生成4升5年纪孩子的学习视频，逐页讲解，并能与生活中的现象融会贯通。---**Textual Information & Chart/Diagram Descriptions**
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**Page Header:**
* **Topic:** 图形与几何 (Shapes and Geometry)
* **Sub-topic/Section:** 专题五 (Topic Five)
* **Series/Brand:** 学之舟 (Xuezizhou - Boat of Learning)
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**Main Content:**
The radius is usually represented by the letter `r`.
③ **Diameter:** A line segment that passes through the center of the circle and has both ends on the circle is called the diameter of the circle. The diameter is generally represented by the letter `d`. As shown in the figure:
**Chart Description (Figure 1: Diagram of Circle with Radius and Diameter)**
* **Type:** Geometric figure.
* **Main Elements:** A circle is depicted.
* A horizontal line segment across the circle, passing through the center, is labeled "直径(d)" (Diameter (d)).
* A line segment from the center to the top edge of the circle is labeled "半径(r)" (Radius (r)).
* The center of the circle is labeled "O".
* **Relative Position:** The radius is shown as extending from the center 'O' to the circumference, and the diameter is shown as extending through the center 'O' connecting two points on the circumference.
**Key Points (重点说明):**
* A circle can have infinitely many radii, and these radii have fixed and equal lengths.
* Three points not on the same straight line determine a circle.
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**2. Basic Properties of a Circle**
(1) A circle has infinitely many diameters and radii, and in the same circle or equal circles, all diameters (or radii) are equal in length.
(2) A circle is both a centrally symmetric figure and an axially symmetric figure. The center of symmetry is the center of the circle, and the axis of symmetry is the line containing any diameter.
→ Circles with equal radii are called equal circles.
(3) In the same circle or equal circles, the radius is half of the diameter, i.e.: `r = 1/2 d` or `d = 2r`.
(4) The position of a circle is determined by its center, and its size is determined by its radius (or diameter).
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**3. How to Draw a Circle**
(1) Spread the two legs of the compass, determine the distance between the two legs, which is the radius of the circle.
(2) Fix the leg with the needle point; this fixed point is the center of the circle.
(3) Rotate the leg with the pencil tip one full turn, and a circle is drawn.
**Chart Description (Figure 2: Compass Usage Illustration)**
* **Type:** Illustration/Diagram (圆规使用示意图 - Schematic Diagram of Compass Usage).
* **Main Elements:**
* A blue drawing compass is shown with its two legs spread.
* One leg, with a needle point, is fixed at a point. This point represents the center of the circle.
* The other leg, with a pencil tip, is shown in motion, drawing a circle on a flat surface.
* A dashed line extends from the fixed needle point to the circle's circumference, indicating the radius.
* **Labels/Annotations:** The dashed line representing the radius is labeled "2厘米" (2 centimeters).
* **Overall:** The illustration demonstrates how to use a compass to draw a circle of a specific radius.
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**4. Circumference and Area of a Circle**
(1) **Pi (Circumference Ratio):** The ratio of a circle's circumference to its diameter is called pi. Pi is represented by the letter "π" (read as pài), it is an infinite non-repeating decimal, `π = 3.1415926535...`, but in practical applications, its approximate value is often taken, i.e., `π ≈ 3.14`.
(2) **Circumference of a Circle:** The length of the curve that forms a circle is called its circumference. Because the circumference of a circle is `π` times its diameter, when we know the diameter or radius of the circle, we can calculate its circumference. That is, Circumference = Diameter × Pi or Circumference = 2 × Radius × Pi.
If `d` represents the diameter of the circle, `r` represents the radius of the circle, and `C` represents the circumference of the circle, the formula for the circumference of a circle is `C = πd` or `C = 2πr`.
(3) **Area of a Circle:** The size of the plane occupied by a circle is called its area.
**Chart Description (Figure 3: Circle with Multiple Radii/Diameters)**
* **Type:** Geometric figure.
* **Main Elements:** A circle is drawn with its center clearly marked. Multiple straight line segments are drawn, all passing through the center and connecting two points on the circumference. These lines divide the circle into several equal sectors.
* **Annotation:** "同圆内有半径、直径无数条，长度都相等。" (Inside the same circle, there are infinitely many radii and diameters, and their lengths are equal.) This text serves as an explanation for the visual representation.
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**Page Footer:**
* 学之舟 97 知识窗 (Xuezizhou 97 Knowledge Window)
**Extracted Content:**
**Top Banner:**
小学生知识通 / 小学数学 (Elementary School Knowledge Guide / Elementary Mathematics)
**Section: 学之舟 (Boat of Learning)**
**重点说明 (Key Explanation)**
半圆的周长包括圆周的一半和一条直径长，不要忘记直径长。
(The perimeter of a semicircle includes half of the circumference and the length of a diameter. Don't forget the diameter length.)
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**Main Content:**
**圆的面积公式 (Area Formula for a Circle):**
如果用 S 表示圆的面积，r 表示圆的半径，那么圆的面积的字母公式是 S = πr².
(If S represents the area of a circle and r represents its radius, then the formula for the area of a circle is S = πr².)
**5. 圆环的面积 (Area of Annulus)**
**(1) 圆环的定义 (Definition of Annulus)**
在半径不相等的两个同心圆中，两个圆之间的部分被称为圆环。
(In two concentric circles with unequal radii, the part between the two circles is called an annulus.)
如图阴影部分是一个圆环。在圆环中，大圆叫外圆，用 R 表示它的半径，小圆叫内圆，用 r 表示它的半径。
(The shaded part in the figure is an annulus. In an annulus, the larger circle is called the outer circle, and its radius is represented by R. The smaller circle is called the inner circle, and its radius is represented by r.)
**Chart Description (Annulus Diagram):**
* **Type:** Geometric figure, specifically a diagram of an annulus (concentric circles with the area between them shaded).
* **Main Elements:**
* Two concentric circles.
* Center point labeled 'O'.
* A line segment from O to the outer circle's circumference, labeled 'R', representing the radius of the outer circle.
* A line segment from O to the inner circle's circumference, labeled 'r', representing the radius of the inner circle.
* The region between the two circles is shaded, indicating the annulus.
**(2) 如图，圆环的面积 = 大圆面积 - 小圆面积。**
(As shown in the figure, Area of Annulus = Area of Large Circle - Area of Small Circle.)
如果用 R 来表示大圆的半径，r 表示小圆的半径，那么公式就可以表示为 S = πR² - πr² = π(R² - r²)。
(If R represents the radius of the large circle and r represents the radius of the small circle, then the formula can be expressed as S = πR² - πr² = π(R² - r²).)
**温馨提示 (Warm Tip):**
组成圆环的两个圆必是同心圆，它们的圆心一致，半径不同。
(The two circles that form an annulus must be concentric circles; their centers are the same, but their radii are different.)
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**考点六 扇形的认识 (Key Point Six: Understanding Sectors)**
**1. 认识扇形 (Understanding Sectors)**
**(1) 弧的定义 (Definition of Arc)**
连接圆上任意两点的部分叫做圆弧，也称为弧，表示符号是 "⌒"，以 A, B 为端点的弧记为 $\overgroup{AB}$，读作弧 AB 或圆弧 AB。
(The part connecting any two points on a circle is called an arc, also known as a circular arc. Its symbol is "⌒". An arc with endpoints A and B is denoted as $\overgroup{AB}$ and read as arc AB or circular arc AB.)
