Unit C1: Surface Areas, Volumes
and Measurement in Circles单元 C1：表面积、体积与圆的度量
The geometry warm-up. Three-dimensional distance, the volumes and surface areas of the standard solids, the angles formed between lines and planes inside a 3D figure, and the radian measure that underwrites every later trigonometry and calculus result. Every section opens with the formula a crammer needs and follows with one or two carefully chosen worked examples.几何热身单元。三维距离、标准立体的体积与表面积、三维图形中直线与平面之间的夹角，以及之后每一节三角与微积分都要用的弧度制度量。每节先给出临阵磨枪所需的公式，再配一两道精心挑选的例题。
3D Distance Formula三维距离公式 SL 3.1
Find the distance between $A = (1, 2, -1)$ and $B = (4, 6, 3)$.求 $A = (1, 2, -1)$ 与 $B = (4, 6, 3)$ 之间的距离。
Apply the formula. Compute each squared difference:
套用公式。逐项算平方差：
$$ (4 - 1)^{2} = 9, \qquad (6 - 2)^{2} = 16, \qquad (3 - (-1))^{2} = 4^{2} = 16. $$Sum and take the square root.
求和取根。
$$ d(A, B) \;=\; \sqrt{9 + 16 + 16} \;=\; \sqrt{41} \;\approx\; 6.40. $$Result. The distance is $\sqrt{41}$ units, approximately $6.40$ to three significant figures.
结果。距离为 $\sqrt{41}$（单位长度），三位有效数字约为 $6.40$。
Going deeper: why two applications of Pythagoras深入：为何要用两次勾股
Place a right-angled box whose opposite corners are $P_{1}$ and $P_{2}$, with edges parallel to the axes. The edge lengths are $|x_{2} - x_{1}|$, $|y_{2} - y_{1}|$, and $|z_{2} - z_{1}|$. The diagonal of the bottom face has length $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ by the 2D Pythagorean theorem. The space diagonal $P_{1} P_{2}$ forms a right triangle with this face diagonal as one leg and the vertical edge as the other leg. Apply Pythagoras a second time: $$ d^{2} \;=\; \bigl[(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}\bigr] + (z_{2} - z_{1})^{2}. $$ Taking the positive square root gives the formula. The proof generalises: the distance in $\mathbb{R}^{n}$ is $\sqrt{\sum_{k=1}^{n} (x_{k}^{(2)} - x_{k}^{(1)})^{2}}$.
作一长方体，以 $P_{1}$、$P_{2}$ 为对顶点，棱平行于坐标轴。棱长为 $|x_{2} - x_{1}|$、$|y_{2} - y_{1}|$、$|z_{2} - z_{1}|$。由二维勾股，底面对角线长 $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$。空间对角线 $P_{1} P_{2}$ 与底面对角线、铅垂棱构成直角三角形，再用一次勾股： $$ d^{2} \;=\; \bigl[(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}\bigr] + (z_{2} - z_{1})^{2}. $$ 取正根即得公式。该证明可推广：$\mathbb{R}^{n}$ 中距离为 $\sqrt{\sum_{k=1}^{n} (x_{k}^{(2)} - x_{k}^{(1)})^{2}}$。
Point $P = (x, 0, 0)$ lies on the $x$-axis and is equidistant from $A = (1, 2, 3)$ and $B = (5, -2, 1)$. Find $x$.点 $P = (x, 0, 0)$ 在 $x$ 轴上，且到 $A = (1, 2, 3)$ 与 $B = (5, -2, 1)$ 的距离相等。求 $x$。
Step 1 — Set the two squared distances equal (avoids unnecessary square roots).
第 1 步 — 令两个距离的平方相等（避免不必要的开方）。
$$ (x-1)^2 + (0-2)^2 + (0-3)^2 \;=\; (x-5)^2 + (0+2)^2 + (0-1)^2 $$Step 2 — Expand both sides.
第 2 步 — 展开两边。
$$ x^2 - 2x + 1 + 4 + 9 \;=\; x^2 - 10x + 25 + 4 + 1 $$Step 3 — Cancel $x^2$ (it appears on both sides) and solve the remaining linear equation.
第 3 步 — 消去两边都有的 $x^2$，解剩下的线性方程。
$$ -2x + 14 = -10x + 30 \;\Longrightarrow\; 8x = 16 \;\Longrightarrow\; x = 2 $$Squaring the distance instead of working with square roots throughout is the key simplification — it turns what looks like a 3D geometry problem into ordinary algebra, and the $x^2$ terms cancelling is a reliable sign the setup is correct.
