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Unit C1 · Geometry & TrigonometryUnit C1 · 几何与三角

Surface Areas, Volumes and Measurement in Circles表面积、体积与圆的度量

IB-Style Practice Questions · Paper 1A · Paper 1B · Paper 2IB 风格练习题 · 第一卷 A 节 · 第一卷 B 节 · 第二卷

EASY MEDIUM HARD Paper 1A Paper 1B Paper 2

Syllabus SL 3.1, 3.2, 3.4考纲 SL 3.1、3.2、3.4AA SL/HL



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PART I  ·  PAPER 1 SECTION A第一部分  ·  第一卷 A 节No calculator · short response · 25 marks不可使用计算器 · 简答题 · 25 分

Section A · Short ResponseA 节 · 简答题

Quote each formula on its own line before substituting numerical values. Leave answers exact (in terms of $\pi$ or as surds) unless the question asks for a decimal. No calculator permitted.代值之前先单独一行写出公式。除题目要求小数外,答案保留精确形式(带 $\pi$ 或带根号)。不可使用计算器。

Q1EASY Paper 1A C1.2 Volume of a Cone [4 marks]

A right circular cone has base radius $r = 3$ and vertical height $h = 4$. Find its volume, leaving the answer in terms of $\pi$.一正圆锥底半径 $r = 3$、铅直高 $h = 4$。求其体积(以 $\pi$ 表示)。

(a) Quote the volume formula for a right circular cone.写出正圆锥体积公式。 [1]
(b) Substitute $r = 3$ and $h = 4$ and simplify.代入 $r = 3$、$h = 4$ 并化简。 [2]
(c) State the answer in the exact form $V = k\pi$ for an integer $k$.以 $V = k\pi$($k$ 为整数)的形式给出答案。 [1]
Q2MEDIUM Paper 1A C1.5 Arc Length and Sector Area [6 marks]

A sector of a circle has radius $r = 10$ and central angle $\theta = \dfrac{\pi}{3}$ radians.一扇形半径 $r = 10$,圆心角 $\theta = \dfrac{\pi}{3}$ 弧度。

(a) Quote the arc-length and sector-area formulas (radian forms).写出弧长与扇形面积公式(弧度形式)。 [2]
(b) Compute the arc length $s$ in exact form.求弧长 $s$ 的精确值。 [2]
(c) Compute the sector area $A$ in exact form.求扇形面积 $A$ 的精确值。 [2]
Q3MEDIUM Paper 1A C1.3 Surface Area of a Closed Cylinder [7 marks]

A closed right circular cylinder has radius $r = 5$ and height $h = 12$.一封闭正圆柱半径 $r = 5$、高 $h = 12$。

(a) State the area of one circular end.写出一个圆形端面的面积。 [1]
(b) Explain why the curved (lateral) surface unrolls into a rectangle of width $h$ and length $2\pi r$, and hence find its area.说明侧面(曲面)展开为宽 $h$、长 $2\pi r$ 的矩形,并求其面积。 [3]
(c) Add the three pieces (two ends plus curved surface) and give the total in the form $S = k\pi$.合并三部分(两端面加侧面),以 $S = k\pi$ 的形式给出总和。 [3]
Q4HARD Paper 1A C1.1 + C1.4 3D Distance and Angle to a Plane [8 marks]

Consider the points $A = (1, 2, -1)$ and $B = (4, 6, 3)$ in three-dimensional space.考虑空间中两点 $A = (1, 2, -1)$ 与 $B = (4, 6, 3)$。

(a) Find the distance $AB$ in exact form.求距离 $AB$ 的精确值。 [3]
(b) Find the length of the projection of $\overrightarrow{AB}$ onto the $xy$-plane (that is, the horizontal distance between $A$ and $B$).求 $\overrightarrow{AB}$ 在 $xy$ 平面上投影的长度(即 $A$ 与 $B$ 的水平距离)。 [2]
(c) Hence show that the angle $\theta$ that the line $AB$ makes with the $xy$-plane satisfies $\tan\theta = \dfrac{4}{5}$.由此证明直线 $AB$ 与 $xy$ 平面所成的角 $\theta$ 满足 $\tan\theta = \dfrac{4}{5}$。 [3]
PART II  ·  PAPER 1 SECTION B第二部分  ·  第一卷 B 节No calculator · extended response · 12 marks不可使用计算器 · 长答题 · 12 分

Section B · Extended ResponseB 节 · 长答题

Decompose composite solids into named pieces (hemisphere, cylinder, cone, disc) before applying any formula. State which faces are present and which are absent. Leave answers in exact form.先把组合立体分解为已命名的部件(半球、圆柱、圆锥、圆盘),再代入公式。指明哪些面存在、哪些面缺失。答案保留精确形式。

Q5HARD Paper 1B C1.2 + C1.3 Composite Solid (Hemisphere on Cylinder) [12 marks]