**(2) 圆心角的定义 (Definition of Central Angle)**
以圆心为顶点的角叫做圆心角。如图所示，∠AOB 是圆心角。
(An angle with its vertex at the center of a circle is called a central angle. As shown in the figure, ∠AOB is a central angle.)
**Chart Description (Central Angle Diagram):**
* **Type:** Geometric figure, specifically a circle with a central angle.
* **Main Elements:**
* A circle with center 'O'.
* Two points 'A' and 'B' on the circumference of the circle.
* Two radii, OA and OB, connecting the center O to points A and B respectively, forming angle AOB.
* The angle at the center is labeled 'O'.
**Section: 实例讲解 (Example Explanation)**
**扇形的认识 (Understanding Sectors)**
**Chart Description (Sector Diagram):**
* **Type:** Geometric figure, specifically a sector of a circle.
* **Main Elements:**
* A circle with center 'O'.
* Two radii, OA and OB, originating from the center O.
* Points 'A' and 'B' on the circumference.
* The arc connecting points A and B on the circumference.
* The region enclosed by the two radii and the arc is shaded, representing the sector.
* Labels: 'O' (center), '半径' (radius) pointing to OA and OB, '圆心角' (central angle) pointing to the angle at O, and '弧' (arc) pointing to the arc AB.
* A dotted line indicates the full circle from which the sector is part.
**重点说明 (Key Explanation)**
扇形是圆的一部分，经常放到圆中解决相关问题。
(A sector is part of a circle, often used to solve related problems within a circle context.)
**(3) 扇形的定义 (Definition of Sector)**
一条弧和经过这条弧两端的两条半径所围成的图形叫做扇形。
(A figure enclosed by an arc and the two radii originating from the ends of the arc is called a sector.)
可以把扇形看作圆的一部分。如图所示，扇形是阴影部分，OA、OB 是它的半径。
(A sector can be regarded as a part of a circle. As shown in the figure, the sector is the shaded part, and OA, OB are its radii.)
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**Page Header:**
* 图形与几何 (Graphics and Geometry)
* 专题五 (Topic Five)
**Diagram 1: Sector Illustration**
* **Type:** Geometric Figure (Circle with a sector highlighted)
* **Main Elements:**
* A circle with center labeled 'O'.
* Two radii originating from 'O' and extending to points 'A' and 'B' on the circumference.
* The region bounded by the two radii OA, OB and the arc AB is shaded green, representing a sector.
* **Labels:** O, A, B.
**2. 扇形的基本性质 (Basic Properties of Sectors)**
* (1) 扇形有且只有一条对称轴，对称轴是弧的中点和圆心确定的直线。
* (A sector has one and only one axis of symmetry. The axis of symmetry is the straight line determined by the midpoint of the arc and the center of the circle.)
* (2) 在同圆或等圆中，扇形的大小由它的圆心角决定。
* (In the same circle or congruent circles, the size of a sector is determined by its central angle.)
**3. 扇形的面积 (Area of Sectors)**
* 扇形是与圆形有关的一种重要图形，其面积与圆心角（顶角）、圆半径、弧长相关，如果用r表示扇形半径，l表示扇形弧长，S表示扇形面积，其面积公式如下：
* (A sector is an important figure related to circles. Its area is related to its central angle (vertex angle), radius, and arc length. If r represents the radius of the sector, l represents the arc length of the sector, and S represents the area of the sector, its area formulas are as follows:)
* 扇形的面积1 = 圆心角的度数 × (π × 半径 × 半径 / 360)
* 扇形的面积2 = (1/2) × 扇形半径 × 扇形弧长
* *Annotation related to arc length:* l = nπr / 180 (This formula is provided as a calculation for the arc length 'l' used in "扇形的面积2".)
* **计算公式 (Calculation Formulas):**
* S₁ = nπr² / 360
* S₂ = (1/2)rl
**考点七 平面图形的认识 (Knowledge Point Seven: Understanding Plane Figures)**
* (Usually perimeter is denoted by C, area by S.)
**1. 平面图形的周长和面积 (Perimeter and Area of Plane Figures)**
* (1) **周长:** 围成一个图形的所有边长的总和叫做这个图形的周长。
* (Perimeter: The sum of the lengths of all sides enclosing a figure is called the perimeter of that figure.)
* (2) **面积:** 物体表面或平面图形所占平面的大小叫做它的面积。
* (Area: The size of the plane occupied by an object's surface or a plane figure is called its area.)
* (3) 周长单位与长度单位相同：千米 (km)、米 (m)、分米 (dm)、厘米 (cm)、毫米 (mm)。
* (Perimeter units are the same as length units: kilometer (km), meter (m), decimeter (dm), centimeter (cm), millimeter (mm).)
* (4) 面积单位有：平方千米 (km²)、公顷 (hm²)、平方米 (m²)、平方分米 (dm²)、平方厘米 (cm²)。
* (Area units include: square kilometer (km²), hectare (hm²), square meter (m²), square decimeter (dm²), square centimeter (cm²).)
**2. 平面组合图形的面积 (Area of Composite Plane Figures)**
* (1) 平面组合图形的定义 (Definition of Composite Plane Figures)
* 三角形、长方形、正方形、平行四边形、梯形、菱形、圆和扇形等图形被称为基本图形或规则图形。
* (Triangles, rectangles, squares, parallelograms, trapezoids, rhombuses, circles, and sectors are called basic figures or regular figures.)
* 由两个或两个以上的基本图形组合而成的比较复杂的图形，叫...
* (More complex figures formed by combining two or more basic figures are called...) (Sentence is incomplete, likely continues on another page.)
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**Right Column Content:**
**学之舟 (Boat of Learning - series title)**
**特别提示 (Special Tip)**
* 半圆和四分之一圆是特殊的扇形。
* (A semicircle and a quarter circle are special types of sectors.)
* **扇形的周长:**
* (Perimeter of a sector:)
* C = 2r + nπr / 180 (n是扇形所对的圆心角)。
* (C = 2r + nπr / 180 (n is the central angle corresponding to the sector).)
**Diagram 2: Semicircle**
* **Type:** Geometric Figure (Semicircle)
* **Main Elements:** A simple representation of a half-circle. No labels are visible.
**重点说明 (Key Explanation)**
* 周长和面积的单位不同，因此它们永远不会相等，不能说边长为2的正方形周长面积相等。
* (Units of perimeter and area are different, so they will never be equal. It cannot be said that the perimeter and area of a square with side length 2 are equal.)
**图解知识 (Illustrated Knowledge)**
**Diagram 3: Football Field**
* **Type:** Image of a football field.
* **Main Elements:**
* A colored illustration of a football field.
* Text label below the image: 足球场面积 = 长 × 宽
* (Football field area = length × width)
**Page Footer:**
* 学之舟 99 知识通 (Boat of Learning 99 Knowledge Express)
**Header:**
小学生知识通 / 小学数学
**Content:**
**作组合图形 (Composing Composite Figures)**
**(2) 求组合图形面积的步骤 (Steps to find the area of composite figures)**
* ① 弄清所求的是组合图形哪些部分的面积。
(Clarify which parts of the composite figure are required to find the area of.)
* ② 根据图中条件联想各种简单图形的特征，看组合图形可以分成几块什么样的图形。能否找到应用公式的条件，能否通过平移、旋转和折叠等方法，使图形化繁为简。
(Based on the conditions in the figure, recall the characteristics of various simple figures, and see how the composite figure can be divided into several types of shapes. Can conditions for applying formulas be found? Can methods such as translation, rotation, and folding be used to simplify the figure?)