用距离的平方而不是全程处理平方根，是这里关键的化简技巧——它把一道看似三维几何的题目变成了普通代数；两边的 $x^2$ 项能相消，正是设置正确的可靠信号。
Volumes of Standard Solids标准立体的体积 SL 3.1
| Solid | Volume |
|---|---|
| Cuboid (length $l$, width $w$, height $h$) | $V = l \, w \, h$ |
| Prism (base area $B$, height $h$) | $V = B \, h$ |
| Cylinder (radius $r$, height $h$) | $V = \pi r^{2} h$ |
| Cone (radius $r$, height $h$) | $V = \tfrac{1}{3} \pi r^{2} h$ |
| Pyramid (base area $B$, height $h$) | $V = \tfrac{1}{3} B \, h$ |
| Sphere (radius $r$) | $V = \tfrac{4}{3} \pi r^{3}$ |
| 立体 | 体积 |
|---|---|
| 长方体（长 $l$、宽 $w$、高 $h$） | $V = l \, w \, h$ |
| 棱柱（底面积 $B$、高 $h$） | $V = B \, h$ |
| 圆柱（半径 $r$、高 $h$） | $V = \pi r^{2} h$ |
| 圆锥（半径 $r$、高 $h$） | $V = \tfrac{1}{3} \pi r^{2} h$ |
| 棱锥（底面积 $B$、高 $h$） | $V = \tfrac{1}{3} B \, h$ |
| 球（半径 $r$） | $V = \tfrac{4}{3} \pi r^{3}$ |
Find the volume of a sphere of radius $6 \text{ cm}$, giving the answer (a) as an exact multiple of $\pi$ and (b) to three significant figures.求半径 $6 \text{ cm}$ 的球的体积，分别 (a) 以 $\pi$ 的精确倍数表示，(b) 取三位有效数字。
Apply the sphere formula.
套用球的体积公式。
$$ V \;=\; \tfrac{4}{3} \pi r^{3} \;=\; \tfrac{4}{3} \pi (6)^{3} \;=\; \tfrac{4}{3} \pi \cdot 216 \;=\; 288 \pi \text{ cm}^{3}. $$(a) Exact. $V = 288 \pi \text{ cm}^{3}$.
(a) 精确值。$V = 288 \pi \text{ cm}^{3}$。
(b) Decimal. $V \approx 288 \cdot 3.14159 \approx 905 \text{ cm}^{3}$ to three significant figures.
(b) 小数。$V \approx 288 \cdot 3.14159 \approx 905 \text{ cm}^{3}$（三位有效数字）。
Going deeper: where the cone factor $\tfrac{1}{3}$ comes from深入：圆锥的 $\tfrac{1}{3}$ 因子从何而来
A cone of radius $r$ and height $h$ can be modelled as the solid of revolution generated by rotating the line $y = (r/h) \, x$ about the $x$-axis between $x = 0$ and $x = h$. By the disk method (developed in Unit E4), $$ V \;=\; \pi \int_{0}^{h} y^{2} \, dx \;=\; \pi \int_{0}^{h} \frac{r^{2}}{h^{2}} \, x^{2} \, dx \;=\; \frac{\pi r^{2}}{h^{2}} \cdot \frac{h^{3}}{3} \;=\; \frac{1}{3} \pi r^{2} h. $$ The factor $\tfrac{1}{3}$ is exactly the antiderivative of $x^{2}$ evaluated from $0$ to $h$. The same argument with the line $y = r$ (a cylinder) integrates to $\pi r^{2} h$ with no third, because $\int r^{2} \, dx = r^{2} x$.
半径 $r$、高 $h$ 的圆锥可视为直线 $y = (r/h) \, x$ 绕 $x$ 轴在 $x = 0$ 到 $x = h$ 上旋转得到的立体。由圆盘法（E4 单元发展）： $$ V \;=\; \pi \int_{0}^{h} y^{2} \, dx \;=\; \pi \int_{0}^{h} \frac{r^{2}}{h^{2}} \, x^{2} \, dx \;=\; \frac{\pi r^{2}}{h^{2}} \cdot \frac{h^{3}}{3} \;=\; \frac{1}{3} \pi r^{2} h. $$ 因子 $\tfrac{1}{3}$ 正是 $x^{2}$ 从 $0$ 到 $h$ 的原函数值。换成 $y = r$（圆柱）同法积出 $\pi r^{2} h$，无 $\tfrac{1}{3}$，因为 $\int r^{2} \, dx = r^{2} x$。
A solid consists of a cylinder of radius $4 \text{ cm}$ and height $10 \text{ cm}$, topped with a hemisphere of the same radius. Find the total volume in terms of $\pi$.某立体由半径 $4 \text{ cm}$、高 $10 \text{ cm}$ 的圆柱，加上同半径的半球顶部组成。求总体积（以 $\pi$ 表示）。
Step 1 — Split into two familiar solids and compute each separately. This is the key move for any composite-solid problem: never guess a combined formula, always build it from the standard-solid list.
