathrm { c m } )$ 表示这擦碗的高度，用 $_ { \mathcal { X } }$ (只)表示这擦碗的数量，请用含有 $\mathcal { X }$ 的代数式表示 $h$ + +(2)若这擦碗的高度为 $1 1 . 2 \ \mathrm { c m }$ ,求这擦碗的数量. + +20.（9分)为了加强公民的节水意识，合理利用水资源，某城市规定用水收费标准如下：每户每月用水量不超过6立方米时，水费按 $a$ 元/立方米收费；每户每月用水量超过6立方米时，不超过的部分每立方米仍按 $a$ 元收费，超过的部分按 $c$ 元/立方米收费.该市某用户今年3,4月份的用水量和水费如下表所示： + +
月份用水量x/立方米收费y/元
3510.5
4921.6
+ +(1)求 $a , c$ 的值，并写出每月用水量不超过6立方米和超过6立方米时，水费$y$ 与用水量 $\mathcal { X }$ 之间的关系式； + +（2)已知某用户5月份的用水量为8立方米,求该用户5月份的水费. + +19.(9分）如图， $B , C , E$ 三点在同一直线上，如果 $\angle D A E = \angle E$ ， $\angle B = \angle D$ ,那么 $A B$ 与 $C D$ 平行吗？请说明理由. + +![](images/8a04bc1e9663d4ad4c993d6cb04b79ae2122b39628050c38286c8122a55ef14e.jpg) + +21.(9分)如图， $F G , E D$ 分别交 $B C$ 于点 $M , N , \angle E N C + \angle C M G = 1 8 0 ^ { \circ } , A B / ,$ /$C D$ + +$( 1 ) \angle 2 = \angle 3$ 吗？为什么？(2)若 $\angle A = \angle 1 + 7 0 ^ { \circ }$ $\angle A C B = 4 2 ^ { \circ }$ ，求 $\angle B$ 的度数. + +![](images/b9c0fcad0708c1213588412537f8fb20d4bf46ad6cfe1db819bf22bbff5e73f6.jpg) + +![](images/fe9d16609172101cb21c32fc87540304e42868534b4455ac0750db3b2e104162.jpg) + +22.(10分)一辆汽车在某次行驶过程中,油箱中的剩余油量y(升)与行驶路程 $\mathcal { X }$ (千米)之间的关系，其部分图象如图所示. + +(1)求 $y$ 与 $\mathcal { X }$ 之间的关系式（不必注明 $_ { \mathcal { X } }$ 的取值范围). + +(2)已知当油箱中的剩余油量为8升时，该汽车会开始提示加油，在此次行驶过程中,行驶了 500千米时,司机发现离前方最近的加油站有 30千米的路程,在开往该加油站的途中,汽车开始提示加油,这时离加油站的路程是多少千米？ + +![](images/2098999b8fea4cc32dbbf4a38299b988eb44a62b98ff276f04648de3362b480f.jpg) + +23.新考法阅读理解题 $\circledcirc$ 视频讲解（10分）我们将 $( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }$ 进行变$a ^ { 2 } + b ^ { 2 } = ( a + b ) ^ { 2 } - 2 a b , a b = { \frac { ( a + b ) ^ { 2 } - ( a ^ { 2 } + b ^ { 2 } ) } { 2 } } \colon$ 2等.根据以上变形解决下列问题： + +(1)已知 $a ^ { 2 } + b ^ { 2 } = 8 , ( a + b ) ^ { 2 } = 4 8$ ，则 $a b { = } \quad \quad$ (2)若 $_ { \mathcal { X } }$ 满足 $( 2 5 - x ) ( x - 1 0 ) = - 1 5$ ，求 $( 2 5 - x ) ^ { 2 } + ( x - 1 0 ) ^ { 2 }$ 的值； + +(3)如图,四边形 $A B E D$ 是梯形， $D A \_ { \perp } A B , E B \_ A B , A D = A C , B E = B C ,$ 连接 $C D , C E$ ，若 $A C \cdot B C { = } 1 0$ ，则图中阴影部分的面积为 + +![](images/be379db5d023ded231c8f17dabd1286c2520191726d5b5efa14c93de75d2d827.jpg) + +
学校
+ +# 第四章 三角形 + +# 基础过关测试卷 + +班级 + +时间：60分钟 满分：100分 + +
题序二三评卷人总分
得分
+ +# 一、选择题(每小题4分，共32分) + +1.小明用三根火柴组成的图形如下图所示,则其中是三角形的是 + +
姓名
+ +A 4 A B C D + +（ ） + +2.下列图形具有稳定性的是 + +
学号
+ +![](images/77d88d42f67f7f25a997b2fc31daecd3c8951c74dbc99ba448b9e0e5b942c737.jpg) + +3.下列各组图形属于全等图形的是 + +![](images/f2493801f002041d056d8f587cd65388731a9d6bb2c3333eb41d38d78d63b75d.jpg) + +4.下列每组数分别是三根木棒的长度，不能用它们摆成三角形的是 + +A $. 5 \ \mathrm { c m } , 8 \ \mathrm { c m } , 1 2 \ \mathrm { c m }$ $\mathrm { B . 6 ~ c m , 8 ~ c m , 1 2 ~ c m }$ $\mathrm { C . 5 ~ c m , 6 ~ c m , 8 ~ c m }$ （204号 $\mathrm { D . 5 ~ c m , 6 ~ c m , 1 2 ~ c m }$ + +5.下列条件中，能够判定 $\triangle A B C { \underline { { \triangle } } } \triangle D E F$ 的是 + +A $A B = D E , B C = E F , \angle A = \angle E$ B $. A B { = } D E , B C { = } E F$ $\angle A = \angle D$ $\scriptstyle \complement , \angle A = \angle E , A B = D F , \angle B = \angle D$ D. $\scriptstyle \angle A = \angle D , \angle B = \angle E , B C = E F$ + +6.下列不一定在三角形内部的线段是 + +A.三角形的角平分线 B.三角形的中线C.三角形的高 D.以上皆不对 + +7.在 $\triangle A B C$ 中， $\angle A = \angle B + \angle C$ 则 $\triangle A B C$ 的形状是 （） + +A.锐角三角形 B.直角三角形 C.钝角三角形 D.不能确定 + +8.如图， $A D$ 是 $\triangle A B C$ 的中线，点 $E , F$ 分别是 $A D$ 及其延长线上的点，且 $D E =$ $D F$ ,连接BF,CE,下列说法： $\begin{array} { r } { \textcircled { 1 } \triangle B D F { \cong } \triangle C D E ; \textcircled { 2 } B F { = } C E ; \textcircled { 3 } S _ { \triangle A B D } = } \end{array}$ $S _ { \triangle A D C }$ $\textcircled { 4 } B F / / C E .$ 其中正确的结论有 （ ） + +A. $\textcircled{1} \textcircled{2} \textcircled{3}$ +B. $\textcircled{2} \textcircled{3} \textcircled{4}$ +C. $\textcircled{1} \textcircled{3} \textcircled{4}$ +D. $\textcircled{1} \textcircled{2} \textcircled{3} \textcircled{4}$ + +![](images/5cd4880fead03b9d7e12cec18f86cb6f1bc30e7417a58aa918f836c476bb3cf7.jpg) + +# 二、填空题(每小题4分，共16分) + +9.如图所示，用直尺和圆规作一个角等于已知角，要说明 $\angle A ^ { \prime } O ^ { \prime } B ^ { \prime } { = } \angle A O B$ ，需要说明 $\triangle C ^ { \prime } O ^ { \prime } D ^ { \prime } { \overset { \sim } { = } } \triangle C O D$ ，则这两个三角形全等的依据是 （写出全等的简写). + +![](images/23765189cea2b4233c4761b39ab77ade1498a4c7503f206c4b7a1dda4d59395b.jpg) +第9题图 + +![](images/561a502250e58c9b4ac2464c5a4e118272d82c4e8bac6a980ddaf78e318f3c58.jpg) +第11题图 + +![](images/ee53651de6ff3445249bdfaa2c9237e7e493dfe0ebe020ac589a53785b84f07f.jpg) +第12题图 + +10.已知一个三角形的两边长分别是1和6,第三边长是整数，则第三边长为 + +11.如图，要测量池塘的宽度 $A B$ ，在池外选取一点 $P$ ，连接 $A P , B P$ 并各自延 长，使 $P C = P A , P D = P B$ 连接 $C D$ ，测得 $C D$ 的长为 $6 0 ~ \mathrm { m }$ ,则池塘的宽 $A B$ 为 m. + +12.如图，已知 $\angle A = \angle 1 = \angle A B C = 7 0 ^ { \circ } , C D \bot B C$ ，则 $\angle 2$ 的度数为 + +# 三、解答题(本大题6小题，共52分） + +13.(6分)如图， $A D \bot B C , \angle B A C$ 为锐角，写出图中所有的锐角三角形、直角三角形，并用符号表示这些三角形. + +![](images/636046c44e8680175c8f0ff4836d84159128bf1165c4ce5de8036d59b5d8def4.jpg) + +14.(8分)如图,画△ABC,使其两边为已知线段 $a , b$ ，夹角为 $\beta .$ （要求：用尺规作图,不写作法，保留作图痕迹;不在已知的线、角上作图) + +![](images/d9f021820db7d90009355b41a909f635796eb1ec2f2deba473891e26f42d0c27.jpg) + +15.(8分)如图，在 $\triangle A C E$ 中，点 $F$ 为 $C E$ 的延长线上一点，点 $B , D$ 分别为 $A C$ ，$C E$ 上的点,且 $A E / / B D$ + +(1)若 $A C { = } 8 , C E { = } 1 0$ ，求 $A E$ 的取值范围; + +(2)若 $\angle C B D = 5 5 ^ { \circ }$ $\angle C = 6 0 ^ { \circ }$ ，求 $\angle A E F$ 的度数. + +![](images/3ac71e8f13e2d85e2e1d6aadb56b009834b0b0a8c7106355d2a92cc255d1cbbf.jpg) + +16.(8分)如图，在 $\triangle A B C$ 中， $A D$ 是 $B C$ 边上的高, $A E$ 是 $\angle B A C$ 的平分线，$\angle E A D = \angle C A D .$ + +（ $1 ) \triangle A E D$ 和 $\triangle A C D$ 全等吗？为什么？(2)若 $\angle B = 4 2 ^ { \circ }$ ，求 $\angle C$ 的度数. + +![](images/bec728e82341bcb07e2a1692762d5ccd6cace728fee0ff42261655edcc13fa71.jpg) + +17.(10分)如图所示，在 $\triangle A B C$ 中,角平分线 $A D , B E , C F$ 相交于点 $O$ + +(1) $\angle B O C$ 与 $\angle B A C$ 的大小有什么关系？为什么？ + +(2) $\angle C O A$ 与 $\angle A B C$ 的大小有什么关系？ $\angle B O A$ 与 $\angle A C B$ 呢？ + +![](images/9f77c08a21df0907eff2f774a31838ea177bd5209edb9cfea1a0e684610958a7.jpg) + +18. $\circledcirc$ 视频讲解（12分)如图，已知 $D E \bot A C , B F \bot A C ,$ 垂足分别是点 $E , F , A E { = }$ CF,DC // AB. + +(1 $) \triangle C D E$ 和 $\triangle A B F$ 全等吗？请说明理由. +(2)试猜想 $A D$ 和 $B C$ 的数量和位置关系,并说明理由. +(3)若连接 $D F , B E$ ，则 $D F$ 和 $B E$ 相等吗？请说明理由. + +![](images/fcd924fc0891b0319ae45e07e9b343a5689ef911992766b7db30939d7527a986.jpg) + +# 能力提优测试卷 + +时间：60分钟 满分：100分 + +
题序二三评卷人总分
得分
+ +# 一、选择题(每小题4分，共32分) + +1.下列图形中,正确画出 $\triangle A B C$ 的AC 边上的高的图形是 + +![](images/3a4f8741717a68ab4052634ce3d3320a57e550eaf826171c9f8e304ef82e27cf.jpg) + +2.如果一个三角形的两边长分别为2和5,其第三边长为正整数,那么此三角形的周长可能为 （ ） + +A.14 B.13 C.10 D.9 + +3.如图， $. A B / / D F , A C \perp C E$ 于点 $C$ ，直线 $B C$ 与 $D F$ 相交于点 $E$ ，若 $\angle A = 2 0 ^ { \circ }$ ，则$\angle C E F$ 等于 （ ） + +A. ${ 1 1 0 } ^ { o }$ （204号 B.100° C.80° D. $7 0 ^ { \circ }$ + +5872Ab + +![](images/c57aead7fda36cacd51a21da49318a0e488eb569e6d590d0151d02940cc70b86.jpg) +第3题图 + +![](images/3034a5ed68ab601ad2e32076fed2847c470f20fbc1cecef924e08dff008232e2.jpg) +第4题图 + +4.如图，已知 $\triangle A B C$ 的六个元素如图所示，则甲、乙、丙三个三角形中与△ABC全等的图形是 （ ） + +A.甲 B.乙 C.丙 D.乙与丙 + +5.如图,AE//DF, $A E { = } D F$ ,则添加下列条件还不能使 $\triangle E A C { \cong } \triangle F D B$ 的是 + +![](images/97acb876afabed35b710e49c4d6bb6bbffb34be4ed24d772b7af8b5d294773ea.jpg) + +A. $A B { = } C D$ B. CE// BF C. $C E { = } B F$ （204号 D. $\angle E { = } \angle F$ （20 + +6.张叔叔用同种材料制成如图所示的金属框架，已知 $\angle B = \angle E , A B = D E , B F$ $= E C$ ，其中△ABC的周长为 $2 4 \ \mathrm { c m } , C F = 3 \ \mathrm { c m }$ ，则制成整个金属框架所需这种材料的长度为 （） + +A. $4 5 ~ \mathrm { c m }$ （204号 B.48 cm C.51 cm D. $5 4 ~ \mathrm { c m }$ + +![](images/4f371b3002d9a9c937d8f0d15ed138e02d5dedf64cd00387d686e4ff4f1823ef.jpg) +第6题图 + +![](images/5164dc0f2d055da67ba5eb940ede95b6832aab21494fce1422d2af9c7764f768.jpg) +第7题图 + +![](images/ab6717f01fd7d06617ed73c1d95bf987298bb82011c0650e07da1adeff46c69a.jpg) +第8题图 + +7.如图，在 $\triangle A B C$ 中，已知点 $D , E , F$ 分别为 $B C , A D , C E$ 的中点，且 ${ \cal S } _ { \triangle A B C } =$ $8 \ \mathrm { c m ^ { 2 } }$ ，则图中阴影部分 $\triangle B E F$ 的面积等于 （ ） + +A. $1 ~ \mathrm { c m ^ { 2 } }$ （204号 $\mathrm { B } . 2 \ \mathrm { c m } ^ { 2 }$ （204号 （204号 $\mathrm { C . 4 \ c m ^ { 2 } }$ （204号 $\mathrm { D } \mathrm { : 6 \ c m ^ { 2 } }$ + +8. $\circledcirc$ 视频讲解如图， $A D$ 是 $\triangle A B C$ 的中线，点 $E , F$ 分别是 $A D$ 和 $A D$ 的延长线上的点，且 $D E { = } D F$ ，连接 $B F , C E$ ，且 $\angle F B D = 3 5 ^ { \circ }$ $\angle B D F { = } 7 5 ^ { \circ }$ .则下列说法：$\textcircled { 1 } \triangle B D F { \cong } \triangle C D E ; \textcircled { 2 } \triangle A B D$ 和 $\triangle A C D$ 的面积相等； $\textcircled { 3 } B F / / C E ; \textcircled { 4 } L E C$ $= 7 0 ^ { \circ }$ .其中正确的有 （） + +A.1个 B.2个 C.3个 D.4个 + +# 二、填空题(每小题5分，共20分） + +9.如图， $A D$ 是 $\triangle A B C$ 的中线，且△ABD比△ACD的周长长 $3 ~ \mathrm { c m }$ ，则 $A B$ 比 $A C$ 长 cm. + +![](images/71f5f574c1f6e1bba4ec1cf81f98fcc382f5799806e39ab020f2643ab042c263.jpg) +第9题图 + +![](images/a15a80010cdb28085e67cdd2d44f28d61bae467ab25371f354c43fa40fb9525a.jpg) +第10题图 + +![](images/944f3f42e8d02ce18054e906e8e10c46ad17a05f0ddaac18330b79bb04b004cc.jpg) +第11题图 + +![](images/5e314ad812638bb1b26e145901bbb4e51e05e6952f974c9f25aa0409c5aa790e.jpg) +第12题图 + +10.如图，这是一个三角形测平架，已知 $A B { = } A C$ ，在 $B C$ 的中点 $D$ 处挂一个重锤，自然下垂.调整架身，使点 $A$ 恰好在重锤线上，则 $A D$ 和 $B C$ 的位置关系为 + +11.小聪做了一个如图所示的风筝，其中 $\angle E D H = \angle F D H , E D = F D = 2 0 ~ \mathrm { c m } ,$ $E H { = } 2 5 \ \mathrm { c m }$ ,则此四边形风筝的周长是 cm. + +12.如图，在正方形网格中， $\triangle A B C$ 的三个顶点及点 $D , E , F , G , H$ 都在格点上,现以 $D , E , F , G , H$ 中的三点为顶点画三角形，则能与 $\triangle A B C$ 全等的三角形是 + +# 三、解答题(本大题5小题，共48分) + +l3.(8分)如图，在 $\triangle A B C$ 中，点 $D$ 是 $B C$ 的中点,点 $E$ 是 $A B$ 边上一点,过点 $C$ （20作 $C F / / A B$ ，交 $E D$ 的延长线于点F.试说明 $\triangle B D E { \cong } \triangle C D F .$ + +![](images/b5fab8081a5e3f0ea8a9840302daa2bb3d72542d81cf3ba302b75d79c04674c5.jpg) + +14.(8分)如图,已知线段 $a$ 和 $\angle \alpha$ ,用尺规作一个 $\triangle A B C$ ，使 $A B = A C = 2 a$ ，$\angle B A C = 1 8 0 ^ { \circ } - \angle \alpha .$ （保留作图痕迹,不写作法） + +![](images/215f7ec1eb45d422cfcfb2d4dd7775577b99bec4ada13e1dd16d0ba4fb6030b9.jpg) + +15.(10 分)在新修的花园中,有一条“ $Z ^ { \prime }$ 字形绿色长廊 $A B C D$ ，如图， $A B / / C D$ ，在AB,BC,CD三段绿色长廊上各修建一凉亭E,M,F,且BE=CF，点M是 $B C$ 的中点，点 $E , M , F$ 在同一条直线上.若凉亭 $M$ 与 $F$ 之间有一池塘，在用皮尺不能直接测量的情况下,你能得出凉亭 $M$ 与 $F$ 之间的距离吗？试说明理由. + +![](images/4ac72d2d57fd533293d50ce1e983796a2d53c61ac56595a04dbdce6e3cd8dca6.jpg) + +16.(10 分)如图所示,在四边形 $A B C D$ 中， $A D / / B C$ ，点 $E$ 为 $C D$ 的中点，连接$A E , B E$ ，延长 $A E$ 交 $B C$ 的延长线于点 $F$ + +(1)判断 $F C$ 与 $A D$ 的数量关系,并说明理由. + +(2)若 $A B { = } B C { + } A D$ ，则 $B E \bot A F$ 吗？为什么？ + +![](images/bb7601968bbeabb9ef1215d29c4ef1109166c74013dbb7c124dee63824609d82.jpg) + +17.新考法综合与实践 $\circledcirc$ 视频讲解（12分）如图， $C D$ 是经过 $\angle B C A$ 顶点 $C$ 的一条直线,CA=CB,点E,F分别是直线CD上两点,且∠BEC=∠CFA=α. + +（1)若直线CD经过∠BCA的内部，且点E,F在射线CD上.如图1,若 $\angle B C A = 9 0 ^ { \circ }$ $\alpha { = } 9 0 ^ { \circ }$ ，则 $B E$ $C F$ （填“>”“ $=$ ”或“ $< ^ { 9 9 }$ + +(2)在(1)的条件下，如图2,若0°<∠BCA<180°，请添加一个关于α与$\angle B C A$ 关系的条件 ,使(1)中的结论仍然成立,并说明理由； + +（3)如图3,若直线CD经过∠BCA的外部,α=∠BCA,请提出关于EF,$B E , A F$ 三条线段数量关系的合理猜想,并说明理由. + +![](images/bdaab4afeb8ddd28affa1b99c3c489c42adb4083ad36ce5a5c9a75ad57a0987b.jpg) +图1 + +![](images/b9dc855fd3169eb89569d47a7449c494e9b3f1e6b72d0c5eda889442f8627ee5.jpg) +图2 + +![](images/d1730998969cf2cbaf218682f1d65f9d0b59e862a64ad8b5c72c6a5e7d4b5100.jpg) +图3 + +# 第五章 生活中的轴对称 + +# 基础过关测试卷 + +时间：60分钟 满分：100分 + +
题序二三评卷人总分
得分
+ +# 一、选择题(每小题4分，共32分） + +1.下列图形中，是轴对称图形的是 + +![](images/5be74c2c8c8ef2bb879676abcc87f0259163169e405bb3c2aa12d1bbf8341d6b.jpg) + +2.等腰三角形是轴对称图形，它的对称轴是 + +A.过顶点的直线 B.底边上的高C.顶角的平分线所在的直线 D.腰上的高所在的直线 + +3.将一张长方形纸片对折，然后用笔尖在纸片上扎出字母B,再把它展开铺平，那么可以看到的图形是 （ ） + +![](images/26af1b8beacf79304a913b2e667fc9bb7c0a7baeffe1e4e8d41d06c4e13bb664.jpg) + +4.如图， $\triangle A O D$ 与△BOC关于直线l成轴对称,若 $\angle D A O = 3 0 ^ { \circ }$ $\angle A D O = 2 5 ^ { \circ }$ ， 则 $\angle B O C$ 的度数为 + +A. $1 2 5 ^ { \circ }$ （20 B.120° C.115° D. $1 1 0 ^ { \circ }$ + +![](images/e6e3bb1ef95a27593ed917f449888f8bf41e1004c162a21c23b848ac273091a7.jpg) +第4题图 + +![](images/3d5fff82e9b5bf58ad10ca22fb2e188a33558b4bb504a54060ccc521eb86ba12.jpg) +第5题图 + +![](images/283b759b53452f19ef4d09d6b60925222028e7ce4e46a9898846ff1bf79cd8d6.jpg) +第6题图 + +![](images/7a4c6b615a7e9bf9e360a339cdf5f2f4e4a68e806108890a45cbe75b29a7bd37.jpg) +第8题图 + +5.如图，在 $\mathrm { R t } \triangle A B C$ 中， $\angle C = 9 0 ^ { \circ }$ ,以顶点 $A$ 为圆心,适当的长为半径画弧,分 + +别交 $A C , A B$ 于点 $M , N$ ，再分别以点 $M , N$ 为圆心，大于 $\cdot \frac { 1 } { 2 } M N$ 的长为半径画弧，两弧交于点 $P$ ，作射线 $A P$ 交边BC于点 $D$ .若 $C D { = } 4$ $A B { = } 1 5$ ，则 $\triangle A B D$ 的面积是 + +A.15 B.30 C.45 D.60 + +6.如图，在 $\triangle A B C$ 中,按以下步骤作图： $\textcircled{1}$ 分别以点 $B , C$ 为圆心，以大于 ${ \frac { 1 } { 2 } } B C$ 的长为半径作弧，两弧交于 $M , N$ 两点； $\textcircled{2}$ 作直线 $M N$ 交 $A B$ 于点 $D$ ，连接CD.若 $A B { = } 1 0 \ \mathrm { c m } , A C { = } 4 \ \mathrm { c m }$ ，则 $\triangle A D C$ 的周长为 （） + +A. $1 4 ~ \mathrm { c m }$ B. $1 3 ~ \mathrm { c m }$ C. 12 cm D.11 cm + +7.如图，将一张正方形纸片沿对角线折叠一次，然后在得到的三角形的三个角上各挖去一个圆洞，最后将正方形纸片展开,得到的图案是 （ ） + +![](images/ab7fc988313964adf35da3960d3c7b1eb28591688a51f851810488b192ea2204.jpg) + +8.如图,在四边形 $A B C D$ 中， $\angle D = 9 0 ^ { \circ }$ ,AE⊥CB,且 $A C$ 平分 $\angle D A E , C D = 4$ ，$A D { = } 6 , B E { = } 2$ ,则四边形 $A B C D$ 的面积为 （ ） + +A.24 B.26 C.28 D.30 + +# 二、填空题(每小题5分，共20分) + +9.如图，这是一个轴对称图形，它的对称轴有 条. + +![](images/6e16085aa9c8ccac501f0ed07fa588a92f97fd0b49fd2fb7f0616f010af46ef8.jpg) +第9题图 + +![](images/b7094c6ce2159c5215655e45e9ea0a5d70196d98266f80e4b6379ec3a0f93890.jpg) +第11题图 + +![](images/e0be3378705c3912bacfa2da131aa4b188efcb5adf72872f70bb39c2fa71059a.jpg) +第12题图 + +10.已知直线l是线段 $A B$ 的垂直平分线，点 $C$ 是直线l上的点，连接 $C A , C B$ ， 若 $\angle B A C = 6 0 ^ { \circ }$ ，则 $\angle A B C$ 的度数为 + +11.如图，这是一个轴对称图形， $B C$ 所在的直线是它的对称轴，则图中共有对全等三角形. + +12. $\circledcirc$ 视讲解在如图所示的正方形网格中画有两条线段，现在要再画一条，使图中的三条线段组成一个轴对称图形，能满足条件的线段有 条. + +# 三、解答题（本大题5小题，共48分） + +13.(6分)下图中的图形都是轴对称图形，请你画出它们的对称轴. + +![](images/fcc7879d0078d9a7e2a6faf7e7fe149fbf281a3e20f16859339a5db7706a1141.jpg) + +14.（8分)如图,已知△ABC和 $\triangle A ^ { \prime } B ^ { \prime } C ^ { \prime }$ 关于直线l对称. + +(1） $\triangle A B C$ 与 $\triangle A ^ { \prime } B ^ { \prime } C ^ { \prime }$ （填“全等”或“不全等”). + +(2)若 $A B { = } 4 \ \mathrm { c m } , A ^ { \prime } C ^ { \prime } { = } 6 \ \mathrm { c m } , B C { = } 3 \ \mathrm { c m } , A C$ 边上的高为 $2 \mathrm { \ c m }$ ，求 $\triangle A ^ { \prime } B ^ { \prime } C ^ { \prime }$ 的 周长和面积. + +![](images/3217f837003877801e6d0c72cecdc99223178b93a5358e3b919841bf9de1af30.jpg) + +15.(10分)两个城镇 $A , B$ 与两条公路 $l _ { 1 } , l _ { 2 }$ 的位置如图所示，电信部门需在 $C$ 处修建一座信号发射塔,要求发射塔到两个城镇 $A , B$ 的距离必须相等，到两条公路 $l _ { 1 } , l _ { 2 }$ 的距离也必须相等,那么点 $C$ 应选在何处？请在图中用尺规作图找出所有符合条件的点C.(不写已知、求作、作法,只保留作图痕迹) + +![](images/571a1d0d252afa4259fc2c93b205e2e7821f894463d909aed53b5a35be727218.jpg) + +16.(10 分)已知等腰三角形一腰上的高与另一腰的夹角的度数为 $5 6 ^ { \circ }$ ,求这个等腰三角形的底角的度数. + +17.新考法阅读理解题视频讲（14分)阅读材料：同学们已经知道，角平分线的性质是角平分线上的点到这个角的两边的距离相等,那么把上述性质反过来就可以得到在一个角的内部,到角的两边的距离相等的点在这个角的平分线上.问题情境：如图，在四边形ABCD中， $C E \bot A B$ 于点 $E$ ${ \vec { \mathfrak { z } } } , C D { = } C B$ $\angle A B C +$ $\angle A D C = 1 8 0 ^ { \circ }$ 问题解决：试说明 $A C$ 平分 $\angle B A D$ 合作探究：(1)在"问题解决”的基础上,若 $A E { = } 3 B E { = } 9$ ，能求出 $A D$ 的长吗？请说明理由.（提示：在两个直角三角形中，一直角边与斜边分别相等的两直角三角形全等)(2)如果△ABC和 $\triangle A C D$ 的面积分别为40和 28,请求出 $\triangle B C E$ 的面积. + +![](images/0e188d8137783bc6629e639ae1bff0fa28217c7f08d8efe5c14c20f1d5a9c72a.jpg) + +# 能力提优测试卷 + +时间：60分钟 满分：100分 + +
题序二三评卷人总分
得分
+ +# 一、选择题(每小题4分，共32分） + +1.第24届冬季奥林匹克运动会于2022年2月在北京举行，下面由冬季奥运会比赛项目图标组成的四个图形中，可以看作轴对称图形的是 （ ） + +![](images/a82b986f0670d8e2218db721caa62146cecc270e09f4abd0cb31c93b4a89505c.jpg) + +2.如图,若△ABC与 $\triangle D E F$ 关于直线l对称， $B E$ 交l于点 $O$ ,则下列说法不一 定正确的是 （ + +A. AB//EF B. $A C { = } D F$ C.AD⊥l D. $B O { = } E O$ + +![](images/d0a27a77bad966163454a995053cafd96d93dd43cb542761c21e88cdb85dc0c0.jpg) +第2题图 + +![](images/bc6bc9b14b927b480b0e18dad583de139280679ec42e9092e7b5d707fd5e6286.jpg) +第3题图 + +![](images/49197a7ae22499de00155631453d375e9e0914bf5b919407a1ba68e71230fbc0.jpg) +第5题图 + +3.如图，在△ABC中, $A B { = } 6$ $A C { = } 4$ ，分别以点 $B$ 和点 $C$ 为圆心,以大于 $B C$ 一半的长为半径画弧，两弧相交于点 $M , N ,$ 作出直线 $M N$ 交 $A B$ 于点 $D$ ，连接$C D$ ，则 $\triangle A D C$ 的周长为 （ ） + +A.8 B.10 C.12 D.14 + +4.若等腰三角形一腰上的高与另一腰的夹角为 $1 5 ^ { \circ }$ ,则顶角的度数为（） + +A $7 5 ^ { \circ }$ （204号 B. $1 5 ^ { \circ }$ C.15°或 $1 6 5 ^ { \circ }$ D. $7 5 ^ { \circ }$ 或 $1 0 5 ^ { \circ }$ + +5.如图，在 $\triangle A B C$ 中， $\angle C = 9 0 ^ { \circ } , A B = 1 0 , A D$ 平分 $\angle B A C .$ 若 $C D = 3$ ，则$\triangle A B D$ 的面积为 （ ） + +A.15 B.24 C.30 D.48 + +6.如图,在△ABC中， $B O$ 平分∠ABC,CO平分△ABC的外角 $\angle A C D , M N$ 经过点 $O$ ，与 $A B , A C$ 相交于点 $M , N$ ，且 $M N / / B C$ ，则 $B M , C N , M N$ 之间的关系是 （） + +A.1 $B M { + } C N { = } M N$ E $3 . B M { - } C N { = } M N$ C.CN- $B M { = } M N$ D.BM-CN=2MN + +![](images/756cab4a12ac935b2f5295cc2b49d0e301eb4414842effb6cc84e030a9df918c.jpg) +第6题图 + +![](images/78bb3167099bf4f0599f65795c273315d7651052d3bd808d0d826fdb6d530e14.jpg) +第7题图 + +![](images/8e947cd9284752797e35f8ffd5135f308216d0f5db08f61365d1fc6422c828f5.jpg) +第8题图 + +7.如图,将长方形纸片ABCD折叠,使顶点 $B$ 落在边 $A D$ 的点 $E$ 上，折痕FG交$B C$ 于点 $G$ ，交 $A B$ 于点 $F$ ，若 $\angle A E F { = } 2 0 ^ { \circ }$ ，则 $\angle F G B$ 的度数为 （ ） + +A. $2 { 0 } ^ { \circ }$ （20 （20 $\mathrm { B } . 2 5 ^ { \circ }$ （20 $ { \mathrm { C } } . 3 0 ^ { \circ }$ D.35° + +8. $\circledcirc$ 视频讲解如图，等腰三角形 $A B C$ 的底边 $B C$ 长为4,面积是16,腰 $A C$ 的垂直平分线EF分别交 $A C , A B$ 边于点 $E , F .$ 若点 $D$ 为 $B C$ 边的中点，点 $M$ 为线段EF上一动点，则 $\triangle C D M$ 周长的最小值为 （） + +A.6 B.8 C.10 D.12 + +# 二、填空题(每小题5分，共20分) + +9.等腰三角形的周长为 $1 4 ~ \mathrm { c m }$ ,其中一边长为 $4 \mathrm { \ c m }$ ，则它的底边长为 cm. + +10.如图，在 $\triangle A B C$ 中， $\angle B A C { > } 9 0 ^ { \circ } ,$ AB的垂直平分线交BC于点 $E , A C$ 的垂直平分线交BC于点 $F$ .若 $\triangle A E F$ 的周长为 $6 ~ \mathrm { c m }$ ，则 $B C =$ cm. + +![](images/a5e5a29632a5a0563a8e5dd2fe7d736d0da7d47c8fab34ebae773687f127879e.jpg) +第11题图 + +![](images/2eaf507a4687831260eb1f32ca43dd616ea37fae98a93f690cc4b1dc88ebbe7e.jpg) +第10题图 + +![](images/0940a76b6725a159f58ae59917532acfc1bde723ace05c9a817a3c60713a6364.jpg) +第12题图 + +11.如图， $O E$ 是 $\angle A O B$ 的平分线， $B D \bot O A$ 于点 $D , A C \bot B O$ 于点 $C$ ,则关于直线OE对称的三角形有 对. + +12.如图，在 $\triangle A B C$ 中， $A B { = } A C$ ，点 $D$ 为线段BC上一动点（不与点 $B , C$ 重合)，连接 $A D$ ，作 $\angle D A E { = } \angle B A C$ ，且 $A D { = } A E$ ，连接 $D E , C E .$ 当 $C E / / A B$ 时，若 $\angle B A D = 3 5 ^ { \circ }$ ，则 $\angle D E C$ 的度数为 + +# 三、解答题(本大题5小题，共48分） + +l3.(8分)如图，在等边△ABC中,过 $B C$ 边上一点 $P$ ，作 $\angle D P E = 6 0 ^ { \circ }$ ,分别与边$A B , A C$ 相交于点D,E.在图中找出与 $\angle E P C$ 始终相等的角,并说明理由. + +![](images/3657d08511585f5b2dbfc81c5b96765aec95c72dfa21a5aee90f91e60e9b4919.jpg) + +14.(8分)如图,在 $1 0 \times 1 0$ 的正方形网格中,每个小正方形的边长都为1,网格中有一个格点 $\triangle A B C$ (即三角形的顶点都在格点上). + +(1)在图中作出 $\triangle A B C$ 关于直线l对称的 $\triangle A _ { 1 } B _ { 1 } C _ { 1 }$ （要求：点 $A$ 与点 $A _ { 1 }$ ，点$B$ 与点 $B _ { 1 }$ ，点 $C$ 与点 $C _ { 1 }$ 相对应)； + +(2)在(1)的结果下，连接 $B B _ { 1 }$ ， $C C _ { 1 }$ ,求四边形 $B B _ { 1 } C _ { 1 } C$ 的面积. + +![](images/8895bfd8964a5faf6816a6b464187de20f6606d70f228bcc772ae0952437dfe2.jpg) + +15.(10分)如图，在 $\triangle A B C$ 中， $A D \perp B C , E F$ 垂直平分 $A C$ ，交 $A C$ 于点 $F$ ，交BC于点 $E$ ，且 $B D { = } D E$ ，连接 $A E$ + +(1)若 $\angle B A E { = } 4 0 ^ { \circ }$ ，求 $\angle C$ 的度数； +(2)若 $\triangle A B C$ 的周长为 $1 4 { \mathrm { ~ c m } } , A C { = } 6 { \mathrm { ~ c m } }$ ，求 $D C$ 的长. + +![](images/b138907b52911cd1f053242869cca2b24be6699636df370ea034fc7afecd0175.jpg) + +16.(10分)如图，点 $P$ 关于 $O A , O B$ 对称的对称点分别为点 $C , D$ ，连接 $C D$ ，交$O A$ 于点 $M$ ，交 $O B$ 于点 $N$ + +(1)若 $C D$ 的长为 $1 8 ~ \mathrm { c m }$ ，求 $\triangle P M N$ 的周长； + +(2)若 $\angle C = 2 1 ^ { \circ }$ $\angle D = 2 8 ^ { \circ }$ ，求 $\angle M P N$ 的度数. + +![](images/cd8e3aa78fc5077f3a1f2d8305db79772a2c56fc1e402a7ab3af8e25e8087cb4.jpg) + +17.视频讲解（12分）如图，在 $\triangle A B C$ 中， $A B { = } A C$ $\angle B A C = 9 0 ^ { \circ }$ ，点 $O$ 为BC的中点. + +(1)写出点 $O$ 到 $\triangle A B C$ 的三个顶点 $A , B , C$ 的距离的大小关系（不要求说明理由). + +(2)如果点 $M , N$ 分别在线段 $A B , A C$ 上移动,在移动过程中保持 $A N { = } B M$ ， 请判断 $\triangle { O M N }$ 的形状,并说明理由. + +(3)在(2)的条件下,四边形 AMON的面积是否发生变化？请说明理由. + +![](images/f20572db92dd6feb97613235a0e983f70dca245cbc375e79cc01b4582ec4c024.jpg) + +# 月考测试卷（二） + +时间：90分钟 满分：120分考试范围:第一章～第五章第2节 + +
题序二三评卷人总分
得分
+ +# 一、选择题(本大题共10小题，每小题3分，共30 分) + +1.下列运算正确的是 + +A. $a ^ { 2 } \cdot a ^ { 4 } = a ^ { 8 }$ B $( - 2 a ^ { 2 } ) ^ { 3 } = - 6 a ^ { 5 }$ $\complement , a ^ { 5 } \div a ^ { 3 } = a ^ { 2 }$ $\mathrm { D } . ~ ( - a - b ) ^ { 2 } = a ^ { 2 } - 2 a b + b ^ { 2 }$ + +2.在 $\triangle A B C$ 中， $A B { = } 3 ~ \mathrm { c m }$ $B C { = } 7 ~ \mathrm { c m }$ ，若 $A C$ 的长为整数,则 $A C$ 的长可能是（ + +$$ +\begin{array} { r l r l r l r l r l r l } { ) } & { { } \mathrm { c m } } & { \qquad } & { } & { { } \mathrm { B } , 5 \ \mathrm { c m } } & { \qquad } & { } & { { } \mathrm { C } , 4 \ \mathrm { c m } } & { \qquad } & { } & { { } \mathrm { D } , 2 \ \mathrm { c m } } \end{array} +$$ + +3.如图，在 $\triangle A B C$ 中，点 $D , E , F$ 分别在边 $B C , A B , A C$ 上，下列能判定 $D E$ //$A C$ 的条件是 （ ） + +A. $\angle 1 = \angle 3$ +B. $\angle 3 { = } \angle C$ +C. $\angle 2 = \angle 4$ +D. $\angle 1 + \angle 2 = 1 8 0 ^ { \circ }$ + +![](images/524454d8a00b015069dd8c70689d491f11bead7516196d20326c206820b76038.jpg) + +4.“漏壶"是一种古代计时器，如图，在壶内盛一定量的水，水从壶底的小孔漏出，壶内壁画有刻度，人们根据壶中水面的位置计算时间.用 $\mathcal { X }$ 表示漏水时间， $y$ 表示壶底到水面的高度，不考虑水量变化对压力的影响，下列图象能表示 $y$ 与 $\mathcal { X }$ 的对应关系的是 （ ） + +![](images/f7acc8a9605ecd5a42d55d41fce99aef1e0b9d960699eff7287001bce05b479c.jpg) + +5.如图，直线 $A B , C D$ 相交于点 $O , O E \bot C D , O F$ 平分∠BOD, $\angle A O E = 2 4 ^ { \circ }$ ，则$\angle C O F$ 的度数是 （ ） + +A. $1 4 6 ^ { \circ }$ B.147° C.157° D.136° + +![](images/4b26281e697478683a0af18d041d0dbba7e032862d919ec31e9590adc1b8c555.jpg) +第5题图 + +![](images/c6a1f1de3c8a7659ebef9e5898c92fe9fb208c9ec47ccc3493ef8984968b44af.jpg) +第7题图 + +![](images/330366f53c0a12db4914b3f38ff8daaeca544950e74f86df47512dba18b681c9.jpg) +第8题图 + +6.如果 $x ^ { 2 } + x = 3$ ,那么代数式 $( x + 1 ) ( x - 1 ) + x ( x + 2 )$ 的值是 + +A.2 B.3 C.5 D. 6 + +7.如图，点 $D , E$ 分别是 $B C , A D$ 的中点， $\triangle C E F$ 与 $\triangle C E D$ 关于直线 $C E$ 对称，若 $\triangle A B C$ 的面积是8,则 $\triangle C E F$ 的面积为 （ ） + +A.8 B.6 C.4 D.2 + +8.如图，已知 $A B { = } A C , A F { = } A E$ $\angle E A F { = } \angle B A C$ ，点 $C , D , E , F$ 共线.则下列结论：①△AFB≌△AEC; $\textcircled { 2 } B F { = } C E$ $\textcircled{3}$ ∠BFC=∠EAF; $\textcircled { 4 } A B { = } B C .$ 其中正确的是 （） + +$$ +\mathrm { ~ B _ { \cdot } ~ } \textcircled { 1 } \textcircled { 2 } \textcircled { 4 } \qquad \mathrm { C . } \textcircled { 1 } \textcircled { 2 } \qquad \mathrm { D . } \textcircled { 1 } \textcircled { 2 } \textcircled { 3 } \textcircled { 4 } +$$ + +9. $\circledcirc$ 视频讲解如图， $\triangle A B C$ 的内部有一点 $P$ ，且点 $D , E , F$ 是点 $P$ 分别以 $A B , B C$ $A C$ 为对称轴的对称点.若 $\triangle A B C$ 的内角 $\angle A = 7 0 ^ { \circ }$ $\angle B = 6 0 ^ { \circ }$ $\angle C = 5 0 ^ { \circ }$ ，则$\angle A D B + \angle B E C + \angle C F A$ 的度数是 （） + +A. $1 8 0 ^ { \circ }$ B.270° C.360° D. $4 8 0 ^ { \circ }$ + +![](images/0e05d7f21eaac115da4e2ec694f61b04d93142aa3f442eb16aa28b1edf5612bb.jpg) +第9题图 + +![](images/112ca1d9080b20187f0b65ffda02dd8faf836bddcaee7b21330071724270627f.jpg) +第10题图 + +10.如图，在 $\triangle A B C$ 中， $B D , C E$ 分别是 $A C , A B$ 边上的中线，分别延长 $B D , C E$ 到点 $F , G$ 使 $D F = B D , E G = C E$ ，则下列结论： $ G A { = } A F , \mathcal { O } G A / / B C$ $\textcircled { 3 } A F / / B C , \textcircled { 4 }$ 点 $G , A , F$ 在同一条直线上， $\textcircled{5}$ 点 $A$ 是线段 $G F$ 的中点.其中正确的有 （） + +A.2个 B.3个 C.4个 D.5个 + +# 二、填空题(本大题共5小题，每小题3分，共15分) + +11.如图,直线 $A B$ 左边是计算器上的数字5,若以 $A B$ 为对称轴，则它的轴对称图形是数字 · + +![](images/d3582c039258099f4fbe11b16f9ce681b157aa48bd8e1522ff43de2ea829da63.jpg) +第11题图 + +![](images/bb9a5ebe28efdf0133e60b99d7119975fb4a64068bed68987ad8924b987764f1.jpg) +第13题图 + +![](images/f949f1e68e271867d75ddc80f33f5354b4ffc0458fc9d0318ea1fb53327a2170.jpg) +第14题图 + +12.某商场将一商品在保持销售价80元/件不变的前提下，规定凡购买超过5件者，超出的部分打5折出售.若顾客购买x(x大于5)件，应付y元,则y与$\mathcal { X }$ 之间的关系式是 + +13.如图,点 $F$ 在直线 $C D$ 上， $F G$ 平分∠EFD,AB// $C D$ $\angle 1 = 5 6 ^ { \circ }$ $\angle 2$ 的度数为 + +14.如图,在△ABC中,AD⊥BC于点D,BE⊥AC于点E,AD,BE相交于点F,如果 $B F { = } A C , B C { = } 8 , C D { = } 2$ ，那么 $A F { = } \_$ + +15. 新考法填空双空题视频讲解两个边长分别为a和b的正方形按如图1所示的方式放置,其未叠合部分(阴影)的面积为Si;若再在图中大正方形的右下角摆放一个边长为b的小正方形(如图2)，两个小正方形叠合部分（阴影）的面积为 $S _ { 2 }$ .若 $a + b = 8 , a b = 1 0$ ，则 $S _ { 1 } + S _ { 2 } =$ ；当 $S _ { 1 } + S _ { 2 } = 4 0$ 时，则图3中阴影部分的面积 $S _ { 3 } =$ + +![](images/77c059b3139a742758e269ca9b1a9cdf2c092ea818eda936a89f9a14d249253b.jpg) +图1 + +![](images/c8e2e49a4a4d3b11a674b492f6e1ecac54a44cab4582113704ce85d24e82486b.jpg) +图2 + +![](images/0a0e9f04c1a3f21824dfcf740e091c798a31aaccca9e9670f55e3faf117391bd.jpg) +图3 + +# 三、解答题（本大题共8小题，满分75分） + +16.(10分)（1)计算： $( \frac { 2 } { 3 } ) ^ { 2 0 2 3 } \times 1 . 5 ^ { 2 0 2 4 } \times ( - 1 ) ^ { 2 0 2 4 } ,$ (2)某种T形零件从正面看到的平面图形是轴对称图形,具体尺寸如图所示,请你计算阴影部分的周长和面积. + +![](images/0d7c272b27b654d93f6d241a7ab43ead77726a01ec04c4889b091108d281b524.jpg) + +17.(9分)如图,AD是一段斜坡,AB 是水平线,现为了测量斜坡上一点D的竖直高度 $D B$ 的长度,欢欢在 $D$ 处立上一竹竿 $C D$ ,并保证 $C D \perp A D$ ,然后在竿顶 $C$ 处垂下一根绳 $C E$ ,与斜坡的交点为点 $E$ ,他调整好绳子 $C E$ 的长度,使得 $C E { = } A D$ ，此时她测得 $D E { = } 2$ 米,求 $D B$ 的长度. + +![](images/fe99707b8d3f3330bd5b21e83fc4fd2cdd7e67668e9e43f189d7f2b1ed6595c5.jpg) + +18.(9分)小明骑车上学,当他骑了一段时间后，想起要买某本书，于是又折回到刚经过的新华书店，买到书后继续骑车去学校.他离家的距离(米)与所用的时间(分钟)的关系如图所示.根据图象回答下列问题： + +(1)小明家到学校的距离是 米. + +(2)小明在书店停留了 分钟. + +(3)本次上学途中，小明一共骑行了 米. + +20.(9分)若 $a$ 为任意自然数，尝试猜想：代数式 $( 3 a + 2 ) ( 2 a + 3 ) - 3 a ( 2 a + 1 )$ 的值是奇数，还是偶数？并说明理由. + +(4)据统计，骑车的速度超过330米/分就超越了安全限度，小明买到书后继续骑车到学校的这段时间内的骑车速度在安全限度内吗？请说明理由. + +学校班级姓名学号考生注意考试内无将学校丶班级、名、 学号填写在指定的位置上° + +![](images/566c7139c56b2495626b1a52c57ce606f954c6b4871ac242b1c09162d295a9f1.jpg) + +21.(9分)如图，将 $\mathrm { R t } \triangle A B C$ 沿某条直线折叠，使斜边的两个端点 $A$ 与 $B$ 重合，折痕为 $D E$ + +19.(9分)如图，在长方形 $A B C D$ 中， $A B { = } 4 , A D { = } 6$ ，延长 $B C$ 到点 $E$ ，使 $C E { = }$ 2,连接 $D E$ ，动点 $P$ 从点 $B$ 出发，以每秒2个单位长度的速度沿BC— $C D$ $D A$ 向终点 $A$ 运动，设点 $P$ 的运动时间为 $t$ 秒，当 $t$ 的值为多少时， $\triangle A B P$ 和△DCE全等？ + +(1)若 $A C { = } 6 \ \mathrm { c m } , B C { = } 8 \ \mathrm { c m }$ ,求△ACD的周长； +(2)若 $\angle C A D : \angle B A D = 1 : 2$ ，求 $\angle B$ 的度数. + +![](images/16e320f6b65049b25d0c5bd656a332695cb1f8952eb2d30b48e79f35660db3db.jpg) + +![](images/82dcd5486379c67e7c00ab338b5c5712233f9680e06e5aaa0af053220cf6d40c.jpg) + +22.(10分)已知直线 $A M , C N$ 和点 $B$ 在同一平面内，且 $A M / / C N , A B \perp B C .$ （20(1)如图 $1 , A M$ 与 $B C$ 交于点 $D$ ，若 $\angle A = 4 0 ^ { \circ }$ ，求 $\angle C$ 的度数； + +(2)如图2,若 $B D \bot A M$ 垂足为点 $D$ ，试说明： $\angle A B D { = } \angle C .$ + +![](images/7c7feb64da9d5d716aa0d8b357c882a31bb9400135839d86f424e1a68e7faec7.jpg) +图1 + +![](images/37df25d37c31f9f1ab3f56438abe30c3cc9b0b5835bb4bb27b4477fbf6e940dc.jpg) +图2 + +23. $\circledcirc$ 视频讲解（10分)如图所示，在锐角 $\triangle A B C$ 中， $B D , C E$ 分别是ABC的两条高，点 $P$ 在 $B D$ 的延长线上, $C A { = } B P$ ，点 $Q$ 在 $C E$ 上， $Q C { = } A B$ + +（1)试判断： $\angle 1$ $\angle 2$ （填 $>$ ”、“ $<$ ”或“ $=$ ”). + +(2)试探究线段 $P A$ 与 $A Q$ 之间的数量关系和位置关系,并说明理由. + +(3)若把(1)中的 $\triangle A B C$ 改为钝角三角形， $A C > A B$ $\angle A$ 是钝角，其他条件不变，试探究线段 $P A$ 与 $A Q$ 之间的数量关系和位置关系.请在备用图中画出图形,并直接写出结论. + +![](images/fadd9cf4e6178a725fb5ebaee613cb959c91931fe98e6243c7069decab5f2b99.jpg) + +![](images/1f0d99982fbc6e5b1bf39cadbbf6311a798cd09f9271f043200963b8111dd73d.jpg) + +# 第六章 概率初步 + +# 基础过关测试卷 + +时间：60分钟 满分：100分 + +
题序二三评卷人总分
得分
+ +# 一、选择题(每小题4分，共32分) + +1.下列事件中，是必然事件的是 + +A.明天一定是晴天 +B.打开手机就有未接电话 +C.通常温度降到 $0 ~ ^ { \circ } C$ 以下，纯净的水会结冰 +D.2024年有370天 + +2.掷一枚质地均匀的正方体骰子，骰子的六个面上分别是1点到6点的点数，则下列说法正确的是 （ ） + +A.掷一枚骰子，朝上的一面的点数一定是4B.掷一枚骰子,朝上的一面的点数一定是奇数C.掷一枚骰子，朝上的一面的点数出现1的可能性最大D.掷一枚骰子，朝上的一面的点数可能出现5 + +3.下列成语描述的事件中，属于随机事件的是 + +A.守株待兔 B.风吹草动 C.一手遮天 D.水中捞月 + +4.从1,2,3,4,5这五个数中,任意取出一个数,是奇数的概率为 + +A $\frac { 4 } { 9 }$ B $\frac { 3 } { 5 }$ C $\frac { 2 } { 5 }$ D $\frac { 1 } { 5 }$ + +5.在抛掷一枚图钉的实验中,某小组一共做了1000 次实验,最后出现钉尖朝上的次数为620，则此时出现钉尖朝上的频率为 （ ） + +A.0.48 B.0.5 C. 0. 62 D.无法确定 + +6.将4个红球、3个白球、2个黑球放入一个不透明的袋子里，从中摸出8个球，恰好红球、白球、黑球都摸到，这件事情 （ + +A.必然发生 B.不可能发生C.很可能发生 D.可能发生 + +7.小明在做一道正确答案是一5的计算题时，由于运算符号（“ $^ +$ ”“-”“ $\times$ ”或“÷")被墨迹污染，看见的算式是“一204”,那么小明还能做对的概率是（ + +A $\frac { 1 } { 6 }$ $\mathrm { B } . { \frac { 1 } { 4 } } \qquad \mathrm { C } . { \frac { 1 } { 3 } } \qquad \mathrm { D } . { \frac { 1 } { 2 } }$ + +8.一个转盘的颜色如图所示，其中 $\angle A O B = 6 0 ^ { \circ } , A C$ 为直径，转动转盘，指针落在红色区域的概率为 （ ） + +A $\frac { 1 } { 6 }$ $\begin{array} { c } { \mathrm { B . } { \frac { 1 } { 3 } } } \\ { \mathrm { D . } { \frac { 2 } { 3 } } } \end{array}$ C $\frac 1 2$ + +![](images/a3cef98fe41c86f6c5613a4fd8c75a6194e8248c0d762b7f91dbdfedaaf67743.jpg) + +# 二、填空题(每小题4分，共16分) + +9.“3个人分成两组，一定有2个人分在一组"的事件是一个 事件.（填"必然”“不可能”或“随机”) + +10.在一个不透明的口袋中装有若干个红球和白球，它们除颜色外其他完全相同，通过多次摸球试验后发现，摸到红球的频率稳定在 $2 5 \%$ 附近，则从口袋中随机地摸出一个球,摸到白球的概率为 + +11.在一个不透明的盒子里装有10 枚大小相同的白色棋子和黑色棋子,若随机从中摸出一枚棋子是白色棋子的概率为 $\frac { 1 } { 5 }$ ,则在设计该游戏时,盒子里应放有白色棋子_枚. + +12.若小球在如图所示的地面上自由滚动,并随机停留在某块方砖上，则它最终停留在黑色区域的概率是 + +![](images/60832e6b3e1a1b0d2105b584819a714f695384a45f3997a6c373842e35d4b7cd.jpg) + +# 三、解答题(本大题5小题，共52分） + +13.(8分)如图,有一些写有数字的卡片,它们的背面都相同,现将它们的背面朝上，从中任意摸出一张. + +![](images/3d0a8a65465dc8a2d4d9627fa1ad85643767eae1edb4030002f47aaf47d14655.jpg) + +(1)摸到哪个数字卡片的可能性最大？摸到哪个数字卡片的可能性最小? + +(2)摸到的数字是奇数和摸到的数字是偶数的可能性哪个大？ + +14.(10分)在一个不透明的袋子中装有仅颜色不同的6个红球和9个黑球. + +(1)若从袋子中随机摸出1个球,分别求摸到红球和摸到白球的概率; (2)若再往袋子中放人 $m$ 个黑球,然后随机摸出1个球，若摸到红球的概率 为 $\frac 1 3$ 求 $m$ 的值. + +15.(10 分)如图，一个转盘被等分成六个扇形,并在上面依次写上数字1,2,3,4,5,6. + +(1)若自由转动转盘，当它停止转动后,指针指向奇数的概率是多少? + +(2)请你用这个转盘设计一个游戏,要求当自由转动的转盘停止时，指针指向的区域的概率为 $\frac { 2 } { 3 }$ + +![](images/21d06646f327de740481946d4d9c1d30fb6f8e3916b4347094fa1c073269cce5.jpg) + +16.（12 分)下面是小明和同学们做"抛掷质地均匀的硬币试验"获得的数据： + +
抛掷次数n100200300400500600
正面朝上的频数m5198153200255306
n 正面朝上的频率m
+ +正面朝上的频率 +0.52 +0.51 +0.50 +0.49 +0.48100 200300400 500600抛掷次数(1)填写表中的空格； +（2)画出折线统计图； +(3)当试验次数很大时，“正面朝上"的频率在 附近摆动. + +17. $\circledcirc$ 视频讲解（12分）已知下列事件： $\textcircled{1}$ 投掷一枚普通的正方体骰子，所得的点数小于7； $\textcircled{2}$ 随机地从一副扑克牌(52张)中摸出一张扑克牌，正好是红桃(大、小王除外). + +(1)事件 $\textcircled{1}$ 是一个 事件，发生的概率为事件 $\textcircled{2}$ 是一个 事件，发生的概率为(2)我们把事件 $A$ 记作"投掷一枚普通的正方体骰子，所得的点数为3的倍数”；事件 $B$ 记作"随机地从一副扑克牌(52张)中摸出一张扑克牌,扑克牌上的数字正好是5的倍数”.请你把这两个事件 $A , B$ 发生的概率用“<"连接起来. + +# 能力提优测试卷 + +时间：60分钟 满分：100分 + +
题序二三评卷人总分
得分
+ +# 一、选择题(每小题5分，共30分) + +1.下列事件为必然事件的是 + +A.从两个班级中任选三名学生，至少有两名学生来自同一个班级 +B.明天会下雨 +C.打开电视机,CCTV1正在播放新闻 +D.购买一张彩票中奖一百万元 + +2.如图，从四张印有品牌标志图案的卡片中任取一张，取出印有品牌标志图案的是轴对称图形的卡片的概率是 ） + +![](images/c8a6fcf5dfc8ced182d8e3d34a8888d730050ed306566e24188ea5d72c064666.jpg) + +A.1 B $\frac { 3 } { 4 }$ $\frac 1 2$ D $\frac { 1 } { 4 }$ + +3.在一个不透明的布袋中装有红色、白色球共40个，除颜色外其他完全相同.小明通过多次摸球试验后发现,其中摸到红色球的频率稳定在 $1 5 \%$ 左右，则布袋中白色球可能有 （ ） + +A.4个 B.6个 C.34个 D.36个 + +4.一个小球在如图所示的地板上自由滚动，并随机停在某块方砖上,如果每一块方砖除颜色外其他完全相同，那么小球最终停留在黑砖上的概率是（ ） + +A $\frac { 5 } { 9 }$ B $\frac { 4 } { 9 }$ C $\frac { 4 } { 5 }$ D.1 + +![](images/75afb139f7e16060bd39aeb9e18be49165f6fd6db5237d96efb6d5427d14531f.jpg) +第4题图 + +![](images/92e858056a5386c619a279a22c5e245aa3e2cffb33334000c996b39e5b966bb5.jpg) +第5题图 + +![](images/6715cdd201acae8a9f9fa63b401933d8226e49552f68e85daff69fbbd7d3de41.jpg) +第6题图 + +5.某小组做"用频率估计概率”的试验时，统计了某一结果出现的频率,绘制了如图所示的折线统计图，则符合这一结果的试验最有可能的是 （ ） + +A.在“石头、剪刀、布"的游戏中，小明随机出的是“剪刀” +B.一副去掉大、小王的普通扑克牌洗匀后，从中任抽一张牌的花色是红桃 +C.暗箱中有1个红球和2个黄球，它们只有颜色上的区别，从中任取一个球是黄球 +D.掷一枚质地均匀的正六面体骰子,向上一面的点数是4 + +6.如图，在两个同心圆中，四条直径把圆分成八等份，若往圆面投掷飞镖，则飞镖落在黑色区域的概率是 （ ） + +A $\frac 1 2$ B $\frac 1 3$ $\frac { 1 } { 4 }$ D $\frac { 1 } { 6 }$ + +# 二、填空题(每小题5分，共20分） + +7.某电视台在2023年5月举办的青年歌手大奖赛活动中，得奖选手由观众发短信投票产生，并对发短信者进行抽奖活动.一万条短信为一个开奖组，设一等奖1名，二等奖3名，三等奖6名.小静同学发了一条短信，那么她获奖的概率是 + +8.在一个不透明的口袋里装有只有颜色不同的黑、白两种颜色的球共20个，某学习小组做摸球试验，将球搅匀后从中随机摸出一个球记下颜色，再把它放回袋中，不断重复，试验数据如下表： + +
摸球的次数n1001502005008001000
摸到白球的次数m5896116295484601
摸到白球的频率 n0.580.640.580.590.6050.601
+ +根据以上数据，估计口袋中黑球有 个 + +9.有八张大小、形状完全相同的卡片，卡片上分别写有数字3,4,5,6,7,8,9,10,从中随机抽取一张，抽出的卡片上的数字恰好为3的倍数的概率是 + +10.在一个不透明的袋子中装有除颜色外其余均相同的7个小球,其中红球2个，黑球5个，若再放入 $m$ 个一样的黑球并摇匀，此时,随机摸出一个球是黑球的概率等于 $\frac { 4 } { 5 }$ ，则 $m$ 的值为 + +# 三、解答题(本大题5小题，共50分） + +11.(8分)如图，一个游戏转盘中,红、黄、蓝三个扇形的圆心角度数分别为 $4 0 ^ { \circ }$ ，${ 1 2 0 } ^ { \circ } , { 2 0 0 } ^ { \circ }$ ，若让转盘自由转动. + +（1)求指针停止后在蓝色区域的概率； +(2)求指针停止后在黄色或红色区域的概率. + +![](images/f0ad503afeecce06892f325b65c33e6f4b7cc3da3fb15dc47fcd4c36349cc8dc.jpg) + +14.(10 分)某信息兴趣小组利用电脑成功设计了一个运算程序,这个程序可用如图所示的框图表示.小明同学任取一个自然数 $\mathcal { X }$ 输入求值. + +(1)试写出与输出的数 $y$ 有关的一个必然事件； + +(2)若输入的数是2至9这八个连续正整数中的一个，求输出的数是3的倍数的概率. + +输人x → 平方 -x ÷2 输出y + +12.（10分)如图,可以随机在图中取点. + +(1)这个点取在阴影部分的概率是(2)在保留原阴影部分的情况下，请你重新设计图案(直接在图上涂阴影)，使得这个点取在阴影部分的概率为. + +![](images/3209ce4912cf9de0700baf0b89ba9309fc3c1fae786b13114f4d72efbfcab6eb.jpg) + +13.(10 分)在一个不透明的袋子里装有红、黄、白三种颜色的球共50个,它们除了颜色不同外其余都相同,其中黄球比白球少5个,已知从袋子里随机摸出一个球是红球的概率是 $\frac { 3 } { 1 0 }$ + +（1)求袋子里红球的个数； +(2)求从袋子里随机摸出一个球是白球的概率. + +15.视讲(12分)如图，端午节期间，某商场为了吸引顾客，设立了一个可以自由转动的转盘，并规定顾客每购买200元的商品,就能获得一次转动转盘的机会,如果转盘停止后，指针上对准红、黄、绿的区域，顾客就可以分别获得50元、20元、10元的奖金，对准无色区域则无奖金(转盘等分成16份). + +（1)小明购物180元，他获得奖金的概率是多少？（2)小德购物210元，他获得奖金的概率是多少？ + +(3)现商场想调整获得10元奖金的概率为,其他金额的获奖率不变，,则需要将多少个无色区域涂上绿色？ + +![](images/75b87daacef404e34c6367ce9a5747477e65b83b60bbcc7387dda5bc74885d2b.jpg) + +# 专项训练卷（一） 整式的乘除运算 + +时间：60分钟 满分：100分 + +
题序二三评卷人总分
得分
+ +# 一、选择题(每小题4分，共32分) + +1.计算 $a ^ { 5 } \cdot a \div a ^ { 4 }$ 的结果为 + +A.a B. $a ^ { 2 }$ （204号 C. $a ^ { 3 }$ （20 D. $a ^ { 4 }$ （204号 + +2.下列运算中，正确的是 + +A. $x ^ { 2 } + x ^ { 5 } = x ^ { 7 }$ B. $( - 3 x ) ^ { 3 } { = } 2 7 x ^ { 3 }$ C. $( \boldsymbol { \mathscr { x } } ^ { 4 } ) ^ { 2 } = \boldsymbol { \mathscr { x } } ^ { 8 }$ D $( x - 1 ) ^ { 2 } = x ^ { 2 } - 2 x - 1$ + +3.可乐中含有大量的咖啡因，世界卫生组织建议青少年每天咖啡因的摄入量不能超过 $0 . 0 0 0 0 8 5 ~ \mathrm { k g }$ ，则数据0.000085用科学记数法表示为 （ ） + +$$ +3 . 5 { \times } 1 0 ^ { 5 } \qquad \quad \mathrm { B } . 8 . 5 { \times } 1 0 ^ { - 4 } \qquad \mathrm { C . ~ 8 . ~ 5 { \times } 1 0 ^ { - 5 } } \qquad \quad \mathrm { D . ~ 8 . ~ 5 { \times } 1 0 ^ { - 6 } } +$$ + +4.若单项式 ${ \frac { 2 } { 7 } } x ^ { m + 2 } y ^ { 2 }$ 与单项式 $7 x y z$ 的乘积是一个9次单项式,则 $m$ 的值为（ + +A.2 B.3 C.4 D.5 + +5.设整式 $M { = } x ( x { - } 5 ) , N { = } ( x { + } 1 ) ( x { - } 6 )$ ，则 $M$ 和 $N$ 的大小关系是（） + +A $M { > } N$ B. $M { = } N$ C. $M { < } N$ D. $M { = } 2 N$ + +6.如图，有一块边长为 $_ { \mathcal { X } }$ 米的正方形草地，现将该正方形草地的南北方向减少3米，东西方向增加3米，则得到一块长为 $( x { + 3 } )$ 米，宽为 $( x { - } 3 )$ 米的长方形草地，那么长方形草地的面积比原来正方形草地的面积 ） + +A.增加了18平方米 B.增加了9平方米C.保持不变 D.减少了9平方米 + +![](images/55d5c281f2ea4971093b79602bb3e15f6186a7e135c1296554642fc566ca5317.jpg) +第6题图 + +![](images/1b2f4418dcecc0e79c12d5df1b966537fdec84f6e1d841b4393c487492c36c1c.jpg) +第7题图 + +7.如图所示，该图是用边长分别为 $\mathcal { X }$ 与 $y$ 的一个大正方形和四个小正方形以及四个长为 $\mathcal { X }$ ，宽为 $y$ 的长方形组成的 $( x > y )$ ，小明利用这个图形很容易解决如下问题：已知 $x ^ { 2 } + 4 y ^ { 2 } = 8 , x y = 2$ ，则 $x { + 2 y }$ 的值为4.那么这样的解决数学问题时所体现的数学思想是 （） + +A.数形结合思想B.逆向思维思想 C.分类思想 D.公理化思想 + +8. $\circledcirc$ 视讲解在一节数学课上，熊老师设计了一个如图所示的计算程序，则最后输出的结果为 （ ） + +输人有理数a $\sqrt { + ( a - 1 ) ^ { 2 } }$ （1+a)（1-a-a→输出 + +A.2a-3 B.2a-1 C.2a+1 D.2a+3 + +# 二、填空题(每小题4分，共16分) + +9.若 $2 m + n { = } - 1$ ，则 $2 ^ { 2 m } \bullet 2 ^ { n } = \underline { { { \vphantom { \sum } 2 ^ { n } } } }$ + +10.已知 $\scriptstyle x ^ { 2 } - 3 x = 1$ ，则代数式 $2 x ( x - 1 ) - 4 x + 3$ 的值为 + +11.若一个三角形的面积为 $x ( 2 x ^ { 2 } - 5 x )$ ，它的一条边长为 $( 2 x ) ^ { 2 }$ ,则这条边上的高为 + +12.若关于 $a$ 的代数式 $( \frac { 1 } { 2 } a + k ) ^ { 2 } - ( 2 - a ) \left( 2 + a \right) ( k$ 是常数)的结果中不含有常数项，则 $k ^ { 2 }$ 的值为 + +# 三、解答题(本大题6小题，共52分) + +13.(6分)计算： $( 1 ) ( x ^ { 2 n + 1 } ) ^ { 3 } \div x ^ { 5 n } \bullet x ; ( 2 ) - 1 ^ { 2 0 2 4 } - 3 2 ^ { 0 } \div ( - 3 ) ^ { - 1 } \times ( - \frac { 1 } { 2 } ) ^ { - 2 } .$ + +14.(8分)先化简，再求值： + +$( 1 ) x ( x + 7 ) + ( x - 5 ) ( x + 5 ) - ( x - 1 ) ( x + 6 )$ ,其中 $x { = } - 3$ $( 2 ) x ( 2 x + 5 ) - ( x ^ { 3 } - 2 x ^ { 2 } ) \div ( - \frac { 1 } { 2 } x )$ ，其中 $x { = } 5$ + +15.(8分）已知 $a = ( - \frac { 1 } { 8 } \times \frac { 1 } { 9 } ) ^ { 3 4 } \times ( 9 \times 8 ) ^ { 3 4 } , b = 9 4 ^ { 2 } - 9 3 \times 9 5 .$ 试化简代数式 $- ( 6 x y ^ { 2 } ) ^ { 2 } \div ( - 3 x y )$ ,并求当 $\scriptstyle x = a - 2 , y = b$ 时,该代数式的值. + +16.(8分)我们经常利用图形描述问题和分析问题,借助于直观的几何图形,把问题变得简明、形象,有助于探索解决问题的思路. + +(1)如图,小明构造了一个几何图形,请你用两种不同的方法计算图中阴影部分的面积 ， ，利用它们可以验证的公式为 + +(2)根据(1)中你得到的公式计算： $\textcircled { 1 } 2 9 9 ^ { 2 } ; \textcircled { 2 } 4 7 ^ { 2 } - 4 7 \times 8 6 + 4 3 ^ { 2 }$ (3)已知 $a ^ { 2 } + b ^ { 2 } = 1 0 , a b = 3$ ，求 $( a - b ) ^ { 2 }$ 的值. + +17.（10分)一条防洪堤坝,其横断面是梯形，上底宽为a米，下底宽为(a十2b)米，坝高为 $( a + 2 b )$ 米. + +(1)试用含 $a , b$ 的代数式表示该防洪堤坝的横断面的面积. + +(2)当a=2°÷2-1,b=（－2024)°－6×（－1)-9时,求该防洪堤坝的横断面的面积. + +(3)在(2)的条件下,若该防洪堤坝的长为() $( \frac { 1 } { 1 0 } ) ^ { - 2 } \div 1 0 ^ { - 1 }$ 米,则修建该防洪堤坝需要多少方土(用科学记数法表示)？（提示：一方土等于一立方米) + +18.新考法综合与实践视频讲解（12分)如图1，这是2023年12月份的日历，现按照图中所示的方式框住日历中的5个数,可以发现，框住的5个数具有下列特点：如 $8 \times 2 4 + 1 0 \times 2 2 - 2 \times 1 6 ^ { 2 } = 1 9 2 + 2 2 0 - 5 1 2 = - 1 0 0$ ，又如 $1 1 \times$ $2 7 + 1 3 \times 2 5 - 2 \times 1 9 ^ { 2 } = 2 9 7 + 3 2 5 - 7 2 2 = - 1 0 0$ ,不难发现,结果都是—100. + +(1)我们可以看出，上述框住的5个数满足一定的规律，请你用语言叙述这个规律： + +（2)设框住的5个数的中间数为x,请你用含x的代数式表示你发现的规律：·并利用整式的运算对你发现的规律加以说明. + +(3)如果像图2那样利用"十字框"框住5个数,那么(1)中的规律还成立吗？先利用图中两个框住的5个数验证一下，如果不成立，写出你发现的规律,然后利用整式的运算加以说明你的结论. + +
二日三2023年12月 四五六日
010203
04050607080910
1121314151617
18192021222324
25262728293031
+ +
2023年12月
一三三四五六日
010203
04Q50607080910
11121314151617
18192021222324
2s262728293031
+ +![](images/c1bddd9060ae242ae56c763b6c74e7a5725fad3df7d4d4ee5a606bdfdfdf24a8.jpg) +图1 +图2 + +# 专项训练卷（二） 几何题的计算 + +时间：60分钟 满分：100分 + +
题序二三评卷人总分
得分
+ +# 一、选择题(每小题4分，共32分) + +1.若 $\angle A = 3 0 ^ { \circ } 2 0 ^ { \prime }$ ，则 $\angle A$ 的余角的度数为 ） + +A. $6 0 ^ { \circ } 4 0 ^ { \prime }$ （204号 B. $6 0 ^ { \circ } 2 0 ^ { \prime }$ （204号 $\mathrm { C } . 5 9 ^ { \circ } 4 0 ^ { \prime }$ $\mathrm { D } . 5 9 ^ { \circ } 2 0 ^ { \prime }$ + +2.若等腰三角形顶角的度数比底角的度数大 $3 0 ^ { \circ }$ ,则这个等腰三角形底角的度数为 （ ） + +A $4 0 ^ { \circ }$ （20 $\mathrm { B } , 4 3 ^ { \circ }$ C.47° D.50° + +3.如图，为了估计池塘两端 $A , B$ 两点之间的距离，小康采取了如下方法：在池塘的一侧选取了一点 $P$ ，连接 $P A , P B$ ，测得 $P A = 2 0$ 米， $P B { = } 1 3$ 米,那么 $A$ ，$B$ 两点之间的距离可能是 （） + +A.7米 B.20米 C.33米 D.35米 + +![](images/a93735101569c9240e659854d78a8ce5dd3c961c91e834296caaf545c727e282.jpg) +第3题图 + +![](images/413cf56a578a078a516cdaccadba439f00461d47cb608a144e6e98c756847213.jpg) +第4题图 + +![](images/8c75b54141bb4cbd9a794aa09e383664adf84effd6313f0a95ba26c8d053dffd.jpg) +第5题图 + +4.如图,在 $\triangle A B C$ 中， $\angle C = 9 0 ^ { \circ }$ ， $A D$ 平分 $\angle B A C$ ，若 $S _ { \triangle A B D } = 1 5 ~ \mathrm { c m ^ { 2 } }$ $A B =$ $1 0 ~ \mathrm { c m }$ ，则线段 $C D$ 的长为 （ ） + +$\mathrm { A } . 3 \ \mathrm { c m }$ （20 （204号 $\mathrm { B . 4 \ c m }$ （204号 （20 $\mathrm { C } . 5 \ \mathrm { c m }$ （20 $\mathrm { D } , 6 \ \mathrm { c m }$ + +5.如图，为了测量河两岸 $A , B$ 两点之间的距离,小明设计了如下方案： $\textcircled{1}$ 在点 $B$ 的右侧取一点 $C$ ,利用精密测角仪测得 $\angle A B C = 7 4 ^ { \circ }$ $\angle A C B = 3 6 ^ { \circ }$ $\textcircled{2}$ 在点 $M$ 处立一根标杆,利用精密测角仪测得 $\angle M B C = 7 4 ^ { \circ }$ $\angle M C B = 3 6 ^ { \circ }$ .若测得 $B M$ 的长为58米，则 $A , B$ 两点之间的距离为 （ ） + +A.74米 B.58米 C.36米 D.28米 + +6.如图,已知直线 $A B , C D$ 与直线 $E F$ 分别相交于点 $G , H$ ，点 $M , N$ 分别是直线$E F , C D$ 上的一点,若 $A B / / C D , \angle 1 = 5 0 ^ { \circ } , \angle 3 = 8 0 ^ { \circ }$ ，则 $\angle 2$ 的度数为（ ） + +A $3 0 ^ { \circ }$ （20 B. $2 8 ^ { \circ }$ （204号 C.25° D.22° + +![](images/d2cceead6a5269f46eb0b5a705acf58c178e0b6bd59bbeddf7eed5157fde9779.jpg) +第6题图 + +![](images/1210227d4bd3dbcea8860402135f54e52ed237a09fa8513dbc34cb7a9f136941.jpg) +第8题图 + +7.已知△ABC≌△DEF, $\triangle A B C$ 的三边长分别为4,6,9,若 $\triangle D E F$ 与△ABC的三边长对应边长分别为 $4 , 3 x { - } 6 , 2 y { + } 1$ ，则 $( \boldsymbol { \mathcal { x } } - \boldsymbol { y } ) ^ { 2 0 2 4 }$ 的值为 （ ） + +A.6 B.4 C.2 D.0 + +8 $\circledcirc$ 视频讲解如图，在 $\triangle A B C$ 中,分别以点 $B$ 和点 $C$ 为圆心，大于 $\cdot { \frac { 1 } { 2 } } B C$ 的长为半径画弧,两弧相交于点 $M , N$ ，作直线 $M N$ ，交 $A C$ 于点 $D$ ，交 $B C$ 于点 $E$ ，连接BD,若△ABD 的周长为 $1 9 , \triangle A B C$ 的周长为25,则 $B C$ 的长为 （ ） + +A.4 B.5 C.6 D.7 + +# 二、填空题(每小题4分，共16分) + +9.如图，直线 $A B , C D$ 相交于点 $O , E O \bot A B$ 于点 $O$ ,若∠BOD： $\angle D O E { = } 5$ ：13,则 $\angle A O C$ 的度数为 + +![](images/0f32dfb46de48bf0dfd4a2ce42bc3e086dd28da6f9f64c6beeb59300e717a4da.jpg) +第9题图 + +![](images/4a1074b73840dc7410abbcb14038b6270e4b3059e977bbab007f9585b3d6cb7d.jpg) +图2 + +![](images/149e83e52222403bea85f9c63414409219e1ad3e3f03a88c8035c4b3a4f20237.jpg) +第10题图 + +![](images/db11fc3ba6bf5fb5feac032dc3cd314c98631a38fc5aa32cfeab6de8e2f4efd7.jpg) +图1 +第11题图 + +10.如图，在等边 $\triangle A B C$ 中， $A D , B E$ 分别是角平分线， $A D$ 与 $B E$ 相交于点 $O$ ，则 $\displaystyle \frac { O E } { O D } =$ + +11.新考法真实问题情境如图1,这是一辆山地自行车的实物图，图2是其抽象出来的部分示意图，已知直线 $E F$ 与 $B D$ 相交于点 $P , A B / / C D , \angle P = 1 8 ^ { \circ }$ ，$\angle C F P = \angle \alpha$ $\angle A B P { = } 9 2 ^ { \circ }$ ，则 $\angle \alpha$ 的补角的度数为 + +12.如图，在 $\triangle A B C$ 纸片中， $\angle A C B = 9 0 ^ { \circ }$ ，将该三角形纸片折叠，使得点 $A$ 落在边 $B C$ 上的点 $E$ 处， $C A$ 与 $C E$ 重合，折痕为 $C D$ ，若 $2 \angle A + 3 \angle B = 2 2 2 ^ { \circ }$ ，则$\angle B D E$ 的度数为 + +![](images/323bbc0cf663a250ad0a9bb6ffd46f6b7a6c4ae564515ffd28cb58c87d58c328.jpg) + +# 三、解答题(本大题5小题，共52分) + +13.(8分)如图，点 $D$ 在 $\angle A B C$ 的边 $A B$ 上，作 $D E \bot B C$ 于点 $E$ ，过 $\angle A B C$ 内的一点 $F$ 作 $F G \bot B C$ 于点 $G$ ，连接 $E F$ ，且 $E F / / A B$ ，若 $\angle B = 3 1 ^ { \circ }$ ，求 $\angle A D E$ 和$\angle F$ 的度数. + +![](images/7ac1e7a7c12ae1777fc2881a31d795e743a7166e5d42e17ac6f776cffc6b8c8b.jpg) + +14.(9分)如图,已知直线 $A B , C D$ 相交于点O,OM⊥AB,ON⊥CD. + +(1)若 $\angle 1 = 2 9 ^ { \circ }$ ，求 $\angle 2$ 的度数; +(2)若 $\angle B O D { = } 6 2 ^ { \circ }$ ，求 $\angle 1$ 和 $\angle M O N$ 的度数. + +![](images/8a3a83e68b5e4e46143c1d0d9862e4c82b5a6c4f90f27005cb423cbcb1aa4d43.jpg) + +15.(9 分)如图,在四边形 $A B C D$ 中， $A D / / B C$ ，点 $E$ 是 $C D$ 的中点，连接 $A E$ 并延长交 $B C$ 的延长线于点 $F , B E \bot A F$ + +(1)若 $B E { = } 4 , A E { = } 2$ ,求四边形 $A B C D$ 的面积； +(2)若 $\angle D A E = 5 9 ^ { \circ }$ ，求 $\angle A B F$ 的度数. + +![](images/44d148d333301abf6cfa5b7f953352410c887e78ff58936ecc6dfed5d6c2a1c4.jpg) + +16.（12分)阅读材料，完成任务： + +材料：我们知道，在 $\triangle A B C$ 中，若 $A B { = } A C$ ，则 $\angle B { = } \angle C$ ，这就是等腰三角形的性质，用语言叙述就是等腰三角形的两底角相等，即等边对等角；反之，可以得到等腰三角形的判定：在 $\triangle A B C$ 中，若 $\angle B = \angle C$ ，则 $A B { = } A C$ ，即两角相等的三角形是等腰三角形，即等角对等边. +任务：如图，在△ABC中， $A B { = } A C , A D \bot B C$ 于点 $D$ ，点 $N$ 是 $A D$ 上一点，且 $N M / / A C$ ,连接BN.(1)若 $\angle A B C = 6 5 ^ { \circ }$ ，求 $\angle A N M$ 的度数; +(2)若 $A C { = } 8$ ，点 $M$ 为AB 的中点， $B N { = } 3$ ,求MN的长和△BMN的周长. + +![](images/023d0d97e297b4ede745092bf496cd318192ed642103f9f05c8a2b55dbd6f18d.jpg) + +17. $\circledcirc$ 视频讲解（14分）综合与实践. + +问题情境：在 $\triangle A B C$ 中,已知边 $A B , B C$ 的垂直平分线分别交AB,BC于点 $D , F$ ，与 $A C$ 分别交于点 $E , G .$ +初步探究： $( 1 ) \textcircled{ 1}$ 如图1,若 $\angle A B C = 1 0 0 ^ { \circ }$ ，求 $\angle E B G$ 的度数； +$\textcircled{2}$ 如图2，若 $\angle A B C = 7 0 ^ { \circ }$ ，求 $\angle E B G$ 的度数； +归纳猜想：(2)若 $\angle A B C = \alpha$ $\angle A B C = 9 0 ^ { \circ }$ ,用含 $\alpha$ 的代数式直接写出 $\angle E B G$ 的度数. + +![](images/ccf3d8da3e2d3441588efd2a6e81a924e091878a81ed7624e347581c1c09ad99.jpg) +图1 + +![](images/37080da8dc8ef8c3e1da643977c172d3d809efda4ddfac6a5e8542cec970fa6a.jpg) +图2 + +# 专项训练卷(三) 几何题的说理 + +时间：60分钟 满分：100分 + +
题序二三评卷人总分
得分
+ +# 一、选择题(每小题5分，共30分) + +1.如图,已知直线l上有一点 $A$ ，直线 $m \perp n$ ，则 $\angle 1$ 与 $\angle 2$ 的关系是 + +A. $\angle 1$ 与 $\angle 2$ 相等 B. $\angle 1$ 与 $\angle 2$ 互余C. $\angle 1$ 与 $\angle 2$ 互补 D. $\angle 1$ 与 $\angle 2$ 是对顶角 + +![](images/b853effb10fe22d338ebfbb902f846525683ec03bf9b27c58ccbcaee7d2f362f.jpg) +第1题图 + +![](images/12839f8809656798cf434b57b4e3c52f532f976d48fe604d31c88ee8ef9c4e5d.jpg) +第2题图 + +![](images/6d8b3514358d7de0e59d33b99d64f63a5e651a134461e8f3f5025d563cf9dc30.jpg) +第3题图 + +2.如图，已知点 $D , B$ 在线段 $A E$ 上,若 $\angle A B C = \angle E$ ，则下列结论正确的是 （ + +A. AC// DF B. BD // AC $\complement . B C / / E F$ （204号 （204号 $\mathrm { D } , D G / / A C$ （20 + +3.将一副三角板和一张对边平行的纸条按如图所示的方式摆放,两把三角板的一条直角边重合,则 $\angle 1$ 与 $\angle 2$ 之间的数量关系是 （ ） + +A. $\angle 1 + \angle 2 = 3 0 ^ { \circ }$ B. $\angle 1 + \angle 2 = 4 5 ^ { \circ }$ $\mathrm { { C } } _ { \cdot } \angle 1 + \angle 2 = 6 0 ^ { \circ }$ $D _ { \bullet } 2 \angle 1 + \angle 2 = 7 5 ^ { \circ }$ + +4.在 $\triangle A B C$ 中，若 $\angle A { = } 2 \angle B$ $\angle A = \angle C + 2 0 ^ { \circ }$ ，则 $\triangle A B C$ 的形状是 （ ） + +A.锐角三角形 B.钝角三角形C.直角三角形 D.无法确定 + +5.若△ABC的三边长分别为 $a , b , c$ ，则 $| a + b + c | + | a - b - c |$ 的值为（) + +A. $a + 2 b$ B. $2 b + 2 c$ C. $2 a + 2 c$ （204号 D. $2 b { + } c$ （20 + +6. $\circledcirc$ 视频讲解如图，在△ABC中， $A B { = } A C$ ，以点 $C$ 为圆心， $C B$ 长为半径画弧，交 $A B$ 于点 $B$ 和点 $D$ ,再分别以点 $B , D$ 为圆心，大于 $\frac { 1 } { 2 } B D$ 长为半径画弧,两弧相交于点 $M$ ，作射线CM交 $A B$ 于点 $E$ ，连接 $C D , D M , B M ,$ 则下列结论正确的有（ + +![](images/ab45dbe2363ddf7573709e8d1b4997443e283d15a47dcee3dbdca5b1fdc06e04.jpg) + +$$ +\textcircled { 1 } \angle C B D = \angle C D B ; \textcircled { 2 } C M \perp B D ; \textcircled { 3 } \angle C D B + \frac { 1 } { 2 } \angle A = 9 0 ^ { \circ } . +$$ + +A $\textcircled{1} \textcircled{2}$ B. $\textcircled{1} \textcircled{3}$ C. $\textcircled{2} \textcircled{3}$ D. $\textcircled{1} \textcircled{2} \textcircled{3}$ + +# 二、填空题(每小题5分，共20分) + +7.如图,已知点 $C$ 是 $A E$ 的中点， $\scriptstyle \angle A = \angle E C D , A B = C D$ 那么△ABC≌△CDE 的根据是 .（填"SSS"、“SAS"“ASA"或"AAS") + +![](images/02b911f3500ed03b123c1fb0c18d96a4302e02a34de19825e26cd66ec7323a2a.jpg) +第7题图 + +![](images/b9c8c0263bb16eff7deaa1d8013b898b4ab41980adce426e5a674f74569434b1.jpg) +第8题图 + +![](images/c506603b27fbf2dcdbb844c3ee104ddfff9bbe7c0e35eb21870c9f9ce435dbeb.jpg) +第9题图 + +![](images/e924382215435b7549d74138301e8eb6285f659972e9f3be06896fdac5ae8c49.jpg) +第10题图 + +8.如图所示,这是一个正五边形,它是一个轴对称图形,它有 条对称轴. + +9.如图，在“ $4$ ”字图中， $B C$ 与 $E F$ 相交于点 $D$ ,若图中有 $a$ 对同位角, $b$ 对内错角， $c$ 对同旁内角， $d$ 对对顶角，则 $( a + b - c + d ) ^ { 3 }$ 的值为 + +10.如图,已知 $A B / / C D , A E \bot E F ,$ 则 $\angle A , \angle C , \angle F$ 之间的数量关系是 + +# 三、解答题(本大题5小题，共50分) + +11.(8分)如图，直线 $A B$ 与 $E F$ 相交于点 $O$ (1)若 $\angle A O E { = } 4 5 ^ { \circ }$ $\angle A O F { = } 3 \angle C O F$ ，试说明 $A B \bot C O$ (2)在(1)的条件下,找出所有与 $\angle G O F$ 互余的角和互补的角. + +![](images/b022f649616ba4ab894144c0fb012fd50b4e9653d3e64d99df3d0193d415632b.jpg) + +12.(8分)如图，点 $D , E$ 在边BC上, $E F$ 垂直平分 $A C$ $A B { = } A E$ ,AD是△ABE 的中线. + +（1）试说明△ABD△AED； +(2）试探究 $\angle B$ 与 $\angle C$ 之间的数量关系，并说明理由. + +![](images/ee71b1afc7d9faa9f527eb244b5fba79e15b4b0aff20e91c8c5de225309ee0b0.jpg) + +13.(8分)如图，在 $\mathrm { R t } \triangle A B C$ 中， $\angle B = 9 0 ^ { \circ } , A D$ 是 $\angle B A C$ 的平分线,DE⊥AC于点 $E$ ，点 $F$ 是 $A B$ 上一点， $B { \cal F } { = } C E$ + +(1)△BDF和 $\triangle E D C$ 全等吗？请说明理由. +(2)试猜想 $A C , A B , B F$ 之间的数量关系，并说明理由. + +![](images/92cfdd907ea2dbb12469e29fd543e96d1b3229598ff69190db96dc5a97899a8c.jpg) + +14.（12分)如图，已知点 $P , Q$ 分别在直线 $A B$ ， $C D$ 上， $\angle A P N + \angle P N Q +$ $\angle C Q N = 3 6 0 ^ { \circ }$ + +(1)试说明 $A B / / C D$ + +(2)若 $P M , Q M$ 分别平分 $\angle B P N$ $\angle D Q N$ ，试探究 $\angle M$ 与 $\angle N$ 之间的数量关系，并说明理由. + +![](images/80eb1cac2a021ada699403d6b752ada3a3f25a43b456d387f25b05b1ba63a578.jpg) + +15.新考法阅读理解题 $\circledcirc$ 视频讲解（14分)阅读材料，完成后面的任务： + +材料：如图1，在凹四边形ABOC中，试探究$\angle B O C$ 与 $\angle A , \angle B , \angle C$ 之间的数量关系．小明发现并提供了如下方法：解： $\angle B O C = \angle A + \angle B + \angle C .$ 如图2,连接 + +![](images/304f0c613589c47f939e2e92674324f1a0800101ee8c8199b00f084ea5e0b278.jpg) +图1 + +![](images/5840df022ed06d95be097ba8aca5fa44b87423c1e8040ce4257099a96fa77f6a.jpg) +图2 + +$A O$ 并延长，因为 $\angle 1 + \angle B + \angle A O B = 1 8 0 ^ { \circ } , \angle 3 + \angle A O B = 1 8 0 ^ { \circ }$ ，所以 $\angle 3 =$ $\angle 1 + \angle B$ ，同理 $\angle 4 = \angle 2 + \angle C$ 所以 $\angle 3 + \angle 4 = \angle 1 + \angle 2 + \angle B + \angle C ,$ 所以 $\angle B O C = \angle B A C + \angle B + \angle C .$ + +任务：(1)你有与小明不同的方法吗？请写出你的解答过程. + +(2)如图3所示，根据图中标注的角的度数，请说明 $\angle A + \angle C + \angle D + \angle F =$ $2 3 0 ^ { \circ }$ + +![](images/b1119104638303808606f85458112bdc1a66a31f00be9753d74199ccb58b24e1.jpg) + +![](images/1e620d511f484cdbdd6b112371bf4acbea44ef03189714cca65626e94c784e9f.jpg) +图4 + +(3)如图4所示，在 $\triangle A B C$ 中， $A B { = } A C$ $\angle A B C , \angle A C B$ 的平分线 $B D , C E$ 交于点 $O , \angle E O D + \angle O B F = 1 8 0 ^ { \circ } , B F$ 交 $A C$ 的延长线于点 $F$ $, \angle C D G =$ $\angle F , D G$ 交 $B C$ 的延长线于点 $G$ ,找出图中的平行线，并说明理由. + +(4)在(3)的条件下，若 $\angle A = 5 8 ^ { \circ }$ ，直接写出 $\angle A D G$ 的度数： + +![](images/357b43173835f003d78a1c038762b3596da146dea3499e09342b9895aa80f747.jpg) + +# 期末测试卷 + +时间：90分钟 满分：120分考试范围：第一章～第六章 + +
题序二三评卷人总分
得分