* ③ 一题能否多解，选择最简便的算法。
(Can a problem have multiple solutions? Choose the simplest algorithm.)
**3. 计算组合图形面积的思路和方法 (Ideas and methods for calculating the area of composite figures)**
**(1) 思路一：图形变换 (Idea One: Figure Transformation)**
* **① 平移 (Translation)**
* a. 点的移动 (等积变形)：根据“同底等高的三角形面积相等”和“平行线的一条直线上的任何一点到另一条直线的距离相等”可以把三角形的一个顶点在平行于所对边的平行线上移动，面积不变，从而把原来复杂的图形简化。
(Point movement (equi-areal transformation): According to "triangles with the same base and height have equal areas" and "the distance from any point on one parallel line to another parallel line is equal", a vertex of a triangle can be moved along a line parallel to its opposite side without changing its area, thus simplifying the original complex figure.)
* b. 面的移动 (平移法)：把某个图形沿直线进行平移，面积不发生变化，从而简化原来复杂的图形。
(Face movement (translation method): Translate a certain figure along a straight line without changing its area, thereby simplifying the original complex figure.)
* **② 旋转 (Rotation)**
* a. 以点为旋转中心 (旋转法)：将复杂图形的某一部分以一个固定点 (一般为定点) 为旋转中心，旋转适当的角度，从而简化原来复杂的图形。
(With a point as the center of rotation (rotation method): Rotate a part of a complex figure around a fixed point (generally a fixed point) by an appropriate angle, thereby simplifying the original complex figure.)
* b. 以直线为轴旋转 (折叠法)：将复杂图形的某一部分以一条直线 (一般为边或对角线) 为对称轴折叠，从而简化原来复杂的图形。
(With a line as the axis of rotation (folding method): Fold a part of a complex figure along a line (generally a side or diagonal) as the axis of symmetry, thereby simplifying the original complex figure.)
* **③ 对称 (Symmetry)**
* a. 对称添加 (扩大法)：将复杂的原图形以一条直线 (一般为边) 为对称轴翻折，把原图形和翻折后的图形看成一个整体，求出其面积，然后再求原图形的面积。
(Symmetrical addition (enlargement method): Fold the original complex figure along a straight line (generally a side) as the axis of symmetry, treat the original figure and the folded figure as a whole, find its area, and then find the area of the original figure.)
* b. 对称等分 (缩小法)：找到复杂图形的对称轴 (或对称中心)，把它平均分成若干个全等的小图形，再求出一个小图形的面积，最后再乘以份数得出总面积。
(Symmetrical division (reduction method): Find the axis of symmetry (or center of symmetry) of a complex figure, divide it into several congruent small figures, find the area of one small figure, and finally multiply by the number of parts to get the total area.)
**(2) 思路二：等量替换 (Idea Two: Equivalent Substitution)**
* 应用等价替换思想，把题中的问题或条件转换成其他的问题或条件，使问题或条件更直观明了。
(Apply the idea of equivalent substitution to convert the problem or conditions in the question into other problems or conditions, making the problem or conditions more intuitive and clear.)
**重点说明 (Key Points/Important Notes)**
* **分割法 (Division Method):**
* 把一个组合图形根据它的特征和已知条件分割成几个简单的规则图形。
(Divide a composite figure into several simple regular figures according to its characteristics and known conditions.)
* **重叠法 (Overlap Method):**
* 就是把所求阴影部分的面积问题转化为可求面积的规则图形的重叠部分的方法。
(It is a method of transforming the problem of finding the area of the shaded part into finding the overlapping part of regular figures whose areas can be found.)
**Footer:**
学之舟 100 知识通
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**General Information**
* **Topic Title:** 图形与几何 专题五 (Graphics and Geometry - Special Topic 5)
* **Publisher/Section Label (Right Banner):** 学之舟 (Boat of Learning)
* **Page Footer:** 学之舟 101 知识通 (Boat of Learning 101 Knowledge Express)
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**Section: Problem-Solving Strategies for Geometry**
**1. 割补法 (Cutting and Patching Method)**
* **Explanation:** 把图形的某一部分分割下来补到另一部分上，使它变成一个我们已学过的几何图形，然后再进行计算。(Cut a part of the figure and patch it onto another part, making it a geometric figure we have already learned, and then calculate.)
**2. (3) 思路三: 从整体看问题 (Thought Process 3: Viewing the Problem Holistically)**
* **Explanation:** 从整体上分析条件和问题，不纠结题目中的细枝末节，抓主要矛盾，以特殊情况验证一般情况，从而找准切入点，确定合适的方法，解决问题。(Analyze the conditions and problems from a holistic perspective, without getting bogged down in the minor details of the problem. Focus on the main contradictions, verify general cases with specific situations, thereby finding the right entry point and determining appropriate methods to solve the problem.)
**3. 挖空法 (Hollowing Out Method)**
* **Explanation:** 把多边形看成是一个完整的规则图形，计算它的面积以后，再减去空缺部分的面积。(Consider the polygon as a complete regular figure, calculate its area, and then subtract the area of the missing part.)
**4. (4) 思路四: 灵活运用公式 (Thought Process 4: Flexible Use of Formulas)**
* **Explanation:** 当不能直接用公式解决问题时，就要考虑对公式进行恒等变形，灵活运用公式解决问题。(When a problem cannot be solved directly with a formula, consider performing equivalent transformations on the formula and flexibly applying it to solve the problem.)
**5. 计算阴影部分的面积思路 (Ideas for Calculating the Area of Shaded Regions)**
* **Introduction:** 计算阴影部分的面积常用的思路是两种，一种是直接求面积，一种是间接求面积。常用的方法有以下几种: (There are two common approaches for calculating the area of shaded regions: one is direct calculation, and the other is indirect calculation. The common methods are as follows:)
* **(1) 直接求法 (Direct Method):** 这种方法是根据已知条件，从整体出发直接求出阴影部分的面积。(This method directly calculates the area of the shaded region based on given conditions, starting from the whole.)
* **(2) 相减法 (Subtraction Method):** 这种方法是将所求的阴影部分的面积看成是若干个基本规则图形的面积之差。(This method views the area of the shaded region as the difference in areas of several basic regular geometric figures.)
* **(3) 重新组合法 (Recombination Method):** 这种方法是将阴影部分图形拆开，根据具体情况重新组合成一个新的图形，设法求出这个新图形的面积即可。(This method disassembles the shaded figure and recombines it into a new figure based on the specific situation. Then, one tries to find the area of this new figure.)
* **(4) 辅助线法 (Auxiliary Line Method):** 这种方法是根据具体情况在图形中加一条或若干条辅助线，使阴影部分转化成若干个基本规则图形，然后再采用相加、相减法解决就可以。(This method involves adding one or more auxiliary lines to the figure based on the specific situation, transforming the shaded region into several basic regular geometric figures, and then solving it by addition or subtraction.)
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**Section: 重点说明 (Key Explanation)**
**Title:** 对称图形的解题方法 (Problem-solving Methods for Symmetric Figures)
**Diagram 1 Description:**
* **Type:** Geometric figure within a square.
* **Main Elements:**
* A square with side length labeled as '4'.
* Inside the square, there are four petal-like shapes formed by arcs, arranged symmetrically around the center, resembling an 'X' pattern. These petal shapes are implied to be the shaded area for calculation.
* Dimensions are indicated by arrows: A vertical arrow on the left side pointing upwards, spanning the height, labeled '4'. A horizontal arrow on the bottom side pointing rightwards, spanning the width, labeled '4'.