第 1 步 — 拆成两个熟悉的立体，分别计算。这是处理任何组合体问题的关键一步：绝不要凭空猜一个合并公式，而是从标准立体清单出发逐一搭建。
$$ V_{\text{cylinder}} = \pi r^2 h = \pi (4)^2(10) = 160\pi $$Step 2 — Hemisphere volume is half a sphere.
第 2 步 — 半球体积是球体积的一半。
$$ V_{\text{hemisphere}} = \tfrac{1}{2}\left(\tfrac{4}{3}\pi r^3\right) = \tfrac{2}{3}\pi (4)^3 = \tfrac{128}{3}\pi $$Step 3 — Add the two volumes (common denominator 3).
第 3 步 — 两体积相加（通分为 3）。
$$ V_{\text{total}} = 160\pi + \tfrac{128}{3}\pi = \tfrac{480\pi + 128\pi}{3} = \tfrac{608\pi}{3} \text{ cm}^3 $$The radius of the hemisphere must match the cylinder's radius for the two pieces to join seamlessly — always check this compatibility before adding volumes, since composite-solid questions sometimes deliberately give a mismatched dimension to test this.
半球的半径必须与圆柱的半径相同，两部分才能无缝拼接——加总体积之前务必检查这一相容性，因为组合体题目有时会故意给出不匹配的尺寸来考察这一点。
Surface Areas表面积 SL 3.1
| Solid | Surface area |
|---|---|
| Cylinder (closed, radius $r$, height $h$) | $S = 2 \pi r^{2} + 2 \pi r h$ |
| Cone (closed, radius $r$, slant $l$) | $S = \pi r^{2} + \pi r l$ |
| Sphere (radius $r$) | $S = 4 \pi r^{2}$ |
Cone anatomy. The circular base contributes $\pi r^{2}$. The curved surface unrolls into a sector of a disk of radius $l$ (the slant), with arc length equal to the circumference $2 \pi r$; the sector area $\tfrac{1}{2} l \cdot 2 \pi r = \pi r l$.
Slant height. For a right circular cone, the slant $l$, the radius $r$, and the vertical height $h$ form a right triangle: $l^{2} = r^{2} + h^{2}$.
| 立体 | 表面积 |
|---|---|
| 圆柱（封闭，半径 $r$、高 $h$） | $S = 2 \pi r^{2} + 2 \pi r h$ |
| 圆锥（封闭，半径 $r$、斜高 $l$） | $S = \pi r^{2} + \pi r l$ |
| 球（半径 $r$） | $S = 4 \pi r^{2}$ |
圆锥结构。底面圆贡献 $\pi r^{2}$。侧面展开是半径 $l$（斜高）的圆盘扇形，弧长等于周长 $2 \pi r$；扇形面积 $\tfrac{1}{2} l \cdot 2 \pi r = \pi r l$。
斜高。正圆锥中，斜高 $l$、半径 $r$、铅直高 $h$ 构成直角三角形：$l^{2} = r^{2} + h^{2}$。
A closed right circular cone has base radius $3 \text{ cm}$ and vertical height $4 \text{ cm}$. Find the total surface area, leaving the answer in terms of $\pi$.一封闭正圆锥底半径 $3 \text{ cm}$，铅直高 $4 \text{ cm}$。求总表面积（以 $\pi$ 表示）。
Step 1. Find the slant height. The slant, radius, and height form a right triangle:
第 1 步：求斜高。斜高、半径、铅直高构成直角三角形：
$$ l \;=\; \sqrt{r^{2} + h^{2}} \;=\; \sqrt{3^{2} + 4^{2}} \;=\; \sqrt{9 + 16} \;=\; \sqrt{25} \;=\; 5 \text{ cm}. $$Step 2. Curved surface. $S_{\text{curved}} = \pi r l = \pi (3)(5) = 15 \pi$.