+ +# 一、选择题(本大题共10小题,每小题3分，共30 分) + +1.下列"表情”中,属于轴对称图形的是 + +![](images/8855ee32514d781c76c28b250c5e304683f2d8320d5d1d3abe000adc8c603582.jpg) + +2.下列说法中，正确的是 + +A.若 $\angle \alpha$ 与 $\angle \beta$ 是同位角，则 $\angle \alpha { = } \angle \beta$ +B.若 $\angle 1 + \angle 2 = 9 0 ^ { \circ }$ ，则 $\angle 1$ 与 $\angle 2$ 互余 +C.两条边和一个角分别对应相等的两个三角形全等 +D.若一个事件发生的概率为0,则这个事件是不确定事件 + +3.若△ABC是等腰三角形,其中两边长分别为 $3 \ \mathrm { c m } , 6 \ \mathrm { c m }$ ，则第三边长可能为（ ） + +A.3 cm B.4 cm C. 6 cm D. $3 ~ \mathrm { c m }$ 或 $6 ~ \mathrm { c m }$ + +4.如图，在 $\triangle A B C$ 和 $\triangle D E C$ 中，已知 $C B = C E$ ，还需添加两个条件才能使$\triangle A B C { \cong } \triangle D E C$ ，不能添加的一组条件是 （ ） + +A. $A B { = } D E$ $\angle B = \angle E$ $3 . A B { = } D E , A C { = } D C$ $\scriptstyle \mathrm { C } , A B = D E , \angle A = \angle D$ D. $\angle A = \angle D , \angle B = \angle E$ + +![](images/969be655392a420e822bf85a33aecdb9fe7b9b40c7c809e7744a986245fea9dd.jpg) + +5.“四时花竞巧，九子粽争新”,端午节吃粽子是我国的传统习俗.小颖妈妈在超市购买了豆沙粽和蛋黄粽共15个，这些粽子除了内部馅料不同外，其他均相同.小颖从中任选了一个粽子,若她选到蛋黄粽的概率为 $\frac { 3 } { 5 }$ ，则购买的豆沙粽的个数是 + +A.5 B.6 C.8 D.9 + +6.如图,将长方形纸片 $A B C D$ 沿 $E F$ 折叠后，点 $D , C$ 分别落在点 $D _ { 1 } , C _ { 1 }$ 的位 置， $E D _ { 1 }$ 的延长线交 $B C$ 于点 $G$ ，若 $\angle E F G { = 6 4 } ^ { \circ }$ ，则 $\angle E G B$ 等于 （ + +A. $1 2 8 ^ { \circ }$ （204号 B.130° （204号 $C . 1 3 2 ^ { \circ }$ （20 D. $1 3 6 ^ { \circ }$ + +![](images/b1c805e02f145869410a7422b960077e565fe44873be81f7af66669d97c1baee.jpg) +第6题图 + +![](images/bd2864a40ffcc6b99019ae0e96898793e4668ccc92b595f3d9e3a37d4d549e07.jpg) +第7题图 + +![](images/4864dfe3aa14d74b74375855177d93d10361f90a5c3f8199ddba7708d954ed61.jpg) +第8题图 + +7.为增强居民的节水意识,某自来水公司采用以户为单位分段计费的办法收费,即每月用水不超过10吨,每吨收费 $a$ 元;若超过10吨,则10吨水按每吨$a$ 元收费,超过10吨的部分按每吨 $^ { b }$ 元收费，公司为居民绘制的水费y(元)随当月用水量 $_ { \mathcal { X } }$ （吨)变化而变化的图象如图所示，则下列结论错误的是（ ） + +A. $a = 1$ .5 +B. $b { = } 2$ +C.若小明家7月份缴水费30元，则当月用水18.5吨D.若小明家3月份用水14吨，则应缴水费23元 + +8.如图，在四边形 $A B C D$ 中， $A E \bot C D$ ，点 $B$ 关于 $A C$ 的对称点 $B ^ { \prime }$ 恰好落在 $C D$ 上，点 $B ^ { \prime }$ 关于 $A E$ 的对称点为点 $D$ ，若 $\angle B A D = 1 0 0 ^ { \circ }$ ，则 $\angle A C B$ 的度数为（ ） + +A. $4 0 ^ { \circ }$ （20 B.45° C.60° D.80° + +9.如果一个正整数能表示为两个连续偶数的平方差，那么称这个正整数为“和平数”如 $4 = 2 ^ { 2 } - 0 ^ { 2 } , 1 2 = 4 ^ { 2 } - 2 ^ { 2 }$ ,因此4,12 这两个数都是“和平数”.介于1到301之间的所有“和平数"之和为 （） + +A.5776 B.4096 C.2022 D.108 + +10. $\circledcirc$ 视频讲解如图，在等腰 $\triangle A B C$ 中， $A B = A C , \angle B A C = 5 0 ^ { \circ } , \angle B A$ 的平分线与线段 $A B$ 的垂直平分线交于点 $O$ ，点 $C$ 沿 $E F$ 折叠后与点 $O$ 重合，则$\angle C E F$ 的度数是 （ ） + +A. $4 5 ^ { \circ }$ B. $5 0 ^ { \circ }$ （20 $\mathrm { C } . 5 5 ^ { \circ }$ （20 D. $6 0 ^ { \circ }$ + +![](images/d39326d8d354618f7c808e9fd0fd13f3da982f6738437174b453704ae1922063.jpg) + +# 二、填空题(本大题共5小题,每小题3分，共15分) + +11.一个游戏转盘如图所示，自由转动转盘，当转盘停止转动后,指针落在字母“C"所示区域内的概率是 + +（2)如图，在△ABC中， $\angle A = 9 0 ^ { \circ }$ ，点 $E$ 为 $B C$ 上一点，点 $A$ 和点 $E$ 关于 $B D$ 对称,点 $B$ 和点 $C$ 关于 $D E$ 对称，求 $\angle A B C$ 和 $\angle C$ 的度数. + +![](images/5307ce51a1b88d363cc71f860079ed72ac05d6631ddec358a7999f38bbc77bd2.jpg) +第11题图 + +![](images/3d3ddb0d0294ce77ad79ea9bd2bdd76f82a88cbaf4adf0f753a17f22efcd60ca.jpg) +第12题图 + +![](images/60e5a9b64348db6ef279682835f6ee2bd4e15acbf79deb03e374e137ac4d8b11.jpg) +第14题图 + +![](images/d16f88fabcce8dd837e6737fd2a4885fc1478f94f5bea4d8d6e5d7c6e49f8a24.jpg) +第15题图 + +12.如图， $\scriptstyle \cdot B E = B A , A B / / D E , B C = D E$ ，若 $\angle B A C = 4 0 ^ { \circ }$ $\angle E { = } 2 5 ^ { \circ }$ ，则 $\angle B D E$ 的 度数为 + +13.如果 $x ^ { 2 } + 3 x { = } 2 0 2 4$ ,那么代数式 $x ( 2 x + 1 ) - ( x - 1 ) ^ { 2 }$ 的值为 + +![](images/ab49604f8ffa3913d9b38e61212e10652ec86b7e3adde3c5cb2f83a39798d99f.jpg) + +14.如图，在 $4 \times 4$ 的正方形网格中，已有4个小方格涂成了灰色，现在要从其余白色小方格中选出一个也涂成灰色，使整个灰色部分的图形构成轴对称图形,这样的白色小方格有 个. + +15. $\circledast$ 视频讲解如图，在 $\triangle A B C$ 中， $A B { = } A C$ ，点 $D , E , F$ 分别是边 $A B \circ B C , C A$ 上的点， $D E$ 与 $B F$ 相交于点 $G , B D { = } B C , B E { = } C F$ 若 $\angle A = 4 0 ^ { \circ }$ ，则 $\angle D G F$ 的度数为 + +# 三、解答题(本大题共8小题，满分75分） + +16.(10分）（1)计算： $( - \frac { 1 } { 4 } ) ^ { - 1 } + ( \pi - 2 ) ^ { 0 } - ( - 3 ) ^ { - 3 } \div ( - 3 ) ^ { - 4 } + | - 2 | .$ + +17.(9分)公安人员在破案时常常根据案发现场作案人员留下的脚印推断犯人的身高，如果用 $a$ 表示脚印长度， $b$ 表示身高，它们的关系接近于 $b { = } 7 a -$ 3.07. + +(1)若某人的脚印长度为 $2 4 . 5 ~ \mathrm { c m }$ ，则他的身高约为多少厘米？ +(2)在某次案件中，抓获了两名可疑人员，一个身高为 $1 . 8 7 \mathrm { ~ m ~ }$ ，另一个身高为$1 . 7 5 \mathrm { ~ m ~ }$ ,现场测量的脚印长度为 $2 6 . 7 \ \mathrm { c m }$ ,请你帮助侦查一下，哪个可疑人员作案的可能性更大，并说明理由. + +![](images/6dfe0827c875877d90d7d3b3b98ba9cc4d0496e2988cb7e6043b070807d97311.jpg) + +18.(9分)已知 $( m + n ) ^ { 2 } = 1 1 ^ { 2 }$ ， $m n { = } 1$ ，求 $( m - n ) ^ { 2 }$ 的值. + +19.(9分)如图，在 $\triangle A B C$ 中， $\angle A = 6 8 ^ { \circ }$ ，点 $D$ 是 $B C$ 上一点， $B D , C D$ 的垂直平分线分别交 $A B$ $A C$ 于点 $E , F$ ，求 $\angle E D F$ 的度数. + +20.(9分)如图，点 $C$ 在线段 $A E$ 上， $B C / / D E , A C = D E , B C = C E$ ，延长 $A B$ 分别交 $C D , E D$ 于点 $G , F$ + +（1)试说明： $A B { = } C D$ + +![](images/b4df81009446f5179bff5f5252e2bedddf3a8d8b0f2c88db7e6877c5891a607b.jpg) + +(2)若 $\angle D = 3 0 ^ { \circ }$ $\angle E { = } 6 5 ^ { \circ }$ ，求 $\angle F G C$ 的度数. + +![](images/c492cb1542614cbef2ce27c7387c4214b37bf2798c7d96b59a1c790c31e1b737.jpg) + +21.(9分)如图，点 $D$ 是 $\angle B A C$ 外一点，过点 $D$ 作 $D E / / A B$ 交 $A C$ 于点 $F$ ，以$D E$ 为边作 $\angle E D G .$ 若 $D G / / A C$ ,请画出图形,判断 $\angle B A C$ 与 $\angle E D G$ 的数量关系，并说明理由. + +![](images/58468dc02d99cebfff32b239baedfde7a5598e2c741a3712157b344badb5f76e.jpg) + +22.(10分)如图，张达设计了一个被均匀地分成了8份的圆形转盘，该转盘分别标有1,2,3,4,5,6,7,8这8个数,转动转盘，当转盘停止时，指针指向的数即为转出的数.(当指针恰好指在分界线上时，无效重转) + +(1)求小彬转出的数是4的倍数的概率. + +(2)现有两张分别写有3和5的卡片，随机转动转盘，转盘停止后记下转出的数，与两张卡片上的数分别作为三条线段的长度.这三条线段能构成三角形的概率是多少？ + +![](images/2f65936634cb4ad55eaeac91faf25bd2b9ffe6a6d5602fa19d326e16fc2424f2.jpg) + +23. $\circledcirc$ 视频讲(10分)现有一笔直的公路连接 $M , N$ 两地,甲车从 $M$ 地驶往 $N$ 地，速度为 $8 0 ~ \mathrm { k m / h }$ ，同时乙车从 $N$ 地驶往 $M$ 地，速度为 $1 0 0 ~ \mathrm { k m / h } .$ 途中甲车发生故障,于是停车修理了 $2 . 5 \mathrm { ~ h ~ }$ ，修好后立即开车驶往 $N$ 地.设乙车行驶的时间为 $^ { t \mathrm { ~ h ~ } }$ ，两车之间的距离为 $s \ \mathrm { k m } .$ 已知 $s$ 与 $t$ 之间的关系的部分图象如图所示. + +$( 1 ) M , N$ 两地的实际距离为 km. (2)图象中点 $C$ 的实际意义是 （3)甲车出发几小时后发生故障？ (4)乙车出发几小时后两车相距 $2 0 0 ~ \mathrm { k m } ?$ （20 + +![](images/1a9cb12234edcd732bb7dbfcd7629189427dc354c1bc8a5b0e816d896e0a89e1.jpg) + +# 答案和解析 + +# 第一章整式的乘除 + +# 卷中悟法 + +# 1.正确计算幂的乘方 + +1.幂的乘方是变乘方为指数的乘 法，即底数不变，指数相乘.而同底数幂 的乘法是变乘法为指数的加法，即底数 不变，指数相加. + +2.幂的乘方不能和同底数幂的乘法相混淆.例如不能把 $( a ^ { 5 } ) ^ { 2 }$ 的计算结果错误地写成 $a ^ { 7 }$ ，也不能把 $a ^ { 5 } \cdot a ^ { 2 }$ 的计算结果错误地写成 $a ^ { 1 0 }$ + +2.利用转化的思想解幂的有关计算 + +解幂的有关计算时，一般有以下两种转化方法： + +(1)把不同底数的幂转化为相同底数的幂；(2)把不同指数的幂转化为相同指数的幂. + +# 3.多项式乘多项式的解题技巧 + +对于多项式乘多项式，由于项可能较多，所以计算比较复杂、繁琐，对于此类题有以下几种解题技巧： + +(1)可以将一个多项式拆开，转化为 先单项式乘多项式，再求和的形式. + +(2)确定积中每项的符号时，按“同号得正，异号得负”的法则进行。 + +(3)观察多项式的组成形式，合理利用平方差公式及完全平方公式解题. + +(4)利用公式 $\left( x + a \right) \left( x + b \right) = x ^ { 2 } +$ $( a + b ) x + a b$ ，能提高运算效率. + +(5)最终结果中，若有同类项要合并同类项. + +4.单项式除以单项式的解题技巧 + +单项式除以单项式可以按以下三个步骤进行： + +(1)将系数相除，作为商的系数。(2)对于被除式和除式均含有的字母按照同底数幂的除法法则相除.(3)被除式中单独含有的字母，连同它的指数一起作为商的因式. + +对于商中符号的确定可按照“同号得正，异号得负”的法则，对于多项式除以单项式，可将多项式拆开，将其转化为单项式除以单项式的问题. + +# 基础过关参考答案 + +# 一、选择题 + +
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CDBACBAD
+ +# 二、填空题 + +
9101112
1.89×10-55x62m4n+2m²n² -2m²n-1
+ +# 三、解答题 + +13.解：（1）原式 $= - b ^ { 5 } \cdot b ^ { 1 2 } = - b ^ { 1 7 }$ ；… · (3分) (2）原式 $= - x ^ { 3 } \cdot x ^ { 4 } - 4 x ^ { 6 } \cdot x = - x ^ { 7 }$ $- 4 x ^ { 7 } = - 5 x ^ { 7 }$ ……· (6分) + +14.解：(1)原式 $= x ^ { 2 } - x + 2 x ^ { 2 } = 3 x ^ { 2 } - x$ · (4分) (2）原式 $= 7 x + 7 - x ^ { 2 } - x - x ^ { 2 } + 3 x$ $= - 2 x ^ { 2 } + 9 x + 7$ ……· (8分) + +15.解：原式 $= 4 ( x ^ { 2 } - 2 x + 1 ) - ( 4 x ^ { 2 } - 9 )$ $= 4 x ^ { 2 } - 8 x + 4 - 4 x ^ { 2 } + 9 = - 8 x + 1 3 .$ …(5分) 当 $x { = } - 2$ 时，原式 $= - 8 \times ( - 2 ) +$ $1 3 { = } 2 9$ …（8分) + +16.解：(1)因为 $a \ast b { = } 2 ^ { a } \times 2 ^ { b }$ ，所以 $1 \ast 3$ $= 2 ^ { 1 } \times 2 ^ { 3 } = 2 \times 8 = 1 6$ ;…（3分)(2)因为 $2 * ( x + 1 ) = 2 5 6$ ，所以 $2 ^ { 2 } \times$ $2 ^ { x + 1 } = 2 ^ { 8 }$ ，则 $2 + x + 1 = 8$ ，解得 $x = \mathrm { ~ i ~ }$ 5． ··(8分) + +17. $\begin{array} { r l r } & { } & { \hbar \pmb { \mathscr { G } } : ( 1 ) ( a x - 1 ) ( x + 7 ) - x ^ { 2 } + 7 = a x ^ { 2 } \left| \right. } \\ & { } & { \left. + 7 a x - x - 7 - x ^ { 2 } + 7 = ( a - 1 ) x ^ { 2 } + \right. } \end{array}$ $( 7 a - 1 ) x$ ，因为它不含有 $x ^ { 2 }$ 的项，所以 $a = 1$ … (4分) + +$$ +\begin{array} { c } { { ( 2 ) ( a - 5 ) ( a + 5 ) + ( a - 5 ) ^ { 2 } - a ( 2 a } } \\ { { + 1 ) = a ^ { 2 } - 2 5 + 2 5 - 1 0 a + a ^ { 2 } - 2 a ^ { 2 } } } \end{array} +$$ + +$- a { = } - 1 1 a$ ， …· (8分)当 $a = 1$ 时，原式 $= - 1 1 \times 1 = - 1 1$ …… (10分) + +18.解: $( 1 ) P = ( x - y ) ^ { 5 } \div ( y - x ) ^ { 5 } \cdot ( x$ $- y ) ^ { 2 } = - ( x - y ) ^ { 5 - 5 + 2 } = - ( x - y ) ^ { 2 }$ $= - { x ^ { 2 } } + 2 x y - { y ^ { 2 } }$ ； …· (3分)$Q = [ ( 2 x - y ) ^ { 2 } + ( 2 x + y ) ( 2 x - y ) +$ $8 x y ] \div 2 x = ( 4 x ^ { 2 } - 4 x y + y ^ { 2 } + 4 x ^ { 2 } -$ $y ^ { 2 } + 8 x y ) \div 2 x$ $= ( 8 x ^ { 2 } + 4 x y ) \div 2 x = 4 x + 2 y .$ （20…· (8分)(2)由题意，得 $x = ( 3 - \pi ) ^ { 0 } = 1 , \ ;$ $y =$ $( - { \frac { 1 } { 2 } } ) ^ { - 1 } { = } { - 2 } , { \frac { \phantom { - } } { \cdots \cdots \cdots \cdots \cdots } }$ （10分）所以 $P { = } { - } ( 1 { + } 2 ) ^ { 2 } { = } { - } 9 , Q { = } 4 \times 1$ $+ 2 \times ( - 2 ) = 0 .$ ·（12分） + +·能力提优参考答案· + +# 一、选择题 + +
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CBAABDCC
+ +# 二、填空题 + +
9101112
9-2ab，（a-b）²(a+6）²= (a-b)²+4ab2
+ +# 三、解答题 + +13.(1)解：原式 $= a ^ { 6 } - 9 a ^ { 6 } - 6 4 a ^ { 6 }$ （2分） $= - 7 2 a ^ { 6 }$ （3分） (2)解：原式 $= 5 - 1 - 3$ ……· (2分) + +$= 1$ · (3分) + +14.解： $( x - 3 ) ^ { 2 } + ( x + 3 ) ( x - 3 ) + 2 x ( 2$ $- x ) = x ^ { 2 } - 6 x + 9 + x ^ { 2 } - 9 + 4 x -$ $2 x ^ { 2 } = - 2 x$ … (6分) 当 $x { = } - \frac { 1 } { 2 }$ 时，原式 $= - 2 \times ( - \frac { 1 } { 2 } ) =$ …· (8分) + +15.解：一； …· (4分) 正确的解答过程： $\alpha ( 1 + a ) - ( a - 1 ) ^ { 2 }$ $= a + a ^ { 2 } - ( a ^ { 2 } - 2 a + 1 ) = a + a ^ { 2 } - a ^ { 2 }$ $+ 2 a - 1 = 3 a - 1$ ……(8分) + +16.解：根据题意得, $( 3 x ^ { 2 } - 5 x y ) \div \frac { x } { 2 } =$ $6 x - 1 0 y$ ，即第一个多项式是 $6 x -$ $1 0 y$ …· (5分)则正确的结果应为 $( 6 x - 1 0 y$ $\frac { x + y } { 2 } = 3 x ^ { 2 } + 3 x y - 5 x y - 5 y ^ { 2 } = 3 x ^ { 2 }$ $- 2 x y - 5 y ^ { 2 }$ …· (8分) + +17.解： $( x + a ) ( x + b ) = x ^ { 2 } + ( a + b ) x +$ $a b = x ^ { 2 } + m x + 2 8$ ，对比可得 $a + b =$ $m , a b { = } 2 8$ ： … (6分)当 $a = 1 , b = 2 8$ 时， $m { = } 2 9$ ；当 $a = 2 , b$ $= 1 4$ 时， $m { = } 1 6$ ；当 $a = 4 , b = 7$ 时， $m$ $= 1 1$ …(10分) + +18.解：(1)原长方形铁皮的面积是 $( 4 a +$ 60) $( 3 a + 6 0 ) = 1 2 a ^ { 2 } + 4 2 0 a + 3 6 0 0$ $( \mathrm { c m } ^ { 2 }$ ）； …（5分）(2)油漆这个铁盒的面积是 $3 a \cdot 4 a$ $+ 2 \times 3 0 \times 4 a + 2 \times 3 0 \times 3 a = 1 2 a ^ { 2 } +$ $4 2 0 a ( \mathrm { c m } ^ { 2 } )$ + +$( 1 2 a ^ { 2 } + 4 2 0 a ) \div { \frac { a } { 5 0 } } = 6 0 0 a + 2 1 0 0 0$ （元)．… …（10分）答：油漆这个铁盒需要（ $6 0 0 a \ +$ 21000）元. …（12分） + +# 第二章相交线与平行线 + +# 卷中悟法 + +# 一、同位角、内错角与同旁内角的识别 + +1.定义法 + +根据定义，两个角共涉及三条直线（或射线或线段），两角的一边分别在两条直线上，而另一边在同一直线上，两角有“共线边”是定义的实质，抓住“一边共线”便不难识别，如图1中的 $\angle 1$ 和 $\angle 2$ 涉及 $E F , M G , N D$ 三条直线，且它们都有一边在直线 $E F$ 上，故 $\angle 1$ 与 $\angle 2$ 是同位角.又如图2中的 $\angle 1$ 和 $\angle 2$ 是否为同位角？因其涉及 $A D , A C , A B , B C$ 四条直线，无共线边，故 $\angle 1$ 和 $\angle 2$ 不是同位角. + +角（图 $\textcircled{2}$ 中的 $\angle 1$ 和 $\angle 2 )$ + +(3)将两个角的两边延长，如果能构成“U"型（或反置或倒置)，那么这两个角为同旁内角（图 $\textcircled{3}$ 中的 $\angle 1$ 和 $\angle 2 )$ + +要注意这三种基本结构的变式.这种方法的好处是有助于记忆和理解，在判断两个角的关系时也不会出错.对顶角是有公共顶点的两个角，而同位角、内错角和同旁内角是没有公共顶点的两个角，学习时要注意它们的区别. + +![](images/baeaf0d81820417d7932e234b8fd65462016b8eb6ede2456b9b502253a98506e.jpg) +图1 + +# 二、平行线的判定与性质的应用 + +![](images/21512e0c928ef40af9018a1142ae82f4674d679b657e79b0221eaf60826a1ddd.jpg) +图3 + +![](images/5ed7c91d88194399cf1cade422e98e51a285d0accef222fa80432215e4a49b02.jpg) +图2 + +2.描粗相关线条法 + +(1)将两个角的两边延长，如果能构 +成“ $F ^ { \prime }$ 型(或倒置或反置)，那么这两个角 +为同位角（图 $\textcircled{1}$ 中的 $\angle 1$ 和 $\angle 2 )$ (2)将两个角的两边延长，如果能构 +成“Z"型(或反置)，那么这两个角为内错 + +1.判定两直线平行的方法有五种：(1)平行线的定义；(2)平行公理的推论：如果两条直线都与第三条直线平行，那么这两条直线平行；(3)同位角相等，两直线平行；(4)内错角相等，两直线平行；(5)同旁内角互补，两直线平行.判定两直线平行时，定义一般不常用，其他四种方法要灵活运用，证明时要注意书写格式. + +2.由两条直线平行得到同位角相等、内错角相等和同旁内角互补，解题时应结合图形先确认所成的角是否为两平行线被第三条直线所截得的同位角或内错角或同旁内角，同时要学会简单的几何说理，做到每一步有理有据. + +# 一、选择题 + +
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+ +# 二、填空题 + +
9101112
72°60°1270°
+ +# 三、解答题 + +13.解： $( 1 ) \angle 1$ 与 $\angle 2$ 是内错角， $\angle 2$ 与 $\angle 3$ 是同旁内角；… · (3分) (2) $\angle 2$ 与 $\angle 5$ 是同旁内角； $\angle 4$ 与 $\angle 5$ 是同位角； $\angle 5$ 与 $\angle 6$ 是对顶角. …· (6分) + +14.解：如图所示． …(6分)因为 $\angle D A C = \angle A C B$ ，根据内错角相等，两直线平行，所以 $A D / / C B$ ·· (8分) + +![](images/d9f040b426c55809068a26bd0e3c913c4cbf34f8cedc2a385cd215770096f8a8.jpg) + +15.解：因为点 $O$ 为直线 $B D$ 上一点，所以 $\angle C O D + \angle B O C = 1 8 0 ^ { \circ }$ 将 $\angle C O D = 2 \angle B O C$ 代人，得$2 \angle B O C + \angle B O C = 1 8 0 ^ { \circ }$ ，解得$\angle B O C = 6 0 ^ { \circ }$ ，： ……· (5分)因为 $O C \bot O A$ ，所以 $\angle A O C = 9 0 ^ { \circ }$ ，所以 $\angle A O B = \angle C O A - \angle B O C = 9 0 ^ { \circ } -$ $6 0 ^ { \circ } = 3 0 ^ { \circ }$ ： …(8分) + +16.解：（1)因为 $A D / / B C$ ，所以 $\angle D +$ $\angle C = 1 8 0 ^ { \circ }$ 因为 $\angle E A D = \angle C$ ，所以 $\angle E A D +$ $\angle D = 1 8 0 ^ { \circ }$ ，所以 $A E / / C D$ ；……· (4分) + +(2）因为 $A E / / C D$ ，所以 $\angle A E B = { \bf \Phi }$ $\angle C$ ，因为 $\angle F E C = \angle B A E$ ，所以 $\angle B$ $= \angle E F C = 5 0 ^ { \circ }$ ： ·（8分) + +17.解：（1)如图，由题意，得 $\angle F A B = \ ;$ $4 5 ^ { \circ }$ ${ } ^ { \circ } , \angle E B C = 8 0 ^ { \circ }$ 因为 $A F / / B E$ ，所以 $\angle F A B = { \bf \left[ \begin{array} { l } { { \begin{array} { r l r } \end{array} } } \end{array} \right] }$ $\angle A B E { = } 4 5 ^ { \circ }$ ，因为 $\angle E B C = 8 0 ^ { \circ }$ ，所以 $\angle A B C = { \mathrm { : } }$ （20 $3 5 ^ { \circ }$ ·· (5分)$( 2 ) D$ 处应在 $C$ 处的南偏西 $4 5 ^ { \circ }$ 方向. (6分)理由：要使 $C D \ / / \ A B$ ，即就要使$\angle A B C = \angle B C D = 3 5 ^ { \circ }$ 又因为 $C G / / B E$ ，所以 $\angle G C B = $ $\angle E B C = 8 0 ^ { \circ }$ ，所以 $\angle G C D = 4 5 ^ { \circ }$ 即 $D$ 处在 $C$ 处的南偏西 $4 5 ^ { \circ }$ 方向.·（10分） + +![](images/9e5ae854b5d526bd733ef31a45cc3376bb883a492c8dbd290df911910010891d.jpg) + +18.解： $( 1 ) \textcircled{ 1}$ 两直线平行，内错角相等60· （2分）$\textcircled{2} 3 0$ （4分）$\textcircled{3} 6 0$ （6分） + +（2）因为 $A B ~ / / ~ C D$ ，所以 $\angle B +$ +$\angle B C E = 1 8 0 ^ { \circ }$ +又因为 $\angle B = 4 0 ^ { \circ }$ ，所以 $\angle B C E = 1 8 0 ^ { \circ }$ +$- \angle B = 1 8 0 ^ { \circ } - 4 0 ^ { \circ } = 1 4 0 ^ { \circ } .$ （8分） +又因为 $C N$ 是 $\angle B C E$ 的平分线，所 +以 $. \angle B C N = 1 4 0 ^ { \circ } \div 2 = 7 0 ^ { \circ } ,$ +因为 $C N \bot C M$ ，所以 $\angle B C M = 9 0 ^ { \circ } -$ +$\angle B C N = 9 0 ^ { \circ } - 7 0 ^ { \circ } = 2 0 ^ { \circ }$ ：（12分） + +# 能力提优参考答案 + +# 一、选择题 + +
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+ +# 二、填空题 + +
9101112
15°答案不唯一，如∠A =∠EBC或∠D= ∠DCF或∠A+ ∠ABC=180°或∠D +∠BCD=180°等30°(180- 2a)°
+ +$9 . 1 5 ^ { \circ }$ 【解析】由尺规作图的痕迹，得$\angle B O C = \angle \alpha = 6 0 ^ { \circ }$ 则 $\angle A O C = 7 5 ^ { \circ } -$ $6 0 ^ { \circ } = 1 5 ^ { \circ }$ + +10.答案不唯一,如 $\angle A = \angle E B C$ 或 $\angle D$ $= \angle D C F$ 或 $\angle A + \angle A B C = 1 8 0 ^ { \circ }$ 或$\angle D + \angle B C D = 1 8 0 ^ { \circ }$ 等【解析】添加条件 $\angle A = \angle E B C$ 或 $\angle D =$ $\angle D C F$ 或 $\angle A + \angle A B C = 1 8 0 ^ { \circ }$ 或$\angle D + \angle B C D = 1 8 0 ^ { \circ }$ 均可以得出 $A D$ $/ / B F$ + +11. $3 0 ^ { \circ }$ 【解析】因为 $\angle B A C = 1 0 0 ^ { \circ }$ $\angle D A E = 5 0 ^ { \circ }$ ，所以 $\angle C A E = 1 8 0 ^ { \circ } -$ $1 0 0 ^ { \circ } - 5 0 ^ { \circ } = 3 0 ^ { \circ }$ .因为 $A E / / B C$ ，所以$\angle C = \angle C A E = 3 0 ^ { \circ } .$ + +12.( $\ : 1 8 0 - 2 \alpha \ : ) ^ { \circ } \ :$ 【解析】由 $C E$ 平分$\angle A C D$ ， $\angle E C A = \alpha ^ { \circ }$ ，得 $\angle A C D =$ $( 2 \alpha ) ^ { \circ } , \angle B C D = ( 1 8 0 - 2 \alpha ) ^ { \circ }$ .由 $F G$ $/ / C D$ ，可得 $\angle G F B = \angle B C D = ( 1 8 0 $ $- 2 \alpha ) ^ { \circ }$ + +# 三、解答题 + +13.解：设这个角为 $x ^ { \circ }$ ,则它的余角、补角分别为 $9 0 ^ { \circ } - x ^ { \circ }$ ${ 1 8 0 } ^ { \circ } - { x } ^ { \circ }$ ，（2分）根据题意列方程,得 $\frac { 1 } { 2 } \left( 1 8 0 - x \right) -$ $( 9 0 - x ) { = } 2 0$ ，解得 $x { = } 4 0$ 即这个角的度数为 $4 0 ^ { \circ }$ …(6分) + +14.解：因为 $\angle 1 = \angle 2$ ，所以 $A B / / C D$ …(4分)因为 $\angle A = 5 5 ^ { \circ } 1 6 ^ { \prime }$ ，所以 $\angle A D C =$ $1 8 0 ^ { \circ } - 5 5 ^ { \circ } 1 6 ^ { \prime } = 1 2 4 ^ { \circ } 4 4 ^ { \prime }$ … (8分) + +15.解：因为 $A B / / C D , \angle E D F = 7 0 ^ { \circ }$ ，所以 $\angle A B D = \angle E D F = 7 0 ^ { \circ }$ …(4分)因为 $B G$ 平分 $\angle A B D$ ，所以 $\angle D B G =$ ∠ABD=35°,所以∠FBG=180°$- \angle D B G = 1 4 5 ^ { \circ }$ · (8分) + +解： $C D$ 是 $\angle A C B$ 的平分线.（2分） +理由：因为 $C D \perp A B , E F \perp A B$ ，所以 +$E F / / C D$ ， +所以 $\angle E = \angle B C D$ ， $\angle A C D =$ +$\angle E M C$ … · (5分) +因为 $\angle E = \angle E M C$ ，所以 $\angle B C D =$ +$\angle A C D$ ，即 $C D$ 是 $\angle A C B$ 的平分线.…… (8分) + +17.解： $( 1 ) A F / / B E$ ·· (2分)理由：因为 $A D / / B C$ ，所以 $\angle B =$ $\angle D O E$ ··(3分)因为 $\angle A = \angle B$ ，所以 $\angle A = \angle D O E$ 所以 $A F / / B E$ ··· (5分)（2）因为 $A D / / B C$ ，所以 $\angle B +$ $\angle B O D { = } 1 8 0 ^ { \circ }$ …· (7分)因为 $\angle B O D = 3 \angle B$ ，所以 $\angle B +$ $3 \angle B = 1 8 0 ^ { \circ }$ ，可得 $\angle B = 4 5 ^ { \circ }$ ，故 $\angle A$ $= \angle B = 4 5 ^ { \circ }$ ·…· (10分) + +18.解：（1）因为 $D E / / O B$ ，所以 $\angle O =$ $\angle A C E$ 因为 $\angle A O B = 4 0 ^ { \circ }$ ，所以 $\angle A C E =$ （20 $4 0 ^ { \circ }$ 。因为 $\angle A C D + \angle A C E = 1 8 0 ^ { \circ }$ ，所以$\angle A C D { = } 1 4 0 ^ { \circ }$ 因为 $C F$ 平分 $\angle A C D$ ，所以 $\angle A C F { = }$ $7 0 ^ { \circ }$ ，所以 $\angle E C F = 7 0 ^ { \circ } + 4 0 ^ { \circ } = 1 1 0 ^ { \circ }$ ·（4分） + +(2）因为 $C G \bot C F$ ，所以 $\angle F C G =$ $9 0 ^ { \circ }$ ，所以 $\angle D C G + \angle D C F = 9 0 ^ { \circ }$ + +因为 $\angle A C O = 1 8 0 ^ { \circ }$ ，所以 $\angle G C O + $ $\angle F C A = 9 0 ^ { \circ }$ +因为 $\angle A C F = \angle D C F$ ，所以 $\angle G C O$ $= \angle G C D$ ，即 $C G$ 平分 $\angle O C D$ +…·（7分)(3)结论：当 $\angle A O B = 6 0 ^ { \circ }$ 时， $C D$ 平分 $\angle O C F$ （8分）理由：因为 $D E / / O B , \angle O = 6 0 ^ { \circ }$ ，所以$\angle D C O = \angle O = 6 0 ^ { \circ }$ ，所以 $\angle A C D = $ （20 ${ 1 2 0 } ^ { \circ }$ +因为 $C F$ 平分 $\angle A C D$ ，所以 $\angle D C F =$ $6 0 ^ { \circ }$ ，所以 $\angle D C O = \angle D C F$ ，即 $C D$ 平分 $\angle O C F$ （12分） + +·月考测试卷（一）参考答案 以 $( a ^ { m } ) ^ { 2 } \div ( a ^ { n } ) ^ { 3 } = a ^ { 2 m } \div a ^ { 3 n } = a ^ { 2 m - 3 n } .$ ·(4分) 因为 $a ^ { m } = 4 , a ^ { n } = \frac { 1 } { 2 }$ ，所以 $a ^ { 2 m - 3 n } = a ^ { 2 m }$ $\div a ^ { 3 n } = ( a ^ { m } ) ^ { 2 } \div ( a ^ { n } ) ^ { 3 } = 4 ^ { 2 } \div ( \frac { 1 } { 2 } ) ^ { 3 } =$ $1 6 \times 8 { = } 1 2 8$ + +# 一、选择题 + +18.解： $( 2 a + b ) ( 2 a - b ) - ( 4 a b ^ { 3 } - 8 a ^ { 3 } b )$ $\div 2 a b - 4 a ^ { 2 } - b ^ { 2 } - ( 2 b ^ { 2 } - 4 a ^ { 2 } ) = 4 a ^ { 2 }$ $- b ^ { 2 } - 2 b ^ { 2 } + 4 a ^ { 2 } = 8 a ^ { 2 } - 3 b ^ { 2 }$ ·… (5分) 当 $a = - 1$ $b = - 2$ 时，原式 $= 8 \times$ $( - 1 ) ^ { 2 } - 3 \times ( - 2 ) ^ { 2 } = - 4 .$ … (9分) + +
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+ +19.解： $\displaystyle ( x + 7 ) ( x - 3 ) = x ^ { 2 } + 4 x - 2 1 ,$ 与$x ^ { 2 } + m x - n$ 对比，可得 $m = 4 , n = 2 1$ ··(6分)则 $( 5 m - n ) ^ { 2 0 2 4 } = ( 5 \times 4 - 2 1 ) ^ { 2 0 2 4 } =$ $( - 1 ) ^ { 2 0 2 4 } = 1$ ： …· （9分) + +# 二、填空题 + +
1112131415
166.x²y²+3x²y²-3xy²y+1 2.2²2-3+i1-x2025
+ +# 三、解答题 + +20.解 $: x ( x - 2 ) ^ { 2 } - x ^ { 2 } ( x - 6 ) - 3 = x ( x ^ { 2 } -$ $4 x + 4 ) - x ^ { 3 } + 6 x ^ { 2 } - 3 = x ^ { 3 } - 4 x ^ { 2 } + 4 x$ $- x ^ { 3 } + 6 x ^ { 2 } - 3 = 2 x ^ { 2 } + 4 x - 3 .$ ·· (5分) 因为 $x ^ { 2 } + 2 x = 4$ ，所以 $2 x ^ { 2 } + 4 x =$ $2 ( x ^ { 2 } + 2 x ) = 2 \times 4 = 8$ ，所以原式 $= 8$ $- 3 { = } 5$ ： … (9分) + +16.(1)解：原式 $= a ^ { 4 } - a ^ { 6 } \div a ^ { 2 }$ （204号（4分）$= a ^ { 4 } - a ^ { 4 } = 0$ （5分）(2）解：原式 $= 1 - 4 + 4 \times 1$ ·(4分）$= 1 - 4 + 4 = 1$ （5分） + +17.解：因为 $a ^ { 2 m } = ( a ^ { m } ) ^ { 2 }$ ， $a ^ { 3 n } = ( a ^ { n } ) ^ { 3 }$ ，所 + +21.解： $( x ^ { 2 } - 4 x + 1 ) - ( - 3 x ^ { 2 } ) = x ^ { 2 } - 4 x$ $+ 1 + 3 x ^ { 2 } = 4 x ^ { 2 } - 4 x + 1$ …(4分)$( 4 x ^ { 2 } - 4 x + 1 ) \bullet ( - 3 x ^ { 2 } ) = - 1 2 x ^ { 4 } +$ $1 2 x ^ { 3 } - 3 x ^ { 2 }$ ，即正确的结果是一 $1 2 x ^ { 4 } + 1 2 x ^ { 3 } -$ （204号 $3 \mathcal { X } ^ { 2 }$ …· (9分) + +22.解：设长方形的长为 $\mathcal { X }$ ，宽为 $y .$ 根据题意,得 $2 x + 2 y = 1 6 , 2 x ^ { 2 } + 2 y ^ { 2 }$ $= 6 8$ ，即 $x + y = 8 , x ^ { 2 } + y ^ { 2 } = 3 4 ,$ … (2分）将 $x + y = 8$ 两边同时平方，得 $x ^ { 2 } +$ $2 x y + y ^ { 2 } = 6 4$ ···(4分)把 $x ^ { 2 } + y ^ { 2 } = 3 4$ 代人 $x ^ { 2 } + 2 x y + y ^ { 2 } =$ 64中，得 $3 4 + 2 x y = 6 4$ ，解得 $x y =$ 15,即长方形 $A B C D$ 的面积为15.（10分） + +23.解： $\mathrm { ( 1 ) } \mathrm { \textlangle } { { \mathbb { D } } } S _ { 1 } = a ( a + 4 ) = a ^ { 2 } + 4 a , S _ { 2 }$ $= ( a + 2 ) ^ { 2 } = a ^ { 2 } + 4 a + 4 .$ (2分)$\textcircled{2}$ 因为 $S _ { 1 } - S _ { 2 } = ( a ^ { 2 } + 4 a ) - ( a ^ { 2 } + 4 a$ $+ 4 ) = a ^ { 2 } + 4 a - a ^ { 2 } - 4 a - 4 = - 4 <$ 0,所以 $S _ { 1 } { < } S _ { 2 }$ ·…(6分)(2)由 $M { = } N$ ，可得 $M - N { = } 0$ ，则 $a ^ { 2 }$ $- 4 + ( a + 1 ) ^ { 2 } = 0$ ,整理得 $2 a ^ { 2 } + 2 a -$ $3 { = } 0$ ，即 $2 a ^ { 2 } + 2 a = 3$ 则 $a ( a + 1 ) = a ^ { 2 } + a = { \frac { 1 } { 2 } } ( 2 a ^ { 2 } + 2 a ) =$ ${ \frac { 1 } { 2 } } \times 3 = { \frac { 3 } { 2 } }$ · (10分) + +# 第三章变量之间的关系 + +# 卷中悟法 + +# 一、自变量与因变量的确定 + +1.自变量是先发生变化的量，因变量是后发生变化的量.2.自变量是主动发生变化的量，因 + +变量是随着自变量的变化而发生变化 +的量.3.常量(不发生变化的量).4.在一个变化的关系式中，只有一 +个自变量和一个因变量，且因变量需要 +写在等号左边. + +# 二、速度(或路程)图象 + +1.弄清哪一条轴(通常是纵轴)表示速度(或路程)，哪一条轴(通常是横轴）表示时间；2.准确读懂不同走向的线所表示的意义：(1)上升的线：从左向右呈上升状的线，其代表速度增加（或匀速远离起点）；(2)水平的线：与水平轴(横轴)平行的线，其代表匀速行驶(或静止)；(3)下降的线：从左向右呈下降状的线，其代表速度减小（或反向运动返回起点). + +# 三、三种变量之间关系的表达方法与特点： + +
表达方法特点
表格法多个变量可以同时出现在同一 张表格中
关系式法准确地反映了因变量与自变量 的数值关系
图象法直观、形象地给出了因变量随自 变量的变化趋势
+ +# ·基础过关参考答案· + +# 一、选择题 + +
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+ +# 二、填空题 + +
9101112
时间 温度y=10x+20y=8.5x①②④
+ +# 三、解答题 + +13.解：(1)当 $f { = } 6 8$ 时， $c = \frac { 5 } { 9 } ( f - 3 2 ) = 1$ 20； （3分）当 $f { = } { - } 4$ 时， $c { = } \frac { 5 } { 9 } ( f { - } 3 2 ) { = } { - } 2 0$ （6分）(2)当 $c = 1 0$ 时， $\frac { 5 } { 9 } ( f - 3 2 ) = 1 0$ ，解得 $f { = } 5 0$ …· (10分) + +14.解：(1)根据图象可以看出这一天的最高温度是 $3 7 \ ^ { \circ } C$ ,是在15 时到达的；最低温度是 $2 3 \ ^ { \circ } \mathrm { C }$ ,是在3时到达的. …· (5分)(2）这一天的温差为 $3 7 - 2 3 = { \mathrm { ? } }$ $1 4 ( ^ { \circ } \mathrm { C } )$ ，经过的时间为 $1 5 - 3 = { \mathrm { ? } }$ 12(时)．·… ·（10分) + +15.解：(1)由题意可知，小正方形的边长是自变量，图中阴影部分的面积为因变量. …（2分）(2)由题意可得 $y = 1 2 ^ { 2 } - 4 x ^ { 2 } = 1 4 4 -$ $4 \mathcal { x } ^ { 2 }$ … (6分)(3)由(2)可知 $y = 1 4 4 - 4 x ^ { 2 }$ ， + +当 $x = 1 ~ \mathrm { c m }$ 时， $y$ 有最大值， $y _ { \# } ^ { \scriptscriptstyle \perp } { + } =$ $1 4 4 - 4 \times 1 ^ { 2 } = 1 4 0 ( \mathrm { c m } ^ { 2 } ) ;$ +当 $x = 5 ~ \mathrm { c m }$ 时， $y$ 有最小值, $y _ { \mathrm { \scriptscriptstyle H Z } } , _ { \mathrm { \scriptscriptstyle I } } =$ $1 4 4 - 4 \times 5 ^ { 2 } = 4 4 ( { \mathrm { c m } } ^ { 2 } ) .$ +所以当小正方形的边长由 $1 ~ \mathrm { c m }$ 变化到 $5 ~ \mathrm { c m }$ 时，图中阴影部分的面积由$1 4 0 ~ \mathrm { c m } ^ { 2 }$ 变到 $4 4 ~ \mathrm { c m } ^ { 2 }$ ．…… (13分) + +16.解：(1)机动车行驶5个小时后加油；··· (3分)(2)因为 $3 6 - 1 2 = 2 4 ( \mathrm { L } )$ ，所以途中加油 $2 4 \mathrm { ~ L ~ }$ （7分）(3)够用. … (8分)理由：因为每小时耗油 $( 4 2 - 1 2 ) \div 5$ $= 6 ( \mathrm { L } )$ ，所以加油后可行驶 $3 6 \div 6 =$ $6 ( \mathrm { { h } ) }$ 又因为 $4 0 \times 6 = 2 4 0 ( \mathrm { k m } ) > 2 3 0 \mathrm { ~ k m }$ 所以油箱中的油够用．（15 分) + +# ·能力提优参考答案 + +# 一、选择题 + +
123456
DDCDAA
+ +# 二、填空题 + +
78910
y=1.2x+580y=2t+991350
+ +# 三、解答题 + +11.解： $x = 1 ~ { \mathrm { k m } }$ 时， $y = 3 5 \times 1 + 2 =$ $3 7 \ { ^ { \circ } } \mathrm { C } \ ; x = 5 \ \mathrm { k m }$ 时， $y = 3 5 \times 5 + 2 =$ $1 7 7 \ { ^ { \circ } C } : x = 1 0 \ { \mathrm { k m } }$ 时， $y = 3 5 \times 1 0 + 2$ $= 3 5 2 \mathrm { ~ } ^ { \circ } \mathrm { C }$ $x = 2 0 \ \mathrm { k m }$ 时， $y = 3 5 \times 2 0$ $+ 2 { = } 7 0 2 \ \mathrm { ^ { \circ } C } \ ; x { = } 3 0 \ \mathrm { k m }$ 时， $y = 3 5 \times$ $3 0 + 2 { = } 1 0 5 2 \mathrm { ~ \textdegree ~ }$ ： …(8分) + +12.解：(1)24; (3分)(2)4； （6分）(3)由图象得，每升高1千米，气温下降 $2 4 \div 4 = 6 \mathrm { ~ \textdegree C }$ ，则 $t { = } 2 4 { - } 6 h$ （10分） + +13.解：(1)若购进 $A$ 型电脑 $\mathcal { X }$ 台，则购进 $B$ 型电脑 $( 3 5 - x )$ 台.$y$ 与 $_ { \mathcal { X } }$ 之间的关系式为 $y = 0 . 6 x ^ { + }$ $0 . 4 ( 3 5 - x )$ ，即 $y = 0 . 2 x + 1 4$ … (5分)(2)由题意得， $3 5 - x = 2 . 5 x$ ，解得 $x$ $= 1 0$ ，则 $\scriptstyle y = 0 , 2 \times 1 0 + 1 4 = 1 6$ 万元.答：该公司需要投入资金16万元.··（10分） + +14.解：(1)由题意得， $y _ { \mathbb { H } } = 3 0 \times 4 + 5 ( x$ $- 4 ) = 5 x + 1 0 0 ,$ $y _ { \mathrm { Z } } = 3 0 \times 4 \times 0 . ~ 9 + 5 x \times 0 . ~ 9 = 4 . ~ 5 x$ $+ 1 0 8$ …（5分）(2)当 $y _ { \mathbb { H } } = y _ { Z }$ 时， $1 0 0 + 5 x { = } 4 . 5 x +$ 108,解得 $x { = } 1 6$ 故购买16 盒乒乓球时，在两家商店购买的付款金额一样多.（10分） + +15.解：(1）由图象可得，点 $A$ 表示充满电后行驶150千米时，剩余电量为35千瓦·时. …(2分）(2)在 $0 \sim 1 5 0$ 千米范围内，汽车行驶1千米的平均耗电量是 $( 6 0 - 3 5 ) \div$ + +$1 5 0 { = } \frac { 1 } { 6 }$ 千瓦·时.当汽车行驶了120千米时，蓄电池的剩余电量为 $6 0 - { \frac { 1 } { 6 } } \times 1 2 0 { = } 4 0$ 千瓦·时.…（7分） + +(3)在 $1 5 0 \sim 2 0 0$ 千米范围内，汽车行驶1千米的平均耗电量是(35－10)$\div ( 2 0 0 - 1 5 0 ) = \frac { 1 } { 2 }$ 千瓦·时.设当汽车行驶了 $_ { \mathcal { X } }$ 千米时,剩余电量降至20千瓦·时，根据题意得， $3 5 -$ $\frac { 1 } { 2 } ( x - 1 5 0 ) = 2 0$ ，解得 $\scriptstyle x = 1 8 0$ 答：当汽车行驶了180千米时，剩余电量降至20千瓦·时．（12分） + +# 期中测试卷 参考答案 + +# 一、选择题 + +
12345678910
ACDDCBADBD
+ +# 二、填空题 + +
11 6012131415
148°45°-192
+ +# 三、解答题 + +16.(1)解：原式 $= 4 x ^ { 2 } - x + 8 x - 2 - 4 x ^ { 2 } +$ $2 x$ … （3分）$= 9 x - 2$ ：（5分） + +(2）解： $\bullet \bullet A B / / C D , \angle A B F = 4 0 ^ { \circ } , \bullet \bullet$ $\angle C F B + \angle B = 1 8 0 ^ { \circ } , \therefore \angle C F B =$ $1 8 0 ^ { \circ } - \angle B = 1 4 0 ^ { \circ } .$ 又： $\angle C F E$ ： $\angle E F B = 3 ~ : ~ 4 , ~ \cdot ^ { . }$ $\begin{array} { r l } & { \angle C F E { = } \frac { 3 } { 7 } \angle C F B { = } 6 0 ^ { \circ } . } \\ & { \mathrel { \mathop { : } } A B \mathbin { / } C D , \mathrel { \mathop { : } } \angle B E F { = } \angle C F E { = } } \end{array}$ $6 0 ^ { \circ }$ ： …… (5分) + +$2 . 1 + 3 ( x - 6 ) = 3 x - 5 . 4$ ，即 $y = 3 x$ $- 5 . 4$ ；… …· (6分)(2)当 $x = 8$ 立方米时， $y = 3 x - 5 . 4$ $= 3 \times 8 - 5 . 4 { = } 1 8 . 6 $ 元.答：该用户5月份的水费是18.6元.·· (9分) + +17.解： $( 3 x + 2 ) ( 3 x - 2 ) + x ( x - 2 ) =$ $9 x ^ { 2 } - 4 + x ^ { 2 } - 2 x = 1 0 x ^ { 2 } - 2 x - 4 .$ （20·（4分)因为 $5 x ^ { 2 } - x - 2 = 0$ ，所以 $5 x ^ { 2 } - x = 2$ 当 $5 x ^ { 2 } - x = 2$ 时，原式 $= 2 \left( 5 x ^ { 2 } - x \right)$ $- 4 { = } 2 \times 2 - 4 { = } 0$ …（9分) + +18.解：(1)由表格可得， $h { = } 4 { + } 1 . 2 ( x { - }$ $1 ) = 1 . 2 x + 2 . 8 $ …· (4分)(2)当 $h { = } 1 1 . 2 ~ \mathrm { c m }$ 时，即 $1 . 2 x + 2 . 8 \ :$ $= 1 1 . 2$ ，解得 $x = 7 .$ ··（7分）答：当这擦碗的高度为 $1 1 . ~ 2 ~ \mathrm { c m }$ 时，碗的数量为7只. … (9分) + +21.解： $( 1 ) \angle 2 = \angle 3$ … (1分)理由： $\angle E N C + \angle C M G = 1 8 0 ^ { \circ }$ $\angle C M G = \angle F M N$ $\angle E N C +$ $\angle F M N { = } 1 8 0 ^ { \circ } , \therefore F G / / E D ,$ ·… (4分)$\cdot \angle 2 = \angle D , \because { A B } / / { C D } , \therefore \angle 3 =$ $\angle D , \therefore \angle 2 = \angle 3$ ；（6分)$( 2 ) \because A B / / C D , \therefore \angle A + \angle A C D =$ $1 8 0 ^ { \circ }$ …(7分)$\cdot \angle A = \angle 1 + 7 0 ^ { \circ } , \angle A C B = 4 2 ^ { \circ } , \therefore$ $( \angle 1 + 7 0 ^ { \circ } ) + ( \angle 1 + 4 2 ^ { \circ } ) = 1 8 0 ^ { \circ } , \ :$ $\angle 1 = 3 4 ^ { \circ }$ ，$\bullet A B / / C D , \bullet \bullet \angle B = \angle 1 = 3 4 ^ { \circ } .$ ·· (9分) + +19.解： $A B / / C D$ … (2分)理由：因为 $\angle D A E = \angle E$ ，所以 $A D / /$ $B C$ ，所以 $\angle B + \angle B A D = 1 8 0 ^ { \circ }$ ·· (5分)因为 $\angle B = \angle D$ ，所以 $\angle D + \angle B A D$ $= 1 8 0 ^ { \circ }$ ，所以 $A B / / C D$ …· (9分) + +22.解：(1)由图象可得,汽车的耗油量为$( 6 0 - 4 5 ) \div 1 5 0 = 0 . 1$ 升/千米,则 $y$ $= 6 0 - 0 . \ : 1 x$ ·…·(4分)(2)当 $y = 8$ 时， $6 0 - 0 . 1 x { = } 8$ ，解得 $\mathcal { X }$ $= 5 2 0$ ,即行驶520千米时，油箱中的剩余油量为8升.$5 0 0 + 3 0 - 5 2 0 = 1 0$ 千米，故在开往该加油站的途中，汽车开始提示加油，这时离加油站的路程是10千米.…·（10分） + +20.解：(1)由表格可得,3月份的用水量为5立方米，没有超过6立方米，则 $a$ $= 1 0 . 5 \div 5 = 2 . 1$ 元/立方米.4月份的用水量为9立方米，超过了6立方米，则 $6 \times 2 . 1 + ( 9 - 6 ) c =$ 21.6,解得 $c { = } 3$ 元/立方米.故每月用水量不超过6立方米时， $y$ $= 2 . 1 x$ 每月用水量超过6立方米时， $y = 6 \times \colon 2 3 .$ + +解：(1)因为 $a ^ { 2 } + b ^ { 2 } = 8$ ， $( a + b ) ^ { 2 } =$ + +48,所以ab=（a+6)²-(a²+6²） ${ \frac { 4 8 - 8 } { 2 } } = 2 0$ 故填20．（2分) + +别为 $a , b ,$ 设第三边为 $c$ ，那么有 $\mid a - b \mid$ $< c < a + b$ 3.可证明线段之间的不等关系. + +(2)设 $2 5 - x = a , x - 1 0 = b$ ，由 $( a +$ $b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }$ 进行变形得， $a ^ { 2 } + 1$ $b ^ { 2 } = ( a + b ) ^ { 2 } - 2 a b .$ +所以 $( 2 5 - x ) ^ { 2 } + ( x - 1 0 ) ^ { 2 } = [ ( 2 5 -$ $x ) + ( x - 1 0 ) ] ^ { 2 } - 2 ( 2 5 - x ) ( x - 1 0 )$ $= 1 5 ^ { 2 } - 2 \times ( - 1 5 ) = 2 2 5 + 3 0 = 2 5 5 .$ （20·…· (7分)(3)设 $A D { = } A C { = } a$ $B E { = } B C { = } b$ ，则图中阴影部分的面积为 ${ \frac { 1 } { 2 } } \left( a + b \right) \left( a \right.$ $+ b ) - { \frac { 1 } { 2 } } ( a ^ { 2 } + b ^ { 2 } ) = { \frac { 1 } { 2 } } [ ( a + b ) ^ { 2 } - ( a ^ { 2 } ) ^ { 2 }$ $+ b ^ { 2 } ) ] = \frac { 1 } { 2 } \times 2 a b = a b = 1 0 .$ 故填10.：(10分） + +# 二、利用全等三角形的性质证明边、角相等 + +利用全等三角形的性质证明边、角相等的关键在于找出边、角所在的三角形，然后证明其全等，从而得出结论。一般步骤是： + +(1)找出所要证明的边或角所在的三角形；(2)根据已知条件或结论证明三角形全等；(3)利用全等三角形的性质得出所证结论. + +# 第四章 三角形 + +# 卷中悟法 + +# 一、三角形三边关系的应用 + +1.判断已知的三条线段 $a , b , c$ 能否构成一个三角形，判断的方法有三种：(1)当 $a + b > c , b + c > a , a + c > b$ 都成立时， $a , b , c$ 可构成三角形；(2)当 $| a - b | <$ $c { < a + b }$ 时， $a , b , c$ 可构成三角形；(3)当$a$ 最长，且 $b + c > a$ 时， $a , b , c$ 可构成三角形. + +若不能通过全等三角形直接证明，可通过线段(角)的和或差将所证结论转化为全等三角形中的线段(角).此外证明边相等也可通过线段的中点，角平分线的性质、定理等证明；角相等可通过角平分线的定义、对顶角相等等证明. + +2.已知三角形的两边，确定第三边的取值范围.如果已知三角形的两边分 + +# 三、合理选择全等三角形的判定方法 + +从判定两个三角形全等的方法可知，要判定两个三角形全等，需要知道这两个三角形分别有三个元素（其中至少一个元素是边)对应相等，这样就可以利用题目中的已知边(角)迅速、准确地确定要补充的边（角)，有目的地完善三角形全等的条件，从而得到判定两个三角形全等的思路. + +找夹角→SAS +已知两边 找第三边→SSS找直角→H边为角的对边-找任一角→AAS +已知一边一角 找夹角的另一边→SAS找夹边的另一角→ASA找边的对角→AAS找夹边→ASA +已知两角找其中一个已知角的对边→AAS基础过关参考答案 + +# 一、选择题 + +
12345678
CADDDCBD
+ +# 二、填空题 + +
9101112
sss66060°
+ +# 三、解答题 + +13.解：锐角三角形有2个，分别是△ABE,△BAC;,. ·(3分)直角三角形有3个，分别是Rt△ABD,Rt△AED,Rt△ADC.·· (6分) + +14.解：如图所示. …… (8分) + +![](images/4a743fd1ae54400b3178886b8be5eb39c333648347ad287bca3828468767942d.jpg) + +15.解：(1)设 $A E$ 的长为 $\mathcal { X }$ ，则 $A E$ 的取值范围是 $1 0 - 8 < x < 1 0 + 8$ ，即 $2 { < } x$ ${ < } 1 8$ …·(3分) + +（2）因为 $A E / / B D$ ，所以 $\angle A =$ $\angle C B D = 5 5 ^ { \circ }$ 所以 $\angle A E F = 1 8 0 ^ { \circ } - \angle A E C = \angle A$ $+ \angle C = 5 5 ^ { \circ } + 6 0 ^ { \circ } = 1 1 5 ^ { \circ }$ … (8分) + +16.解： $( 1 ) \triangle A E D { \cong } \triangle A C D .$ … (1分)理由：因为 $A D$ 是 $B C$ 边上的高，所以 $\angle A D E { = } \angle A D C { = } 9 0 ^ { \circ }$ 因为 $\angle E A D = \angle C A D , A D = A D$ ，所以△AED≌△ACD(ASA).…(3分) + +(2)因为 $\angle B = 4 2 ^ { \circ }$ ，所以 $\angle B A D = 9 0 ^ { \circ }$ $- 4 2 ^ { \circ } = 4 8 ^ { \circ }$ ，因为 $\angle E A D = \angle C A D$ ， $\angle B A E =$ $\angle C A E$ ，所以 $\angle B A E = \angle C A E =$ $2 \angle E A D$ ，所以 $\angle C A D = \frac { 1 } { 3 } \angle B A D = \frac { 1 } { 3 } \times 4 8 ^ { \circ } =$ $1 6 ^ { \circ }$ ，所以 $\angle C = 9 0 ^ { \circ } - 1 6 ^ { \circ } = 7 4 ^ { \circ }$ （8分） + +17.解： $\mathrm { ( 1 ) } \angle B O C { = } 9 0 ^ { \circ } { + } \frac { 1 } { 2 } \angle B A C .$ （2分）理由：因为 BE平分 $\angle A B C$ ，所以$\angle O B C { = } { \frac { 1 } { 2 } } \angle A B C .$ （204号同理可得 $\angle O C D = \frac { 1 } { 2 } \angle A C B$ 所以 $\angle O B C + \angle O C D = \frac { 1 } { 2 } ( \angle A B C +$ $\angle A C B )$ ， ： (4分)又因为 $\angle B A C + \angle A B C + \angle A C B =$ $1 8 0 ^ { \circ }$ ，所以 $\angle A B C + \angle A C B = 1 8 0 ^ { \circ } -$ + +$\angle B A C$ ， +所以 $\angle O B C + \angle O C D = { \frac { 1 } { 2 } } \left( 1 8 0 ^ { \circ } - \right.$ $\angle B A C )$ ，… …· (6分)又因为 $\angle B O C = 1 8 0 ^ { \circ } - ( \angle O B C +$ $\angle O C D )$ ， +所以 $\angle B O C = 1 8 0 ^ { \circ } - { \frac { 1 } { 2 } }$ (180°$\angle B A C = 9 0 ^ { \circ } + { \frac { 1 } { 2 } } \angle B A C .$ +： (8分) + +(2)同理可得 $\angle C O A = 9 0 ^ { \circ } + { \frac { 1 } { 2 } } \angle A B C$ $\angle B O A = 9 0 ^ { \circ } + \frac { 1 } { 2 } \angle A C B$ … (10分） + +18.解： $\triangle C D E { \cong } \triangle A B F$ …(1分)理由：因为 $D E \bot A C , B F \bot A C .$ 所以$\angle D E C = \angle B F A = 9 0 ^ { \circ }$ ，因为 $D C / /$ $A B$ ，所以 $\angle A C D = \angle B A C$ ，因为 $A E$ ${ = } C F$ ，所以 $A F { = } C E$ ，所以 $\triangle C D E { \cong }$ $\triangle A B F$ · (4分)$\mathrm { ( 2 ) } A D { = } B C , A D / / B C .$ …(5分)理由：因为 $\triangle C D E \cong \triangle A B F$ ，所以$D E { = } B F$ ，因为 $\angle A E D = \angle C F B = 9 0 ^ { \circ }$ $A E =$ $C F$ ，所以 $\triangle A D E { \cong } \triangle C B F$ 所以 $A D { = } B C$ $\angle D A E = \angle B C F$ ，所以 $A D / / B C .$ · (8分)$( 3 ) D F { = } B E$ … (9分)理由：如图，在 $\triangle A D F$ 和 $\triangle C B E$ 中，$A D = B C$ ， $\angle D A F = \angle B C E$ $A F =$ + +$C E$ ，所以 $\triangle A D F { \cong } \triangle C B E$ ，所以DF$= B E .$ … ：(12分） + +D CB·能力提优参考答案· + +# 一、选择题 + +
12345678
CBADCABD
+ +# 二、填空题 + +
9 3AD⊥BC101112
90△EHD和△EGF
+ +# 三、解答题 + +13.解：因为 $C F / / A B$ ，所以 $\angle B =$ $\angle F C D , \angle B E D = \angle F .$ 因为点 $D$ 是 $B C$ 的中点，所以 $B D =$ CD.·在 $\triangle B D E$ 与 $\triangle C D F$ 中，$\angle B E D = \angle F , \angle B = \angle F C D , B D =$ $C D$ ，所以△BDE≌△CDF(AAS).· + +14.解：具体作图如下： + +![](images/593019b817191593a62636f39b6c0a802245709246b48a62ba28b1afe627538d.jpg) + +则△ABC是所求作的三角形．…· (8分) + +15.解：只要测量出 $E M$ 的长度即为凉亭 + +$M$ 与 $F$ 之间的距离. · (3分)理由：因为 $A B / / C D$ ，所以 $\angle B = \mathrm { : }$ $\angle C .$ +因为点 $M$ 是 $B C$ 的中点，所以 $B M =$ CM. ·· · (6分)在△BEM和 $\triangle C F M$ 中， +$\angle B = \angle C$ ， $B M = C M$ ， $\angle B M E = { \mathrm { : } }$ $\angle C M F$ ， +所以 $\triangle B E M { \cong } \triangle C F M ( \mathrm { A S A } )$ ，所以$M F { = } M E$ … (10分) + +16.解： $( 1 ) F C { = } A D .$ · (1分)理由：因为 $A D / / B C$ ，所以 $\angle A D C = { \mathrm { ? } }$ $\angle E C F .$ 因为点 $E$ 是 $C D$ 的中点，所以 $D E = \ \mathrm { i }$ EC.在 $\triangle A D E$ 和 $\triangle F C E$ 中，$\angle A D C = \angle E C F , D E = E C , \angle A E D$ $= \angle C E F$ 所以 $\triangle A D E { \cong } \triangle F C E$ （ASA)，所以$F C { = } A D$ ;· …· (4分) + +(2)由(1)可知 $\triangle A D E { \cong } \triangle F C E$ ，所以 $A E { = } E F , A D { = } C F .$ +因为 $A B { = } B C { + } A D { = } B C { + } C F$ ，所以$A B { = } B F$ (7分)在△ABE和△FBE中， +$\scriptstyle A B = B F , A E = E F , B E = B E ,$ +所以 $\triangle A B E { \cong } \triangle F B E ( \mathrm { S S S } )$ +所以 $\angle A E B = \angle F E B = 9 0 ^ { \circ }$ ，所以 $B E$ ⊥AF. （10分） + +17.解：(1)因为 $\angle B E C = \angle C F A = \alpha =$ $9 0 ^ { \circ }$ ，所以 $\angle B C E + \angle C B E = 1 8 0 ^ { \circ } -$ （20 $9 0 ^ { \circ } = 9 0 ^ { \circ } .$ 因为 $\angle B C A = \angle B C E + \angle A C F =$ $9 0 ^ { \circ }$ ，所以 $\angle C B E { = } \angle A C F .$ 在 $\triangle B C E$ 和 $\triangle C A F$ 中，$\angle B E C = \angle C F A , \angle C B E = \angle A C F ,$ $B C { = } A C$ 所以 $\triangle B C E { \cong } \triangle C A F$ （AAS)，所以$B E { = } C F$ …·（4分)$( 2 ) \alpha + \angle B C A = 1 8 0 ^ { \circ } ($ 或 $\alpha$ 与 $\angle B C A$ 互补)．…… …· (5分)理由：因为 $\angle B E C = \angle C F A = \alpha$ ，所以$\angle B E F = 1 8 0 ^ { \circ } - \alpha .$ （20因为 $\alpha + \angle B C A = 1 8 0 ^ { \circ }$ ，所以 $\angle B C A$ $= 1 8 0 ^ { \circ } - \alpha .$ 因为 $\angle E B C + \angle B C E = 1 8 0 ^ { \circ } - \alpha$ ，$\angle B C A = \angle B C E + \angle A C F = 1 8 0 ^ { \circ } -$ （20 $\alpha$ ，所以 $\angle E B C = \angle A C F$ … (6分)在△BCE和△ $C A F$ 中，$\angle C B E = \angle A C F , \angle B E C = \angle C F A ,$ $B C { = } C A$ ，所以 $\triangle B C E { \cong } \triangle C A F$ (AAS)，所以$B E { = } C F .$ ： … (（7分)$( 3 ) E F { = } B E { + } A F .$ …（8分)理由：因为 $\angle B C A = \alpha$ ，所以 $\angle B C E +$ $\angle A C F = 1 8 0 ^ { \circ } - \alpha .$ 因为 $\angle B E C = \alpha$ ，所以 $\angle E B C +$ （20$\angle B C E = 1 8 0 ^ { \circ } - \alpha$ ，所以 $\angle E B C =$ + +$\angle A C F$ …· (10分)在 $\triangle B E C$ 和△CFA中， +$\angle E B C = \angle F C A , \angle B E C = \angle C F A ,$ $B C { = } C A$ ， +所以 $\triangle B E C { \cong } \triangle C F A ( \mathrm { A A S } )$ ，所以$B E { = } C F , E C { = } F A ,$ +所以 $E F { = } E C { + } C F { = } B E { + } A F .$ （2 +·（12分） + +# 第五章生活中的轴对称 + +# 卷中悟法 + +# 一、轴对称图形及其对称轴的识别方法 + +1.判断一个图形是不是轴对称图形，可以用折纸的方法按照轴对称图形的概念，看是否能找到一条直线，将图形沿其折叠，使直线两旁的部分能够完全重合（即用轴对称图形的定义进行识别). + +2.识别轴对称图形的关键是要找到作为对称轴的直线，沿直线折叠后两边的部分能够完全重合，有时这样的直线能找到多条，说明这个轴对称图形有多条对称轴。 + +# 二、线段垂直平分线与角平分线性质的综合应用 + +1.线段垂直平分线的性质一般与全等三角形、角平分线的知识综合命题，线段垂直平分线上的点到线段两个端点的距离相等，而角平分线上的点到角两边的距离相等，两者是不同的. + +2.解题时，主要用线段垂直平分线上的点到线段两个端点的距离相等及角平分线上的点到角两边的距离相等来得出线段相等或三角形全等的结论，再利用全等三角形的性质解决问题. + +# 基础过关参考答案 + +# 一、选择题 + +
12345678
BCCABAAD
+ +# 二、填空题 + +
9101112
无数60°34
+ +# 三、解答题 + +13.解：如图所示. （6分） + +![](images/d3f2ed627f97afc5a1d991a5e3cf1235d5897c85f9c4ae59ed94fd12c2654a52.jpg) + +14.解：(1)全等. ：(2分)(2)因为 $\triangle A B C$ 和 $\triangle A ^ { \prime } B ^ { \prime } C ^ { \prime }$ 关于直线 $l$ 对称，所以 $A ^ { \prime } B ^ { \prime } { = } A B { = } 4 ~ ($ m，$B ^ { \prime } C ^ { \prime } { = } B C { = } 3 \ \mathrm { c m } , A ^ { \prime } C$ 边上的高 $=$ $2 \mathrm { \ c m }$ ，所以 $\triangle A ^ { \prime } B ^ { \prime } C ^ { \prime }$ 的周长 $= 4 + 6 +$ $3 { = } 1 3 ( \mathrm { c m } )$ ： (6分)面积= $[ = \frac { 1 } { 2 } \times 6 \times 2 = 6 ( c m ^ { 2 } )$ …· (8分) + +15.解：如图所示，(1)作出线段 $A B$ 的垂直平分线； （3分）(2)作出角的平分线. （6分）它们的交点即为所求作的点 $C$ （有2个). …· (10分) + +![](images/400dda2c4d0c7a5a728045e745ef50ed76b56ae545825aa14ef21ccb163fe340.jpg) + +16.解： $\textcircled{1}$ 当顶角 $\angle A { < } 9 0 ^ { \circ }$ 时，如图 $\textcircled{1}$ 所示，因为 $B D \perp A C$ ，所以 $\angle A +$ $\angle A B D { = } 9 0 ^ { \circ }$ ，因为 $\angle A B D = 5 6 ^ { \circ }$ ，所以 $\angle A = 9 0 ^ { \circ } -$ $5 6 ^ { \circ } = 3 4 ^ { \circ }$ ，因为 $A B { = } A C$ ，所以 $\angle A B C = \angle C =$ ${ \frac { 1 } { 2 } } \times ( 1 8 0 ^ { \circ } - 3 4 ^ { \circ } ) = 7 3 ^ { \circ } .$ $\textcircled{2}$ 当顶角 $\angle A { > } 9 0 ^ { \circ }$ 时，如图 $\textcircled{2}$ 所示，因为 $B D \perp A C$ ，所以 $\angle D A B +$ $\angle A B D { = } 9 0 ^ { \circ }$ 因为 $\angle A B D = 5 6 ^ { \circ }$ ，所以 $\angle D A B = 9 0 ^ { \circ }$ $- 5 6 ^ { \circ } = 3 4 ^ { \circ }$ ，所以 $\angle B A C = 1 4 6 ^ { \circ }$ 因为 $A B { = } A C$ ，所以 $\angle A B C = \angle C =$ $\frac { 1 } { 2 } \times ( 1 8 0 ^ { \circ } - 1 4 6 ^ { \circ } ) = 1 7 ^ { \circ } .$ 综上所述，这个等腰三角形的底角的度数为 $7 3 ^ { \circ }$ 或 $1 7 ^ { \circ }$ ·……（10分) + +![](images/4ad7e7c76a6f71cef5ef9269dbea16437bcc96c8dc599a116705c12e22269a34.jpg) +图 $\textcircled{1}$ + +![](images/6f1edcb1a8d5fbcc6f195c377cfa598efd9db4ae0da1742d5030821c0c546d64.jpg) +图 $\textcircled{2}$ + +，解：问题解决：如图，作 $C F \bot A D$ 的延长线于点 $F$ ，所以 $\angle F { = } 9 0 ^ { \circ }$ +因为 $C E \bot A B$ ，所以 $\angle F { = } \angle C E A { = }$ $\angle C E B = 9 0 ^ { \circ }$ +因为 $\angle A D C + \angle C D F = 1 8 0 ^ { \circ }$ ，且$\angle A B C + \angle A D C = 1 8 0 ^ { \circ }$ ，所以 $\angle C D F$ $= \angle B$ +在 $\triangle C D F$ 和 $\triangle C B E$ 中， $\angle F =$ $\angle C E B , \angle C D F = \angle B , C D = C B$ ，所以 $\triangle C D F { \cong } \triangle C B E ( \mathrm { A A S } )$ ，所以 $C F$ ${ } = { C E } .$ +因为 $C F \bot A D , C E \bot A B$ ，所以 $A C$ 平分 $\angle B A D$ …（4分）合作探究：（1）在 $\mathrm { R t } \bigtriangleup C A F$ 和$\mathrm { R t } \triangle C A E$ 中， $C F = C E , A C = A C$ ，所以 $\triangle C A F { \cong } \triangle C A E .$ +所以 $A F { = } A E$ +因为 $\triangle C D F { \cong } \triangle C B E$ ，所以 $D F =$ BE. +因为 $3 B E { = } 9$ ，所以 $B E { = } 3$ ，所以 $D F$ $= 3$ +因为 $A D { = } A F { - } D F$ ，所以 $A D { = } A E$ $- D F { = } 9 - 3 { = } 6 .$ …（9分) + +(2)因为 $\triangle C A F { \cong } \triangle C A E , \triangle C D F { \cong }$ $\triangle C B E$ ，所以 $S _ { \triangle C A F } { = } S _ { \triangle C A E }$ ， $S _ { \triangle C D F } { = } S _ { \triangle C B E }$ ， + +设△BCE的面积为 $\mathcal { X }$ ，则 $\triangle C D F$ 的 +面积为 $\mathcal { X }$ ， +由题意，得 $2 8 + x { = } 4 0 - x$ ，解得 $x { = } 6 .$ +所以 $\triangle B C E$ 的面积为6.（14分） + +![](images/60df6d2ca892777369bdf09433249fb81e47d10bf83ee1d8f5d7abf4c77d6685.jpg) + +# 一、选择题 + +
12345678
DABDABDC
+ +# 二、填空题 + +
9101112
4或66425°
+ +# 三、解答题 + +13.解： $\angle B D P = \angle E P C .$ …… (1分) 理由： $\triangle A B C$ 为等边三角形， $\angle D P E = 6 0 ^ { \circ } , \therefore \angle D P E = \angle B =$ $6 0 ^ { \circ }$ · (3分) $\cdot \angle D P E + \angle E P C = 1 8 0 ^ { \circ } - \angle B P D ,$ （20 $\angle B + \angle B D P = 1 8 0 ^ { \circ } - \angle B P D ,$ $\therefore \angle D P E + \angle E P C = \angle B + \angle B D P ,$ $\cdot \angle E P C = \angle B D P$ ·…· (8分) + +14.解：(1)如图， $\triangle A _ { 1 } B _ { 1 } C _ { 1 }$ 是 $\triangle A B C$ 关于直线的对称图形;（4分) + +![](images/3507aab1d126833d0b9e8b1c4fa6cffe6b72ff565e84cc57b7a0d6d04d0c505c.jpg) + +(2)由图形可得四边形 $B B _ { 1 } C _ { 1 } C$ 是等 腰梯形， $B B _ { 1 } { = } 4 , C C _ { 1 } { = } 2$ ，高是4, $S _ { \perp \perp \perp \perp \perp \perp } \ L _ { \mathcal { \perp } B B _ { 1 } C _ { 1 } C } = \frac { 1 } { 2 } \times ( 4 + 2 ) \times 4 =$ 12. ·…· (8分) + +15.解： $( 1 ) \because A D$ 垂直平分 $B E , E F$ 垂直平分 $A C , \therefore A B = A E = E C , \therefore \angle C =$ $\angle C A E .$ ： $\cdot \angle B A E = 4 0 ^ { \circ } , \therefore \angle A E D = 7 0 ^ { \circ } , \therefore$ $\angle C = \frac { 1 } { 2 } \angle A E D = 3 5 ^ { \circ }$ （20 $( 2 ) { \because } \triangle A B C$ 的周长为 $1 4 \ { \mathrm { c m } } , A C =$ $6 ~ \mathrm { c m }$ ，$\scriptstyle \cdot . A B + B C = A B + B E + E C = 1 4 - 6$ （20$= 8 \mathrm { c m }$ ，即 $2 D E + 2 E C = 2 \left( D E + \right.$ $E C ) = 8 \mathrm { ~ \ c m } , \therefore D E + E C = D C =$ $4 \ \mathrm { c m } .$ …（10分） + +16.解：(1)由轴对称的性质可得 $P M =$ $C M , N D { = } N P .$ $\cdot C D { = } 1 8 \ \mathrm { c m } , \therefore \triangle P M N$ 的周长 $=$ $P N + P M + M N = D N + M N + C M$ $= C D { = } 1 8 ~ \mathrm { c m }$ … (3分)(2)∵点 $P$ 关于 $O A , O B$ 对称的对称点分别为 $C , D$ ， + +∴OA垂直平分 $P C , O B$ 垂直平分$P D$ ， +∴ $C M = P M$ ， $P N = D N$ $\angle C =$ $\angle M P C = 2 1 ^ { \circ }$ $\angle D = \angle N P D = 2 8 ^ { \circ } .$ ·… (6分)$\cdot \angle C M P = 1 8 0 ^ { \circ } - \angle C - \angle M P C =$ $1 3 8 ^ { \circ } , \angle P N D = 1 8 0 ^ { \circ } - \angle D - \angle N P D$ $= 1 2 4 ^ { \circ }$ 。 +： $\angle P M N = 4 2 ^ { \circ }$ $\angle P N M = 5 6 ^ { \circ }$ +$\therefore \angle M P N { = } 1 8 0 ^ { \circ } { - } \angle P M N { - } \angle P N M$ $= 8 2 ^ { \circ }$ …· (10分) + +17.解：（1）点 $O$ 到 $\triangle A B C$ 的三个顶点$A , B , C$ 的距离的大小关系是 $O A =$ $O B { = } O C$ ；： …(2分）（ $2 ) \triangle { O M N }$ 是等腰直角三角形.(3分)理由： $\cdot A B { = } A C$ $\angle B A C = 9 0 ^ { \circ }$ ，点 $O$ 为 $B C$ 的中点，： $A O$ 平分 $\angle B A C , A O \bot B C ,$ ∴ $\angle A O B = 9 0 ^ { \circ }$ ， $\angle B = \angle C = 4 5 ^ { \circ }$ $\angle B A O = \angle C A O = 4 5 ^ { \circ } , \therefore \angle C A O =$ $\angle B , O A { = } O B$ …· (5分)在 $\triangle B O M$ 和 $\triangle A O N$ 中，$A N { = } B M$ ,∠CAO=∠B,OA=OB,:△BOM≌△AON(SAS),:OM=ON,∠AON=∠BOM.： $\cdot \angle A O B = \angle B O M + \angle A O M = 9 0 ^ { \circ }$ 。 $\angle A O N + \angle A O M = 9 0 ^ { \circ }$ ，即 $\angle M O N$ $= 9 0 ^ { \circ }$ ， + +$\triangle { O M N }$ 是等腰直角三角形.… (8分）(3)四边形AMON的面积不发生变化 …· (9分)理由：由(2)可得 $\triangle A O N { \underline { { \underline { { \circ } } } } } \triangle B O M .$ 故 $S _ { \textsc { p q j j k A M O N } } = S _ { \triangle A M O } + S _ { \triangle M B O } =$ $S _ { \triangle A B O } = \frac { 1 } { 2 } S _ { \triangle A B C } .$ （12分） + +·月考测试卷（二）参考答案· + +# 一、选择题 + +
12345678910
CBBCBCDACD
+ +# 二、填空题 + +
1112131415
2y=40x+20062°43420
+ +# 三、解答题 + +16.(1)解：原式 $= ( \frac { 2 } { 3 } ) ^ { 2 0 2 3 } \times 1 . 5 ^ { 2 0 2 3 } \times 1 . 5$ $\times 1$ （204号 （2分）$= ( { \frac { 2 } { 3 } } \times { \frac { 3 } { 2 } } ) ^ { 2 0 2 3 } \times 1 . 5 { = } 1 . 5 .$ （5分） + +(2)解：由图形可得，阴影部分的周长$= 2 ( y + x + x + 0 . 5 x + 3 y ) = 2 ( 2 . 5 x$ $+ 4 y ) = 5 x + 8 y$ ； ···(2分)阴影部分的面积 $= y ( 2 x + 0 . ~ 5 x ) +$ $0 . 5 x \bullet 3 y = 2 x y + 0 . 5 x y + 1 . 5 x y =$ $4 \mathit { x y } .$ ·(5分) + +17.解：如图，延长 $C E$ 交 $A B$ 于点 $F$ ，则 CF⊥AB. ··· …(1分) + +![](images/519a91d0f6ae04b62a520082e2a64d59bed69b4508a3306d82b37763c1a620b9.jpg) + +$1 2 ) = 4 5 0$ 米/分.因为 $4 5 0 > 3 3 0$ ，所以小明买到书后继续骑车到学校的这段时间内的骑车速度不在安全限度内. ：（9分） + +因为 $\angle A + \angle 1 = 9 0 ^ { \circ }$ $\angle C + \angle 2 = { \ : }$ +$9 0 ^ { \circ }$ $\angle 1 = \angle 2$ ，所以 $\angle A = \angle C$ …·（4分) +在△ABD和 $\triangle C D E$ 中， +因为 $\angle A = \angle C , \angle A B D = \angle C D E =$ +$9 0 ^ { \circ }$ $\scriptstyle { \mathrm { ) } } ^ { \circ } , A D = C E$ +所以 $\triangle A B D { \cong } \triangle C D E$ （AAS)，所以 +$B D { = } D E .$ +因为 $D E { = } 2$ 米，所以 $D B { = } 2$ 米.… (9分) + +19.解：由已知可得 $A B { = } C D { = } 4$ ,分两种情况讨论：$\textcircled{1}$ 若点 $P$ 在 $B C$ 边上时，即 $\angle A B P { = }$ $\angle D C E = 9 0 ^ { \circ }$ 时，若满足 $B P { = } C E { = } 2$ ,根据"SAS”可得出 $\triangle A B P { \overset { \circ } { = } } \triangle D C E .$ 由题意得 $2 t = 2$ ，解得 $t { = } 1$ （3分） + +$\textcircled{2}$ 若点 $P$ 在 $A D$ 边上时，即 $\angle B A P$ $= \angle D C E = 9 0 ^ { \circ }$ 时， +若满足 $A P = C E = 2$ ,根据“SAS”可得出 $\triangle B A P { \overset { \subset } { = } } \triangle D C E .$ （20 +由题意得 $1 6 - 2 t = 2$ ，解得 $t { = } 7$ +…· （7分)综上所述，当 $t$ 的值为1或7时，$\triangle A B P$ 和 $\triangle D C E$ 全等·（9分) + +18.解：(1)由图象可得，小明家到学校的距离是1500米.故填1500．·…（1分）（2)小明在书店停留的时间为从8分钟到12分钟,共停留了 $1 2 - 8 = 4$ 分钟.故填4. …· (3分)(3)骑行总路程 $= 1 2 0 0 + ( 1 2 0 0 -$ $6 0 0 ) + ( 1 5 0 0 - 6 0 0 ) = 1 2 0 0 + 6 0 0 +$ $9 0 0 { = } 2 7 0 0$ 米.故填 2700．…… (6分) + +(4)买到书后继续骑车到学校的这段时间是 $1 2 { \sim } 1 4$ 分钟，在 $1 2 \sim 1 4$ 分钟时平均速度 $= ( 1 5 0 0 - 6 0 0 ) \div ( 1 4 -$ + +20.解：代数式 $\left( 3 a + 2 \right) \left( 2 a + 3 \right) - 3 a \left( 2 a \right.$ $+ 1 )$ 的值是偶数. …·(1分)理由： $( 3 a + 2 ) ( 2 a + 3 ) - 3 a ( 2 a + 1 )$ $= 6 a ^ { 2 } + 9 a + 4 a + 6 - 6 a ^ { 2 } - 3 a = 1 0 a$ $+ 6 = 2 ( 5 a + 3 )$ …（6分)因为 $a$ 为自然数，所以 $5 a + 3$ 是整数，所以代数式 $\left( 3 a + 2 \right) \left( 2 a + 3 \right) -$ $3 a ( 2 a + 1 )$ 的值是偶数．（9分) + +21.解：（1）由折叠的性质可得， $D A =$ $D B _ { \ast }$ 因为 $A C { = } 6 \ \mathrm { c m } , B C { = } 8 \ \mathrm { c m }$ ， + +所以 $\triangle A C D$ 的周长 $= D A + D C +$ $A C = D B + D C + A C = B C + A C =$ $1 4 ~ \mathrm { c m }$ ；： …（4分）(2）设 $\angle C A D { = } { _ { x } }$ ，则 $\angle B A D { = } 2 x$ 由图形的折叠可得， $\scriptstyle \angle B = \angle B A D =$ $2 x$ 在 $\mathrm { R t } \triangle A B C$ 中， $\angle B + \angle B A C = { \bf \Phi } _ { \bf \Phi }$ $9 0 ^ { \circ }$ ，可得 $2 x + 2 x + x = 9 0 ^ { \circ }$ ，解得 $x =$ $1 8 ^ { \circ }$ ，则 $\angle B { = } 2 x { = } 3 6 ^ { \circ }$ ． (9分) + +22.解：(1）因为 $\angle A = 4 0 ^ { \circ }$ $A B \bot B C$ ，所以 $\angle A D B = 9 0 ^ { \circ } - 4 0 ^ { \circ } = 5 0 ^ { \circ }$ 因为 $A M / / C N$ ，所以 $\angle C = \angle A D B$ $= 5 0 ^ { \circ }$ · (3分)(2)如图，延长 $D B$ 交 $C N$ 于点 $H$ 因为 $M D / / C N , B D \perp A M ,$ 所以 $B D$ $\bot C N$ ，所以 $\angle C H B = 9 0 ^ { \circ }$ …· (5分)因为 $B D \perp A M$ ，所以 $\angle D A B +$ $\angle A B D { = } 9 0 ^ { \circ }$ 因为 $A B \perp A M$ ，所以 $\angle A B D +$ $\angle C B H = 1 8 0 ^ { \circ } - 9 0 ^ { \circ } = 9 0 ^ { \circ }$ ，所以$\angle D A B { = } \angle C B H .$ · (8分)因为 $\angle A B D = 9 0 ^ { \circ } - \angle B A D , \angle C =$ $9 0 ^ { \circ } - \angle C B H$ ，所以 $\angle A B D = \angle C$ ·· (10分) + +![](images/9edfc832e116364adb349487830ef11282928617594f913adfa4278cd44d0f7c.jpg) + +23.解：(1)因为 $B D , C E$ 都是 $\triangle A B C$ 的高，所以 $\angle A D B = \angle A E C = 9 0 ^ { \circ }$ + +因为 $\angle 1 = 9 0 ^ { \circ } - \angle C A E , \angle 2 = 9 0 ^ { \circ } -$ $\angle B A D$ ，所以 $\angle 1 = \angle 2$ 故填 $=$ 。 (2分) + +(2)结论： $A P { = } A Q , A P \bot A Q .$ +… (3分)理由：由(1)可知 $\angle 1 = \angle 2$ +在 $\triangle Q A C$ 和 $\triangle A P B$ 中， +$Q C = A B , \angle 1 = \angle 2 , C A = B P ,$ +所以 $\triangle Q A C { \cong } \triangle A P B$ （SAS)，所以$A Q { = } A P$ $\angle Q A C = \angle P$ +因为 $\angle D A P + \angle P = 9 0 ^ { \circ }$ ，所以$\angle D A P + \angle Q A C = 9 0 ^ { \circ }$ ，即 $\angle Q A P =$ $9 0 ^ { \circ }$ ，所以 $A Q \bot A P$ ；…(7分)$( 3 ) A P { = } A Q , A P \bot A Q .$ … (10分)理由：如图所示.因为 $B D , C E$ 都是 $\triangle A B C$ 的高，所以$B D \perp A C , C E \perp A B .$ +所以 $\angle 1 + \angle C A E = 9 0 ^ { \circ }$ ， $\angle 2 +$ $\angle D A B = 9 0 ^ { \circ }$ +因为 $\angle C A E = \angle D A B$ ，所以 $\angle 1 =$ $\angle 2$ +在 $\triangle Q A C$ 和 $\triangle A P B$ 中， +$Q C = A B , \angle 1 = \angle 2 , C A = B P ,$ +所以 $\triangle Q A C { \cong } \triangle A P B ( \mathrm { S A S } )$ ，所以$A Q { = } A P , \angle Q A C { = } \angle P .$ +因为 $\angle P D A = 9 0 ^ { \circ }$ ，所以 $\angle P +$ $\angle P A D = 9 0 ^ { \circ }$ ，所以 $\angle Q A C + \angle P A D$ + +![](images/49350b8aa84418c1421cb7b363edaf8a28da1c74176c955ef8973ff3dc9c1dcd.jpg) + +$= 9 0 ^ { \circ }$ ，即 $\angle Q A P { = } 9 0 ^ { \circ }$ 所以 $A Q \bot A P$ 故 $A P { = } A Q { , } A P \bot A G$ + +# 第六章概率初步 + +# 卷中悟法 + +1.用面积法求概率 + +对于受几何图形的面积影响的随机事件，在一个平面区域内的每个点，事件发生的可能性都是相等的，如果所有可能发生的区域的面积为 $S$ ，所求事件 $A$ 发生的区域的面积为 $S ^ { \prime }$ ，那么 $P ( A ) = { \frac { S ^ { \prime } } { S } }$ + +2.概率的计算方法 + +概率是用来刻画随机事件发生的可能性大小的一个 $0 \sim 1$ 的常数，概率小则事件发生的可能性小，概率大则事件发生的可能性大，如果一次试验中所有可能的结果是有限多个，并且一次试验中各种可能结果发生的可能性都相等，那么可用列举法求概率. + +用列举法求概率的一般步骤： + +(1)列举出一次试验的所有可能的结果；(2)数出 $m , n$ (3)代入概率的计算公式 $P ( A ) { = } { \frac { m } { n } }$ + +# 基础过关参考答案 + +# 一、选择题 + +
12345678
CDABCABB
+ +# 二、填空题 + +
9101112
必然0.75238
+ +# 三、解答题 + +13.解：(1)因为总共有6张卡片，其中数字是1的有2张，数字是2的有3张，数字是3的有1张，所以摸到数字2卡片的可能性最大，摸到数字3卡片的可能性最小. …（4分）(2)因为总共有6张卡片，其中数字是奇数的有3张，数字是偶数的有3张，所以摸到的数字是奇数和摸到的数字是偶数的可能性一样大.… (8分) + +14.解：(1)根据题意，得 $P$ (摸到红球) $=$ ${ \frac { 6 } { 6 + 9 } } { = } { \frac { 6 } { 1 5 } } { = } { \frac { 2 } { 5 } }$ ,P(摸到白球) ${ \mathfrak { o } } = 0$ （5分） + +(2)根据题意，得 ${ \frac { 6 } { 6 + 9 + m } } = { \frac { 1 } { 3 } }$ ，所以$1 5 + m { = } 1 8$ ，解得 $m { = } 3$ …(10分） + +15.解：(1)一个转盘被等分成六个扇形，并在上面依次写上数字 $1 , 2 , 3 , 4 , 5$ 6,其中有3个扇形是奇数，所以自由转动转盘，当它停止转动后，指针指向奇数的概率是 $\frac { 3 } { 6 } = \frac { 1 } { 2 }$ （6分）(2)答案不唯一，如当自由转动的转盘停止时，指针指向数字大于2的区域．· …·（10分） +16.解：(1)如下表： ·· (6分) + +# ·能力提优参考答案· + +# 一、选择题 + +