* **Textual Explanation:** 根据所求图形的对称性，将所求图形面积平均分成若干份，先求出其中的一份面积，然后求出总面积。(Based on the symmetry of the desired figure, divide the area of the figure into several equal parts, first find the area of one part, and then find the total area.)
**Diagram 2 Description:**
* **Type:** Right-angled triangle.
* **Main Elements:**
* A right-angled triangle with its right angle at the bottom-left vertex.
* The vertical side (height) is labeled '2'.
* The horizontal side (base) is labeled '2'.
* Dimensions are indicated by arrows: A vertical arrow on the left side pointing upwards, spanning the height, labeled '2'. A horizontal arrow on the bottom side pointing rightwards, spanning the width, labeled '2'.
* **Textual Explanation:** 分析: 把原图平均分成八分,就得到上图。(Analysis: Divide the original figure into eight equal parts to get the figure above.)
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**Section: 真题解析 (Real Problem Analysis)**
**Question:**
* **(海南小考)** 在一条线段中间另有6个点，则这8个点可以构成( )条线段。( (Hainan Primary School Exam) There are 6 other points between two endpoints of a line segment. How many line segments can these 8 points form?)
**Options:**
* A. 21
* B. 28
* C. 36
**Answer:**
* [答案]B ([Answer]B)
---
**Section: 真题回放 (Real Problem Review)**
**Question (Example 1):**
* **Title:** 例1 (龙华小考) (Example 1 (Longhua Primary School Exam))
* **Question Stem:** 下图中有 ( ) 条线段。(How many line segments are there in the figure below?)
**Diagram 3 Description (for Example 1):**
* **Type:** Line segment with marked points.
* **Main Elements:**
* A single horizontal straight line segment.
* Four distinct points are marked on this line segment. These points include the two endpoints and two points in between.
* Small vertical tick marks indicate the precise positions of these four points.
**Options:**
* A. 4
* B. 6
* C. 10
* D. 12
Here is the extracted content from the image:
**General Information:**
* **Subject:** 小学生知识通 / 小学数学 (Elementary Student Knowledge Guide / Elementary Mathematics)
* **Page Number:** 102
---
**Section 1: Analysis for a Previous Question (Question Stem not visible)**
* **Section Title:** 解析 (Analysis)
* **Textual Information:**
独立的线段有4条，两条线段合并的线段有3条，三条线段合并的线段有2条，四条线段合并的线段有1条，据此列加法计算即可。
* **Mathematical Formula:**
4 + 3 + 2 + 1 = 10 (条)
* **Answer:**
【答案】C
---
**Section 2: Question 2 (例2)**
* **Question Stem:**
例2 (龙华小考) 下列说法中，正确的是( )。
* **Options:**
* A. 直角三角形的两条直角边互相垂直
* B. 三角形三个内角中可以有一个钝角，一个直角
* C. 两个锐角的和一定比直角大
* D. 周角的大小是平角的4倍
* **Annotation for Option D:**
→周角是360°，平角是180°。
* **Section Title:** 解析 (Analysis)
* **Explanation:**
在直角三角形中，直角三角形的两条直角边互相垂直；三角形的内角和是180°，三角形三个内角中最多只有1个钝角或1个直角；在钝角三角形中，两个锐角的和小于直角；在直角三角形中，两个锐角的和等于直角；周角=360°，平角=180°，一个周角=2个平角，据此判断。
选项A，直角三角形的两条直角边互相垂直，原题说法正确；
选项B，三角形三个内角中最多只有1个钝角或1个直角，原题说法错误；
选项C，在钝角三角形中，两个锐角的和小于直角，在直角三角形中，两个锐角的和等于直角，原题说法错误；
选项D，周角的大小是平角的2倍，原题说法错误。
* **Answer:**
【答案】A
---
**Section 3: Example Explanation (实例讲解)**
* **Section Title:** 实例讲解 (Example Explanation)
* **Textual Information (Highlighted Box):**
所有三角形的内角和是等大的
* **Textual Information:**
无论是锐角三角形、直角三角形还是钝角三角形，它们的内角和都是180°。
* **Chart/Diagram Description:**
* **Type:** Geometric figures - Three triangles.
* **Main Elements:**
* Three distinct triangles are displayed side-by-side.
* The leftmost triangle appears to be an acute triangle (all angles less than 90 degrees).
* The middle triangle appears to be an obtuse triangle (one angle greater than 90 degrees).
* The rightmost triangle is explicitly shown as a right-angled triangle, with a small square symbol indicating a 90-degree angle.
* **Labels and Annotations:** No specific vertex labels (e.g., A, B, C) or angle values are provided on the triangles themselves, apart from the right angle symbol.
---
**Section 4: Question from "真题解析" (True Question Analysis)**
* **Section Title:** 真题解析 (True Question Analysis)
* **Question Stem:**
(广东小考) 把一段圆柱木料锯成三段，增加了 ( )个底面积。
* **Options:**
* A.3
* B.4
* C.6
* D.2
* **Answer:**
【答案】B
---
**Section 5: Question 3 (例3 - Judgment Questions)**
* **Section Title:** 例3 判断题 (Example 3 Judgment Questions)
* **Sub-question 1:**
1. (保定小考) 两个三角形拼在一起组成一个四边形，它的内角和是360°。( )
* **Sub-question 2:**
2. (吉水小考) 角的大小与角的两边叉开的大小有关，与角的两边的长短无关。( )
* **Sub-question 3:**
3. (京山小考) 一个扇形的圆心角是120°，它的面积是所在圆 (Content is cut off and incomplete.)
**Extracted Content:**
**Section 1: 图形与几何 (Graphics and Geometry) - 专题五 (Special Topic 5)**
**Problem 4:**
* **Question Stem:** (商丘小考)在同一平面内的两条直线不是相交，就是平行。( )
* **Other Relevant Text:** 垂直也是相交的一种特殊情况。
* **Analysis (解析):**
* 4. 在同一平面内的两条直线不是相交，就是平行，此题说法正确。
**Problem 5:**
* **Question Stem:** (历下小考)任何面积相等的三角形一定能拼成一个平行四边形。( )
* **Other Relevant Text:** 面积相等不代表完全相同。
* **Analysis (解析):**
* 5. 两个完全一样的三角形能拼成一个平行四边形，而面积相等的三角形不一定完全一样。面积相等的三角形不一定能拼成一个平行四边形。
**General Analysis (解析) for Problems 1-3 (partial content provided in image):**
* **Analysis for Problem 1:**
* 1. 三角形的内角和是180°，四边形的内角和是360°。据此判断，两个三角形拼在一起组成一个四边形，它的内角和是360°，原题说法正确。
* **Analysis for Problem 2:**
* 2. 此题主要考查了角的大小比较，角的大小与角的两边叉开的大小有关，与角的两边的长短无关，据此判断此题说法正确。
* **Analysis for Problem 3:**
* 3. 扇形的面积 = (扇形圆心角的度数 ÷ 360) × 圆的面积，据此作答即可。一个扇形的圆心角是120°，它的面积是所在圆面积的 1/3。
**Overall Answers for Problems 1-5 (from Analysis section):**
* **【答案】** 1.√; 2.√; 3.√; 4.√; 5.×
---
**Section 2: 真题解析 (Real Exam Analysis)**
**Problem 1 (Guangdong Exam):**
* **Question Stem:** (广东小考)把一个长6厘米，宽4厘米的长方形按2∶1放大后，得到的图形的面积是( )平方厘米。
* **Options:**
* A.48
* B.24
* C.96
* D.72
* **Answer:** 【答案】C
**Problem 2 (Beijing Exam - 例4):**
* **Question Stem:** 例4 (北京小考) 如图，∠AOB的顶点O在直线l上，已知图中所有小于平角的角之和是400度，则∠AOB=____度。
* **Other Relevant Text:** 所有小于平角的角有5个。
* **Chart/Diagram Description:**
* **Type:** Geometric figure illustrating angles formed by rays originating from a point on a straight line.