第 2 步：侧面（曲面）。$S_{\text{侧}} = \pi r l = \pi (3)(5) = 15 \pi$。
Step 3. Base. $S_{\text{base}} = \pi r^{2} = \pi (3)^{2} = 9 \pi$.
第 3 步：底面。$S_{\text{底}} = \pi r^{2} = \pi (3)^{2} = 9 \pi$。
Step 4. Total.
第 4 步：求和。
$$ S \;=\; 15 \pi + 9 \pi \;=\; 24 \pi \text{ cm}^{2}. $$Result. The total surface area is $24 \pi \text{ cm}^{2}$, approximately $75.4 \text{ cm}^{2}$.
结果。总表面积为 $24 \pi \text{ cm}^{2}$，约为 $75.4 \text{ cm}^{2}$。
A solid is a cylinder of radius $4 \text{ cm}$ and height $10 \text{ cm}$, topped with a hemisphere of the same radius (as in C1.2b). Find the exposed surface area.某立体由半径 $4 \text{ cm}$、高 $10 \text{ cm}$ 的圆柱，加上同半径的半球顶部组成（同 C1.2b）。求其外露表面积。
Step 1 — Identify which faces are actually exposed. The top circular face of the cylinder is covered by the hemisphere, so it contributes nothing. The flat circular face of the hemisphere is glued to the cylinder, so it isn't exposed either. Only three surfaces count: the cylinder's curved side, the cylinder's base, and the hemisphere's curved surface.
第 1 步 — 确定哪些面真正外露。圆柱的顶面被半球覆盖，不贡献面积；半球的平面一侧粘在圆柱上，同样不外露。只有三个面需要计算：圆柱侧面、圆柱底面、半球曲面。
Step 2 — Cylinder's curved surface and base only (no top).
第 2 步 — 只算圆柱侧面与底面（无顶面）。
$$ S_{\text{cyl}} = 2\pi r h + \pi r^2 = 2\pi(4)(10) + \pi(4)^2 = 80\pi + 16\pi = 96\pi $$Step 3 — Hemisphere's curved surface is half a sphere's surface (not half of $4\pi r^2$ plus the flat circle, since that circle is hidden).
第 3 步 — 半球的曲面是球面积的一半（不是 $4\pi r^2$ 的一半再加平面圆，因为那个圆被遮住了）。
$$ S_{\text{hemi, curved}} = \tfrac{1}{2}(4\pi r^2) = 2\pi(4)^2 = 32\pi $$Step 4 — Add the three exposed pieces.
第 4 步 — 三部分外露面积相加。
$$ S_{\text{total}} = 96\pi + 32\pi = 128\pi \text{ cm}^2 $$This is the single most common error type in composite-solid surface area questions: including a face that's actually hidden inside the join, or forgetting the flat circle really is covered. Always sketch the solid and physically cross out every hidden face before adding.
这是组合体表面积题最常见的错误类型：把实际藏在拼接处内部的面也算了进去，或忘记那个平面圆确实被遮住了。求和之前，务必先画出立体草图，把每个被遮住的面都实际划掉。
Angles in 3D: Line and Plane三维夹角：直线与平面 SL 3.2
Angle between a line and a plane. Project the line onto the plane (drop a perpendicular from a point on the line to the plane and join the foot to the original point). The angle the line makes with the plane is the angle between the line and its projection. Equivalently, it is the complement of the angle the line makes with the normal to the plane.
Recipe for cuboid / pyramid problems. (1) Draw the figure and label the corners. (2) Identify the right triangle that contains the angle you want. (3) Compute the two legs of that triangle from the cuboid edge lengths, often via the 2D Pythagorean theorem on the floor. (4) Apply $\tan \theta = \text{opposite} / \text{adjacent}$.