抛掷 次数 n100200300400500600
正面朝 上的频 数m5198153200255306
正面朝 上的频0.510.490.510.500.510.51
+ +
123456
ABCBDA
+ +# 二、填空题 + +正面朝上的频率 +0.52 +0.51 +0.50 +0.49 +0.48100 200300400500600抛掷次数 + +（3)由折线统计图可知，“正面朝上”的频率在0.51附近摆动.（12分） + +17.解： $( 1 ) \textcircled{ 1}$ 是必然事件，发生的概率为1； （3分）$\textcircled{2}$ 是随机事件，发生的概率为 ${ \frac { 1 3 } { 5 2 } } =$ $\frac { 1 } { 4 }$ · (6分) + +11.解：（1)蓝色区域的圆心角度数为$2 { 0 0 } ^ { \circ }$ ，则 $P$ (指针停止后在蓝色区域)$= \frac { 2 0 0 ^ { \circ } } { 3 6 0 ^ { \circ } } = \frac { 5 } { 9 }$ （4分） + +(2)如下图： （9分) + +(2)红、黄两个扇形的圆心角度数分别为 $4 0 ^ { \circ } , 1 2 0 ^ { \circ }$ ，则 $P$ （指针停止后在黄色或红色区域） $\tan { \frac { 4 0 ^ { \circ } + 1 2 0 ^ { \circ } } { 3 6 0 ^ { \circ } } } = { \frac { 4 } { 9 } }$ （8分） + +(2)事件 $A$ 发生的概率为 ${ \frac { 2 } { 6 } } = { \frac { 1 } { 3 } }$ ，事件 $B$ 发生的概率为 $\frac { 8 } { 5 2 } { = } \frac { 2 } { 1 3 }$ 因为 $\frac { 2 } { 1 3 } < \frac { 1 } { 3 }$ ，所以 $P ( B ) { \ < } P ( A )$ …· (12分) + +![](images/07b7572baf4bd6597005af10e6396a616af100bee5fcc01583f86ebb50f7c95a.jpg) + +12.解：(1)设阴影部分的面积是 $a$ ，则整个图形的面积是 $7 a$ ，可得 $P$ （这个点取在阴影部分） $= \frac { a } { 7 a } = \frac { 1 } { 7 }$ 故答案为$\frac { 1 } { 7 }$ …· (5分) + +# 三、解答题 + +
78910
100083183
+ +# 专项训练卷（一）整式的乘除运算 参考答案· + +（2)如图所示(答案不唯一）：…… (10分) + +13.解：(1)袋子里红球的个数为 $5 0 \times \frac { 3 } { 1 0 }$ $= 1 5$ （4分） + +(2)设白球有 $\mathcal { X }$ 个，由题意得 $x + x -$ $5 + 1 5 = 5 0$ ,解得 $x { = } 2 0$ ,·(8分) 则 $P$ （随机摸出一个球是白球） $= \frac { 2 0 } { 5 0 }$ $= \frac { 2 } { 5 }$ · (10分) + +# 一、选择题 + +
12345678
BCCAADAB
+ +14.解：（1)根据题意得 $y = { \frac { x ^ { 2 } - x } { 2 } } =$ ${ \frac { 1 } { 2 } } x ( x - 1 ) .$ 写出的必然事件不唯一，如输出的数$y$ 是整数. …(5分)(2)当输入的数是2至9这八个连续正整数中的一个时，输出的结果分别为 $1 , 3 , 6 , 1 0 , 1 5 , 2 1 , 2 8 , 3 6$ ，则 $P$ （输出的数是3的倍数) $= \frac { 5 } { 8 }$ （10分） + +
9101112
125x-524
+ +15.解：（1)因为 $1 8 0 < 2 0 0$ ，所以小明购物180元，不能获得转动转盘的机会，故小明获得奖金的概率为0;… (4分) + +# 二、填空题 + +(2)小德购物210元，能获得一次转动转盘的机会， $P$ （获得奖金） $= { \frac { 6 } { 1 6 } } =$ $\frac { 3 } { 8 }$ · (8分) + +(3)设需要将 $\mathcal { X }$ 个无色区域涂上绿色，可得 ${ \frac { x + 3 } { 1 6 } } = { \frac { 1 } { 4 } }$ ，解得 $x = 1$ ，所以需要将1个无色区域涂上绿色.（12分） + +# 三、解答题 + +13.解：（1）原式 $= x ^ { 6 n + 3 } \ \div \ x ^ { 5 n } \cdot x =$ $\boldsymbol { x } ^ { 6 n + 3 - 5 n + 1 } = \boldsymbol { x } ^ { n + 4 }$ （3分） + +（2）原式 $= - 1 - 1 \div ( - \frac { 1 } { 3 } ) \times 4 = - 1$ $- 1 \times ( - 3 ) \times 4 = - 1 + 1 2 = 1 1 .$ （20号（6分） + +14.解：(1)原式 $= x ^ { 2 } + 7 x + x ^ { 2 } - 2 5 - ( x ^ { 2 }$ $- x + 6 x - 6 )$ $= x ^ { 2 } + 7 x + x ^ { 2 } - 2 5 - x ^ { 2 } + x - 6 x + 6$ $= x ^ { 2 } + 2 x - 1 9$ …(3分) 当 $x = - 3$ 时，原式 $= ( - 3 ) ^ { 2 } + 2 \times$ $( - 3 ) - 1 9 = 9 - 6 - 1 9 = - 1 6 .$ …（4分) (2）原式 $= 2 x ^ { 2 } + 5 x - ( - 2 x ^ { 2 } + 4 x ) =$ $2 x ^ { 2 } + 5 x + 2 x ^ { 2 } - 4 x = 4 x ^ { 2 } + x .$ ·…(7分) 当 $x = 5$ 时，原式 $= 4 \times 5 ^ { 2 } + 5 = 1 0 5 .$ …(8分) + +15.解 ${ \mathrm { } } _ { ; a = ( - { \frac { 1 } { 8 } } \times 8 \times { \frac { 1 } { 9 } } \times 9 ) ^ { 3 4 } = ( - 1 ) ^ { 3 4 } }$ $= 1$ ， … (2分) $b { = } 9 4 ^ { 2 } - ( 9 4 { - } 1 ) ( 9 4 { + } 1 ) { = } 9 4 ^ { 2 } { - } 9 4 ^ { 2 }$ $+ 1 { = } 1$ · (4分) $- ( 6 x y ^ { 2 } ) ^ { 2 } \div ( - 3 x y ) = - 3 6 x ^ { 2 } y ^ { 4 } \div \frac { 1 } { 2 }$ $( - 3 x y ) = 1 2 x y ^ { 3 }$ ， ·(6分) 当 $x = a - 2 = 1 - 2 = - 1$ $y = b = 1$ 时，原式 $= 1 2 \times ( - 1 ) \times 1 ^ { 3 } = - 1 2$ ·… (8分) + +16.解:(1) $( a - b ) ^ { 2 }$ $a ^ { 2 } - 2 a b + b ^ { 2 }$ (a$b ) ^ { 2 } = a ^ { 2 } - 2 a b + b ^ { 2 }$ .…(2分) $( 2 ) \textcircled { 1 } 2 9 9 ^ { 2 } = ( 3 0 0 - 1 ) ^ { 2 } = 3 0 0 ^ { 2 } - 2 \times$ $3 0 0 \times 1 + 1 = 9 0 0 0 0 - 6 0 0 + 1 =$ 89401；···· …·（4分） $\textcircled { 2 } 4 7 ^ { 2 } - 4 7 \times 8 6 + 4 3 ^ { 2 } = 4 7 ^ { 2 } - 2 \times 4 7 \times$ $4 3 + 4 3 ^ { 2 } = ( 4 7 - 4 3 ) ^ { 2 } = 4 ^ { 2 } = 1 6 .$ …· (6分) $\begin{array} { l } { { ( 3 ) ( a - b ) ^ { 2 } = a ^ { 2 } - 2 a b + b ^ { 2 } = a ^ { 2 } + b ^ { 2 } } } \\ { { \phantom { - } } } \\ { { - 2 a b = 1 0 - 2 \times 3 = 4 , \phantom { - } \cdots \ ( 8 \ \rlap / ) } } \end{array}$ + +17.解： $( 1 ) \frac { 1 } { 2 } \left( a + a + 2 b \right) \times \left( a + 2 b \right) =$ ${ \frac { 1 } { 2 } } ( 2 a + 2 b ) \times ( a + 2 b ) = ( a + b ) ( a +$ $2 b ) = a ^ { 2 } + 3 a b + 2 b ^ { 2 } ,$ 所以该防洪堤坝的横断面的面积为$( a ^ { 2 } + 3 a b + 2 b ^ { 2 } )$ 平方米．（3分)$( 2 ) a { = } 1 \div 2 ^ { - 1 } { = } 2 , b { = } 1 { - } 6 \times ( { - } 1 )$ $= 1 + 6 = 7$ ，： …(5分）所以该防洪堤坝的横断面的面积为$a ^ { 2 } + 3 a b + 2 b ^ { 2 } = 2 ^ { 2 } + 3 \times 2 \times 7 + 2 \times 7 ^ { 2 }$ $= 4 + 4 2 + 9 8 = 1 4 4$ 平方米.…· (7分) + +(3)修建该防洪堤坝需要() $( \frac { 1 } { 1 0 } ) ^ { - 2 } \div 1$ $1 0 ^ { - 1 } \times 1 4 4 = 1 0 ^ { 2 } \div 1 0 ^ { - 1 } \times 1 4 4 = 1 4 4$ $\times 1 0 0 0 { = } 1 . 4 4 \times 1 0 ^ { 5 }$ 立方米 $= 1 . 4 4 \times$ （20 $1 0 ^ { 5 }$ 方土. · (10分) + +18.解：(1)每个框四个角上的数交叉相乘后求和，再与中间的数的平方的2倍作差，结果都等于－100.…· (2分) + +(2)设中间的数为 $\mathcal { X }$ ,则其他四个数依次是 $x ^ { - } 8 , x ^ { - } 6 , x + 6 , x + 8$ ，所以规律为 $\left( x + 6 \right) \left( x - 6 \right) + \left( x + 8 \right) \left( x - 1 \right)$ $8 ) - 2 x ^ { 2 } = - 1 0 0 .$ +理由：左边 $= x ^ { 2 } - 3 6 + x ^ { 2 } - 6 4 - 2 x ^ { 2 }$ $= x ^ { 2 } + x ^ { 2 } - 2 x ^ { 2 } - 1 0 0 = - 1 0 0 =$ 右边,所以结论成立．…（6分)(3)(1)中的规律不成立．（7分)验证： $9 \times 2 3 + 1 5 \times 1 7 - 2 \times 1 6 ^ { 2 } =$ —50， $1 2 \times 2 6 + 1 8 \times 2 0 - 2 \times 1 9 ^ { 2 } =$ 一50,所以(1)中的规律不成立. +·（9分）发现的规律：设中间的数为 $\mathcal { X }$ ,则其他四个数依次是 $x ^ { - 7 } , x ^ { - 1 } , x ^ { + 1 } , x$ $+ 7$ ，所以规律为 $( x + 7 ) ( x - 7 ) + ( x$ $+ 1 ) ( x - 1 ) - 2 x ^ { 2 } = - 5 0 .$ +…·（10分）理由：左边 $= x ^ { 2 } - 4 9 + x ^ { 2 } - 1 - 2 x ^ { 2 } =$ $x ^ { 2 } + x ^ { 2 } - 2 x ^ { 2 } - 5 0 = - 5 0 =$ 右边. +……· (12分) + +专项训练卷（二）几何题的计算参考答案 + +# 一、选择题 + +
1235678
CDBABADC
+ +# 二、填空题 + +
9101112
25°170°6°
+ +# 三、解答题 + +13.解：因为 $D E \bot B C$ ，所以 $\angle D E B =$ $9 0 ^ { \circ }$ ，所以 $\angle B + \angle E D B = 9 0 ^ { \circ }$ ，因为$\angle B = 3 1 ^ { \circ }$ ，所以 $\angle E D B = 9 0 ^ { \circ } - 3 1 ^ { \circ } =$ $5 9 ^ { \circ }$ ，所以 $\angle A D E = 1 8 0 ^ { \circ } - \angle E D B =$ $1 8 0 ^ { \circ } - 5 9 ^ { \circ } = 1 2 1 ^ { \circ }$ ·· (4分)因为 $E F / / A B$ ，所以 $\angle F E D = \vdots$ $\angle E D B = 5 9 ^ { \circ }$ ，因为 $F G \bot B C , D E \bot$ BC，所以 $F G / / D E$ ，所以 $\angle F = \vdots$ $\angle F E D = 5 9 ^ { \circ }$ (8分) + +14.解：(1)因为 $O M \perp A B , O N \perp C D$ 所以 $\angle 1 + \angle A O C = 9 0 ^ { \circ }$ $\angle 2 + \angle A O C$ $= 9 0 ^ { \circ }$ ，所以 $\angle 1 = \angle 2$ ，因为 $\angle 1 = 2 9 ^ { \circ }$ ，所以 $\angle 2 = \angle 1 = 2 9 ^ { \circ }$ ·……(3分)(2)因为 $\angle B O D { = } 6 2 ^ { \circ }$ ，所以 $\angle A O C =$ $\angle B O D { = } 6 2 ^ { \circ }$ ，因为 $\angle A O M = 9 0 ^ { \circ }$ ，所以 $\angle 1 = \vdots$ $\angle A O M - \angle A O C = 9 0 ^ { \circ } - 6 2 ^ { \circ } = 2 8 ^ { \circ }$ … (6分)因为 $\angle C O N = 9 0 ^ { \circ }$ ，所以 $\angle M O N = \mathrm { : }$ $\angle 1 + \angle C O N = 2 8 ^ { \circ } + 9 0 ^ { \circ } = 1 1 8 ^ { \circ } .$ …· (9分) + +15.解：(1)因为点 $E$ 是 $\it C D$ 的中点，所以$D E = C E$ ，因为 $A D / / B C$ ，所以$\angle D A E = \angle F$ ， $\angle D = \angle E C F$ ，所以$\triangle A D E { \cong } \triangle F C E .$ ，所以 $A E = F E$ $S _ { \triangle A D E } { = } S _ { \triangle F C E }$ ，所以 $A F { = } 2 A E { = } 2 { \times } 2 { = } 4$ ，因为 $B E$ + +$\perp A F$ ，所以 $S _ { \triangle A B F } = \frac 1 2 B E \cdot A F = \frac 1 2$ +$\times 4 \times 4 = 8$ ， +所以 $S _ { \substack { \scriptscriptstyle { [ \mathscr { A } _ { 1 } \mathscr { A } _ { 2 } ] \mathscr { H } _ { 4 } } \mathscr { A } B C D } } = S _ { \triangle \mathscr { A } B F } = 8 ,$ … (5分) + +（2）因为 $A D / / B C$ ，所以 $\angle F =$ $\angle D A E = 5 9 ^ { \circ }$ ，因为 $\triangle A D E \cong$ $\triangle F C E$ ，所以 $A E { = } F E$ ，因为 $B E \bot$ $A F$ ，所以 $B E$ 是线段 $A F$ 的垂直平分线，所以 $A B = B F$ ，所以 $\angle B A F =$ $\angle F = 5 9 ^ { \circ }$ ，所以 $\angle A B F = 1 8 0 ^ { \circ } - 5 9 ^ { \circ }$ $- 5 9 ^ { \circ } = 6 2 ^ { \circ }$ ·… (9分) + +16.解：(1)因为 $A B { = } A C$ ，所以 $\angle A B C =$ $\angle C = 6 5 ^ { \circ }$ ，因为 $A D \perp B C$ ，所以$\angle C A D = 9 0 ^ { \circ } - \angle C = 9 0 ^ { \circ } - 6 5 ^ { \circ } = 2 5 ^ { \circ }$ 因为 $M N / / \ A C$ ，所以 $\angle A N M =$ $\angle D A C = 2 5 ^ { \circ }$ …(3分)(2)因为 $A B = A C = 8$ ，点 $M$ 为 $A B$ 的中点，所以 $A M { = } B M { = } \frac { 1 } { 2 } { \times } 8 { = } 4$ ，… (5分)因为MN//AC，所以 $\angle A N M =$ $\angle D A C$ ，因为 $A B { = } A C , A D \bot B C ,$ 所以 $\angle B A D = \angle C A D$ ，所以 $\angle M A N =$ $\angle A N M$ ，所以 $A M { = } M N$ ，所以 $M N$ $= A M { = } 4$ ·…· (10分)因为 $B N { = } 3$ ，所以 $\triangle B N M$ 的周长为$B M + M N + B N { = } 4 { + } 4 { + } 3 { = } 1 1 .$ …（12分) + +17.解： $( 1 ) \textcircled{ 1 }$ 因为 $D E$ 垂直平分 $A B$ ，所以 $A E { = } B E$ ，所以 $\angle A = \angle A B E$ ，因为 $F G$ 垂直平分 $B C$ ，所以 $C G { = } B G$ + +所以 $\angle G B C = \angle C$ ，因为 $\angle A B C = { \mathrm { : } }$ $1 0 0 ^ { \circ }$ ，所以 $\angle A + \angle C = 1 8 0 ^ { \circ } - 1 0 0 ^ { \circ } =$ $8 0 ^ { \circ }$ ，所以 $\angle A B E + \angle C B G = \angle A +$ $\angle C = 8 0 ^ { \circ }$ ，所以 $\angle E B G = 1 0 0 ^ { \circ } - 8 0 ^ { \circ }$ $= 2 0 ^ { \circ }$ （4分）$\textcircled{2}$ 因为 $D E$ 垂直平分 $A B$ ，所以 $A E =$ $B E$ ，所以 $\angle A = \angle A B E .$ ，因为 $F G$ 垂直平分 $B C$ ，所以 $C G = B G$ ，所以$\angle G B C = \angle C$ ，因为 $\angle A B C = 7 0 ^ { \circ }$ ，所以 $\angle A + \angle C = 1 8 0 ^ { \circ } - 7 0 ^ { \circ } = 1 1 0 ^ { \circ }$ +所以 $\angle A B E + \angle C B G = 1 1 0 ^ { \circ }$ ，所以$\angle A B G + \angle E B G + \angle E B G + \angle C B E$ $= 1 1 0 ^ { \circ }$ +所以 $\angle A B C + \angle E B G = 1 1 0 ^ { \circ }$ ，所以$\angle E B G = 1 1 0 ^ { \circ } - 7 0 ^ { \circ } = 4 0 ^ { \circ } .$ ： +…· (10分) + +11.解：（1）因为 $\angle A O E = 4 5 ^ { \circ }$ ，所以$\angle B O F { = } \angle A O E { = } 4 5 ^ { \circ }$ ，所以 $\angle A O F$ $= 1 8 0 ^ { \circ } - 4 5 ^ { \circ } = 1 3 5 ^ { \circ }$ ，因为 $\angle A O F =$ $3 \angle C O F$ ，所以 $3 \angle C O F = 1 3 5 ^ { \circ }$ ，所以$\angle C O F { = } 4 5 ^ { \circ }$ ，所以 $\angle C O B { = } 9 0 ^ { \circ }$ ，所以 $A B \bot C O .$ ·(4分)(2）与 $\angle C O F$ 互余的角： $\angle B O F$ $\angle A O E$ ；与 $\angle C O F$ 互补的角：$\angle C O E$ …(8分) + +# 三、解答题 + +(2)当 $0 ^ { \circ } < \alpha < 9 0 ^ { \circ }$ 时， $\angle E B G = 1 8 0 ^ { \circ }$ $- 2 \alpha$ ； …（12分）当 $9 0 ^ { \circ } < \alpha < 1 8 0 ^ { \circ }$ 时， $\angle E B G = 2 \alpha -$ $1 8 0 ^ { \circ }$ ·(14分) + +专项训练卷（三）几何题的说理 参考答案· + +12.解：（1）因为 $A D$ 是 $\triangle A B E$ 的中线，所以 $B D { = } D E$ 在 $\triangle A B D$ 和 $\triangle A E D$ 中， $A B = A E$ $A D { = } A D , B D { = } D E$ ，所以 $\triangle A B D { \underline { { \underline { { \circ } } } } }$ $\triangle A E D$ …(3分)$( 2 ) \angle B = 2 \angle C .$ （4分）理由：因为 $A B = A E$ ，所以 $\angle B =$ $\angle A E B$ ，因为 $E F$ 垂直平分 $A C$ ，所以$A E { = } C E$ ，所以 $\angle E A C = \angle C$ ， ·· (6分)因为 $\angle A E B = 1 8 0 ^ { \circ } - \angle A E C =$ $\angle E A C + \angle C ,$ 所以 $\angle A E B = 2 \angle C$ ，所以 $\angle B { = } 2 \angle C .$ ·· (8分) + +# 一、选择题 + +
123456
BCBABD
+ +
78910
SAS527∠A+∠C+ ∠EFC=270°
+ +# 二、填空题 + +13.解： $( 1 ) \triangle B D F { \cong } \triangle E D C .$ …(1分)理由：因为 $\angle B = 9 0 ^ { \circ } , A D$ 是 $\angle B A C$ 的平分线， $D E \bot A C$ 所以 $B D = D E$ ，因为 $B F { = } C E , \angle B$ $= \angle C E D = 9 0 ^ { \circ }$ ，所以 $\triangle B D F \cong$ $\triangle E D C$ ：$( 2 ) A C { = } A B { + } B F .$ ：(5分)理由：易得 $\triangle A B D \cong \triangle A E D$ ，所以$A B { = } A E$ ，因为 $B F { = } C E$ ，所以 $A C =$ + +$A E + C E { = } A B { + } B F ,$ 即 $\scriptstyle A C = A B + B F$ （8分） + +14.解：(1)如图，过点 $N$ 作 $N E / / A B$ ，所以 $\angle A P N + \angle P N E { = } 1 8 0 ^ { \circ }$ 因为 $\angle A P N + \angle P N Q + \angle C Q N =$ $3 6 0 ^ { \circ }$ ，所以 $\angle A P N + \angle P N E +$ $\angle E N Q + \angle C Q N = 3 6 0 ^ { \circ } ,$ 所以 $1 8 0 ^ { \circ } + \angle E N Q + \angle C Q N = 3 6 0 ^ { \circ }$ 所以 $\angle E N Q + \angle C Q N = 1 8 0 ^ { \circ }$ 所以 $E N / / C D$ ，… ：（4分）因为 $N E / / A B$ ，所以 $A B / / C D$ …· (5分) + +![](images/8af7f4aec5432f7c3c80018398d422ea40723dafcd9358b2cbfb59f9c2740066.jpg) + +(2） $\angle M { = } \frac { 1 } { 2 } \angle P N Q .$ …… (6分) + +理由：因为 $A B / / C D / / N E$ ，所以$\angle B P N ~ = ~ \angle P N E ,$ $\angle D Q N =$ $\angle Q N E$ +所以 $\angle B P N + \angle D Q N = \angle P N Q$ 同理可得 $\angle B P M + \angle D Q M = \angle M$ +因为PM,QM分别平分 $\angle B P N$ $\angle D Q N$ ， +所以 $\angle B P N = 2 \angle B P M , \angle D Q N =$ $2 \angle D Q D$ ，所以 $\angle B P N + \angle D Q N =$ $2 ( \angle B P M + \angle D Q M )$ +所以 $\angle P N Q = 2 \angle M ,$ 即 $\angle M = \vdots$ $\frac { 1 } { 2 } \angle P N Q .$ … (12分) + +15.解：(1)有. …(1分）如图，连接BC,在 $\triangle A B C$ 中， $\angle A +$ $\angle A B C + \angle A C B = 1 8 0 ^ { \circ }$ ，在 $\triangle C B C$ 中， $\angle B O C + \angle O B C + \angle O C B =$ $1 8 0 ^ { \circ }$ ，所以 $\angle A + \angle A B C + \angle A C B =$ $\angle B O C + \angle O B C + \angle O C B$ ，所以 $\angle A$ $+ \angle A B C - \angle O B C + \angle A C B -$ $\angle O C B = \angle B O C$ ，所以 $\angle B O C = \angle A$ $+ \angle A B O + \angle A C O .$ ……(4分) + +![](images/a71e08cbb80f53e828d8278c6cc62286159ae8e9a36fc915792628d4d46b6b01.jpg) + +(2)如图，连接 $A D$ ，由(1)可知 $\angle F +$ $\angle 2 + \angle 3 = 1 3 0 ^ { \circ } , \angle C + \angle 4 + \angle 1 =$ ${ 1 0 0 } ^ { \circ }$ ，所以 $\angle F + \angle 2 + \angle 3 + \angle C +$ $\angle 4 + \angle 1 = 2 3 0 ^ { \circ }$ ，所以 $\angle F + ( \angle 1 +$ $\angle 2 ) + \angle C + ( \angle 3 + \angle 4 ) = 2 3 0 ^ { \circ } ,$ 所以 $\angle F A B + \angle C + \angle C D E + \angle F =$ $2 3 0 ^ { \circ }$ ……(7分) + +AE130° 100B C D + +$( 3 ) E C / / B F / / D G .$ ·(8分)理由：因为 $\angle E O D + \angle O B F = 1 8 0 ^ { \circ }$ $\angle E O D = \angle B O C .$ ，所以 $\angle B O C +$ $\angle O B F { = } 1 8 0 ^ { \circ }$ ，所以 $E C / / B F$ 因为 $\angle C D G = \angle F$ ，所以 $D G / / B F$ 所以 $E C / / B F / / D G .$ ·…（10分）(4)149.5° …（14分)提示：因为 $A B { = } A C$ ，所以 $\angle A B C =$ $\angle A C B = \frac { 1 } { 2 } \left( 1 8 0 ^ { \circ } - 5 8 ^ { \circ } \right) = 6 1 ^ { \circ } ,$ 又因为 $C E$ 平分 $\angle A C B$ ，所以 $\angle E C D = :$ $\angle E C B = 3 0 . 5 ^ { \circ }$ ,又因为 $D G / / E C$ ，所以 $\angle C D G = \angle D C E = 3 0 . ~ 5 ^ { \circ }$ ，即$\angle A D G = 1 4 9 . 5 ^ { \circ } .$ + +期末测试卷参考答案· + +# 一、选择题 + +
12345678910
DBCCBACAAB
+ +# 二、填空题 + +
1112131415
13115°2023370°
+ +# 三、解答题 + +16.(1)解：原式 $= - 4 + 1 - ( - 3 ) + 2$ ·（4分）$= 2$ （5分） + +(2)解：因为点 $A$ 和点 $E$ 关于 $B D$ 对称，所以 $\angle A B D = \angle E B D$ ，即 $\angle A B C$ $= 2 \angle A B D = 2 \angle E B D .$ ……(1分)又因为点 $B$ 和点 $C$ 关于 $D E$ 对称，所以 $\angle D B E = \angle C$ ，所以 $\angle A B C = \left\{ \begin{array} { l l } { \begin{array} { r l r } \end{array} } \end{array} \right.$ $2 \angle C$ ·…· (3分)因为 $\angle A = 9 0 ^ { \circ }$ ，所以 $\angle A B C + \angle C =$ $9 0 ^ { \circ }$ ，即 $3 \angle C = 9 0 ^ { \circ }$ 解得 $\angle C = 3 0 ^ { \circ }$ ，则 $\angle A B C = 2 \angle C =$ （20 $6 0 ^ { \circ }$ ·… (5分) + +17.解：(1)当 $a = 2 4 . 5 ~ \mathrm { c m }$ 时， $b = 7 a -$ $3 . 0 7 = 7 \times 2 4 . 5 - 3 . 0 7 = 1 6 8 . 4 3$ (cm).答：他的身高约为168.43厘米；……· (4分) + +(2)当 $a = 2 6 . 7 \ \mathrm { c m }$ 时， $b { = } 7 a { - } 3 . 0 7$ $= 7 \times 2 6 . 7 \mathrm { - } 3 . 0 7 \mathrm { = } 1 8 3 . 8 3 ~ \mathrm { c m }$ ，显然身高为 $1 . 8 7 \mathrm { ~ m ~ }$ 的比较接近，因此身高为 $1 . 8 7 \mathrm { ~ m ~ }$ 的人作案的可能性更大. … (9分) + +18.解：由 $( m + n ) ^ { 2 } = 1 1 ^ { 2 }$ ，得 $m ^ { 2 } + 2 m n +$ $n ^ { 2 } = 1 2 1$ …·(3分)因为 $m n { = } 1$ ，所以 $m ^ { 2 } + 2 + n ^ { 2 } = 1 2 1$ 整理得 $m ^ { 2 } + n ^ { 2 } = 1 1 9$ ,·（7分)所以 $( m - n ) ^ { 2 } = m ^ { 2 } - 2 m n + n ^ { 2 } = 1 1 9$ $- 2 { = } 1 1 7$ …· (9分) + +19.解：因为 $B D , C D$ 的垂直平分线分别交 $A B , A C$ 于点 $E , F$ ，所以 $E B =$ $E D , F D = F C$ 所以 $\angle E D B = \angle B , \angle F D C = \angle C$ …· (4分)所以 $\angle E D B + \angle F D C = \angle B + \angle C$ 因为 $\angle E D F = 1 8 0 ^ { \circ } - ( \angle E D B +$ $\angle F D C ) , \angle A = 1 8 0 ^ { \circ } - ( \angle B + \angle C )$ 所以 $\angle E D F = \angle A = 6 8 ^ { \circ }$ （9分） + +20.解：(1)因为 $B C / / D E$ ，所以 $\angle A C B =$ $\angle E .$ 在 $\triangle A B C$ 和 $\triangle D C E$ 中，$A C { = } D E$ $= D E , \angle A C B = \angle E , B C = C E$ 所以 $\triangle A B C \cong \triangle D C E ( \mathrm { S A S } )$ ，所以$A B { = } C D$ ;·.. ·…(4分)（2）因为 $\triangle A B C { \cong } \triangle D C E$ $\angle D =$ $3 0 ^ { \circ }$ ，所以 $\angle A = \angle D = 3 0 ^ { \circ }$ 因为 $\angle E { = } 6 5 ^ { \circ }$ ，所以 $\angle A F E = 1 8 0 ^ { \circ } -$ $3 0 ^ { \circ } - 6 5 ^ { \circ } = 8 5 ^ { \circ }$ ，所以 $\angle D F A = 1 8 0 ^ { \circ } -$ $8 5 ^ { \circ } = 9 5 ^ { \circ }$ ， + +所以 $\angle D G F = 1 8 0 ^ { \circ } - 9 5 ^ { \circ } - 3 0 ^ { \circ } = 5 5 ^ { \circ }$ 所以 $\angle F G C = 1 8 0 ^ { \circ } - 5 5 ^ { \circ } = 1 2 5 ^ { \circ }$ …· (9分) + +21.解：相等或互补. …（1分）理由：如图所示. + +![](images/f2beb444fe95ce8522b41d1460aa36419706105898518696946facab36e4af11.jpg) + +分情况讨论： +当 $D G$ 边位于 $D E$ 上方时， +因为 $D E ~ / / ~ A B$ ，所以 $\angle B A C =$ $\angle E F C$ +因为 $D G / / A C$ ，所以 $\angle E F C =$ $\angle E D G$ ，所以 $\angle B A C = \angle E D G$ +· (4分)当 $D G$ 边位于 $D E$ 下方时， +因为 $D E ~ / / ~ A B$ ，所以 $\angle B A C = { \vdots }$ $\angle A F D$ +因为 $D G / / A C$ ，所以 $\angle E D G ^ { \prime } +$ $\angle A F D = 1 8 0 ^ { \circ }$ ，所以 $\angle E D G ^ { \prime } ~ + ~$ $\angle B A C = 1 8 0 ^ { \circ }$ +即 $\angle B A C$ 与 $\angle E D G$ 的数量关系是相等或互补. （9分） + +22.解：(1)在8种等可能结果中，转出的数是4的倍数的结果有2种：4,8.则$P$ （小彬转出的数是4的倍数） $\scriptstyle \displaystyle { \frac { 2 } { 8 } } =$ $\frac { 1 } { 4 }$ （4分） + +（2)有两张分别写有3和5的卡片，设转盘停止后记下转出的数为 $\mathcal { X }$ + +要想组成三角形，则 $5 - 3 < x < 5 +$ 3，即 $2 { < } x { < } 8$ 在8种等可能结果中，能构成三角形的结果有5种： $3 , 3 , 5 ; 3 , 4 , 5 ; 3 , 5 , 5$ 3,5,6;3,5,7.则 $P$ （这三条线段能构成三角形) $= \frac { 5 } { 8 }$ （10分） + +23.解：（1)由图象得， $M , N$ 两地的实际距离为 $6 0 0 ~ \mathrm { k m } .$ 故填 600.···… (1分)(2)点 $C$ 的实际意义是乙车行驶 $4 \textrm { h }$ 后两车相遇.故填乙车行驶 $4 \textrm { h }$ 后两车相遇. ··(2分）(3)设甲车出发 $\mathcal { X }$ 小时后发生故障，根据题意得， $4 \times 1 0 0 + 8 0 x = 6 0 0$ ，解得 $x { = } 2 . 5$ 答：甲车出发2.5小时后发生故障.（4分） + +(4)设乙车出发 $a$ 小时后两车相距$2 0 0 \ \mathrm { k m } .$ 分情况讨论： +$\textcircled{1}$ 当甲、乙两车相遇前相距 $2 0 0 \ \mathrm { k m }$ 时，可得 $1 0 0 a + 8 0 a = 6 0 0 - 2 0 0$ ，解得 $a { = } \frac { 2 0 } { 9 }$ （2 +$\textcircled{2}$ 当甲、乙两车相遇后相距 $2 0 0 \ \mathrm { k m }$ 时，可得 $8 0 ( a - 2 . 5 ) + 1 0 0 a = 6 0 0 +$ 200,解得 $a { = } \frac { 5 0 } { 9 }$ （20 +综上所述，当乙车出发 $\frac { 2 0 } { 9 }$ h或 $\frac { 5 0 } { 9 }$ h后，两车相距 $2 0 0 \ \mathrm { k m } ,$ （10分） + +# 考前练卷霸 + +# 成绩进步大 + +精编试题 卷中悟法 叁考答案 详细解析梳理重点知识 设置精巧点拨 规范学生答题 拓展学习思维掌握解题技巧 培育关键能力 养成良好习惯 提升学科素养 \ No newline at end of file diff --git a/juanba/juanba/auto/juanba_content_list.json b/juanba/juanba/auto/juanba_content_list.json new file mode 100644 index 0000000000000000000000000000000000000000..205d43e255a26f20b5a822d0c3c9370b5406fb38 --- /dev/null +++ b/juanba/juanba/auto/juanba_content_list.json @@ -0,0 +1,8427 @@ +[ + { + "type": "text", + "text": "初中同步测试卷", + "text_level": 1, + "page_idx": 0 + }, + { + "type": "text", + "text": "金太阳教育研究院编", + "page_idx": 0 + }, + { + "type": "text", + "text": "19套 ", + "page_idx": 0 + }, + { + "type": "text", + "text": "卷+月考卷+期中卷+专项卷+期末卷", + "page_idx": 0 + }, + { + "type": "text", + "text": "图书在版编目(CIP)数据", + "text_level": 1, + "page_idx": 2 + }, + { + "type": "text", + "text": "卷霸．初中同步测试卷 数学北师大版七年级下册／金太阳教育研究院编．--南昌：江西高校出版社，2023．2(2023.12重印）", + "page_idx": 2 + }, + { + "type": "text", + "text": "ISBN 978-7-5762-3604-0 ", + "page_idx": 2 + }, + { + "type": "text", + "text": "I. $\\textcircled{1}$ 卷…Ⅱ. $\\textcircled{1}$ 金…Ⅲ. $\\textcircled{1}$ 中学数学课一初中一教 学参考资料IV. $\\textcircled{1} G 6 3 4$ ", + "page_idx": 2 + }, + { + "type": "text", + "text": "中国国家版本馆CIP数据核字（2023)第017754号", + "page_idx": 2 + }, + { + "type": "text", + "text": "卷霸初中同步测试卷 数学 北师大版七年级下册JUANBA CHUZHONG TONGBU CESHIJUANSHUXUEBEISHIDA BANQI NIANJIXIACE", + "page_idx": 2 + }, + { + "type": "table", + "img_path": "images/1cc13d84eb7e58f710e9a1c71d7b63f5da3e3e6b463c7f8842d6b9afd2f1c798.jpg", + "table_caption": [], + "table_footnote": [ + "赣版权登字-07-2023-116版权所有侵权必究" + ], + "table_body": "
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2023年2月第1版
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书定 号价ISBN 978-7-5762-3604-0
28.00元
", + "page_idx": 2 + }, + { + "type": "text", + "text": "初中同步测试卷", + "text_level": 1, + "page_idx": 3 + }, + { + "type": "text", + "text": "CHUZHONG TONGBU CESHIJUAN", + "page_idx": 3 + }, + { + "type": "text", + "text": "金太阳教育研究院编", + "page_idx": 3 + }, + { + "type": "text", + "text": "梳理重点知识 拓展学习思维 ", + "text_level": 1, + "page_idx": 3 + }, + { + "type": "text", + "text": "本套试卷围绕新一轮基础教育课程改革的新思想、新理念，充分考虑学生的学习体验，以提高学生素质、促进学生学习方式的转变、促进学生积极主动全面的发展为目标，对学生的学习进行同步指导、检测及评价，帮助学生发现优势和不足，促使学生制订改进计划，调整学习策略，从而提高学习效率。", + "page_idx": 3 + }, + { + "type": "text", + "text": "主编：刘美平副主编：刘黎明熊林华编委:(按姓氏笔画排序)刘宗政　许　清　宋丽革　杨建余燕婷 　梁东强 　谭 　薇 　熊三勇", + "page_idx": 3 + }, + { + "type": "text", + "text": "", + "page_idx": 3 + }, + { + "type": "text", + "text": "", + "page_idx": 3 + }, + { + "type": "text", + "text": "数学北师大族", + "text_level": 1, + "page_idx": 3 + }, + { + "type": "text", + "text": "七年级下册", + "page_idx": 3 + }, + { + "type": "text", + "text": "亮点一 结构合理", + "text_level": 1, + "page_idx": 3 + }, + { + "type": "text", + "text": "采用“单元卷+阶段卷+专项卷”的模式编写，思路清晰，结构完整", + "page_idx": 3 + }, + { + "type": "text", + "text": "亮点", + "text_level": 1, + "page_idx": 3 + }, + { + "type": "text", + "text": "亮点二 内容新颖", + "text_level": 1, + "page_idx": 3 + }, + { + "type": "text", + "text": "紧紧围绕教材编创新题、好题，难易适中，内容丰富，特色鲜明", + "page_idx": 3 + }, + { + "type": "text", + "text": "亮点三 卷中悟法", + "page_idx": 3 + }, + { + "type": "text", + "text": "参考答案中含重难知识总结、方法归纳、解题技巧等，重点突出", + "page_idx": 3 + }, + { + "type": "text", + "text": "亮点四 视频讲解优秀教师详细讲解答题技巧，快速提升学生解题、得分能力", + "page_idx": 3 + }, + { + "type": "text", + "text": "本卷特别注重科学性和实用性，在编写过程中，得到了很多一线教师的热心帮助。我们精心设计，细致核对，力求做到尽善尽美。虽然我们竭尽心智，但疏漏之处在所难免，恳请广大师生在使用中提出宝贵的意见，以便我们修改完善。", + "page_idx": 3 + }, + { + "type": "text", + "text": "第一章 整式的乘除", + "page_idx": 4 + }, + { + "type": "text", + "text": "第五章 生活中的轴对称", + "text_level": 1, + "page_idx": 4 + }, + { + "type": "text", + "text": "基础过关测试卷能力提优测试卷", + "page_idx": 4 + }, + { + "type": "text", + "text": "$\\textcircled{4}$ 阅读理解题", + "text_level": 1, + "page_idx": 4 + }, + { + "type": "text", + "text": "第二章 相交线与平行线基础过关测试卷 L能力提优测试卷· 13月考测试卷（一） 17第三章 变量之间的关系基础过关测试卷· 25能力提优测试卷· 29期中测试卷 33", + "page_idx": 4 + }, + { + "type": "text", + "text": "", + "page_idx": 4 + }, + { + "type": "text", + "text": "", + "page_idx": 4 + }, + { + "type": "text", + "text": "第四章 三角形", + "page_idx": 4 + }, + { + "type": "text", + "text": "基础过关测试卷 41", + "page_idx": 4 + }, + { + "type": "text", + "text": "基础过关测试卷 49 \n能力提优测试卷 53", + "page_idx": 4 + }, + { + "type": "image", + "img_path": "", + "img_caption": [], + "img_footnote": [], + "page_idx": 4 + }, + { + "type": "text", + "text": "月考测试卷（二） 57", + "page_idx": 4 + }, + { + "type": "text", + "text": "六章 概率初步", + "page_idx": 4 + }, + { + "type": "text", + "text": "基础过关测试卷 65 \n能力提优测试卷 69 \n专项训练卷（一） 整式的乘除运算 ： 73 \n专项训练卷（二） 几何题的计算 77 \n专项训练卷（三） 几何题的说理 81 \n期末测试卷 85 \n答案和解析 93 \n新定义/3/18/34 \n解题方法/40/52/84", + "page_idx": 4 + }, + { + "type": "text", + "text": "", + "page_idx": 4 + }, + { + "type": "text", + "text": "", + "page_idx": 4 + }, + { + "type": "text", + "text": "代数推理", + "text_level": 1, + "page_idx": 4 + }, + { + "type": "text", + "text": "等式规律探索/6/19 ", + "page_idx": 4 + }, + { + "type": "text", + "text": "$\\circledcirc$ 真实问题情境", + "text_level": 1, + "page_idx": 4 + }, + { + "type": "text", + "text": "汽车探照灯/14 ", + "page_idx": 4 + }, + { + "type": "text", + "text": "山地自行车/78", + "page_idx": 4 + }, + { + "type": "text", + "text": "$\\textcircled{4}$ 数学文化", + "text_level": 1, + "page_idx": 4 + }, + { + "type": "text", + "text": "《详解九章算法》/35", + "page_idx": 4 + }, + { + "type": "text", + "text": "$\\circledcirc$ 综合与实践", + "text_level": 1, + "page_idx": 4 + }, + { + "type": "text", + "text": "拓展探究型/48/76 ", + "page_idx": 4 + }, + { + "type": "text", + "text": "$\\textcircled{6}$ 填空双空题/59 ", + "page_idx": 4 + }, + { + "type": "text", + "text": "第一章 整式的乘除", + "text_level": 1, + "page_idx": 5 + }, + { + "type": "text", + "text": "基础过关测试卷", + "text_level": 1, + "page_idx": 5 + }, + { + "type": "text", + "text": "时间：60分钟 满分：100分", + "page_idx": 5 + }, + { + "type": "table", + "img_path": "images/36ca64a80cdb73b3ca108c79e90360d4078548f57d0d726001d654e7877b2e1f.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 5 + }, + { + "type": "text", + "text": "一、选择题(每小题4分，共32分)", + "text_level": 1, + "page_idx": 5 + }, + { + "type": "text", + "text": "1.计算 $a ^ { 6 } \\bullet a ^ { 2 }$ 的结果是", + "page_idx": 5 + }, + { + "type": "text", + "text": "A $a ^ { 3 }$ B. $a ^ { 4 }$ C.a8 D. $a ^ { 1 2 }$ ", + "page_idx": 5 + }, + { + "type": "text", + "text": "2.计算 $2 0 2 4 ^ { \\circ } \\times 2 ^ { - 1 }$ 的结果是", + "page_idx": 5 + }, + { + "type": "text", + "text": "A.-2024 B.-2 C.0 D $\\frac 1 2$ ", + "page_idx": 5 + }, + { + "type": "text", + "text": "3.计算 $( 2 a ) \\ \\bullet \\ ( a b )$ 的结果是", + "page_idx": 5 + }, + { + "type": "text", + "text": "A. 2ab B. $2 a ^ { 2 } b$ （20 C.3ab D. $3 a ^ { 2 } b$ ", + "page_idx": 5 + }, + { + "type": "text", + "text": "4.若 $( x - 2 ) ( x + a ) = x ^ { 2 } + b x - 6$ ，则 ", + "page_idx": 5 + }, + { + "type": "text", + "text": "A. $a = 3 , b = 1$ $\\mathrm { B } _ { \\cdot } a { = } 3 , b { = } - 5$ （20 $\\stackrel { \\cdot } { \\scriptscriptstyle \\mathscr { O } } . a = - 3 , b = - 1$ I $) . a { = } { - 3 , b { = } { - 5 } }$ ", + "page_idx": 5 + }, + { + "type": "text", + "text": "5.下列运算中，正确的是", + "page_idx": 5 + }, + { + "type": "text", + "text": "A. $x ^ { 2 } + 5 x ^ { 2 } = 6 x ^ { 4 } \\quad \\mathrm { ~ B . ~ } x ^ { 3 } \\bullet x ^ { 2 } = x ^ { 6 } \\qquad \\mathrm { ~ C . ~ } ( x ^ { 2 } ) ^ { 3 } = x ^ { 6 }$ D. $( \\boldsymbol { \\mathscr { x y } } ) ^ { 3 } = \\boldsymbol { \\mathscr { x y } } ^ { 3 }$ （204号", + "page_idx": 5 + }, + { + "type": "text", + "text": "6.若 $M \\bullet ( 3 x - y ^ { 2 } ) = y ^ { 4 } - 9 x ^ { 2 }$ ，则多项式 $M$ 为 ", + "page_idx": 5 + }, + { + "type": "text", + "text": "$ \\mathrm { A } . - 3 x + y ^ { 2 } \\qquad \\mathrm { B } . - y ^ { 2 } - 3 x \\qquad \\mathrm { C } . ~ 3 x + y ^ { 2 }$ $\\mathrm { D } . 3 x - y ^ { 2 }$ ", + "page_idx": 5 + }, + { + "type": "text", + "text": "7.当 $a { = } \\frac { 1 } { 3 }$ 时，代数式 $( a - 4 ) ( a - 3 ) - a ( a + 2 )$ 的值为", + "page_idx": 5 + }, + { + "type": "text", + "text": "A.9 B.-9 C.3 D $\\frac 1 3$ ", + "page_idx": 5 + }, + { + "type": "text", + "text": "8.已知 $\\displaystyle a + b = - 4 , a b = - 1 2$ ，则 $a ^ { 2 } + b ^ { 2 }$ 的值为", + "page_idx": 5 + }, + { + "type": "text", + "text": "A.46 B.44 C.42 D.40", + "page_idx": 5 + }, + { + "type": "text", + "text": "二、填空题(每小题4分，共16分)", + "text_level": 1, + "page_idx": 5 + }, + { + "type": "text", + "text": "9.一种桑蚕丝的直径约为0.0000189米，数据0.0000189用科学记数法表示为", + "page_idx": 5 + }, + { + "type": "text", + "text": "10.计算： $1 0 x ^ { 8 } \\div 2 x ^ { 2 } =$ ", + "page_idx": 5 + }, + { + "type": "text", + "text": "11.计算： $2 m ^ { 2 } n \\bullet ( m ^ { 2 } + n - 1 ) =$ ", + "page_idx": 5 + }, + { + "type": "text", + "text": "12. $\\circledcirc$ 视频讲解在化简求 $( a + 3 b ) ^ { 2 } + ( 2 a + 3 b ) ( 2 a - 3 b ) + a ( 5 a - 6 b )$ 的值时，小明把 $a$ 的值看错后代入得结果为10,而小红代人正确的 $\\boldsymbol { a }$ 的值得正确的结果也是10,经探究后，发现所求代数式的值与 $b$ 的值无关，则他们俩代入的 $a$ 的值的商为", + "page_idx": 5 + }, + { + "type": "text", + "text": "三、解答题(本大题6小题，共52分)", + "text_level": 1, + "page_idx": 5 + }, + { + "type": "text", + "text": "13.(6分)计算： $( 1 ) ( - b ^ { 5 } ) \\bullet ( - b ^ { 3 } ) ^ { 4 } ; ( 2 ) ( - x ) ^ { 3 } \\bullet ( - x ^ { 2 } ) ^ { 2 } - ( - 2 x ^ { 3 } ) ^ { 2 } \\bullet x .$ ", + "page_idx": 5 + }, + { + "type": "text", + "text": "14.（8分)计算： $( 1 ) x ^ { 2 } - { \\frac { 1 } { 2 } } x ( 2 - 4 x ) , ( 2 ) ( x + 1 ) ( 7 - x ) - x ( x - 3 ) .$ ", + "page_idx": 5 + }, + { + "type": "text", + "text": "15.(8分)先化简，再求值： $4 ( x - 1 ) ^ { 2 } - ( 2 x + 3 ) ( 2 x - 3 )$ ,其中 $x { = } - 2$ ", + "page_idx": 6 + }, + { + "type": "text", + "text": "16.新考法阅读理解题（8分）规定 $a \\ast b = 2 ^ { a } \\times 2 ^ { b }$ ", + "page_idx": 6 + }, + { + "type": "text", + "text": "(1)求 $1 \\ast 3$ 的值； \n(2)若 $2 * ( x + 1 ) = 2 5 6$ ，求 $_ { \\mathcal { X } }$ 的值. ", + "page_idx": 6 + }, + { + "type": "text", + "text": "17.(10 分)如果代数式 $\\left( a x ^ { - } 1 \\right) \\left( x ^ { + } 7 \\right) - x ^ { 2 } + 7$ 化简后,不含有 $x ^ { 2 }$ 的项.", + "page_idx": 6 + }, + { + "type": "text", + "text": "(1)求 $a$ 的值；", + "page_idx": 6 + }, + { + "type": "text", + "text": "(2）求 $( a - 5 ) ( a + 5 ) + ( a - 5 ) ^ { 2 } - a ( 2 a + 1 )$ 的值. ", + "page_idx": 6 + }, + { + "type": "text", + "text": "18. $\\circledcirc$ 视频讲解（12分）已知 $P { = } ( x { - } y ) ^ { 5 } \\div ( y { - } x ) ^ { 5 } { \\bullet } ( x { - } y ) ^ { 2 } , Q { = } [ ( 2 x { - } y ) ^ { 2 } { + }$ $( 2 x + y ) ( 2 x - y ) + 8 x y ] \\div 2 x .$ （20 ", + "page_idx": 6 + }, + { + "type": "text", + "text": "(1)化简 $P$ 和 $Q$ ", + "page_idx": 6 + }, + { + "type": "text", + "text": "(2)当 $x = ( 3 - \\pi ) ^ { 0 }$ $y = ( - \\frac { 1 } { 2 } ) ^ { - 1 }$ 时，分别求 $P$ 和 $Q$ 的值.", + "page_idx": 6 + }, + { + "type": "text", + "text": "能力提优测试卷", + "text_level": 1, + "page_idx": 7 + }, + { + "type": "text", + "text": "时间：60分钟 满分：100分", + "page_idx": 7 + }, + { + "type": "table", + "img_path": "images/3b8bd4a46f5aefb1f7d4e32cd14874580f965f01be206cbaff8f45d6d7a34ed2.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 7 + }, + { + "type": "text", + "text": "一、选择题(每小题4分，共32分)", + "text_level": 1, + "page_idx": 7 + }, + { + "type": "text", + "text": "1.下列运算正确的是", + "page_idx": 7 + }, + { + "type": "text", + "text": "A. $a ^ { 2 } \\cdot a ^ { 3 } = a ^ { 6 }$ （20 B. $a ^ { 8 } \\div a ^ { 4 } = a ^ { 2 }$ （204号$\\mathrm { C } , 5 a - 3 a = 2 a$ $\\operatorname { D } . ( - 2 a b ^ { 2 } ) ^ { 2 } = - 4 a ^ { 2 } b ^ { 4 }$ ", + "page_idx": 7 + }, + { + "type": "text", + "text": "2.熔喷布，俗称口罩的“心脏”,是口罩中间的过滤层，能过滤细菌，防止病菌传播.经测量，医用外科口罩的熔喷布厚度约为0.000156米，数据0.000156用科学记数法表示为 （ ）", + "page_idx": 7 + }, + { + "type": "text", + "text": "A. $0 . 1 5 6 \\times 1 0 ^ { - 3 }$ （20 $\\mathrm { B } . 1 . 5 6 { \\times } 1 0 ^ { - 4 }$ （204号 $: 1 5 . 6 \\times 1 0 ^ { - 5 }$ （20 I $) . 1 . 5 6 \\times 1 0 ^ { - 3 }$ （20 ", + "page_idx": 7 + }, + { + "type": "text", + "text": "3.一个多项式除以 $4 . x ^ { 2 }$ ,商是一个三次三项式,那么这个多项式是 （）", + "page_idx": 7 + }, + { + "type": "text", + "text": "A.五次三项式 B.四次三项式C.五次四项式 D.四次四项式", + "page_idx": 7 + }, + { + "type": "text", + "text": "4.已知 $( a + b ) ^ { 2 } = 4 9 , a ^ { 2 } + b ^ { 2 } = 2 5$ ，则 $a b$ 的值为", + "page_idx": 7 + }, + { + "type": "text", + "text": "A.12 B.20 C.24 D.48 ", + "page_idx": 7 + }, + { + "type": "text", + "text": "5.若 $x { + } m$ 与 $x - 5$ 的乘积中不含 $\\mathcal { X }$ 的一次项,则 $m$ 的值为", + "page_idx": 7 + }, + { + "type": "text", + "text": "A.-5 B.5 C.0 D.3 ", + "page_idx": 7 + }, + { + "type": "text", + "text": "6 $\\begin{array} { r l r } { . ( - 5 a ^ { 2 } + 4 b ^ { 2 } ) ( } & { { } } & { ) = 2 5 a ^ { 4 } - 1 6 b ^ { 4 } } \\end{array}$ ,则括号内应填 ", + "page_idx": 7 + }, + { + "type": "text", + "text": "A. $5 a ^ { 2 } + 4 b ^ { 2 }$ $3 , 5 a ^ { 2 } - 4 b ^ { 2 }$ $\\mathrm { C } , - 5 a ^ { 2 } + 4 b ^ { 2 }$ $\\mathrm { D } . - 5 a ^ { 2 } - 4 b ^ { 2 }$ ", + "page_idx": 7 + }, + { + "type": "text", + "text": "7.长方体的长、宽、高分别是 $4 x - 3 , x$ 和 $2 x$ ，则它的体积等于", + "page_idx": 7 + }, + { + "type": "text", + "text": "A. $4 x ^ { 3 } - 3 x ^ { 2 }$ （204号 $\\mathrm { B } , 4 x ^ { 2 } - 3 x$ C. $8 x ^ { 3 } - 6 x ^ { 2 }$ D. $2 \\mathcal { x } ^ { 2 }$ （204号", + "page_idx": 7 + }, + { + "type": "text", + "text": "8.新考法代数推理 $\\circledcirc$ 视频讲解观察下列各式及其展开式：", + "page_idx": 7 + }, + { + "type": "equation", + "img_path": "images/1bcdad025af99f86b3be8cddfaa05266dbbf9a1287ace1df960bf1170d3bb920.jpg", + "text": "$$\n\\begin{array} { l } { { ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 } ; } } \\\\ { { \\ } } \\\\ { { ( a + b ) ^ { 3 } = a ^ { 3 } + 3 a ^ { 2 } b + 3 a b ^ { 2 } + b ^ { 3 } ; } } \\\\ { { \\ } } \\\\ { { ( a + b ) ^ { 4 } = a ^ { 4 } + 4 a ^ { 3 } b + 6 a ^ { 2 } b ^ { 2 } + 4 a b ^ { 3 } + b ^ { 4 } ; } } \\\\ { { \\ } } \\\\ { { ( a + b ) ^ { 5 } = a ^ { 5 } + 5 a ^ { 4 } b + 1 0 a ^ { 3 } b ^ { 2 } + 1 0 a ^ { 2 } b ^ { 3 } + 5 a b ^ { 4 } + b ^ { 5 } ; } } \\end{array}\n$$", + "text_format": "latex", + "page_idx": 7 + }, + { + "type": "text", + "text": "请你猜想 $( a + b ) ^ { 1 4 }$ 的展开式第三项的系数是", + "page_idx": 7 + }, + { + "type": "text", + "text": "A.66 B.78 C.91 D.105 ", + "page_idx": 7 + }, + { + "type": "text", + "text": "二、填空题(每小题4分，共16分)", + "text_level": 1, + "page_idx": 7 + }, + { + "type": "text", + "text": "9.已知 $a \\cdot ( - b ^ { 2 } ) ^ { 2 } { = } 3$ ，则 $a ^ { 2 } b ^ { 8 }$ 的值为", + "page_idx": 7 + }, + { + "type": "text", + "text": "10.在一次“学数学，爱数学\"的主题班会上，主持人小明同学亮出了 $A , B , C$ 三张卡片，上面分别写有 $\\begin{array} { r } { \\boxed { - 2 a b } \\boxed { a ^ { 2 } + 2 a b + b ^ { 2 } } \\boxed { ( a - b ) ^ { 2 } } } \\end{array}$ ,其中有两张卡片上的式子相乘,所得的积为一 $2 a ^ { 3 } b + 4 a ^ { 2 } b ^ { 2 } - 2 a b ^ { 3 } .$ 这两张卡片是 和", + "page_idx": 7 + }, + { + "type": "text", + "text": "11.如图，将完全相同的四个长方形纸片拼成一个大的正方形，用两种不同的方法表示这个大正方形的面积,则可以得出的一个等式为", + "page_idx": 7 + }, + { + "type": "image", + "img_path": "images/ecdef52c2a8dc6203035fd1b01da09e2ebb0ab20a8580483e15c2230bdc3388e.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 7 + }, + { + "type": "text", + "text": "12.若规定 $\\left| { \\begin{array} { l l } { a } & { b } \\\\ { c } & { d } \\end{array} } \\right| = a d - b c$ ,当 $\\left| { \\begin{array} { l l } { x - 2 } & { x - 2 } \\\\ { x + 2 } & { x - 2 } \\end{array} } \\right| = 0$ 时， $_ { \\mathcal { X } }$ 的值为", + "page_idx": 7 + }, + { + "type": "text", + "text": "三、解答题(本大题6小题，共52分)", + "text_level": 1, + "page_idx": 7 + }, + { + "type": "text", + "text": "13.（6分）计算： $( 1 ) a ^ { 4 } { \\bf \\lambda } { \\bf \\dot { \\alpha } } a ^ { 2 } - ( 3 a ^ { 3 } ) ^ { 2 } + ( - 4 a ^ { 2 } ) ^ { 3 } { \\bf \\dot { \\beta } } ( 2 ) \\left| - 5 \\right| - ( 1 - \\pi ) ^ { 0 } +$ （20 $( - { \\frac { 1 } { 3 } } ) ^ { - 1 } .$ ", + "page_idx": 7 + }, + { + "type": "text", + "text": "14.(8分)先化简，再求值： $( x - 3 ) ^ { 2 } + ( x + 3 ) ( x - 3 ) + 2 x ( 2 - x )$ ,其中 $x { = } - \\frac { 1 } { 2 }$ ", + "page_idx": 8 + }, + { + "type": "text", + "text": "15.(8分)小红在计算 $\\alpha ( 1 + a ) - ( a - 1 ) ^ { 2 }$ 时,解答过程如下：", + "page_idx": 8 + }, + { + "type": "text", + "text": "a(1+a)-(a-1)² \n=a+a²-(a²-1)·第一步$| = a + a ^ { 2 } - a ^ { 2 } - 1 \\cdots$ ·第二步=a-1……第三步", + "page_idx": 8 + }, + { + "type": "text", + "text": "小红的解答从第 步开始出错，请写出正确的解答过程.", + "page_idx": 8 + }, + { + "type": "text", + "text": "16.（8 分)小明在进行两个多项式的乘法运算时,不小心把乘 $\\frac { x + y } { 2 }$ 错抄成乘 $\\cdot \\frac { x } { 2 }$ 结果得到 $3 x ^ { 2 } - 5 x y$ ,则第一个多项式是多少？正确的结果应该是多少？", + "page_idx": 8 + }, + { + "type": "text", + "text": "17.(10分)已知等式 $\\left( x + a \\right) \\left( x + b \\right) = x ^ { 2 } + m x + 2 8$ 其中 $a , b , m$ 均为正整数，你认为 $m$ 可取哪些值？它与 $a , b$ 的取值有关吗？请你写出所有满足题意的 $m$ （20的值.", + "page_idx": 8 + }, + { + "type": "text", + "text": "18. $\\circledcirc$ 视讲（12分)一张长方形铁皮如图1所示，四个角都剪去边长为30（cm）的正方形，再四周折起，做成一个有底无盖的铁盒如图2所示,铁盒底面长方形的长是 $4 a ( \\mathrm { c m } )$ ，宽是 $3 a ( \\mathrm { c m } )$ ，这个无盖铁盒各个面的面积之和称为铁盒的表面积.", + "page_idx": 8 + }, + { + "type": "text", + "text": "(1)请用含 $a$ 的代数式表示图中原长方形铁皮的面积.", + "page_idx": 8 + }, + { + "type": "text", + "text": "(2)若要在铁盒的各个外表面漆上某种油漆，每元钱可漆的面积为 ${ \\frac { a } { 5 0 } } ( \\mathrm { c m } ^ { 2 } )$ ，则油漆这个铁盒需要多少钱？（用含 $a$ 的代数式表示）", + "page_idx": 8 + }, + { + "type": "image", + "img_path": "images/4ac5e58a9e1d159b053e76e8c0b1b24106f6a1fcab84cea3d4665b5abfe54cb9.jpg", + "img_caption": [ + "图1" + ], + "img_footnote": [], + "page_idx": 8 + }, + { + "type": "image", + "img_path": "images/471d45e26a1d2b0f9e1d3455956d33f1779c774a984bca221b90d10b7147b7e2.jpg", + "img_caption": [ + "图2" + ], + "img_footnote": [], + "page_idx": 8 + }, + { + "type": "text", + "text": "第二章 相交线与平行线", + "text_level": 1, + "page_idx": 9 + }, + { + "type": "text", + "text": "基础过关测试卷", + "text_level": 1, + "page_idx": 9 + }, + { + "type": "text", + "text": "时间：60分钟 满分：100分", + "page_idx": 9 + }, + { + "type": "table", + "img_path": "images/7676a0561c284f3e00526e49a8cf1ef5c1d674aadb9104050618227d3b9c669c.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 9 + }, + { + "type": "text", + "text": "一、选择题(每小题4分，共32分）", + "text_level": 1, + "page_idx": 9 + }, + { + "type": "text", + "text": "1.下面的四个图形中， $\\angle 1$ 与 $\\angle 2$ 是对顶角的是", + "page_idx": 9 + }, + { + "type": "image", + "img_path": "images/09bfb80391d7cada3c9ec9b3a50bcc972fe2be22482e9edb69b73cbcc4dd3341.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 9 + }, + { + "type": "text", + "text": "2.如果一个角的度数为 $6 0 ^ { \\circ }$ ,那么这个角的补角的度数是", + "page_idx": 9 + }, + { + "type": "text", + "text": "A.30° $\\mathrm { B } , 6 0 ^ { \\circ }$ （204号 C.90° D.120° ", + "page_idx": 9 + }, + { + "type": "text", + "text": "3.如图，在立定跳远中，体育老师是这样测量运动员的成绩的，将一块直角三角板的一边附在起跳线上，另一边与拉直的皮尺重合，这样做的理由是（ ）", + "page_idx": 9 + }, + { + "type": "text", + "text": "A.垂线段最短 B.过两点有且只有一条直线C.两点之间，线段最短 D.过一点可以作无数条直线", + "page_idx": 9 + }, + { + "type": "image", + "img_path": "images/fc6396eb4338900f4250315d3b56eefad5a7e9d27f45773d6194e94ab89b1e1f.jpg", + "img_caption": [ + "第3题图" + ], + "img_footnote": [], + "page_idx": 9 + }, + { + "type": "image", + "img_path": "images/b6551db5d54d147f026bdbe8e65a5ae41b948e73a18b43a33013d6bd04c360ec.jpg", + "img_caption": [ + "第4题图" + ], + "img_footnote": [], + "page_idx": 9 + }, + { + "type": "image", + "img_path": "images/172f6ccad68b806503983e360dcb7e6280b3d054388784683e1d2d20527bb09d.jpg", + "img_caption": [ + "第5题图" + ], + "img_footnote": [], + "page_idx": 9 + }, + { + "type": "text", + "text": "4.如图，直线 $a$ 和直线 $b$ 被直线 $c$ 所截，则 $\\angle 1$ 与 $\\angle 2$ 是", + "page_idx": 9 + }, + { + "type": "text", + "text": "A.同位角 B.内错角 C.同旁内角 D.对顶角", + "page_idx": 9 + }, + { + "type": "text", + "text": "5.如图，直线 $l _ { 1 } / / l _ { 2 }$ $\\angle 1 = 5 5 ^ { \\circ }$ ，则 $\\angle 2$ 的度数为", + "page_idx": 9 + }, + { + "type": "text", + "text": "A. $3 5 ^ { \\circ }$ （204号 B.45° C. $5 5 ^ { \\circ }$ （20 D.125° ", + "page_idx": 9 + }, + { + "type": "text", + "text": "6.如图，直线 $a$ 和直线 $b$ 被直线 $c$ 所截，下列条件不能判定直线 $a$ 与 $b$ 平行的是（", + "page_idx": 9 + }, + { + "type": "text", + "text": "A. $\\angle 1 = \\angle 3$ （204号 ${ \\mathrm { B } } _ { \\cdot } \\angle 2 + \\angle 4 = 1 8 0 ^ { \\circ }$ C. $\\angle 1 = \\angle 4$ （204号 D. $\\angle 3 = \\angle 4$ （20 ", + "page_idx": 9 + }, + { + "type": "image", + "img_path": "images/91164ed58178049018dbc2f1d9b78c460d93442e77e103f840bb4930a4d29bd4.jpg", + "img_caption": [ + "第6题图" + ], + "img_footnote": [], + "page_idx": 9 + }, + { + "type": "image", + "img_path": "images/5fbf38418b9e922932108778435b4529a39ba2195adc5abe2cc04689eb1a0165.jpg", + "img_caption": [ + "第7题图" + ], + "img_footnote": [], + "page_idx": 9 + }, + { + "type": "image", + "img_path": "images/35309b3e10db9ba87ed1b981dce53a1e4a08e26526cf858e4d0d225d5866ba5d.jpg", + "img_caption": [ + "第8题图" + ], + "img_footnote": [], + "page_idx": 9 + }, + { + "type": "text", + "text": "7.如图，直线 $l _ { 1 } , l _ { 2 } , l _ { 3 }$ 相交于点 $O ,$ 直线 $l _ { 4 } / / l _ { 1 }$ ，若 $\\angle 1 = 1 2 4 ^ { \\circ }$ $\\angle 2 = 8 8 ^ { \\circ }$ ，则 $\\angle 3$ 的 度数为 ", + "page_idx": 9 + }, + { + "type": "text", + "text": "A $2 6 ^ { \\circ }$ （204号 B. $3 6 ^ { \\circ }$ C.46° D.56° ", + "page_idx": 9 + }, + { + "type": "text", + "text": "8. $\\circledcirc$ 视频讲解如图所示，与 $\\angle \\alpha$ 构成同位角的角的个数为", + "page_idx": 9 + }, + { + "type": "text", + "text": "A.1 B.2 C.3 D.4 ", + "page_idx": 9 + }, + { + "type": "text", + "text": "二、填空题(每小题4分，共16分)", + "text_level": 1, + "page_idx": 9 + }, + { + "type": "text", + "text": "9.如图所示，直线AB 和 $C D$ 相交于点 $O$ ，若 $\\angle A O D$ 与 $\\angle B O C$ 的度数之和为$2 1 6 ^ { \\circ }$ ，则 $\\angle A O C$ 的度数为", + "page_idx": 9 + }, + { + "type": "image", + "img_path": "images/a7c545ac432eacbb1f095732f2ac012aa58a6fcf0a4b3a545bc7c4cf5571f578.jpg", + "img_caption": [ + "第9题图" + ], + "img_footnote": [], + "page_idx": 9 + }, + { + "type": "image", + "img_path": "images/89a47e96a23454043b98d7de694711411966dd8e19f7ce38df84c205e94e76c9.jpg", + "img_caption": [ + "第10题图" + ], + "img_footnote": [], + "page_idx": 9 + }, + { + "type": "image", + "img_path": "images/0703d9e526f150120fafab88f4c90b2d747dd0cec779a9a0e6bc6f09554eb308.jpg", + "img_caption": [ + "第11题图" + ], + "img_footnote": [], + "page_idx": 9 + }, + { + "type": "image", + "img_path": "images/7f5b9c17c03d868a3191d665c09d5f4945b02ea8b2dc82041decb5ad56fdcdc9.jpg", + "img_caption": [ + "第12题图" + ], + "img_footnote": [], + "page_idx": 9 + }, + { + "type": "text", + "text": "10.如图，直线 $a / / b$ ，直线 $c$ 与直线 $a , b$ 分别相交于点 $A , B ,$ 若 $\\angle 1 = 6 0 ^ { \\circ }$ ，则 $\\angle 2$ 的度数为 ", + "page_idx": 9 + }, + { + "type": "text", + "text": "11.如图, $\\angle A = 7 0 ^ { \\circ }$ ，点 $O$ 是 $A B$ 上一点，直线 $O D$ 与AB的夹角 $\\angle B O D { = } 8 2 ^ { \\circ }$ ，要 使 $O D / / A C$ ，则直线 $O D$ 绕点 $O$ 按逆时针方向至少旋转__.度. ", + "page_idx": 9 + }, + { + "type": "text", + "text": "12.如图,AB//CD, $\\angle D = 7 5 ^ { \\circ }$ $\\angle C A D : \\angle B A C = 2 : 1 ,$ 则 $\\angle C A D$ 的度数为", + "page_idx": 9 + }, + { + "type": "text", + "text": "三、解答题(本大题6小题，共52分)", + "text_level": 1, + "page_idx": 10 + }, + { + "type": "text", + "text": "13.(6 分)(1)如果把下图看成是直线 $A B , E F$ 被直线 $C D$ 所截,那么 $\\angle 1$ 与 $\\angle 2$ 是一对什么角？ $\\angle 2$ 与 $\\angle 3$ 呢？", + "page_idx": 10 + }, + { + "type": "text", + "text": "(2)如果把下图看成是直线 $A B , C D$ 被直线 $E F$ 所截,那么 $\\angle 2$ 与 $\\angle 5$ 是一对什么角？ $\\angle 4$ 与 $\\angle 5$ 呢？ $\\angle 5$ 与 $\\angle 6$ 呢？", + "page_idx": 10 + }, + { + "type": "image", + "img_path": "images/8db2e574dc67bba88b40bc4bd6c72772a8ca950c03ebfb22f8650375bee08dc3.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 10 + }, + { + "type": "text", + "text": "14.(8分)如图,利用尺规作图,在 $A C$ 的上方作 $\\angle C A D = \\angle A C B$ ,并说明AD// CB.(尺规作图要求保留作图痕迹,不写作法) ", + "page_idx": 10 + }, + { + "type": "image", + "img_path": "images/8375a430c0dc71ec1d71a2d66fefd88eb325471d8b9eeeb14b16c40146bc402a.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 10 + }, + { + "type": "text", + "text": "15.(8分)如图所示，点 $O$ 为直线 $B D$ 上一点， $O C \\bot O A$ ，垂足为点 $\\quad O , \\angle C O D =$ $2 \\angle B O C$ ，求 $\\angle A O B$ 的度数.", + "page_idx": 10 + }, + { + "type": "image", + "img_path": "images/47510441fb6eba02ad4fe86fd36e42f373e75f6162192f5d537aa58ad91540f7.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 10 + }, + { + "type": "text", + "text": "16.(8分)如图 $, A D / / B C , \\angle E A D = \\angle C , \\angle F E C = \\angle B A E ,$ $\\angle E F C = 5 0 ^ { \\circ }$ ", + "page_idx": 10 + }, + { + "type": "text", + "text": "(1)试说明线段 $A E$ 与线段 $C D$ 平行；", + "page_idx": 10 + }, + { + "type": "text", + "text": "(2)求 $\\angle B$ 的度数. ", + "page_idx": 10 + }, + { + "type": "image", + "img_path": "images/c717d5c11b67ec92f2ac5f0ef8ce816efa0d52e5e4beb39e003fd4e179e8c887.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 10 + }, + { + "type": "text", + "text": "17.(10分)如图， $B$ 处在 $A$ 处的南偏西 $4 5 ^ { \\circ }$ 方向， $C$ 处在 $B$ 处的北偏东 $8 0 ^ { \\circ }$ 方向.", + "page_idx": 10 + }, + { + "type": "text", + "text": "(1)求 $\\angle A B C$ 的度数. ", + "page_idx": 10 + }, + { + "type": "text", + "text": "(2)要使 $C D / / A B , D$ 处应在 $C$ 处的什么方向？", + "page_idx": 10 + }, + { + "type": "image", + "img_path": "images/d68deb47892e9e184f3fd8876399a1a5286deb2b82416cab43b0208f02b3969d.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 10 + }, + { + "type": "text", + "text": "18. $\\circledcirc$ 视频讲解（12分）（1） $\\textcircled{1}$ 如图1,已知 $A B / / C D$ $\\angle A B C = 6 0 ^ { \\circ }$ ，根据，可得 $B C D =$ 。；", + "page_idx": 10 + }, + { + "type": "text", + "text": "$\\textcircled{2}$ 如图2,在 $\\textcircled{1}$ 的条件下,若 $C M$ 平分 $\\angle B C D$ ，则 $\\ _ { - } B C M =$ 。$\\textcircled{3}$ 如图3，在 $\\textcircled{1}$ 和 $\\textcircled{2}$ 的条件下,若 $C N \\bot C M$ 则 $\\angle B C N =$ ", + "page_idx": 10 + }, + { + "type": "text", + "text": "(2)尝试解决下面的问题：如图 $4 , A B / / C D$ $\\angle B = 4 0 ^ { \\circ }$ ,CN是 $\\angle B C E$ 的平分线， $C N \\bot C M ,$ 求 $\\angle B C M$ 的度数.", + "page_idx": 10 + }, + { + "type": "image", + "img_path": "images/59f4f05a35ec3e2742ee566a14c2141eb76caf3765da9206ceb645aaabeecc9e.jpg", + "img_caption": [ + "图1 图2" + ], + "img_footnote": [], + "page_idx": 10 + }, + { + "type": "image", + "img_path": "images/3b4311782887b174853f5596e1952c5e6430b135cfd859f9661ed93b7b5dfdcc.jpg", + "img_caption": [ + "图3 " + ], + "img_footnote": [], + "page_idx": 10 + }, + { + "type": "image", + "img_path": "images/3b3b7271fae7ce7bab8dd69f5296b2ba9a3764cba158e476b000b79930d11b4d.jpg", + "img_caption": [ + "图4" + ], + "img_footnote": [], + "page_idx": 10 + }, + { + "type": "table", + "img_path": "images/19a73c9c596d299c66e117109116374967a92ff83f948eac34ba211ca92719fb.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
学校
", + "page_idx": 11 + }, + { + "type": "text", + "text": "能力提优测试卷", + "text_level": 1, + "page_idx": 11 + }, + { + "type": "text", + "text": "时间：60分钟 满分：100分", + "page_idx": 11 + }, + { + "type": "table", + "img_path": "images/28eb579dc350af3c0b04deccd4a76a269a1ea3cf88a8918a8ed0595c988a4368.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 11 + }, + { + "type": "text", + "text": "一、选择题(每小题4分，共32分)", + "text_level": 1, + "page_idx": 11 + }, + { + "type": "text", + "text": "1.下列说法不正确的是", + "page_idx": 11 + }, + { + "type": "table", + "img_path": "images/213f0e39b25bb6f5fc53e19186ad97d9b3a9f4b458866814691d96d4efc483bd.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
姓名
", + "page_idx": 11 + }, + { + "type": "table", + "img_path": "images/7b3c18203ae8bc6fc7f6d30d48a08fc3b99036fb8d80a9b4a116c0492e3fa3c2.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
学号