* **Main Elements:**
* **Point O:** A central point, labeled 'O', located on line 'l'.
* **Line l:** A straight line passing through point 'O'.
* **Rays:**
* Ray OA: Extends upwards and to the left from O, labeled 'A' at its end.
* Ray OB: Extends upwards and to the right from O, labeled 'B' at its end.
* **Labels:** Points A, B, O, and line l are labeled.
* **Relative Position:** Ray OA and Ray OB originate from O and are on one side of line l (above it), forming an angle ∠AOB. The line l itself forms two straight angles (or 180-degree angles) with rays extending from O along the line. Implicitly, there are angles formed by combinations of rays OA, OB, and the line l (e.g., angle between OA and the left part of l, angle between OB and the right part of l, and angle AOB).
* **Analysis (解析):**
* 三个较小的角的和是一个平角，则用“总度数-180° = 剩余两个角的度数”；剩余两个角的度数比平角多了一个∠AOB, 故∠AOB的度数=剩余两个角的度数-180°。
* 400 - 180 - 180
* = 220 - 180
* = 40 (度)
**Problem 3 (Gansu Exam):**
* **Question Stem:** (甘肃小考)下面图形中，只有一条对称轴的是 ( )。
* **Options:**
* A.等腰三角形 (Isosceles triangle)
* B.长方形 (Rectangle)
* C.正方形 (Square)
* **Answer:** 【答案】A
Here is the extracted content from the image:
---
**Page Header:**
* 小学生知识通 (Primary School Knowledge Pass)
* 小学数学 (Primary School Mathematics)
---
**Section: 真题解析 (Real Question Analysis)**
**Problem 1:**
* **Question Stem:** (甘肃小学考) 甲圆半径是乙圆半径的2倍，则甲圆的面积是乙圆面积的 ( ) 倍。
* *(Translation: (Gansu Primary School Exam) The radius of circle A is 2 times the radius of circle B, then the area of circle A is ( ) times the area of circle B.)*
* **Options:**
* A. 2
* B. 4
* C. 8
* **Answer:**
* 【答案】B
---
**Section: 真题解析 (Real Question Analysis)**
**Problem 2:**
* **Question Stem:** (甘肃小学考) 一个圆柱的底面半径2厘米，高3厘米。它的表面积是( )平方厘米。(π取3.14)
* *(Translation: (Gansu Primary School Exam) A cylinder has a base radius of 2 cm and a height of 3 cm. Its surface area is ( ) square cm. (Take π as 3.14))*
* **Options:**
* A. 62.8
* B. 31.4
* C. 78.5
* D. 42.8
* **Answer:**
* 【答案】A
---
**Section: 例5 (Example 5)**
* **Context:** 【答案】40 (This seems to be the answer to a previous question not fully shown on the page.)
* **Question Stem:** (房县小学考) 一个三角形的三条边长度和是35厘米，三条边长度之比是2 : 3 : 2, 这个三角形最长的边是____厘米。按____边分类，它是____三角形。
* *(Translation: (Fangxian Primary School Exam) The sum of the lengths of the three sides of a triangle is 35 cm, and the ratio of the lengths of the three sides is 2 : 3 : 2. The longest side of this triangle is ____ cm. According to side classification, it is a ____ triangle.)*
* **Annotation/Hint:** →按照比例有两个边长度相等，为等腰三角形。(According to the ratio, two sides have equal length, so it's an isosceles triangle.)
* **Analysis (解析):**
* **Textual Explanation:** 已知三角形的周长与三边的长度比，要求三角形最长的边的长度，用三角形的周长×最长边占三角形周长的分率=三角形最长的边的长度；要求这个三角形按边分是什么三角形，比较三边的长度，两边相等的三角形是等腰三角形，三边相等的三角形是等边三角形，据此列式解答。
* *(Translation: Given the perimeter of the triangle and the ratio of its three sides, to find the length of the longest side, use: perimeter of the triangle × (fraction of longest side in perimeter) = length of the longest side; to find what kind of triangle it is by side classification, compare the lengths of the three sides. A triangle with two equal sides is an isosceles triangle, and a triangle with three equal sides is an equilateral triangle. Solve using these principles.)*
* **Mathematical Formulas/Calculations:**
* $35 \times \frac{3}{2+3+2} = 15 (\text{厘米})$
* $35 \times \frac{2}{2+3+2} = 10 (\text{厘米})$
* **Conclusion:** 按边分类，它是等腰三角形。
* *(Translation: Classified by sides, it is an isosceles triangle.)*
* **Answer:**
* 【答案】15; 等腰 (15; isosceles)
---
**Section: 例6 (Example 6)**
* **Question Stem:** (北京小学考) 用10根火柴棒首尾顺次连接成一个三角形，能接成不同的三角形有____个。
* *(Translation: (Beijing Primary School Exam) Using 10 matchsticks connected end-to-end to form a triangle, how many different triangles can be formed?)*
* **Analysis (解析):**
* **Textual Explanation:** 10根小棒可以分成:
* *(Translation: 10 small sticks can be divided into:)*
* **List of Combinations and Outcomes:**
* 1, 1, 8, 不能组成三角形; (1, 1, 8, cannot form a triangle;)
* 1, 2, 7, 不能组成三角形; (1, 2, 7, cannot form a triangle;)
* 1, 3, 6, 不能组成三角形; (1, 3, 6, cannot form a triangle;)
* 1, 4, 5, 不能组成三角形; (1, 4, 5, cannot form a triangle;)
* 2, 2, 6, 不能组成三角形; (2, 2, 6, cannot form a triangle;)
* 2, 3, 5, 不能组成三角形; (2, 3, 5, cannot form a triangle;)
* 2, 4, 4, 能组成三角形; (2, 4, 4, can form a triangle;)
* 3, 3, 4, 能组成三角形。 (3, 3, 4, can form a triangle.)
* **Answer:**
* 【答案】2
---
**Section: 例7 (Example 7)**
* **Question Stem:** (龙华小学考) 下图中，∠1+∠2+∠3____180°，a+b____co。(括号里填 “>” “<” 或 “=” )
* *(Translation: (Longhua Primary School Exam) In the figure below, ∠1+∠2+∠3____180°, a+b____c. (Fill in ">", "<" or "=" in the parentheses))*
* **Chart/Diagram Description:**
* **Type:** Geometric figure (implied, as the problem refers to "the figure below"). The specific diagram is not visible in the provided image.
* **Main Elements:**
* **Angles:** ∠1, ∠2, ∠3. Based on common mathematical problems of this type and the "重点说明" sections on triangles, these are highly likely to be the interior angles of a triangle.
* **Variables/Sides:** a, b, c. These likely represent the lengths of the sides of a triangle, corresponding to the angles.
* **Labels and Annotations:** ∠1, ∠2, ∠3, a, b, c.
* **Relative Position and Direction:** Not discernible from the partial image, but implied to be part of a single geometric figure, most probably a triangle.