直线与平面的夹角。把直线投影到平面（从直线上一点向平面作垂线，再连垂足与原点）。直线与平面的夹角即直线与其投影之间的角；等价地，是直线与平面法线夹角的余角。
长方体/棱锥题套路。(1) 画图并标顶点。(2) 找含所求角的直角三角形。(3) 由长方体的棱长算出该三角形的两直角边，通常先用二维勾股算底面对角线。(4) 用 $\tan \theta = \text{对边} / \text{邻边}$。
A cuboid has dimensions $4 \text{ cm} \times 3 \text{ cm} \times 2 \text{ cm}$ (length, width, height). Find the acute angle that the space diagonal makes with the base.一长方体尺寸为 $4 \text{ cm} \times 3 \text{ cm} \times 2 \text{ cm}$（长、宽、高）。求空间对角线与底面所成的锐角。
Step 1. Diagonal of the base. The base is a $4 \times 3$ rectangle. By the 2D Pythagorean theorem,
第 1 步：底面对角线。底面为 $4 \times 3$ 矩形。由二维勾股，
$$ d_{\text{base}} \;=\; \sqrt{4^{2} + 3^{2}} \;=\; \sqrt{16 + 9} \;=\; \sqrt{25} \;=\; 5 \text{ cm}. $$Step 2. Space diagonal. Adding the vertical edge of length $2$,
第 2 步：空间对角线。加上长度 $2$ 的铅直棱，
$$ d_{\text{space}} \;=\; \sqrt{d_{\text{base}}^{2} + 2^{2}} \;=\; \sqrt{25 + 4} \;=\; \sqrt{29} \text{ cm}. $$Step 3. Identify the right triangle containing the angle. The space diagonal, the base diagonal (its projection on the base), and the vertical edge form a right triangle. Let $\theta$ be the angle between the space diagonal and the base. The opposite side is the vertical edge (length $2$); the adjacent side is the base diagonal (length $5$).
第 3 步：找含所求角的直角三角形。空间对角线、底面对角线（其在底面的投影）、铅直棱构成直角三角形。设 $\theta$ 为空间对角线与底面的夹角，对边为铅直棱（长 $2$），邻边为底面对角线（长 $5$）。
Step 4. Compute the angle.
第 4 步：算角。
$$ \tan \theta \;=\; \frac{2}{5}, \qquad \theta \;=\; \arctan \tfrac{2}{5} \;\approx\; 21.8^{\circ}. $$Result. The space diagonal makes an angle of approximately $21.8^{\circ}$ with the base.
结果。空间对角线与底面所成的角约为 $21.8^{\circ}$。
Going deeper: angle between a line and a plane, formal definition深入：直线与平面夹角的形式化定义
Let $\ell$ be a line not contained in a plane $\Pi$ and not perpendicular to $\Pi$. The orthogonal projection of $\ell$ on $\Pi$ is the unique line $\ell'$ in $\Pi$ obtained by dropping a perpendicular from each point of $\ell$ to $\Pi$ and joining the feet. The angle between $\ell$ and $\Pi$ is defined as the angle between $\ell$ and $\ell'$. It satisfies $0^{\circ} \le \theta \le 90^{\circ}$.
If $\mathbf{n}$ is a normal vector to $\Pi$ and $\mathbf{d}$ is a direction vector of $\ell$, then the angle $\varphi$ between $\mathbf{d}$ and $\mathbf{n}$ satisfies $$ \cos \varphi \;=\; \frac{|\mathbf{d} \cdot \mathbf{n}|}{\|\mathbf{d}\| \, \|\mathbf{n}\|}, \qquad \theta \;=\; 90^{\circ} - \varphi. $$ The absolute value ensures $\varphi \in [0^{\circ}, 90^{\circ}]$, so $\theta$ is non-negative. The vector form belongs to Unit C3 (HL); for SL applications, the right-triangle method shown in C1.4 suffices.