", + "page_idx": 11 + }, + { + "type": "text", + "text": "A.对顶角相等 \nB.若 $\\angle \\alpha = 1 4 ^ { \\circ } 2 6 ^ { \\prime }$ ，则 $\\angle \\alpha$ 的余角为 $7 5 ^ { \\circ } 3 4 ^ { \\prime }$ \nC.如果 $a / / b , a / / c$ ,那么 $b / / c$ \nD.如果 $a \\perp b , a \\perp c$ ,那么 $b \\perp c$ ", + "page_idx": 11 + }, + { + "type": "text", + "text": "2.如图，与 $\\angle 1$ 是同旁内角的是", + "page_idx": 11 + }, + { + "type": "text", + "text": "A. $\\angle 2$ B. $\\angle 3$ （20 C.∠4 D.∠5 ", + "page_idx": 11 + }, + { + "type": "image", + "img_path": "images/332e60154b6f3e43a3bbb5388b06a36d38ee5df1e5fbbf090a393e27f3028460.jpg", + "img_caption": [ + "第2题图" + ], + "img_footnote": [], + "page_idx": 11 + }, + { + "type": "image", + "img_path": "images/07bb78fa0f38d2df5b32bafe41f2587c5e9f9f317e7ffdb7b6c4ef32956216b5.jpg", + "img_caption": [ + "第3题图" + ], + "img_footnote": [], + "page_idx": 11 + }, + { + "type": "image", + "img_path": "images/18a8baa75a52fd0254186604a494678fca22efafb70779e7bdea77dfcf8d7f1f.jpg", + "img_caption": [ + "第4题图" + ], + "img_footnote": [], + "page_idx": 11 + }, + { + "type": "image", + "img_path": "images/fc050aec774b96e9a21717f13e3e7dde74ac3eae765a30d8319d135c05ff9238.jpg", + "img_caption": [ + "第5题图" + ], + "img_footnote": [], + "page_idx": 11 + }, + { + "type": "text", + "text": "3.如图，直线 $A B , C D$ 相交于点 $O , O E \\bot A B$ 于点 $O , \\angle C O E { = 5 5 ^ { \\circ } }$ ，则 $\\angle B O D$ 的 度数是 ", + "page_idx": 11 + }, + { + "type": "text", + "text": "A. $3 5 ^ { \\circ }$ （20 B.45° C.30° D.40° ", + "page_idx": 11 + }, + { + "type": "text", + "text": "4.如图， $A B / / C D$ $\\angle F E B = 7 0 ^ { \\circ }$ $\\angle E F D$ 的平分线 $F G$ 交 $A B$ 于点 $G$ ，则 $\\angle E F G$ 的度数为 （ ", + "page_idx": 11 + }, + { + "type": "text", + "text": "A. $6 3 ^ { \\circ }$ （20 B.53° C.65° D.55° ", + "page_idx": 11 + }, + { + "type": "text", + "text": "5.如图，直线 $A C / / B D , A O , B O$ 分别是∠BAC, $\\angle A B D$ 的平分线，则 $\\angle B A O$ 与$\\angle A B O$ 的关系一定为 （ ）", + "page_idx": 11 + }, + { + "type": "text", + "text": "A.互余 B.相等 C.互补 D.无法判断 ", + "page_idx": 11 + }, + { + "type": "text", + "text": "6.如图， $A B / / C D$ ，直线 $E F$ 与 $A B , C D$ 分别交于点 $G , H , I G \\bot E F$ 于点 $G$ ，$\\angle A G I = 4 3 ^ { \\circ }$ ，则 $\\angle E H D$ 的度数为 （ ）", + "page_idx": 11 + }, + { + "type": "text", + "text": "A. $5 7 ^ { \\circ }$ （204号 B.53° C.47° D. $4 3 ^ { \\circ }$ ", + "page_idx": 11 + }, + { + "type": "image", + "img_path": "images/bc1fe50ae877da38d6391aa2cc376cc1c465d26dafe2ceadb65d11fe414732b7.jpg", + "img_caption": [ + "第6题图" + ], + "img_footnote": [], + "page_idx": 11 + }, + { + "type": "image", + "img_path": "images/60eee65a70c665e81791f2c056c7087cf32c545520c3eb0c1de6e3be16a18662.jpg", + "img_caption": [ + "第7题图" + ], + "img_footnote": [], + "page_idx": 11 + }, + { + "type": "image", + "img_path": "images/14f1642520409e0e620e9b695a564541f244d266cada4fbc8fb64c1ab22b8a0a.jpg", + "img_caption": [ + "第8题图" + ], + "img_footnote": [], + "page_idx": 11 + }, + { + "type": "text", + "text": "7.如图， $A B / / C D$ ,则根据图中标注的角，下列关系中,成立的是", + "page_idx": 11 + }, + { + "type": "text", + "text": "A $\\angle 1 = \\angle 3$ $3 . \\angle 2 + \\angle 3 = 1 8 0 ^ { \\circ }$ C. $\\angle 2 + \\angle 4 < 1 8 0 ^ { \\circ }$ $\\mathrm { D } . \\angle 3 + \\angle 5 = 1 8 0 ^ { \\circ }$ ", + "page_idx": 11 + }, + { + "type": "text", + "text": "8. 新考法真实问题情境视频讲解如图，这是一汽车探照灯纵剖面，从位于点$O$ 的灯泡发出的两束光线 $O B , O C$ 经过反射以后平行射出，若 $\\angle A B O = \\alpha$ ，$\\angle D C O { = } \\beta$ 则 $\\angle B O C$ 的度数是 （）", + "page_idx": 11 + }, + { + "type": "equation", + "img_path": "images/f088e8baaf1c90cfefafc4e8b350f2f8cfd3a168f6d44203837cadc9e212cea7.jpg", + "text": "$$\n\\mathrm { A } . \\alpha + \\beta \\qquad \\mathrm { B } . 1 8 0 ^ { \\circ } - \\alpha \\qquad \\mathrm { C } . \\frac { 1 } { 2 } ( \\alpha + \\beta ) \\qquad \\mathrm { D } . 9 0 ^ { \\circ } + \\alpha + \\beta\n$$", + "text_format": "latex", + "page_idx": 11 + }, + { + "type": "text", + "text": "二、填空题(每小题4分，共16分)", + "text_level": 1, + "page_idx": 11 + }, + { + "type": "text", + "text": "9.如图，已知 $\\angle \\alpha = 6 0 ^ { \\circ }$ $\\angle \\beta = 7 5 ^ { \\circ }$ ,根据尺规作图的痕迹，可得 $\\angle A O C$ 的度数为", + "page_idx": 11 + }, + { + "type": "image", + "img_path": "images/88c1fda811c8bf79a9007645cc017af0f7f52a851cc3b3eaab1df757afbbac25.jpg", + "img_caption": [ + "第9题图" + ], + "img_footnote": [], + "page_idx": 11 + }, + { + "type": "image", + "img_path": "images/bcb1a86fa5f96dceded4306f00855951fa8cf1dbd48a1ee4e1e525590985fd6c.jpg", + "img_caption": [ + "第10题图" + ], + "img_footnote": [], + "page_idx": 11 + }, + { + "type": "text", + "text": "10.如图,要使得 $A D / / B F$ ，则需要添加的条件是 (写出一个即可).", + "page_idx": 11 + }, + { + "type": "text", + "text": "11.如图,已知AE//BC, $\\angle B A C = 1 0 0 ^ { \\circ }$ $\\angle D A E = 5 0 ^ { \\circ }$ ，则 $\\angle C =$ ", + "page_idx": 11 + }, + { + "type": "image", + "img_path": "images/a216bdbc67b8b1a80bce392e2beee3cb6f3539d673a1ac866a2607398b044152.jpg", + "img_caption": [ + "第11题图" + ], + "img_footnote": [], + "page_idx": 11 + }, + { + "type": "image", + "img_path": "images/56dd743c1c148a11b2c243aea3431640a12631ecdd9e67b15f43ac32de026cc7.jpg", + "img_caption": [ + "第12题图" + ], + "img_footnote": [], + "page_idx": 11 + }, + { + "type": "text", + "text": "12如图，点 $A , C , F , B$ 在同一直线上, $C E$ 平分 $\\angle A C D , F G / / C D$ ，若 $\\angle E C A = \\alpha ^ { \\circ }$ ， 则 $\\angle G F B$ 的度数为 （用含 $\\alpha$ 的代数式表示). ", + "page_idx": 11 + }, + { + "type": "text", + "text": "三、解答题(本大题6小题，共52分)", + "text_level": 1, + "page_idx": 12 + }, + { + "type": "text", + "text": "13.(6分)若一个角的余角比它的补角的 $\\frac 1 2$ 小 $2 0 ^ { \\circ }$ ,求这个角的度数.", + "page_idx": 12 + }, + { + "type": "text", + "text": "14.(8分)如图,在四边形 $A B C D$ 中， $B D$ 是连接两个顶点的线段,若 $\\angle 1 = \\angle 2$ ，$\\angle A = 5 5 ^ { \\circ } 1 6 ^ { \\prime }$ ，求 $\\angle A D C$ 的度数.", + "page_idx": 12 + }, + { + "type": "image", + "img_path": "images/c5c0f0993a8fa7f653f8d2a4adc1640ee61714313fd0ca568f607bfc32f36aad.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 12 + }, + { + "type": "text", + "text": "15.(8分)如图 $, A B / / C D , B G$ 平分∠ABD, $\\angle E D F { = } 7 0 ^ { \\circ }$ ，求 $\\angle F B G$ 的度数. ", + "page_idx": 12 + }, + { + "type": "text", + "text": "A FBGC D E", + "page_idx": 12 + }, + { + "type": "text", + "text": "16.(8分)如图， $C D \\perp A B , E F \\perp A B , \\angle E = \\angle E M C , C D$ 是 $\\angle A C B$ 的平分线吗? 请说明理由. ", + "page_idx": 12 + }, + { + "type": "image", + "img_path": "images/755f858ea92f95d87afe6a0237eb29ed31baa404119df315fbc46caa85ad399a.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 12 + }, + { + "type": "text", + "text": "17.(10 分)如图,已知 $A D / / B C , \\angle A = \\angle B .$ （20", + "page_idx": 12 + }, + { + "type": "text", + "text": "(1)试判断 $A F$ 与 $B E$ 的位置关系,并说明理由；", + "page_idx": 12 + }, + { + "type": "text", + "text": "(2)若 $\\angle B O D { = } 3 \\angle B$ ，求 $\\angle A$ 的度数. ", + "page_idx": 12 + }, + { + "type": "image", + "img_path": "images/b2a7e854eb7ea0fed18167c8c1bc2d43e6da8b69dbb6be2fc8e4d613dff394b2.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 12 + }, + { + "type": "text", + "text": "18. $\\circledcirc$ 视频讲解（12分）如图，点 $C$ 在 $\\angle A O B$ 的一边 $O A$ 上，过点 $C$ 的直线 $D E$ /$O B , C F$ 平分 $\\_ A C D , C G \\bot C F$ ，垂足为点 $C .$ ", + "page_idx": 12 + }, + { + "type": "text", + "text": "(1)若 $\\angle A O B = 4 0 ^ { \\circ }$ ，求 $\\angle E C F$ 的度数. ", + "page_idx": 12 + }, + { + "type": "text", + "text": "(2)试说明： $C G$ 平分 $\\angle O C D$ ", + "page_idx": 12 + }, + { + "type": "text", + "text": "(3）当 $\\angle A O B$ 为多少度时， $C D$ 平分 $\\angle O C F ?$ 请说明理由.", + "page_idx": 12 + }, + { + "type": "image", + "img_path": "images/a93e0a7d161995dd4ca25e09ea8a0f9156650a6357b1377fec9adc11b2da533c.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 12 + }, + { + "type": "text", + "text": "月考测试卷（一）", + "text_level": 1, + "page_idx": 13 + }, + { + "type": "text", + "text": "时间：90分钟 满分：120分考试范围：整式的乘除", + "page_idx": 13 + }, + { + "type": "table", + "img_path": "images/cf1d7489f00b0516db5668e4a7aee5fe273808b814fcf65e368d43e624d105fa.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 13 + }, + { + "type": "text", + "text": "一、选择题(本大题共10小题，每小题3分，共30分)", + "text_level": 1, + "page_idx": 13 + }, + { + "type": "text", + "text": "1.下列运算结果为 $a ^ { 6 }$ 的是", + "page_idx": 13 + }, + { + "type": "text", + "text": "A. $a ^ { 2 } \\cdot a ^ { 3 }$ B. $a ^ { 1 2 } \\div a ^ { 2 }$ （20 C. $( a ^ { 3 } ) ^ { 2 }$ （204号 D. $( \\frac { 1 } { 2 } a ^ { 3 } ) ^ { 2 }$ ", + "page_idx": 13 + }, + { + "type": "text", + "text": "2.某种病毒的平均直径大约在 $8 0 \\sim 1 4 0$ 纳米之间，已知1纳米 $= 1 0 ^ { - 9 }$ 米，则140纳米用科学记数法可表示为 （", + "page_idx": 13 + }, + { + "type": "text", + "text": "A. $1 4 0 \\times 1 0 ^ { - 9 }$ 米 B. 1. $4 \\times 1 0 ^ { - 7 }$ 米C. $1 4 \\times 1 0 ^ { - 8 }$ 米 D. 1. $4 \\times 1 0 ^ { - 8 }$ 米", + "page_idx": 13 + }, + { + "type": "text", + "text": "3.如图,阴影部分的面积是${ \\frac { 7 } { 2 } } x y$ B ${ \\frac { 9 } { 2 } } x y$ C.4xy D.2xy", + "page_idx": 13 + }, + { + "type": "image", + "img_path": "images/55c7078cd429e2c982ad8e76aa7baa8cf6a53417ce7bb7dc538d5c1195669578.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 13 + }, + { + "type": "text", + "text": "", + "page_idx": 13 + }, + { + "type": "text", + "text": "4.计算 $2 8 x ^ { 4 } y ^ { 2 } \\div ( - 7 x ^ { 3 } y )$ 的正确结果是", + "page_idx": 13 + }, + { + "type": "text", + "text": "A.4xy B.-4xy C.4x²y D.4xy² ", + "page_idx": 13 + }, + { + "type": "text", + "text": "5.化简 $x ( 2 x - 1 ) - x ^ { 2 } ( 2 - x )$ 的结果是 $\\mathrm { A } , - x ^ { 3 } - x \\qquad \\mathrm { B } , x ^ { 3 } - x \\qquad \\mathrm { C } , - x ^ { 2 } - 1$ D. $x ^ { 3 } - 1$ （20 ", + "page_idx": 13 + }, + { + "type": "text", + "text": "", + "page_idx": 13 + }, + { + "type": "text", + "text": "6.已知 $m + n { = } 2 , m n { = } - 2$ ，则 $( 1 + m ) ( 1 + n )$ 的值为", + "page_idx": 13 + }, + { + "type": "text", + "text": "A.6 B.-2 C.0 D.1 ", + "page_idx": 13 + }, + { + "type": "text", + "text": "7.如图，从边长为 $( a + 4 ) { \\mathrm { ~ c m } }$ 的正方形纸片中剪去一个边长为 $( a + 1 ) \\ \\mathrm { c m }$ 的小正方形（ $_ { ( a > 0 ) }$ ，剩余部分沿虚线又剪拼成一个长方形(不重叠、无缝隙)，则长方形的面积为 （ ）", + "page_idx": 13 + }, + { + "type": "text", + "text": "A. $( 2 a ^ { 2 } + 5 a$ ）cm² B.( $3 a + 1 5$ ）cm² C. $\\left( 6 a + 1 5 \\right) ~ \\mathrm { c m } ^ { 2 }$ D $\\ : ( 8 a + 1 5 ) \\ : \\mathrm { ~ c m ^ { 2 } }$ ", + "page_idx": 13 + }, + { + "type": "image", + "img_path": "images/4e0e9cd28098091d41436ac2d2af9075c51c928ee09515918460739a3df88a2f.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 13 + }, + { + "type": "text", + "text": "8.豆豆认为下列括号内都可以填 $a ^ { 4 }$ ,你认为使等式成立的只能是", + "page_idx": 13 + }, + { + "type": "text", + "text": "A $\\mathrm { ~ ~ \\cdot ~ } a ^ { 1 2 } = ( \\mathrm { ~ ~ \\rho ~ } ) ^ { 2 }$ $\\mathrm { B } , a ^ { 1 2 } = ( \\phantom { a } ) ^ { 3 }$ $\\mathrm { ~ \\cal ~ C ~ } _ { \\bullet } a ^ { 1 2 } = ( \\mathrm { ~ ~ { ~ \\partial ~ } ~ } ) ^ { 4 }$ （204号 $\\mathrm { D } . { a } ^ { 1 2 } = ( \\phantom { a } ) ^ { 6 }$ ", + "page_idx": 13 + }, + { + "type": "text", + "text": "9.若 $9 ^ { x + 2 } 3 ^ { x + 5 } = 2 7 ^ { 2 x - 1 }$ ，则 $\\mathcal { X }$ 等于 ", + "page_idx": 13 + }, + { + "type": "text", + "text": "A.3 B.-3 C.4 D.-4 ", + "page_idx": 13 + }, + { + "type": "text", + "text": "10. $\\circledcirc$ 视频讲解对于等式 $( a + b ) ^ { 2 } = a ^ { 2 } + b ^ { 2 }$ ,甲、乙、丙三人有不同看法，则下列说法正确的是 （ ）", + "page_idx": 13 + }, + { + "type": "table", + "img_path": "images/f89b975cd6715b71ec85b09dd27930003bb0c5819bd947d857cf605b3a016bb6.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
甲：无论a和b取何值， 等式均不能成立.乙：只有当a=0时，丙：当a=0或b=0 等式才成立.时,等式成立.
", + "page_idx": 13 + }, + { + "type": "text", + "text": "A.只有甲正确 B.只有乙正确C.只有丙正确 D.三人说法均不正确", + "page_idx": 13 + }, + { + "type": "text", + "text": "二、填空题(本大题共5小题，每小题3分，共15分)", + "text_level": 1, + "page_idx": 13 + }, + { + "type": "text", + "text": "11.一颗人造地球卫星的速度为 $2 . 8 8 \\times 1 0 ^ { 7 }$ 米/时，一架喷气式飞机的速度为$1 . 8 \\times 1 0 ^ { 6 }$ 米/时，则这颗人造地球卫星的速度是这架喷气式飞机的速度的倍.", + "page_idx": 13 + }, + { + "type": "text", + "text": "12.如果一个三角形的底边长为 $2 x ^ { 2 } y + x y - y ^ { 2 }$ ，这条边上的高为 $6 x y$ ,那么这个三角形的面积为", + "page_idx": 13 + }, + { + "type": "text", + "text": "13.新考法阅读理解题规定一种新运算“ $\\divideontimes$ ”,对于任意实数 $a$ 和 $b$ ，有 $a \\times b$ $= a \\div b + 1$ ，则 $( 6 x ^ { 3 } y - 3 x y ^ { 2 } ) * 3 x y =$ ", + "page_idx": 13 + }, + { + "type": "text", + "text": "14.随着数学学习的深人,数系不断扩充,引入新数 $i$ ，规定 $i ^ { 2 } = - 1$ ，并且新数 $i$ 的加、减、乘、除运算和整式的加、减、乘、除运算类似，则 $\\left( 1 + i \\right) \\left( 2 - i \\right)$ 的运算结果是", + "page_idx": 14 + }, + { + "type": "text", + "text": "15.新考法代数推理 $\\circledcirc$ 视频讲解请你计算： $( 1 - x ) ( 1 + x ) , ( 1 - x ) ( 1 + x + x ^ { 2 } ) , \\cdots ,$ 猜想 $( 1 - x ) ( 1 + x + x ^ { 2 } + \\cdots + x ^ { 2 0 2 4 } )$ 的结果是 ", + "page_idx": 14 + }, + { + "type": "text", + "text": "三、解答题(本大题共8小题，满分75分)", + "text_level": 1, + "page_idx": 14 + }, + { + "type": "text", + "text": "16.(10分)（1)计算： $a ^ { 3 } \\bullet a + ( - a ^ { 2 } ) ^ { 3 } \\div a ^ { 2 }$ (2）计算： $( \\pi - 3 . 1 4 ) ^ { \\circ } - \\Bigl ( - \\frac { 1 } { 2 } \\Bigr ) ^ { - 2 } + ( - 2 ) ^ { 2 } \\times ( - 1 ) ^ { 2 0 2 4 } .$ ", + "page_idx": 14 + }, + { + "type": "text", + "text": "", + "page_idx": 14 + }, + { + "type": "text", + "text": "17.(9分)已知 $a ^ { m } = 4 , a ^ { n } = \\frac { 1 } { 2 }$ ，求 $a ^ { 2 m - 3 n }$ 的值. ", + "page_idx": 14 + }, + { + "type": "text", + "text": "18.(9分)先化简,再求值： $( 2 a + b ) ( 2 a - b ) - ( 4 a b ^ { 3 } - 8 a ^ { 3 } b ) \\div 2 a b \\cdot$ 其中 $a = - 1$ $b { = } - 2 .$ （204号", + "page_idx": 14 + }, + { + "type": "image", + "img_path": "images/c754a00b53adf7d98dd422ef604f3c9d1e2c30444d0237522348b5ba7a369518.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 15 + }, + { + "type": "text", + "text": "19.(9分)已知 $( x { + } 7 )$ 和 $( x { - } 3 )$ 的积为 $x ^ { 2 } + m x ^ { - } n$ ，求 $( 5 m - n ) ^ { 2 0 2 4 }$ 的值. ", + "page_idx": 15 + }, + { + "type": "text", + "text": "20.（9分)已知 $\\ x ^ { 2 } + 2 x { = } 4$ ，求代数式 $x ( x - 2 ) ^ { 2 } - x ^ { 2 } ( x - 6 ) - 3$ 的值. ", + "page_idx": 15 + }, + { + "type": "text", + "text": "21.(9分)小颖在计算某一个多项式乘一 $3 \\mathcal { x } ^ { 2 }$ 时，因抄错运算符号，写成了加上$- 3 x ^ { 2 }$ ，结果得到 $x ^ { 2 } - 4 x + 1$ ,那么原题正确的计算结果是什么？请计算出正确的结果.", + "page_idx": 15 + }, + { + "type": "text", + "text": "22.（10分)如图，长方形ABCD的周长为16,以长方形的四条边为边长向外作四个正方形，若这四个正方形的面积之和为68,求长方形ABCD的面积.", + "page_idx": 16 + }, + { + "type": "image", + "img_path": "images/fd0b649c896175f81415e08d645f3e92a3e847d55cfb5c7c73e3166f1e87da47.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 16 + }, + { + "type": "text", + "text": "23. $\\circledcirc$ 视讲(10分)我们通常用作差法比较代数式的大小.例如：已知 $M { = } 2 x +$ $3 , N { = } 2 { x } { + } 1$ ，比较 $M$ 和 $N$ 的大小.先求 $M - N$ ，若 $M { - } N { > } 0$ ，则 $M { > } N$ ；若$M { - } N { < } 0$ ，则 $M { < } N$ ；若 $M { - } N { = } 0$ ，则 $M { = } N$ ,反之亦成立.本题中,因为 $M$ $- N { = } 2 x { + } 3 { - } ( 2 x { + } 1 ) { = } 2 { > } 0$ ，所以 $M { \\ > } N .$ ", + "page_idx": 16 + }, + { + "type": "text", + "text": "(1)如图1,这是一个边长为 $a$ 的正方形,将正方形一边不变,另一边增加4,得到如图2所示的新长方形，此长方形的面积为 $S _ { 1 }$ ;将图1中正方形边长增加2得到如图3所示的新正方形,此正方形的面积为 $S _ { 2 }$ $\\textcircled{1}$ 用含 $a$ 的代数式分别表示 $S _ { 1 }$ 和 $S _ { 2 }$ （结果需要化简）；$\\textcircled{2}$ 请用作差法比较 $S _ { 1 }$ 与 $S _ { 2 }$ 的大小.", + "page_idx": 16 + }, + { + "type": "text", + "text": "(2）若 $M { = } a ^ { 2 }$ ， $N { = } 4 { - } ( a { + } 1 ) ^ { 2 }$ ，且 $M { = } N$ 求 $\\alpha ( a { + } 1 )$ 的值. ", + "page_idx": 16 + }, + { + "type": "image", + "img_path": "images/c6829e9bc8ea0bef99a2be237868c01d2f9fc2926798f18bb8ece99391e16b6f.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 16 + }, + { + "type": "text", + "text": "第三章 变量之间的关系", + "text_level": 1, + "page_idx": 17 + }, + { + "type": "text", + "text": "基础过关测试卷", + "text_level": 1, + "page_idx": 17 + }, + { + "type": "text", + "text": "时间：60分钟 满分：100分", + "page_idx": 17 + }, + { + "type": "table", + "img_path": "images/c1560e3c8ddaa1753fa4b6bd45be77d9b71d026d4ea52327fe317f224ed4c253.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 17 + }, + { + "type": "text", + "text": "一、选择题(每小题4分，共32分)", + "text_level": 1, + "page_idx": 17 + }, + { + "type": "text", + "text": "1.某超市某种商品的单价为70元/件，若买 $_ { \\mathcal { X } }$ 件该商品的总价为 $y$ 元,则其中的常量是 （ ）", + "page_idx": 17 + }, + { + "type": "text", + "text": "A.70 B. $_ { \\mathcal { X } }$ C.y D.不确定", + "page_idx": 17 + }, + { + "type": "text", + "text": "2.在利用太阳能热水器来加热水的过程中，热水器里的水温随所晒时间的长短而变化，这个问题中因变量是 （ ）", + "page_idx": 17 + }, + { + "type": "text", + "text": "A.太阳光的强弱B.水的温度 C.所晒时间 D.热水器的容积 ", + "page_idx": 17 + }, + { + "type": "text", + "text": "3.在用图象表示变量之间的关系时，下列说法最恰当的是", + "page_idx": 17 + }, + { + "type": "text", + "text": "A.用水平方向的数轴上的点表示因变量B.用竖直方向的数轴上的点表示自变量C.用横轴上的点表示自变量·D.用横轴或纵轴上的点表示自变量", + "page_idx": 17 + }, + { + "type": "text", + "text": "4.变量 $y$ 与 $\\mathcal { X }$ 之间的关系式是 $\\scriptstyle { y = 2 x - 3 }$ ,当因变量 $y = 6$ 时，自变量 $\\mathcal { X }$ 的值是（", + "page_idx": 17 + }, + { + "type": "text", + "text": "A.15 B.9 C.4.5 D.1. 5 ", + "page_idx": 17 + }, + { + "type": "text", + "text": "5.为了加强爱国主义教育，每周一学校都要举行庄严的升旗仪式，同学们凝视着冉冉上升的国旗，下列哪个函数图象能近似地刻画上升的国旗离旗杆顶端的距离与时间的关系 （ ）", + "page_idx": 17 + }, + { + "type": "image", + "img_path": "images/2501e7cb5a037909c2831198f8f79bec0e5f06e9987b438dc4e641928241582f.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 17 + }, + { + "type": "text", + "text": "6.根据如图所示的程序计算变量 $y$ 的对应值，若输入变量 $\\mathcal { X }$ 的值为一1,则输出 （20 $y$ 的值为 （ ", + "page_idx": 17 + }, + { + "type": "text", + "text": "A.-2 B.2 C.-1 D.0 ", + "page_idx": 17 + }, + { + "type": "image", + "img_path": "images/7b3b272397e842106fd0d1d45d2d9552309372f280068667b34554d54ca2ed4c.jpg", + "img_caption": [ + "第6题图" + ], + "img_footnote": [], + "page_idx": 17 + }, + { + "type": "image", + "img_path": "images/f882ee50ed053feaf7c27ff63e62c33b654001f1acef23d3e15f23d47359bdcc.jpg", + "img_caption": [ + "第8题图" + ], + "img_footnote": [], + "page_idx": 17 + }, + { + "type": "text", + "text": "7.某大剧场地面的一部分为扇形，观众席的座位数按下表所示的方式设置：", + "page_idx": 17 + }, + { + "type": "table", + "img_path": "images/45a27427a9333c9f41c5f394db2b29e28e1130144ad5b64e40141c00c9b5790a.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
排数x1234
座位数y50535659
", + "page_idx": 17 + }, + { + "type": "text", + "text": "有下列结论： $\\textcircled{1}$ 排数 $\\mathcal { X }$ 是自变量，座位数 $y$ 是因变量； $\\textcircled{2}$ 排数 $\\mathcal { X }$ 是因变量，座位数 $y$ 是自变量； $\\textcircled { 3 } y = 5 0 + 3 x$ $\\textcircled { 4 } y = 4 7 + 3 x .$ 其中正确的结论有 ）", + "page_idx": 17 + }, + { + "type": "text", + "text": "A.1个 B.2个 C.3个 D.4个 ", + "page_idx": 17 + }, + { + "type": "text", + "text": "8. $\\circledcirc$ 视讲如图，该图反映的是小明从家先去食堂吃早餐，接着去图书馆读报，然后回家的过程，其中 $\\mathcal { X }$ 表示时间， $y$ 表示小明离家的距离，小明家、食堂、图书馆在同一直线上，根据图中提供的信息，下列说法正确的有 （）$\\textcircled{1}$ 食堂离小明家 $0 . 4 \\ \\mathrm { k m }$ $\\textcircled{2}$ 小明从食堂到图书馆用了 $3 ~ \\mathrm { m i n }$ $\\textcircled{3}$ 图书馆在小明家和食堂之间； $\\textcircled{4}$ 小明从图书馆回家的平均速度是 $0 . 0 4 ~ \\mathrm { k m / m i n }$ ", + "page_idx": 17 + }, + { + "type": "text", + "text": "A.4个 B.3个 C.2个 D.1个 ", + "page_idx": 17 + }, + { + "type": "text", + "text": "二、填空题(每小题5分，共20分)", + "text_level": 1, + "page_idx": 17 + }, + { + "type": "text", + "text": "9.“早穿皮祅午穿纱，围着火炉吃西瓜\"这句谚语反映了我国新疆地区一天中，温度随时间的变化而变化，其中自变量是 ，因变量是", + "page_idx": 17 + }, + { + "type": "text", + "text": "10.李老师带领 $\\mathcal { X }$ 名学生到某动物园参观，已知成人票每张20元，学生票每张10元.设门票总费用为 $y$ 元，则 $y$ 与 $\\mathcal { X }$ 之间的关系式为", + "page_idx": 17 + }, + { + "type": "text", + "text": "11.某人购进一批苹果到市场上零售，已知卖出苹果数量 $\\mathcal { X }$ 与售价 $y$ 的关系如下表.", + "page_idx": 17 + }, + { + "type": "table", + "img_path": "images/c051a02d37ded3198c5ce003a8b8d2530dd3758fe4b255a6f50436cc48e2dafa.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
数量x/kg12345
售价y/元8.51725.53442.5
", + "page_idx": 17 + }, + { + "type": "text", + "text": "写出用 $\\mathcal { X }$ 表示 $y$ 的关系式：", + "page_idx": 17 + }, + { + "type": "text", + "text": "12.如图,这是小明从学校到家行进的路程s(米)与时间t(分)的函数图象.观察图象，从中得到如下信息：$\\textcircled{1}$ 学校离小明家1000米； $\\textcircled{2}$ 小明用了20分钟到家；$\\textcircled{3}$ 小明前10分钟走了路程的一半； $\\textcircled{4}$ 小明后10分钟比前 10 分钟走得快.其中正确的是 . (填序号)", + "page_idx": 18 + }, + { + "type": "image", + "img_path": "images/54572e5cbe3e37f31b117863c30a7a3aedb56a5454dfe05a88cf39770bc4a75a.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 18 + }, + { + "type": "text", + "text": "三、解答题(本大题4小题，共48分)", + "text_level": 1, + "page_idx": 18 + }, + { + "type": "text", + "text": "13.(10分)若用c表示摄氏温度，f表示华氏温度，则c与f之间的关系式为c$= \\frac { 5 } { 9 } ( f - 3 2 )$ .试分别求：", + "page_idx": 18 + }, + { + "type": "text", + "text": "(1)当 $f { = } 6 8$ 和 $f { = } { - } 4$ 时， $c$ 的值; \n(2)当 $c { = } 1 0$ 时， $f$ 的值.", + "page_idx": 18 + }, + { + "type": "text", + "text": "14.(10分)温度的变化是人们经常谈论的话题,请根据下面的图象与同伴讨论这一天温度变化的情况.", + "page_idx": 18 + }, + { + "type": "text", + "text": "（1)这一天的最高温度是多少？是在几时到达的？最低温度呢？（2)这一天的温差是多少？从最低温度到最高温度经过多长时间?", + "page_idx": 18 + }, + { + "type": "image", + "img_path": "images/a39cd68baf6c32d935fb6ec24d8ee727c56e70cfafa71e05ab881ada43bb741f.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 18 + }, + { + "type": "text", + "text": "15.(13分)如图所示,将一个边长为 $1 2 \\ \\mathrm { c m }$ 的正方形的四个角都剪去一个大小相等的小正方形，当小正方形的边长由小到大变化时，图中阴影部分的面积也随之发生变化.", + "page_idx": 18 + }, + { + "type": "text", + "text": "(1)在这个变化过程中，自变量、因变量各是什么？ \n(2)如果小正方形的边长为xcm,图中阴影部分的面积为ycm²,请写出y与 $\\mathcal { X }$ 之间的关系式. \n(3)当小正方形的边长由 $1 ~ \\mathrm { c m }$ 变化到 $5 ~ \\mathrm { c m }$ 时,图中阴影部分的面积是怎样变化的？", + "page_idx": 18 + }, + { + "type": "image", + "img_path": "images/9c785ebffdea8a32ab4f53393056256e9576d2053466b11a9560c14e910e66eb.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 18 + }, + { + "type": "text", + "text": "16.视濒讲（15分)某机动车出发前油箱内有油42L,行驶若干小时后，途中在加油站加油若干升,油箱中剩余油量 $Q ( \\mathrm { L } )$ 与行驶时间 $t ( \\mathrm { h } )$ 之间的关系如图所示.根据图象回答下列问题：", + "page_idx": 18 + }, + { + "type": "text", + "text": "（1)机动车行驶多少小时后加油？ \n(2)途中加油多少升？ \n(3)已知加油站距目的地还有 $2 3 0 ~ \\mathrm { k m }$ ，车速为 $4 0 ~ \\mathrm { k m / h }$ ,若要到达目的地,油箱中的油是否够用？请说明理由.", + "page_idx": 18 + }, + { + "type": "text", + "text": "Q/L 4236302418126 01234567891011t/h ", + "page_idx": 18 + }, + { + "type": "text", + "text": "能力提优测试卷", + "text_level": 1, + "page_idx": 19 + }, + { + "type": "text", + "text": "时间：60分钟 满分：100分", + "page_idx": 19 + }, + { + "type": "text", + "text": "班级", + "page_idx": 19 + }, + { + "type": "table", + "img_path": "images/c2cd25c74811c40a3cbc110a1fd3a6a9491201a37efce38defc6c1575783366b.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
学校
", + "page_idx": 19 + }, + { + "type": "table", + "img_path": "images/50bcae0f4a52005642482e010b01fa48d1f21e30ab3375a5ad08c6b4a9e8fb66.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 19 + }, + { + "type": "text", + "text": "一、选择题(每小题5分，共30分)", + "text_level": 1, + "page_idx": 19 + }, + { + "type": "table", + "img_path": "images/3c7b33f3a16c98d92ab08f2463a5137afe03015d6162b7eb24cd4c8774029ed4.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
姓名
", + "page_idx": 19 + }, + { + "type": "text", + "text": "1.2023 年10月19日，广东省茂名市遭受台风\"三巴\"和弱冷空气共同影响，大部分地区发生强降雨，某条河受暴雨袭击，一天的水位记录如下：", + "page_idx": 19 + }, + { + "type": "table", + "img_path": "images/a1c546c862a4fe9ef4d1eac773de6f20a63fd6aa4c18dcded6bc955b37bf6759.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
学号
", + "page_idx": 19 + }, + { + "type": "table", + "img_path": "images/c8ebe6fd5ae4cc05be54088aff77b869f7f06e128aebad5a8ef42e75e36fd7b5.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
时间/时04812162024
水位/米22.534.5568
", + "page_idx": 19 + }, + { + "type": "text", + "text": "观察表中的数据，水位上升最快的时段是", + "page_idx": 19 + }, + { + "type": "text", + "text": "A. $8 \\sim 1 2$ 时 B. $1 2 { \\sim } 1 6$ 时 $C . 1 6 { \\sim } 2 0$ 时 D. $2 0 { \\sim } 2 4$ 时", + "page_idx": 19 + }, + { + "type": "text", + "text": "2.自动测温仪记录的图象如图所示，它反映了齐齐哈尔市的春季某天气温 $T$ 如何随时间 $t$ 的变化而变化，下列从图象中得到的信息正确的是 （ ）", + "page_idx": 19 + }, + { + "type": "text", + "text": "A.0点时气温达到最低 B.最低气温是零下 $4 ~ ^ { \\circ } C$ C.0点到14点之间气温持续上升 D.最高气温是 $8 ~ ^ { \\circ } C$ ", + "page_idx": 19 + }, + { + "type": "image", + "img_path": "images/81b3700ab667276ebded8d3e2de13cc85cbb1dd67f5cc4cde677cd9d66480076.jpg", + "img_caption": [ + "第2题图" + ], + "img_footnote": [], + "page_idx": 19 + }, + { + "type": "image", + "img_path": "images/2910ef549cd54d24ceafe5cdf94a7a6027caeb5867d867509e637bbf8556c3b5.jpg", + "img_caption": [ + "第6题图" + ], + "img_footnote": [], + "page_idx": 19 + }, + { + "type": "text", + "text": "3.下面的表格列出了一个试验的统计数据，表示皮球从高处落下时，弹跳高度 $b$ （ $\\mathrm { c m } )$ 与下降高度 $d ( \\mathrm { c m } )$ 的关系，下列式子中，能表示这种关系的是 （ ）", + "page_idx": 19 + }, + { + "type": "table", + "img_path": "images/fbb0e4eb6d591b713921e56ff96542d23c792fbbb55161bc29a64ad73110c801.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
考生注意
", + "page_idx": 19 + }, + { + "type": "table", + "img_path": "images/64ca6d9545c8503f7784c69dccfeff5ffb0315ab07dff99a2ba2f9859bc78cf8.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
d/cm5080100150
b/cm25405075
", + "page_idx": 19 + }, + { + "type": "text", + "text": "A. $b { = } d ^ { 2 }$ （204号 B. $b { = } 2 d$ $\\mathrm { C } . \\mathit { b } \\mathrm { = } \\frac { \\mathit { d } } { 2 }$ $\\mathrm { D } , b { = } d { + } 2 5$ ", + "page_idx": 19 + }, + { + "type": "text", + "text": "4.某工厂有甲、乙两个大小相同的蓄水池，如图所示，且中间有管道连通.现要向甲池中注水,若单位时间内的注水量不变，则从注水开始，乙池水面上升的高度 $h$ 与注水时间 $t$ 之间的关系的图象可能是 （ ）", + "page_idx": 19 + }, + { + "type": "image", + "img_path": "images/cfc7cf376608c06fbf1ee069161c5e40282fc2ca2816867a2f193058976fe29e.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 19 + }, + { + "type": "text", + "text": "5.向一个容器内均匀地注入水，液面升高的高度 $y$ 与注水时间 $x$ 满足如图所示的图象，则符合图象条件的容器为 （ ）", + "page_idx": 19 + }, + { + "type": "image", + "img_path": "images/a309d4823d11785bf0b72388263392fdf0cefc512bebd43e52c5ed3246627b5c.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 19 + }, + { + "type": "text", + "text": "6. $\\circledcirc$ 视频讲解如图1，在矩形 $M N P Q$ 中，动点 $R$ 从点 $N$ 出发，沿 $N { } P { } Q { } M$ 方向运动至点 $M$ 处停止,设点 $R$ 运动的路程为 $x , \\triangle M N R$ 的面积为 $y$ ,如果 $y$ 关于 $\\mathcal { X }$ 的函数图象如图2所示，那么当点 $R$ 运动到 $M Q$ 的中点时， $\\triangle M N R$ 的面积为 （）", + "page_idx": 19 + }, + { + "type": "text", + "text": "A.5 B.9 C.10 D.不可确定", + "page_idx": 19 + }, + { + "type": "text", + "text": "二、填空题(每小题5分，共20分)", + "text_level": 1, + "page_idx": 19 + }, + { + "type": "text", + "text": "7.气象观测小组进行活动，一号探测气球从海拔 $5 ~ \\mathrm { m }$ 处出发，以 $1 . 2 ~ \\mathrm { m / m i n }$ 的速度上升，气球所在位置的海拔 $y$ （单位： $\\mathrm { m }$ 与上升时间 $_ { \\mathcal { X } }$ （单位： $\\operatorname* { m i n }$ 之间的关系式为", + "page_idx": 19 + }, + { + "type": "text", + "text": "8.小明从家跑步到学校，接着马上原路步行回家.如图，这是小明离家的路程 $y$ （米)与时间 $t$ （分)的函数图象，则小明回家的速度是每分钟步行 米.", + "page_idx": 19 + }, + { + "type": "text", + "text": "→y/米 800 0 5 15分 ", + "page_idx": 19 + }, + { + "type": "text", + "text": "9.共享单车为市民的出行带来了方便.某共享单车公司规定，首次骑行需交99元押金，第一次骑行收费标准如下表所示：(不足半小时的按半小时计算)", + "page_idx": 19 + }, + { + "type": "table", + "img_path": "images/88f2b5a199cc8aaa74723a58230bff3912aaaad541124fb0a47b9a3615f50e27.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
骑行时间t/小时0.511.52
骑行费用y/元99+199+299+399+4
", + "page_idx": 19 + }, + { + "type": "text", + "text": "则第一次骑行费用 $y$ (元)与骑行时间 $t$ （小时)之间的关系式为", + "page_idx": 19 + }, + { + "type": "text", + "text": "10.小明步行，小颖骑车，他们同时从小颖家出发，以各自的速度 y/米匀速到公园游玩，小颖先到并停留了8分钟，发现相机忘在了家里，于是沿原路以同样的速度回家去取.若小明的步行可i0 20 1020/分钟速度为180米/分，他们各自距离出发点的路程 $y$ 与出发时间 $x$ 之间的关系图象如图所示，则当小明到达公园时小颖离家 米.", + "page_idx": 19 + }, + { + "type": "text", + "text": "三、解答题(本大题5小题，共50分)", + "text_level": 1, + "page_idx": 20 + }, + { + "type": "text", + "text": "11.(8分)地壳的厚度约为 $8 ~ \\mathrm { k m }$ 到 $4 0 ~ \\mathrm { k m }$ ，在地表以下不太深的地方，温度可按关系式 $\\scriptstyle { y = 3 5 x + t }$ 计算,其中 $_ { \\mathcal { X } }$ 是深度 $( \\mathrm { k m } ) , t$ 是地球表面温度， $y$ 是所达深度的温度.当地表温度为2℃时，分别计算当x为1km,5km,10 km,$2 0 ~ \\mathrm { k m } , 3 0 ~ \\mathrm { k m }$ 时地壳的温度.", + "page_idx": 20 + }, + { + "type": "text", + "text": "12.(10 分)在空中，自地面算起,每升高1千米,气温下降若干 $\\mathrm { { } ^ { \\circ } C }$ .某地空中气温 $t ( ^ { \\circ } \\mathrm { C } )$ 与高度 $h$ (千米)之间的图象如图所示.", + "page_idx": 20 + }, + { + "type": "text", + "text": "(1)该地地面气温为 $\\mathrm { { } ^ { \\circ } C }$ (2)当高度为 千米时,气温为 $0 ~ ^ { \\circ } C$ (3)求出空中气温 $t ( ^ { \\circ } \\mathrm { C } )$ 与高度 $h$ (千米)之间的关系式. ", + "page_idx": 20 + }, + { + "type": "text", + "text": "t/C241680 1234h千米", + "page_idx": 20 + }, + { + "type": "text", + "text": "13.（10分)某公司计划购买A,B两种型号的电脑，已知购买一台A型电脑需0.6万元，购买一台 $B$ 型电脑需0.4万元，该公司准备投人资金 $y$ 万元，全部用于购进35台这两种型号的电脑,设购进 $A$ 型电脑 $\\mathcal { X }$ 台.", + "page_idx": 20 + }, + { + "type": "text", + "text": "(1)求 $y$ 与 $\\mathcal { X }$ 之间的关系式. \n(2)若购进B型电脑的数量恰好为A型电脑数量的2.5倍，则该公司需要投人资金多少万元？", + "page_idx": 20 + }, + { + "type": "text", + "text": "14.(10 分)甲、乙两家体育用品商店出售同样的乒乓球拍和乒乓球,乒乓球拍每副定价30元,乒乓球每盒定价5元.现两家商店搞促销活动，甲店：每买一副球拍赠一盒乒乓球.乙店：按定价的9折优惠.某班级需购买球拍4副，乒乓球若干盒(不少于4盒).", + "page_idx": 20 + }, + { + "type": "text", + "text": "(1)设购买乒乓球为 $\\mathcal { X }$ 盒，在甲店购买的付款金额为 $y _ { \\mathbb { H } }$ 元,在乙店购买的付款金额为 $y _ { Z }$ 元,分别写出在两家商店购买的付款金额与乒乓球盒数 $\\mathcal { X }$ 之间的关系式.", + "page_idx": 20 + }, + { + "type": "text", + "text": "（2)购买多少盒乒乓球时，在两家商店购买的付款金额一样多?", + "page_idx": 20 + }, + { + "type": "text", + "text": "15.视频讲（12分)新能源纯电动汽车的不断普及让很多人感受到了它的好处，其中最重要的一点就是对环境的保护.如图，这是某型号新能源纯电动汽车充满电后,蓄电池剩余电量y(千瓦·时)与已行驶路程 $\\mathcal { X }$ (千米)之间的关系图象.", + "page_idx": 20 + }, + { + "type": "text", + "text": "(1)图中点 $A$ 表示的实际意义是什么？(2)当汽车行驶了120千米时，求蓄电池的剩余电量.(3)当汽车行驶了多少千米时，剩余电量降至20千瓦·时？", + "page_idx": 20 + }, + { + "type": "text", + "text": "↑y千瓦·时 \n60 A \n35 \n10 \n0 150200x/干米 ", + "page_idx": 20 + }, + { + "type": "text", + "text": "期中测试卷", + "text_level": 1, + "page_idx": 21 + }, + { + "type": "text", + "text": "时间：90分钟 满分：120分考试范围：第一章～第三章", + "page_idx": 21 + }, + { + "type": "table", + "img_path": "images/c5a6364bfe53238b3fba4b44a6c4fedf7200054a7e5442151d4cb0b81ff8ab1f.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 21 + }, + { + "type": "text", + "text": "一、选择题(本大题共10小题，每小题3分，共30分）", + "text_level": 1, + "page_idx": 21 + }, + { + "type": "text", + "text": "1.下列运算正确的是", + "page_idx": 21 + }, + { + "type": "text", + "text": "A. $( a ^ { 3 } ) ^ { 4 } = a ^ { 1 2 }$ B $( - 2 a ) ^ { 2 } = - 4 a ^ { 2 }$ $\\mathrm { C } . a ^ { 3 } \\cdot a ^ { 3 } = a ^ { 9 }$ $\\operatorname { D } . a ^ { 6 } \\div a ^ { 2 } = a ^ { 3 }$ ", + "page_idx": 21 + }, + { + "type": "text", + "text": "2.如图， $A C \\bot B C$ ，直线 $E F$ 经过点 $C$ ，若 $\\angle 1 = 3 5 ^ { \\circ }$ ，则 $\\angle 2$ 的度数为 （）", + "page_idx": 21 + }, + { + "type": "text", + "text": "A $3 5 ^ { \\circ }$ （20 B.45° C.55° D. 65° ", + "page_idx": 21 + }, + { + "type": "image", + "img_path": "images/e58b9b8f0d5b3b107d85b633a189613f9cdd05e3f3678dd6b6412a0b97abcad5.jpg", + "img_caption": [ + "第2题图" + ], + "img_footnote": [], + "page_idx": 21 + }, + { + "type": "image", + "img_path": "images/30f13a80499e1dd8be7addf79730ba9a13bd52821386add2efe896d309bd4d12.jpg", + "img_caption": [ + "第6题图" + ], + "img_footnote": [], + "page_idx": 21 + }, + { + "type": "text", + "text": "3.据专家介绍，某种病毒的直径大约为0.000000125米，它与飞沫等体液结合后体积会变大，正确佩戴口罩能有效预防.数据0.000000125用科学记数法表示为 （ ）", + "page_idx": 21 + }, + { + "type": "text", + "text": "A. $0 . 1 2 5 { \\times } 1 0 ^ { - 8 }$ $\\mathrm { B } . 1 2 . 5 { \\times } 1 0 ^ { - 6 }$ （204号 $\\mathrm { C } . 1 . 2 5 { \\times } 1 0 ^ { - 6 }$ （20 （204号 $\\mathrm { D } . 1 . 2 5 { \\times } 1 0 ^ { - 7 }$ ", + "page_idx": 21 + }, + { + "type": "text", + "text": "4.一根弹簧长 $8 ~ \\mathrm { c m }$ ，它所挂物体的质量不能超过 $5 ~ \\mathrm { k g }$ ，并且所挂物体的质量每增加 $1 ~ \\mathrm { k g }$ ，弹簧就伸长 $0 . 5 ~ \\mathrm { c m }$ ,则挂上物体后弹簧的长度 $y ( \\mathrm { c m } )$ 与所挂物体的质量 $x ( \\mathrm { k g } ) ( 0 { \\leqslant } x { \\leqslant } 5 )$ 之间的关系式为 C ）", + "page_idx": 21 + }, + { + "type": "text", + "text": "A. $y { = } 0 . 5 ( x { + } 8 )$ $\\mathrm { B } . \\mathrm { { } } y { = } 0 . 5 x { - } 8 $ \nC. $y { = } 0 . 5 ( x { - } 8 )$ $\\mathrm { D } . { \\it y } = 0 . 5 x ^ { + } 8$ ", + "page_idx": 21 + }, + { + "type": "text", + "text": "5.一个长方体的高为 $x \\ \\mathrm { c m }$ ,长比高的3倍少 $4 \\mathrm { \\ c m }$ ，宽是高的2倍，那么这个长方体的体积是 （ ）", + "page_idx": 21 + }, + { + "type": "text", + "text": "A. $( 3 x ^ { 3 } - 4 x ^ { 2 } ) \\mathrm { { c m } ^ { 3 } }$ B $\\dot { \\cdot } ( 6 x ^ { 3 } + 8 x ^ { 2 } ) \\mathrm { c m } ^ { 3 }$ C. $( 6 x ^ { 3 } - 8 x ^ { 2 } ) \\mathrm { c m } ^ { 3 }$ $\\mathrm { D } . ( 6 x ^ { 2 } - 8 x ) \\mathrm { c m } ^ { 3 }$ ", + "page_idx": 21 + }, + { + "type": "text", + "text": "6.如图，下列条件中,能判定 $A B / / C D$ 的是", + "page_idx": 21 + }, + { + "type": "equation", + "img_path": "images/fb0a3ea2462ccfa186816d0c0542352d5d6f0f73c26fa66891792060003d89a1.jpg", + "text": "$$\n\\Delta . \\angle 1 = \\angle 2 \\qquad \\mathrm { B } . \\angle 3 = \\angle 4 \\qquad \\mathrm { C } . \\angle B = \\angle D \\qquad \\mathrm { D } . \\angle D = \\angle 5\n$$", + "text_format": "latex", + "page_idx": 21 + }, + { + "type": "text", + "text": "7.当 $a = - 1$ 时,式子 $( 2 8 a ^ { 3 } b - 1 4 a ^ { 2 } b + 7 a b ) \\div ( - 7 a b )$ 的值是", + "page_idx": 21 + }, + { + "type": "equation", + "img_path": "images/9cd64e98e5cec098884c6ffe2ec14f44ed1dccbae6d6172dccb30c77136ccc8e.jpg", + "text": "$$\n\\mathrm { A } , - 7 \\mathrm { B } , - 6 \\mathrm { C } , - 3 \\mathrm { D } , - 2\n$$", + "text_format": "latex", + "page_idx": 21 + }, + { + "type": "text", + "text": "8. $\\circledcirc$ 视频讲解如图，直线 $A B / / C D / / E F$ ，点 $O$ 在直线 $E F$ 上，下列结论正确的是（", + "page_idx": 21 + }, + { + "type": "text", + "text": "A. $\\angle \\alpha + \\angle \\beta - \\angle \\gamma = 9 0 ^ { \\circ }$ B $. \\angle \\alpha + \\angle \\beta + \\angle \\gamma = 1 8 0 ^ { \\circ }$ C.∠x+∠β-∠α=180° D.∠a+∠γ-∠β=180° ", + "page_idx": 21 + }, + { + "type": "image", + "img_path": "images/56a0385115498cecfdbd7e93fd3f31282b8bea3e63526f27793646bc5af918cc.jpg", + "img_caption": [ + "第8题图" + ], + "img_footnote": [], + "page_idx": 21 + }, + { + "type": "image", + "img_path": "images/a991c53c7a4836cc578c6967bea3ccde4ae581755849596f0c4aac971f9129f5.jpg", + "img_caption": [ + "第10题图" + ], + "img_footnote": [], + "page_idx": 21 + }, + { + "type": "text", + "text": "9. 新考法阅读理解题定义一种新的运算：若 $a \\neq 0$ ，则有 $a \\mathbf { A } b = a ^ { - 2 } + a b +$ |-b|.那么 $( - \\frac { 1 } { 2 } ) \\pmb { \\Delta } 2$ 的值是 ", + "page_idx": 21 + }, + { + "type": "text", + "text": "A.-3 B.5 C.13 D $\\frac { 3 } { 2 }$ ", + "page_idx": 21 + }, + { + "type": "text", + "text": "10.小明和小华是同班同学，也是邻居.某日早晨,小明7：40 先出发去学校，走了一段路后，在途中停下吃了早餐，后来发现上学时间快到了，就跑步到学校；小华离家后直接乘公共汽车到了学校.他们从家到学校已走的路程s(米)和所用时间t(分钟)的关系图象如图所示.则下列说法中，错误的是 （ ）", + "page_idx": 21 + }, + { + "type": "text", + "text": "A.小明吃早餐用时5分钟B.小华到学校的平均速度是240米/分C.小明跑步的平均速度是100米/分D.小华到学校的时间是7：55", + "page_idx": 21 + }, + { + "type": "text", + "text": "二、填空题(本大题共5小题，每小题3分，共15分)", + "text_level": 1, + "page_idx": 22 + }, + { + "type": "text", + "text": "11.某人购进一批葡萄到市场上零售，已知卖出葡萄质量 $\\mathcal { X }$ 与销售额 $y$ 之间的关系如下表：", + "page_idx": 22 + }, + { + "type": "table", + "img_path": "images/41eb0d9f665d2a7a8855507517daf2ec2c8dbdb0d35e84ff8f6a9461c7917113.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
质量x/千克12345
销售额y/元612182430
", + "page_idx": 22 + }, + { + "type": "text", + "text": "则当卖出葡萄质量 $_ { \\mathcal { X } }$ 为10千克时,销售额 $y$ 为 元", + "page_idx": 22 + }, + { + "type": "text", + "text": "12.已知 $x ^ { + } y { = } 2 x y$ ，则 $( 2 x - 1 ) ( 2 y - 1 )$ 的值为", + "page_idx": 22 + }, + { + "type": "text", + "text": "13.如图,已知直线 $E F \\bot M N$ 垂足为点 $F$ ，且 $\\angle 1 = 1 3 8 ^ { \\circ }$ ，则当 $\\angle 2 =$ 时， $A B / / C D$ ", + "page_idx": 22 + }, + { + "type": "image", + "img_path": "images/caa23f6889a06123a156d3a4016accbafbce5e2197abc3c12cfcfb3d31cb3378.jpg", + "img_caption": [ + "第13题图" + ], + "img_footnote": [], + "page_idx": 22 + }, + { + "type": "image", + "img_path": "images/4a1afd0387d10b5e51a2da8da622db3945bf77e731d5f4cb88bf281dd9c32d4f.jpg", + "img_caption": [ + "第14题图" + ], + "img_footnote": [], + "page_idx": 22 + }, + { + "type": "image", + "img_path": "", + "img_caption": [ + "图1" + ], + "img_footnote": [], + "page_idx": 22 + }, + { + "type": "image", + "img_path": "images/25e90cfea0d3fa9ef4299f4d0e3f7d76e149c168355275316bb73dff4ae70512.jpg", + "img_caption": [ + "第15题图" + ], + "img_footnote": [], + "page_idx": 22 + }, + { + "type": "text", + "text": "14.如图,AB//DE, $\\angle 1 = 1 3 5 ^ { \\circ }$ $A C \\bot C D$ ，则 $\\angle D$ 的度数为", + "page_idx": 22 + }, + { + "type": "text", + "text": "15.新考法数学文化 $\\circledcirc$ 靓频讲我国宋朝数学家杨辉在他的著作《详解九章算法》中提出的“杨辉三角\"如图1所示，并观察如图2所示的等式.根据前面各式的规律，请你猜想 $( 2 x - 1 ) ^ { 6 }$ 的展开式中含 $x ^ { 5 }$ 项的系数是", + "page_idx": 22 + }, + { + "type": "image", + "img_path": "images/ee7768ebb741ee407069dbb2ecdd824b00d93e87e1e3292a714bc3f50cac3bcf.jpg", + "img_caption": [ + "图2" + ], + "img_footnote": [], + "page_idx": 22 + }, + { + "type": "text", + "text": "三、解答题(本大题共8小题，满分75分）", + "text_level": 1, + "page_idx": 22 + }, + { + "type": "text", + "text": "16.(10分)（1)计算： $( x + 2 ) ( 4 x - 1 ) - 2 x ( 2 x - 1 ) .$ ", + "page_idx": 22 + }, + { + "type": "text", + "text": "(2)如图， $A B / / C D$ ，点 $E$ 在 $A B$ 上，点 $F$ 在 $C D$ 上，如果 $\\angle C F E : \\angle E F B =$ （204号 $3 : 4$ $\\angle A B F { = } 4 0 ^ { \\circ }$ ，求 $\\angle B E F$ 的度数.", + "page_idx": 22 + }, + { + "type": "text", + "text": "17.(9分)已知 $5 x ^ { 2 } - x - 2 = 0$ ，求代数式 $( 3 x + 2 ) ( 3 x - 2 ) + x ( x - 2 )$ 的值. ", + "page_idx": 22 + }, + { + "type": "text", + "text": "18.(9分)如图，把一些相同规格的碗整齐地叠放在水平桌面上，这擦碗的高度随着碗的数量变化而变化的情况如下表所示：", + "page_idx": 23 + }, + { + "type": "table", + "img_path": "images/63a08f6355b4875c22bb389075681934d7f4e23c4ac271d460bd0c10ac9842b7.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
碗的数量/只12345：
高度/cm45.26.47.68.8：
", + "page_idx": 23 + }, + { + "type": "image", + "img_path": "images/b77c674c3552af08200c0107c317ddee910577b2c1ae9325d7d7639e74b3bfd0.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 23 + }, + { + "type": "text", + "text": "(1)若用 $h ( \\mathrm { c m } )$ 表示这擦碗的高度，用 $_ { \\mathcal { X } }$ (只)表示这擦碗的数量，请用含有 $\\mathcal { X }$ 的代数式表示 $h$ ", + "page_idx": 23 + }, + { + "type": "text", + "text": "(2)若这擦碗的高度为 $1 1 . 2 \\ \\mathrm { c m }$ ,求这擦碗的数量. ", + "page_idx": 23 + }, + { + "type": "text", + "text": "20.（9分)为了加强公民的节水意识，合理利用水资源，某城市规定用水收费标准如下：每户每月用水量不超过6立方米时，水费按 $a$ 元/立方米收费；每户每月用水量超过6立方米时，不超过的部分每立方米仍按 $a$ 元收费，超过的部分按 $c$ 元/立方米收费.该市某用户今年3,4月份的用水量和水费如下表所示：", + "page_idx": 23 + }, + { + "type": "table", + "img_path": "images/2a2f27ea494fa29ab1c45267aa02b847882fd6e98cf865ad5bd170ba6d33cacd.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
月份用水量x/立方米收费y/元
3510.5
4921.6
", + "page_idx": 23 + }, + { + "type": "text", + "text": "(1)求 $a , c$ 的值，并写出每月用水量不超过6立方米和超过6立方米时，水费$y$ 与用水量 $\\mathcal { X }$ 之间的关系式；", + "page_idx": 23 + }, + { + "type": "text", + "text": "（2)已知某用户5月份的用水量为8立方米,求该用户5月份的水费.", + "page_idx": 23 + }, + { + "type": "text", + "text": "19.(9分）如图， $B , C , E$ 三点在同一直线上，如果 $\\angle D A E = \\angle E$ ， $\\angle B = \\angle D$ ,那么 $A B$ 与 $C D$ 平行吗？请说明理由.", + "page_idx": 23 + }, + { + "type": "image", + "img_path": "images/8a04bc1e9663d4ad4c993d6cb04b79ae2122b39628050c38286c8122a55ef14e.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 23 + }, + { + "type": "text", + "text": "21.(9分)如图， $F G , E D$ 分别交 $B C$ 于点 $M , N , \\angle E N C + \\angle C M G = 1 8 0 ^ { \\circ } , A B / ,$ /$C D$ ", + "page_idx": 23 + }, + { + "type": "text", + "text": "$( 1 ) \\angle 2 = \\angle 3$ 吗？为什么？(2)若 $\\angle A = \\angle 1 + 7 0 ^ { \\circ }$ $\\angle A C B = 4 2 ^ { \\circ }$ ，求 $\\angle B$ 的度数.", + "page_idx": 23 + }, + { + "type": "image", + "img_path": "images/b9c0fcad0708c1213588412537f8fb20d4bf46ad6cfe1db819bf22bbff5e73f6.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 23 + }, + { + "type": "image", + "img_path": "images/fe9d16609172101cb21c32fc87540304e42868534b4455ac0750db3b2e104162.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 23 + }, + { + "type": "text", + "text": "22.(10分)一辆汽车在某次行驶过程中,油箱中的剩余油量y(升)与行驶路程 $\\mathcal { X }$ (千米)之间的关系，其部分图象如图所示.", + "page_idx": 24 + }, + { + "type": "text", + "text": "(1)求 $y$ 与 $\\mathcal { X }$ 之间的关系式（不必注明 $_ { \\mathcal { X } }$ 的取值范围). ", + "page_idx": 24 + }, + { + "type": "text", + "text": "(2)已知当油箱中的剩余油量为8升时，该汽车会开始提示加油，在此次行驶过程中,行驶了 500千米时,司机发现离前方最近的加油站有 30千米的路程,在开往该加油站的途中,汽车开始提示加油,这时离加油站的路程是多少千米？", + "page_idx": 24 + }, + { + "type": "image", + "img_path": "images/2098999b8fea4cc32dbbf4a38299b988eb44a62b98ff276f04648de3362b480f.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 24 + }, + { + "type": "text", + "text": "23.新考法阅读理解题 $\\circledcirc$ 视频讲解（10分）我们将 $( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }$ 进行变$a ^ { 2 } + b ^ { 2 } = ( a + b ) ^ { 2 } - 2 a b , a b = { \\frac { ( a + b ) ^ { 2 } - ( a ^ { 2 } + b ^ { 2 } ) } { 2 } } \\colon$ 2等.根据以上变形解决下列问题：", + "page_idx": 24 + }, + { + "type": "text", + "text": "(1)已知 $a ^ { 2 } + b ^ { 2 } = 8 , ( a + b ) ^ { 2 } = 4 8$ ，则 $a b { = } \\quad \\quad$ (2)若 $_ { \\mathcal { X } }$ 满足 $( 2 5 - x ) ( x - 1 0 ) = - 1 5$ ，求 $( 2 5 - x ) ^ { 2 } + ( x - 1 0 ) ^ { 2 }$ 的值；", + "page_idx": 24 + }, + { + "type": "text", + "text": "", + "page_idx": 24 + }, + { + "type": "text", + "text": "(3)如图,四边形 $A B E D$ 是梯形， $D A \\_ { \\perp } A B , E B \\_ A B , A D = A C , B E = B C ,$ 连接 $C D , C E$ ，若 $A C \\cdot B C { = } 1 0$ ，则图中阴影部分的面积为", + "page_idx": 24 + }, + { + "type": "image", + "img_path": "images/be379db5d023ded231c8f17dabd1286c2520191726d5b5efa14c93de75d2d827.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 24 + }, + { + "type": "table", + "img_path": "images/5fbd3a57eee74e895003584f50096787e9981f63b824a04ae23abe8e6fb7e232.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
学校
", + "page_idx": 25 + }, + { + "type": "text", + "text": "第四章 三角形", + "text_level": 1, + "page_idx": 25 + }, + { + "type": "text", + "text": "基础过关测试卷", + "text_level": 1, + "page_idx": 25 + }, + { + "type": "text", + "text": "班级", + "page_idx": 25 + }, + { + "type": "text", + "text": "时间：60分钟 满分：100分", + "page_idx": 25 + }, + { + "type": "table", + "img_path": "images/c52b3e7b27c8f3eed5f466d3d6f7383710835f1921531edf52a2e316a61742b3.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 25 + }, + { + "type": "text", + "text": "一、选择题(每小题4分，共32分)", + "text_level": 1, + "page_idx": 25 + }, + { + "type": "text", + "text": "1.小明用三根火柴组成的图形如下图所示,则其中是三角形的是", + "page_idx": 25 + }, + { + "type": "table", + "img_path": "images/3834f721f2e5595714cd0b0f58cba162ad2adfe54d0644e0e6137c56a4c4ffbe.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
姓名
", + "page_idx": 25 + }, + { + "type": "text", + "text": "A 4 A B C D ", + "page_idx": 25 + }, + { + "type": "text", + "text": "（ ）", + "page_idx": 25 + }, + { + "type": "text", + "text": "2.下列图形具有稳定性的是", + "page_idx": 25 + }, + { + "type": "table", + "img_path": "images/28cf184fd6220cf4fd2feee4b53944ab3becbae6cb790ca2f2d624f833ef04b8.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
学号
", + "page_idx": 25 + }, + { + "type": "image", + "img_path": "images/77d88d42f67f7f25a997b2fc31daecd3c8951c74dbc99ba448b9e0e5b942c737.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 25 + }, + { + "type": "text", + "text": "3.下列各组图形属于全等图形的是", + "page_idx": 25 + }, + { + "type": "image", + "img_path": "images/f2493801f002041d056d8f587cd65388731a9d6bb2c3333eb41d38d78d63b75d.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 25 + }, + { + "type": "text", + "text": "4.下列每组数分别是三根木棒的长度，不能用它们摆成三角形的是", + "page_idx": 25 + }, + { + "type": "text", + "text": "A $. 5 \\ \\mathrm { c m } , 8 \\ \\mathrm { c m } , 1 2 \\ \\mathrm { c m }$ $\\mathrm { B . 6 ~ c m , 8 ~ c m , 1 2 ~ c m }$ $\\mathrm { C . 5 ~ c m , 6 ~ c m , 8 ~ c m }$ （204号 $\\mathrm { D . 5 ~ c m , 6 ~ c m , 1 2 ~ c m }$ ", + "page_idx": 25 + }, + { + "type": "text", + "text": "5.下列条件中，能够判定 $\\triangle A B C { \\underline { { \\triangle } } } \\triangle D E F$ 的是", + "page_idx": 25 + }, + { + "type": "text", + "text": "A $A B = D E , B C = E F , \\angle A = \\angle E$ B $. A B { = } D E , B C { = } E F$ $\\angle A = \\angle D$ $\\scriptstyle \\complement , \\angle A = \\angle E , A B = D F , \\angle B = \\angle D$ D. $\\scriptstyle \\angle A = \\angle D , \\angle B = \\angle E , B C = E F$ ", + "page_idx": 25 + }, + { + "type": "text", + "text": "6.下列不一定在三角形内部的线段是", + "page_idx": 25 + }, + { + "type": "text", + "text": "A.三角形的角平分线 B.三角形的中线C.三角形的高 D.以上皆不对", + "page_idx": 25 + }, + { + "type": "text", + "text": "7.在 $\\triangle A B C$ 中， $\\angle A = \\angle B + \\angle C$ 则 $\\triangle A B C$ 的形状是 （）", + "page_idx": 25 + }, + { + "type": "text", + "text": "A.锐角三角形 B.直角三角形 C.钝角三角形 D.不能确定", + "page_idx": 25 + }, + { + "type": "text", + "text": "8.如图， $A D$ 是 $\\triangle A B C$ 的中线，点 $E , F$ 分别是 $A D$ 及其延长线上的点，且 $D E =$ $D F$ ,连接BF,CE,下列说法： $\\begin{array} { r } { \\textcircled { 1 } \\triangle B D F { \\cong } \\triangle C D E ; \\textcircled { 2 } B F { = } C E ; \\textcircled { 3 } S _ { \\triangle A B D } = } \\end{array}$ $S _ { \\triangle A D C }$ $\\textcircled { 4 } B F / / C E .$ 其中正确的结论有 （ ）", + "page_idx": 25 + }, + { + "type": "text", + "text": "A. $\\textcircled{1} \\textcircled{2} \\textcircled{3}$ \nB. $\\textcircled{2} \\textcircled{3} \\textcircled{4}$ \nC. $\\textcircled{1} \\textcircled{3} \\textcircled{4}$ \nD. $\\textcircled{1} \\textcircled{2} \\textcircled{3} \\textcircled{4}$ ", + "page_idx": 25 + }, + { + "type": "image", + "img_path": "images/5cd4880fead03b9d7e12cec18f86cb6f1bc30e7417a58aa918f836c476bb3cf7.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 25 + }, + { + "type": "text", + "text": "二、填空题(每小题4分，共16分)", + "text_level": 1, + "page_idx": 25 + }, + { + "type": "text", + "text": "9.如图所示，用直尺和圆规作一个角等于已知角，要说明 $\\angle A ^ { \\prime } O ^ { \\prime } B ^ { \\prime } { = } \\angle A O B$ ，需要说明 $\\triangle C ^ { \\prime } O ^ { \\prime } D ^ { \\prime } { \\overset { \\sim } { = } } \\triangle C O D$ ，则这两个三角形全等的依据是 （写出全等的简写).", + "page_idx": 25 + }, + { + "type": "image", + "img_path": "images/23765189cea2b4233c4761b39ab77ade1498a4c7503f206c4b7a1dda4d59395b.jpg", + "img_caption": [ + "第9题图" + ], + "img_footnote": [], + "page_idx": 25 + }, + { + "type": "image", + "img_path": "images/561a502250e58c9b4ac2464c5a4e118272d82c4e8bac6a980ddaf78e318f3c58.jpg", + "img_caption": [ + "第11题图" + ], + "img_footnote": [], + "page_idx": 25 + }, + { + "type": "image", + "img_path": "images/ee53651de6ff3445249bdfaa2c9237e7e493dfe0ebe020ac589a53785b84f07f.jpg", + "img_caption": [ + "第12题图" + ], + "img_footnote": [], + "page_idx": 25 + }, + { + "type": "text", + "text": "10.已知一个三角形的两边长分别是1和6,第三边长是整数，则第三边长为", + "page_idx": 25 + }, + { + "type": "text", + "text": "11.如图，要测量池塘的宽度 $A B$ ，在池外选取一点 $P$ ，连接 $A P , B P$ 并各自延 长，使 $P C = P A , P D = P B$ 连接 $C D$ ，测得 $C D$ 的长为 $6 0 ~ \\mathrm { m }$ ,则池塘的宽 $A B$ 为 m. ", + "page_idx": 25 + }, + { + "type": "text", + "text": "12.如图，已知 $\\angle A = \\angle 1 = \\angle A B C = 7 0 ^ { \\circ } , C D \\bot B C$ ，则 $\\angle 2$ 的度数为", + "page_idx": 25 + }, + { + "type": "text", + "text": "三、解答题(本大题6小题，共52分）", + "text_level": 1, + "page_idx": 25 + }, + { + "type": "text", + "text": "13.(6分)如图， $A D \\bot B C , \\angle B A C$ 为锐角，写出图中所有的锐角三角形、直角三角形，并用符号表示这些三角形.", + "page_idx": 25 + }, + { + "type": "image", + "img_path": "images/636046c44e8680175c8f0ff4836d84159128bf1165c4ce5de8036d59b5d8def4.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 25 + }, + { + "type": "text", + "text": "14.(8分)如图,画△ABC,使其两边为已知线段 $a , b$ ，夹角为 $\\beta .$ （要求：用尺规作图,不写作法，保留作图痕迹;不在已知的线、角上作图)", + "page_idx": 26 + }, + { + "type": "image", + "img_path": "images/d9f021820db7d90009355b41a909f635796eb1ec2f2deba473891e26f42d0c27.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 26 + }, + { + "type": "text", + "text": "15.(8分)如图，在 $\\triangle A C E$ 中，点 $F$ 为 $C E$ 的延长线上一点，点 $B , D$ 分别为 $A C$ ，$C E$ 上的点,且 $A E / / B D$ ", + "page_idx": 26 + }, + { + "type": "text", + "text": "(1)若 $A C { = } 8 , C E { = } 1 0$ ，求 $A E$ 的取值范围; ", + "page_idx": 26 + }, + { + "type": "text", + "text": "(2)若 $\\angle C B D = 5 5 ^ { \\circ }$ $\\angle C = 6 0 ^ { \\circ }$ ，求 $\\angle A E F$ 的度数. ", + "page_idx": 26 + }, + { + "type": "image", + "img_path": "images/3ac71e8f13e2d85e2e1d6aadb56b009834b0b0a8c7106355d2a92cc255d1cbbf.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 26 + }, + { + "type": "text", + "text": "16.(8分)如图，在 $\\triangle A B C$ 中， $A D$ 是 $B C$ 边上的高, $A E$ 是 $\\angle B A C$ 的平分线，$\\angle E A D = \\angle C A D .$ ", + "page_idx": 26 + }, + { + "type": "text", + "text": "（ $1 ) \\triangle A E D$ 和 $\\triangle A C D$ 全等吗？为什么？(2)若 $\\angle B = 4 2 ^ { \\circ }$ ，求 $\\angle C$ 的度数.", + "page_idx": 26 + }, + { + "type": "image", + "img_path": "images/bec728e82341bcb07e2a1692762d5ccd6cace728fee0ff42261655edcc13fa71.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 26 + }, + { + "type": "text", + "text": "17.(10分)如图所示，在 $\\triangle A B C$ 中,角平分线 $A D , B E , C F$ 相交于点 $O$ ", + "page_idx": 26 + }, + { + "type": "text", + "text": "(1) $\\angle B O C$ 与 $\\angle B A C$ 的大小有什么关系？为什么？", + "page_idx": 26 + }, + { + "type": "text", + "text": "(2) $\\angle C O A$ 与 $\\angle A B C$ 的大小有什么关系？ $\\angle B O A$ 与 $\\angle A C B$ 呢？", + "page_idx": 26 + }, + { + "type": "image", + "img_path": "images/9f77c08a21df0907eff2f774a31838ea177bd5209edb9cfea1a0e684610958a7.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 26 + }, + { + "type": "text", + "text": "18. $\\circledcirc$ 视频讲解（12分)如图，已知 $D E \\bot A C , B F \\bot A C ,$ 垂足分别是点 $E , F , A E { = }$ CF,DC // AB.", + "page_idx": 26 + }, + { + "type": "text", + "text": "(1 $) \\triangle C D E$ 和 $\\triangle A B F$ 全等吗？请说明理由. \n(2)试猜想 $A D$ 和 $B C$ 的数量和位置关系,并说明理由. \n(3)若连接 $D F , B E$ ，则 $D F$ 和 $B E$ 相等吗？请说明理由.", + "page_idx": 26 + }, + { + "type": "image", + "img_path": "images/fcd924fc0891b0319ae45e07e9b343a5689ef911992766b7db30939d7527a986.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 26 + }, + { + "type": "text", + "text": "能力提优测试卷", + "text_level": 1, + "page_idx": 27 + }, + { + "type": "text", + "text": "时间：60分钟 满分：100分", + "page_idx": 27 + }, + { + "type": "table", + "img_path": "images/4aa6982eb6b990bc36d960faf67375edcf9e3332f51b3d3ea440e8b2d041da43.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 27 + }, + { + "type": "text", + "text": "一、选择题(每小题4分，共32分)", + "text_level": 1, + "page_idx": 27 + }, + { + "type": "text", + "text": "1.下列图形中,正确画出 $\\triangle A B C$ 的AC 边上的高的图形是", + "page_idx": 27 + }, + { + "type": "image", + "img_path": "images/3a4f8741717a68ab4052634ce3d3320a57e550eaf826171c9f8e304ef82e27cf.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 27 + }, + { + "type": "text", + "text": "2.如果一个三角形的两边长分别为2和5,其第三边长为正整数,那么此三角形的周长可能为 （ ）", + "page_idx": 27 + }, + { + "type": "text", + "text": "A.14 B.13 C.10 D.9 ", + "page_idx": 27 + }, + { + "type": "text", + "text": "3.如图， $. A B / / D F , A C \\perp C E$ 于点 $C$ ，直线 $B C$ 与 $D F$ 相交于点 $E$ ，若 $\\angle A = 2 0 ^ { \\circ }$ ，则$\\angle C E F$ 等于 （ ）", + "page_idx": 27 + }, + { + "type": "text", + "text": "A. ${ 1 1 0 } ^ { o }$ （204号 B.100° C.80° D. $7 0 ^ { \\circ }$ ", + "page_idx": 27 + }, + { + "type": "text", + "text": "5872Ab", + "page_idx": 27 + }, + { + "type": "image", + "img_path": "images/c57aead7fda36cacd51a21da49318a0e488eb569e6d590d0151d02940cc70b86.jpg", + "img_caption": [ + "第3题图" + ], + "img_footnote": [], + "page_idx": 27 + }, + { + "type": "image", + "img_path": "images/3034a5ed68ab601ad2e32076fed2847c470f20fbc1cecef924e08dff008232e2.jpg", + "img_caption": [ + "第4题图" + ], + "img_footnote": [], + "page_idx": 27 + }, + { + "type": "text", + "text": "4.如图，已知 $\\triangle A B C$ 的六个元素如图所示，则甲、乙、丙三个三角形中与△ABC全等的图形是 （ ）", + "page_idx": 27 + }, + { + "type": "text", + "text": "A.甲 B.乙 C.丙 D.乙与丙 ", + "page_idx": 27 + }, + { + "type": "text", + "text": "5.如图,AE//DF, $A E { = } D F$ ,则添加下列条件还不能使 $\\triangle E A C { \\cong } \\triangle F D B$ 的是", + "page_idx": 27 + }, + { + "type": "image", + "img_path": "images/97acb876afabed35b710e49c4d6bb6bbffb34be4ed24d772b7af8b5d294773ea.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 27 + }, + { + "type": "text", + "text": "A. $A B { = } C D$ B. CE// BF C. $C E { = } B F$ （204号 D. $\\angle E { = } \\angle F$ （20 ", + "page_idx": 27 + }, + { + "type": "text", + "text": "6.