* **Missing Information:** The actual diagram is not present in the image.
---
**Section: 重点说明 (Key Points)**
* **Point 1 (Triangle Inequality):** 根据三角形的三边关系判断，任意两边之和大于第三边，任意两边之差小于第三边。
* *(Translation: Judge based on the triangle inequality theorem: the sum of any two sides is greater than the third side, and the difference of any two sides is less than the third side.)*
* **Point 2 (Sum of Angles in a Triangle):** 三角形三个内角的度数之和是180°，与三角形的形状没有关系。
* *(Translation: The sum of the three interior angles of a triangle is 180°, which is independent of the shape of the triangle.)*
---
Here is the extracted content from the image, structured as requested:
---
**Section 1: Triangle Properties**
**Chart Description:**
* **Type:** Geometric figure (triangle).
* **Main Elements:**
* A triangle with three interior angles labeled '1', '2', and '3'.
* The sides opposite angles '1', '2', and '3' are labeled 'a', 'c', and 'b' respectively. (Angle '1' is opposite side 'a', Angle '2' is opposite side 'c', Angle '3' is opposite side 'b').
**Other Relevant Text (Explanation/Concept Review):**
* **解析 (Analysis):**
* 在三角形中，三个内角的和是180°，∠1、∠2、∠3分别是这个三角形的三个内角，所以它们的和是180°；三角形中，任意两边之和大于第三边，任意两边之差小于第三边，据此解答。
* 图中，∠1+∠2+∠3=180°, a+b>c。
* **【答案】:** =; >
---
**Section 2: Largest Square from Rectangle**
**Question Stem:**
(安徽小考)用一块长是10厘米，宽是8厘米的长方形厚纸板，剪出一个最大的正方形，这个正方形的面积是( )平方厘米。
**Options:**
A.80
B.40
C.64
**Other Relevant Text (Solution Hint):**
* **真题解析 (Authentic Problem Analysis):** (No detailed text provided, only the question and options)
**Answer:**
【答案】C
---
**Section 3: Folded Rectangle Perimeter**
**Question Stem:**
例8 (安国小考)一个长6厘米，宽2.4厘米的长方形，沿对角线折叠后，得到如图所示几何图，阴影部分的周长是_________厘米。
**Chart Description:**
* **Type:** Geometric figure (rectangle with internal line and shading).
* **Main Elements:**
* A rectangle with vertices labeled A (top right), B (bottom right), C (bottom left), D (top left).
* A point E is located on the side CD (the top side of the rectangle based on labeling).
* A line segment BE connects point B (bottom right) to point E (on top left side).
* The region defined by vertices B, C, E is shaded green, forming a triangle BCE.
* *Note on potential inconsistency:* The diagram's vertex labeling (A top right, D top left) is unconventional for a rectangle, and the description "沿对角线折叠后" (folded along diagonal) is hard to reconcile directly with the depicted shaded triangle BCE without further assumptions or a different initial figure. However, the solution below calculates the perimeter of the original rectangle.
**Other Relevant Text (Explanation):**
* **解析 (Analysis):**
* 观察图形, BE和CD都是长方形的长, 都是6厘米, CE和BD都是长方形的宽, 都是2.4厘米, 阴影部分的周长就是两个长加上两个宽。
* (6+6) + (2.4+2.4) = 12 + 4.8 = 16.8 (厘米)
**Answer:**
【答案】16.8
---
**Section 4: Counting Triangles**
**Question Stem:**
(广东小考)下图一共有______个三角形。
**Chart Description:**
* **Type:** Geometric figure (fan-like arrangement of lines).
* **Main Elements:**
* A central point labeled O.
* A horizontal line segment below O with points A01, A02, A03, A04, A05, A06 marked from left to right.
* A vertical line segment to the right of O with points A07, A08, A09, A10, A11, A12 marked from bottom to top.
* Lines radiate from point O to all points A01 through A06, forming triangles with the horizontal base.
* Lines radiate from point O to all points A07 through A12, forming triangles with the vertical line segment.
**Other Relevant Text (Explanation):**
* **【考点】** 组合图形的计数。(Knowledge Point: Counting combined figures)
* **【解析】** 先看顶点O在上面的三角形，一共有5+4+3+2+1=15个三角形，再看顶点O在左边的三角形一共共有6+5+4+3+2+1=21个，据此加起来，再加上大三角形即可解答问题。
* **【解答】** 根据题干分析可得: 顶点O在上面的三角形，一共有5+4+3+2+1=15(个)顶点O在左边的三角形一共有6+5+4+3+2+1=21(个)15+21+1=37(个)。故答案为:37。
**Answer:**
【答案】37
---
**Section 5: Circle and Rectangle Area/Perimeter**
**Question Stem:**
例9 (南通小考)如图, 圆和长方形的面积相等，则阴影部分的面积与圆面积的比为_________。如果阴影部分的周长为30厘米，则圆周长为_________厘米。
**Chart Description:**
* **Type:** Geometric figure (circle and rectangle).
* **Main Elements:**
* A circle with its center labeled O.
* A rectangle is positioned to the right of the circle.
* The rectangle's left vertical side aligns with the circle's vertical radius extending downwards from O.
* The rectangle's top horizontal side aligns with the circle's horizontal radius extending rightwards from O.
* The rectangle itself is shaded green. This indicates it is the "阴影部分" (shaded part) as visually depicted.
**Other Relevant Text (Explanation):**
* **解析 (Analysis):**
* 由题意知: 圆的面积 = 长方形的面积, 所以阴影部分的面积等于圆的面积的1-1/4=3/4。若设圆的面积是单位“1”，则阴影部分的面积就是3/4，所以阴影部分的面积与圆面积的比是: 3/4。
* *Note on potential inconsistency:* The problem statement says "圆和长方形的面积相等" (Area of circle = Area of rectangle). However, the explanation then states "阴影部分的面积等于圆的面积的1-1/4=3/4" (Area of shaded part = 3/4 of circle's area). This implies the unshaded part (which appears to be a quarter circle occupied by the rectangle if the rectangle's dimensions are r x r) represents 1/4 of the circle's area. If the rectangle's area is 1/4 of the circle's area, it contradicts the initial premise that the circle's area and rectangle's area are equal (unless both are zero). The explanation seems to define "shaded part" as the circle minus a quarter, rather than the rectangle itself (which is visually shaded).
* The solution for the second part of the question (circumference based on shaded part's perimeter) is incomplete in the provided image.
---
Here is the extracted content from the image, organized as requested:
---
**1. Problem Example/Solution Snippet (Partial Question)**
* **Other Relevant Text (Solution Explanation):**
÷ 1 = 3 : 4; 圆的半径是r, 则圆的面积 = πr² = 长方形的面积, 又因为长方形的宽 = 圆的半径 = r, 所以长方形的长 = πr² ÷ r = πr。 阴影部分的周长 = 长 + 宽 + (长 - 半径) + 1/4 圆周长 = 2πr + 1/2 πr = 30 厘米, 得 πr = 12 厘米, 圆周长 = 2πr = 24 厘米。
* **Answer:**
【答案】3 : 4; 24
---
**2. Problem: Calculating Shaded Area**
* **Question Stem:**
(甘肃小考)计算阴影部分的面积。
(Gansu Primary School Exam) Calculate the area of the shaded part.
* **Chart/Diagram Description:**
* **Type:** Geometric figure (composite shape).