设 $\ell$ 是不在平面 $\Pi$ 内且不与 $\Pi$ 垂直的直线。$\ell$ 在 $\Pi$ 上的正交投影是从 $\ell$ 上每点向 $\Pi$ 作垂线、连接垂足而得的唯一直线 $\ell'$（在 $\Pi$ 内）。$\ell$ 与 $\Pi$ 的夹角定义为 $\ell$ 与 $\ell'$ 的夹角，满足 $0^{\circ} \le \theta \le 90^{\circ}$。
若 $\mathbf{n}$ 是 $\Pi$ 的法向量、$\mathbf{d}$ 是 $\ell$ 的方向向量，则 $\mathbf{d}$ 与 $\mathbf{n}$ 的夹角 $\varphi$ 满足 $$ \cos \varphi \;=\; \frac{|\mathbf{d} \cdot \mathbf{n}|}{\|\mathbf{d}\| \, \|\mathbf{n}\|}, \qquad \theta \;=\; 90^{\circ} - \varphi. $$ 绝对值保证 $\varphi \in [0^{\circ}, 90^{\circ}]$，故 $\theta$ 非负。向量形式属于 C3（HL）；SL 题用 C1.4 的直角三角形法即可。
A square-based pyramid has base side $6 \text{ cm}$ and apex directly above the centre of the base at height $4 \text{ cm}$. Find the angle between two adjacent slant edges.一正四棱锥底边长 $6 \text{ cm}$，顶点在底面中心正上方，高 $4 \text{ cm}$。求两条相邻侧棱之间的夹角。
Step 1 — Find the slant edge length. The horizontal distance from the base centre to a base vertex is half the base diagonal: $\tfrac{1}{2}\sqrt{6^2+6^2} = 3\sqrt{2}$. The slant edge is the hypotenuse of a right triangle with this horizontal leg and the vertical height:
第 1 步 — 求侧棱长。底面中心到底面顶点的水平距离是底面对角线的一半：$\tfrac{1}{2}\sqrt{6^2+6^2} = 3\sqrt{2}$。侧棱是以此水平边和铅直高为直角边的直角三角形的斜边：
$$ e = \sqrt{(3\sqrt{2})^2 + 4^2} = \sqrt{18+16} = \sqrt{34} \text{ cm} $$Step 2 — Identify the triangle formed by two adjacent slant edges and the base edge between their feet. This triangle is isosceles: two sides of length $\sqrt{34}$ (the slant edges) and included base of length $6$ (the base edge itself).
第 2 步 — 找出由两条相邻侧棱及其底端之间的底边构成的三角形。这是一个等腰三角形：两条腰长 $\sqrt{34}$（侧棱），底边长 $6$（正是底面边）。
Step 3 — Use the cosine rule to find the angle between the two equal sides.
第 3 步 — 用余弦定理求两腰之间的夹角。
$$ 6^2 = (\sqrt{34})^2 + (\sqrt{34})^2 - 2(\sqrt{34})(\sqrt{34})\cos\theta \;\Longrightarrow\; 36 = 68 - 68\cos\theta $$ $$ \cos\theta = \frac{32}{68} = \frac{8}{17} \;\Longrightarrow\; \theta \approx 61.9^{\circ} $$This problem needed a tool from outside the "drop a perpendicular, use tan" recipe in the cram-cheat above — the cosine rule (Unit B4) — because the angle sits between two slant edges rather than between a line and the base. Recognizing when the simple right-triangle recipe doesn't apply is itself part of the skill being tested.
这道题需要用到上面速记框中"作垂线、用 tan"套路之外的工具——余弦定理（第 B4 单元）——因为所求角在两条侧棱之间，而不是直线与底面之间。识别出简单直角三角形套路何时不适用，本身就是这道题考察的能力之一。
Radian Measure, Arc Length and Sector Area弧度制、弧长与扇形面积 SL 3.4
A sector of a circle has radius $r = 10 \text{ cm}$ and central angle $\theta = \tfrac{\pi}{3}$ radians. Find (a) the arc length and (b) the sector area, both in exact form.一扇形半径 $r = 10 \text{ cm}$，圆心角 $\theta = \tfrac{\pi}{3}$ 弧度。求 (a) 弧长，(b) 扇形面积，均以精确值表示。
(a) Arc length.
(a) 弧长。
$$ s \;=\; r \theta \;=\; 10 \cdot \frac{\pi}{3} \;=\; \frac{10 \pi}{3} \text{ cm}. $$(b) Sector area.
(b) 扇形面积。
$$ A \;=\; \tfrac{1}{2} r^{2} \theta \;=\; \tfrac{1}{2} (10)^{2} \cdot \frac{\pi}{3} \;=\; \tfrac{1}{2} \cdot 100 \cdot \frac{\pi}{3} \;=\; \frac{50 \pi}{3} \text{ cm}^{2}. $$Check the conversion. An angle of $\pi/3$ radians equals $60^{\circ}$. The arc is one sixth of the full circumference $2 \pi r = 20 \pi$, namely $\tfrac{20 \pi}{6} = \tfrac{10 \pi}{3}$. The sector is one sixth of the full disk area $\pi r^{2} = 100 \pi$, namely $\tfrac{100 \pi}{6} = \tfrac{50 \pi}{3}$. Both checks agree.