张叔叔用同种材料制成如图所示的金属框架，已知 $\\angle B = \\angle E , A B = D E , B F$ $= E C$ ，其中△ABC的周长为 $2 4 \\ \\mathrm { c m } , C F = 3 \\ \\mathrm { c m }$ ，则制成整个金属框架所需这种材料的长度为 （）", + "page_idx": 27 + }, + { + "type": "text", + "text": "A. $4 5 ~ \\mathrm { c m }$ （204号 B.48 cm C.51 cm D. $5 4 ~ \\mathrm { c m }$ ", + "page_idx": 27 + }, + { + "type": "image", + "img_path": "images/4f371b3002d9a9c937d8f0d15ed138e02d5dedf64cd00387d686e4ff4f1823ef.jpg", + "img_caption": [ + "第6题图" + ], + "img_footnote": [], + "page_idx": 27 + }, + { + "type": "image", + "img_path": "images/5164dc0f2d055da67ba5eb940ede95b6832aab21494fce1422d2af9c7764f768.jpg", + "img_caption": [ + "第7题图" + ], + "img_footnote": [], + "page_idx": 27 + }, + { + "type": "image", + "img_path": "images/ab6717f01fd7d06617ed73c1d95bf987298bb82011c0650e07da1adeff46c69a.jpg", + "img_caption": [ + "第8题图" + ], + "img_footnote": [], + "page_idx": 27 + }, + { + "type": "text", + "text": "7.如图，在 $\\triangle A B C$ 中，已知点 $D , E , F$ 分别为 $B C , A D , C E$ 的中点，且 ${ \\cal S } _ { \\triangle A B C } =$ $8 \\ \\mathrm { c m ^ { 2 } }$ ，则图中阴影部分 $\\triangle B E F$ 的面积等于 （ ）", + "page_idx": 27 + }, + { + "type": "text", + "text": "A. $1 ~ \\mathrm { c m ^ { 2 } }$ （204号 $\\mathrm { B } . 2 \\ \\mathrm { c m } ^ { 2 }$ （204号 （204号 $\\mathrm { C . 4 \\ c m ^ { 2 } }$ （204号 $\\mathrm { D } \\mathrm { : 6 \\ c m ^ { 2 } }$ ", + "page_idx": 27 + }, + { + "type": "text", + "text": "8. $\\circledcirc$ 视频讲解如图， $A D$ 是 $\\triangle A B C$ 的中线，点 $E , F$ 分别是 $A D$ 和 $A D$ 的延长线上的点，且 $D E { = } D F$ ，连接 $B F , C E$ ，且 $\\angle F B D = 3 5 ^ { \\circ }$ $\\angle B D F { = } 7 5 ^ { \\circ }$ .则下列说法：$\\textcircled { 1 } \\triangle B D F { \\cong } \\triangle C D E ; \\textcircled { 2 } \\triangle A B D$ 和 $\\triangle A C D$ 的面积相等； $\\textcircled { 3 } B F / / C E ; \\textcircled { 4 } L E C$ $= 7 0 ^ { \\circ }$ .其中正确的有 （）", + "page_idx": 27 + }, + { + "type": "text", + "text": "A.1个 B.2个 C.3个 D.4个 ", + "page_idx": 27 + }, + { + "type": "text", + "text": "二、填空题(每小题5分，共20分）", + "text_level": 1, + "page_idx": 27 + }, + { + "type": "text", + "text": "9.如图， $A D$ 是 $\\triangle A B C$ 的中线，且△ABD比△ACD的周长长 $3 ~ \\mathrm { c m }$ ，则 $A B$ 比 $A C$ 长 cm. ", + "page_idx": 27 + }, + { + "type": "image", + "img_path": "images/71f5f574c1f6e1bba4ec1cf81f98fcc382f5799806e39ab020f2643ab042c263.jpg", + "img_caption": [ + "第9题图" + ], + "img_footnote": [], + "page_idx": 27 + }, + { + "type": "image", + "img_path": "images/a15a80010cdb28085e67cdd2d44f28d61bae467ab25371f354c43fa40fb9525a.jpg", + "img_caption": [ + "第10题图" + ], + "img_footnote": [], + "page_idx": 27 + }, + { + "type": "image", + "img_path": "images/944f3f42e8d02ce18054e906e8e10c46ad17a05f0ddaac18330b79bb04b004cc.jpg", + "img_caption": [ + "第11题图" + ], + "img_footnote": [], + "page_idx": 27 + }, + { + "type": "image", + "img_path": "images/5e314ad812638bb1b26e145901bbb4e51e05e6952f974c9f25aa0409c5aa790e.jpg", + "img_caption": [ + "第12题图" + ], + "img_footnote": [], + "page_idx": 27 + }, + { + "type": "text", + "text": "10.如图，这是一个三角形测平架，已知 $A B { = } A C$ ，在 $B C$ 的中点 $D$ 处挂一个重锤，自然下垂.调整架身，使点 $A$ 恰好在重锤线上，则 $A D$ 和 $B C$ 的位置关系为", + "page_idx": 27 + }, + { + "type": "text", + "text": "11.小聪做了一个如图所示的风筝，其中 $\\angle E D H = \\angle F D H , E D = F D = 2 0 ~ \\mathrm { c m } ,$ $E H { = } 2 5 \\ \\mathrm { c m }$ ,则此四边形风筝的周长是 cm.", + "page_idx": 27 + }, + { + "type": "text", + "text": "12.如图，在正方形网格中， $\\triangle A B C$ 的三个顶点及点 $D , E , F , G , H$ 都在格点上,现以 $D , E , F , G , H$ 中的三点为顶点画三角形，则能与 $\\triangle A B C$ 全等的三角形是", + "page_idx": 27 + }, + { + "type": "text", + "text": "三、解答题(本大题5小题，共48分)", + "text_level": 1, + "page_idx": 28 + }, + { + "type": "text", + "text": "l3.(8分)如图，在 $\\triangle A B C$ 中，点 $D$ 是 $B C$ 的中点,点 $E$ 是 $A B$ 边上一点,过点 $C$ （20作 $C F / / A B$ ，交 $E D$ 的延长线于点F.试说明 $\\triangle B D E { \\cong } \\triangle C D F .$ ", + "page_idx": 28 + }, + { + "type": "image", + "img_path": "images/b5fab8081a5e3f0ea8a9840302daa2bb3d72542d81cf3ba302b75d79c04674c5.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 28 + }, + { + "type": "text", + "text": "14.(8分)如图,已知线段 $a$ 和 $\\angle \\alpha$ ,用尺规作一个 $\\triangle A B C$ ，使 $A B = A C = 2 a$ ，$\\angle B A C = 1 8 0 ^ { \\circ } - \\angle \\alpha .$ （保留作图痕迹,不写作法）", + "page_idx": 28 + }, + { + "type": "image", + "img_path": "images/215f7ec1eb45d422cfcfb2d4dd7775577b99bec4ada13e1dd16d0ba4fb6030b9.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 28 + }, + { + "type": "text", + "text": "15.(10 分)在新修的花园中,有一条“ $Z ^ { \\prime }$ 字形绿色长廊 $A B C D$ ，如图， $A B / / C D$ ，在AB,BC,CD三段绿色长廊上各修建一凉亭E,M,F,且BE=CF，点M是 $B C$ 的中点，点 $E , M , F$ 在同一条直线上.若凉亭 $M$ 与 $F$ 之间有一池塘，在用皮尺不能直接测量的情况下,你能得出凉亭 $M$ 与 $F$ 之间的距离吗？试说明理由.", + "page_idx": 28 + }, + { + "type": "image", + "img_path": "images/4ac72d2d57fd533293d50ce1e983796a2d53c61ac56595a04dbdce6e3cd8dca6.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 28 + }, + { + "type": "text", + "text": "16.(10 分)如图所示,在四边形 $A B C D$ 中， $A D / / B C$ ，点 $E$ 为 $C D$ 的中点，连接$A E , B E$ ，延长 $A E$ 交 $B C$ 的延长线于点 $F$ ", + "page_idx": 28 + }, + { + "type": "text", + "text": "(1)判断 $F C$ 与 $A D$ 的数量关系,并说明理由. ", + "page_idx": 28 + }, + { + "type": "text", + "text": "(2)若 $A B { = } B C { + } A D$ ，则 $B E \\bot A F$ 吗？为什么？", + "page_idx": 28 + }, + { + "type": "image", + "img_path": "images/bb7601968bbeabb9ef1215d29c4ef1109166c74013dbb7c124dee63824609d82.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 28 + }, + { + "type": "text", + "text": "17.新考法综合与实践 $\\circledcirc$ 视频讲解（12分）如图， $C D$ 是经过 $\\angle B C A$ 顶点 $C$ 的一条直线,CA=CB,点E,F分别是直线CD上两点,且∠BEC=∠CFA=α.", + "page_idx": 28 + }, + { + "type": "text", + "text": "（1)若直线CD经过∠BCA的内部，且点E,F在射线CD上.如图1,若 $\\angle B C A = 9 0 ^ { \\circ }$ $\\alpha { = } 9 0 ^ { \\circ }$ ，则 $B E$ $C F$ （填“>”“ $=$ ”或“ $< ^ { 9 9 }$ ", + "page_idx": 28 + }, + { + "type": "text", + "text": "(2)在(1)的条件下，如图2,若0°<∠BCA<180°，请添加一个关于α与$\\angle B C A$ 关系的条件 ,使(1)中的结论仍然成立,并说明理由；", + "page_idx": 28 + }, + { + "type": "text", + "text": "（3)如图3,若直线CD经过∠BCA的外部,α=∠BCA,请提出关于EF,$B E , A F$ 三条线段数量关系的合理猜想,并说明理由.", + "page_idx": 28 + }, + { + "type": "image", + "img_path": "images/bdaab4afeb8ddd28affa1b99c3c489c42adb4083ad36ce5a5c9a75ad57a0987b.jpg", + "img_caption": [ + "图1" + ], + "img_footnote": [], + "page_idx": 28 + }, + { + "type": "image", + "img_path": "images/b9dc855fd3169eb89569d47a7449c494e9b3f1e6b72d0c5eda889442f8627ee5.jpg", + "img_caption": [ + "图2" + ], + "img_footnote": [], + "page_idx": 28 + }, + { + "type": "image", + "img_path": "images/d1730998969cf2cbaf218682f1d65f9d0b59e862a64ad8b5c72c6a5e7d4b5100.jpg", + "img_caption": [ + "图3 " + ], + "img_footnote": [], + "page_idx": 28 + }, + { + "type": "text", + "text": "第五章 生活中的轴对称", + "text_level": 1, + "page_idx": 29 + }, + { + "type": "text", + "text": "基础过关测试卷", + "text_level": 1, + "page_idx": 29 + }, + { + "type": "text", + "text": "时间：60分钟 满分：100分", + "page_idx": 29 + }, + { + "type": "table", + "img_path": "images/9d4fe14d1981a905ba6ccfb92a93ebac997de52c7e853710858b421178602719.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 29 + }, + { + "type": "text", + "text": "一、选择题(每小题4分，共32分）", + "text_level": 1, + "page_idx": 29 + }, + { + "type": "text", + "text": "1.下列图形中，是轴对称图形的是", + "page_idx": 29 + }, + { + "type": "image", + "img_path": "images/5be74c2c8c8ef2bb879676abcc87f0259163169e405bb3c2aa12d1bbf8341d6b.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 29 + }, + { + "type": "text", + "text": "2.等腰三角形是轴对称图形，它的对称轴是", + "page_idx": 29 + }, + { + "type": "text", + "text": "A.过顶点的直线 B.底边上的高C.顶角的平分线所在的直线 D.腰上的高所在的直线", + "page_idx": 29 + }, + { + "type": "text", + "text": "3.将一张长方形纸片对折，然后用笔尖在纸片上扎出字母B,再把它展开铺平，那么可以看到的图形是 （ ）", + "page_idx": 29 + }, + { + "type": "image", + "img_path": "images/26af1b8beacf79304a913b2e667fc9bb7c0a7baeffe1e4e8d41d06c4e13bb664.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 29 + }, + { + "type": "text", + "text": "4.如图， $\\triangle A O D$ 与△BOC关于直线l成轴对称,若 $\\angle D A O = 3 0 ^ { \\circ }$ $\\angle A D O = 2 5 ^ { \\circ }$ ， 则 $\\angle B O C$ 的度数为 ", + "page_idx": 29 + }, + { + "type": "text", + "text": "A. $1 2 5 ^ { \\circ }$ （20 B.120° C.115° D. $1 1 0 ^ { \\circ }$ ", + "page_idx": 29 + }, + { + "type": "image", + "img_path": "images/e6e3bb1ef95a27593ed917f449888f8bf41e1004c162a21c23b848ac273091a7.jpg", + "img_caption": [ + "第4题图" + ], + "img_footnote": [], + "page_idx": 29 + }, + { + "type": "image", + "img_path": "images/3d5fff82e9b5bf58ad10ca22fb2e188a33558b4bb504a54060ccc521eb86ba12.jpg", + "img_caption": [ + "第5题图" + ], + "img_footnote": [], + "page_idx": 29 + }, + { + "type": "image", + "img_path": "images/283b759b53452f19ef4d09d6b60925222028e7ce4e46a9898846ff1bf79cd8d6.jpg", + "img_caption": [ + "第6题图" + ], + "img_footnote": [], + "page_idx": 29 + }, + { + "type": "image", + "img_path": "images/7a4c6b615a7e9bf9e360a339cdf5f2f4e4a68e806108890a45cbe75b29a7bd37.jpg", + "img_caption": [ + "第8题图" + ], + "img_footnote": [], + "page_idx": 29 + }, + { + "type": "text", + "text": "5.如图，在 $\\mathrm { R t } \\triangle A B C$ 中， $\\angle C = 9 0 ^ { \\circ }$ ,以顶点 $A$ 为圆心,适当的长为半径画弧,分", + "page_idx": 29 + }, + { + "type": "text", + "text": "别交 $A C , A B$ 于点 $M , N$ ，再分别以点 $M , N$ 为圆心，大于 $\\cdot \\frac { 1 } { 2 } M N$ 的长为半径画弧，两弧交于点 $P$ ，作射线 $A P$ 交边BC于点 $D$ .若 $C D { = } 4$ $A B { = } 1 5$ ，则 $\\triangle A B D$ 的面积是", + "page_idx": 29 + }, + { + "type": "text", + "text": "A.15 B.30 C.45 D.60 ", + "page_idx": 29 + }, + { + "type": "text", + "text": "6.如图，在 $\\triangle A B C$ 中,按以下步骤作图： $\\textcircled{1}$ 分别以点 $B , C$ 为圆心，以大于 ${ \\frac { 1 } { 2 } } B C$ 的长为半径作弧，两弧交于 $M , N$ 两点； $\\textcircled{2}$ 作直线 $M N$ 交 $A B$ 于点 $D$ ，连接CD.若 $A B { = } 1 0 \\ \\mathrm { c m } , A C { = } 4 \\ \\mathrm { c m }$ ，则 $\\triangle A D C$ 的周长为 （）", + "page_idx": 29 + }, + { + "type": "text", + "text": "A. $1 4 ~ \\mathrm { c m }$ B. $1 3 ~ \\mathrm { c m }$ C. 12 cm D.11 cm ", + "page_idx": 29 + }, + { + "type": "text", + "text": "7.如图，将一张正方形纸片沿对角线折叠一次，然后在得到的三角形的三个角上各挖去一个圆洞，最后将正方形纸片展开,得到的图案是 （ ）", + "page_idx": 29 + }, + { + "type": "image", + "img_path": "images/ab7fc988313964adf35da3960d3c7b1eb28591688a51f851810488b192ea2204.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 29 + }, + { + "type": "text", + "text": "8.如图,在四边形 $A B C D$ 中， $\\angle D = 9 0 ^ { \\circ }$ ,AE⊥CB,且 $A C$ 平分 $\\angle D A E , C D = 4$ ，$A D { = } 6 , B E { = } 2$ ,则四边形 $A B C D$ 的面积为 （ ）", + "page_idx": 29 + }, + { + "type": "text", + "text": "A.24 B.26 C.28 D.30 ", + "page_idx": 29 + }, + { + "type": "text", + "text": "二、填空题(每小题5分，共20分)", + "text_level": 1, + "page_idx": 29 + }, + { + "type": "text", + "text": "9.如图，这是一个轴对称图形，它的对称轴有 条.", + "page_idx": 29 + }, + { + "type": "image", + "img_path": "images/6e16085aa9c8ccac501f0ed07fa588a92f97fd0b49fd2fb7f0616f010af46ef8.jpg", + "img_caption": [ + "第9题图" + ], + "img_footnote": [], + "page_idx": 29 + }, + { + "type": "image", + "img_path": "images/b7094c6ce2159c5215655e45e9ea0a5d70196d98266f80e4b6379ec3a0f93890.jpg", + "img_caption": [ + "第11题图" + ], + "img_footnote": [], + "page_idx": 29 + }, + { + "type": "image", + "img_path": "images/e0be3378705c3912bacfa2da131aa4b188efcb5adf72872f70bb39c2fa71059a.jpg", + "img_caption": [ + "第12题图" + ], + "img_footnote": [], + "page_idx": 29 + }, + { + "type": "text", + "text": "10.已知直线l是线段 $A B$ 的垂直平分线，点 $C$ 是直线l上的点，连接 $C A , C B$ ， 若 $\\angle B A C = 6 0 ^ { \\circ }$ ，则 $\\angle A B C$ 的度数为 ", + "page_idx": 29 + }, + { + "type": "text", + "text": "11.如图，这是一个轴对称图形， $B C$ 所在的直线是它的对称轴，则图中共有对全等三角形.", + "page_idx": 29 + }, + { + "type": "text", + "text": "12. $\\circledcirc$ 视讲解在如图所示的正方形网格中画有两条线段，现在要再画一条，使图中的三条线段组成一个轴对称图形，能满足条件的线段有 条.", + "page_idx": 29 + }, + { + "type": "text", + "text": "三、解答题（本大题5小题，共48分）", + "text_level": 1, + "page_idx": 30 + }, + { + "type": "text", + "text": "13.(6分)下图中的图形都是轴对称图形，请你画出它们的对称轴.", + "page_idx": 30 + }, + { + "type": "image", + "img_path": "images/fcc7879d0078d9a7e2a6faf7e7fe149fbf281a3e20f16859339a5db7706a1141.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 30 + }, + { + "type": "text", + "text": "14.（8分)如图,已知△ABC和 $\\triangle A ^ { \\prime } B ^ { \\prime } C ^ { \\prime }$ 关于直线l对称. ", + "page_idx": 30 + }, + { + "type": "text", + "text": "(1） $\\triangle A B C$ 与 $\\triangle A ^ { \\prime } B ^ { \\prime } C ^ { \\prime }$ （填“全等”或“不全等”). ", + "page_idx": 30 + }, + { + "type": "text", + "text": "(2)若 $A B { = } 4 \\ \\mathrm { c m } , A ^ { \\prime } C ^ { \\prime } { = } 6 \\ \\mathrm { c m } , B C { = } 3 \\ \\mathrm { c m } , A C$ 边上的高为 $2 \\mathrm { \\ c m }$ ，求 $\\triangle A ^ { \\prime } B ^ { \\prime } C ^ { \\prime }$ 的 周长和面积. ", + "page_idx": 30 + }, + { + "type": "image", + "img_path": "images/3217f837003877801e6d0c72cecdc99223178b93a5358e3b919841bf9de1af30.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 30 + }, + { + "type": "text", + "text": "15.(10分)两个城镇 $A , B$ 与两条公路 $l _ { 1 } , l _ { 2 }$ 的位置如图所示，电信部门需在 $C$ 处修建一座信号发射塔,要求发射塔到两个城镇 $A , B$ 的距离必须相等，到两条公路 $l _ { 1 } , l _ { 2 }$ 的距离也必须相等,那么点 $C$ 应选在何处？请在图中用尺规作图找出所有符合条件的点C.(不写已知、求作、作法,只保留作图痕迹)", + "page_idx": 30 + }, + { + "type": "image", + "img_path": "images/571a1d0d252afa4259fc2c93b205e2e7821f894463d909aed53b5a35be727218.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 30 + }, + { + "type": "text", + "text": "16.(10 分)已知等腰三角形一腰上的高与另一腰的夹角的度数为 $5 6 ^ { \\circ }$ ,求这个等腰三角形的底角的度数.", + "page_idx": 30 + }, + { + "type": "text", + "text": "17.新考法阅读理解题视频讲（14分)阅读材料：同学们已经知道，角平分线的性质是角平分线上的点到这个角的两边的距离相等,那么把上述性质反过来就可以得到在一个角的内部,到角的两边的距离相等的点在这个角的平分线上.问题情境：如图，在四边形ABCD中， $C E \\bot A B$ 于点 $E$ ${ \\vec { \\mathfrak { z } } } , C D { = } C B$ $\\angle A B C +$ $\\angle A D C = 1 8 0 ^ { \\circ }$ 问题解决：试说明 $A C$ 平分 $\\angle B A D$ 合作探究：(1)在\"问题解决”的基础上,若 $A E { = } 3 B E { = } 9$ ，能求出 $A D$ 的长吗？请说明理由.（提示：在两个直角三角形中，一直角边与斜边分别相等的两直角三角形全等)(2)如果△ABC和 $\\triangle A C D$ 的面积分别为40和 28,请求出 $\\triangle B C E$ 的面积.", + "page_idx": 30 + }, + { + "type": "text", + "text": "", + "page_idx": 30 + }, + { + "type": "image", + "img_path": "images/0e188d8137783bc6629e639ae1bff0fa28217c7f08d8efe5c14c20f1d5a9c72a.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 30 + }, + { + "type": "text", + "text": "能力提优测试卷", + "text_level": 1, + "page_idx": 31 + }, + { + "type": "text", + "text": "时间：60分钟 满分：100分", + "page_idx": 31 + }, + { + "type": "table", + "img_path": "images/f8d108db6ba3f5eb1575540a6cd46c33dbcd65c370e79e6dc45ebdd638eab893.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 31 + }, + { + "type": "text", + "text": "一、选择题(每小题4分，共32分）", + "text_level": 1, + "page_idx": 31 + }, + { + "type": "text", + "text": "1.第24届冬季奥林匹克运动会于2022年2月在北京举行，下面由冬季奥运会比赛项目图标组成的四个图形中，可以看作轴对称图形的是 （ ）", + "page_idx": 31 + }, + { + "type": "image", + "img_path": "images/a82b986f0670d8e2218db721caa62146cecc270e09f4abd0cb31c93b4a89505c.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 31 + }, + { + "type": "text", + "text": "2.如图,若△ABC与 $\\triangle D E F$ 关于直线l对称， $B E$ 交l于点 $O$ ,则下列说法不一 定正确的是 （ ", + "page_idx": 31 + }, + { + "type": "text", + "text": "A. AB//EF B. $A C { = } D F$ C.AD⊥l D. $B O { = } E O$ ", + "page_idx": 31 + }, + { + "type": "image", + "img_path": "images/d0a27a77bad966163454a995053cafd96d93dd43cb542761c21e88cdb85dc0c0.jpg", + "img_caption": [ + "第2题图" + ], + "img_footnote": [], + "page_idx": 31 + }, + { + "type": "image", + "img_path": "images/bc6bc9b14b927b480b0e18dad583de139280679ec42e9092e7b5d707fd5e6286.jpg", + "img_caption": [ + "第3题图" + ], + "img_footnote": [], + "page_idx": 31 + }, + { + "type": "image", + "img_path": "images/49197a7ae22499de00155631453d375e9e0914bf5b919407a1ba68e71230fbc0.jpg", + "img_caption": [ + "第5题图" + ], + "img_footnote": [], + "page_idx": 31 + }, + { + "type": "text", + "text": "3.如图，在△ABC中, $A B { = } 6$ $A C { = } 4$ ，分别以点 $B$ 和点 $C$ 为圆心,以大于 $B C$ 一半的长为半径画弧，两弧相交于点 $M , N ,$ 作出直线 $M N$ 交 $A B$ 于点 $D$ ，连接$C D$ ，则 $\\triangle A D C$ 的周长为 （ ）", + "page_idx": 31 + }, + { + "type": "text", + "text": "A.8 B.10 C.12 D.14 ", + "page_idx": 31 + }, + { + "type": "text", + "text": "4.若等腰三角形一腰上的高与另一腰的夹角为 $1 5 ^ { \\circ }$ ,则顶角的度数为（）", + "page_idx": 31 + }, + { + "type": "text", + "text": "A $7 5 ^ { \\circ }$ （204号 B. $1 5 ^ { \\circ }$ C.15°或 $1 6 5 ^ { \\circ }$ D. $7 5 ^ { \\circ }$ 或 $1 0 5 ^ { \\circ }$ ", + "page_idx": 31 + }, + { + "type": "text", + "text": "5.如图，在 $\\triangle A B C$ 中， $\\angle C = 9 0 ^ { \\circ } , A B = 1 0 , A D$ 平分 $\\angle B A C .$ 若 $C D = 3$ ，则$\\triangle A B D$ 的面积为 （ ）", + "page_idx": 31 + }, + { + "type": "text", + "text": "A.15 B.24 C.30 D.48 ", + "page_idx": 31 + }, + { + "type": "text", + "text": "6.如图,在△ABC中， $B O$ 平分∠ABC,CO平分△ABC的外角 $\\angle A C D , M N$ 经过点 $O$ ，与 $A B , A C$ 相交于点 $M , N$ ，且 $M N / / B C$ ，则 $B M , C N , M N$ 之间的关系是 （）", + "page_idx": 31 + }, + { + "type": "text", + "text": "A.1 $B M { + } C N { = } M N$ E $3 . B M { - } C N { = } M N$ C.CN- $B M { = } M N$ D.BM-CN=2MN ", + "page_idx": 31 + }, + { + "type": "image", + "img_path": "images/756cab4a12ac935b2f5295cc2b49d0e301eb4414842effb6cc84e030a9df918c.jpg", + "img_caption": [ + "第6题图" + ], + "img_footnote": [], + "page_idx": 31 + }, + { + "type": "image", + "img_path": "images/78bb3167099bf4f0599f65795c273315d7651052d3bd808d0d826fdb6d530e14.jpg", + "img_caption": [ + "第7题图" + ], + "img_footnote": [], + "page_idx": 31 + }, + { + "type": "image", + "img_path": "images/8e947cd9284752797e35f8ffd5135f308216d0f5db08f61365d1fc6422c828f5.jpg", + "img_caption": [ + "第8题图" + ], + "img_footnote": [], + "page_idx": 31 + }, + { + "type": "text", + "text": "7.如图,将长方形纸片ABCD折叠,使顶点 $B$ 落在边 $A D$ 的点 $E$ 上，折痕FG交$B C$ 于点 $G$ ，交 $A B$ 于点 $F$ ，若 $\\angle A E F { = } 2 0 ^ { \\circ }$ ，则 $\\angle F G B$ 的度数为 （ ）", + "page_idx": 31 + }, + { + "type": "text", + "text": "A. $2 { 0 } ^ { \\circ }$ （20 （20 $\\mathrm { B } . 2 5 ^ { \\circ }$ （20 $ { \\mathrm { C } } . 3 0 ^ { \\circ }$ D.35° ", + "page_idx": 31 + }, + { + "type": "text", + "text": "8. $\\circledcirc$ 视频讲解如图，等腰三角形 $A B C$ 的底边 $B C$ 长为4,面积是16,腰 $A C$ 的垂直平分线EF分别交 $A C , A B$ 边于点 $E , F .$ 若点 $D$ 为 $B C$ 边的中点，点 $M$ 为线段EF上一动点，则 $\\triangle C D M$ 周长的最小值为 （）", + "page_idx": 31 + }, + { + "type": "text", + "text": "A.6 B.8 C.10 D.12 ", + "page_idx": 31 + }, + { + "type": "text", + "text": "二、填空题(每小题5分，共20分)", + "text_level": 1, + "page_idx": 31 + }, + { + "type": "text", + "text": "9.等腰三角形的周长为 $1 4 ~ \\mathrm { c m }$ ,其中一边长为 $4 \\mathrm { \\ c m }$ ，则它的底边长为 cm.", + "page_idx": 31 + }, + { + "type": "text", + "text": "10.如图，在 $\\triangle A B C$ 中， $\\angle B A C { > } 9 0 ^ { \\circ } ,$ AB的垂直平分线交BC于点 $E , A C$ 的垂直平分线交BC于点 $F$ .若 $\\triangle A E F$ 的周长为 $6 ~ \\mathrm { c m }$ ，则 $B C =$ cm.", + "page_idx": 31 + }, + { + "type": "image", + "img_path": "images/a5e5a29632a5a0563a8e5dd2fe7d736d0da7d47c8fab34ebae773687f127879e.jpg", + "img_caption": [ + "第11题图" + ], + "img_footnote": [], + "page_idx": 31 + }, + { + "type": "image", + "img_path": "images/2eaf507a4687831260eb1f32ca43dd616ea37fae98a93f690cc4b1dc88ebbe7e.jpg", + "img_caption": [ + "第10题图" + ], + "img_footnote": [], + "page_idx": 31 + }, + { + "type": "image", + "img_path": "images/0940a76b6725a159f58ae59917532acfc1bde723ace05c9a817a3c60713a6364.jpg", + "img_caption": [ + "第12题图" + ], + "img_footnote": [], + "page_idx": 31 + }, + { + "type": "text", + "text": "11.如图， $O E$ 是 $\\angle A O B$ 的平分线， $B D \\bot O A$ 于点 $D , A C \\bot B O$ 于点 $C$ ,则关于直线OE对称的三角形有 对.", + "page_idx": 31 + }, + { + "type": "text", + "text": "12.如图，在 $\\triangle A B C$ 中， $A B { = } A C$ ，点 $D$ 为线段BC上一动点（不与点 $B , C$ 重合)，连接 $A D$ ，作 $\\angle D A E { = } \\angle B A C$ ，且 $A D { = } A E$ ，连接 $D E , C E .$ 当 $C E / / A B$ 时，若 $\\angle B A D = 3 5 ^ { \\circ }$ ，则 $\\angle D E C$ 的度数为", + "page_idx": 31 + }, + { + "type": "text", + "text": "三、解答题(本大题5小题，共48分）", + "text_level": 1, + "page_idx": 32 + }, + { + "type": "text", + "text": "l3.(8分)如图，在等边△ABC中,过 $B C$ 边上一点 $P$ ，作 $\\angle D P E = 6 0 ^ { \\circ }$ ,分别与边$A B , A C$ 相交于点D,E.在图中找出与 $\\angle E P C$ 始终相等的角,并说明理由.", + "page_idx": 32 + }, + { + "type": "image", + "img_path": "images/3657d08511585f5b2dbfc81c5b96765aec95c72dfa21a5aee90f91e60e9b4919.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 32 + }, + { + "type": "text", + "text": "14.(8分)如图,在 $1 0 \\times 1 0$ 的正方形网格中,每个小正方形的边长都为1,网格中有一个格点 $\\triangle A B C$ (即三角形的顶点都在格点上).", + "page_idx": 32 + }, + { + "type": "text", + "text": "(1)在图中作出 $\\triangle A B C$ 关于直线l对称的 $\\triangle A _ { 1 } B _ { 1 } C _ { 1 }$ （要求：点 $A$ 与点 $A _ { 1 }$ ，点$B$ 与点 $B _ { 1 }$ ，点 $C$ 与点 $C _ { 1 }$ 相对应)；", + "page_idx": 32 + }, + { + "type": "text", + "text": "(2)在(1)的结果下，连接 $B B _ { 1 }$ ， $C C _ { 1 }$ ,求四边形 $B B _ { 1 } C _ { 1 } C$ 的面积. ", + "page_idx": 32 + }, + { + "type": "image", + "img_path": "images/8895bfd8964a5faf6816a6b464187de20f6606d70f228bcc772ae0952437dfe2.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 32 + }, + { + "type": "text", + "text": "15.(10分)如图，在 $\\triangle A B C$ 中， $A D \\perp B C , E F$ 垂直平分 $A C$ ，交 $A C$ 于点 $F$ ，交BC于点 $E$ ，且 $B D { = } D E$ ，连接 $A E$ ", + "page_idx": 32 + }, + { + "type": "text", + "text": "(1)若 $\\angle B A E { = } 4 0 ^ { \\circ }$ ，求 $\\angle C$ 的度数； \n(2)若 $\\triangle A B C$ 的周长为 $1 4 { \\mathrm { ~ c m } } , A C { = } 6 { \\mathrm { ~ c m } }$ ，求 $D C$ 的长. ", + "page_idx": 32 + }, + { + "type": "image", + "img_path": "images/b138907b52911cd1f053242869cca2b24be6699636df370ea034fc7afecd0175.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 32 + }, + { + "type": "text", + "text": "16.(10分)如图，点 $P$ 关于 $O A , O B$ 对称的对称点分别为点 $C , D$ ，连接 $C D$ ，交$O A$ 于点 $M$ ，交 $O B$ 于点 $N$ ", + "page_idx": 32 + }, + { + "type": "text", + "text": "(1)若 $C D$ 的长为 $1 8 ~ \\mathrm { c m }$ ，求 $\\triangle P M N$ 的周长； ", + "page_idx": 32 + }, + { + "type": "text", + "text": "(2)若 $\\angle C = 2 1 ^ { \\circ }$ $\\angle D = 2 8 ^ { \\circ }$ ，求 $\\angle M P N$ 的度数. ", + "page_idx": 32 + }, + { + "type": "image", + "img_path": "images/cd8e3aa78fc5077f3a1f2d8305db79772a2c56fc1e402a7ab3af8e25e8087cb4.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 32 + }, + { + "type": "text", + "text": "17.视频讲解（12分）如图，在 $\\triangle A B C$ 中， $A B { = } A C$ $\\angle B A C = 9 0 ^ { \\circ }$ ，点 $O$ 为BC的中点.", + "page_idx": 32 + }, + { + "type": "text", + "text": "(1)写出点 $O$ 到 $\\triangle A B C$ 的三个顶点 $A , B , C$ 的距离的大小关系（不要求说明理由).", + "page_idx": 32 + }, + { + "type": "text", + "text": "(2)如果点 $M , N$ 分别在线段 $A B , A C$ 上移动,在移动过程中保持 $A N { = } B M$ ， 请判断 $\\triangle { O M N }$ 的形状,并说明理由. ", + "page_idx": 32 + }, + { + "type": "text", + "text": "(3)在(2)的条件下,四边形 AMON的面积是否发生变化？请说明理由.", + "page_idx": 32 + }, + { + "type": "image", + "img_path": "images/f20572db92dd6feb97613235a0e983f70dca245cbc375e79cc01b4582ec4c024.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 32 + }, + { + "type": "text", + "text": "月考测试卷（二）", + "text_level": 1, + "page_idx": 33 + }, + { + "type": "text", + "text": "时间：90分钟 满分：120分考试范围:第一章～第五章第2节", + "page_idx": 33 + }, + { + "type": "table", + "img_path": "images/193cc2f6f11ee2b44c0adfbf788d5a064e6b685fed213409e9b5fbbba453f6e8.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 33 + }, + { + "type": "text", + "text": "一、选择题(本大题共10小题，每小题3分，共30 分)", + "text_level": 1, + "page_idx": 33 + }, + { + "type": "text", + "text": "1.下列运算正确的是", + "page_idx": 33 + }, + { + "type": "text", + "text": "A. $a ^ { 2 } \\cdot a ^ { 4 } = a ^ { 8 }$ B $( - 2 a ^ { 2 } ) ^ { 3 } = - 6 a ^ { 5 }$ $\\complement , a ^ { 5 } \\div a ^ { 3 } = a ^ { 2 }$ $\\mathrm { D } . ~ ( - a - b ) ^ { 2 } = a ^ { 2 } - 2 a b + b ^ { 2 }$ ", + "page_idx": 33 + }, + { + "type": "text", + "text": "2.在 $\\triangle A B C$ 中， $A B { = } 3 ~ \\mathrm { c m }$ $B C { = } 7 ~ \\mathrm { c m }$ ，若 $A C$ 的长为整数,则 $A C$ 的长可能是（", + "page_idx": 33 + }, + { + "type": "equation", + "img_path": "images/b0ae1f3f663e99d76fdfaf6df2b77bf6c5ffc448ce264c4cd537b86a0c0897c5.jpg", + "text": "$$\n\\begin{array} { r l r l r l r l r l r l } { ) } & { { } \\mathrm { c m } } & { \\qquad } & { } & { { } \\mathrm { B } , 5 \\ \\mathrm { c m } } & { \\qquad } & { } & { { } \\mathrm { C } , 4 \\ \\mathrm { c m } } & { \\qquad } & { } & { { } \\mathrm { D } , 2 \\ \\mathrm { c m } } \\end{array}\n$$", + "text_format": "latex", + "page_idx": 33 + }, + { + "type": "text", + "text": "3.如图，在 $\\triangle A B C$ 中，点 $D , E , F$ 分别在边 $B C , A B , A C$ 上，下列能判定 $D E$ //$A C$ 的条件是 （ ）", + "page_idx": 33 + }, + { + "type": "text", + "text": "A. $\\angle 1 = \\angle 3$ \nB. $\\angle 3 { = } \\angle C$ \nC. $\\angle 2 = \\angle 4$ \nD. $\\angle 1 + \\angle 2 = 1 8 0 ^ { \\circ }$ ", + "page_idx": 33 + }, + { + "type": "image", + "img_path": "images/524454d8a00b015069dd8c70689d491f11bead7516196d20326c206820b76038.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 33 + }, + { + "type": "text", + "text": "4.“漏壶\"是一种古代计时器，如图，在壶内盛一定量的水，水从壶底的小孔漏出，壶内壁画有刻度，人们根据壶中水面的位置计算时间.用 $\\mathcal { X }$ 表示漏水时间， $y$ 表示壶底到水面的高度，不考虑水量变化对压力的影响，下列图象能表示 $y$ 与 $\\mathcal { X }$ 的对应关系的是 （ ）", + "page_idx": 33 + }, + { + "type": "image", + "img_path": "images/f7acc8a9605ecd5a42d55d41fce99aef1e0b9d960699eff7287001bce05b479c.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 33 + }, + { + "type": "text", + "text": "5.如图，直线 $A B , C D$ 相交于点 $O , O E \\bot C D , O F$ 平分∠BOD, $\\angle A O E = 2 4 ^ { \\circ }$ ，则$\\angle C O F$ 的度数是 （ ）", + "page_idx": 33 + }, + { + "type": "text", + "text": "A. $1 4 6 ^ { \\circ }$ B.147° C.157° D.136° ", + "page_idx": 33 + }, + { + "type": "image", + "img_path": "images/4b26281e697478683a0af18d041d0dbba7e032862d919ec31e9590adc1b8c555.jpg", + "img_caption": [ + "第5题图" + ], + "img_footnote": [], + "page_idx": 33 + }, + { + "type": "image", + "img_path": "images/c6a1f1de3c8a7659ebef9e5898c92fe9fb208c9ec47ccc3493ef8984968b44af.jpg", + "img_caption": [ + "第7题图" + ], + "img_footnote": [], + "page_idx": 33 + }, + { + "type": "image", + "img_path": "images/330366f53c0a12db4914b3f38ff8daaeca544950e74f86df47512dba18b681c9.jpg", + "img_caption": [ + "第8题图" + ], + "img_footnote": [], + "page_idx": 33 + }, + { + "type": "text", + "text": "6.如果 $x ^ { 2 } + x = 3$ ,那么代数式 $( x + 1 ) ( x - 1 ) + x ( x + 2 )$ 的值是", + "page_idx": 33 + }, + { + "type": "text", + "text": "A.2 B.3 C.5 D. 6", + "page_idx": 33 + }, + { + "type": "text", + "text": "7.如图，点 $D , E$ 分别是 $B C , A D$ 的中点， $\\triangle C E F$ 与 $\\triangle C E D$ 关于直线 $C E$ 对称，若 $\\triangle A B C$ 的面积是8,则 $\\triangle C E F$ 的面积为 （ ）", + "page_idx": 33 + }, + { + "type": "text", + "text": "A.8 B.6 C.4 D.2 ", + "page_idx": 33 + }, + { + "type": "text", + "text": "8.如图，已知 $A B { = } A C , A F { = } A E$ $\\angle E A F { = } \\angle B A C$ ，点 $C , D , E , F$ 共线.则下列结论：①△AFB≌△AEC; $\\textcircled { 2 } B F { = } C E$ $\\textcircled{3}$ ∠BFC=∠EAF; $\\textcircled { 4 } A B { = } B C .$ 其中正确的是 （）", + "page_idx": 33 + }, + { + "type": "equation", + "img_path": "images/0f7aeee023eb873bb82f657d499a7eb6fc0615dcb73f447f398ac7a383dbd0b7.jpg", + "text": "$$\n\\mathrm { ~ B _ { \\cdot } ~ } \\textcircled { 1 } \\textcircled { 2 } \\textcircled { 4 } \\qquad \\mathrm { C . } \\textcircled { 1 } \\textcircled { 2 } \\qquad \\mathrm { D . } \\textcircled { 1 } \\textcircled { 2 } \\textcircled { 3 } \\textcircled { 4 }\n$$", + "text_format": "latex", + "page_idx": 33 + }, + { + "type": "text", + "text": "9. $\\circledcirc$ 视频讲解如图， $\\triangle A B C$ 的内部有一点 $P$ ，且点 $D , E , F$ 是点 $P$ 分别以 $A B , B C$ $A C$ 为对称轴的对称点.若 $\\triangle A B C$ 的内角 $\\angle A = 7 0 ^ { \\circ }$ $\\angle B = 6 0 ^ { \\circ }$ $\\angle C = 5 0 ^ { \\circ }$ ，则$\\angle A D B + \\angle B E C + \\angle C F A$ 的度数是 （）", + "page_idx": 33 + }, + { + "type": "text", + "text": "A. $1 8 0 ^ { \\circ }$ B.270° C.360° D. $4 8 0 ^ { \\circ }$ ", + "page_idx": 33 + }, + { + "type": "image", + "img_path": "images/0e05d7f21eaac115da4e2ec694f61b04d93142aa3f442eb16aa28b1edf5612bb.jpg", + "img_caption": [ + "第9题图" + ], + "img_footnote": [], + "page_idx": 33 + }, + { + "type": "image", + "img_path": "images/112ca1d9080b20187f0b65ffda02dd8faf836bddcaee7b21330071724270627f.jpg", + "img_caption": [ + "第10题图" + ], + "img_footnote": [], + "page_idx": 33 + }, + { + "type": "text", + "text": "10.如图，在 $\\triangle A B C$ 中， $B D , C E$ 分别是 $A C , A B$ 边上的中线，分别延长 $B D , C E$ 到点 $F , G$ 使 $D F = B D , E G = C E$ ，则下列结论： $ G A { = } A F , \\mathcal { O } G A / / B C$ $\\textcircled { 3 } A F / / B C , \\textcircled { 4 }$ 点 $G , A , F$ 在同一条直线上， $\\textcircled{5}$ 点 $A$ 是线段 $G F$ 的中点.其中正确的有 （）", + "page_idx": 33 + }, + { + "type": "text", + "text": "A.2个 B.3个 C.4个 D.5个 ", + "page_idx": 33 + }, + { + "type": "text", + "text": "二、填空题(本大题共5小题，每小题3分，共15分)", + "text_level": 1, + "page_idx": 34 + }, + { + "type": "text", + "text": "11.如图,直线 $A B$ 左边是计算器上的数字5,若以 $A B$ 为对称轴，则它的轴对称图形是数字 ·", + "page_idx": 34 + }, + { + "type": "image", + "img_path": "images/d3582c039258099f4fbe11b16f9ce681b157aa48bd8e1522ff43de2ea829da63.jpg", + "img_caption": [ + "第11题图" + ], + "img_footnote": [], + "page_idx": 34 + }, + { + "type": "image", + "img_path": "images/bb9a5ebe28efdf0133e60b99d7119975fb4a64068bed68987ad8924b987764f1.jpg", + "img_caption": [ + "第13题图" + ], + "img_footnote": [], + "page_idx": 34 + }, + { + "type": "image", + "img_path": "images/f949f1e68e271867d75ddc80f33f5354b4ffc0458fc9d0318ea1fb53327a2170.jpg", + "img_caption": [ + "第14题图" + ], + "img_footnote": [], + "page_idx": 34 + }, + { + "type": "text", + "text": "12.某商场将一商品在保持销售价80元/件不变的前提下，规定凡购买超过5件者，超出的部分打5折出售.若顾客购买x(x大于5)件，应付y元,则y与$\\mathcal { X }$ 之间的关系式是", + "page_idx": 34 + }, + { + "type": "text", + "text": "13.如图,点 $F$ 在直线 $C D$ 上， $F G$ 平分∠EFD,AB// $C D$ $\\angle 1 = 5 6 ^ { \\circ }$ $\\angle 2$ 的度数为", + "page_idx": 34 + }, + { + "type": "text", + "text": "14.如图,在△ABC中,AD⊥BC于点D,BE⊥AC于点E,AD,BE相交于点F,如果 $B F { = } A C , B C { = } 8 , C D { = } 2$ ，那么 $A F { = } \\_$ ", + "page_idx": 34 + }, + { + "type": "text", + "text": "15. 新考法填空双空题视频讲解两个边长分别为a和b的正方形按如图1所示的方式放置,其未叠合部分(阴影)的面积为Si;若再在图中大正方形的右下角摆放一个边长为b的小正方形(如图2)，两个小正方形叠合部分（阴影）的面积为 $S _ { 2 }$ .若 $a + b = 8 , a b = 1 0$ ，则 $S _ { 1 } + S _ { 2 } =$ ；当 $S _ { 1 } + S _ { 2 } = 4 0$ 时，则图3中阴影部分的面积 $S _ { 3 } =$ ", + "page_idx": 34 + }, + { + "type": "image", + "img_path": "images/77c059b3139a742758e269ca9b1a9cdf2c092ea818eda936a89f9a14d249253b.jpg", + "img_caption": [ + "图1" + ], + "img_footnote": [], + "page_idx": 34 + }, + { + "type": "image", + "img_path": "images/c8e2e49a4a4d3b11a674b492f6e1ecac54a44cab4582113704ce85d24e82486b.jpg", + "img_caption": [ + "图2" + ], + "img_footnote": [], + "page_idx": 34 + }, + { + "type": "image", + "img_path": "images/0a0e9f04c1a3f21824dfcf740e091c798a31aaccca9e9670f55e3faf117391bd.jpg", + "img_caption": [ + "图3 " + ], + "img_footnote": [], + "page_idx": 34 + }, + { + "type": "text", + "text": "三、解答题（本大题共8小题，满分75分）", + "text_level": 1, + "page_idx": 34 + }, + { + "type": "text", + "text": "16.(10分)（1)计算： $( \\frac { 2 } { 3 } ) ^ { 2 0 2 3 } \\times 1 . 5 ^ { 2 0 2 4 } \\times ( - 1 ) ^ { 2 0 2 4 } ,$ (2)某种T形零件从正面看到的平面图形是轴对称图形,具体尺寸如图所示,请你计算阴影部分的周长和面积.", + "page_idx": 34 + }, + { + "type": "text", + "text": "", + "page_idx": 34 + }, + { + "type": "image", + "img_path": "images/0d7c272b27b654d93f6d241a7ab43ead77726a01ec04c4889b091108d281b524.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 34 + }, + { + "type": "text", + "text": "17.(9分)如图,AD是一段斜坡,AB 是水平线,现为了测量斜坡上一点D的竖直高度 $D B$ 的长度,欢欢在 $D$ 处立上一竹竿 $C D$ ,并保证 $C D \\perp A D$ ,然后在竿顶 $C$ 处垂下一根绳 $C E$ ,与斜坡的交点为点 $E$ ,他调整好绳子 $C E$ 的长度,使得 $C E { = } A D$ ，此时她测得 $D E { = } 2$ 米,求 $D B$ 的长度.", + "page_idx": 34 + }, + { + "type": "image", + "img_path": "images/fe99707b8d3f3330bd5b21e83fc4fd2cdd7e67668e9e43f189d7f2b1ed6595c5.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 34 + }, + { + "type": "text", + "text": "18.(9分)小明骑车上学,当他骑了一段时间后，想起要买某本书，于是又折回到刚经过的新华书店，买到书后继续骑车去学校.他离家的距离(米)与所用的时间(分钟)的关系如图所示.根据图象回答下列问题：", + "page_idx": 35 + }, + { + "type": "text", + "text": "(1)小明家到学校的距离是 米.", + "page_idx": 35 + }, + { + "type": "text", + "text": "(2)小明在书店停留了 分钟. ", + "page_idx": 35 + }, + { + "type": "text", + "text": "(3)本次上学途中，小明一共骑行了 米.", + "page_idx": 35 + }, + { + "type": "text", + "text": "20.(9分)若 $a$ 为任意自然数，尝试猜想：代数式 $( 3 a + 2 ) ( 2 a + 3 ) - 3 a ( 2 a + 1 )$ 的值是奇数，还是偶数？并说明理由.", + "page_idx": 35 + }, + { + "type": "text", + "text": "(4)据统计，骑车的速度超过330米/分就超越了安全限度，小明买到书后继续骑车到学校的这段时间内的骑车速度在安全限度内吗？请说明理由.", + "page_idx": 35 + }, + { + "type": "text", + "text": "学校班级姓名学号考生注意考试内无将学校丶班级、名、 学号填写在指定的位置上°", + "page_idx": 35 + }, + { + "type": "image", + "img_path": "images/566c7139c56b2495626b1a52c57ce606f954c6b4871ac242b1c09162d295a9f1.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 35 + }, + { + "type": "text", + "text": "21.(9分)如图，将 $\\mathrm { R t } \\triangle A B C$ 沿某条直线折叠，使斜边的两个端点 $A$ 与 $B$ 重合，折痕为 $D E$ ", + "page_idx": 35 + }, + { + "type": "text", + "text": "19.(9分)如图，在长方形 $A B C D$ 中， $A B { = } 4 , A D { = } 6$ ，延长 $B C$ 到点 $E$ ，使 $C E { = }$ 2,连接 $D E$ ，动点 $P$ 从点 $B$ 出发，以每秒2个单位长度的速度沿BC— $C D$ $D A$ 向终点 $A$ 运动，设点 $P$ 的运动时间为 $t$ 秒，当 $t$ 的值为多少时， $\\triangle A B P$ 和△DCE全等？", + "page_idx": 35 + }, + { + "type": "text", + "text": "(1)若 $A C { = } 6 \\ \\mathrm { c m } , B C { = } 8 \\ \\mathrm { c m }$ ,求△ACD的周长； \n(2)若 $\\angle C A D : \\angle B A D = 1 : 2$ ，求 $\\angle B$ 的度数. ", + "page_idx": 35 + }, + { + "type": "image", + "img_path": "images/16e320f6b65049b25d0c5bd656a332695cb1f8952eb2d30b48e79f35660db3db.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 35 + }, + { + "type": "image", + "img_path": "images/82dcd5486379c67e7c00ab338b5c5712233f9680e06e5aaa0af053220cf6d40c.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 35 + }, + { + "type": "text", + "text": "22.(10分)已知直线 $A M , C N$ 和点 $B$ 在同一平面内，且 $A M / / C N , A B \\perp B C .$ （20(1)如图 $1 , A M$ 与 $B C$ 交于点 $D$ ，若 $\\angle A = 4 0 ^ { \\circ }$ ，求 $\\angle C$ 的度数；", + "page_idx": 36 + }, + { + "type": "text", + "text": "", + "page_idx": 36 + }, + { + "type": "text", + "text": "(2)如图2,若 $B D \\bot A M$ 垂足为点 $D$ ，试说明： $\\angle A B D { = } \\angle C .$ ", + "page_idx": 36 + }, + { + "type": "image", + "img_path": "images/7c7feb64da9d5d716aa0d8b357c882a31bb9400135839d86f424e1a68e7faec7.jpg", + "img_caption": [ + "图1" + ], + "img_footnote": [], + "page_idx": 36 + }, + { + "type": "image", + "img_path": "images/37df25d37c31f9f1ab3f56438abe30c3cc9b0b5835bb4bb27b4477fbf6e940dc.jpg", + "img_caption": [ + "图2" + ], + "img_footnote": [], + "page_idx": 36 + }, + { + "type": "text", + "text": "23. $\\circledcirc$ 视频讲解（10分)如图所示，在锐角 $\\triangle A B C$ 中， $B D , C E$ 分别是ABC的两条高，点 $P$ 在 $B D$ 的延长线上, $C A { = } B P$ ，点 $Q$ 在 $C E$ 上， $Q C { = } A B$ ", + "page_idx": 36 + }, + { + "type": "text", + "text": "（1)试判断： $\\angle 1$ $\\angle 2$ （填 $>$ ”、“ $<$ ”或“ $=$ ”).", + "page_idx": 36 + }, + { + "type": "text", + "text": "(2)试探究线段 $P A$ 与 $A Q$ 之间的数量关系和位置关系,并说明理由.", + "page_idx": 36 + }, + { + "type": "text", + "text": "(3)若把(1)中的 $\\triangle A B C$ 改为钝角三角形， $A C > A B$ $\\angle A$ 是钝角，其他条件不变，试探究线段 $P A$ 与 $A Q$ 之间的数量关系和位置关系.请在备用图中画出图形,并直接写出结论.", + "page_idx": 36 + }, + { + "type": "image", + "img_path": "images/fadd9cf4e6178a725fb5ebaee613cb959c91931fe98e6243c7069decab5f2b99.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 36 + }, + { + "type": "image", + "img_path": "images/1f0d99982fbc6e5b1bf39cadbbf6311a798cd09f9271f043200963b8111dd73d.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 36 + }, + { + "type": "text", + "text": "第六章 概率初步", + "text_level": 1, + "page_idx": 37 + }, + { + "type": "text", + "text": "基础过关测试卷", + "text_level": 1, + "page_idx": 37 + }, + { + "type": "text", + "text": "时间：60分钟 满分：100分", + "page_idx": 37 + }, + { + "type": "table", + "img_path": "images/4e149ce47f162f0ebab595efb01a327c1edd74f837f5d0fe299c93abe5798adc.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 37 + }, + { + "type": "text", + "text": "一、选择题(每小题4分，共32分)", + "text_level": 1, + "page_idx": 37 + }, + { + "type": "text", + "text": "1.下列事件中，是必然事件的是", + "page_idx": 37 + }, + { + "type": "text", + "text": "A.明天一定是晴天 \nB.打开手机就有未接电话 \nC.通常温度降到 $0 ~ ^ { \\circ } C$ 以下，纯净的水会结冰 \nD.2024年有370天", + "page_idx": 37 + }, + { + "type": "text", + "text": "2.掷一枚质地均匀的正方体骰子，骰子的六个面上分别是1点到6点的点数，则下列说法正确的是 （ ）", + "page_idx": 37 + }, + { + "type": "text", + "text": "A.掷一枚骰子，朝上的一面的点数一定是4B.掷一枚骰子,朝上的一面的点数一定是奇数C.掷一枚骰子，朝上的一面的点数出现1的可能性最大D.掷一枚骰子，朝上的一面的点数可能出现5", + "page_idx": 37 + }, + { + "type": "text", + "text": "3.下列成语描述的事件中，属于随机事件的是", + "page_idx": 37 + }, + { + "type": "text", + "text": "A.守株待兔 B.风吹草动 C.一手遮天 D.水中捞月", + "page_idx": 37 + }, + { + "type": "text", + "text": "4.从1,2,3,4,5这五个数中,任意取出一个数,是奇数的概率为", + "page_idx": 37 + }, + { + "type": "text", + "text": "A $\\frac { 4 } { 9 }$ B $\\frac { 3 } { 5 }$ C $\\frac { 2 } { 5 }$ D $\\frac { 1 } { 5 }$ ", + "page_idx": 37 + }, + { + "type": "text", + "text": "5.在抛掷一枚图钉的实验中,某小组一共做了1000 次实验,最后出现钉尖朝上的次数为620，则此时出现钉尖朝上的频率为 （ ）", + "page_idx": 37 + }, + { + "type": "text", + "text": "A.0.48 B.0.5 C. 0. 62 D.无法确定", + "page_idx": 37 + }, + { + "type": "text", + "text": "6.将4个红球、3个白球、2个黑球放入一个不透明的袋子里，从中摸出8个球，恰好红球、白球、黑球都摸到，这件事情 （", + "page_idx": 37 + }, + { + "type": "text", + "text": "A.必然发生 B.不可能发生C.很可能发生 D.可能发生", + "page_idx": 37 + }, + { + "type": "text", + "text": "7.小明在做一道正确答案是一5的计算题时，由于运算符号（“ $^ +$ ”“-”“ $\\times$ ”或“÷\")被墨迹污染，看见的算式是“一204”,那么小明还能做对的概率是（", + "page_idx": 37 + }, + { + "type": "text", + "text": "A $\\frac { 1 } { 6 }$ $\\mathrm { B } . { \\frac { 1 } { 4 } } \\qquad \\mathrm { C } . { \\frac { 1 } { 3 } } \\qquad \\mathrm { D } . { \\frac { 1 } { 2 } }$ ", + "page_idx": 37 + }, + { + "type": "text", + "text": "8.一个转盘的颜色如图所示，其中 $\\angle A O B = 6 0 ^ { \\circ } , A C$ 为直径，转动转盘，指针落在红色区域的概率为 （ ）", + "page_idx": 37 + }, + { + "type": "text", + "text": "A $\\frac { 1 } { 6 }$ $\\begin{array} { c } { \\mathrm { B . } { \\frac { 1 } { 3 } } } \\\\ { \\mathrm { D . } { \\frac { 2 } { 3 } } } \\end{array}$ C $\\frac 1 2$ ", + "page_idx": 37 + }, + { + "type": "image", + "img_path": "images/a3cef98fe41c86f6c5613a4fd8c75a6194e8248c0d762b7f91dbdfedaaf67743.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 37 + }, + { + "type": "text", + "text": "二、填空题(每小题4分，共16分)", + "text_level": 1, + "page_idx": 37 + }, + { + "type": "text", + "text": "9.“3个人分成两组，一定有2个人分在一组\"的事件是一个 事件.（填\"必然”“不可能”或“随机”)", + "page_idx": 37 + }, + { + "type": "text", + "text": "10.在一个不透明的口袋中装有若干个红球和白球，它们除颜色外其他完全相同，通过多次摸球试验后发现，摸到红球的频率稳定在 $2 5 \\%$ 附近，则从口袋中随机地摸出一个球,摸到白球的概率为", + "page_idx": 37 + }, + { + "type": "text", + "text": "11.在一个不透明的盒子里装有10 枚大小相同的白色棋子和黑色棋子,若随机从中摸出一枚棋子是白色棋子的概率为 $\\frac { 1 } { 5 }$ ,则在设计该游戏时,盒子里应放有白色棋子_枚.", + "page_idx": 37 + }, + { + "type": "text", + "text": "12.若小球在如图所示的地面上自由滚动,并随机停留在某块方砖上，则它最终停留在黑色区域的概率是", + "page_idx": 37 + }, + { + "type": "image", + "img_path": "images/60832e6b3e1a1b0d2105b584819a714f695384a45f3997a6c373842e35d4b7cd.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 37 + }, + { + "type": "text", + "text": "三、解答题(本大题5小题，共52分）", + "text_level": 1, + "page_idx": 38 + }, + { + "type": "text", + "text": "13.(8分)如图,有一些写有数字的卡片,它们的背面都相同,现将它们的背面朝上，从中任意摸出一张.", + "page_idx": 38 + }, + { + "type": "image", + "img_path": "images/3d0a8a65465dc8a2d4d9627fa1ad85643767eae1edb4030002f47aaf47d14655.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 38 + }, + { + "type": "text", + "text": "(1)摸到哪个数字卡片的可能性最大？摸到哪个数字卡片的可能性最小?", + "page_idx": 38 + }, + { + "type": "text", + "text": "(2)摸到的数字是奇数和摸到的数字是偶数的可能性哪个大？ ", + "page_idx": 38 + }, + { + "type": "text", + "text": "14.(10分)在一个不透明的袋子中装有仅颜色不同的6个红球和9个黑球.", + "page_idx": 38 + }, + { + "type": "text", + "text": "(1)若从袋子中随机摸出1个球,分别求摸到红球和摸到白球的概率; (2)若再往袋子中放人 $m$ 个黑球,然后随机摸出1个球，若摸到红球的概率 为 $\\frac 1 3$ 求 $m$ 的值. ", + "page_idx": 38 + }, + { + "type": "text", + "text": "15.(10 分)如图，一个转盘被等分成六个扇形,并在上面依次写上数字1,2,3,4,5,6.", + "page_idx": 38 + }, + { + "type": "text", + "text": "(1)若自由转动转盘，当它停止转动后,指针指向奇数的概率是多少?", + "page_idx": 38 + }, + { + "type": "text", + "text": "(2)请你用这个转盘设计一个游戏,要求当自由转动的转盘停止时，指针指向的区域的概率为 $\\frac { 2 } { 3 }$ ", + "page_idx": 38 + }, + { + "type": "image", + "img_path": "images/21d06646f327de740481946d4d9c1d30fb6f8e3916b4347094fa1c073269cce5.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 38 + }, + { + "type": "text", + "text": "16.（12 分)下面是小明和同学们做\"抛掷质地均匀的硬币试验\"获得的数据：", + "page_idx": 38 + }, + { + "type": "table", + "img_path": "images/0b45f459a5f9cf3ee490c293b6d5acd169f90d3dae212e2bc8aaf5f700211bf9.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
抛掷次数n100200300400500600
正面朝上的频数m5198153200255306
n 正面朝上的频率m
", + "page_idx": 38 + }, + { + "type": "text", + "text": "正面朝上的频率 \n0.52 \n0.51 \n0.50 \n0.49 \n0.48100 200300400 500600抛掷次数(1)填写表中的空格； \n（2)画出折线统计图； \n(3)当试验次数很大时，“正面朝上\"的频率在 附近摆动.", + "page_idx": 38 + }, + { + "type": "text", + "text": "", + "page_idx": 38 + }, + { + "type": "text", + "text": "17. $\\circledcirc$ 视频讲解（12分）已知下列事件： $\\textcircled{1}$ 投掷一枚普通的正方体骰子，所得的点数小于7； $\\textcircled{2}$ 随机地从一副扑克牌(52张)中摸出一张扑克牌，正好是红桃(大、小王除外).", + "page_idx": 38 + }, + { + "type": "text", + "text": "(1)事件 $\\textcircled{1}$ 是一个 事件，发生的概率为事件 $\\textcircled{2}$ 是一个 事件，发生的概率为(2)我们把事件 $A$ 记作\"投掷一枚普通的正方体骰子，所得的点数为3的倍数”；事件 $B$ 记作\"随机地从一副扑克牌(52张)中摸出一张扑克牌,扑克牌上的数字正好是5的倍数”.请你把这两个事件 $A , B$ 发生的概率用“<\"连接起来.", + "page_idx": 38 + }, + { + "type": "text", + "text": "", + "page_idx": 38 + }, + { + "type": "text", + "text": "能力提优测试卷", + "text_level": 1, + "page_idx": 39 + }, + { + "type": "text", + "text": "时间：60分钟 满分：100分", + "page_idx": 39 + }, + { + "type": "table", + "img_path": "images/39f96013a79d2f356d742fc068cb72a00627c32767b146998f10091606e39838.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 39 + }, + { + "type": "text", + "text": "一、选择题(每小题5分，共30分)", + "text_level": 1, + "page_idx": 39 + }, + { + "type": "text", + "text": "1.下列事件为必然事件的是", + "page_idx": 39 + }, + { + "type": "text", + "text": "A.从两个班级中任选三名学生，至少有两名学生来自同一个班级 \nB.明天会下雨 \nC.打开电视机,CCTV1正在播放新闻 \nD.购买一张彩票中奖一百万元", + "page_idx": 39 + }, + { + "type": "text", + "text": "2.如图，从四张印有品牌标志图案的卡片中任取一张，取出印有品牌标志图案的是轴对称图形的卡片的概率是 ）", + "page_idx": 39 + }, + { + "type": "image", + "img_path": "images/c8a6fcf5dfc8ced182d8e3d34a8888d730050ed306566e24188ea5d72c064666.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 39 + }, + { + "type": "text", + "text": "A.1 B $\\frac { 3 } { 4 }$ $\\frac 1 2$ D $\\frac { 1 } { 4 }$ ", + "page_idx": 39 + }, + { + "type": "text", + "text": "3.在一个不透明的布袋中装有红色、白色球共40个，除颜色外其他完全相同.小明通过多次摸球试验后发现,其中摸到红色球的频率稳定在 $1 5 \\%$ 左右，则布袋中白色球可能有 （ ）", + "page_idx": 39 + }, + { + "type": "text", + "text": "A.4个 B.6个 C.34个 D.36个 ", + "page_idx": 39 + }, + { + "type": "text", + "text": "4.一个小球在如图所示的地板上自由滚动，并随机停在某块方砖上,如果每一块方砖除颜色外其他完全相同，那么小球最终停留在黑砖上的概率是（ ）", + "page_idx": 39 + }, + { + "type": "text", + "text": "A $\\frac { 5 } { 9 }$ B $\\frac { 4 } { 9 }$ C $\\frac { 4 } { 5 }$ D.1 ", + "page_idx": 39 + }, + { + "type": "image", + "img_path": "images/75afb139f7e16060bd39aeb9e18be49165f6fd6db5237d96efb6d5427d14531f.jpg", + "img_caption": [ + "第4题图" + ], + "img_footnote": [], + "page_idx": 39 + }, + { + "type": "image", + "img_path": "images/92e858056a5386c619a279a22c5e245aa3e2cffb33334000c996b39e5b966bb5.jpg", + "img_caption": [ + "第5题图" + ], + "img_footnote": [], + "page_idx": 39 + }, + { + "type": "image", + "img_path": "images/6715cdd201acae8a9f9fa63b401933d8226e49552f68e85daff69fbbd7d3de41.jpg", + "img_caption": [ + "第6题图" + ], + "img_footnote": [], + "page_idx": 39 + }, + { + "type": "text", + "text": "5.某小组做\"用频率估计概率”的试验时，统计了某一结果出现的频率,绘制了如图所示的折线统计图，则符合这一结果的试验最有可能的是 （ ）", + "page_idx": 39 + }, + { + "type": "text", + "text": "A.在“石头、剪刀、布\"的游戏中，小明随机出的是“剪刀” \nB.一副去掉大、小王的普通扑克牌洗匀后，从中任抽一张牌的花色是红桃 \nC.暗箱中有1个红球和2个黄球，它们只有颜色上的区别，从中任取一个球是黄球 \nD.掷一枚质地均匀的正六面体骰子,向上一面的点数是4", + "page_idx": 39 + }, + { + "type": "text", + "text": "6.如图，在两个同心圆中，四条直径把圆分成八等份，若往圆面投掷飞镖，则飞镖落在黑色区域的概率是 （ ）", + "page_idx": 39 + }, + { + "type": "text", + "text": "A $\\frac 1 2$ B $\\frac 1 3$ $\\frac { 1 } { 4 }$ D $\\frac { 1 } { 6 }$ ", + "page_idx": 39 + }, + { + "type": "text", + "text": "二、填空题(每小题5分，共20分）", + "text_level": 1, + "page_idx": 39 + }, + { + "type": "text", + "text": "7.某电视台在2023年5月举办的青年歌手大奖赛活动中，得奖选手由观众发短信投票产生，并对发短信者进行抽奖活动.一万条短信为一个开奖组，设一等奖1名，二等奖3名，三等奖6名.小静同学发了一条短信，那么她获奖的概率是", + "page_idx": 39 + }, + { + "type": "text", + "text": "8.在一个不透明的口袋里装有只有颜色不同的黑、白两种颜色的球共20个，某学习小组做摸球试验，将球搅匀后从中随机摸出一个球记下颜色，再把它放回袋中，不断重复，试验数据如下表：", + "page_idx": 39 + }, + { + "type": "table", + "img_path": "images/0e7b718805ddef278fc24ef1362a39b3057ef6ace24d2de36a78e45bf58ba584.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
摸球的次数n1001502005008001000
摸到白球的次数m5896116295484601
摸到白球的频率 n0.580.640.580.590.6050.601
", + "page_idx": 39 + }, + { + "type": "text", + "text": "根据以上数据，估计口袋中黑球有 个", + "page_idx": 39 + }, + { + "type": "text", + "text": "9.有八张大小、形状完全相同的卡片，卡片上分别写有数字3,4,5,6,7,8,9,10,从中随机抽取一张，抽出的卡片上的数字恰好为3的倍数的概率是", + "page_idx": 39 + }, + { + "type": "text", + "text": "10.在一个不透明的袋子中装有除颜色外其余均相同的7个小球,其中红球2个，黑球5个，若再放入 $m$ 个一样的黑球并摇匀，此时,随机摸出一个球是黑球的概率等于 $\\frac { 4 } { 5 }$ ，则 $m$ 的值为", + "page_idx": 40 + }, + { + "type": "text", + "text": "三、解答题(本大题5小题，共50分）", + "text_level": 1, + "page_idx": 40 + }, + { + "type": "text", + "text": "11.(8分)如图，一个游戏转盘中,红、黄、蓝三个扇形的圆心角度数分别为 $4 0 ^ { \\circ }$ ，${ 1 2 0 } ^ { \\circ } , { 2 0 0 } ^ { \\circ }$ ，若让转盘自由转动.", + "page_idx": 40 + }, + { + "type": "text", + "text": "（1)求指针停止后在蓝色区域的概率； \n(2)求指针停止后在黄色或红色区域的概率.", + "page_idx": 40 + }, + { + "type": "image", + "img_path": "images/f0ad503afeecce06892f325b65c33e6f4b7cc3da3fb15dc47fcd4c36349cc8dc.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 40 + }, + { + "type": "text", + "text": "14.(10 分)某信息兴趣小组利用电脑成功设计了一个运算程序,这个程序可用如图所示的框图表示.小明同学任取一个自然数 $\\mathcal { X }$ 输入求值.", + "page_idx": 40 + }, + { + "type": "text", + "text": "(1)试写出与输出的数 $y$ 有关的一个必然事件；", + "page_idx": 40 + }, + { + "type": "text", + "text": "(2)若输入的数是2至9这八个连续正整数中的一个，求输出的数是3的倍数的概率.", + "page_idx": 40 + }, + { + "type": "text", + "text": "输人x → 平方 -x ÷2 输出y", + "page_idx": 40 + }, + { + "type": "text", + "text": "12.（10分)如图,可以随机在图中取点.", + "page_idx": 40 + }, + { + "type": "text", + "text": "(1)这个点取在阴影部分的概率是(2)在保留原阴影部分的情况下，请你重新设计图案(直接在图上涂阴影)，使得这个点取在阴影部分的概率为.", + "page_idx": 40 + }, + { + "type": "text", + "text": "", + "page_idx": 40 + }, + { + "type": "image", + "img_path": "images/3209ce4912cf9de0700baf0b89ba9309fc3c1fae786b13114f4d72efbfcab6eb.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 40 + }, + { + "type": "text", + "text": "13.(10 分)在一个不透明的袋子里装有红、黄、白三种颜色的球共50个,它们除了颜色不同外其余都相同,其中黄球比白球少5个,已知从袋子里随机摸出一个球是红球的概率是 $\\frac { 3 } { 1 0 }$ ", + "page_idx": 40 + }, + { + "type": "text", + "text": "（1)求袋子里红球的个数； \n(2)求从袋子里随机摸出一个球是白球的概率.", + "page_idx": 40 + }, + { + "type": "text", + "text": "15.视讲(12分)如图，端午节期间，某商场为了吸引顾客，设立了一个可以自由转动的转盘，并规定顾客每购买200元的商品,就能获得一次转动转盘的机会,如果转盘停止后，指针上对准红、黄、绿的区域，顾客就可以分别获得50元、20元、10元的奖金，对准无色区域则无奖金(转盘等分成16份).", + "page_idx": 40 + }, + { + "type": "text", + "text": "（1)小明购物180元，他获得奖金的概率是多少？（2)小德购物210元，他获得奖金的概率是多少？", + "page_idx": 40 + }, + { + "type": "text", + "text": "(3)现商场想调整获得10元奖金的概率为,其他金额的获奖率不变，,则需要将多少个无色区域涂上绿色？", + "page_idx": 40 + }, + { + "type": "image", + "img_path": "images/75b87daacef404e34c6367ce9a5747477e65b83b60bbcc7387dda5bc74885d2b.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 40 + }, + { + "type": "text", + "text": "专项训练卷（一） 整式的乘除运算 ", + "text_level": 1, + "page_idx": 41 + }, + { + "type": "text", + "text": "时间：60分钟 满分：100分", + "page_idx": 41 + }, + { + "type": "table", + "img_path": "images/d90f366932995163b01bba40175884a040d82ca5e94fd15565d4abbf190b2995.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 41 + }, + { + "type": "text", + "text": "一、选择题(每小题4分，共32分)", + "text_level": 1, + "page_idx": 41 + }, + { + "type": "text", + "text": "1.计算 $a ^ { 5 } \\cdot a \\div a ^ { 4 }$ 的结果为", + "page_idx": 41 + }, + { + "type": "text", + "text": "A.a B. $a ^ { 2 }$ （204号 C. $a ^ { 3 }$ （20 D. $a ^ { 4 }$ （204号 ", + "page_idx": 41 + }, + { + "type": "text", + "text": "2.下列运算中，正确的是", + "page_idx": 41 + }, + { + "type": "text", + "text": "A. $x ^ { 2 } + x ^ { 5 } = x ^ { 7 }$ B. $( - 3 x ) ^ { 3 } { = } 2 7 x ^ { 3 }$ C. $( \\boldsymbol { \\mathscr { x } } ^ { 4 } ) ^ { 2 } = \\boldsymbol { \\mathscr { x } } ^ { 8 }$ D $( x - 1 ) ^ { 2 } = x ^ { 2 } - 2 x - 1$ ", + "page_idx": 41 + }, + { + "type": "text", + "text": "3.可乐中含有大量的咖啡因，世界卫生组织建议青少年每天咖啡因的摄入量不能超过 $0 . 0 0 0 0 8 5 ~ \\mathrm { k g }$ ，则数据0.000085用科学记数法表示为 （ ）", + "page_idx": 41 + }, + { + "type": "equation", + "img_path": "images/842a8adcd5a9dc5d7189a50e84a6a6cb72f324d3db54d4b9a4849a4e223ae421.jpg", + "text": "$$\n3 . 5 { \\times } 1 0 ^ { 5 } \\qquad \\quad \\mathrm { B } . 8 . 5 { \\times } 1 0 ^ { - 4 } \\qquad \\mathrm { C . ~ 8 . ~ 5 { \\times } 1 0 ^ { - 5 } } \\qquad \\quad \\mathrm { D . ~ 8 . ~ 5 { \\times } 1 0 ^ { - 6 } }\n$$", + "text_format": "latex", + "page_idx": 41 + }, + { + "type": "text", + "text": "4.若单项式 ${ \\frac { 2 } { 7 } } x ^ { m + 2 } y ^ { 2 }$ 与单项式 $7 x y z$ 的乘积是一个9次单项式,则 $m$ 的值为（", + "page_idx": 41 + }, + { + "type": "text", + "text": "A.2 B.3 C.4 D.5", + "page_idx": 41 + }, + { + "type": "text", + "text": "5.设整式 $M { = } x ( x { - } 5 ) , N { = } ( x { + } 1 ) ( x { - } 6 )$ ，则 $M$ 和 $N$ 的大小关系是（）", + "page_idx": 41 + }, + { + "type": "text", + "text": "A $M { > } N$ B. $M { = } N$ C. $M { < } N$ D. $M { = } 2 N$ ", + "page_idx": 41 + }, + { + "type": "text", + "text": "6.如图，有一块边长为 $_ { \\mathcal { X } }$ 米的正方形草地，现将该正方形草地的南北方向减少3米，东西方向增加3米，则得到一块长为 $( x { + 3 } )$ 米，宽为 $( x { - } 3 )$ 米的长方形草地，那么长方形草地的面积比原来正方形草地的面积 ）", + "page_idx": 41 + }, + { + "type": "text", + "text": "A.增加了18平方米 B.增加了9平方米C.保持不变 D.减少了9平方米", + "page_idx": 41 + }, + { + "type": "image", + "img_path": "images/55d5c281f2ea4971093b79602bb3e15f6186a7e135c1296554642fc566ca5317.jpg", + "img_caption": [ + "第6题图" + ], + "img_footnote": [], + "page_idx": 41 + }, + { + "type": "image", + "img_path": "images/1b2f4418dcecc0e79c12d5df1b966537fdec84f6e1d841b4393c487492c36c1c.jpg", + "img_caption": [ + "第7题图" + ], + "img_footnote": [], + "page_idx": 41 + }, + { + "type": "text", + "text": "7.如图所示，该图是用边长分别为 $\\mathcal { X }$ 与 $y$ 的一个大正方形和四个小正方形以及四个长为 $\\mathcal { X }$ ，宽为 $y$ 的长方形组成的 $( x > y )$ ，小明利用这个图形很容易解决如下问题：已知 $x ^ { 2 } + 4 y ^ { 2 } = 8 , x y = 2$ ，则 $x { + 2 y }$ 的值为4.那么这样的解决数学问题时所体现的数学思想是 （）", + "page_idx": 41 + }, + { + "type": "text", + "text": "A.数形结合思想B.逆向思维思想 C.分类思想 D.公理化思想", + "page_idx": 41 + }, + { + "type": "text", + "text": "8. $\\circledcirc$ 视讲解在一节数学课上，熊老师设计了一个如图所示的计算程序，则最后输出的结果为 （ ）", + "page_idx": 41 + }, + { + "type": "text", + "text": "输人有理数a $\\sqrt { + ( a - 1 ) ^ { 2 } }$ （1+a)（1-a-a→输出", + "page_idx": 41 + }, + { + "type": "text", + "text": "A.2a-3 B.2a-1 C.2a+1 D.2a+3 ", + "page_idx": 41 + }, + { + "type": "text", + "text": "二、填空题(每小题4分，共16分)", + "text_level": 1, + "page_idx": 41 + }, + { + "type": "text", + "text": "9.若 $2 m + n { = } - 1$ ，则 $2 ^ { 2 m } \\bullet 2 ^ { n } = \\underline { { { \\vphantom { \\sum } 2 ^ { n } } } }$ ", + "page_idx": 41 + }, + { + "type": "text", + "text": "10.已知 $\\scriptstyle x ^ { 2 } - 3 x = 1$ ，则代数式 $2 x ( x - 1 ) - 4 x + 3$ 的值为 ", + "page_idx": 41 + }, + { + "type": "text", + "text": "11.若一个三角形的面积为 $x ( 2 x ^ { 2 } - 5 x )$ ，它的一条边长为 $( 2 x ) ^ { 2 }$ ,则这条边上的高为", + "page_idx": 41 + }, + { + "type": "text", + "text": "12.若关于 $a$ 的代数式 $( \\frac { 1 } { 2 } a + k ) ^ { 2 } - ( 2 - a ) \\left( 2 + a \\right) ( k$ 是常数)的结果中不含有常数项，则 $k ^ { 2 }$ 的值为", + "page_idx": 41 + }, + { + "type": "text", + "text": "三、解答题(本大题6小题，共52分)", + "text_level": 1, + "page_idx": 41 + }, + { + "type": "text", + "text": "13.(6分)计算： $( 1 ) ( x ^ { 2 n + 1 } ) ^ { 3 } \\div x ^ { 5 n } \\bullet x ; ( 2 ) - 1 ^ { 2 0 2 4 } - 3 2 ^ { 0 } \\div ( - 3 ) ^ { - 1 } \\times ( - \\frac { 1 } { 2 } ) ^ { - 2 } .$ ", + "page_idx": 41 + }, + { + "type": "text", + "text": "14.(8分)先化简，再求值：", + "page_idx": 41 + }, + { + "type": "text", + "text": "$( 1 ) x ( x + 7 ) + ( x - 5 ) ( x + 5 ) - ( x - 1 ) ( x + 6 )$ ,其中 $x { = } - 3$ $( 2 ) x ( 2 x + 5 ) - ( x ^ { 3 } - 2 x ^ { 2 } ) \\div ( - \\frac { 1 } { 2 } x )$ ，其中 $x { = } 5$ ", + "page_idx": 41 + }, + { + "type": "text", + "text": "", + "page_idx": 41 + }, + { + "type": "text", + "text": "15.(8分）已知 $a = ( - \\frac { 1 } { 8 } \\times \\frac { 1 } { 9 } ) ^ { 3 4 } \\times ( 9 \\times 8 ) ^ { 3 4 } , b = 9 4 ^ { 2 } - 9 3 \\times 9 5 .$ 试化简代数式 $- ( 6 x y ^ { 2 } ) ^ { 2 } \\div ( - 3 x y )$ ,并求当 $\\scriptstyle x = a - 2 , y = b$ 时,该代数式的值. ", + "page_idx": 42 + }, + { + "type": "text", + "text": "16.(8分)我们经常利用图形描述问题和分析问题,借助于直观的几何图形,把问题变得简明、形象,有助于探索解决问题的思路.", + "page_idx": 42 + }, + { + "type": "text", + "text": "(1)如图,小明构造了一个几何图形,请你用两种不同的方法计算图中阴影部分的面积 ， ，利用它们可以验证的公式为", + "page_idx": 42 + }, + { + "type": "text", + "text": "(2)根据(1)中你得到的公式计算： $\\textcircled { 1 } 2 9 9 ^ { 2 } ; \\textcircled { 2 } 4 7 ^ { 2 } - 4 7 \\times 8 6 + 4 3 ^ { 2 }$ (3)已知 $a ^ { 2 } + b ^ { 2 } = 1 0 , a b = 3$ ，求 $( a - b ) ^ { 2 }$ 的值.", + "page_idx": 42 + }, + { + "type": "text", + "text": "", + "page_idx": 42 + }, + { + "type": "text", + "text": "17.（10分)一条防洪堤坝,其横断面是梯形，上底宽为a米，下底宽为(a十2b)米，坝高为 $( a + 2 b )$ 米.", + "page_idx": 42 + }, + { + "type": "text", + "text": "(1)试用含 $a , b$ 的代数式表示该防洪堤坝的横断面的面积.", + "page_idx": 42 + }, + { + "type": "text", + "text": "(2)当a=2°÷2-1,b=（－2024)°－6×（－1)-9时,求该防洪堤坝的横断面的面积.", + "page_idx": 42 + }, + { + "type": "text", + "text": "(3)在(2)的条件下,若该防洪堤坝的长为() $( \\frac { 1 } { 1 0 } ) ^ { - 2 } \\div 1 0 ^ { - 1 }$ 米,则修建该防洪堤坝需要多少方土(用科学记数法表示)？（提示：一方土等于一立方米)", + "page_idx": 42 + }, + { + "type": "text", + "text": "18.新考法综合与实践视频讲解（12分)如图1，这是2023年12月份的日历，现按照图中所示的方式框住日历中的5个数,可以发现，框住的5个数具有下列特点：如 $8 \\times 2 4 + 1 0 \\times 2 2 - 2 \\times 1 6 ^ { 2 } = 1 9 2 + 2 2 0 - 5 1 2 = - 1 0 0$ ，又如 $1 1 \\times$ $2 7 + 1 3 \\times 2 5 - 2 \\times 1 9 ^ { 2 } = 2 9 7 + 3 2 5 - 7 2 2 = - 1 0 0$ ,不难发现,结果都是—100.", + "page_idx": 42 + }, + { + "type": "text", + "text": "(1)我们可以看出，上述框住的5个数满足一定的规律，请你用语言叙述这个规律：", + "page_idx": 42 + }, + { + "type": "text", + "text": "（2)设框住的5个数的中间数为x,请你用含x的代数式表示你发现的规律：·并利用整式的运算对你发现的规律加以说明.", + "page_idx": 42 + }, + { + "type": "text", + "text": "(3)如果像图2那样利用\"十字框\"框住5个数,那么(1)中的规律还成立吗？先利用图中两个框住的5个数验证一下，如果不成立，写出你发现的规律,然后利用整式的运算加以说明你的结论.", + "page_idx": 42 + }, + { + "type": "table", + "img_path": "images/745ec034aebe844c186b1fde40fa6ef95baef3c93a88a8c7b6932e42f1f05ee2.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