* **Main Elements:** The diagram shows a shape composed of a right trapezoid and a quarter-circle. The base of the trapezoid is also the diameter of the quarter-circle. The shaded area is the region of the composite shape *excluding* the quarter-circle.
* **Labels and Annotations:** A dimension `d=20 cm` is labeled below the common base/diameter.
* **Other Relevant Text (Analysis):**
【解析】阴影部分的面积 = 右侧梯形的面积 - 四分之一圆的面积, 梯形的下底和高、以及圆的半径都是 20 ÷ 2 厘米, 上底是 20 厘米, 然后根据梯形和圆的面积公式解答即可。
(The area of the shaded part = Area of the right trapezoid - Area of the quarter circle. The lower base and height of the trapezoid, as well as the radius of the circle, are all 20 ÷ 2 cm. The upper base is 20 cm. Then, calculate according to the area formulas for trapezoids and circles.)
* **Other Relevant Text (Solution):**
【解答】
20 ÷ 2 = 10 (厘米)
(10+20) × 10 ÷ 2 - 3.14 × 10² ÷ 4
= 150 - 78.5
= 71.5 (平方厘米)
* **Answer:**
【答案】阴影部分的面积是 71.5 平方厘米。
(The area of the shaded part is 71.5 square centimeters.)
---
**3. Problem Example 10: Vegetable Plot Area Increase**
* **Question Stem:**
例 10 (江口小考) 王红家有一块边长 15 米的正方形菜地, 今年她把这块菜地的一组对边分别增加了 3 米, 另一组对边长度不变。这块菜地的面积增加了多少平方米?
(Example 10 (Jiangkou Primary School Exam) Wang Hong's family has a square vegetable plot with a side length of 15 meters. This year, she increased one pair of opposite sides of this plot by 3 meters each, while the other pair of opposite sides remained unchanged. By how many square meters did the area of this vegetable plot increase?)
* **Chart/Diagram Description:**
* **Type:** Geometric figure (square).
* **Main Elements:** A square shape is shown, representing the original vegetable plot.
* **Labels and Annotations:** The top side is labeled "15米" (15 meters) and the right side is labeled "15米" (15 meters), indicating its dimensions.
* **Other Relevant Text (Analysis):**
【解析】根据题意可知, 先求出原来的正方形菜地的面积, 用 边长 × 边长 = 正方形的面积, 然后求出一组对边增加 3 米后的长方形面积, 用 长 × 宽 = 长方形的面积, 最后用现在的长方形面积 - 原来的正方形面积 = 这块菜地的面积增加的部分, 据此列式解答。
(According to the problem, first find the area of the original square vegetable plot using side length × side length = area of a square. Then find the area of the rectangle after one pair of opposite sides increased by 3 meters, using length × width = area of a rectangle. Finally, use the current rectangular area - the original square area = the increased part of the vegetable plot's area, and solve accordingly.)
* **Other Relevant Text (Solution):**
【答案】
正方形面积: 15 × 15 = 225 (平方米)
长方形面积: 15 × (15+3) = 15 × 18 = 270 (平方米)
增加面积: 270 - 225 = 45 (平方米)
答: 这块菜地的面积增加了 45 平方米。
(Square area: 15 × 15 = 225 (square meters))
(Rectangular area: 15 × (15+3) = 15 × 18 = 270 (square meters))
(Increased area: 270 - 225 = 45 (square meters))
(Answer: The area of this vegetable plot increased by 45 square meters.)
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**4. Problem: Trapezoid Area Formula**
* **Question Stem:**
(甘肃小考)一个梯形, 上底长 a 厘米, 下底 b 厘米, 高 h 厘米。它的面积是( )平方厘米。如果 a = b, 那么这个图形就是一个( )形。
(Gansu Primary School Exam) A trapezoid has an upper base of 'a' cm, a lower base of 'b' cm, and a height of 'h' cm. Its area is ( ) square centimeters. If a = b, then this figure is a ( ) shape.
* **Answer:**
【答案】1/2 (a+b)h; 长方形
(1/2 (a+b)h; rectangle)
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**5. Problem Example 11: Swimming Pool Choice**
* **Question Stem:**
例 11 (宁波小考)某游泳馆有大小两个游泳池, 小聪来到游泳馆游泳, 这时游泳池中的游泳人数情况如下图所示。根据当时的情况, 管理员应将小聪安排在哪个游泳池中? 说说你的理由。
(Example 11 (Ningbo Primary School Exam) A swimming center has two swimming pools, one large and one small. Xiao Cong came to the swimming center to swim. The number of people swimming in the pools at that time is shown in the figure below. Based on the current situation, which swimming pool should the manager arrange Xiao Cong to be in? Explain your reason.)
* **Chart/Diagram Description:**
* **Type:** Two geometric figures (rectangles) representing swimming pools.
* **Main Elements:**
* **Left Pool (Smaller):** A rectangle.
* **Dimensions:** Labeled "25米" (25 meters) for the width and "40米" (40 meters) for the length.
* **Content:** Text inside "有200人在游泳" (200 people are swimming).
* **Right Pool (Larger):** A rectangle.
* **Dimensions:** Labeled "35米" (35 meters) for the width and "60米" (60 meters) for the length.
* **Content:** Text inside "有350人在游泳" (350 people are swimming).
* **Note:** The image does not contain the solution or explanation for this problem.
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**Section: 图形与几何 专题五 (Geometry and Shapes - Topic 5)**
**Problem Context (Implied): Swimming Pool Allocation**
**Other Relevant Text (Analysis/Solution for Swimming Pool Problem):**
* **解析 (Analysis):** 因为大游泳池平均每人占6平方米，小游泳池平均每人占5平方米，所以管理员应将小聪安排在大游泳池。
* *Translation:* Because the large swimming pool on average occupies 6 square meters per person, and the small swimming pool on average occupies 5 square meters per person, the manager should arrange Xiao Cong to the large swimming pool.
* **【答案】 (Answer):**
* 小 (Small): 40 × 25 ÷ 200 = 1000 ÷ 200 = 5 (平方米)
* 大 (Large): 60 × 35 ÷ 350 = 2100 ÷ 350 = 6 (平方米)
* 答 (Answer): 应将小聪安排在大游泳池。因为大游泳池人均占的面积大。
* *Translation:* Xiao Cong should be arranged to the large swimming pool. Because the large swimming pool occupies a larger area per person.
* **Annotation:** 根据每个人的平均面积来判断去哪个游泳池。
* *Translation:* Determine which swimming pool to go to based on the average area per person.
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**Question 12**
* **Question Stem:** 例 12 (汕头小考) 求下图阴影部分的面积。
* *Translation:* Example 12 (Shantou Primary School Exam) Find the area of the shaded part in the figure below.
* **Chart/Diagram Description:**
* **Type:** Geometric figure - a semi-annulus (half of a circular ring).
* **Main Elements:**
* Two concentric semicircles.
* The region between the two semicircles is shaded.
* **Dimensions:**
* The outer diameter is labeled as "10 cm".
* The inner diameter is labeled as "6 cm".
* **Other Relevant Text (Analysis/Solution):**
* **解析 (Analysis):** 从图中可以看出，阴影部分是圆环的一半，其中圆环的面积 = (大圆半径的平方 - 小圆半径的平方) × π，所以阴影部分的面积 = 圆环的面积 ÷ 2。
* *Translation:* From the figure, it can be seen that the shaded part is half of a circular ring, where the area of the circular ring = (square of large radius - square of small radius) × π. Therefore, the area of the shaded part = area of circular ring ÷ 2.
