📚 Mensuration: Volume and Surface Area of Pyramids, Cones and Spheres | 测量：棱锥、圆锥和球体的体积与表面积
Mensuration questions on pyramids, cones and spheres are a very common part of the IGCSE mathematics exam. You need to be confident in applying the correct formula, identifying the right radius or height, and presenting your answer in the required units. These topics appear in both calculator and non-calculator papers, and they can be tested as straightforward 'plug-in' questions or as part of longer, multi-step problems involving compound solids, density or unit conversion.
关于棱锥、圆锥和球体的测量题是 IGCSE 数学考试中非常常见的部分。你需要能够自信地套用正确公式、识别对应的半径或高度，并按要求的单位写出答案。这些主题既会出现在计算器试卷中，也会出现在非计算器试卷中；它们既可能以直接代入公式的简单题形式考查，也可能作为涉及组合体、密度或单位换算的多步骤长题的一部分。
1. Understanding Mensuration in IGCSE | 理解 IGCSE 测量
Mensuration covers the measurement of lengths, areas and volumes of two-dimensional and three-dimensional shapes. In the IGCSE syllabus, pyramids, cones and spheres appear under the topic of shape, space and measures, and they are examined in both the Core and Extended tiers. The examiner is not only testing whether you can recall a formula, but whether you can read a diagram, decide which length is which, and then communicate a final answer with a sensible degree of accuracy. A large proportion of marks in this topic are method marks, so writing down the formula you are using is always worthwhile.
测量（mensuration）涵盖二维图形与三维图形的长度、面积和体积的度量。在 IGCSE 考纲中，棱锥、圆锥和球体属于"图形、空间与度量"这一主题，在 Core（基础）和 Extended（拓展）两个层级都会考查。考官考查的不只是你能否背出公式，还包括你能否读懂图形、判断哪条长度是哪一个量，并最终以合理的精度写出答案。这个主题中很大一部分分数是方法分，所以先写出所用公式永远是值得的。
- Vertex / apex 顶点 —— the single point at the top of a pyramid or cone. 棱锥或圆锥顶部的那个点。
- Perpendicular height 垂直高 —— the height measured at 90° to the base. 与底面成 90° 测得的高。
- Slant height 斜高 —— the height along a sloping face or edge. 沿着斜侧面或斜棱量出的高。
- Radius / diameter 半径 / 直径 —— remember d = 2r and r = d ÷ 2. 记住 d = 2r，r = d ÷ 2。
2. Key Formulae at a Glance | 核心公式速览
Before attempting any question, make sure you can reproduce the following formulae from memory. Notice the pattern: every solid that tapers to a point carries a factor of ⅓ compared with the prism or cylinder that just fits around it. The sphere is different, and its volume formula uses the fraction (4/3).
在做任何题之前，请确保你能凭记忆默写下列公式。注意其中的规律：所有收拢到一个顶点的立体，都比恰好包住它的棱柱或圆柱多一个 ⅓ 的系数。球体则不同，它的体积公式使用的是分数 (4/3)。
| Shape 立体 | Volume 体积 | Surface area 表面积 |
|---|---|---|
| Pyramid 棱锥 | V = ⅓ × base area × h | base area + area of all triangular faces |
| Cone 圆锥 | V = ⅓πr²h | curved = πrl; total = πrl + πr² |
| Sphere 球体 | V = (4/3)πr³ | A = 4πr² |
| Hemisphere 半球 | V = (2/3)πr³ | curved = 2πr²; total = 3πr² |
l = √r² + h² for a cone, and l = √(h² + (a ÷ 2)²) for a square-based pyramid
3. Volume of a Pyramid | 棱锥的体积
The volume of any pyramid, whatever shape its base has, is one third of the base area multiplied by the perpendicular height. The ⅓ exists because exactly three congruent pyramids can be packed into a prism with the same base and the same height. A very common error is to use the slant edge as the height; the height must be perpendicular to the base, measured straight up from the base to the apex.
无论底面是什么形状，棱锥的体积都等于底面积乘以垂直高的三分之一。这个 ⅓ 之所以存在，是因为恰好三个全等的棱锥可以拼满一个与它同底同高的棱柱。非常常见的错误是把斜棱当作高；高必须垂直于底面，从底面竖直量到顶点。
V = ⅓ × base area × perpendicular height
se is a triangle instead, calculate the triangular area first, then apply the same ⅓ rule. For a cone the base is a circle, so the base area is πr², giving V = ⅓πr²h.
例题：一个棱锥底面为边长 6 cm 的正方形，垂直高为 10 cm。底面积为 6 × 6 = 36 cm²，所以 V = ⅓ × 36 × 10 = 120 cm³。如果底面是三角形，先算出三角形面积，再套用同样的 ⅓ 规则。对于圆锥，底面是圆，底面积为 πr²，于是 V = ⅓πr²h。
4. Surface Area of a Pyramid | 棱锥的表面积
pyramid with base edge a and slant height l, the four triangular faces each have area ½ × a × l, so the total surface area is a² + 2al. To find l when it is not given, use Pythagoras on the right-angled triangle formed by the vertical height, half the base edge and the slant height.