换算验算。$\pi/3$ 弧度即 $60^{\circ}$。弧长为整周长 $2 \pi r = 20 \pi$ 的 $\tfrac{1}{6}$，即 $\tfrac{20 \pi}{6} = \tfrac{10 \pi}{3}$。扇形面积为整圆面积 $\pi r^{2} = 100 \pi$ 的 $\tfrac{1}{6}$，即 $\tfrac{100 \pi}{6} = \tfrac{50 \pi}{3}$。两项检验一致。
Going deeper: why radians make calculus tidy深入：为何弧度让微积分整洁
In radian measure, the small-angle limit reads $$ \lim_{\theta \to 0} \frac{\sin \theta}{\theta} \;=\; 1. $$ In degree measure, the same limit reads $\lim_{\theta \to 0} (\sin \theta) / \theta = \pi / 180$, because $\sin \theta$ in degrees equals $\sin(\pi \theta / 180)$ in radians. The clean derivative $\tfrac{d}{dx} \sin x = \cos x$ holds only when $x$ is in radians; in degrees, an extra factor of $\pi / 180$ appears. Every calculus formula in Units E1 through E4 silently assumes radians.
在弧度制下，小角度极限为 $$ \lim_{\theta \to 0} \frac{\sin \theta}{\theta} \;=\; 1. $$ 若用角度，则 $\lim_{\theta \to 0} (\sin \theta) / \theta = \pi / 180$，因为角度制的 $\sin \theta$ 等于弧度制的 $\sin(\pi \theta / 180)$。简洁的导数 $\tfrac{d}{dx} \sin x = \cos x$ 仅在 $x$ 取弧度时成立；用角度则多出因子 $\pi / 180$。E1 至 E4 单元的每个微积分公式都默认弧度。
A sector has radius $8 \text{ cm}$ and central angle $1.2$ radians. Find its perimeter.一扇形半径 $8 \text{ cm}$，圆心角 $1.2$ 弧度。求其周长。
Step 1 — Identify all three edges of a sector: the arc, plus two straight radii. A common error is to compute only the arc length and call it the perimeter.
第 1 步 — 确认扇形有三条边：一段弧，加两条半径。常见错误是只算弧长，就把它当成周长。
Step 2 — Arc length.
第 2 步 — 弧长。
$$ s = r\theta = 8(1.2) = 9.6 \text{ cm} $$Step 3 — Add the two straight radii.
第 3 步 — 加上两条直边（半径）。
$$ P = s + 2r = 9.6 + 2(8) = 9.6 + 16 = 25.6 \text{ cm} $$Perimeter and area questions are often mixed in the same problem specifically to test whether you remember that a sector's boundary is not just its arc — the two straight edges are easy to forget once the arc-length formula is fresh in mind.
周长与面积问题常常出现在同一道题中，就是为了考察你是否记得扇形的边界不仅仅是那段弧——一旦刚用完弧长公式，两条直边很容易被忘记。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- Quote the formula before substituting. Writing $V = \tfrac{4}{3} \pi r^{3}$ on its own line earns the M1 even if the arithmetic later slips.
- 先写公式再代值。单独一行写 $V = \tfrac{4}{3} \pi r^{3}$ 可拿 M1，即便后续算错。
- Decide exact form vs three significant figures. If the question says "in terms of $\pi$", leave the answer as $288 \pi$, not $904.78$. If it asks for a numerical value to 3sf, do not leave $\pi$ in the answer.
- 区分精确值与三位有效数字。题目说"以 $\pi$ 表示"则留 $288 \pi$，不写 $904.78$。若要三位有效数字的数值，则不要留 $\pi$。
- Compute the slant before the cone surface. The slip "$\pi r h + \pi r^{2}$" with $h$ in place of $l$ is the canonical wrong setup.
- 圆锥表面积前先算斜高。把 $h$ 当 $l$ 写成"$\pi r h + \pi r^{2}$"是标志性错误。
- Always draw the figure. A labelled diagram with the right triangle picked out earns the M1 mark for setting up; an unlabelled formula does not.
- 必须画图。标注顶点并标出所用直角三角形可拿建模 M1；无图的公式拿不到。
- For "angle between line and base", drop the perpendicular from the upper point to the base. The right triangle has the perpendicular (vertical edge) opposite the angle and the projection on the base adjacent.
- "直线与底面的夹角"：从上方点向底面作垂线。所得直角三角形中，垂线（铅直棱）为对边，底面投影为邻边。
- Check calculator mode before any trigonometric evaluation. RAD for $s = r \theta$ and $A = \tfrac{1}{2} r^{2} \theta$. DEG only if the angle is explicitly given in degrees and not used inside a calculus expression.