二日三2023年12月 四五六日
010203
04050607080910
1121314151617
18192021222324
25262728293031
", + "page_idx": 42 + }, + { + "type": "table", + "img_path": "images/8ea1a428b12def23806c0bbef651d2a52029e1b33d1a3b861ccd384dbe45126f.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
2023年12月
一三三四五六日
010203
04Q50607080910
11121314151617
18192021222324
2s262728293031
", + "page_idx": 42 + }, + { + "type": "image", + "img_path": "images/c1bddd9060ae242ae56c763b6c74e7a5725fad3df7d4d4ee5a606bdfdfdf24a8.jpg", + "img_caption": [ + "图1", + "图2" + ], + "img_footnote": [], + "page_idx": 42 + }, + { + "type": "text", + "text": "专项训练卷（二） 几何题的计算", + "text_level": 1, + "page_idx": 43 + }, + { + "type": "text", + "text": "时间：60分钟 满分：100分", + "page_idx": 43 + }, + { + "type": "table", + "img_path": "images/b1342baf5d563659efc57bc87106534b3f2bd3c7c3b3d86248a46b26b6370cbb.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 43 + }, + { + "type": "text", + "text": "一、选择题(每小题4分，共32分)", + "text_level": 1, + "page_idx": 43 + }, + { + "type": "text", + "text": "1.若 $\\angle A = 3 0 ^ { \\circ } 2 0 ^ { \\prime }$ ，则 $\\angle A$ 的余角的度数为 ）", + "page_idx": 43 + }, + { + "type": "text", + "text": "A. $6 0 ^ { \\circ } 4 0 ^ { \\prime }$ （204号 B. $6 0 ^ { \\circ } 2 0 ^ { \\prime }$ （204号 $\\mathrm { C } . 5 9 ^ { \\circ } 4 0 ^ { \\prime }$ $\\mathrm { D } . 5 9 ^ { \\circ } 2 0 ^ { \\prime }$ ", + "page_idx": 43 + }, + { + "type": "text", + "text": "2.若等腰三角形顶角的度数比底角的度数大 $3 0 ^ { \\circ }$ ,则这个等腰三角形底角的度数为 （ ）", + "page_idx": 43 + }, + { + "type": "text", + "text": "A $4 0 ^ { \\circ }$ （20 $\\mathrm { B } , 4 3 ^ { \\circ }$ C.47° D.50° ", + "page_idx": 43 + }, + { + "type": "text", + "text": "3.如图，为了估计池塘两端 $A , B$ 两点之间的距离，小康采取了如下方法：在池塘的一侧选取了一点 $P$ ，连接 $P A , P B$ ，测得 $P A = 2 0$ 米， $P B { = } 1 3$ 米,那么 $A$ ，$B$ 两点之间的距离可能是 （）", + "page_idx": 43 + }, + { + "type": "text", + "text": "A.7米 B.20米 C.33米 D.35米", + "page_idx": 43 + }, + { + "type": "image", + "img_path": "images/a93735101569c9240e659854d78a8ce5dd3c961c91e834296caaf545c727e282.jpg", + "img_caption": [ + "第3题图" + ], + "img_footnote": [], + "page_idx": 43 + }, + { + "type": "image", + "img_path": "images/413cf56a578a078a516cdaccadba439f00461d47cb608a144e6e98c756847213.jpg", + "img_caption": [ + "第4题图" + ], + "img_footnote": [], + "page_idx": 43 + }, + { + "type": "image", + "img_path": "images/8c75b54141bb4cbd9a794aa09e383664adf84effd6313f0a95ba26c8d053dffd.jpg", + "img_caption": [ + "第5题图" + ], + "img_footnote": [], + "page_idx": 43 + }, + { + "type": "text", + "text": "4.如图,在 $\\triangle A B C$ 中， $\\angle C = 9 0 ^ { \\circ }$ ， $A D$ 平分 $\\angle B A C$ ，若 $S _ { \\triangle A B D } = 1 5 ~ \\mathrm { c m ^ { 2 } }$ $A B =$ $1 0 ~ \\mathrm { c m }$ ，则线段 $C D$ 的长为 （ ）", + "page_idx": 43 + }, + { + "type": "text", + "text": "$\\mathrm { A } . 3 \\ \\mathrm { c m }$ （20 （204号 $\\mathrm { B . 4 \\ c m }$ （204号 （20 $\\mathrm { C } . 5 \\ \\mathrm { c m }$ （20 $\\mathrm { D } , 6 \\ \\mathrm { c m }$ ", + "page_idx": 43 + }, + { + "type": "text", + "text": "5.如图，为了测量河两岸 $A , B$ 两点之间的距离,小明设计了如下方案： $\\textcircled{1}$ 在点 $B$ 的右侧取一点 $C$ ,利用精密测角仪测得 $\\angle A B C = 7 4 ^ { \\circ }$ $\\angle A C B = 3 6 ^ { \\circ }$ $\\textcircled{2}$ 在点 $M$ 处立一根标杆,利用精密测角仪测得 $\\angle M B C = 7 4 ^ { \\circ }$ $\\angle M C B = 3 6 ^ { \\circ }$ .若测得 $B M$ 的长为58米，则 $A , B$ 两点之间的距离为 （ ）", + "page_idx": 43 + }, + { + "type": "text", + "text": "A.74米 B.58米 C.36米 D.28米", + "page_idx": 43 + }, + { + "type": "text", + "text": "6.如图,已知直线 $A B , C D$ 与直线 $E F$ 分别相交于点 $G , H$ ，点 $M , N$ 分别是直线$E F , C D$ 上的一点,若 $A B / / C D , \\angle 1 = 5 0 ^ { \\circ } , \\angle 3 = 8 0 ^ { \\circ }$ ，则 $\\angle 2$ 的度数为（ ）", + "page_idx": 43 + }, + { + "type": "text", + "text": "A $3 0 ^ { \\circ }$ （20 B. $2 8 ^ { \\circ }$ （204号 C.25° D.22° ", + "page_idx": 43 + }, + { + "type": "image", + "img_path": "images/d2cceead6a5269f46eb0b5a705acf58c178e0b6bd59bbeddf7eed5157fde9779.jpg", + "img_caption": [ + "第6题图" + ], + "img_footnote": [], + "page_idx": 43 + }, + { + "type": "image", + "img_path": "images/1210227d4bd3dbcea8860402135f54e52ed237a09fa8513dbc34cb7a9f136941.jpg", + "img_caption": [ + "第8题图" + ], + "img_footnote": [], + "page_idx": 43 + }, + { + "type": "text", + "text": "7.已知△ABC≌△DEF, $\\triangle A B C$ 的三边长分别为4,6,9,若 $\\triangle D E F$ 与△ABC的三边长对应边长分别为 $4 , 3 x { - } 6 , 2 y { + } 1$ ，则 $( \\boldsymbol { \\mathcal { x } } - \\boldsymbol { y } ) ^ { 2 0 2 4 }$ 的值为 （ ）", + "page_idx": 43 + }, + { + "type": "text", + "text": "A.6 B.4 C.2 D.0 ", + "page_idx": 43 + }, + { + "type": "text", + "text": "8 $\\circledcirc$ 视频讲解如图，在 $\\triangle A B C$ 中,分别以点 $B$ 和点 $C$ 为圆心，大于 $\\cdot { \\frac { 1 } { 2 } } B C$ 的长为半径画弧,两弧相交于点 $M , N$ ，作直线 $M N$ ，交 $A C$ 于点 $D$ ，交 $B C$ 于点 $E$ ，连接BD,若△ABD 的周长为 $1 9 , \\triangle A B C$ 的周长为25,则 $B C$ 的长为 （ ）", + "page_idx": 43 + }, + { + "type": "text", + "text": "A.4 B.5 C.6 D.7 ", + "page_idx": 43 + }, + { + "type": "text", + "text": "二、填空题(每小题4分，共16分)", + "text_level": 1, + "page_idx": 43 + }, + { + "type": "text", + "text": "9.如图，直线 $A B , C D$ 相交于点 $O , E O \\bot A B$ 于点 $O$ ,若∠BOD： $\\angle D O E { = } 5$ ：13,则 $\\angle A O C$ 的度数为", + "page_idx": 43 + }, + { + "type": "image", + "img_path": "images/0f32dfb46de48bf0dfd4a2ce42bc3e086dd28da6f9f64c6beeb59300e717a4da.jpg", + "img_caption": [ + "第9题图" + ], + "img_footnote": [], + "page_idx": 43 + }, + { + "type": "image", + "img_path": "images/4a1074b73840dc7410abbcb14038b6270e4b3059e977bbab007f9585b3d6cb7d.jpg", + "img_caption": [ + "图2" + ], + "img_footnote": [], + "page_idx": 43 + }, + { + "type": "image", + "img_path": "images/149e83e52222403bea85f9c63414409219e1ad3e3f03a88c8035c4b3a4f20237.jpg", + "img_caption": [ + "第10题图" + ], + "img_footnote": [], + "page_idx": 43 + }, + { + "type": "image", + "img_path": "images/db11fc3ba6bf5fb5feac032dc3cd314c98631a38fc5aa32cfeab6de8e2f4efd7.jpg", + "img_caption": [ + "图1", + "第11题图" + ], + "img_footnote": [], + "page_idx": 43 + }, + { + "type": "text", + "text": "10.如图，在等边 $\\triangle A B C$ 中， $A D , B E$ 分别是角平分线， $A D$ 与 $B E$ 相交于点 $O$ ，则 $\\displaystyle \\frac { O E } { O D } =$ ", + "page_idx": 43 + }, + { + "type": "text", + "text": "11.新考法真实问题情境如图1,这是一辆山地自行车的实物图，图2是其抽象出来的部分示意图，已知直线 $E F$ 与 $B D$ 相交于点 $P , A B / / C D , \\angle P = 1 8 ^ { \\circ }$ ，$\\angle C F P = \\angle \\alpha$ $\\angle A B P { = } 9 2 ^ { \\circ }$ ，则 $\\angle \\alpha$ 的补角的度数为", + "page_idx": 43 + }, + { + "type": "text", + "text": "12.如图，在 $\\triangle A B C$ 纸片中， $\\angle A C B = 9 0 ^ { \\circ }$ ，将该三角形纸片折叠，使得点 $A$ 落在边 $B C$ 上的点 $E$ 处， $C A$ 与 $C E$ 重合，折痕为 $C D$ ，若 $2 \\angle A + 3 \\angle B = 2 2 2 ^ { \\circ }$ ，则$\\angle B D E$ 的度数为", + "page_idx": 44 + }, + { + "type": "image", + "img_path": "images/323bbc0cf663a250ad0a9bb6ffd46f6b7a6c4ae564515ffd28cb58c87d58c328.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 44 + }, + { + "type": "text", + "text": "三、解答题(本大题5小题，共52分)", + "text_level": 1, + "page_idx": 44 + }, + { + "type": "text", + "text": "13.(8分)如图，点 $D$ 在 $\\angle A B C$ 的边 $A B$ 上，作 $D E \\bot B C$ 于点 $E$ ，过 $\\angle A B C$ 内的一点 $F$ 作 $F G \\bot B C$ 于点 $G$ ，连接 $E F$ ，且 $E F / / A B$ ，若 $\\angle B = 3 1 ^ { \\circ }$ ，求 $\\angle A D E$ 和$\\angle F$ 的度数.", + "page_idx": 44 + }, + { + "type": "image", + "img_path": "images/7ac1e7a7c12ae1777fc2881a31d795e743a7166e5d42e17ac6f776cffc6b8c8b.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 44 + }, + { + "type": "text", + "text": "14.(9分)如图,已知直线 $A B , C D$ 相交于点O,OM⊥AB,ON⊥CD. ", + "page_idx": 44 + }, + { + "type": "text", + "text": "(1)若 $\\angle 1 = 2 9 ^ { \\circ }$ ，求 $\\angle 2$ 的度数; \n(2)若 $\\angle B O D { = } 6 2 ^ { \\circ }$ ，求 $\\angle 1$ 和 $\\angle M O N$ 的度数. ", + "page_idx": 44 + }, + { + "type": "image", + "img_path": "images/8a3a83e68b5e4e46143c1d0d9862e4c82b5a6c4f90f27005cb423cbcb1aa4d43.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 44 + }, + { + "type": "text", + "text": "15.(9 分)如图,在四边形 $A B C D$ 中， $A D / / B C$ ，点 $E$ 是 $C D$ 的中点，连接 $A E$ 并延长交 $B C$ 的延长线于点 $F , B E \\bot A F$ ", + "page_idx": 44 + }, + { + "type": "text", + "text": "(1)若 $B E { = } 4 , A E { = } 2$ ,求四边形 $A B C D$ 的面积； \n(2)若 $\\angle D A E = 5 9 ^ { \\circ }$ ，求 $\\angle A B F$ 的度数. ", + "page_idx": 44 + }, + { + "type": "image", + "img_path": "images/44d148d333301abf6cfa5b7f953352410c887e78ff58936ecc6dfed5d6c2a1c4.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 44 + }, + { + "type": "text", + "text": "16.（12分)阅读材料，完成任务：", + "page_idx": 44 + }, + { + "type": "text", + "text": "材料：我们知道，在 $\\triangle A B C$ 中，若 $A B { = } A C$ ，则 $\\angle B { = } \\angle C$ ，这就是等腰三角形的性质，用语言叙述就是等腰三角形的两底角相等，即等边对等角；反之，可以得到等腰三角形的判定：在 $\\triangle A B C$ 中，若 $\\angle B = \\angle C$ ，则 $A B { = } A C$ ，即两角相等的三角形是等腰三角形，即等角对等边. \n任务：如图，在△ABC中， $A B { = } A C , A D \\bot B C$ 于点 $D$ ，点 $N$ 是 $A D$ 上一点，且 $N M / / A C$ ,连接BN.(1)若 $\\angle A B C = 6 5 ^ { \\circ }$ ，求 $\\angle A N M$ 的度数; \n(2)若 $A C { = } 8$ ，点 $M$ 为AB 的中点， $B N { = } 3$ ,求MN的长和△BMN的周长.", + "page_idx": 44 + }, + { + "type": "text", + "text": "", + "page_idx": 44 + }, + { + "type": "image", + "img_path": "images/023d0d97e297b4ede745092bf496cd318192ed642103f9f05c8a2b55dbd6f18d.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 44 + }, + { + "type": "text", + "text": "17. $\\circledcirc$ 视频讲解（14分）综合与实践.", + "page_idx": 44 + }, + { + "type": "text", + "text": "问题情境：在 $\\triangle A B C$ 中,已知边 $A B , B C$ 的垂直平分线分别交AB,BC于点 $D , F$ ，与 $A C$ 分别交于点 $E , G .$ \n初步探究： $( 1 ) \\textcircled{ 1}$ 如图1,若 $\\angle A B C = 1 0 0 ^ { \\circ }$ ，求 $\\angle E B G$ 的度数； \n$\\textcircled{2}$ 如图2，若 $\\angle A B C = 7 0 ^ { \\circ }$ ，求 $\\angle E B G$ 的度数； \n归纳猜想：(2)若 $\\angle A B C = \\alpha$ $\\angle A B C = 9 0 ^ { \\circ }$ ,用含 $\\alpha$ 的代数式直接写出 $\\angle E B G$ 的度数. ", + "page_idx": 44 + }, + { + "type": "image", + "img_path": "images/ccf3d8da3e2d3441588efd2a6e81a924e091878a81ed7624e347581c1c09ad99.jpg", + "img_caption": [ + "图1" + ], + "img_footnote": [], + "page_idx": 44 + }, + { + "type": "image", + "img_path": "images/37080da8dc8ef8c3e1da643977c172d3d809efda4ddfac6a5e8542cec970fa6a.jpg", + "img_caption": [ + "图2" + ], + "img_footnote": [], + "page_idx": 44 + }, + { + "type": "text", + "text": "专项训练卷(三) 几何题的说理", + "text_level": 1, + "page_idx": 45 + }, + { + "type": "text", + "text": "时间：60分钟 满分：100分", + "page_idx": 45 + }, + { + "type": "table", + "img_path": "images/422d077658d327569d996bcdd69df3a8fa247e993a8fd438d34a7f8fcb9c0b2d.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 45 + }, + { + "type": "text", + "text": "一、选择题(每小题5分，共30分)", + "text_level": 1, + "page_idx": 45 + }, + { + "type": "text", + "text": "1.如图,已知直线l上有一点 $A$ ，直线 $m \\perp n$ ，则 $\\angle 1$ 与 $\\angle 2$ 的关系是 ", + "page_idx": 45 + }, + { + "type": "text", + "text": "A. $\\angle 1$ 与 $\\angle 2$ 相等 B. $\\angle 1$ 与 $\\angle 2$ 互余C. $\\angle 1$ 与 $\\angle 2$ 互补 D. $\\angle 1$ 与 $\\angle 2$ 是对顶角", + "page_idx": 45 + }, + { + "type": "image", + "img_path": "images/b853effb10fe22d338ebfbb902f846525683ec03bf9b27c58ccbcaee7d2f362f.jpg", + "img_caption": [ + "第1题图" + ], + "img_footnote": [], + "page_idx": 45 + }, + { + "type": "image", + "img_path": "images/12839f8809656798cf434b57b4e3c52f532f976d48fe604d31c88ee8ef9c4e5d.jpg", + "img_caption": [ + "第2题图" + ], + "img_footnote": [], + "page_idx": 45 + }, + { + "type": "image", + "img_path": "images/6d8b3514358d7de0e59d33b99d64f63a5e651a134461e8f3f5025d563cf9dc30.jpg", + "img_caption": [ + "第3题图" + ], + "img_footnote": [], + "page_idx": 45 + }, + { + "type": "text", + "text": "2.如图，已知点 $D , B$ 在线段 $A E$ 上,若 $\\angle A B C = \\angle E$ ，则下列结论正确的是 （ ", + "page_idx": 45 + }, + { + "type": "text", + "text": "A. AC// DF B. BD // AC $\\complement . B C / / E F$ （204号 （204号 $\\mathrm { D } , D G / / A C$ （20 ", + "page_idx": 45 + }, + { + "type": "text", + "text": "3.将一副三角板和一张对边平行的纸条按如图所示的方式摆放,两把三角板的一条直角边重合,则 $\\angle 1$ 与 $\\angle 2$ 之间的数量关系是 （ ）", + "page_idx": 45 + }, + { + "type": "text", + "text": "A. $\\angle 1 + \\angle 2 = 3 0 ^ { \\circ }$ B. $\\angle 1 + \\angle 2 = 4 5 ^ { \\circ }$ $\\mathrm { { C } } _ { \\cdot } \\angle 1 + \\angle 2 = 6 0 ^ { \\circ }$ $D _ { \\bullet } 2 \\angle 1 + \\angle 2 = 7 5 ^ { \\circ }$ ", + "page_idx": 45 + }, + { + "type": "text", + "text": "4.在 $\\triangle A B C$ 中，若 $\\angle A { = } 2 \\angle B$ $\\angle A = \\angle C + 2 0 ^ { \\circ }$ ，则 $\\triangle A B C$ 的形状是 （ ）", + "page_idx": 45 + }, + { + "type": "text", + "text": "A.锐角三角形 B.钝角三角形C.直角三角形 D.无法确定", + "page_idx": 45 + }, + { + "type": "text", + "text": "5.若△ABC的三边长分别为 $a , b , c$ ，则 $| a + b + c | + | a - b - c |$ 的值为（) ", + "page_idx": 45 + }, + { + "type": "text", + "text": "A. $a + 2 b$ B. $2 b + 2 c$ C. $2 a + 2 c$ （204号 D. $2 b { + } c$ （20 ", + "page_idx": 45 + }, + { + "type": "text", + "text": "6. $\\circledcirc$ 视频讲解如图，在△ABC中， $A B { = } A C$ ，以点 $C$ 为圆心， $C B$ 长为半径画弧，交 $A B$ 于点 $B$ 和点 $D$ ,再分别以点 $B , D$ 为圆心，大于 $\\frac { 1 } { 2 } B D$ 长为半径画弧,两弧相交于点 $M$ ，作射线CM交 $A B$ 于点 $E$ ，连接 $C D , D M , B M ,$ 则下列结论正确的有（", + "page_idx": 45 + }, + { + "type": "image", + "img_path": "images/ab45dbe2363ddf7573709e8d1b4997443e283d15a47dcee3dbdca5b1fdc06e04.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 45 + }, + { + "type": "equation", + "img_path": "images/e92c4f7222ca5571a88a3a7894993f7da495113d488c508b7cfc80fb4af2b395.jpg", + "text": "$$\n\\textcircled { 1 } \\angle C B D = \\angle C D B ; \\textcircled { 2 } C M \\perp B D ; \\textcircled { 3 } \\angle C D B + \\frac { 1 } { 2 } \\angle A = 9 0 ^ { \\circ } .\n$$", + "text_format": "latex", + "page_idx": 45 + }, + { + "type": "text", + "text": "A $\\textcircled{1} \\textcircled{2}$ B. $\\textcircled{1} \\textcircled{3}$ C. $\\textcircled{2} \\textcircled{3}$ D. $\\textcircled{1} \\textcircled{2} \\textcircled{3}$ ", + "page_idx": 45 + }, + { + "type": "text", + "text": "二、填空题(每小题5分，共20分)", + "text_level": 1, + "page_idx": 45 + }, + { + "type": "text", + "text": "7.如图,已知点 $C$ 是 $A E$ 的中点， $\\scriptstyle \\angle A = \\angle E C D , A B = C D$ 那么△ABC≌△CDE 的根据是 .（填\"SSS\"、“SAS\"“ASA\"或\"AAS\") ", + "page_idx": 45 + }, + { + "type": "image", + "img_path": "images/02b911f3500ed03b123c1fb0c18d96a4302e02a34de19825e26cd66ec7323a2a.jpg", + "img_caption": [ + "第7题图" + ], + "img_footnote": [], + "page_idx": 45 + }, + { + "type": "image", + "img_path": "images/b9c8c0263bb16eff7deaa1d8013b898b4ab41980adce426e5a674f74569434b1.jpg", + "img_caption": [ + "第8题图" + ], + "img_footnote": [], + "page_idx": 45 + }, + { + "type": "image", + "img_path": "images/c506603b27fbf2dcdbb844c3ee104ddfff9bbe7c0e35eb21870c9f9ce435dbeb.jpg", + "img_caption": [ + "第9题图" + ], + "img_footnote": [], + "page_idx": 45 + }, + { + "type": "image", + "img_path": "images/e924382215435b7549d74138301e8eb6285f659972e9f3be06896fdac5ae8c49.jpg", + "img_caption": [ + "第10题图" + ], + "img_footnote": [], + "page_idx": 45 + }, + { + "type": "text", + "text": "8.如图所示,这是一个正五边形,它是一个轴对称图形,它有 条对称轴.", + "page_idx": 45 + }, + { + "type": "text", + "text": "9.如图，在“ $4$ ”字图中， $B C$ 与 $E F$ 相交于点 $D$ ,若图中有 $a$ 对同位角, $b$ 对内错角， $c$ 对同旁内角， $d$ 对对顶角，则 $( a + b - c + d ) ^ { 3 }$ 的值为", + "page_idx": 45 + }, + { + "type": "text", + "text": "10.如图,已知 $A B / / C D , A E \\bot E F ,$ 则 $\\angle A , \\angle C , \\angle F$ 之间的数量关系是 ", + "page_idx": 45 + }, + { + "type": "text", + "text": "三、解答题(本大题5小题，共50分)", + "text_level": 1, + "page_idx": 45 + }, + { + "type": "text", + "text": "11.(8分)如图，直线 $A B$ 与 $E F$ 相交于点 $O$ (1)若 $\\angle A O E { = } 4 5 ^ { \\circ }$ $\\angle A O F { = } 3 \\angle C O F$ ，试说明 $A B \\bot C O$ (2)在(1)的条件下,找出所有与 $\\angle G O F$ 互余的角和互补的角.", + "page_idx": 45 + }, + { + "type": "text", + "text": "", + "page_idx": 45 + }, + { + "type": "text", + "text": "", + "page_idx": 45 + }, + { + "type": "image", + "img_path": "images/b022f649616ba4ab894144c0fb012fd50b4e9653d3e64d99df3d0193d415632b.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 45 + }, + { + "type": "text", + "text": "12.(8分)如图，点 $D , E$ 在边BC上, $E F$ 垂直平分 $A C$ $A B { = } A E$ ,AD是△ABE 的中线. ", + "page_idx": 46 + }, + { + "type": "text", + "text": "（1）试说明△ABD△AED； \n(2）试探究 $\\angle B$ 与 $\\angle C$ 之间的数量关系，并说明理由.", + "page_idx": 46 + }, + { + "type": "image", + "img_path": "images/ee71b1afc7d9faa9f527eb244b5fba79e15b4b0aff20e91c8c5de225309ee0b0.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 46 + }, + { + "type": "text", + "text": "13.(8分)如图，在 $\\mathrm { R t } \\triangle A B C$ 中， $\\angle B = 9 0 ^ { \\circ } , A D$ 是 $\\angle B A C$ 的平分线,DE⊥AC于点 $E$ ，点 $F$ 是 $A B$ 上一点， $B { \\cal F } { = } C E$ ", + "page_idx": 46 + }, + { + "type": "text", + "text": "(1)△BDF和 $\\triangle E D C$ 全等吗？请说明理由. \n(2)试猜想 $A C , A B , B F$ 之间的数量关系，并说明理由.", + "page_idx": 46 + }, + { + "type": "image", + "img_path": "images/92cfdd907ea2dbb12469e29fd543e96d1b3229598ff69190db96dc5a97899a8c.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 46 + }, + { + "type": "text", + "text": "14.（12分)如图，已知点 $P , Q$ 分别在直线 $A B$ ， $C D$ 上， $\\angle A P N + \\angle P N Q +$ $\\angle C Q N = 3 6 0 ^ { \\circ }$ ", + "page_idx": 46 + }, + { + "type": "text", + "text": "(1)试说明 $A B / / C D$ ", + "page_idx": 46 + }, + { + "type": "text", + "text": "(2)若 $P M , Q M$ 分别平分 $\\angle B P N$ $\\angle D Q N$ ，试探究 $\\angle M$ 与 $\\angle N$ 之间的数量关系，并说明理由.", + "page_idx": 46 + }, + { + "type": "image", + "img_path": "images/80eb1cac2a021ada699403d6b752ada3a3f25a43b456d387f25b05b1ba63a578.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 46 + }, + { + "type": "text", + "text": "15.新考法阅读理解题 $\\circledcirc$ 视频讲解（14分)阅读材料，完成后面的任务：", + "page_idx": 46 + }, + { + "type": "text", + "text": "材料：如图1，在凹四边形ABOC中，试探究$\\angle B O C$ 与 $\\angle A , \\angle B , \\angle C$ 之间的数量关系．小明发现并提供了如下方法：解： $\\angle B O C = \\angle A + \\angle B + \\angle C .$ 如图2,连接", + "page_idx": 46 + }, + { + "type": "image", + "img_path": "images/304f0c613589c47f939e2e92674324f1a0800101ee8c8199b00f084ea5e0b278.jpg", + "img_caption": [ + "图1" + ], + "img_footnote": [], + "page_idx": 46 + }, + { + "type": "image", + "img_path": "images/5840df022ed06d95be097ba8aca5fa44b87423c1e8040ce4257099a96fa77f6a.jpg", + "img_caption": [ + "图2" + ], + "img_footnote": [], + "page_idx": 46 + }, + { + "type": "text", + "text": "$A O$ 并延长，因为 $\\angle 1 + \\angle B + \\angle A O B = 1 8 0 ^ { \\circ } , \\angle 3 + \\angle A O B = 1 8 0 ^ { \\circ }$ ，所以 $\\angle 3 =$ $\\angle 1 + \\angle B$ ，同理 $\\angle 4 = \\angle 2 + \\angle C$ 所以 $\\angle 3 + \\angle 4 = \\angle 1 + \\angle 2 + \\angle B + \\angle C ,$ 所以 $\\angle B O C = \\angle B A C + \\angle B + \\angle C .$ ", + "page_idx": 46 + }, + { + "type": "text", + "text": "任务：(1)你有与小明不同的方法吗？请写出你的解答过程.", + "page_idx": 46 + }, + { + "type": "text", + "text": "(2)如图3所示，根据图中标注的角的度数，请说明 $\\angle A + \\angle C + \\angle D + \\angle F =$ $2 3 0 ^ { \\circ }$ ", + "page_idx": 46 + }, + { + "type": "image", + "img_path": "images/b1119104638303808606f85458112bdc1a66a31f00be9753d74199ccb58b24e1.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 46 + }, + { + "type": "image", + "img_path": "images/1e620d511f484cdbdd6b112371bf4acbea44ef03189714cca65626e94c784e9f.jpg", + "img_caption": [ + "图4" + ], + "img_footnote": [], + "page_idx": 46 + }, + { + "type": "text", + "text": "(3)如图4所示，在 $\\triangle A B C$ 中， $A B { = } A C$ $\\angle A B C , \\angle A C B$ 的平分线 $B D , C E$ 交于点 $O , \\angle E O D + \\angle O B F = 1 8 0 ^ { \\circ } , B F$ 交 $A C$ 的延长线于点 $F$ $, \\angle C D G =$ $\\angle F , D G$ 交 $B C$ 的延长线于点 $G$ ,找出图中的平行线，并说明理由.", + "page_idx": 46 + }, + { + "type": "text", + "text": "", + "page_idx": 46 + }, + { + "type": "text", + "text": "(4)在(3)的条件下，若 $\\angle A = 5 8 ^ { \\circ }$ ，直接写出 $\\angle A D G$ 的度数：", + "page_idx": 46 + }, + { + "type": "image", + "img_path": "images/357b43173835f003d78a1c038762b3596da146dea3499e09342b9895aa80f747.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 47 + }, + { + "type": "text", + "text": "期末测试卷", + "text_level": 1, + "page_idx": 47 + }, + { + "type": "text", + "text": "时间：90分钟 满分：120分考试范围：第一章～第六章", + "page_idx": 47 + }, + { + "type": "table", + "img_path": "images/6572b6d9b33c75a13a7a635040223d610cfdb4623488ee091864b02a035cefa1.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
题序二三评卷人总分
得分
", + "page_idx": 47 + }, + { + "type": "text", + "text": "一、选择题(本大题共10小题,每小题3分，共30 分)", + "text_level": 1, + "page_idx": 47 + }, + { + "type": "text", + "text": "1.下列\"表情”中,属于轴对称图形的是", + "page_idx": 47 + }, + { + "type": "image", + "img_path": "images/8855ee32514d781c76c28b250c5e304683f2d8320d5d1d3abe000adc8c603582.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 47 + }, + { + "type": "text", + "text": "2.下列说法中，正确的是", + "page_idx": 47 + }, + { + "type": "text", + "text": "A.若 $\\angle \\alpha$ 与 $\\angle \\beta$ 是同位角，则 $\\angle \\alpha { = } \\angle \\beta$ \nB.若 $\\angle 1 + \\angle 2 = 9 0 ^ { \\circ }$ ，则 $\\angle 1$ 与 $\\angle 2$ 互余 \nC.两条边和一个角分别对应相等的两个三角形全等 \nD.若一个事件发生的概率为0,则这个事件是不确定事件", + "page_idx": 47 + }, + { + "type": "text", + "text": "3.若△ABC是等腰三角形,其中两边长分别为 $3 \\ \\mathrm { c m } , 6 \\ \\mathrm { c m }$ ，则第三边长可能为（ ）", + "page_idx": 47 + }, + { + "type": "text", + "text": "A.3 cm B.4 cm C. 6 cm D. $3 ~ \\mathrm { c m }$ 或 $6 ~ \\mathrm { c m }$ ", + "page_idx": 47 + }, + { + "type": "text", + "text": "4.如图，在 $\\triangle A B C$ 和 $\\triangle D E C$ 中，已知 $C B = C E$ ，还需添加两个条件才能使$\\triangle A B C { \\cong } \\triangle D E C$ ，不能添加的一组条件是 （ ）", + "page_idx": 47 + }, + { + "type": "text", + "text": "A. $A B { = } D E$ $\\angle B = \\angle E$ $3 . A B { = } D E , A C { = } D C$ $\\scriptstyle \\mathrm { C } , A B = D E , \\angle A = \\angle D$ D. $\\angle A = \\angle D , \\angle B = \\angle E$ ", + "page_idx": 47 + }, + { + "type": "image", + "img_path": "images/969be655392a420e822bf85a33aecdb9fe7b9b40c7c809e7744a986245fea9dd.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 47 + }, + { + "type": "text", + "text": "5.“四时花竞巧，九子粽争新”,端午节吃粽子是我国的传统习俗.小颖妈妈在超市购买了豆沙粽和蛋黄粽共15个，这些粽子除了内部馅料不同外，其他均相同.小颖从中任选了一个粽子,若她选到蛋黄粽的概率为 $\\frac { 3 } { 5 }$ ，则购买的豆沙粽的个数是", + "page_idx": 47 + }, + { + "type": "text", + "text": "", + "page_idx": 47 + }, + { + "type": "text", + "text": "A.5 B.6 C.8 D.9 ", + "page_idx": 47 + }, + { + "type": "text", + "text": "6.如图,将长方形纸片 $A B C D$ 沿 $E F$ 折叠后，点 $D , C$ 分别落在点 $D _ { 1 } , C _ { 1 }$ 的位 置， $E D _ { 1 }$ 的延长线交 $B C$ 于点 $G$ ，若 $\\angle E F G { = 6 4 } ^ { \\circ }$ ，则 $\\angle E G B$ 等于 （ ", + "page_idx": 47 + }, + { + "type": "text", + "text": "A. $1 2 8 ^ { \\circ }$ （204号 B.130° （204号 $C . 1 3 2 ^ { \\circ }$ （20 D. $1 3 6 ^ { \\circ }$ ", + "page_idx": 47 + }, + { + "type": "image", + "img_path": "images/b1c805e02f145869410a7422b960077e565fe44873be81f7af66669d97c1baee.jpg", + "img_caption": [ + "第6题图" + ], + "img_footnote": [], + "page_idx": 47 + }, + { + "type": "image", + "img_path": "images/bd2864a40ffcc6b99019ae0e96898793e4668ccc92b595f3d9e3a37d4d549e07.jpg", + "img_caption": [ + "第7题图" + ], + "img_footnote": [], + "page_idx": 47 + }, + { + "type": "image", + "img_path": "images/4864dfe3aa14d74b74375855177d93d10361f90a5c3f8199ddba7708d954ed61.jpg", + "img_caption": [ + "第8题图" + ], + "img_footnote": [], + "page_idx": 47 + }, + { + "type": "text", + "text": "7.为增强居民的节水意识,某自来水公司采用以户为单位分段计费的办法收费,即每月用水不超过10吨,每吨收费 $a$ 元;若超过10吨,则10吨水按每吨$a$ 元收费,超过10吨的部分按每吨 $^ { b }$ 元收费，公司为居民绘制的水费y(元)随当月用水量 $_ { \\mathcal { X } }$ （吨)变化而变化的图象如图所示，则下列结论错误的是（ ）", + "page_idx": 47 + }, + { + "type": "text", + "text": "A. $a = 1$ .5 \nB. $b { = } 2$ \nC.若小明家7月份缴水费30元，则当月用水18.5吨D.若小明家3月份用水14吨，则应缴水费23元", + "page_idx": 47 + }, + { + "type": "text", + "text": "8.如图，在四边形 $A B C D$ 中， $A E \\bot C D$ ，点 $B$ 关于 $A C$ 的对称点 $B ^ { \\prime }$ 恰好落在 $C D$ 上，点 $B ^ { \\prime }$ 关于 $A E$ 的对称点为点 $D$ ，若 $\\angle B A D = 1 0 0 ^ { \\circ }$ ，则 $\\angle A C B$ 的度数为（ ）", + "page_idx": 47 + }, + { + "type": "text", + "text": "A. $4 0 ^ { \\circ }$ （20 B.45° C.60° D.80° ", + "page_idx": 47 + }, + { + "type": "text", + "text": "9.如果一个正整数能表示为两个连续偶数的平方差，那么称这个正整数为“和平数”如 $4 = 2 ^ { 2 } - 0 ^ { 2 } , 1 2 = 4 ^ { 2 } - 2 ^ { 2 }$ ,因此4,12 这两个数都是“和平数”.介于1到301之间的所有“和平数\"之和为 （）", + "page_idx": 47 + }, + { + "type": "text", + "text": "A.5776 B.4096 C.2022 D.108 ", + "page_idx": 47 + }, + { + "type": "text", + "text": "10. $\\circledcirc$ 视频讲解如图，在等腰 $\\triangle A B C$ 中， $A B = A C , \\angle B A C = 5 0 ^ { \\circ } , \\angle B A$ 的平分线与线段 $A B$ 的垂直平分线交于点 $O$ ，点 $C$ 沿 $E F$ 折叠后与点 $O$ 重合，则$\\angle C E F$ 的度数是 （ ）", + "page_idx": 48 + }, + { + "type": "text", + "text": "A. $4 5 ^ { \\circ }$ B. $5 0 ^ { \\circ }$ （20 $\\mathrm { C } . 5 5 ^ { \\circ }$ （20 D. $6 0 ^ { \\circ }$ ", + "page_idx": 48 + }, + { + "type": "image", + "img_path": "images/d39326d8d354618f7c808e9fd0fd13f3da982f6738437174b453704ae1922063.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 48 + }, + { + "type": "text", + "text": "二、填空题(本大题共5小题,每小题3分，共15分)", + "text_level": 1, + "page_idx": 48 + }, + { + "type": "text", + "text": "11.一个游戏转盘如图所示，自由转动转盘，当转盘停止转动后,指针落在字母“C\"所示区域内的概率是", + "page_idx": 48 + }, + { + "type": "text", + "text": "（2)如图，在△ABC中， $\\angle A = 9 0 ^ { \\circ }$ ，点 $E$ 为 $B C$ 上一点，点 $A$ 和点 $E$ 关于 $B D$ 对称,点 $B$ 和点 $C$ 关于 $D E$ 对称，求 $\\angle A B C$ 和 $\\angle C$ 的度数.", + "page_idx": 48 + }, + { + "type": "image", + "img_path": "images/5307ce51a1b88d363cc71f860079ed72ac05d6631ddec358a7999f38bbc77bd2.jpg", + "img_caption": [ + "第11题图" + ], + "img_footnote": [], + "page_idx": 48 + }, + { + "type": "image", + "img_path": "images/3d3ddb0d0294ce77ad79ea9bd2bdd76f82a88cbaf4adf0f753a17f22efcd60ca.jpg", + "img_caption": [ + "第12题图" + ], + "img_footnote": [], + "page_idx": 48 + }, + { + "type": "image", + "img_path": "images/60e5a9b64348db6ef279682835f6ee2bd4e15acbf79deb03e374e137ac4d8b11.jpg", + "img_caption": [ + "第14题图" + ], + "img_footnote": [], + "page_idx": 48 + }, + { + "type": "image", + "img_path": "images/d16f88fabcce8dd837e6737fd2a4885fc1478f94f5bea4d8d6e5d7c6e49f8a24.jpg", + "img_caption": [ + "第15题图" + ], + "img_footnote": [], + "page_idx": 48 + }, + { + "type": "text", + "text": "12.如图， $\\scriptstyle \\cdot B E = B A , A B / / D E , B C = D E$ ，若 $\\angle B A C = 4 0 ^ { \\circ }$ $\\angle E { = } 2 5 ^ { \\circ }$ ，则 $\\angle B D E$ 的 度数为 ", + "page_idx": 48 + }, + { + "type": "text", + "text": "13.如果 $x ^ { 2 } + 3 x { = } 2 0 2 4$ ,那么代数式 $x ( 2 x + 1 ) - ( x - 1 ) ^ { 2 }$ 的值为", + "page_idx": 48 + }, + { + "type": "image", + "img_path": "images/ab49604f8ffa3913d9b38e61212e10652ec86b7e3adde3c5cb2f83a39798d99f.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 48 + }, + { + "type": "text", + "text": "14.如图，在 $4 \\times 4$ 的正方形网格中，已有4个小方格涂成了灰色，现在要从其余白色小方格中选出一个也涂成灰色，使整个灰色部分的图形构成轴对称图形,这样的白色小方格有 个.", + "page_idx": 48 + }, + { + "type": "text", + "text": "15. $\\circledast$ 视频讲解如图，在 $\\triangle A B C$ 中， $A B { = } A C$ ，点 $D , E , F$ 分别是边 $A B \\circ B C , C A$ 上的点， $D E$ 与 $B F$ 相交于点 $G , B D { = } B C , B E { = } C F$ 若 $\\angle A = 4 0 ^ { \\circ }$ ，则 $\\angle D G F$ 的度数为", + "page_idx": 48 + }, + { + "type": "text", + "text": "三、解答题(本大题共8小题，满分75分）", + "text_level": 1, + "page_idx": 48 + }, + { + "type": "text", + "text": "16.(10分）（1)计算： $( - \\frac { 1 } { 4 } ) ^ { - 1 } + ( \\pi - 2 ) ^ { 0 } - ( - 3 ) ^ { - 3 } \\div ( - 3 ) ^ { - 4 } + | - 2 | .$ ", + "page_idx": 48 + }, + { + "type": "text", + "text": "17.(9分)公安人员在破案时常常根据案发现场作案人员留下的脚印推断犯人的身高，如果用 $a$ 表示脚印长度， $b$ 表示身高，它们的关系接近于 $b { = } 7 a -$ 3.07.", + "page_idx": 48 + }, + { + "type": "text", + "text": "(1)若某人的脚印长度为 $2 4 . 5 ~ \\mathrm { c m }$ ，则他的身高约为多少厘米？ \n(2)在某次案件中，抓获了两名可疑人员，一个身高为 $1 . 8 7 \\mathrm { ~ m ~ }$ ，另一个身高为$1 . 7 5 \\mathrm { ~ m ~ }$ ,现场测量的脚印长度为 $2 6 . 7 \\ \\mathrm { c m }$ ,请你帮助侦查一下，哪个可疑人员作案的可能性更大，并说明理由.", + "page_idx": 48 + }, + { + "type": "image", + "img_path": "images/6dfe0827c875877d90d7d3b3b98ba9cc4d0496e2988cb7e6043b070807d97311.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 49 + }, + { + "type": "text", + "text": "18.(9分)已知 $( m + n ) ^ { 2 } = 1 1 ^ { 2 }$ ， $m n { = } 1$ ，求 $( m - n ) ^ { 2 }$ 的值. ", + "page_idx": 49 + }, + { + "type": "text", + "text": "19.(9分)如图，在 $\\triangle A B C$ 中， $\\angle A = 6 8 ^ { \\circ }$ ，点 $D$ 是 $B C$ 上一点， $B D , C D$ 的垂直平分线分别交 $A B$ $A C$ 于点 $E , F$ ，求 $\\angle E D F$ 的度数.", + "page_idx": 49 + }, + { + "type": "text", + "text": "20.(9分)如图，点 $C$ 在线段 $A E$ 上， $B C / / D E , A C = D E , B C = C E$ ，延长 $A B$ 分别交 $C D , E D$ 于点 $G , F$ ", + "page_idx": 49 + }, + { + "type": "text", + "text": "（1)试说明： $A B { = } C D$ ", + "page_idx": 49 + }, + { + "type": "image", + "img_path": "images/b4df81009446f5179bff5f5252e2bedddf3a8d8b0f2c88db7e6877c5891a607b.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 49 + }, + { + "type": "text", + "text": "(2)若 $\\angle D = 3 0 ^ { \\circ }$ $\\angle E { = } 6 5 ^ { \\circ }$ ，求 $\\angle F G C$ 的度数. ", + "page_idx": 49 + }, + { + "type": "image", + "img_path": "images/c492cb1542614cbef2ce27c7387c4214b37bf2798c7d96b59a1c790c31e1b737.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 49 + }, + { + "type": "text", + "text": "21.(9分)如图，点 $D$ 是 $\\angle B A C$ 外一点，过点 $D$ 作 $D E / / A B$ 交 $A C$ 于点 $F$ ，以$D E$ 为边作 $\\angle E D G .$ 若 $D G / / A C$ ,请画出图形,判断 $\\angle B A C$ 与 $\\angle E D G$ 的数量关系，并说明理由.", + "page_idx": 49 + }, + { + "type": "image", + "img_path": "images/58468dc02d99cebfff32b239baedfde7a5598e2c741a3712157b344badb5f76e.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 49 + }, + { + "type": "text", + "text": "22.(10分)如图，张达设计了一个被均匀地分成了8份的圆形转盘，该转盘分别标有1,2,3,4,5,6,7,8这8个数,转动转盘，当转盘停止时，指针指向的数即为转出的数.(当指针恰好指在分界线上时，无效重转)", + "page_idx": 50 + }, + { + "type": "text", + "text": "(1)求小彬转出的数是4的倍数的概率. ", + "page_idx": 50 + }, + { + "type": "text", + "text": "(2)现有两张分别写有3和5的卡片，随机转动转盘，转盘停止后记下转出的数，与两张卡片上的数分别作为三条线段的长度.这三条线段能构成三角形的概率是多少？", + "page_idx": 50 + }, + { + "type": "image", + "img_path": "images/2f65936634cb4ad55eaeac91faf25bd2b9ffe6a6d5602fa19d326e16fc2424f2.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 50 + }, + { + "type": "text", + "text": "23. $\\circledcirc$ 视频讲(10分)现有一笔直的公路连接 $M , N$ 两地,甲车从 $M$ 地驶往 $N$ 地，速度为 $8 0 ~ \\mathrm { k m / h }$ ，同时乙车从 $N$ 地驶往 $M$ 地，速度为 $1 0 0 ~ \\mathrm { k m / h } .$ 途中甲车发生故障,于是停车修理了 $2 . 5 \\mathrm { ~ h ~ }$ ，修好后立即开车驶往 $N$ 地.设乙车行驶的时间为 $^ { t \\mathrm { ~ h ~ } }$ ，两车之间的距离为 $s \\ \\mathrm { k m } .$ 已知 $s$ 与 $t$ 之间的关系的部分图象如图所示.", + "page_idx": 50 + }, + { + "type": "text", + "text": "$( 1 ) M , N$ 两地的实际距离为 km. (2)图象中点 $C$ 的实际意义是 （3)甲车出发几小时后发生故障？ (4)乙车出发几小时后两车相距 $2 0 0 ~ \\mathrm { k m } ?$ （20 ", + "page_idx": 50 + }, + { + "type": "image", + "img_path": "images/1a9cb12234edcd732bb7dbfcd7629189427dc354c1bc8a5b0e816d896e0a89e1.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 50 + }, + { + "type": "text", + "text": "答案和解析 ", + "text_level": 1, + "page_idx": 51 + }, + { + "type": "text", + "text": "第一章整式的乘除", + "text_level": 1, + "page_idx": 51 + }, + { + "type": "text", + "text": "卷中悟法", + "text_level": 1, + "page_idx": 51 + }, + { + "type": "text", + "text": "1.正确计算幂的乘方 ", + "text_level": 1, + "page_idx": 51 + }, + { + "type": "text", + "text": "1.幂的乘方是变乘方为指数的乘 法，即底数不变，指数相乘.而同底数幂 的乘法是变乘法为指数的加法，即底数 不变，指数相加. ", + "page_idx": 51 + }, + { + "type": "text", + "text": "2.幂的乘方不能和同底数幂的乘法相混淆.例如不能把 $( a ^ { 5 } ) ^ { 2 }$ 的计算结果错误地写成 $a ^ { 7 }$ ，也不能把 $a ^ { 5 } \\cdot a ^ { 2 }$ 的计算结果错误地写成 $a ^ { 1 0 }$ ", + "page_idx": 51 + }, + { + "type": "text", + "text": "2.利用转化的思想解幂的有关计算", + "page_idx": 51 + }, + { + "type": "text", + "text": "解幂的有关计算时，一般有以下两种转化方法：", + "page_idx": 51 + }, + { + "type": "text", + "text": "(1)把不同底数的幂转化为相同底数的幂；(2)把不同指数的幂转化为相同指数的幂.", + "page_idx": 51 + }, + { + "type": "text", + "text": "3.多项式乘多项式的解题技巧 ", + "text_level": 1, + "page_idx": 51 + }, + { + "type": "text", + "text": "对于多项式乘多项式，由于项可能较多，所以计算比较复杂、繁琐，对于此类题有以下几种解题技巧：", + "page_idx": 51 + }, + { + "type": "text", + "text": "(1)可以将一个多项式拆开，转化为 先单项式乘多项式，再求和的形式. ", + "page_idx": 51 + }, + { + "type": "text", + "text": "(2)确定积中每项的符号时，按“同号得正，异号得负”的法则进行。", + "page_idx": 51 + }, + { + "type": "text", + "text": "(3)观察多项式的组成形式，合理利用平方差公式及完全平方公式解题.", + "page_idx": 51 + }, + { + "type": "text", + "text": "(4)利用公式 $\\left( x + a \\right) \\left( x + b \\right) = x ^ { 2 } +$ $( a + b ) x + a b$ ，能提高运算效率.", + "page_idx": 51 + }, + { + "type": "text", + "text": "(5)最终结果中，若有同类项要合并同类项.", + "page_idx": 51 + }, + { + "type": "text", + "text": "4.单项式除以单项式的解题技巧 ", + "page_idx": 51 + }, + { + "type": "text", + "text": "单项式除以单项式可以按以下三个步骤进行：", + "page_idx": 51 + }, + { + "type": "text", + "text": "(1)将系数相除，作为商的系数。(2)对于被除式和除式均含有的字母按照同底数幂的除法法则相除.(3)被除式中单独含有的字母，连同它的指数一起作为商的因式.", + "page_idx": 51 + }, + { + "type": "text", + "text": "对于商中符号的确定可按照“同号得正，异号得负”的法则，对于多项式除以单项式，可将多项式拆开，将其转化为单项式除以单项式的问题.", + "page_idx": 51 + }, + { + "type": "text", + "text": "基础过关参考答案", + "text_level": 1, + "page_idx": 51 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 51 + }, + { + "type": "table", + "img_path": "images/73ba0793e157883be177185652ba53789749da1961c023628422cc06482b940c.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
12345678
CDBACBAD
", + "page_idx": 51 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 51 + }, + { + "type": "table", + "img_path": "images/84bc0e36299239acde6b85eac70440973a4fb6472d240fb956df1b015fe54257.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
9101112
1.89×10-55x62m4n+2m²n² -2m²n-1
", + "page_idx": 51 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 51 + }, + { + "type": "text", + "text": "13.解：（1）原式 $= - b ^ { 5 } \\cdot b ^ { 1 2 } = - b ^ { 1 7 }$ ；… · (3分) (2）原式 $= - x ^ { 3 } \\cdot x ^ { 4 } - 4 x ^ { 6 } \\cdot x = - x ^ { 7 }$ $- 4 x ^ { 7 } = - 5 x ^ { 7 }$ ……· (6分) ", + "page_idx": 51 + }, + { + "type": "text", + "text": "14.解：(1)原式 $= x ^ { 2 } - x + 2 x ^ { 2 } = 3 x ^ { 2 } - x$ · (4分) (2）原式 $= 7 x + 7 - x ^ { 2 } - x - x ^ { 2 } + 3 x$ $= - 2 x ^ { 2 } + 9 x + 7$ ……· (8分) ", + "page_idx": 51 + }, + { + "type": "text", + "text": "15.解：原式 $= 4 ( x ^ { 2 } - 2 x + 1 ) - ( 4 x ^ { 2 } - 9 )$ $= 4 x ^ { 2 } - 8 x + 4 - 4 x ^ { 2 } + 9 = - 8 x + 1 3 .$ …(5分) 当 $x { = } - 2$ 时，原式 $= - 8 \\times ( - 2 ) +$ $1 3 { = } 2 9$ …（8分) ", + "page_idx": 51 + }, + { + "type": "text", + "text": "16.解：(1)因为 $a \\ast b { = } 2 ^ { a } \\times 2 ^ { b }$ ，所以 $1 \\ast 3$ $= 2 ^ { 1 } \\times 2 ^ { 3 } = 2 \\times 8 = 1 6$ ;…（3分)(2)因为 $2 * ( x + 1 ) = 2 5 6$ ，所以 $2 ^ { 2 } \\times$ $2 ^ { x + 1 } = 2 ^ { 8 }$ ，则 $2 + x + 1 = 8$ ，解得 $x = \\mathrm { ~ i ~ }$ 5． ··(8分)", + "page_idx": 51 + }, + { + "type": "text", + "text": "17. $\\begin{array} { r l r } & { } & { \\hbar \\pmb { \\mathscr { G } } : ( 1 ) ( a x - 1 ) ( x + 7 ) - x ^ { 2 } + 7 = a x ^ { 2 } \\left| \\right. } \\\\ & { } & { \\left. + 7 a x - x - 7 - x ^ { 2 } + 7 = ( a - 1 ) x ^ { 2 } + \\right. } \\end{array}$ $( 7 a - 1 ) x$ ，因为它不含有 $x ^ { 2 }$ 的项，所以 $a = 1$ … (4分)", + "page_idx": 51 + }, + { + "type": "equation", + "img_path": "images/58a52f364829b0b2ba4ef9beb61ea702c34c85283a4bcf67f4d127e78bb38450.jpg", + "text": "$$\n\\begin{array} { c } { { ( 2 ) ( a - 5 ) ( a + 5 ) + ( a - 5 ) ^ { 2 } - a ( 2 a } } \\\\ { { + 1 ) = a ^ { 2 } - 2 5 + 2 5 - 1 0 a + a ^ { 2 } - 2 a ^ { 2 } } } \\end{array}\n$$", + "text_format": "latex", + "page_idx": 51 + }, + { + "type": "text", + "text": "$- a { = } - 1 1 a$ ， …· (8分)当 $a = 1$ 时，原式 $= - 1 1 \\times 1 = - 1 1$ …… (10分)", + "page_idx": 51 + }, + { + "type": "text", + "text": "18.解: $( 1 ) P = ( x - y ) ^ { 5 } \\div ( y - x ) ^ { 5 } \\cdot ( x$ $- y ) ^ { 2 } = - ( x - y ) ^ { 5 - 5 + 2 } = - ( x - y ) ^ { 2 }$ $= - { x ^ { 2 } } + 2 x y - { y ^ { 2 } }$ ； …· (3分)$Q = [ ( 2 x - y ) ^ { 2 } + ( 2 x + y ) ( 2 x - y ) +$ $8 x y ] \\div 2 x = ( 4 x ^ { 2 } - 4 x y + y ^ { 2 } + 4 x ^ { 2 } -$ $y ^ { 2 } + 8 x y ) \\div 2 x$ $= ( 8 x ^ { 2 } + 4 x y ) \\div 2 x = 4 x + 2 y .$ （20…· (8分)(2)由题意，得 $x = ( 3 - \\pi ) ^ { 0 } = 1 , \\ ;$ $y =$ $( - { \\frac { 1 } { 2 } } ) ^ { - 1 } { = } { - 2 } , { \\frac { \\phantom { - } } { \\cdots \\cdots \\cdots \\cdots \\cdots } }$ （10分）所以 $P { = } { - } ( 1 { + } 2 ) ^ { 2 } { = } { - } 9 , Q { = } 4 \\times 1$ $+ 2 \\times ( - 2 ) = 0 .$ ·（12分）", + "page_idx": 51 + }, + { + "type": "text", + "text": "·能力提优参考答案· ", + "page_idx": 51 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 51 + }, + { + "type": "table", + "img_path": "images/b9ba00df8f5850ee661028fe1ad9dab48a1c1d526d24d5a976cd7c0cdf0edd40.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
12345678
CBAABDCC
", + "page_idx": 51 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 51 + }, + { + "type": "table", + "img_path": "images/581b4c08e47fad205623ab4a0c7f07c59d9776299b9d57d16982f1030edaa07a.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
9101112
9-2ab，（a-b）²(a+6）²= (a-b)²+4ab2
", + "page_idx": 51 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 51 + }, + { + "type": "text", + "text": "13.(1)解：原式 $= a ^ { 6 } - 9 a ^ { 6 } - 6 4 a ^ { 6 }$ （2分） $= - 7 2 a ^ { 6 }$ （3分） (2)解：原式 $= 5 - 1 - 3$ ……· (2分) ", + "page_idx": 51 + }, + { + "type": "text", + "text": "$= 1$ · (3分) ", + "page_idx": 52 + }, + { + "type": "text", + "text": "14.解： $( x - 3 ) ^ { 2 } + ( x + 3 ) ( x - 3 ) + 2 x ( 2$ $- x ) = x ^ { 2 } - 6 x + 9 + x ^ { 2 } - 9 + 4 x -$ $2 x ^ { 2 } = - 2 x$ … (6分) 当 $x { = } - \\frac { 1 } { 2 }$ 时，原式 $= - 2 \\times ( - \\frac { 1 } { 2 } ) =$ …· (8分) ", + "page_idx": 52 + }, + { + "type": "text", + "text": "15.解：一； …· (4分) 正确的解答过程： $\\alpha ( 1 + a ) - ( a - 1 ) ^ { 2 }$ $= a + a ^ { 2 } - ( a ^ { 2 } - 2 a + 1 ) = a + a ^ { 2 } - a ^ { 2 }$ $+ 2 a - 1 = 3 a - 1$ ……(8分) ", + "page_idx": 52 + }, + { + "type": "text", + "text": "16.解：根据题意得, $( 3 x ^ { 2 } - 5 x y ) \\div \\frac { x } { 2 } =$ $6 x - 1 0 y$ ，即第一个多项式是 $6 x -$ $1 0 y$ …· (5分)则正确的结果应为 $( 6 x - 1 0 y$ $\\frac { x + y } { 2 } = 3 x ^ { 2 } + 3 x y - 5 x y - 5 y ^ { 2 } = 3 x ^ { 2 }$ $- 2 x y - 5 y ^ { 2 }$ …· (8分)", + "page_idx": 52 + }, + { + "type": "text", + "text": "17.解： $( x + a ) ( x + b ) = x ^ { 2 } + ( a + b ) x +$ $a b = x ^ { 2 } + m x + 2 8$ ，对比可得 $a + b =$ $m , a b { = } 2 8$ ： … (6分)当 $a = 1 , b = 2 8$ 时， $m { = } 2 9$ ；当 $a = 2 , b$ $= 1 4$ 时， $m { = } 1 6$ ；当 $a = 4 , b = 7$ 时， $m$ $= 1 1$ …(10分)", + "page_idx": 52 + }, + { + "type": "text", + "text": "18.解：(1)原长方形铁皮的面积是 $( 4 a +$ 60) $( 3 a + 6 0 ) = 1 2 a ^ { 2 } + 4 2 0 a + 3 6 0 0$ $( \\mathrm { c m } ^ { 2 }$ ）； …（5分）(2)油漆这个铁盒的面积是 $3 a \\cdot 4 a$ $+ 2 \\times 3 0 \\times 4 a + 2 \\times 3 0 \\times 3 a = 1 2 a ^ { 2 } +$ $4 2 0 a ( \\mathrm { c m } ^ { 2 } )$ ", + "page_idx": 52 + }, + { + "type": "text", + "text": "$( 1 2 a ^ { 2 } + 4 2 0 a ) \\div { \\frac { a } { 5 0 } } = 6 0 0 a + 2 1 0 0 0$ （元)．… …（10分）答：油漆这个铁盒需要（ $6 0 0 a \\ +$ 21000）元. …（12分）", + "page_idx": 52 + }, + { + "type": "text", + "text": "第二章相交线与平行线", + "text_level": 1, + "page_idx": 52 + }, + { + "type": "text", + "text": "卷中悟法", + "text_level": 1, + "page_idx": 52 + }, + { + "type": "text", + "text": "一、同位角、内错角与同旁内角的识别", + "text_level": 1, + "page_idx": 52 + }, + { + "type": "text", + "text": "1.定义法", + "page_idx": 52 + }, + { + "type": "text", + "text": "根据定义，两个角共涉及三条直线（或射线或线段），两角的一边分别在两条直线上，而另一边在同一直线上，两角有“共线边”是定义的实质，抓住“一边共线”便不难识别，如图1中的 $\\angle 1$ 和 $\\angle 2$ 涉及 $E F , M G , N D$ 三条直线，且它们都有一边在直线 $E F$ 上，故 $\\angle 1$ 与 $\\angle 2$ 是同位角.又如图2中的 $\\angle 1$ 和 $\\angle 2$ 是否为同位角？因其涉及 $A D , A C , A B , B C$ 四条直线，无共线边，故 $\\angle 1$ 和 $\\angle 2$ 不是同位角.", + "page_idx": 52 + }, + { + "type": "text", + "text": "角（图 $\\textcircled{2}$ 中的 $\\angle 1$ 和 $\\angle 2 )$ ", + "page_idx": 52 + }, + { + "type": "text", + "text": "(3)将两个角的两边延长，如果能构成“U\"型（或反置或倒置)，那么这两个角为同旁内角（图 $\\textcircled{3}$ 中的 $\\angle 1$ 和 $\\angle 2 )$ ", + "page_idx": 52 + }, + { + "type": "text", + "text": "要注意这三种基本结构的变式.这种方法的好处是有助于记忆和理解，在判断两个角的关系时也不会出错.对顶角是有公共顶点的两个角，而同位角、内错角和同旁内角是没有公共顶点的两个角，学习时要注意它们的区别.", + "page_idx": 52 + }, + { + "type": "image", + "img_path": "images/baeaf0d81820417d7932e234b8fd65462016b8eb6ede2456b9b502253a98506e.jpg", + "img_caption": [ + "图1" + ], + "img_footnote": [], + "page_idx": 52 + }, + { + "type": "text", + "text": "二、平行线的判定与性质的应用", + "text_level": 1, + "page_idx": 52 + }, + { + "type": "image", + "img_path": "images/21512e0c928ef40af9018a1142ae82f4674d679b657e79b0221eaf60826a1ddd.jpg", + "img_caption": [ + "图3 " + ], + "img_footnote": [], + "page_idx": 52 + }, + { + "type": "image", + "img_path": "images/5ed7c91d88194399cf1cade422e98e51a285d0accef222fa80432215e4a49b02.jpg", + "img_caption": [ + "图2" + ], + "img_footnote": [], + "page_idx": 52 + }, + { + "type": "text", + "text": "2.描粗相关线条法", + "page_idx": 52 + }, + { + "type": "text", + "text": "(1)将两个角的两边延长，如果能构 \n成“ $F ^ { \\prime }$ 型(或倒置或反置)，那么这两个角 \n为同位角（图 $\\textcircled{1}$ 中的 $\\angle 1$ 和 $\\angle 2 )$ (2)将两个角的两边延长，如果能构 \n成“Z\"型(或反置)，那么这两个角为内错", + "page_idx": 52 + }, + { + "type": "text", + "text": "1.判定两直线平行的方法有五种：(1)平行线的定义；(2)平行公理的推论：如果两条直线都与第三条直线平行，那么这两条直线平行；(3)同位角相等，两直线平行；(4)内错角相等，两直线平行；(5)同旁内角互补，两直线平行.判定两直线平行时，定义一般不常用，其他四种方法要灵活运用，证明时要注意书写格式.", + "page_idx": 52 + }, + { + "type": "text", + "text": "2.由两条直线平行得到同位角相等、内错角相等和同旁内角互补，解题时应结合图形先确认所成的角是否为两平行线被第三条直线所截得的同位角或内错角或同旁内角，同时要学会简单的几何说理，做到每一步有理有据.", + "page_idx": 52 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 52 + }, + { + "type": "table", + "img_path": "images/56a4955ff233c96b8ce8f988a9bc1dc5bac5b990504262d98430e854f8019d3a.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
12345678
CDABCDBC
", + "page_idx": 52 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 52 + }, + { + "type": "table", + "img_path": "images/29b9f978d7fc4d33743ea6669c83116a60d14b55c8461cc01f67ffbeedfcf6c5.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
9101112
72°60°1270°
", + "page_idx": 52 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 52 + }, + { + "type": "text", + "text": "13.解： $( 1 ) \\angle 1$ 与 $\\angle 2$ 是内错角， $\\angle 2$ 与 $\\angle 3$ 是同旁内角；… · (3分) (2) $\\angle 2$ 与 $\\angle 5$ 是同旁内角； $\\angle 4$ 与 $\\angle 5$ 是同位角； $\\angle 5$ 与 $\\angle 6$ 是对顶角. …· (6分) ", + "page_idx": 52 + }, + { + "type": "text", + "text": "14.解：如图所示． …(6分)因为 $\\angle D A C = \\angle A C B$ ，根据内错角相等，两直线平行，所以 $A D / / C B$ ·· (8分)", + "page_idx": 52 + }, + { + "type": "image", + "img_path": "images/d9f040b426c55809068a26bd0e3c913c4cbf34f8cedc2a385cd215770096f8a8.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 52 + }, + { + "type": "text", + "text": "15.解：因为点 $O$ 为直线 $B D$ 上一点，所以 $\\angle C O D + \\angle B O C = 1 8 0 ^ { \\circ }$ 将 $\\angle C O D = 2 \\angle B O C$ 代人，得$2 \\angle B O C + \\angle B O C = 1 8 0 ^ { \\circ }$ ，解得$\\angle B O C = 6 0 ^ { \\circ }$ ，： ……· (5分)因为 $O C \\bot O A$ ，所以 $\\angle A O C = 9 0 ^ { \\circ }$ ，所以 $\\angle A O B = \\angle C O A - \\angle B O C = 9 0 ^ { \\circ } -$ $6 0 ^ { \\circ } = 3 0 ^ { \\circ }$ ： …(8分)", + "page_idx": 52 + }, + { + "type": "text", + "text": "16.解：（1)因为 $A D / / B C$ ，所以 $\\angle D +$ $\\angle C = 1 8 0 ^ { \\circ }$ 因为 $\\angle E A D = \\angle C$ ，所以 $\\angle E A D +$ $\\angle D = 1 8 0 ^ { \\circ }$ ，所以 $A E / / C D$ ；……· (4分)", + "page_idx": 53 + }, + { + "type": "text", + "text": "(2）因为 $A E / / C D$ ，所以 $\\angle A E B = { \\bf \\Phi }$ $\\angle C$ ，因为 $\\angle F E C = \\angle B A E$ ，所以 $\\angle B$ $= \\angle E F C = 5 0 ^ { \\circ }$ ： ·（8分)", + "page_idx": 53 + }, + { + "type": "text", + "text": "17.解：（1)如图，由题意，得 $\\angle F A B = \\ ;$ $4 5 ^ { \\circ }$ ${ } ^ { \\circ } , \\angle E B C = 8 0 ^ { \\circ }$ 因为 $A F / / B E$ ，所以 $\\angle F A B = { \\bf \\left[ \\begin{array} { l } { { \\begin{array} { r l r } \\end{array} } } \\end{array} \\right] }$ $\\angle A B E { = } 4 5 ^ { \\circ }$ ，因为 $\\angle E B C = 8 0 ^ { \\circ }$ ，所以 $\\angle A B C = { \\mathrm { : } }$ （20 $3 5 ^ { \\circ }$ ·· (5分)$( 2 ) D$ 处应在 $C$ 处的南偏西 $4 5 ^ { \\circ }$ 方向. (6分)理由：要使 $C D \\ / / \\ A B$ ，即就要使$\\angle A B C = \\angle B C D = 3 5 ^ { \\circ }$ 又因为 $C G / / B E$ ，所以 $\\angle G C B = $ $\\angle E B C = 8 0 ^ { \\circ }$ ，所以 $\\angle G C D = 4 5 ^ { \\circ }$ 即 $D$ 处在 $C$ 处的南偏西 $4 5 ^ { \\circ }$ 方向.·（10分）", + "page_idx": 53 + }, + { + "type": "image", + "img_path": "images/9e5ae854b5d526bd733ef31a45cc3376bb883a492c8dbd290df911910010891d.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 53 + }, + { + "type": "text", + "text": "18.解： $( 1 ) \\textcircled{ 1}$ 两直线平行，内错角相等60· （2分）$\\textcircled{2} 3 0$ （4分）$\\textcircled{3} 6 0$ （6分）", + "page_idx": 53 + }, + { + "type": "text", + "text": "（2）因为 $A B ~ / / ~ C D$ ，所以 $\\angle B +$ \n$\\angle B C E = 1 8 0 ^ { \\circ }$ \n又因为 $\\angle B = 4 0 ^ { \\circ }$ ，所以 $\\angle B C E = 1 8 0 ^ { \\circ }$ \n$- \\angle B = 1 8 0 ^ { \\circ } - 4 0 ^ { \\circ } = 1 4 0 ^ { \\circ } .$ （8分） \n又因为 $C N$ 是 $\\angle B C E$ 的平分线，所 \n以 $. \\angle B C N = 1 4 0 ^ { \\circ } \\div 2 = 7 0 ^ { \\circ } ,$ \n因为 $C N \\bot C M$ ，所以 $\\angle B C M = 9 0 ^ { \\circ } -$ \n$\\angle B C N = 9 0 ^ { \\circ } - 7 0 ^ { \\circ } = 2 0 ^ { \\circ }$ ：（12分）", + "page_idx": 53 + }, + { + "type": "text", + "text": "能力提优参考答案", + "text_level": 1, + "page_idx": 53 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 53 + }, + { + "type": "table", + "img_path": "images/c1c4afbe1bfeeaca89d915d370b785382ac6a8ffbc310a42805db0eef8d7c863.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