* **解 (Solution):**
* 10 ÷ 2 = 5 (cm)
* 6 ÷ 2 = 3 (cm)
* (5² - 3²) × 3.14 ÷ 2 = 25.12 (cm²)
* **答 (Answer):** 阴影部分的面积是 25.12 cm²。
* *Translation:* The area of the shaded part is 25.12 cm².
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**Real Problem Analysis 1**
* **Title:** 真题解析 (Real Problem Analysis)
* **Question Stem:** (北京小考) 如图, A、B 两个平行四边形纸片部分重叠, 所占面积为 160 cm², A 的面积为 120 cm², B 的面积为 74 cm², 求重叠部分 (图中阴影部分) 的面积。
* *Translation:* (Beijing Primary School Exam) As shown in the figure, two parallelogram paper sheets A and B partially overlap. The total area occupied is 160 cm², the area of A is 120 cm², and the area of B is 74 cm². Find the area of the overlapping part (the shaded part in the figure).
* **Chart/Diagram Description:**
* **Type:** Geometric figure - two overlapping parallelograms.
* **Main Elements:**
* One parallelogram labeled "A".
* Another parallelogram labeled "B".
* The two parallelograms overlap. The overlapping region is shaded.
* **Other Relevant Text (Solution):**
* **【解答】 (Solution):**
* 120 + 74 - 160
* = 194 - 160
* = 34 (cm²)
* **答 (Answer):** 重叠部分的面积是 34 cm²。
* *Translation:* The area of the overlapping part is 34 cm².
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**Question 13**
* **Question Stem:** 例 13 (防城港小考) 下图空白部分的面积是 800 平方厘米, 阴影部分的面积是多少平方厘米?
* *Translation:* Example 13 (Fangchenggang Primary School Exam) The area of the white part in the figure below is 800 square centimeters. What is the area of the shaded part?
* **Chart/Diagram Description:**
* **Type:** Geometric figure - a semicircle with an inscribed triangle.
* **Main Elements:**
* A semicircle.
* An isosceles right-angled triangle inscribed within the semicircle. The base of the triangle is the diameter of the semicircle. The apex of the triangle is on the circumference.
* A right angle symbol is present at the base of the triangle, indicating the base is perpendicular to the radius drawn to the apex.
* The central triangle region is white.
* The two regions between the triangle and the semicircle's arc are shaded.
* **Other Relevant Text (Analysis/Solution):**
* **解析 (Analysis):** 观察图可知，空白三角形的底是圆的直径 2r，高是圆的半径 r，空白三角形的面积 = 2r × r ÷ 2 = r²，要求半圆的面积，用圆的面积 ÷ 2，然后用半圆的面积 - 空白部分的面积 = 阴影部分面积，据此列式解答。
* *Translation:* Observe the figure. The base of the white triangle is the diameter of the circle, 2r, and the height is the radius of the circle, r. The area of the white triangle = 2r × r ÷ 2 = r². To find the area of the semicircle, use (area of semicircle - area of white part = area of shaded part) to solve according to this formula.
* **Solution:**
* 3.14 × 800 ÷ 2 - 800
* = 1256 - 800
* = 456 (平方厘米)
* **答 (Answer):** 阴影部分的面积是 456 平方厘米。
* *Translation:* The area of the shaded part is 456 square centimeters.
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**Real Problem Analysis 2**
* **Title:** 真题解析 (Real Problem Analysis)
* **Question Stem:** (甘肃小考) 如图梯形中, 有 ( ) 对面积相等的三角形。
* *Translation:* (Gansu Primary School Exam) In the trapezoid shown in the figure, there are ( ) pairs of triangles with equal areas.
* **Chart/Diagram Description:**
* **Type:** Geometric figure - a trapezoid with diagonals.
* **Main Elements:**
* A trapezoid.
* Both diagonals of the trapezoid are drawn, intersecting inside the trapezoid.
* **Options:**
* A. 1
* B. 2
* C. 3
* D. 4
* **Answer:** 【答案】C
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**Problem 1**
**Question Stem:**
(北京小考)
如图，在直角三角形 ABC 中，AB 垂直于 BC，ND 垂直于 AC，NE 垂直于 AB，NF 垂直于 BC，四边形 BFNE 是正方形，AB = 4, BC = 3, AC = 5, ND = 1, 则正方形 BFNE 的边长等于 ______。
**Chart/Diagram Description:**
* **Type:** Geometric figure (Right-angled triangle with an inscribed square and additional lines).
* **Main Elements:**
* A right-angled triangle labeled ABC. The right angle is at B (since AB ⊥ BC).
* Vertices are A (top), B (bottom left), C (bottom right).
* A square labeled BFNE is inscribed within the triangle. E is on AB, F is on BC.
* Point N is a vertex of the square, located inside the triangle.
* A line segment ND is drawn from N to a point D on AC. ND is perpendicular to AC.
* Dashed line segments AN, BN, and CN are drawn, connecting vertex N to the vertices A, B, and C of the triangle, respectively.
* **Labels and Annotations:**
* Vertices: A, B, C, D, E, F, N.
* Lines: NE ⊥ AB, NF ⊥ BC, ND ⊥ AC.
* Given lengths: AB = 4, BC = 3, AC = 5, ND = 1.
* Shape: BFNE is a square.
**Other Relevant Text (Solution/Explanation):**
【解答】
设正方形 BFNE 的边长为 x, 连接 AN、BN、CN, 则
S_△ABC = S_△ANC + S_△ANB + S_△BNC
即 4x/2 + 3x/2 + 5×1/2 = 3×4/2
解得 x=1, 所以正方形 BFNE 的边长为 1。
**Answer:**
【答案】1
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**Problem 2**
**Question Stem:**
例 14 (黄冈小考) 求出如图阴影部分的面积。(单位：cm)
**Chart/Diagram Description:**
* **Type:** Composite geometric figure (a trapezoid with an arc inside).
* **Main Elements:**
* An overall shape resembling a trapezoid.
* The top horizontal side of the trapezoid has a length of 5.
* The bottom horizontal side of the trapezoid has a length of 8.
* A dashed vertical line segment indicates the height of the trapezoid, labeled as 5. This line connects the midpoint of the top side to the bottom side, indicating the height perpendicular to the bases.
* An arc is drawn within the trapezoid, connecting points on the slanted sides and curving downwards, nearly forming a semi-circle.
* Lines connect the endpoints of the top segment (length 5) to the points where the arc meets the bottom segment.
* Lines also connect the midpoint of the top segment to the two points where the arc meets the bottom segment, forming two triangles below the arc, suggesting the arc might be a semicircle or part of a larger circle.
* The region above the arc, within the overall trapezoidal boundary, is shaded. The region below the arc (appearing as a semi-circle) is unshaded.
* **Labels and Annotations:**
* Lengths: "5" (top base), "8" (bottom base), "5" (height).
* Units are specified in the question stem: cm.
**Other Relevant Text (Solution/Explanation):**
解析
图中弧形阴影部分和空白弧形部分的面积是相等的，所以阴影部分的面积可转化为梯形的面积来进行计算，已知梯形的上底是 5 厘米，下底是 8 厘米，高是 5 厘米，根据梯形的面积公式：S = (a+b)h ÷ 2 进行计算即可。
(5 + 8) × 5 ÷ 2
= 13 × 5 ÷ 2
= 32.5 (平方厘米)
**Answer:**
答：阴影部分的面积是 32.5 平方厘米。
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