A solid trophy consists of a solid hemisphere of radius $r$ joined to the top of a solid right circular cylinder of the same radius $r$ and height $r$. The bottom face of the cylinder is a closed disc; the hemisphere sits flush on the top face of the cylinder (no gap, and the shared circular face is internal).一立体奖杯由一实心半球与一同半径的实心正圆柱组合而成,半球安置于圆柱顶面、与之齐平(无空隙,公共圆面为内部,不计入表面);圆柱底面为封闭圆盘。半球与圆柱半径均为 $r$,圆柱高为 $r$。

(a) Show that the total volume of the trophy is $V = \dfrac{5}{3}\pi r^{3}$.证明奖杯的总体积为 $V = \dfrac{5}{3}\pi r^{3}$。 [3]
(b) List the external faces (curved hemisphere, curved cylinder lateral, bottom disc) and show that the total external surface area is $S = 5\pi r^{2}$.列出外表面(半球曲面、圆柱侧面、底面圆盘),并证明总外表面积为 $S = 5\pi r^{2}$。 [4]
(c) Evaluate $V$ and $S$ when $r = 3$, giving both answers in the form $k\pi$ for an integer $k$.取 $r = 3$,分别求 $V$ 与 $S$,并以 $k\pi$($k$ 为整数)的形式给出。 [2]
(d) Show that $V/S = r/3$ for every $r > 0$, and verify that this implies $V = S$ at $r = 3$ (consistent with part (c)). State the geometric meaning of the ratio $V/S$.证明对每个 $r > 0$ 有 $V/S = r/3$,并验证由此得 $r = 3$ 时 $V = S$(与 (c) 一致)。说明比值 $V/S$ 的几何意义。 [3]
PART III  ·  PAPER 2第三部分  ·  第二卷Calculator · mixed response · 23 marks可使用计算器 · 混合题型 · 23 分

Paper 2 · Calculator Permitted第二卷 · 允许使用计算器

A graphing calculator is required. Round final numerical answers to three significant figures unless an exact form is requested. Show every formula before substituting; the method marks live on those lines.需要图形计算器(GDC)。除题目要求精确形式外,最终数值答案取三位有效数字。代值前每个公式都要写出来 — 方法分就在这些行上。

Q6MEDIUM Paper 2 C1.2 + C1.3 Cone Modelling (Slant, Volume, Equivalent Sphere) [10 marks]

A right circular cone has slant height $l = 13 \text{ cm}$ and base radius $r = 5 \text{ cm}$.一正圆锥斜高 $l = 13 \text{ cm}$,底半径 $r = 5 \text{ cm}$。

(a) Find the vertical height $h$ of the cone.求圆锥的铅直高 $h$。 [2]
(b) Find the total surface area of the closed cone (base plus curved surface), giving the answer in exact form.求该封闭圆锥的总表面积(底面加侧面),结果取精确形式。 [3]
(c) Find the volume of the cone in exact form, then give the decimal value to three significant figures.求圆锥体积的精确值,再给出三位有效数字的小数值。 [2]
(d) A solid sphere has the same volume as the cone. Find the radius of the sphere, giving the answer to three significant figures.某实心球的体积与该圆锥相等。求该球的半径,结果取三位有效数字。 [3]
Q7HARD Paper 2 C1.2 + C1.3 Optimisation (Open-Top Cylinder, Fixed Volume) [13 marks]

An open-top cylindrical can has radius $r$, height $h$, and fixed interior volume $V_{0} = 1000 \text{ cm}^{3}$ (so $V_{0}$ is a constant). The can has a closed circular base and a curved lateral surface, but no top.一无盖圆柱形罐半径为 $r$、高为 $h$、内体积固定为 $V_{0} = 1000 \text{ cm}^{3}$($V_{0}$ 为常数)。罐有封闭的圆形底面与曲面侧壁,但无顶。

(a) Use the volume constraint to express $h$ in terms of $r$ and $V_{0}$.由体积约束,将 $h$ 用 $r$ 与 $V_{0}$ 表示。 [1]
(b) Show that the total external surface area is $S(r) = \pi r^{2} + \dfrac{2 V_{0}}{r}$.证明外表面积为 $S(r) = \pi r^{2} + \dfrac{2 V_{0}}{r}$。 [2]
(c) Differentiate $S(r)$ with respect to $r$, set the derivative to zero, and show that the critical radius is $r^{*} = \left(\dfrac{V_{0}}{\pi}\right)^{1/3}$.对 $r$ 求导 $S(r)$,令导数为 $0$,证明临界半径为 $r^{*} = \left(\dfrac{V_{0}}{\pi}\right)^{1/3}$。 [4]
(d) Show that, at this critical radius, the optimal height satisfies $h^{*} = r^{*}$.证明在该临界半径处,最优高度满足 $h^{*} = r^{*}$。 [2]
(e) Use the second-derivative test to confirm that $r^{*}$ gives a minimum (not a maximum) of $S$.用二阶导数检验确认 $r^{*}$ 给出 $S$ 的极小值(非极大值)。 [2]
(f) For $V_{0} = 1000 \text{ cm}^{3}$, compute $r^{*}$, $h^{*}$, and the minimum surface area $S(r^{*})$, each to three significant figures.取 $V_{0} = 1000 \text{ cm}^{3}$,分别求 $r^{*}$、$h^{*}$、$S(r^{*})$,各取三位有效数字。 [2]