12345678
DDADACDA
", + "page_idx": 53 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 53 + }, + { + "type": "table", + "img_path": "images/ec4e19f3136b155bdca5d7b4a51a3b18c309c2efa38a63dec05bd74975923975.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
9101112
15°答案不唯一，如∠A =∠EBC或∠D= ∠DCF或∠A+ ∠ABC=180°或∠D +∠BCD=180°等30°(180- 2a)°
", + "page_idx": 53 + }, + { + "type": "text", + "text": "$9 . 1 5 ^ { \\circ }$ 【解析】由尺规作图的痕迹，得$\\angle B O C = \\angle \\alpha = 6 0 ^ { \\circ }$ 则 $\\angle A O C = 7 5 ^ { \\circ } -$ $6 0 ^ { \\circ } = 1 5 ^ { \\circ }$ ", + "page_idx": 53 + }, + { + "type": "text", + "text": "10.答案不唯一,如 $\\angle A = \\angle E B C$ 或 $\\angle D$ $= \\angle D C F$ 或 $\\angle A + \\angle A B C = 1 8 0 ^ { \\circ }$ 或$\\angle D + \\angle B C D = 1 8 0 ^ { \\circ }$ 等【解析】添加条件 $\\angle A = \\angle E B C$ 或 $\\angle D =$ $\\angle D C F$ 或 $\\angle A + \\angle A B C = 1 8 0 ^ { \\circ }$ 或$\\angle D + \\angle B C D = 1 8 0 ^ { \\circ }$ 均可以得出 $A D$ $/ / B F$ ", + "page_idx": 53 + }, + { + "type": "text", + "text": "", + "page_idx": 53 + }, + { + "type": "text", + "text": "11. $3 0 ^ { \\circ }$ 【解析】因为 $\\angle B A C = 1 0 0 ^ { \\circ }$ $\\angle D A E = 5 0 ^ { \\circ }$ ，所以 $\\angle C A E = 1 8 0 ^ { \\circ } -$ $1 0 0 ^ { \\circ } - 5 0 ^ { \\circ } = 3 0 ^ { \\circ }$ .因为 $A E / / B C$ ，所以$\\angle C = \\angle C A E = 3 0 ^ { \\circ } .$ ", + "page_idx": 53 + }, + { + "type": "text", + "text": "12.( $\\ : 1 8 0 - 2 \\alpha \\ : ) ^ { \\circ } \\ :$ 【解析】由 $C E$ 平分$\\angle A C D$ ， $\\angle E C A = \\alpha ^ { \\circ }$ ，得 $\\angle A C D =$ $( 2 \\alpha ) ^ { \\circ } , \\angle B C D = ( 1 8 0 - 2 \\alpha ) ^ { \\circ }$ .由 $F G$ $/ / C D$ ，可得 $\\angle G F B = \\angle B C D = ( 1 8 0 $ $- 2 \\alpha ) ^ { \\circ }$ ", + "page_idx": 53 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 53 + }, + { + "type": "text", + "text": "13.解：设这个角为 $x ^ { \\circ }$ ,则它的余角、补角分别为 $9 0 ^ { \\circ } - x ^ { \\circ }$ ${ 1 8 0 } ^ { \\circ } - { x } ^ { \\circ }$ ，（2分）根据题意列方程,得 $\\frac { 1 } { 2 } \\left( 1 8 0 - x \\right) -$ $( 9 0 - x ) { = } 2 0$ ，解得 $x { = } 4 0$ 即这个角的度数为 $4 0 ^ { \\circ }$ …(6分)", + "page_idx": 53 + }, + { + "type": "text", + "text": "14.解：因为 $\\angle 1 = \\angle 2$ ，所以 $A B / / C D$ …(4分)因为 $\\angle A = 5 5 ^ { \\circ } 1 6 ^ { \\prime }$ ，所以 $\\angle A D C =$ $1 8 0 ^ { \\circ } - 5 5 ^ { \\circ } 1 6 ^ { \\prime } = 1 2 4 ^ { \\circ } 4 4 ^ { \\prime }$ … (8分)", + "page_idx": 53 + }, + { + "type": "text", + "text": "15.解：因为 $A B / / C D , \\angle E D F = 7 0 ^ { \\circ }$ ，所以 $\\angle A B D = \\angle E D F = 7 0 ^ { \\circ }$ …(4分)因为 $B G$ 平分 $\\angle A B D$ ，所以 $\\angle D B G =$ ∠ABD=35°,所以∠FBG=180°$- \\angle D B G = 1 4 5 ^ { \\circ }$ · (8分)", + "page_idx": 53 + }, + { + "type": "text", + "text": "解： $C D$ 是 $\\angle A C B$ 的平分线.（2分） \n理由：因为 $C D \\perp A B , E F \\perp A B$ ，所以 \n$E F / / C D$ ， \n所以 $\\angle E = \\angle B C D$ ， $\\angle A C D =$ \n$\\angle E M C$ … · (5分) \n因为 $\\angle E = \\angle E M C$ ，所以 $\\angle B C D =$ \n$\\angle A C D$ ，即 $C D$ 是 $\\angle A C B$ 的平分线.…… (8分)", + "page_idx": 53 + }, + { + "type": "text", + "text": "17.解： $( 1 ) A F / / B E$ ·· (2分)理由：因为 $A D / / B C$ ，所以 $\\angle B =$ $\\angle D O E$ ··(3分)因为 $\\angle A = \\angle B$ ，所以 $\\angle A = \\angle D O E$ 所以 $A F / / B E$ ··· (5分)（2）因为 $A D / / B C$ ，所以 $\\angle B +$ $\\angle B O D { = } 1 8 0 ^ { \\circ }$ …· (7分)因为 $\\angle B O D = 3 \\angle B$ ，所以 $\\angle B +$ $3 \\angle B = 1 8 0 ^ { \\circ }$ ，可得 $\\angle B = 4 5 ^ { \\circ }$ ，故 $\\angle A$ $= \\angle B = 4 5 ^ { \\circ }$ ·…· (10分)", + "page_idx": 53 + }, + { + "type": "text", + "text": "18.解：（1）因为 $D E / / O B$ ，所以 $\\angle O =$ $\\angle A C E$ 因为 $\\angle A O B = 4 0 ^ { \\circ }$ ，所以 $\\angle A C E =$ （20 $4 0 ^ { \\circ }$ 。因为 $\\angle A C D + \\angle A C E = 1 8 0 ^ { \\circ }$ ，所以$\\angle A C D { = } 1 4 0 ^ { \\circ }$ 因为 $C F$ 平分 $\\angle A C D$ ，所以 $\\angle A C F { = }$ $7 0 ^ { \\circ }$ ，所以 $\\angle E C F = 7 0 ^ { \\circ } + 4 0 ^ { \\circ } = 1 1 0 ^ { \\circ }$ ·（4分）", + "page_idx": 53 + }, + { + "type": "text", + "text": "(2）因为 $C G \\bot C F$ ，所以 $\\angle F C G =$ $9 0 ^ { \\circ }$ ，所以 $\\angle D C G + \\angle D C F = 9 0 ^ { \\circ }$ ", + "page_idx": 53 + }, + { + "type": "text", + "text": "因为 $\\angle A C O = 1 8 0 ^ { \\circ }$ ，所以 $\\angle G C O + $ $\\angle F C A = 9 0 ^ { \\circ }$ \n因为 $\\angle A C F = \\angle D C F$ ，所以 $\\angle G C O$ $= \\angle G C D$ ，即 $C G$ 平分 $\\angle O C D$ \n…·（7分)(3)结论：当 $\\angle A O B = 6 0 ^ { \\circ }$ 时， $C D$ 平分 $\\angle O C F$ （8分）理由：因为 $D E / / O B , \\angle O = 6 0 ^ { \\circ }$ ，所以$\\angle D C O = \\angle O = 6 0 ^ { \\circ }$ ，所以 $\\angle A C D = $ （20 ${ 1 2 0 } ^ { \\circ }$ \n因为 $C F$ 平分 $\\angle A C D$ ，所以 $\\angle D C F =$ $6 0 ^ { \\circ }$ ，所以 $\\angle D C O = \\angle D C F$ ，即 $C D$ 平分 $\\angle O C F$ （12分）", + "page_idx": 54 + }, + { + "type": "text", + "text": "·月考测试卷（一）参考答案 以 $( a ^ { m } ) ^ { 2 } \\div ( a ^ { n } ) ^ { 3 } = a ^ { 2 m } \\div a ^ { 3 n } = a ^ { 2 m - 3 n } .$ ·(4分) 因为 $a ^ { m } = 4 , a ^ { n } = \\frac { 1 } { 2 }$ ，所以 $a ^ { 2 m - 3 n } = a ^ { 2 m }$ $\\div a ^ { 3 n } = ( a ^ { m } ) ^ { 2 } \\div ( a ^ { n } ) ^ { 3 } = 4 ^ { 2 } \\div ( \\frac { 1 } { 2 } ) ^ { 3 } =$ $1 6 \\times 8 { = } 1 2 8$ ", + "page_idx": 54 + }, + { + "type": "text", + "text": "", + "page_idx": 54 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 54 + }, + { + "type": "text", + "text": "18.解： $( 2 a + b ) ( 2 a - b ) - ( 4 a b ^ { 3 } - 8 a ^ { 3 } b )$ $\\div 2 a b - 4 a ^ { 2 } - b ^ { 2 } - ( 2 b ^ { 2 } - 4 a ^ { 2 } ) = 4 a ^ { 2 }$ $- b ^ { 2 } - 2 b ^ { 2 } + 4 a ^ { 2 } = 8 a ^ { 2 } - 3 b ^ { 2 }$ ·… (5分) 当 $a = - 1$ $b = - 2$ 时，原式 $= 8 \\times$ $( - 1 ) ^ { 2 } - 3 \\times ( - 2 ) ^ { 2 } = - 4 .$ … (9分) ", + "page_idx": 54 + }, + { + "type": "table", + "img_path": "images/fd5cd8d6aee33fcd46cd5c38e62e3c2d713abd73036e71697087c00c9e3cc5a6.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
12345678910
CBABBDCBCC
", + "page_idx": 54 + }, + { + "type": "text", + "text": "19.解： $\\displaystyle ( x + 7 ) ( x - 3 ) = x ^ { 2 } + 4 x - 2 1 ,$ 与$x ^ { 2 } + m x - n$ 对比，可得 $m = 4 , n = 2 1$ ··(6分)则 $( 5 m - n ) ^ { 2 0 2 4 } = ( 5 \\times 4 - 2 1 ) ^ { 2 0 2 4 } =$ $( - 1 ) ^ { 2 0 2 4 } = 1$ ： …· （9分)", + "page_idx": 54 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 54 + }, + { + "type": "table", + "img_path": "images/5f1a42f33c3780265d7527f843f3c5d55760a33125e7164070cd819b7e1e4c82.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
1112131415
166.x²y²+3x²y²-3xy²y+1 2.2²2-3+i1-x2025
", + "page_idx": 54 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 54 + }, + { + "type": "text", + "text": "20.解 $: x ( x - 2 ) ^ { 2 } - x ^ { 2 } ( x - 6 ) - 3 = x ( x ^ { 2 } -$ $4 x + 4 ) - x ^ { 3 } + 6 x ^ { 2 } - 3 = x ^ { 3 } - 4 x ^ { 2 } + 4 x$ $- x ^ { 3 } + 6 x ^ { 2 } - 3 = 2 x ^ { 2 } + 4 x - 3 .$ ·· (5分) 因为 $x ^ { 2 } + 2 x = 4$ ，所以 $2 x ^ { 2 } + 4 x =$ $2 ( x ^ { 2 } + 2 x ) = 2 \\times 4 = 8$ ，所以原式 $= 8$ $- 3 { = } 5$ ： … (9分) ", + "page_idx": 54 + }, + { + "type": "text", + "text": "16.(1)解：原式 $= a ^ { 4 } - a ^ { 6 } \\div a ^ { 2 }$ （204号（4分）$= a ^ { 4 } - a ^ { 4 } = 0$ （5分）(2）解：原式 $= 1 - 4 + 4 \\times 1$ ·(4分）$= 1 - 4 + 4 = 1$ （5分）", + "page_idx": 54 + }, + { + "type": "text", + "text": "17.解：因为 $a ^ { 2 m } = ( a ^ { m } ) ^ { 2 }$ ， $a ^ { 3 n } = ( a ^ { n } ) ^ { 3 }$ ，所", + "page_idx": 54 + }, + { + "type": "text", + "text": "21.解： $( x ^ { 2 } - 4 x + 1 ) - ( - 3 x ^ { 2 } ) = x ^ { 2 } - 4 x$ $+ 1 + 3 x ^ { 2 } = 4 x ^ { 2 } - 4 x + 1$ …(4分)$( 4 x ^ { 2 } - 4 x + 1 ) \\bullet ( - 3 x ^ { 2 } ) = - 1 2 x ^ { 4 } +$ $1 2 x ^ { 3 } - 3 x ^ { 2 }$ ，即正确的结果是一 $1 2 x ^ { 4 } + 1 2 x ^ { 3 } -$ （204号 $3 \\mathcal { X } ^ { 2 }$ …· (9分)", + "page_idx": 54 + }, + { + "type": "text", + "text": "22.解：设长方形的长为 $\\mathcal { X }$ ，宽为 $y .$ 根据题意,得 $2 x + 2 y = 1 6 , 2 x ^ { 2 } + 2 y ^ { 2 }$ $= 6 8$ ，即 $x + y = 8 , x ^ { 2 } + y ^ { 2 } = 3 4 ,$ … (2分）将 $x + y = 8$ 两边同时平方，得 $x ^ { 2 } +$ $2 x y + y ^ { 2 } = 6 4$ ···(4分)把 $x ^ { 2 } + y ^ { 2 } = 3 4$ 代人 $x ^ { 2 } + 2 x y + y ^ { 2 } =$ 64中，得 $3 4 + 2 x y = 6 4$ ，解得 $x y =$ 15,即长方形 $A B C D$ 的面积为15.（10分）", + "page_idx": 54 + }, + { + "type": "text", + "text": "23.解： $\\mathrm { ( 1 ) } \\mathrm { \\textlangle } { { \\mathbb { D } } } S _ { 1 } = a ( a + 4 ) = a ^ { 2 } + 4 a , S _ { 2 }$ $= ( a + 2 ) ^ { 2 } = a ^ { 2 } + 4 a + 4 .$ (2分)$\\textcircled{2}$ 因为 $S _ { 1 } - S _ { 2 } = ( a ^ { 2 } + 4 a ) - ( a ^ { 2 } + 4 a$ $+ 4 ) = a ^ { 2 } + 4 a - a ^ { 2 } - 4 a - 4 = - 4 <$ 0,所以 $S _ { 1 } { < } S _ { 2 }$ ·…(6分)(2)由 $M { = } N$ ，可得 $M - N { = } 0$ ，则 $a ^ { 2 }$ $- 4 + ( a + 1 ) ^ { 2 } = 0$ ,整理得 $2 a ^ { 2 } + 2 a -$ $3 { = } 0$ ，即 $2 a ^ { 2 } + 2 a = 3$ 则 $a ( a + 1 ) = a ^ { 2 } + a = { \\frac { 1 } { 2 } } ( 2 a ^ { 2 } + 2 a ) =$ ${ \\frac { 1 } { 2 } } \\times 3 = { \\frac { 3 } { 2 } }$ · (10分)", + "page_idx": 54 + }, + { + "type": "text", + "text": "第三章变量之间的关系", + "text_level": 1, + "page_idx": 54 + }, + { + "type": "text", + "text": "卷中悟法", + "text_level": 1, + "page_idx": 54 + }, + { + "type": "text", + "text": "一、自变量与因变量的确定", + "text_level": 1, + "page_idx": 54 + }, + { + "type": "text", + "text": "1.自变量是先发生变化的量，因变量是后发生变化的量.2.自变量是主动发生变化的量，因", + "page_idx": 54 + }, + { + "type": "text", + "text": "变量是随着自变量的变化而发生变化 \n的量.3.常量(不发生变化的量).4.在一个变化的关系式中，只有一 \n个自变量和一个因变量，且因变量需要 \n写在等号左边.", + "page_idx": 54 + }, + { + "type": "text", + "text": "二、速度(或路程)图象", + "text_level": 1, + "page_idx": 54 + }, + { + "type": "text", + "text": "1.弄清哪一条轴(通常是纵轴)表示速度(或路程)，哪一条轴(通常是横轴）表示时间；2.准确读懂不同走向的线所表示的意义：(1)上升的线：从左向右呈上升状的线，其代表速度增加（或匀速远离起点）；(2)水平的线：与水平轴(横轴)平行的线，其代表匀速行驶(或静止)；(3)下降的线：从左向右呈下降状的线，其代表速度减小（或反向运动返回起点).", + "page_idx": 54 + }, + { + "type": "text", + "text": "三、三种变量之间关系的表达方法与特点：", + "text_level": 1, + "page_idx": 54 + }, + { + "type": "table", + "img_path": "images/dc3396ebdc8af57253f6aae97c7858506232e377a8ed426219f79bc87aac889b.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
表达方法特点
表格法多个变量可以同时出现在同一 张表格中
关系式法准确地反映了因变量与自变量 的数值关系
图象法直观、形象地给出了因变量随自 变量的变化趋势
", + "page_idx": 54 + }, + { + "type": "text", + "text": "·基础过关参考答案· ", + "text_level": 1, + "page_idx": 55 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 55 + }, + { + "type": "table", + "img_path": "images/d6863459a808f930bf416522b99df10d8034d759133569f9eac5563c0e61b363.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
12345678
ABCCABBC
", + "page_idx": 55 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 55 + }, + { + "type": "table", + "img_path": "images/d4a666bc0b9e5276a2c8344e4a2234872e537e2898d946d3b5be723171d5f297.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
9101112
时间 温度y=10x+20y=8.5x①②④
", + "page_idx": 55 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 55 + }, + { + "type": "text", + "text": "13.解：(1)当 $f { = } 6 8$ 时， $c = \\frac { 5 } { 9 } ( f - 3 2 ) = 1$ 20； （3分）当 $f { = } { - } 4$ 时， $c { = } \\frac { 5 } { 9 } ( f { - } 3 2 ) { = } { - } 2 0$ （6分）(2)当 $c = 1 0$ 时， $\\frac { 5 } { 9 } ( f - 3 2 ) = 1 0$ ，解得 $f { = } 5 0$ …· (10分)", + "page_idx": 55 + }, + { + "type": "text", + "text": "14.解：(1)根据图象可以看出这一天的最高温度是 $3 7 \\ ^ { \\circ } C$ ,是在15 时到达的；最低温度是 $2 3 \\ ^ { \\circ } \\mathrm { C }$ ,是在3时到达的. …· (5分)(2）这一天的温差为 $3 7 - 2 3 = { \\mathrm { ? } }$ $1 4 ( ^ { \\circ } \\mathrm { C } )$ ，经过的时间为 $1 5 - 3 = { \\mathrm { ? } }$ 12(时)．·… ·（10分)", + "page_idx": 55 + }, + { + "type": "text", + "text": "15.解：(1)由题意可知，小正方形的边长是自变量，图中阴影部分的面积为因变量. …（2分）(2)由题意可得 $y = 1 2 ^ { 2 } - 4 x ^ { 2 } = 1 4 4 -$ $4 \\mathcal { x } ^ { 2 }$ … (6分)(3)由(2)可知 $y = 1 4 4 - 4 x ^ { 2 }$ ，", + "page_idx": 55 + }, + { + "type": "text", + "text": "当 $x = 1 ~ \\mathrm { c m }$ 时， $y$ 有最大值， $y _ { \\# } ^ { \\scriptscriptstyle \\perp } { + } =$ $1 4 4 - 4 \\times 1 ^ { 2 } = 1 4 0 ( \\mathrm { c m } ^ { 2 } ) ;$ \n当 $x = 5 ~ \\mathrm { c m }$ 时， $y$ 有最小值, $y _ { \\mathrm { \\scriptscriptstyle H Z } } , _ { \\mathrm { \\scriptscriptstyle I } } =$ $1 4 4 - 4 \\times 5 ^ { 2 } = 4 4 ( { \\mathrm { c m } } ^ { 2 } ) .$ \n所以当小正方形的边长由 $1 ~ \\mathrm { c m }$ 变化到 $5 ~ \\mathrm { c m }$ 时，图中阴影部分的面积由$1 4 0 ~ \\mathrm { c m } ^ { 2 }$ 变到 $4 4 ~ \\mathrm { c m } ^ { 2 }$ ．…… (13分)", + "page_idx": 55 + }, + { + "type": "text", + "text": "16.解：(1)机动车行驶5个小时后加油；··· (3分)(2)因为 $3 6 - 1 2 = 2 4 ( \\mathrm { L } )$ ，所以途中加油 $2 4 \\mathrm { ~ L ~ }$ （7分）(3)够用. … (8分)理由：因为每小时耗油 $( 4 2 - 1 2 ) \\div 5$ $= 6 ( \\mathrm { L } )$ ，所以加油后可行驶 $3 6 \\div 6 =$ $6 ( \\mathrm { { h } ) }$ 又因为 $4 0 \\times 6 = 2 4 0 ( \\mathrm { k m } ) > 2 3 0 \\mathrm { ~ k m }$ 所以油箱中的油够用．（15 分)", + "page_idx": 55 + }, + { + "type": "text", + "text": "·能力提优参考答案", + "text_level": 1, + "page_idx": 55 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 55 + }, + { + "type": "table", + "img_path": "images/ab567175f374f7eb91bfc8ee9bd36e759f5bcd2e9a2723f9516987924939cb31.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
123456
DDCDAA
", + "page_idx": 55 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 55 + }, + { + "type": "table", + "img_path": "images/dd9e1f83e35d5b8bb7181215c8316105eda8e6368881239a07e4b398f4295b7d.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
78910
y=1.2x+580y=2t+991350
", + "page_idx": 55 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 55 + }, + { + "type": "text", + "text": "11.解： $x = 1 ~ { \\mathrm { k m } }$ 时， $y = 3 5 \\times 1 + 2 =$ $3 7 \\ { ^ { \\circ } } \\mathrm { C } \\ ; x = 5 \\ \\mathrm { k m }$ 时， $y = 3 5 \\times 5 + 2 =$ $1 7 7 \\ { ^ { \\circ } C } : x = 1 0 \\ { \\mathrm { k m } }$ 时， $y = 3 5 \\times 1 0 + 2$ $= 3 5 2 \\mathrm { ~ } ^ { \\circ } \\mathrm { C }$ $x = 2 0 \\ \\mathrm { k m }$ 时， $y = 3 5 \\times 2 0$ $+ 2 { = } 7 0 2 \\ \\mathrm { ^ { \\circ } C } \\ ; x { = } 3 0 \\ \\mathrm { k m }$ 时， $y = 3 5 \\times$ $3 0 + 2 { = } 1 0 5 2 \\mathrm { ~ \\textdegree ~ }$ ： …(8分)", + "page_idx": 55 + }, + { + "type": "text", + "text": "", + "page_idx": 55 + }, + { + "type": "text", + "text": "12.解：(1)24; (3分)(2)4； （6分）(3)由图象得，每升高1千米，气温下降 $2 4 \\div 4 = 6 \\mathrm { ~ \\textdegree C }$ ，则 $t { = } 2 4 { - } 6 h$ （10分）", + "page_idx": 55 + }, + { + "type": "text", + "text": "13.解：(1)若购进 $A$ 型电脑 $\\mathcal { X }$ 台，则购进 $B$ 型电脑 $( 3 5 - x )$ 台.$y$ 与 $_ { \\mathcal { X } }$ 之间的关系式为 $y = 0 . 6 x ^ { + }$ $0 . 4 ( 3 5 - x )$ ，即 $y = 0 . 2 x + 1 4$ … (5分)(2)由题意得， $3 5 - x = 2 . 5 x$ ，解得 $x$ $= 1 0$ ，则 $\\scriptstyle y = 0 , 2 \\times 1 0 + 1 4 = 1 6$ 万元.答：该公司需要投入资金16万元.··（10分）", + "page_idx": 55 + }, + { + "type": "text", + "text": "14.解：(1)由题意得， $y _ { \\mathbb { H } } = 3 0 \\times 4 + 5 ( x$ $- 4 ) = 5 x + 1 0 0 ,$ $y _ { \\mathrm { Z } } = 3 0 \\times 4 \\times 0 . ~ 9 + 5 x \\times 0 . ~ 9 = 4 . ~ 5 x$ $+ 1 0 8$ …（5分）(2)当 $y _ { \\mathbb { H } } = y _ { Z }$ 时， $1 0 0 + 5 x { = } 4 . 5 x +$ 108,解得 $x { = } 1 6$ 故购买16 盒乒乓球时，在两家商店购买的付款金额一样多.（10分）", + "page_idx": 55 + }, + { + "type": "text", + "text": "15.解：(1）由图象可得，点 $A$ 表示充满电后行驶150千米时，剩余电量为35千瓦·时. …(2分）(2)在 $0 \\sim 1 5 0$ 千米范围内，汽车行驶1千米的平均耗电量是 $( 6 0 - 3 5 ) \\div$ ", + "page_idx": 55 + }, + { + "type": "text", + "text": "$1 5 0 { = } \\frac { 1 } { 6 }$ 千瓦·时.当汽车行驶了120千米时，蓄电池的剩余电量为 $6 0 - { \\frac { 1 } { 6 } } \\times 1 2 0 { = } 4 0$ 千瓦·时.…（7分）", + "page_idx": 55 + }, + { + "type": "text", + "text": "(3)在 $1 5 0 \\sim 2 0 0$ 千米范围内，汽车行驶1千米的平均耗电量是(35－10)$\\div ( 2 0 0 - 1 5 0 ) = \\frac { 1 } { 2 }$ 千瓦·时.设当汽车行驶了 $_ { \\mathcal { X } }$ 千米时,剩余电量降至20千瓦·时，根据题意得， $3 5 -$ $\\frac { 1 } { 2 } ( x - 1 5 0 ) = 2 0$ ，解得 $\\scriptstyle x = 1 8 0$ 答：当汽车行驶了180千米时，剩余电量降至20千瓦·时．（12分）", + "page_idx": 55 + }, + { + "type": "text", + "text": "期中测试卷 参考答案", + "text_level": 1, + "page_idx": 55 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 55 + }, + { + "type": "table", + "img_path": "images/c7c32ec783f7f6bcdca8e13375f5ca9cbe448b5cf260e52a8175486036b8dc24.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
12345678910
ACDDCBADBD
", + "page_idx": 55 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 55 + }, + { + "type": "table", + "img_path": "images/3d7efd8dea224d7ae9659e532e3e57fa5943bf650e8b3c405c131f5d784a64b8.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
11 6012131415
148°45°-192
", + "page_idx": 55 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 55 + }, + { + "type": "text", + "text": "16.(1)解：原式 $= 4 x ^ { 2 } - x + 8 x - 2 - 4 x ^ { 2 } +$ $2 x$ … （3分）$= 9 x - 2$ ：（5分）", + "page_idx": 55 + }, + { + "type": "text", + "text": "(2）解： $\\bullet \\bullet A B / / C D , \\angle A B F = 4 0 ^ { \\circ } , \\bullet \\bullet$ $\\angle C F B + \\angle B = 1 8 0 ^ { \\circ } , \\therefore \\angle C F B =$ $1 8 0 ^ { \\circ } - \\angle B = 1 4 0 ^ { \\circ } .$ 又： $\\angle C F E$ ： $\\angle E F B = 3 ~ : ~ 4 , ~ \\cdot ^ { . }$ $\\begin{array} { r l } & { \\angle C F E { = } \\frac { 3 } { 7 } \\angle C F B { = } 6 0 ^ { \\circ } . } \\\\ & { \\mathrel { \\mathop { : } } A B \\mathbin { / } C D , \\mathrel { \\mathop { : } } \\angle B E F { = } \\angle C F E { = } } \\end{array}$ $6 0 ^ { \\circ }$ ： …… (5分) ", + "page_idx": 55 + }, + { + "type": "text", + "text": "", + "page_idx": 56 + }, + { + "type": "text", + "text": "$2 . 1 + 3 ( x - 6 ) = 3 x - 5 . 4$ ，即 $y = 3 x$ $- 5 . 4$ ；… …· (6分)(2)当 $x = 8$ 立方米时， $y = 3 x - 5 . 4$ $= 3 \\times 8 - 5 . 4 { = } 1 8 . 6 $ 元.答：该用户5月份的水费是18.6元.·· (9分)", + "page_idx": 56 + }, + { + "type": "text", + "text": "17.解： $( 3 x + 2 ) ( 3 x - 2 ) + x ( x - 2 ) =$ $9 x ^ { 2 } - 4 + x ^ { 2 } - 2 x = 1 0 x ^ { 2 } - 2 x - 4 .$ （20·（4分)因为 $5 x ^ { 2 } - x - 2 = 0$ ，所以 $5 x ^ { 2 } - x = 2$ 当 $5 x ^ { 2 } - x = 2$ 时，原式 $= 2 \\left( 5 x ^ { 2 } - x \\right)$ $- 4 { = } 2 \\times 2 - 4 { = } 0$ …（9分)", + "page_idx": 56 + }, + { + "type": "text", + "text": "18.解：(1)由表格可得， $h { = } 4 { + } 1 . 2 ( x { - }$ $1 ) = 1 . 2 x + 2 . 8 $ …· (4分)(2)当 $h { = } 1 1 . 2 ~ \\mathrm { c m }$ 时，即 $1 . 2 x + 2 . 8 \\ :$ $= 1 1 . 2$ ，解得 $x = 7 .$ ··（7分）答：当这擦碗的高度为 $1 1 . ~ 2 ~ \\mathrm { c m }$ 时，碗的数量为7只. … (9分)", + "page_idx": 56 + }, + { + "type": "text", + "text": "21.解： $( 1 ) \\angle 2 = \\angle 3$ … (1分)理由： $\\angle E N C + \\angle C M G = 1 8 0 ^ { \\circ }$ $\\angle C M G = \\angle F M N$ $\\angle E N C +$ $\\angle F M N { = } 1 8 0 ^ { \\circ } , \\therefore F G / / E D ,$ ·… (4分)$\\cdot \\angle 2 = \\angle D , \\because { A B } / / { C D } , \\therefore \\angle 3 =$ $\\angle D , \\therefore \\angle 2 = \\angle 3$ ；（6分)$( 2 ) \\because A B / / C D , \\therefore \\angle A + \\angle A C D =$ $1 8 0 ^ { \\circ }$ …(7分)$\\cdot \\angle A = \\angle 1 + 7 0 ^ { \\circ } , \\angle A C B = 4 2 ^ { \\circ } , \\therefore$ $( \\angle 1 + 7 0 ^ { \\circ } ) + ( \\angle 1 + 4 2 ^ { \\circ } ) = 1 8 0 ^ { \\circ } , \\ :$ $\\angle 1 = 3 4 ^ { \\circ }$ ，$\\bullet A B / / C D , \\bullet \\bullet \\angle B = \\angle 1 = 3 4 ^ { \\circ } .$ ·· (9分)", + "page_idx": 56 + }, + { + "type": "text", + "text": "19.解： $A B / / C D$ … (2分)理由：因为 $\\angle D A E = \\angle E$ ，所以 $A D / /$ $B C$ ，所以 $\\angle B + \\angle B A D = 1 8 0 ^ { \\circ }$ ·· (5分)因为 $\\angle B = \\angle D$ ，所以 $\\angle D + \\angle B A D$ $= 1 8 0 ^ { \\circ }$ ，所以 $A B / / C D$ …· (9分)", + "page_idx": 56 + }, + { + "type": "text", + "text": "22.解：(1)由图象可得,汽车的耗油量为$( 6 0 - 4 5 ) \\div 1 5 0 = 0 . 1$ 升/千米,则 $y$ $= 6 0 - 0 . \\ : 1 x$ ·…·(4分)(2)当 $y = 8$ 时， $6 0 - 0 . 1 x { = } 8$ ，解得 $\\mathcal { X }$ $= 5 2 0$ ,即行驶520千米时，油箱中的剩余油量为8升.$5 0 0 + 3 0 - 5 2 0 = 1 0$ 千米，故在开往该加油站的途中，汽车开始提示加油，这时离加油站的路程是10千米.…·（10分）", + "page_idx": 56 + }, + { + "type": "text", + "text": "20.解：(1)由表格可得,3月份的用水量为5立方米，没有超过6立方米，则 $a$ $= 1 0 . 5 \\div 5 = 2 . 1$ 元/立方米.4月份的用水量为9立方米，超过了6立方米，则 $6 \\times 2 . 1 + ( 9 - 6 ) c =$ 21.6,解得 $c { = } 3$ 元/立方米.故每月用水量不超过6立方米时， $y$ $= 2 . 1 x$ 每月用水量超过6立方米时， $y = 6 \\times \\colon 2 3 .$ ", + "page_idx": 56 + }, + { + "type": "text", + "text": "解：(1)因为 $a ^ { 2 } + b ^ { 2 } = 8$ ， $( a + b ) ^ { 2 } =$ ", + "page_idx": 56 + }, + { + "type": "text", + "text": "48,所以ab=（a+6)²-(a²+6²） ${ \\frac { 4 8 - 8 } { 2 } } = 2 0$ 故填20．（2分) ", + "page_idx": 56 + }, + { + "type": "text", + "text": "别为 $a , b ,$ 设第三边为 $c$ ，那么有 $\\mid a - b \\mid$ $< c < a + b$ 3.可证明线段之间的不等关系.", + "page_idx": 56 + }, + { + "type": "text", + "text": "(2)设 $2 5 - x = a , x - 1 0 = b$ ，由 $( a +$ $b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }$ 进行变形得， $a ^ { 2 } + 1$ $b ^ { 2 } = ( a + b ) ^ { 2 } - 2 a b .$ \n所以 $( 2 5 - x ) ^ { 2 } + ( x - 1 0 ) ^ { 2 } = [ ( 2 5 -$ $x ) + ( x - 1 0 ) ] ^ { 2 } - 2 ( 2 5 - x ) ( x - 1 0 )$ $= 1 5 ^ { 2 } - 2 \\times ( - 1 5 ) = 2 2 5 + 3 0 = 2 5 5 .$ （20·…· (7分)(3)设 $A D { = } A C { = } a$ $B E { = } B C { = } b$ ，则图中阴影部分的面积为 ${ \\frac { 1 } { 2 } } \\left( a + b \\right) \\left( a \\right.$ $+ b ) - { \\frac { 1 } { 2 } } ( a ^ { 2 } + b ^ { 2 } ) = { \\frac { 1 } { 2 } } [ ( a + b ) ^ { 2 } - ( a ^ { 2 } ) ^ { 2 }$ $+ b ^ { 2 } ) ] = \\frac { 1 } { 2 } \\times 2 a b = a b = 1 0 .$ 故填10.：(10分）", + "page_idx": 56 + }, + { + "type": "text", + "text": "二、利用全等三角形的性质证明边、角相等", + "text_level": 1, + "page_idx": 56 + }, + { + "type": "text", + "text": "利用全等三角形的性质证明边、角相等的关键在于找出边、角所在的三角形，然后证明其全等，从而得出结论。一般步骤是：", + "page_idx": 56 + }, + { + "type": "text", + "text": "(1)找出所要证明的边或角所在的三角形；(2)根据已知条件或结论证明三角形全等；(3)利用全等三角形的性质得出所证结论.", + "page_idx": 56 + }, + { + "type": "text", + "text": "第四章 三角形", + "text_level": 1, + "page_idx": 56 + }, + { + "type": "text", + "text": "卷中悟法", + "text_level": 1, + "page_idx": 56 + }, + { + "type": "text", + "text": "一、三角形三边关系的应用", + "text_level": 1, + "page_idx": 56 + }, + { + "type": "text", + "text": "1.判断已知的三条线段 $a , b , c$ 能否构成一个三角形，判断的方法有三种：(1)当 $a + b > c , b + c > a , a + c > b$ 都成立时， $a , b , c$ 可构成三角形；(2)当 $| a - b | <$ $c { < a + b }$ 时， $a , b , c$ 可构成三角形；(3)当$a$ 最长，且 $b + c > a$ 时， $a , b , c$ 可构成三角形.", + "page_idx": 56 + }, + { + "type": "text", + "text": "若不能通过全等三角形直接证明，可通过线段(角)的和或差将所证结论转化为全等三角形中的线段(角).此外证明边相等也可通过线段的中点，角平分线的性质、定理等证明；角相等可通过角平分线的定义、对顶角相等等证明.", + "page_idx": 56 + }, + { + "type": "text", + "text": "2.已知三角形的两边，确定第三边的取值范围.如果已知三角形的两边分", + "page_idx": 56 + }, + { + "type": "text", + "text": "三、合理选择全等三角形的判定方法", + "text_level": 1, + "page_idx": 56 + }, + { + "type": "text", + "text": "从判定两个三角形全等的方法可知，要判定两个三角形全等，需要知道这两个三角形分别有三个元素（其中至少一个元素是边)对应相等，这样就可以利用题目中的已知边(角)迅速、准确地确定要补充的边（角)，有目的地完善三角形全等的条件，从而得到判定两个三角形全等的思路.", + "page_idx": 56 + }, + { + "type": "text", + "text": "找夹角→SAS \n已知两边 找第三边→SSS找直角→H边为角的对边-找任一角→AAS \n已知一边一角 找夹角的另一边→SAS找夹边的另一角→ASA找边的对角→AAS找夹边→ASA \n已知两角找其中一个已知角的对边→AAS基础过关参考答案", + "page_idx": 57 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 57 + }, + { + "type": "table", + "img_path": "images/13a5e5621cde84d77607add20c05b5b06ccfbc1cf65822c50ab1cf8d3c009911.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
12345678
CADDDCBD
", + "page_idx": 57 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 57 + }, + { + "type": "table", + "img_path": "images/259167fc0881a8ac0778dc341948de8cc4d7854a34b7feda4ee8996a2c60c0e5.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
9101112
sss66060°
", + "page_idx": 57 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 57 + }, + { + "type": "text", + "text": "13.解：锐角三角形有2个，分别是△ABE,△BAC;,. ·(3分)直角三角形有3个，分别是Rt△ABD,Rt△AED,Rt△ADC.·· (6分)", + "page_idx": 57 + }, + { + "type": "text", + "text": "14.解：如图所示. …… (8分) ", + "page_idx": 57 + }, + { + "type": "image", + "img_path": "images/4a743fd1ae54400b3178886b8be5eb39c333648347ad287bca3828468767942d.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 57 + }, + { + "type": "text", + "text": "15.解：(1)设 $A E$ 的长为 $\\mathcal { X }$ ，则 $A E$ 的取值范围是 $1 0 - 8 < x < 1 0 + 8$ ，即 $2 { < } x$ ${ < } 1 8$ …·(3分)", + "page_idx": 57 + }, + { + "type": "text", + "text": "（2）因为 $A E / / B D$ ，所以 $\\angle A =$ $\\angle C B D = 5 5 ^ { \\circ }$ 所以 $\\angle A E F = 1 8 0 ^ { \\circ } - \\angle A E C = \\angle A$ $+ \\angle C = 5 5 ^ { \\circ } + 6 0 ^ { \\circ } = 1 1 5 ^ { \\circ }$ … (8分)", + "page_idx": 57 + }, + { + "type": "text", + "text": "16.解： $( 1 ) \\triangle A E D { \\cong } \\triangle A C D .$ … (1分)理由：因为 $A D$ 是 $B C$ 边上的高，所以 $\\angle A D E { = } \\angle A D C { = } 9 0 ^ { \\circ }$ 因为 $\\angle E A D = \\angle C A D , A D = A D$ ，所以△AED≌△ACD(ASA).…(3分)", + "page_idx": 57 + }, + { + "type": "text", + "text": "(2)因为 $\\angle B = 4 2 ^ { \\circ }$ ，所以 $\\angle B A D = 9 0 ^ { \\circ }$ $- 4 2 ^ { \\circ } = 4 8 ^ { \\circ }$ ，因为 $\\angle E A D = \\angle C A D$ ， $\\angle B A E =$ $\\angle C A E$ ，所以 $\\angle B A E = \\angle C A E =$ $2 \\angle E A D$ ，所以 $\\angle C A D = \\frac { 1 } { 3 } \\angle B A D = \\frac { 1 } { 3 } \\times 4 8 ^ { \\circ } =$ $1 6 ^ { \\circ }$ ，所以 $\\angle C = 9 0 ^ { \\circ } - 1 6 ^ { \\circ } = 7 4 ^ { \\circ }$ （8分）", + "page_idx": 57 + }, + { + "type": "text", + "text": "17.解： $\\mathrm { ( 1 ) } \\angle B O C { = } 9 0 ^ { \\circ } { + } \\frac { 1 } { 2 } \\angle B A C .$ （2分）理由：因为 BE平分 $\\angle A B C$ ，所以$\\angle O B C { = } { \\frac { 1 } { 2 } } \\angle A B C .$ （204号同理可得 $\\angle O C D = \\frac { 1 } { 2 } \\angle A C B$ 所以 $\\angle O B C + \\angle O C D = \\frac { 1 } { 2 } ( \\angle A B C +$ $\\angle A C B )$ ， ： (4分)又因为 $\\angle B A C + \\angle A B C + \\angle A C B =$ $1 8 0 ^ { \\circ }$ ，所以 $\\angle A B C + \\angle A C B = 1 8 0 ^ { \\circ } -$ ", + "page_idx": 57 + }, + { + "type": "text", + "text": "$\\angle B A C$ ， \n所以 $\\angle O B C + \\angle O C D = { \\frac { 1 } { 2 } } \\left( 1 8 0 ^ { \\circ } - \\right.$ $\\angle B A C )$ ，… …· (6分)又因为 $\\angle B O C = 1 8 0 ^ { \\circ } - ( \\angle O B C +$ $\\angle O C D )$ ， \n所以 $\\angle B O C = 1 8 0 ^ { \\circ } - { \\frac { 1 } { 2 } }$ (180°$\\angle B A C = 9 0 ^ { \\circ } + { \\frac { 1 } { 2 } } \\angle B A C .$ \n： (8分)", + "page_idx": 57 + }, + { + "type": "text", + "text": "(2)同理可得 $\\angle C O A = 9 0 ^ { \\circ } + { \\frac { 1 } { 2 } } \\angle A B C$ $\\angle B O A = 9 0 ^ { \\circ } + \\frac { 1 } { 2 } \\angle A C B$ … (10分）", + "page_idx": 57 + }, + { + "type": "text", + "text": "18.解： $\\triangle C D E { \\cong } \\triangle A B F$ …(1分)理由：因为 $D E \\bot A C , B F \\bot A C .$ 所以$\\angle D E C = \\angle B F A = 9 0 ^ { \\circ }$ ，因为 $D C / /$ $A B$ ，所以 $\\angle A C D = \\angle B A C$ ，因为 $A E$ ${ = } C F$ ，所以 $A F { = } C E$ ，所以 $\\triangle C D E { \\cong }$ $\\triangle A B F$ · (4分)$\\mathrm { ( 2 ) } A D { = } B C , A D / / B C .$ …(5分)理由：因为 $\\triangle C D E \\cong \\triangle A B F$ ，所以$D E { = } B F$ ，因为 $\\angle A E D = \\angle C F B = 9 0 ^ { \\circ }$ $A E =$ $C F$ ，所以 $\\triangle A D E { \\cong } \\triangle C B F$ 所以 $A D { = } B C$ $\\angle D A E = \\angle B C F$ ，所以 $A D / / B C .$ · (8分)$( 3 ) D F { = } B E$ … (9分)理由：如图，在 $\\triangle A D F$ 和 $\\triangle C B E$ 中，$A D = B C$ ， $\\angle D A F = \\angle B C E$ $A F =$ ", + "page_idx": 57 + }, + { + "type": "text", + "text": "$C E$ ，所以 $\\triangle A D F { \\cong } \\triangle C B E$ ，所以DF$= B E .$ … ：(12分）", + "page_idx": 57 + }, + { + "type": "text", + "text": "D CB·能力提优参考答案·", + "page_idx": 57 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 57 + }, + { + "type": "table", + "img_path": "images/061a4142c175a0bd54d844e94cbb0e40c04a0a260a80e2474087456b8305b597.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
12345678
CBADCABD
", + "page_idx": 57 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 57 + }, + { + "type": "table", + "img_path": "images/4566febab7a843ece8c76533d5d500cc94bf9f824a63eada9a0581ac69badab3.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
9 3AD⊥BC101112
90△EHD和△EGF
", + "page_idx": 57 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 57 + }, + { + "type": "text", + "text": "13.解：因为 $C F / / A B$ ，所以 $\\angle B =$ $\\angle F C D , \\angle B E D = \\angle F .$ 因为点 $D$ 是 $B C$ 的中点，所以 $B D =$ CD.·在 $\\triangle B D E$ 与 $\\triangle C D F$ 中，$\\angle B E D = \\angle F , \\angle B = \\angle F C D , B D =$ $C D$ ，所以△BDE≌△CDF(AAS).·", + "page_idx": 57 + }, + { + "type": "text", + "text": "14.解：具体作图如下：", + "page_idx": 57 + }, + { + "type": "image", + "img_path": "images/593019b817191593a62636f39b6c0a802245709246b48a62ba28b1afe627538d.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 57 + }, + { + "type": "text", + "text": "则△ABC是所求作的三角形．…· (8分)", + "page_idx": 57 + }, + { + "type": "text", + "text": "15.解：只要测量出 $E M$ 的长度即为凉亭", + "page_idx": 57 + }, + { + "type": "text", + "text": "$M$ 与 $F$ 之间的距离. · (3分)理由：因为 $A B / / C D$ ，所以 $\\angle B = \\mathrm { : }$ $\\angle C .$ \n因为点 $M$ 是 $B C$ 的中点，所以 $B M =$ CM. ·· · (6分)在△BEM和 $\\triangle C F M$ 中， \n$\\angle B = \\angle C$ ， $B M = C M$ ， $\\angle B M E = { \\mathrm { : } }$ $\\angle C M F$ ， \n所以 $\\triangle B E M { \\cong } \\triangle C F M ( \\mathrm { A S A } )$ ，所以$M F { = } M E$ … (10分)", + "page_idx": 58 + }, + { + "type": "text", + "text": "16.解： $( 1 ) F C { = } A D .$ · (1分)理由：因为 $A D / / B C$ ，所以 $\\angle A D C = { \\mathrm { ? } }$ $\\angle E C F .$ 因为点 $E$ 是 $C D$ 的中点，所以 $D E = \\ \\mathrm { i }$ EC.在 $\\triangle A D E$ 和 $\\triangle F C E$ 中，$\\angle A D C = \\angle E C F , D E = E C , \\angle A E D$ $= \\angle C E F$ 所以 $\\triangle A D E { \\cong } \\triangle F C E$ （ASA)，所以$F C { = } A D$ ;· …· (4分)", + "page_idx": 58 + }, + { + "type": "text", + "text": "(2)由(1)可知 $\\triangle A D E { \\cong } \\triangle F C E$ ，所以 $A E { = } E F , A D { = } C F .$ \n因为 $A B { = } B C { + } A D { = } B C { + } C F$ ，所以$A B { = } B F$ (7分)在△ABE和△FBE中， \n$\\scriptstyle A B = B F , A E = E F , B E = B E ,$ \n所以 $\\triangle A B E { \\cong } \\triangle F B E ( \\mathrm { S S S } )$ \n所以 $\\angle A E B = \\angle F E B = 9 0 ^ { \\circ }$ ，所以 $B E$ ⊥AF. （10分）", + "page_idx": 58 + }, + { + "type": "text", + "text": "17.解：(1)因为 $\\angle B E C = \\angle C F A = \\alpha =$ $9 0 ^ { \\circ }$ ，所以 $\\angle B C E + \\angle C B E = 1 8 0 ^ { \\circ } -$ （20 $9 0 ^ { \\circ } = 9 0 ^ { \\circ } .$ 因为 $\\angle B C A = \\angle B C E + \\angle A C F =$ $9 0 ^ { \\circ }$ ，所以 $\\angle C B E { = } \\angle A C F .$ 在 $\\triangle B C E$ 和 $\\triangle C A F$ 中，$\\angle B E C = \\angle C F A , \\angle C B E = \\angle A C F ,$ $B C { = } A C$ 所以 $\\triangle B C E { \\cong } \\triangle C A F$ （AAS)，所以$B E { = } C F$ …·（4分)$( 2 ) \\alpha + \\angle B C A = 1 8 0 ^ { \\circ } ($ 或 $\\alpha$ 与 $\\angle B C A$ 互补)．…… …· (5分)理由：因为 $\\angle B E C = \\angle C F A = \\alpha$ ，所以$\\angle B E F = 1 8 0 ^ { \\circ } - \\alpha .$ （20因为 $\\alpha + \\angle B C A = 1 8 0 ^ { \\circ }$ ，所以 $\\angle B C A$ $= 1 8 0 ^ { \\circ } - \\alpha .$ 因为 $\\angle E B C + \\angle B C E = 1 8 0 ^ { \\circ } - \\alpha$ ，$\\angle B C A = \\angle B C E + \\angle A C F = 1 8 0 ^ { \\circ } -$ （20 $\\alpha$ ，所以 $\\angle E B C = \\angle A C F$ … (6分)在△BCE和△ $C A F$ 中，$\\angle C B E = \\angle A C F , \\angle B E C = \\angle C F A ,$ $B C { = } C A$ ，所以 $\\triangle B C E { \\cong } \\triangle C A F$ (AAS)，所以$B E { = } C F .$ ： … (（7分)$( 3 ) E F { = } B E { + } A F .$ …（8分)理由：因为 $\\angle B C A = \\alpha$ ，所以 $\\angle B C E +$ $\\angle A C F = 1 8 0 ^ { \\circ } - \\alpha .$ 因为 $\\angle B E C = \\alpha$ ，所以 $\\angle E B C +$ （20$\\angle B C E = 1 8 0 ^ { \\circ } - \\alpha$ ，所以 $\\angle E B C =$ ", + "page_idx": 58 + }, + { + "type": "text", + "text": "$\\angle A C F$ …· (10分)在 $\\triangle B E C$ 和△CFA中， \n$\\angle E B C = \\angle F C A , \\angle B E C = \\angle C F A ,$ $B C { = } C A$ ， \n所以 $\\triangle B E C { \\cong } \\triangle C F A ( \\mathrm { A A S } )$ ，所以$B E { = } C F , E C { = } F A ,$ \n所以 $E F { = } E C { + } C F { = } B E { + } A F .$ （2 \n·（12分）", + "page_idx": 58 + }, + { + "type": "text", + "text": "第五章生活中的轴对称", + "text_level": 1, + "page_idx": 58 + }, + { + "type": "text", + "text": "卷中悟法", + "text_level": 1, + "page_idx": 58 + }, + { + "type": "text", + "text": "一、轴对称图形及其对称轴的识别方法", + "text_level": 1, + "page_idx": 58 + }, + { + "type": "text", + "text": "1.判断一个图形是不是轴对称图形，可以用折纸的方法按照轴对称图形的概念，看是否能找到一条直线，将图形沿其折叠，使直线两旁的部分能够完全重合（即用轴对称图形的定义进行识别).", + "page_idx": 58 + }, + { + "type": "text", + "text": "2.识别轴对称图形的关键是要找到作为对称轴的直线，沿直线折叠后两边的部分能够完全重合，有时这样的直线能找到多条，说明这个轴对称图形有多条对称轴。", + "page_idx": 58 + }, + { + "type": "text", + "text": "二、线段垂直平分线与角平分线性质的综合应用", + "text_level": 1, + "page_idx": 58 + }, + { + "type": "text", + "text": "1.线段垂直平分线的性质一般与全等三角形、角平分线的知识综合命题，线段垂直平分线上的点到线段两个端点的距离相等，而角平分线上的点到角两边的距离相等，两者是不同的.", + "page_idx": 58 + }, + { + "type": "text", + "text": "", + "page_idx": 58 + }, + { + "type": "text", + "text": "2.解题时，主要用线段垂直平分线上的点到线段两个端点的距离相等及角平分线上的点到角两边的距离相等来得出线段相等或三角形全等的结论，再利用全等三角形的性质解决问题.", + "page_idx": 58 + }, + { + "type": "text", + "text": "基础过关参考答案", + "text_level": 1, + "page_idx": 58 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 58 + }, + { + "type": "table", + "img_path": "images/730006d834656ffb3a51f28f2c41b95dc9a6ff465503c4b20d78363f1b01f5dd.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
12345678
BCCABAAD
", + "page_idx": 58 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 58 + }, + { + "type": "table", + "img_path": "images/52ea89e617cb69003cb9cea4db997c573b7c096ad915296e1e62510d7c4c4d44.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
9101112
无数60°34
", + "page_idx": 58 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 58 + }, + { + "type": "text", + "text": "13.解：如图所示. （6分）", + "page_idx": 58 + }, + { + "type": "image", + "img_path": "images/d3f2ed627f97afc5a1d991a5e3cf1235d5897c85f9c4ae59ed94fd12c2654a52.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 58 + }, + { + "type": "text", + "text": "14.解：(1)全等. ：(2分)(2)因为 $\\triangle A B C$ 和 $\\triangle A ^ { \\prime } B ^ { \\prime } C ^ { \\prime }$ 关于直线 $l$ 对称，所以 $A ^ { \\prime } B ^ { \\prime } { = } A B { = } 4 ~ ($ m，$B ^ { \\prime } C ^ { \\prime } { = } B C { = } 3 \\ \\mathrm { c m } , A ^ { \\prime } C$ 边上的高 $=$ $2 \\mathrm { \\ c m }$ ，所以 $\\triangle A ^ { \\prime } B ^ { \\prime } C ^ { \\prime }$ 的周长 $= 4 + 6 +$ $3 { = } 1 3 ( \\mathrm { c m } )$ ： (6分)面积= $[ = \\frac { 1 } { 2 } \\times 6 \\times 2 = 6 ( c m ^ { 2 } )$ …· (8分)", + "page_idx": 58 + }, + { + "type": "text", + "text": "15.解：如图所示，(1)作出线段 $A B$ 的垂直平分线； （3分）(2)作出角的平分线. （6分）它们的交点即为所求作的点 $C$ （有2个). …· (10分)", + "page_idx": 59 + }, + { + "type": "image", + "img_path": "images/400dda2c4d0c7a5a728045e745ef50ed76b56ae545825aa14ef21ccb163fe340.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 59 + }, + { + "type": "text", + "text": "16.解： $\\textcircled{1}$ 当顶角 $\\angle A { < } 9 0 ^ { \\circ }$ 时，如图 $\\textcircled{1}$ 所示，因为 $B D \\perp A C$ ，所以 $\\angle A +$ $\\angle A B D { = } 9 0 ^ { \\circ }$ ，因为 $\\angle A B D = 5 6 ^ { \\circ }$ ，所以 $\\angle A = 9 0 ^ { \\circ } -$ $5 6 ^ { \\circ } = 3 4 ^ { \\circ }$ ，因为 $A B { = } A C$ ，所以 $\\angle A B C = \\angle C =$ ${ \\frac { 1 } { 2 } } \\times ( 1 8 0 ^ { \\circ } - 3 4 ^ { \\circ } ) = 7 3 ^ { \\circ } .$ $\\textcircled{2}$ 当顶角 $\\angle A { > } 9 0 ^ { \\circ }$ 时，如图 $\\textcircled{2}$ 所示，因为 $B D \\perp A C$ ，所以 $\\angle D A B +$ $\\angle A B D { = } 9 0 ^ { \\circ }$ 因为 $\\angle A B D = 5 6 ^ { \\circ }$ ，所以 $\\angle D A B = 9 0 ^ { \\circ }$ $- 5 6 ^ { \\circ } = 3 4 ^ { \\circ }$ ，所以 $\\angle B A C = 1 4 6 ^ { \\circ }$ 因为 $A B { = } A C$ ，所以 $\\angle A B C = \\angle C =$ $\\frac { 1 } { 2 } \\times ( 1 8 0 ^ { \\circ } - 1 4 6 ^ { \\circ } ) = 1 7 ^ { \\circ } .$ 综上所述，这个等腰三角形的底角的度数为 $7 3 ^ { \\circ }$ 或 $1 7 ^ { \\circ }$ ·……（10分)", + "page_idx": 59 + }, + { + "type": "image", + "img_path": "images/4ad7e7c76a6f71cef5ef9269dbea16437bcc96c8dc599a116705c12e22269a34.jpg", + "img_caption": [ + "图 $\\textcircled{1}$ " + ], + "img_footnote": [], + "page_idx": 59 + }, + { + "type": "image", + "img_path": "images/6f1edcb1a8d5fbcc6f195c377cfa598efd9db4ae0da1742d5030821c0c546d64.jpg", + "img_caption": [ + "图 $\\textcircled{2}$ " + ], + "img_footnote": [], + "page_idx": 59 + }, + { + "type": "text", + "text": "，解：问题解决：如图，作 $C F \\bot A D$ 的延长线于点 $F$ ，所以 $\\angle F { = } 9 0 ^ { \\circ }$ \n因为 $C E \\bot A B$ ，所以 $\\angle F { = } \\angle C E A { = }$ $\\angle C E B = 9 0 ^ { \\circ }$ \n因为 $\\angle A D C + \\angle C D F = 1 8 0 ^ { \\circ }$ ，且$\\angle A B C + \\angle A D C = 1 8 0 ^ { \\circ }$ ，所以 $\\angle C D F$ $= \\angle B$ \n在 $\\triangle C D F$ 和 $\\triangle C B E$ 中， $\\angle F =$ $\\angle C E B , \\angle C D F = \\angle B , C D = C B$ ，所以 $\\triangle C D F { \\cong } \\triangle C B E ( \\mathrm { A A S } )$ ，所以 $C F$ ${ } = { C E } .$ \n因为 $C F \\bot A D , C E \\bot A B$ ，所以 $A C$ 平分 $\\angle B A D$ …（4分）合作探究：（1）在 $\\mathrm { R t } \\bigtriangleup C A F$ 和$\\mathrm { R t } \\triangle C A E$ 中， $C F = C E , A C = A C$ ，所以 $\\triangle C A F { \\cong } \\triangle C A E .$ \n所以 $A F { = } A E$ \n因为 $\\triangle C D F { \\cong } \\triangle C B E$ ，所以 $D F =$ BE. \n因为 $3 B E { = } 9$ ，所以 $B E { = } 3$ ，所以 $D F$ $= 3$ \n因为 $A D { = } A F { - } D F$ ，所以 $A D { = } A E$ $- D F { = } 9 - 3 { = } 6 .$ …（9分)", + "page_idx": 59 + }, + { + "type": "text", + "text": "(2)因为 $\\triangle C A F { \\cong } \\triangle C A E , \\triangle C D F { \\cong }$ $\\triangle C B E$ ，所以 $S _ { \\triangle C A F } { = } S _ { \\triangle C A E }$ ， $S _ { \\triangle C D F } { = } S _ { \\triangle C B E }$ ，", + "page_idx": 59 + }, + { + "type": "text", + "text": "设△BCE的面积为 $\\mathcal { X }$ ，则 $\\triangle C D F$ 的 \n面积为 $\\mathcal { X }$ ， \n由题意，得 $2 8 + x { = } 4 0 - x$ ，解得 $x { = } 6 .$ \n所以 $\\triangle B C E$ 的面积为6.（14分）", + "page_idx": 59 + }, + { + "type": "image", + "img_path": "images/60df6d2ca892777369bdf09433249fb81e47d10bf83ee1d8f5d7abf4c77d6685.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 59 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 59 + }, + { + "type": "table", + "img_path": "images/dcf6c1e1d00926efd60caf8f4335f5d70d3fe0b0865886d253ec46824d6d3dee.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
12345678
DABDABDC
", + "page_idx": 59 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 59 + }, + { + "type": "table", + "img_path": "images/dad611dcbb28d3b862dc878be521e25a7884a6186cdf31fc76b03f211deea420.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
9101112
4或66425°
", + "page_idx": 59 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 59 + }, + { + "type": "text", + "text": "13.解： $\\angle B D P = \\angle E P C .$ …… (1分) 理由： $\\triangle A B C$ 为等边三角形， $\\angle D P E = 6 0 ^ { \\circ } , \\therefore \\angle D P E = \\angle B =$ $6 0 ^ { \\circ }$ · (3分) $\\cdot \\angle D P E + \\angle E P C = 1 8 0 ^ { \\circ } - \\angle B P D ,$ （20 $\\angle B + \\angle B D P = 1 8 0 ^ { \\circ } - \\angle B P D ,$ $\\therefore \\angle D P E + \\angle E P C = \\angle B + \\angle B D P ,$ $\\cdot \\angle E P C = \\angle B D P$ ·…· (8分) ", + "page_idx": 59 + }, + { + "type": "text", + "text": "14.解：(1)如图， $\\triangle A _ { 1 } B _ { 1 } C _ { 1 }$ 是 $\\triangle A B C$ 关于直线的对称图形;（4分)", + "page_idx": 59 + }, + { + "type": "image", + "img_path": "images/3507aab1d126833d0b9e8b1c4fa6cffe6b72ff565e84cc57b7a0d6d04d0c505c.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 59 + }, + { + "type": "text", + "text": "(2)由图形可得四边形 $B B _ { 1 } C _ { 1 } C$ 是等 腰梯形， $B B _ { 1 } { = } 4 , C C _ { 1 } { = } 2$ ，高是4, $S _ { \\perp \\perp \\perp \\perp \\perp \\perp } \\ L _ { \\mathcal { \\perp } B B _ { 1 } C _ { 1 } C } = \\frac { 1 } { 2 } \\times ( 4 + 2 ) \\times 4 =$ 12. ·…· (8分) ", + "page_idx": 59 + }, + { + "type": "text", + "text": "15.解： $( 1 ) \\because A D$ 垂直平分 $B E , E F$ 垂直平分 $A C , \\therefore A B = A E = E C , \\therefore \\angle C =$ $\\angle C A E .$ ： $\\cdot \\angle B A E = 4 0 ^ { \\circ } , \\therefore \\angle A E D = 7 0 ^ { \\circ } , \\therefore$ $\\angle C = \\frac { 1 } { 2 } \\angle A E D = 3 5 ^ { \\circ }$ （20 $( 2 ) { \\because } \\triangle A B C$ 的周长为 $1 4 \\ { \\mathrm { c m } } , A C =$ $6 ~ \\mathrm { c m }$ ，$\\scriptstyle \\cdot . A B + B C = A B + B E + E C = 1 4 - 6$ （20$= 8 \\mathrm { c m }$ ，即 $2 D E + 2 E C = 2 \\left( D E + \\right.$ $E C ) = 8 \\mathrm { ~ \\ c m } , \\therefore D E + E C = D C =$ $4 \\ \\mathrm { c m } .$ …（10分）", + "page_idx": 59 + }, + { + "type": "text", + "text": "16.解：(1)由轴对称的性质可得 $P M =$ $C M , N D { = } N P .$ $\\cdot C D { = } 1 8 \\ \\mathrm { c m } , \\therefore \\triangle P M N$ 的周长 $=$ $P N + P M + M N = D N + M N + C M$ $= C D { = } 1 8 ~ \\mathrm { c m }$ … (3分)(2)∵点 $P$ 关于 $O A , O B$ 对称的对称点分别为 $C , D$ ，", + "page_idx": 59 + }, + { + "type": "text", + "text": "∴OA垂直平分 $P C , O B$ 垂直平分$P D$ ， \n∴ $C M = P M$ ， $P N = D N$ $\\angle C =$ $\\angle M P C = 2 1 ^ { \\circ }$ $\\angle D = \\angle N P D = 2 8 ^ { \\circ } .$ ·… (6分)$\\cdot \\angle C M P = 1 8 0 ^ { \\circ } - \\angle C - \\angle M P C =$ $1 3 8 ^ { \\circ } , \\angle P N D = 1 8 0 ^ { \\circ } - \\angle D - \\angle N P D$ $= 1 2 4 ^ { \\circ }$ 。 \n： $\\angle P M N = 4 2 ^ { \\circ }$ $\\angle P N M = 5 6 ^ { \\circ }$ \n$\\therefore \\angle M P N { = } 1 8 0 ^ { \\circ } { - } \\angle P M N { - } \\angle P N M$ $= 8 2 ^ { \\circ }$ …· (10分)", + "page_idx": 60 + }, + { + "type": "text", + "text": "17.解：（1）点 $O$ 到 $\\triangle A B C$ 的三个顶点$A , B , C$ 的距离的大小关系是 $O A =$ $O B { = } O C$ ；： …(2分）（ $2 ) \\triangle { O M N }$ 是等腰直角三角形.(3分)理由： $\\cdot A B { = } A C$ $\\angle B A C = 9 0 ^ { \\circ }$ ，点 $O$ 为 $B C$ 的中点，： $A O$ 平分 $\\angle B A C , A O \\bot B C ,$ ∴ $\\angle A O B = 9 0 ^ { \\circ }$ ， $\\angle B = \\angle C = 4 5 ^ { \\circ }$ $\\angle B A O = \\angle C A O = 4 5 ^ { \\circ } , \\therefore \\angle C A O =$ $\\angle B , O A { = } O B$ …· (5分)在 $\\triangle B O M$ 和 $\\triangle A O N$ 中，$A N { = } B M$ ,∠CAO=∠B,OA=OB,:△BOM≌△AON(SAS),:OM=ON,∠AON=∠BOM.： $\\cdot \\angle A O B = \\angle B O M + \\angle A O M = 9 0 ^ { \\circ }$ 。 $\\angle A O N + \\angle A O M = 9 0 ^ { \\circ }$ ，即 $\\angle M O N$ $= 9 0 ^ { \\circ }$ ，", + "page_idx": 60 + }, + { + "type": "text", + "text": "$\\triangle { O M N }$ 是等腰直角三角形.… (8分）(3)四边形AMON的面积不发生变化 …· (9分)理由：由(2)可得 $\\triangle A O N { \\underline { { \\underline { { \\circ } } } } } \\triangle B O M .$ 故 $S _ { \\textsc { p q j j k A M O N } } = S _ { \\triangle A M O } + S _ { \\triangle M B O } =$ $S _ { \\triangle A B O } = \\frac { 1 } { 2 } S _ { \\triangle A B C } .$ （12分）", + "page_idx": 60 + }, + { + "type": "text", + "text": "·月考测试卷（二）参考答案· ", + "page_idx": 60 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 60 + }, + { + "type": "table", + "img_path": "images/086350ff18365748882fe4071dbb660ed98f30de84d60ffd90ae8169c7c91126.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
12345678910
CBBCBCDACD
", + "page_idx": 60 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 60 + }, + { + "type": "table", + "img_path": "images/890c48829d96534405738358c60653f2bef660ad656857880a838cb667c18816.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
1112131415
2y=40x+20062°43420
", + "page_idx": 60 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 60 + }, + { + "type": "text", + "text": "16.(1)解：原式 $= ( \\frac { 2 } { 3 } ) ^ { 2 0 2 3 } \\times 1 . 5 ^ { 2 0 2 3 } \\times 1 . 5$ $\\times 1$ （204号 （2分）$= ( { \\frac { 2 } { 3 } } \\times { \\frac { 3 } { 2 } } ) ^ { 2 0 2 3 } \\times 1 . 5 { = } 1 . 5 .$ （5分）", + "page_idx": 60 + }, + { + "type": "text", + "text": "(2)解：由图形可得，阴影部分的周长$= 2 ( y + x + x + 0 . 5 x + 3 y ) = 2 ( 2 . 5 x$ $+ 4 y ) = 5 x + 8 y$ ； ···(2分)阴影部分的面积 $= y ( 2 x + 0 . ~ 5 x ) +$ $0 . 5 x \\bullet 3 y = 2 x y + 0 . 5 x y + 1 . 5 x y =$ $4 \\mathit { x y } .$ ·(5分)", + "page_idx": 60 + }, + { + "type": "text", + "text": "17.解：如图，延长 $C E$ 交 $A B$ 于点 $F$ ，则 CF⊥AB. ··· …(1分) ", + "page_idx": 60 + }, + { + "type": "image", + "img_path": "images/519a91d0f6ae04b62a520082e2a64d59bed69b4508a3306d82b37763c1a620b9.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 60 + }, + { + "type": "text", + "text": "$1 2 ) = 4 5 0$ 米/分.因为 $4 5 0 > 3 3 0$ ，所以小明买到书后继续骑车到学校的这段时间内的骑车速度不在安全限度内. ：（9分）", + "page_idx": 60 + }, + { + "type": "text", + "text": "因为 $\\angle A + \\angle 1 = 9 0 ^ { \\circ }$ $\\angle C + \\angle 2 = { \\ : }$ \n$9 0 ^ { \\circ }$ $\\angle 1 = \\angle 2$ ，所以 $\\angle A = \\angle C$ …·（4分) \n在△ABD和 $\\triangle C D E$ 中， \n因为 $\\angle A = \\angle C , \\angle A B D = \\angle C D E =$ \n$9 0 ^ { \\circ }$ $\\scriptstyle { \\mathrm { ) } } ^ { \\circ } , A D = C E$ \n所以 $\\triangle A B D { \\cong } \\triangle C D E$ （AAS)，所以 \n$B D { = } D E .$ \n因为 $D E { = } 2$ 米，所以 $D B { = } 2$ 米.… (9分)", + "page_idx": 60 + }, + { + "type": "text", + "text": "19.解：由已知可得 $A B { = } C D { = } 4$ ,分两种情况讨论：$\\textcircled{1}$ 若点 $P$ 在 $B C$ 边上时，即 $\\angle A B P { = }$ $\\angle D C E = 9 0 ^ { \\circ }$ 时，若满足 $B P { = } C E { = } 2$ ,根据\"SAS”可得出 $\\triangle A B P { \\overset { \\circ } { = } } \\triangle D C E .$ 由题意得 $2 t = 2$ ，解得 $t { = } 1$ （3分）", + "page_idx": 60 + }, + { + "type": "text", + "text": "$\\textcircled{2}$ 若点 $P$ 在 $A D$ 边上时，即 $\\angle B A P$ $= \\angle D C E = 9 0 ^ { \\circ }$ 时， \n若满足 $A P = C E = 2$ ,根据“SAS”可得出 $\\triangle B A P { \\overset { \\subset } { = } } \\triangle D C E .$ （20 \n由题意得 $1 6 - 2 t = 2$ ，解得 $t { = } 7$ \n…· （7分)综上所述，当 $t$ 的值为1或7时，$\\triangle A B P$ 和 $\\triangle D C E$ 全等·（9分)", + "page_idx": 60 + }, + { + "type": "text", + "text": "18.解：(1)由图象可得，小明家到学校的距离是1500米.故填1500．·…（1分）（2)小明在书店停留的时间为从8分钟到12分钟,共停留了 $1 2 - 8 = 4$ 分钟.故填4. …· (3分)(3)骑行总路程 $= 1 2 0 0 + ( 1 2 0 0 -$ $6 0 0 ) + ( 1 5 0 0 - 6 0 0 ) = 1 2 0 0 + 6 0 0 +$ $9 0 0 { = } 2 7 0 0$ 米.故填 2700．…… (6分)", + "page_idx": 60 + }, + { + "type": "text", + "text": "(4)买到书后继续骑车到学校的这段时间是 $1 2 { \\sim } 1 4$ 分钟，在 $1 2 \\sim 1 4$ 分钟时平均速度 $= ( 1 5 0 0 - 6 0 0 ) \\div ( 1 4 -$ ", + "page_idx": 60 + }, + { + "type": "text", + "text": "20.解：代数式 $\\left( 3 a + 2 \\right) \\left( 2 a + 3 \\right) - 3 a \\left( 2 a \\right.$ $+ 1 )$ 的值是偶数. …·(1分)理由： $( 3 a + 2 ) ( 2 a + 3 ) - 3 a ( 2 a + 1 )$ $= 6 a ^ { 2 } + 9 a + 4 a + 6 - 6 a ^ { 2 } - 3 a = 1 0 a$ $+ 6 = 2 ( 5 a + 3 )$ …（6分)因为 $a$ 为自然数，所以 $5 a + 3$ 是整数，所以代数式 $\\left( 3 a + 2 \\right) \\left( 2 a + 3 \\right) -$ $3 a ( 2 a + 1 )$ 的值是偶数．（9分)", + "page_idx": 60 + }, + { + "type": "text", + "text": "21.解：（1）由折叠的性质可得， $D A =$ $D B _ { \\ast }$ 因为 $A C { = } 6 \\ \\mathrm { c m } , B C { = } 8 \\ \\mathrm { c m }$ ，", + "page_idx": 60 + }, + { + "type": "text", + "text": "所以 $\\triangle A C D$ 的周长 $= D A + D C +$ $A C = D B + D C + A C = B C + A C =$ $1 4 ~ \\mathrm { c m }$ ；： …（4分）(2）设 $\\angle C A D { = } { _ { x } }$ ，则 $\\angle B A D { = } 2 x$ 由图形的折叠可得， $\\scriptstyle \\angle B = \\angle B A D =$ $2 x$ 在 $\\mathrm { R t } \\triangle A B C$ 中， $\\angle B + \\angle B A C = { \\bf \\Phi } _ { \\bf \\Phi }$ $9 0 ^ { \\circ }$ ，可得 $2 x + 2 x + x = 9 0 ^ { \\circ }$ ，解得 $x =$ $1 8 ^ { \\circ }$ ，则 $\\angle B { = } 2 x { = } 3 6 ^ { \\circ }$ ． (9分)", + "page_idx": 61 + }, + { + "type": "text", + "text": "22.解：(1）因为 $\\angle A = 4 0 ^ { \\circ }$ $A B \\bot B C$ ，所以 $\\angle A D B = 9 0 ^ { \\circ } - 4 0 ^ { \\circ } = 5 0 ^ { \\circ }$ 因为 $A M / / C N$ ，所以 $\\angle C = \\angle A D B$ $= 5 0 ^ { \\circ }$ · (3分)(2)如图，延长 $D B$ 交 $C N$ 于点 $H$ 因为 $M D / / C N , B D \\perp A M ,$ 所以 $B D$ $\\bot C N$ ，所以 $\\angle C H B = 9 0 ^ { \\circ }$ …· (5分)因为 $B D \\perp A M$ ，所以 $\\angle D A B +$ $\\angle A B D { = } 9 0 ^ { \\circ }$ 因为 $A B \\perp A M$ ，所以 $\\angle A B D +$ $\\angle C B H = 1 8 0 ^ { \\circ } - 9 0 ^ { \\circ } = 9 0 ^ { \\circ }$ ，所以$\\angle D A B { = } \\angle C B H .$ · (8分)因为 $\\angle A B D = 9 0 ^ { \\circ } - \\angle B A D , \\angle C =$ $9 0 ^ { \\circ } - \\angle C B H$ ，所以 $\\angle A B D = \\angle C$ ·· (10分)", + "page_idx": 61 + }, + { + "type": "image", + "img_path": "images/9edfc832e116364adb349487830ef11282928617594f913adfa4278cd44d0f7c.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 61 + }, + { + "type": "text", + "text": "23.解：(1)因为 $B D , C E$ 都是 $\\triangle A B C$ 的高，所以 $\\angle A D B = \\angle A E C = 9 0 ^ { \\circ }$ ", + "page_idx": 61 + }, + { + "type": "text", + "text": "因为 $\\angle 1 = 9 0 ^ { \\circ } - \\angle C A E , \\angle 2 = 9 0 ^ { \\circ } -$ $\\angle B A D$ ，所以 $\\angle 1 = \\angle 2$ 故填 $=$ 。 (2分)", + "page_idx": 61 + }, + { + "type": "text", + "text": "(2)结论： $A P { = } A Q , A P \\bot A Q .$ \n… (3分)理由：由(1)可知 $\\angle 1 = \\angle 2$ \n在 $\\triangle Q A C$ 和 $\\triangle A P B$ 中， \n$Q C = A B , \\angle 1 = \\angle 2 , C A = B P ,$ \n所以 $\\triangle Q A C { \\cong } \\triangle A P B$ （SAS)，所以$A Q { = } A P$ $\\angle Q A C = \\angle P$ \n因为 $\\angle D A P + \\angle P = 9 0 ^ { \\circ }$ ，所以$\\angle D A P + \\angle Q A C = 9 0 ^ { \\circ }$ ，即 $\\angle Q A P =$ $9 0 ^ { \\circ }$ ，所以 $A Q \\bot A P$ ；…(7分)$( 3 ) A P { = } A Q , A P \\bot A Q .$ … (10分)理由：如图所示.因为 $B D , C E$ 都是 $\\triangle A B C$ 的高，所以$B D \\perp A C , C E \\perp A B .$ \n所以 $\\angle 1 + \\angle C A E = 9 0 ^ { \\circ }$ ， $\\angle 2 +$ $\\angle D A B = 9 0 ^ { \\circ }$ \n因为 $\\angle C A E = \\angle D A B$ ，所以 $\\angle 1 =$ $\\angle 2$ \n在 $\\triangle Q A C$ 和 $\\triangle A P B$ 中， \n$Q C = A B , \\angle 1 = \\angle 2 , C A = B P ,$ \n所以 $\\triangle Q A C { \\cong } \\triangle A P B ( \\mathrm { S A S } )$ ，所以$A Q { = } A P , \\angle Q A C { = } \\angle P .$ \n因为 $\\angle P D A = 9 0 ^ { \\circ }$ ，所以 $\\angle P +$ $\\angle P A D = 9 0 ^ { \\circ }$ ，所以 $\\angle Q A C + \\angle P A D$ ", + "page_idx": 61 + }, + { + "type": "image", + "img_path": "images/49350b8aa84418c1421cb7b363edaf8a28da1c74176c955ef8973ff3dc9c1dcd.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 61 + }, + { + "type": "text", + "text": "", + "page_idx": 61 + }, + { + "type": "text", + "text": "$= 9 0 ^ { \\circ }$ ，即 $\\angle Q A P { = } 9 0 ^ { \\circ }$ 所以 $A Q \\bot A P$ 故 $A P { = } A Q { , } A P \\bot A G$ ", + "page_idx": 61 + }, + { + "type": "text", + "text": "第六章概率初步", + "text_level": 1, + "page_idx": 61 + }, + { + "type": "text", + "text": "卷中悟法", + "text_level": 1, + "page_idx": 61 + }, + { + "type": "text", + "text": "1.用面积法求概率", + "page_idx": 61 + }, + { + "type": "text", + "text": "对于受几何图形的面积影响的随机事件，在一个平面区域内的每个点，事件发生的可能性都是相等的，如果所有可能发生的区域的面积为 $S$ ，所求事件 $A$ 发生的区域的面积为 $S ^ { \\prime }$ ，那么 $P ( A ) = { \\frac { S ^ { \\prime } } { S } }$ ", + "page_idx": 61 + }, + { + "type": "text", + "text": "2.概率的计算方法", + "page_idx": 61 + }, + { + "type": "text", + "text": "概率是用来刻画随机事件发生的可能性大小的一个 $0 \\sim 1$ 的常数，概率小则事件发生的可能性小，概率大则事件发生的可能性大，如果一次试验中所有可能的结果是有限多个，并且一次试验中各种可能结果发生的可能性都相等，那么可用列举法求概率.", + "page_idx": 61 + }, + { + "type": "text", + "text": "用列举法求概率的一般步骤：", + "page_idx": 61 + }, + { + "type": "text", + "text": "(1)列举出一次试验的所有可能的结果；(2)数出 $m , n$ (3)代入概率的计算公式 $P ( A ) { = } { \\frac { m } { n } }$ ", + "page_idx": 61 + }, + { + "type": "text", + "text": "基础过关参考答案", + "text_level": 1, + "page_idx": 61 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 61 + }, + { + "type": "table", + "img_path": "images/1094296becbf2525d5c077acdee244009f0921be5d567bac1473fbae7e9f7929.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