- 动三角函数前先看计算器模式。$s = r \theta$ 与 $A = \tfrac{1}{2} r^{2} \theta$ 用 RAD。仅在角明确以度给出且不进入微积分时才用 DEG。
- Leave $\pi$ exact when the angle is a clean fraction of $\pi$. $\theta = \pi/6$ gives arc $r \pi / 6$ exactly; converting to a decimal loses marks for "form".
- 若角为 $\pi$ 的简分数，保留 $\pi$。$\theta = \pi/6$ 时弧长精确为 $r \pi / 6$；转小数会因"形式"扣分。
Flashcards闪卡
Unit C1 Practice Quiz单元 C1 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt.
每一条都要"裸做"做对（不看笔记、一次过）才打勾。
- Compute the distance between two points in 3D using $\sqrt{\Delta x^{2} + \Delta y^{2} + \Delta z^{2}}$用 $\sqrt{\Delta x^{2} + \Delta y^{2} + \Delta z^{2}}$ 算三维两点距离
- Find the midpoint of a 3D segment by averaging coordinates用坐标平均法求三维线段中点
- Recall the six standard volume formulas (cuboid, prism, cylinder, cone, pyramid, sphere)记住六个标准立体的体积公式（长方体、棱柱、圆柱、圆锥、棱锥、球）
- Recall the three standard surface-area formulas (cylinder, cone, sphere)记住三个标准表面积公式（圆柱、圆锥、球）
- Compute the slant height $l = \sqrt{r^{2} + h^{2}}$ of a right circular cone before applying the surface formula在用表面积公式前先算正圆锥的斜高 $l = \sqrt{r^{2} + h^{2}}$
- Quote a volume or surface formula on its own line before substituting numerical values代入数值前先单行写出体积或表面积公式
- Draw and label the right triangle inside a cuboid or pyramid that contains the requested angle在长方体或棱锥中画出并标注含所求角的直角三角形
- Identify "angle between line and plane" as the angle between the line and its orthogonal projection把"直线与平面的夹角"识别为直线与其正交投影之间的角
- Convert between degrees and radians fluently using $180^{\circ} = \pi$用 $180^{\circ} = \pi$ 在角度与弧度间熟练换算
- Apply $s = r \theta$ and $A = \tfrac{1}{2} r^{2} \theta$ with $\theta$ in radians在 $\theta$ 取弧度的前提下用 $s = r \theta$ 与 $A = \tfrac{1}{2} r^{2} \theta$
- Check calculator mode (RAD vs DEG) before evaluating any trigonometric expression动三角运算前检查计算器模式（RAD 或 DEG）
IB Paper-Style PracticeIB 试卷风格练习
C1 Practice and Solutions are on the roadmap. They will ship under Practice Questions/Unit_C1_*.html with the bilingual built-in pattern.
C1 配套的 Practice 与 Solutions 已在排期，上线后位于 Practice Questions/Unit_C1_*.html，采用双语内嵌格式。
How to use this guide本指南使用说明
How to use this guide本指南使用说明
C1 is the geometry foundation. The whole unit is SL content, so every HL student inherits it. The mistakes here are not conceptual; they are unit slips, forgotten factors of $\pi$, and the perennial habit of working in degrees when the formula expects radians. Drill the formulas, then drill the unit conversions.C1 是几何基础。整单元为 SL 内容，HL 学生全部继承。本单元的错误几乎都不是概念错误，而是单位失误、漏写 $\pi$ 因子，以及公式要求弧度时却用了角度的老毛病。先把公式背熟，再把单位换算练熟。
Memorise the six volume formulas (cuboid, prism, cylinder, cone, pyramid, sphere) and the three surface-area formulas (cylinder, cone, sphere). Memorise $s = r \theta$ and $A = \tfrac{1}{2} r^{2} \theta$ for radians. Practise one cuboid-diagonal angle problem end to end.
背熟六个体积公式（长方体、棱柱、圆柱、圆锥、棱锥、球）与三个表面积公式（圆柱、圆锥、球）。背熟弧度制的 $s = r \theta$ 与 $A = \tfrac{1}{2} r^{2} \theta$。完整做一道长方体对角线夹角题。
Draw the figure for every 3D problem. Identify the foot of the perpendicular before reaching for a trigonometric ratio. Convert every angle to radians when calculus or rotation enters; an arc length computed with degrees plugged into $s = r \theta$ is the canonical wrong answer.
每道三维题都要画图。在动用三角比之前先找到垂足。一旦涉及微积分或旋转，所有角都换成弧度；把角度代入 $s = r \theta$ 算弧长是标志性错答。