12345678
CDABCABB
", + "page_idx": 61 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 61 + }, + { + "type": "table", + "img_path": "images/d49f9e6333d55184ca8d1c28319a0e246c9ebaad6dfa6929ccf07d2345fa44ae.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
9101112
必然0.75238
", + "page_idx": 61 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 61 + }, + { + "type": "text", + "text": "13.解：(1)因为总共有6张卡片，其中数字是1的有2张，数字是2的有3张，数字是3的有1张，所以摸到数字2卡片的可能性最大，摸到数字3卡片的可能性最小. …（4分）(2)因为总共有6张卡片，其中数字是奇数的有3张，数字是偶数的有3张，所以摸到的数字是奇数和摸到的数字是偶数的可能性一样大.… (8分)", + "page_idx": 61 + }, + { + "type": "text", + "text": "14.解：(1)根据题意，得 $P$ (摸到红球) $=$ ${ \\frac { 6 } { 6 + 9 } } { = } { \\frac { 6 } { 1 5 } } { = } { \\frac { 2 } { 5 } }$ ,P(摸到白球) ${ \\mathfrak { o } } = 0$ （5分）", + "page_idx": 61 + }, + { + "type": "text", + "text": "(2)根据题意，得 ${ \\frac { 6 } { 6 + 9 + m } } = { \\frac { 1 } { 3 } }$ ，所以$1 5 + m { = } 1 8$ ，解得 $m { = } 3$ …(10分）", + "page_idx": 61 + }, + { + "type": "text", + "text": "15.解：(1)一个转盘被等分成六个扇形，并在上面依次写上数字 $1 , 2 , 3 , 4 , 5$ 6,其中有3个扇形是奇数，所以自由转动转盘，当它停止转动后，指针指向奇数的概率是 $\\frac { 3 } { 6 } = \\frac { 1 } { 2 }$ （6分）(2)答案不唯一，如当自由转动的转盘停止时，指针指向数字大于2的区域．· …·（10分） \n16.解：(1)如下表： ·· (6分)", + "page_idx": 61 + }, + { + "type": "text", + "text": "", + "page_idx": 61 + }, + { + "type": "text", + "text": "·能力提优参考答案· ", + "text_level": 1, + "page_idx": 62 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 62 + }, + { + "type": "table", + "img_path": "images/f6db4d0895323ae150163b833934a9433dd042a788bf6989e2bdcdd9bdb8db8a.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
抛掷 次数 n100200300400500600
正面朝 上的频 数m5198153200255306
正面朝 上的频0.510.490.510.500.510.51
", + "page_idx": 62 + }, + { + "type": "table", + "img_path": "images/c4f48494f10401ac0b1b77822d2f75988aef04e8e89893397f5205ec0c5ad8df.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
123456
ABCBDA
", + "page_idx": 62 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 62 + }, + { + "type": "text", + "text": "正面朝上的频率 \n0.52 \n0.51 \n0.50 \n0.49 \n0.48100 200300400500600抛掷次数", + "page_idx": 62 + }, + { + "type": "text", + "text": "（3)由折线统计图可知，“正面朝上”的频率在0.51附近摆动.（12分）", + "page_idx": 62 + }, + { + "type": "text", + "text": "17.解： $( 1 ) \\textcircled{ 1}$ 是必然事件，发生的概率为1； （3分）$\\textcircled{2}$ 是随机事件，发生的概率为 ${ \\frac { 1 3 } { 5 2 } } =$ $\\frac { 1 } { 4 }$ · (6分)", + "page_idx": 62 + }, + { + "type": "text", + "text": "11.解：（1)蓝色区域的圆心角度数为$2 { 0 0 } ^ { \\circ }$ ，则 $P$ (指针停止后在蓝色区域)$= \\frac { 2 0 0 ^ { \\circ } } { 3 6 0 ^ { \\circ } } = \\frac { 5 } { 9 }$ （4分）", + "page_idx": 62 + }, + { + "type": "text", + "text": "(2)如下图： （9分) ", + "page_idx": 62 + }, + { + "type": "text", + "text": "(2)红、黄两个扇形的圆心角度数分别为 $4 0 ^ { \\circ } , 1 2 0 ^ { \\circ }$ ，则 $P$ （指针停止后在黄色或红色区域） $\\tan { \\frac { 4 0 ^ { \\circ } + 1 2 0 ^ { \\circ } } { 3 6 0 ^ { \\circ } } } = { \\frac { 4 } { 9 } }$ （8分）", + "page_idx": 62 + }, + { + "type": "text", + "text": "(2)事件 $A$ 发生的概率为 ${ \\frac { 2 } { 6 } } = { \\frac { 1 } { 3 } }$ ，事件 $B$ 发生的概率为 $\\frac { 8 } { 5 2 } { = } \\frac { 2 } { 1 3 }$ 因为 $\\frac { 2 } { 1 3 } < \\frac { 1 } { 3 }$ ，所以 $P ( B ) { \\ < } P ( A )$ …· (12分)", + "page_idx": 62 + }, + { + "type": "image", + "img_path": "images/07b7572baf4bd6597005af10e6396a616af100bee5fcc01583f86ebb50f7c95a.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 62 + }, + { + "type": "text", + "text": "12.解：(1)设阴影部分的面积是 $a$ ，则整个图形的面积是 $7 a$ ，可得 $P$ （这个点取在阴影部分） $= \\frac { a } { 7 a } = \\frac { 1 } { 7 }$ 故答案为$\\frac { 1 } { 7 }$ …· (5分)", + "page_idx": 62 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 62 + }, + { + "type": "table", + "img_path": "images/869e0e4da8095db6a120a803b9e6f1ab050d75fe59570971c0a28a2e8d65664b.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
78910
100083183
", + "page_idx": 62 + }, + { + "type": "text", + "text": "专项训练卷（一）整式的乘除运算 参考答案· ", + "text_level": 1, + "page_idx": 62 + }, + { + "type": "text", + "text": "（2)如图所示(答案不唯一）：…… (10分)", + "page_idx": 62 + }, + { + "type": "text", + "text": "13.解：(1)袋子里红球的个数为 $5 0 \\times \\frac { 3 } { 1 0 }$ $= 1 5$ （4分）", + "page_idx": 62 + }, + { + "type": "text", + "text": "(2)设白球有 $\\mathcal { X }$ 个，由题意得 $x + x -$ $5 + 1 5 = 5 0$ ,解得 $x { = } 2 0$ ,·(8分) 则 $P$ （随机摸出一个球是白球） $= \\frac { 2 0 } { 5 0 }$ $= \\frac { 2 } { 5 }$ · (10分) ", + "page_idx": 62 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 62 + }, + { + "type": "table", + "img_path": "images/fb413d1fd5cf6c6a2d902a702ba150e3604047dab5a5f3ebdde1bb4dd5daf4ae.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
12345678
BCCAADAB
", + "page_idx": 62 + }, + { + "type": "text", + "text": "14.解：（1)根据题意得 $y = { \\frac { x ^ { 2 } - x } { 2 } } =$ ${ \\frac { 1 } { 2 } } x ( x - 1 ) .$ 写出的必然事件不唯一，如输出的数$y$ 是整数. …(5分)(2)当输入的数是2至9这八个连续正整数中的一个时，输出的结果分别为 $1 , 3 , 6 , 1 0 , 1 5 , 2 1 , 2 8 , 3 6$ ，则 $P$ （输出的数是3的倍数) $= \\frac { 5 } { 8 }$ （10分）", + "page_idx": 62 + }, + { + "type": "table", + "img_path": "images/de19c3070a208c21d3e5641c8514f562b8893a9a1ba43833ce7060286f31c33d.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
9101112
125x-524
", + "page_idx": 62 + }, + { + "type": "text", + "text": "15.解：（1)因为 $1 8 0 < 2 0 0$ ，所以小明购物180元，不能获得转动转盘的机会，故小明获得奖金的概率为0;… (4分)", + "page_idx": 62 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 62 + }, + { + "type": "text", + "text": "(2)小德购物210元，能获得一次转动转盘的机会， $P$ （获得奖金） $= { \\frac { 6 } { 1 6 } } =$ $\\frac { 3 } { 8 }$ · (8分)", + "page_idx": 62 + }, + { + "type": "text", + "text": "(3)设需要将 $\\mathcal { X }$ 个无色区域涂上绿色，可得 ${ \\frac { x + 3 } { 1 6 } } = { \\frac { 1 } { 4 } }$ ，解得 $x = 1$ ，所以需要将1个无色区域涂上绿色.（12分）", + "page_idx": 62 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 62 + }, + { + "type": "text", + "text": "13.解：（1）原式 $= x ^ { 6 n + 3 } \\ \\div \\ x ^ { 5 n } \\cdot x =$ $\\boldsymbol { x } ^ { 6 n + 3 - 5 n + 1 } = \\boldsymbol { x } ^ { n + 4 }$ （3分）", + "page_idx": 62 + }, + { + "type": "text", + "text": "（2）原式 $= - 1 - 1 \\div ( - \\frac { 1 } { 3 } ) \\times 4 = - 1$ $- 1 \\times ( - 3 ) \\times 4 = - 1 + 1 2 = 1 1 .$ （20号（6分）", + "page_idx": 62 + }, + { + "type": "text", + "text": "14.解：(1)原式 $= x ^ { 2 } + 7 x + x ^ { 2 } - 2 5 - ( x ^ { 2 }$ $- x + 6 x - 6 )$ $= x ^ { 2 } + 7 x + x ^ { 2 } - 2 5 - x ^ { 2 } + x - 6 x + 6$ $= x ^ { 2 } + 2 x - 1 9$ …(3分) 当 $x = - 3$ 时，原式 $= ( - 3 ) ^ { 2 } + 2 \\times$ $( - 3 ) - 1 9 = 9 - 6 - 1 9 = - 1 6 .$ …（4分) (2）原式 $= 2 x ^ { 2 } + 5 x - ( - 2 x ^ { 2 } + 4 x ) =$ $2 x ^ { 2 } + 5 x + 2 x ^ { 2 } - 4 x = 4 x ^ { 2 } + x .$ ·…(7分) 当 $x = 5$ 时，原式 $= 4 \\times 5 ^ { 2 } + 5 = 1 0 5 .$ …(8分) ", + "page_idx": 62 + }, + { + "type": "text", + "text": "15.解 ${ \\mathrm { } } _ { ; a = ( - { \\frac { 1 } { 8 } } \\times 8 \\times { \\frac { 1 } { 9 } } \\times 9 ) ^ { 3 4 } = ( - 1 ) ^ { 3 4 } }$ $= 1$ ， … (2分) $b { = } 9 4 ^ { 2 } - ( 9 4 { - } 1 ) ( 9 4 { + } 1 ) { = } 9 4 ^ { 2 } { - } 9 4 ^ { 2 }$ $+ 1 { = } 1$ · (4分) $- ( 6 x y ^ { 2 } ) ^ { 2 } \\div ( - 3 x y ) = - 3 6 x ^ { 2 } y ^ { 4 } \\div \\frac { 1 } { 2 }$ $( - 3 x y ) = 1 2 x y ^ { 3 }$ ， ·(6分) 当 $x = a - 2 = 1 - 2 = - 1$ $y = b = 1$ 时，原式 $= 1 2 \\times ( - 1 ) \\times 1 ^ { 3 } = - 1 2$ ·… (8分) ", + "page_idx": 62 + }, + { + "type": "text", + "text": "", + "page_idx": 63 + }, + { + "type": "text", + "text": "16.解:(1) $( a - b ) ^ { 2 }$ $a ^ { 2 } - 2 a b + b ^ { 2 }$ (a$b ) ^ { 2 } = a ^ { 2 } - 2 a b + b ^ { 2 }$ .…(2分) $( 2 ) \\textcircled { 1 } 2 9 9 ^ { 2 } = ( 3 0 0 - 1 ) ^ { 2 } = 3 0 0 ^ { 2 } - 2 \\times$ $3 0 0 \\times 1 + 1 = 9 0 0 0 0 - 6 0 0 + 1 =$ 89401；···· …·（4分） $\\textcircled { 2 } 4 7 ^ { 2 } - 4 7 \\times 8 6 + 4 3 ^ { 2 } = 4 7 ^ { 2 } - 2 \\times 4 7 \\times$ $4 3 + 4 3 ^ { 2 } = ( 4 7 - 4 3 ) ^ { 2 } = 4 ^ { 2 } = 1 6 .$ …· (6分) $\\begin{array} { l } { { ( 3 ) ( a - b ) ^ { 2 } = a ^ { 2 } - 2 a b + b ^ { 2 } = a ^ { 2 } + b ^ { 2 } } } \\\\ { { \\phantom { - } } } \\\\ { { - 2 a b = 1 0 - 2 \\times 3 = 4 , \\phantom { - } \\cdots \\ ( 8 \\ \\rlap / ) } } \\end{array}$ ", + "page_idx": 63 + }, + { + "type": "text", + "text": "17.解： $( 1 ) \\frac { 1 } { 2 } \\left( a + a + 2 b \\right) \\times \\left( a + 2 b \\right) =$ ${ \\frac { 1 } { 2 } } ( 2 a + 2 b ) \\times ( a + 2 b ) = ( a + b ) ( a +$ $2 b ) = a ^ { 2 } + 3 a b + 2 b ^ { 2 } ,$ 所以该防洪堤坝的横断面的面积为$( a ^ { 2 } + 3 a b + 2 b ^ { 2 } )$ 平方米．（3分)$( 2 ) a { = } 1 \\div 2 ^ { - 1 } { = } 2 , b { = } 1 { - } 6 \\times ( { - } 1 )$ $= 1 + 6 = 7$ ，： …(5分）所以该防洪堤坝的横断面的面积为$a ^ { 2 } + 3 a b + 2 b ^ { 2 } = 2 ^ { 2 } + 3 \\times 2 \\times 7 + 2 \\times 7 ^ { 2 }$ $= 4 + 4 2 + 9 8 = 1 4 4$ 平方米.…· (7分)", + "page_idx": 63 + }, + { + "type": "text", + "text": "(3)修建该防洪堤坝需要() $( \\frac { 1 } { 1 0 } ) ^ { - 2 } \\div 1$ $1 0 ^ { - 1 } \\times 1 4 4 = 1 0 ^ { 2 } \\div 1 0 ^ { - 1 } \\times 1 4 4 = 1 4 4$ $\\times 1 0 0 0 { = } 1 . 4 4 \\times 1 0 ^ { 5 }$ 立方米 $= 1 . 4 4 \\times$ （20 $1 0 ^ { 5 }$ 方土. · (10分) ", + "page_idx": 63 + }, + { + "type": "text", + "text": "", + "page_idx": 63 + }, + { + "type": "text", + "text": "18.解：(1)每个框四个角上的数交叉相乘后求和，再与中间的数的平方的2倍作差，结果都等于－100.…· (2分)", + "page_idx": 63 + }, + { + "type": "text", + "text": "(2)设中间的数为 $\\mathcal { X }$ ,则其他四个数依次是 $x ^ { - } 8 , x ^ { - } 6 , x + 6 , x + 8$ ，所以规律为 $\\left( x + 6 \\right) \\left( x - 6 \\right) + \\left( x + 8 \\right) \\left( x - 1 \\right)$ $8 ) - 2 x ^ { 2 } = - 1 0 0 .$ \n理由：左边 $= x ^ { 2 } - 3 6 + x ^ { 2 } - 6 4 - 2 x ^ { 2 }$ $= x ^ { 2 } + x ^ { 2 } - 2 x ^ { 2 } - 1 0 0 = - 1 0 0 =$ 右边,所以结论成立．…（6分)(3)(1)中的规律不成立．（7分)验证： $9 \\times 2 3 + 1 5 \\times 1 7 - 2 \\times 1 6 ^ { 2 } =$ —50， $1 2 \\times 2 6 + 1 8 \\times 2 0 - 2 \\times 1 9 ^ { 2 } =$ 一50,所以(1)中的规律不成立. \n·（9分）发现的规律：设中间的数为 $\\mathcal { X }$ ,则其他四个数依次是 $x ^ { - 7 } , x ^ { - 1 } , x ^ { + 1 } , x$ $+ 7$ ，所以规律为 $( x + 7 ) ( x - 7 ) + ( x$ $+ 1 ) ( x - 1 ) - 2 x ^ { 2 } = - 5 0 .$ \n…·（10分）理由：左边 $= x ^ { 2 } - 4 9 + x ^ { 2 } - 1 - 2 x ^ { 2 } =$ $x ^ { 2 } + x ^ { 2 } - 2 x ^ { 2 } - 5 0 = - 5 0 =$ 右边. \n……· (12分)", + "page_idx": 63 + }, + { + "type": "text", + "text": "专项训练卷（二）几何题的计算参考答案", + "page_idx": 63 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 63 + }, + { + "type": "table", + "img_path": "images/f1576fa69ee6dff3cee6ebc24ad69e4715b43684ebba20a0d1de2a647a01ef8a.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
1235678
CDBABADC
", + "page_idx": 63 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 63 + }, + { + "type": "table", + "img_path": "images/60a59a9c8c8643ee4c9506bd666d67d2592d09f12e2a575d65d728aad61d741c.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
9101112
25°170°6°
", + "page_idx": 63 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 63 + }, + { + "type": "text", + "text": "13.解：因为 $D E \\bot B C$ ，所以 $\\angle D E B =$ $9 0 ^ { \\circ }$ ，所以 $\\angle B + \\angle E D B = 9 0 ^ { \\circ }$ ，因为$\\angle B = 3 1 ^ { \\circ }$ ，所以 $\\angle E D B = 9 0 ^ { \\circ } - 3 1 ^ { \\circ } =$ $5 9 ^ { \\circ }$ ，所以 $\\angle A D E = 1 8 0 ^ { \\circ } - \\angle E D B =$ $1 8 0 ^ { \\circ } - 5 9 ^ { \\circ } = 1 2 1 ^ { \\circ }$ ·· (4分)因为 $E F / / A B$ ，所以 $\\angle F E D = \\vdots$ $\\angle E D B = 5 9 ^ { \\circ }$ ，因为 $F G \\bot B C , D E \\bot$ BC，所以 $F G / / D E$ ，所以 $\\angle F = \\vdots$ $\\angle F E D = 5 9 ^ { \\circ }$ (8分)", + "page_idx": 63 + }, + { + "type": "text", + "text": "14.解：(1)因为 $O M \\perp A B , O N \\perp C D$ 所以 $\\angle 1 + \\angle A O C = 9 0 ^ { \\circ }$ $\\angle 2 + \\angle A O C$ $= 9 0 ^ { \\circ }$ ，所以 $\\angle 1 = \\angle 2$ ，因为 $\\angle 1 = 2 9 ^ { \\circ }$ ，所以 $\\angle 2 = \\angle 1 = 2 9 ^ { \\circ }$ ·……(3分)(2)因为 $\\angle B O D { = } 6 2 ^ { \\circ }$ ，所以 $\\angle A O C =$ $\\angle B O D { = } 6 2 ^ { \\circ }$ ，因为 $\\angle A O M = 9 0 ^ { \\circ }$ ，所以 $\\angle 1 = \\vdots$ $\\angle A O M - \\angle A O C = 9 0 ^ { \\circ } - 6 2 ^ { \\circ } = 2 8 ^ { \\circ }$ … (6分)因为 $\\angle C O N = 9 0 ^ { \\circ }$ ，所以 $\\angle M O N = \\mathrm { : }$ $\\angle 1 + \\angle C O N = 2 8 ^ { \\circ } + 9 0 ^ { \\circ } = 1 1 8 ^ { \\circ } .$ …· (9分)", + "page_idx": 63 + }, + { + "type": "text", + "text": "15.解：(1)因为点 $E$ 是 $\\it C D$ 的中点，所以$D E = C E$ ，因为 $A D / / B C$ ，所以$\\angle D A E = \\angle F$ ， $\\angle D = \\angle E C F$ ，所以$\\triangle A D E { \\cong } \\triangle F C E .$ ，所以 $A E = F E$ $S _ { \\triangle A D E } { = } S _ { \\triangle F C E }$ ，所以 $A F { = } 2 A E { = } 2 { \\times } 2 { = } 4$ ，因为 $B E$ ", + "page_idx": 63 + }, + { + "type": "text", + "text": "$\\perp A F$ ，所以 $S _ { \\triangle A B F } = \\frac 1 2 B E \\cdot A F = \\frac 1 2$ \n$\\times 4 \\times 4 = 8$ ， \n所以 $S _ { \\substack { \\scriptscriptstyle { [ \\mathscr { A } _ { 1 } \\mathscr { A } _ { 2 } ] \\mathscr { H } _ { 4 } } \\mathscr { A } B C D } } = S _ { \\triangle \\mathscr { A } B F } = 8 ,$ … (5分)", + "page_idx": 63 + }, + { + "type": "text", + "text": "（2）因为 $A D / / B C$ ，所以 $\\angle F =$ $\\angle D A E = 5 9 ^ { \\circ }$ ，因为 $\\triangle A D E \\cong$ $\\triangle F C E$ ，所以 $A E { = } F E$ ，因为 $B E \\bot$ $A F$ ，所以 $B E$ 是线段 $A F$ 的垂直平分线，所以 $A B = B F$ ，所以 $\\angle B A F =$ $\\angle F = 5 9 ^ { \\circ }$ ，所以 $\\angle A B F = 1 8 0 ^ { \\circ } - 5 9 ^ { \\circ }$ $- 5 9 ^ { \\circ } = 6 2 ^ { \\circ }$ ·… (9分)", + "page_idx": 63 + }, + { + "type": "text", + "text": "16.解：(1)因为 $A B { = } A C$ ，所以 $\\angle A B C =$ $\\angle C = 6 5 ^ { \\circ }$ ，因为 $A D \\perp B C$ ，所以$\\angle C A D = 9 0 ^ { \\circ } - \\angle C = 9 0 ^ { \\circ } - 6 5 ^ { \\circ } = 2 5 ^ { \\circ }$ 因为 $M N / / \\ A C$ ，所以 $\\angle A N M =$ $\\angle D A C = 2 5 ^ { \\circ }$ …(3分)(2)因为 $A B = A C = 8$ ，点 $M$ 为 $A B$ 的中点，所以 $A M { = } B M { = } \\frac { 1 } { 2 } { \\times } 8 { = } 4$ ，… (5分)因为MN//AC，所以 $\\angle A N M =$ $\\angle D A C$ ，因为 $A B { = } A C , A D \\bot B C ,$ 所以 $\\angle B A D = \\angle C A D$ ，所以 $\\angle M A N =$ $\\angle A N M$ ，所以 $A M { = } M N$ ，所以 $M N$ $= A M { = } 4$ ·…· (10分)因为 $B N { = } 3$ ，所以 $\\triangle B N M$ 的周长为$B M + M N + B N { = } 4 { + } 4 { + } 3 { = } 1 1 .$ …（12分)", + "page_idx": 63 + }, + { + "type": "text", + "text": "17.解： $( 1 ) \\textcircled{ 1 }$ 因为 $D E$ 垂直平分 $A B$ ，所以 $A E { = } B E$ ，所以 $\\angle A = \\angle A B E$ ，因为 $F G$ 垂直平分 $B C$ ，所以 $C G { = } B G$ ", + "page_idx": 63 + }, + { + "type": "text", + "text": "所以 $\\angle G B C = \\angle C$ ，因为 $\\angle A B C = { \\mathrm { : } }$ $1 0 0 ^ { \\circ }$ ，所以 $\\angle A + \\angle C = 1 8 0 ^ { \\circ } - 1 0 0 ^ { \\circ } =$ $8 0 ^ { \\circ }$ ，所以 $\\angle A B E + \\angle C B G = \\angle A +$ $\\angle C = 8 0 ^ { \\circ }$ ，所以 $\\angle E B G = 1 0 0 ^ { \\circ } - 8 0 ^ { \\circ }$ $= 2 0 ^ { \\circ }$ （4分）$\\textcircled{2}$ 因为 $D E$ 垂直平分 $A B$ ，所以 $A E =$ $B E$ ，所以 $\\angle A = \\angle A B E .$ ，因为 $F G$ 垂直平分 $B C$ ，所以 $C G = B G$ ，所以$\\angle G B C = \\angle C$ ，因为 $\\angle A B C = 7 0 ^ { \\circ }$ ，所以 $\\angle A + \\angle C = 1 8 0 ^ { \\circ } - 7 0 ^ { \\circ } = 1 1 0 ^ { \\circ }$ \n所以 $\\angle A B E + \\angle C B G = 1 1 0 ^ { \\circ }$ ，所以$\\angle A B G + \\angle E B G + \\angle E B G + \\angle C B E$ $= 1 1 0 ^ { \\circ }$ \n所以 $\\angle A B C + \\angle E B G = 1 1 0 ^ { \\circ }$ ，所以$\\angle E B G = 1 1 0 ^ { \\circ } - 7 0 ^ { \\circ } = 4 0 ^ { \\circ } .$ ： \n…· (10分)", + "page_idx": 64 + }, + { + "type": "text", + "text": "11.解：（1）因为 $\\angle A O E = 4 5 ^ { \\circ }$ ，所以$\\angle B O F { = } \\angle A O E { = } 4 5 ^ { \\circ }$ ，所以 $\\angle A O F$ $= 1 8 0 ^ { \\circ } - 4 5 ^ { \\circ } = 1 3 5 ^ { \\circ }$ ，因为 $\\angle A O F =$ $3 \\angle C O F$ ，所以 $3 \\angle C O F = 1 3 5 ^ { \\circ }$ ，所以$\\angle C O F { = } 4 5 ^ { \\circ }$ ，所以 $\\angle C O B { = } 9 0 ^ { \\circ }$ ，所以 $A B \\bot C O .$ ·(4分)(2）与 $\\angle C O F$ 互余的角： $\\angle B O F$ $\\angle A O E$ ；与 $\\angle C O F$ 互补的角：$\\angle C O E$ …(8分)", + "page_idx": 64 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 64 + }, + { + "type": "text", + "text": "(2)当 $0 ^ { \\circ } < \\alpha < 9 0 ^ { \\circ }$ 时， $\\angle E B G = 1 8 0 ^ { \\circ }$ $- 2 \\alpha$ ； …（12分）当 $9 0 ^ { \\circ } < \\alpha < 1 8 0 ^ { \\circ }$ 时， $\\angle E B G = 2 \\alpha -$ $1 8 0 ^ { \\circ }$ ·(14分)", + "page_idx": 64 + }, + { + "type": "text", + "text": "专项训练卷（三）几何题的说理 参考答案· ", + "page_idx": 64 + }, + { + "type": "text", + "text": "12.解：（1）因为 $A D$ 是 $\\triangle A B E$ 的中线，所以 $B D { = } D E$ 在 $\\triangle A B D$ 和 $\\triangle A E D$ 中， $A B = A E$ $A D { = } A D , B D { = } D E$ ，所以 $\\triangle A B D { \\underline { { \\underline { { \\circ } } } } }$ $\\triangle A E D$ …(3分)$( 2 ) \\angle B = 2 \\angle C .$ （4分）理由：因为 $A B = A E$ ，所以 $\\angle B =$ $\\angle A E B$ ，因为 $E F$ 垂直平分 $A C$ ，所以$A E { = } C E$ ，所以 $\\angle E A C = \\angle C$ ， ·· (6分)因为 $\\angle A E B = 1 8 0 ^ { \\circ } - \\angle A E C =$ $\\angle E A C + \\angle C ,$ 所以 $\\angle A E B = 2 \\angle C$ ，所以 $\\angle B { = } 2 \\angle C .$ ·· (8分)", + "page_idx": 64 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 64 + }, + { + "type": "table", + "img_path": "images/aab0f10fc802bcb90d2cb3ca55784e6e2662fc844771d0e25bd247052b398512.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
123456
BCBABD
", + "page_idx": 64 + }, + { + "type": "table", + "img_path": "images/29573c5ed323be93da3b4c291e3e7ec9fba633b8493646faed3fe82b197d36cc.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
78910
SAS527∠A+∠C+ ∠EFC=270°
", + "page_idx": 64 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 64 + }, + { + "type": "text", + "text": "13.解： $( 1 ) \\triangle B D F { \\cong } \\triangle E D C .$ …(1分)理由：因为 $\\angle B = 9 0 ^ { \\circ } , A D$ 是 $\\angle B A C$ 的平分线， $D E \\bot A C$ 所以 $B D = D E$ ，因为 $B F { = } C E , \\angle B$ $= \\angle C E D = 9 0 ^ { \\circ }$ ，所以 $\\triangle B D F \\cong$ $\\triangle E D C$ ：$( 2 ) A C { = } A B { + } B F .$ ：(5分)理由：易得 $\\triangle A B D \\cong \\triangle A E D$ ，所以$A B { = } A E$ ，因为 $B F { = } C E$ ，所以 $A C =$ ", + "page_idx": 64 + }, + { + "type": "text", + "text": "$A E + C E { = } A B { + } B F ,$ 即 $\\scriptstyle A C = A B + B F$ （8分）", + "page_idx": 64 + }, + { + "type": "text", + "text": "14.解：(1)如图，过点 $N$ 作 $N E / / A B$ ，所以 $\\angle A P N + \\angle P N E { = } 1 8 0 ^ { \\circ }$ 因为 $\\angle A P N + \\angle P N Q + \\angle C Q N =$ $3 6 0 ^ { \\circ }$ ，所以 $\\angle A P N + \\angle P N E +$ $\\angle E N Q + \\angle C Q N = 3 6 0 ^ { \\circ } ,$ 所以 $1 8 0 ^ { \\circ } + \\angle E N Q + \\angle C Q N = 3 6 0 ^ { \\circ }$ 所以 $\\angle E N Q + \\angle C Q N = 1 8 0 ^ { \\circ }$ 所以 $E N / / C D$ ，… ：（4分）因为 $N E / / A B$ ，所以 $A B / / C D$ …· (5分)", + "page_idx": 64 + }, + { + "type": "image", + "img_path": "images/8af7f4aec5432f7c3c80018398d422ea40723dafcd9358b2cbfb59f9c2740066.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 64 + }, + { + "type": "text", + "text": "(2） $\\angle M { = } \\frac { 1 } { 2 } \\angle P N Q .$ …… (6分) ", + "page_idx": 64 + }, + { + "type": "text", + "text": "理由：因为 $A B / / C D / / N E$ ，所以$\\angle B P N ~ = ~ \\angle P N E ,$ $\\angle D Q N =$ $\\angle Q N E$ \n所以 $\\angle B P N + \\angle D Q N = \\angle P N Q$ 同理可得 $\\angle B P M + \\angle D Q M = \\angle M$ \n因为PM,QM分别平分 $\\angle B P N$ $\\angle D Q N$ ， \n所以 $\\angle B P N = 2 \\angle B P M , \\angle D Q N =$ $2 \\angle D Q D$ ，所以 $\\angle B P N + \\angle D Q N =$ $2 ( \\angle B P M + \\angle D Q M )$ \n所以 $\\angle P N Q = 2 \\angle M ,$ 即 $\\angle M = \\vdots$ $\\frac { 1 } { 2 } \\angle P N Q .$ … (12分)", + "page_idx": 64 + }, + { + "type": "text", + "text": "15.解：(1)有. …(1分）如图，连接BC,在 $\\triangle A B C$ 中， $\\angle A +$ $\\angle A B C + \\angle A C B = 1 8 0 ^ { \\circ }$ ，在 $\\triangle C B C$ 中， $\\angle B O C + \\angle O B C + \\angle O C B =$ $1 8 0 ^ { \\circ }$ ，所以 $\\angle A + \\angle A B C + \\angle A C B =$ $\\angle B O C + \\angle O B C + \\angle O C B$ ，所以 $\\angle A$ $+ \\angle A B C - \\angle O B C + \\angle A C B -$ $\\angle O C B = \\angle B O C$ ，所以 $\\angle B O C = \\angle A$ $+ \\angle A B O + \\angle A C O .$ ……(4分)", + "page_idx": 64 + }, + { + "type": "text", + "text": "", + "page_idx": 64 + }, + { + "type": "image", + "img_path": "images/a71e08cbb80f53e828d8278c6cc62286159ae8e9a36fc915792628d4d46b6b01.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 64 + }, + { + "type": "text", + "text": "(2)如图，连接 $A D$ ，由(1)可知 $\\angle F +$ $\\angle 2 + \\angle 3 = 1 3 0 ^ { \\circ } , \\angle C + \\angle 4 + \\angle 1 =$ ${ 1 0 0 } ^ { \\circ }$ ，所以 $\\angle F + \\angle 2 + \\angle 3 + \\angle C +$ $\\angle 4 + \\angle 1 = 2 3 0 ^ { \\circ }$ ，所以 $\\angle F + ( \\angle 1 +$ $\\angle 2 ) + \\angle C + ( \\angle 3 + \\angle 4 ) = 2 3 0 ^ { \\circ } ,$ 所以 $\\angle F A B + \\angle C + \\angle C D E + \\angle F =$ $2 3 0 ^ { \\circ }$ ……(7分)", + "page_idx": 64 + }, + { + "type": "text", + "text": "AE130° 100B C D ", + "page_idx": 64 + }, + { + "type": "text", + "text": "$( 3 ) E C / / B F / / D G .$ ·(8分)理由：因为 $\\angle E O D + \\angle O B F = 1 8 0 ^ { \\circ }$ $\\angle E O D = \\angle B O C .$ ，所以 $\\angle B O C +$ $\\angle O B F { = } 1 8 0 ^ { \\circ }$ ，所以 $E C / / B F$ 因为 $\\angle C D G = \\angle F$ ，所以 $D G / / B F$ 所以 $E C / / B F / / D G .$ ·…（10分）(4)149.5° …（14分)提示：因为 $A B { = } A C$ ，所以 $\\angle A B C =$ $\\angle A C B = \\frac { 1 } { 2 } \\left( 1 8 0 ^ { \\circ } - 5 8 ^ { \\circ } \\right) = 6 1 ^ { \\circ } ,$ 又因为 $C E$ 平分 $\\angle A C B$ ，所以 $\\angle E C D = :$ $\\angle E C B = 3 0 . 5 ^ { \\circ }$ ,又因为 $D G / / E C$ ，所以 $\\angle C D G = \\angle D C E = 3 0 . ~ 5 ^ { \\circ }$ ，即$\\angle A D G = 1 4 9 . 5 ^ { \\circ } .$ ", + "page_idx": 64 + }, + { + "type": "text", + "text": "", + "page_idx": 65 + }, + { + "type": "text", + "text": "期末测试卷参考答案· ", + "page_idx": 65 + }, + { + "type": "text", + "text": "一、选择题", + "text_level": 1, + "page_idx": 65 + }, + { + "type": "table", + "img_path": "images/f0815b89dd39ef58fbb57a9370582f3fc9f589949603f10bff3065b25d0f1939.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
12345678910
DBCCBACAAB
", + "page_idx": 65 + }, + { + "type": "text", + "text": "二、填空题", + "text_level": 1, + "page_idx": 65 + }, + { + "type": "table", + "img_path": "images/428ba12dfbd7ca39038e229c8d16418ea7854beb27f94479a3d42c9f3b4a00ee.jpg", + "table_caption": [], + "table_footnote": [], + "table_body": "
1112131415
13115°2023370°
", + "page_idx": 65 + }, + { + "type": "text", + "text": "三、解答题", + "text_level": 1, + "page_idx": 65 + }, + { + "type": "text", + "text": "16.(1)解：原式 $= - 4 + 1 - ( - 3 ) + 2$ ·（4分）$= 2$ （5分）", + "page_idx": 65 + }, + { + "type": "text", + "text": "(2)解：因为点 $A$ 和点 $E$ 关于 $B D$ 对称，所以 $\\angle A B D = \\angle E B D$ ，即 $\\angle A B C$ $= 2 \\angle A B D = 2 \\angle E B D .$ ……(1分)又因为点 $B$ 和点 $C$ 关于 $D E$ 对称，所以 $\\angle D B E = \\angle C$ ，所以 $\\angle A B C = \\left\\{ \\begin{array} { l l } { \\begin{array} { r l r } \\end{array} } \\end{array} \\right.$ $2 \\angle C$ ·…· (3分)因为 $\\angle A = 9 0 ^ { \\circ }$ ，所以 $\\angle A B C + \\angle C =$ $9 0 ^ { \\circ }$ ，即 $3 \\angle C = 9 0 ^ { \\circ }$ 解得 $\\angle C = 3 0 ^ { \\circ }$ ，则 $\\angle A B C = 2 \\angle C =$ （20 $6 0 ^ { \\circ }$ ·… (5分)", + "page_idx": 65 + }, + { + "type": "text", + "text": "17.解：(1)当 $a = 2 4 . 5 ~ \\mathrm { c m }$ 时， $b = 7 a -$ $3 . 0 7 = 7 \\times 2 4 . 5 - 3 . 0 7 = 1 6 8 . 4 3$ (cm).答：他的身高约为168.43厘米；……· (4分)", + "page_idx": 65 + }, + { + "type": "text", + "text": "(2)当 $a = 2 6 . 7 \\ \\mathrm { c m }$ 时， $b { = } 7 a { - } 3 . 0 7$ $= 7 \\times 2 6 . 7 \\mathrm { - } 3 . 0 7 \\mathrm { = } 1 8 3 . 8 3 ~ \\mathrm { c m }$ ，显然身高为 $1 . 8 7 \\mathrm { ~ m ~ }$ 的比较接近，因此身高为 $1 . 8 7 \\mathrm { ~ m ~ }$ 的人作案的可能性更大. … (9分)", + "page_idx": 65 + }, + { + "type": "text", + "text": "18.解：由 $( m + n ) ^ { 2 } = 1 1 ^ { 2 }$ ，得 $m ^ { 2 } + 2 m n +$ $n ^ { 2 } = 1 2 1$ …·(3分)因为 $m n { = } 1$ ，所以 $m ^ { 2 } + 2 + n ^ { 2 } = 1 2 1$ 整理得 $m ^ { 2 } + n ^ { 2 } = 1 1 9$ ,·（7分)所以 $( m - n ) ^ { 2 } = m ^ { 2 } - 2 m n + n ^ { 2 } = 1 1 9$ $- 2 { = } 1 1 7$ …· (9分)", + "page_idx": 65 + }, + { + "type": "text", + "text": "19.解：因为 $B D , C D$ 的垂直平分线分别交 $A B , A C$ 于点 $E , F$ ，所以 $E B =$ $E D , F D = F C$ 所以 $\\angle E D B = \\angle B , \\angle F D C = \\angle C$ …· (4分)所以 $\\angle E D B + \\angle F D C = \\angle B + \\angle C$ 因为 $\\angle E D F = 1 8 0 ^ { \\circ } - ( \\angle E D B +$ $\\angle F D C ) , \\angle A = 1 8 0 ^ { \\circ } - ( \\angle B + \\angle C )$ 所以 $\\angle E D F = \\angle A = 6 8 ^ { \\circ }$ （9分）", + "page_idx": 65 + }, + { + "type": "text", + "text": "20.解：(1)因为 $B C / / D E$ ，所以 $\\angle A C B =$ $\\angle E .$ 在 $\\triangle A B C$ 和 $\\triangle D C E$ 中，$A C { = } D E$ $= D E , \\angle A C B = \\angle E , B C = C E$ 所以 $\\triangle A B C \\cong \\triangle D C E ( \\mathrm { S A S } )$ ，所以$A B { = } C D$ ;·.. ·…(4分)（2）因为 $\\triangle A B C { \\cong } \\triangle D C E$ $\\angle D =$ $3 0 ^ { \\circ }$ ，所以 $\\angle A = \\angle D = 3 0 ^ { \\circ }$ 因为 $\\angle E { = } 6 5 ^ { \\circ }$ ，所以 $\\angle A F E = 1 8 0 ^ { \\circ } -$ $3 0 ^ { \\circ } - 6 5 ^ { \\circ } = 8 5 ^ { \\circ }$ ，所以 $\\angle D F A = 1 8 0 ^ { \\circ } -$ $8 5 ^ { \\circ } = 9 5 ^ { \\circ }$ ，", + "page_idx": 65 + }, + { + "type": "text", + "text": "所以 $\\angle D G F = 1 8 0 ^ { \\circ } - 9 5 ^ { \\circ } - 3 0 ^ { \\circ } = 5 5 ^ { \\circ }$ 所以 $\\angle F G C = 1 8 0 ^ { \\circ } - 5 5 ^ { \\circ } = 1 2 5 ^ { \\circ }$ …· (9分)", + "page_idx": 65 + }, + { + "type": "text", + "text": "21.解：相等或互补. …（1分）理由：如图所示.", + "page_idx": 65 + }, + { + "type": "image", + "img_path": "images/f2beb444fe95ce8522b41d1460aa36419706105898518696946facab36e4af11.jpg", + "img_caption": [], + "img_footnote": [], + "page_idx": 65 + }, + { + "type": "text", + "text": "分情况讨论： \n当 $D G$ 边位于 $D E$ 上方时， \n因为 $D E ~ / / ~ A B$ ，所以 $\\angle B A C =$ $\\angle E F C$ \n因为 $D G / / A C$ ，所以 $\\angle E F C =$ $\\angle E D G$ ，所以 $\\angle B A C = \\angle E D G$ \n· (4分)当 $D G$ 边位于 $D E$ 下方时， \n因为 $D E ~ / / ~ A B$ ，所以 $\\angle B A C = { \\vdots }$ $\\angle A F D$ \n因为 $D G / / A C$ ，所以 $\\angle E D G ^ { \\prime } +$ $\\angle A F D = 1 8 0 ^ { \\circ }$ ，所以 $\\angle E D G ^ { \\prime } ~ + ~$ $\\angle B A C = 1 8 0 ^ { \\circ }$ \n即 $\\angle B A C$ 与 $\\angle E D G$ 的数量关系是相等或互补. （9分）", + "page_idx": 65 + }, + { + "type": "text", + "text": "22.解：(1)在8种等可能结果中，转出的数是4的倍数的结果有2种：4,8.则$P$ （小彬转出的数是4的倍数） $\\scriptstyle \\displaystyle { \\frac { 2 } { 8 } } =$ $\\frac { 1 } { 4 }$ （4分）", + "page_idx": 65 + }, + { + "type": "text", + "text": "（2)有两张分别写有3和5的卡片，设转盘停止后记下转出的数为 $\\mathcal { X }$ ", + "page_idx": 65 + }, + { + "type": "text", + "text": "要想组成三角形，则 $5 - 3 < x < 5 +$ 3，即 $2 { < } x { < } 8$ 在8种等可能结果中，能构成三角形的结果有5种： $3 , 3 , 5 ; 3 , 4 , 5 ; 3 , 5 , 5$ 3,5,6;3,5,7.则 $P$ （这三条线段能构成三角形) $= \\frac { 5 } { 8 }$ （10分）", + "page_idx": 65 + }, + { + "type": "text", + "text": "23.解：（1)由图象得， $M , N$ 两地的实际距离为 $6 0 0 ~ \\mathrm { k m } .$ 故填 600.···… (1分)(2)点 $C$ 的实际意义是乙车行驶 $4 \\textrm { h }$ 后两车相遇.故填乙车行驶 $4 \\textrm { h }$ 后两车相遇. ··(2分）(3)设甲车出发 $\\mathcal { X }$ 小时后发生故障，根据题意得， $4 \\times 1 0 0 + 8 0 x = 6 0 0$ ，解得 $x { = } 2 . 5$ 答：甲车出发2.5小时后发生故障.（4分）", + "page_idx": 65 + }, + { + "type": "text", + "text": "(4)设乙车出发 $a$ 小时后两车相距$2 0 0 \\ \\mathrm { k m } .$ 分情况讨论： \n$\\textcircled{1}$ 当甲、乙两车相遇前相距 $2 0 0 \\ \\mathrm { k m }$ 时，可得 $1 0 0 a + 8 0 a = 6 0 0 - 2 0 0$ ，解得 $a { = } \\frac { 2 0 } { 9 }$ （2 \n$\\textcircled{2}$ 当甲、乙两车相遇后相距 $2 0 0 \\ \\mathrm { k m }$ 时，可得 $8 0 ( a - 2 . 5 ) + 1 0 0 a = 6 0 0 +$ 200,解得 $a { = } \\frac { 5 0 } { 9 }$ （20 \n综上所述，当乙车出发 $\\frac { 2 0 } { 9 }$ h或 $\\frac { 5 0 } { 9 }$ h后，两车相距 $2 0 0 \\ \\mathrm { k m } ,$ （10分）", + "page_idx": 65 + }, + { + "type": "text", + "text": "考前练卷霸 ", + "text_level": 1, + "page_idx": 67 + }, + { + "type": "text", + "text": "成绩进步大", + "text_level": 1, + "page_idx": 67 + }, + { + "type": "text", + "text": "精编试题 卷中悟法 叁考答案 详细解析梳理重点知识 设置精巧点拨 规范学生答题 拓展学习思维掌握解题技巧 培育关键能力 养成良好习惯 提升学科素养", + "page_idx": 67 + } +] \ No newline at end of file diff --git a/juanba/juanba/auto/juanba_layout.pdf b/juanba/juanba/auto/juanba_layout.pdf new file mode 100644 index 0000000000000000000000000000000000000000..3d77ad0232e50d14872534055f289420313ff2db --- /dev/null +++ b/juanba/juanba/auto/juanba_layout.pdf @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid 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出版发行江西高校出版社
社 址江西省南昌市洪都北大道96号
总编室电话(0791)88504319，（0791)83829608（编辑部）
销售电话4008842220（全国免费热线），（0791)88175897
www. juacp.com
江西凯顺印务有限公司
址刷销本张数次全国新华书店
787mm×1092mm 1/8
8
网印经开印字版152千字
2023年2月第1版
2023年12月第2次印刷
书定 号价ISBN 978-7-5762-3604-0
28.00元
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题序二三评卷人总分
得分
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甲：无论a和b取何值， 等式均不能成立.乙：只有当a=0时，丙：当a=0或b=0 等式才成立.时,等式成立.
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学校
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题序二三评卷人总分
得分
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题序二三评卷人总分
得分
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学校
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"text": "密封线内不要答题°" + }, + { + "category_id": 15, + "poly": [ + 98.0, + 1360.0, + 130.0, + 1360.0, + 130.0, + 1662.0, + 98.0, + 1662.0 + ], + "score": 0.967, + "text": "学号填写在指定的位置上°" + }, + { + "category_id": 15, + "poly": [ + 17.0, + 1275.0, + 135.0, + 1275.0, + 135.0, + 1313.0, + 17.0, + 1313.0 + ], + "score": 0.999, + "text": "考生注意" + }, + { + "category_id": 16, + "poly": [ + 27.0, + 1331.0, + 52.0, + 1331.0, + 52.0, + 1358.0, + 27.0, + 1358.0 + ], + "score": 0.421, + "text": "D" + }, + { + "category_id": 15, + "poly": [ + 65.0, + 1326.0, + 100.0, + 1326.0, + 100.0, + 1704.0, + 65.0, + 1704.0 + ], + "score": 0.912, + "text": "②考试前，先将学校、班级、姓名" + }, + { + "category_id": 15, + "poly": [ + 22.0, + 1357.0, + 53.0, + 1357.0, + 53.0, + 1585.0, + 22.0, + 1585.0 + ], + "score": 0.959, + "text": "密封线内不要答题°" + }, + { + "category_id": 15, + "poly": [ + 98.0, + 1360.0, + 130.0, + 1360.0, + 130.0, + 1661.0, + 98.0, + 1661.0 + ], + "score": 0.975, + "text": "学号填写在指定的位置上°" + }, + { + "category_id": 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题序二三评卷人总分
得分
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题序二三评卷人总分
得分
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题序二三评卷人总分
得分
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题序二三评卷人总分
得分
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"如图所示的框图表示.小明同学任取一个自然数" + }, + { + "category_id": 15, + "poly": [ + 2251.0, + 209.0, + 2398.0, + 209.0, + 2398.0, + 254.0, + 2251.0, + 254.0 + ], + "score": 0.976, + "text": "输入求值." + }, + { + "category_id": 15, + "poly": [ + 169.0, + 865.0, + 728.0, + 865.0, + 728.0, + 910.0, + 169.0, + 910.0 + ], + "score": 0.96, + "text": "12.（10分)如图,可以随机在图中取点." + }, + { + "category_id": 15, + "poly": [ + 1533.0, + 266.0, + 1856.0, + 266.0, + 1856.0, + 311.0, + 1533.0, + 311.0 + ], + "score": 0.995, + "text": "(1)试写出与输出的数" + }, + { + "category_id": 15, + "poly": [ + 1886.0, + 266.0, + 2206.0, + 266.0, + 2206.0, + 311.0, + 1886.0, + 311.0 + ], + "score": 0.976, + "text": "有关的一个必然事件；" + }, + { + "category_id": 15, + "poly": [ + 173.0, + 145.0, + 1309.0, + 145.0, + 1309.0, + 191.0, + 173.0, + 191.0 + ], + "score": 0.984, + "text": "10.在一个不透明的袋子中装有除颜色外其余均相同的7个小球,其中红球2个，" + }, + { + "category_id": 15, + "poly": [ + 223.0, + 200.0, + 505.0, + 200.0, + 505.0, + 252.0, + 223.0, + 252.0 + ], + "score": 0.95, + "text": "黑球5个，若再放入" + }, + { + "category_id": 15, + "poly": [ + 541.0, + 200.0, + 1312.0, + 200.0, + 1312.0, + 252.0, + 541.0, + 252.0 + ], + "score": 0.965, + "text": "个一样的黑球并摇匀，此时,随机摸出一个球是黑球的" + }, + { + "category_id": 15, + "poly": [ + 217.0, + 266.0, + 352.0, + 266.0, + 352.0, + 331.0, + 217.0, + 331.0 + ], + "score": 0.999, + "text": "概率等于" + }, + { + "category_id": 15, + "poly": [ + 388.0, + 266.0, + 436.0, + 266.0, + 436.0, + 331.0, + 388.0, + 331.0 + ], + "score": 0.967, + "text": "，则" + }, + { + "category_id": 15, + "poly": [ + 472.0, + 266.0, + 584.0, + 266.0, + 584.0, + 331.0, + 472.0, + 331.0 + ], + "score": 1.0, + "text": "的值为" + }, + { + "category_id": 15, + "poly": [ + 1530.0, + 1145.0, + 2218.0, + 1145.0, + 2218.0, + 1186.0, + 1530.0, + 1186.0 + ], + "score": 0.978, + "text": "（1)小明购物180元，他获得奖金的概率是多少？" + }, + { + "category_id": 15, + "poly": [ + 1530.0, + 1202.0, + 2218.0, + 1202.0, + 2218.0, + 1243.0, + 1530.0, + 1243.0 + ], + "score": 0.976, + "text": "（2)小德购物210元，他获得奖金的概率是多少？" + }, + { + "category_id": 15, + "poly": [ + 1533.0, + 323.0, + 2613.0, + 323.0, + 2613.0, + 370.0, + 1533.0, + 370.0 + ], + "score": 0.971, + "text": "(2)若输入的数是2至9这八个连续正整数中的一个，求输出的数是3的倍" + }, + { + "category_id": 15, + "poly": [ + 1578.0, + 379.0, + 1734.0, + 379.0, + 1734.0, + 424.0, + 1578.0, + 424.0 + ], + "score": 0.986, + "text": "数的概率." + }, + { + "category_id": 15, + "poly": [ + 220.0, + 922.0, + 703.0, + 922.0, + 703.0, + 968.0, + 220.0, + 968.0 + ], + "score": 0.994, + "text": "(1)这个点取在阴影部分的概率是" + }, + { + "category_id": 15, + "poly": [ + 1479.0, + 916.0, + 2611.0, + 916.0, + 2611.0, + 961.0, + 1479.0, + 961.0 + ], + "score": 0.986, + "text": "15.视讲(12分)如图，端午节期间，某商场为了吸引顾客，设立了一个可以" + }, + { + "category_id": 15, + "poly": [ + 1532.0, + 971.0, + 2612.0, + 971.0, + 2612.0, + 1021.0, + 1532.0, + 1021.0 + ], + "score": 0.979, + "text": "自由转动的转盘，并规定顾客每购买200元的商品,就能获得一次转动转盘" + }, + { + "category_id": 15, + "poly": [ + 1528.0, + 1030.0, + 2611.0, + 1030.0, + 2611.0, + 1075.0, + 1528.0, + 1075.0 + ], + "score": 0.978, + "text": "的机会,如果转盘停止后，指针上对准红、黄、绿的区域，顾客就可以分别获" + }, + { + "category_id": 15, + "poly": [ + 1527.0, + 1087.0, + 2577.0, + 1087.0, + 2577.0, + 1133.0, + 1527.0, + 1133.0 + ], + "score": 0.995, + "text": "得50元、20元、10元的奖金，对准无色区域则无奖金(转盘等分成16份)." + }, + { + "category_id": 15, + "poly": [ + 167.0, + 1357.0, + 1306.0, + 1357.0, + 1306.0, + 1403.0, + 167.0, + 1403.0 + ], + "score": 0.972, + "text": "13.(10 分)在一个不透明的袋子里装有红、黄、白三种颜色的球共50个,它们除" + }, + { + "category_id": 15, + "poly": [ + 219.0, + 1412.0, + 1305.0, + 1412.0, + 1305.0, + 1462.0, + 219.0, + 1462.0 + ], + "score": 0.982, + "text": "了颜色不同外其余都相同,其中黄球比白球少5个,已知从袋子里随机摸出" + }, + { + "category_id": 15, + "poly": [ + 216.0, + 1482.0, + 552.0, + 1482.0, + 552.0, + 1546.0, + 216.0, + 1546.0 + ], + "score": 0.951, + "text": "一个球是红球的概率是" + }, + { + "category_id": 16, + "poly": [ + 547.0, + 1507.0, + 552.0, + 1507.0, + 552.0, + 1545.0, + 547.0, + 1545.0 + ], + "score": 0.152, + "text": 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题序二三评卷人总分
得分
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1069.0 + ], + "score": 0.999, + "text": "1.判定两直线平行的方法有五种：" + }, + { + "category_id": 15, + "poly": [ + 1435.0, + 1092.0, + 2004.0, + 1092.0, + 2004.0, + 1127.0, + 1435.0, + 1127.0 + ], + "score": 0.978, + "text": "(1)平行线的定义；(2)平行公理的推论：" + }, + { + "category_id": 15, + "poly": [ + 1435.0, + 1151.0, + 2005.0, + 1151.0, + 2005.0, + 1186.0, + 1435.0, + 1186.0 + ], + "score": 1.0, + "text": "如果两条直线都与第三条直线平行，那" + }, + { + "category_id": 15, + "poly": [ + 1437.0, + 1210.0, + 2005.0, + 1210.0, + 2005.0, + 1245.0, + 1437.0, + 1245.0 + ], + "score": 0.98, + "text": "么这两条直线平行；(3)同位角相等，两" + }, + { + "category_id": 15, + "poly": [ + 1433.0, + 1267.0, + 2005.0, + 1267.0, + 2005.0, + 1306.0, + 1433.0, + 1306.0 + ], + "score": 0.984, + "text": "直线平行；(4)内错角相等，两直线平行；" + }, + { + "category_id": 15, + "poly": [ + 1436.0, + 1329.0, + 2006.0, + 1329.0, + 2006.0, + 1364.0, + 1436.0, + 1364.0 + ], + "score": 0.97, + "text": "(5)同旁内角互补，两直线平行.判定两" + }, + { + "category_id": 15, + "poly": [ + 1434.0, + 1386.0, + 2006.0, + 1386.0, + 2006.0, + 1425.0, + 1434.0, + 1425.0 + ], + "score": 0.995, + "text": "直线平行时，定义一般不常用，其他四种" + }, + { + "category_id": 15, + "poly": [ + 1434.0, + 1446.0, + 2006.0, + 1446.0, + 2006.0, + 1484.0, + 1434.0, + 1484.0 + ], + "score": 0.998, + "text": "方法要灵活运用，证明时要注意书写" + }, + { + "category_id": 15, + "poly": [ + 1429.0, + 1502.0, + 1516.0, + 1502.0, + 1516.0, + 1547.0, + 1429.0, + 1547.0 + ], + "score": 0.986, + "text": "格式." + } + ], + "page_info": { + "page_no": 52, + "width": 2858, + "height": 2042 + } + }, + { + "layout_dets": [ + { + "category_id": 5, + "poly": [ + 854, + 1076, + 1427, + 1076, + 1427, + 1437, + 854, + 1437 + ], + "score": 0.97, + "html": "
9101112
15°答案不唯一，如∠A =∠EBC或∠D= ∠DCF或∠A+ ∠ABC=180°或∠D +∠BCD=180°等30°(180- 2a)°
" + }, + { + "category_id": 5, + "poly": [ + 870, + 880, + 1409, + 880, + 1409, + 997, + 870, + 997 + ], + "score": 0.967, + "html": "
12345678
DDADACDA
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表达方法特点
表格法多个变量可以同时出现在同一 张表格中
关系式法准确地反映了因变量与自变量 的数值关系
图象法直观、形象地给出了因变量随自 变量的变化趋势
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"(1)找出所要证明的边或角所在的" + }, + { + "category_id": 15, + "poly": [ + 2036.0, + 732.0, + 2159.0, + 732.0, + 2159.0, + 775.0, + 2036.0, + 775.0 + ], + "score": 0.994, + "text": "三角形；" + }, + { + "category_id": 15, + "poly": [ + 2104.0, + 793.0, + 2607.0, + 793.0, + 2607.0, + 832.0, + 2104.0, + 832.0 + ], + "score": 0.998, + "text": "(2)根据已知条件或结论证明三角" + }, + { + "category_id": 15, + "poly": [ + 2034.0, + 847.0, + 2159.0, + 847.0, + 2159.0, + 893.0, + 2034.0, + 893.0 + ], + "score": 0.998, + "text": "形全等；" + }, + { + "category_id": 15, + "poly": [ + 2106.0, + 912.0, + 2608.0, + 912.0, + 2608.0, + 947.0, + 2106.0, + 947.0 + ], + "score": 0.997, + "text": "(3)利用全等三角形的性质得出所" + }, + { + "category_id": 15, + "poly": [ + 2034.0, + 964.0, + 2155.0, + 964.0, + 2155.0, + 1012.0, + 2034.0, + 1012.0 + ], + "score": 0.931, + "text": "证结论." + }, + { + "category_id": 15, + "poly": [ + 123.0, + 702.0, + 460.0, + 702.0, + 460.0, + 741.0, + 123.0, + 741.0 + ], + "score": 0.947, + "text": "18.解：(1)由表格可得，" + }, + 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1597.0, + 1567.0, + 1597.0 + ], + "score": 0.961, + "text": "可构成三角形；(2)当" + }, + { + "category_id": 15, + "poly": [ + 1553.0, + 1620.0, + 1602.0, + 1620.0, + 1602.0, + 1655.0, + 1553.0, + 1655.0 + ], + "score": 0.999, + "text": "时，" + }, + { + "category_id": 15, + "poly": [ + 1695.0, + 1620.0, + 2004.0, + 1620.0, + 2004.0, + 1655.0, + 1695.0, + 1655.0 + ], + "score": 0.964, + "text": "可构成三角形；(3)当" + }, + { + "category_id": 15, + "poly": [ + 1454.0, + 1675.0, + 1582.0, + 1675.0, + 1582.0, + 1713.0, + 1454.0, + 1713.0 + ], + "score": 0.998, + "text": "最长，且" + }, + { + "category_id": 15, + "poly": [ + 1717.0, + 1675.0, + 1767.0, + 1675.0, + 1767.0, + 1713.0, + 1717.0, + 1713.0 + ], + "score": 0.999, + "text": "时，" + }, + { + "category_id": 15, + "poly": [ + 1863.0, + 1675.0, + 2006.0, + 1675.0, + 2006.0, + 1713.0, + 1863.0, + 1713.0 + ], + "score": 1.0, + "text": "可构成三" + }, + { + "category_id": 15, + "poly": [ + 1426.0, + 1727.0, + 1517.0, + 1727.0, + 1517.0, + 1775.0, + 1426.0, + 1775.0 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"(2)当" + }, + { + "category_id": 15, + "poly": [ + 949.0, + 1447.0, + 997.0, + 1447.0, + 997.0, + 1490.0, + 949.0, + 1490.0 + ], + "score": 0.994, + "text": "时，" + }, + { + "category_id": 15, + "poly": [ + 1199.0, + 1447.0, + 1280.0, + 1447.0, + 1280.0, + 1490.0, + 1199.0, + 1490.0 + ], + "score": 0.879, + "text": "，解得" + }, + { + "category_id": 15, + "poly": [ + 870.0, + 1505.0, + 1308.0, + 1505.0, + 1308.0, + 1545.0, + 870.0, + 1545.0 + ], + "score": 0.967, + "text": ",即行驶520千米时，油箱中的" + }, + { + "category_id": 15, + "poly": [ + 781.0, + 1561.0, + 1028.0, + 1561.0, + 1028.0, + 1602.0, + 781.0, + 1602.0 + ], + "score": 0.994, + "text": "剩余油量为8升." + }, + { + "category_id": 15, + "poly": [ + 1070.0, + 1620.0, + 1307.0, + 1620.0, + 1307.0, + 1657.0, + 1070.0, + 1657.0 + ], + "score": 0.962, + "text": "千米，故在开往" + }, + { + "category_id": 15, + "poly": [ + 781.0, + 1675.0, + 1307.0, + 1675.0, + 1307.0, + 1714.0, + 781.0, + 1714.0 + ], + "score": 0.999, + "text": "该加油站的途中，汽车开始提示加" + }, + { + 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0.921, + "text": "元." + }, + { + "category_id": 15, + "poly": [ + 781.0, + 374.0, + 1293.0, + 374.0, + 1293.0, + 412.0, + 781.0, + 412.0 + ], + "score": 0.999, + "text": "答：该用户5月份的水费是18.6元." + }, + { + "category_id": 15, + "poly": [ + 1142.0, + 428.0, + 1310.0, + 428.0, + 1310.0, + 473.0, + 1142.0, + 473.0 + ], + "score": 0.821, + "text": "·· (9分)" + }, + { + "category_id": 15, + "poly": [ + 1483.0, + 337.0, + 1571.0, + 337.0, + 1571.0, + 375.0, + 1483.0, + 375.0 + ], + "score": 0.796, + "text": "(2)设" + }, + { + "category_id": 15, + "poly": [ + 1888.0, + 337.0, + 1934.0, + 337.0, + 1934.0, + 375.0, + 1888.0, + 375.0 + ], + "score": 0.773, + "text": "，由" + }, + { + "category_id": 15, + "poly": [ + 1744.0, + 393.0, + 1936.0, + 393.0, + 1936.0, + 436.0, + 1744.0, + 436.0 + ], + "score": 0.936, + "text": "进行变形得，" + }, + { + "category_id": 16, + "poly": [ + 1746.0, + 451.0, + 1757.0, + 451.0, + 1757.0, + 494.0, + 1746.0, + 494.0 + ], + "score": 0.0, + "text": "" + }, + { + "category_id": 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2175.0, + 369.0, + 2084.0, + 369.0 + ], + "score": 0.999, + "text": "度内." + }, + { + "category_id": 15, + "poly": [ + 2479.0, + 323.0, + 2613.0, + 323.0, + 2613.0, + 366.0, + 2479.0, + 366.0 + ], + "score": 0.747, + "text": "：（9分）" + }, + { + "category_id": 15, + "poly": [ + 1433.0, + 153.0, + 1694.0, + 153.0, + 1694.0, + 193.0, + 1433.0, + 193.0 + ], + "score": 0.958, + "text": "17.解：如图，延长" + }, + { + "category_id": 15, + "poly": [ + 1750.0, + 153.0, + 1789.0, + 153.0, + 1789.0, + 193.0, + 1750.0, + 193.0 + ], + "score": 1.0, + "text": "交" + }, + { + "category_id": 15, + "poly": [ + 1847.0, + 153.0, + 1924.0, + 153.0, + 1924.0, + 193.0, + 1847.0, + 193.0 + ], + "score": 1.0, + "text": "于点" + }, + { + "category_id": 15, + "poly": [ + 1957.0, + 153.0, + 2009.0, + 153.0, + 2009.0, + 193.0, + 1957.0, + 193.0 + ], + "score": 0.989, + "text": "，则" + }, + { + "category_id": 15, + "poly": [ + 1483.0, + 210.0, + 1673.0, + 210.0, + 1673.0, + 249.0, + 1483.0, + 249.0 + ], + "score": 0.839, + 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985.0 + ], + "score": 1.0, + "text": "所以" + }, + { + "category_id": 15, + "poly": [ + 1811.0, + 942.0, + 2009.0, + 942.0, + 2009.0, + 985.0, + 1811.0, + 985.0 + ], + "score": 0.889, + "text": "（AAS)，所以" + }, + { + "category_id": 16, + "poly": [ + 1621.0, + 999.0, + 1632.0, + 999.0, + 1632.0, + 1041.0, + 1621.0, + 1041.0 + ], + "score": 0.0, + "text": "" + }, + { + "category_id": 15, + "poly": [ + 1484.0, + 1057.0, + 1554.0, + 1057.0, + 1554.0, + 1098.0, + 1484.0, + 1098.0 + ], + "score": 1.0, + "text": "因为" + }, + { + "category_id": 15, + "poly": [ + 1664.0, + 1057.0, + 1785.0, + 1057.0, + 1785.0, + 1098.0, + 1664.0, + 1098.0 + ], + "score": 0.94, + "text": "米，所以" + }, + { + "category_id": 15, + "poly": [ + 1895.0, + 1057.0, + 1944.0, + 1057.0, + 1944.0, + 1098.0, + 1895.0, + 1098.0 + ], + "score": 0.997, + "text": "米." + }, + { + "category_id": 15, + "poly": [ + 1838.0, + 1113.0, + 2011.0, + 1113.0, + 2011.0, + 1158.0, + 1838.0, + 1158.0 + ], + "score": 0.778, + "text": "… (9分)" + }, 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1031.0, + 264.0, + 1031.0, + 316.0, + 1011.0, + 316.0 + ], + "score": 0.414, + "text": "。" + }, + { + "category_id": 15, + "poly": [ + 1299.0, + 268.0, + 1435.0, + 268.0, + 1435.0, + 314.0, + 1299.0, + 314.0 + ], + "score": 0.801, + "text": " (2分)" + }, + { + "category_id": 15, + "poly": [ + 2158.0, + 1004.0, + 2500.0, + 1004.0, + 2500.0, + 1047.0, + 2158.0, + 1047.0 + ], + "score": 0.963, + "text": "14.解：(1)根据题意，得" + }, + { + "category_id": 15, + "poly": [ + 2534.0, + 1004.0, + 2699.0, + 1004.0, + 2699.0, + 1047.0, + 2534.0, + 1047.0 + ], + "score": 0.922, + "text": "(摸到红球)" + }, + { + "category_id": 15, + "poly": [ + 2417.0, + 1073.0, + 2615.0, + 1073.0, + 2615.0, + 1130.0, + 2417.0, + 1130.0 + ], + "score": 0.956, + "text": ",P(摸到白球)" + }, + { + "category_id": 16, + "poly": [ + 2674.0, + 1073.0, + 2687.0, + 1073.0, + 2687.0, + 1130.0, + 2674.0, + 1130.0 + ], + "score": 0.0, + "text": "" + }, + { + "category_id": 15, + "poly": [ + 2627.0, + 1156.0, + 2740.0, + 1156.0, + 2740.0, + 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1824.0, + 851.0 + ], + "score": 0.971, + "text": "，那么" + }, + { + "category_id": 16, + "poly": [ + 2055.0, + 791.0, + 2066.0, + 791.0, + 2066.0, + 870.0, + 2055.0, + 870.0 + ], + "score": 0.487, + "text": "（" + }, + { + "category_id": 15, + "poly": [ + 2160.0, + 1370.0, + 2735.0, + 1370.0, + 2735.0, + 1405.0, + 2160.0, + 1405.0 + ], + "score": 0.985, + "text": "15.解：(1)一个转盘被等分成六个扇形，" + }, + { + "category_id": 15, + "poly": [ + 2212.0, + 1425.0, + 2557.0, + 1425.0, + 2557.0, + 1463.0, + 2212.0, + 1463.0 + ], + "score": 1.0, + "text": "并在上面依次写上数字" + }, + { + "category_id": 16, + "poly": [ + 2724.0, + 1425.0, + 2736.0, + 1425.0, + 2736.0, + 1463.0, + 2724.0, + 1463.0 + ], + "score": 0.391, + "text": "." + }, + { + "category_id": 15, + "poly": [ + 2209.0, + 1482.0, + 2736.0, + 1482.0, + 2736.0, + 1521.0, + 2209.0, + 1521.0 + ], + "score": 0.99, + "text": "6,其中有3个扇形是奇数，所以自由" + }, + { + "category_id": 15, + "poly": [ + 2211.0, + 1539.0, + 2736.0, + 1539.0, + 2736.0, + 1577.0, + 2211.0, + 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2561.0, + 1843.0 + ], + "score": 0.747, + "text": "…·（10分）" + }, + { + "category_id": 15, + "poly": [ + 2162.0, + 1859.0, + 2414.0, + 1859.0, + 2414.0, + 1888.0, + 2162.0, + 1888.0 + ], + "score": 0.82, + "text": "16（1)加下主" + }, + { + "category_id": 16, + "poly": [ + 2632.0, + 1859.0, + 2736.0, + 1859.0, + 2736.0, + 1889.0, + 2632.0, + 1889.0 + ], + "score": 0.629, + "text": "（6）" + }, + { + "category_id": 15, + "poly": [ + 1622.0, + 943.0, + 2129.0, + 943.0, + 2129.0, + 982.0, + 1622.0, + 982.0 + ], + "score": 1.0, + "text": "概率是用来刻画随机事件发生的可" + }, + { + "category_id": 15, + "poly": [ + 1558.0, + 999.0, + 1793.0, + 999.0, + 1793.0, + 1038.0, + 1558.0, + 1038.0 + ], + "score": 1.0, + "text": "能性大小的一个" + }, + { + "category_id": 15, + "poly": [ + 1871.0, + 999.0, + 2128.0, + 999.0, + 2128.0, + 1038.0, + 1871.0, + 1038.0 + ], + "score": 0.992, + "text": "的常数，概率小则" + }, + { + "category_id": 15, + "poly": [ + 1557.0, + 1055.0, + 2129.0, + 1055.0, + 2129.0, + 1094.0, + 1557.0, + 1094.0 + ], + 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534.0, + 2212.0, + 534.0 + ], + "score": 0.956, + "text": "字是1的有2张，数字是2的有3" + }, + { + "category_id": 15, + "poly": [ + 2213.0, + 553.0, + 2739.0, + 553.0, + 2739.0, + 591.0, + 2213.0, + 591.0 + ], + "score": 0.957, + "text": "张，数字是3的有1张，所以摸到数" + }, + { + "category_id": 15, + "poly": [ + 2213.0, + 609.0, + 2739.0, + 609.0, + 2739.0, + 647.0, + 2213.0, + 647.0 + ], + "score": 0.989, + "text": "字2卡片的可能性最大，摸到数字3" + }, + { + "category_id": 15, + "poly": [ + 2215.0, + 665.0, + 2491.0, + 665.0, + 2491.0, + 703.0, + 2215.0, + 703.0 + ], + "score": 0.995, + "text": "卡片的可能性最小." + }, + { + "category_id": 15, + "poly": [ + 2600.0, + 666.0, + 2738.0, + 666.0, + 2738.0, + 705.0, + 2600.0, + 705.0 + ], + "score": 0.774, + "text": "…（4分）" + }, + { + "category_id": 15, + "poly": [ + 2213.0, + 722.0, + 2738.0, + 722.0, + 2738.0, + 761.0, + 2213.0, + 761.0 + ], + "score": 0.991, + "text": "(2)因为总共有6张卡片，其中数字" + }, + { + "category_id": 15, + "poly": [ + 2213.0, + 780.0, + 2740.0, + 780.0, + 2740.0, + 818.0, + 2213.0, + 818.0 + ], + "score": 0.99, + "text": "是奇数的有3张，数字是偶数的有3" + }, + { + "category_id": 15, + "poly": [ + 2213.0, + 838.0, + 2737.0, + 838.0, + 2737.0, + 873.0, + 2213.0, + 873.0 + ], + "score": 0.994, + "text": "张，所以摸到的数字是奇数和摸到的" + }, + { + "category_id": 15, + "poly": [ + 2210.0, + 892.0, + 2626.0, + 892.0, + 2626.0, + 932.0, + 2210.0, + 932.0 + ], + "score": 0.983, + "text": "数字是偶数的可能性一样大." + }, + { + "category_id": 15, + "poly": [ + 2594.0, + 948.0, + 2739.0, + 948.0, + 2739.0, + 991.0, + 2594.0, + 991.0 + ], + "score": 0.834, + "text": "… (8分)" + }, + { + "category_id": 15, + "poly": [ + 308.0, + 158.0, + 376.0, + 158.0, + 376.0, + 196.0, + 308.0, + 196.0 + ], + "score": 1.0, + "text": "所以" + }, + { + "category_id": 15, + "poly": [ + 491.0, + 158.0, + 607.0, + 158.0, + 607.0, + 196.0, + 491.0, + 196.0 + ], + "score": 1.0, + "text": "的周长" + }, + { + "category_id": 15, + "poly": [ + 397.0, + 272.0, + 440.0, + 272.0, + 440.0, + 311.0, + 397.0, + 311.0 + ], + "score": 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