表面积指的是所有面的面积总和，除非题目只要求侧面（lateral）面积，否则要包括底面。对于底面边长为 a、斜高为 l 的正四棱锥，四个三角形侧面每个面积为 ½ × a × l，因此总表面积为 a² + 2al。若题目没有给出 l，可利用"垂直高、底面半边长、斜高"构成的直角三角形，用勾股定理求出。
Total SA = a² + 2al, where l = √(h² + (a ÷ 2)²)
re-based pyramid has base edge 6 cm and perpendicular height 4 cm. The slant height is l = √4² + 3² = √25 surface area is 6² + 2 × 6 × 5 = 36 + 60 = 96 cm². Read the question carefully: if it says "the outside of the pyramid is painted", the base is included; if it says "the sides are made of glass", it is not.
例题：正四棱锥底面边长 6 cm，垂直高 4 cm。斜高 l = √4² + 3² = √25 = 5 cm。总表面积为 6² + 2 × 6 × 5 = 36 + 60 = 96 cm²。读题时要仔细：如果题目说"棱锥外表面被涂漆"，则包含底面；如果说"侧面由玻璃制成"，则不包含底面。
5. Volume of a Cone | 圆锥的体积
A cone is simply a pyramid with a circular base, so the same ⅓ relationship applies. Three cones of ice cream fit exactly into the cylindrical tub that surrounds them, which is the physical reason behind the formula. You must use the radius, not the diameter, and the perpendicular height, not the slant height.
圆锥本质上就是底面为圆的棱锥，因此同样的 ⅓ 关系成立。三个圆锥形的冰淇淋恰好能装满包住它们的圆柱形桶，这就是公式背后的直观原因。你必须使用半径而不是直径，使用垂直高而不是斜高。
V = ⅓πr²h
In a non-calculator paper you should leave the answer as 21π cm³; in a calculator paper, give three significant figures unless the question says otherwise. If the question gives the diameter as 6 cm, the radius is 3 cm before you even start.
例题：圆锥半径为 3 cm，垂直高为 7 cm。则 V = ⅓ × π × 3² × 7 = ⅓ × π × 9 × 7 = 21π ≈ 65.97 cm³。在非计算器试卷中，答案应写成 21π cm³；在计算器试卷中，除非题目另有说明，一般保留三位有效数字。如果题目给出直径是 6 cm，那么一开始半径就已经是 3 cm。
6. Curved and Total Surface Area of a Cone | 圆锥的侧面积与总表面积
When a cone is cut along a slant edge and opened out flat, the curved surface becomes a sector whose radius is the slant height l and whose arc length is the circumference 2πr. Working out the area of that sector gives the elegant result πrl. Because l is always longer than r, the curved surface area is always greater than the base area πr².
把一个圆锥沿一条斜棱剪开并摊平，侧面就变成一个扇形，其半径是斜高 l，弧长是底面周长 2πr。计算这个扇形的面积，就得到漂亮的结果 πrl。由于 l 总是比 r 长，所以侧面积总大于底面积 πr²。
Curved SA = πrl Total SA = πrl + πr² l = √r² + h²
ed example: a cone has radius 5 cm and perpendicular height 12 cm. The slant height is l = √5² + 12² m². Notice how often 5, 12, 13 and 3, 4, 5 appear in exam questions; recognising these Pythagorean triples saves valuable time.
例题：圆锥半径为 5 cm，垂直高为 12 cm。斜高 l = √5² + 12² = √169 = 13 cm。侧面积 = π × 5 × 13 = 65π ≈ 204.2 cm²。总表面积 = 65π + π × 5² = 65π + 25π = 90π ≈ 282.7 cm²。注意考试题中经常出现 5、12、13 和 3、4、5 这两组数；能识别这些勾股数可以节省宝贵时间。
7. Volume and Surface Area of a Sphere | 球体的体积与表面积
A sphere is the set of all points at a fixed distance r from a centre. It has no edges and no vertices, and every point on its surface is the same distance from the centre, which is why a single formula describes the whole surface. The surface area 4πr² is exactly four times the area of the great circle πr², a fact worth remembering as a memory aid.
球体是所有到球心距离为定值 r 的点的集合。它没有棱也没有顶点，表面上每一点到球心的距离都相同，因此一个公式就能描述整个表面。表面积 4πr² 恰好是"大圆"面积 πr² 的四倍，这一点可以当作记忆辅助来记住。
V = (4/3)πr³ A = 4πr²
². Remember that volume is measured in cubic units (cm³, m³) while surface area is measured in square units (cm², m²). Writing the wrong unit costs marks even when the number is correct.
例题：球体半径为 6 cm。体积 V = (4/3) × π × 6³ = (4/3) × π × 216 = 288π ≈ 904.8 cm³，表面积 A = 4 × π × 6² = 144π ≈ 452.4 cm²。记住体积用立方单位（cm³、m³），表面积用平方单位（cm²、m²）。数字正确但单位写错，同样会丢分。
8. Hemispheres, Frustums and Compound Solids | 半球、圆台与组合体
Half of a sphere is a hemisphere, and all of its formulae are simply halved. Its volume is (2/3)πr³ and its curved surface area is 2πr². If the flat circular face is exposed, add πr², giving a total surface area of 3πr². Compound solids are extremely common in IGCSE papers: a cone on top of a cylinder, a hemisphere on top of a cone, or a sphere with a cylindrical hole drilled through it. The strategy is always the same - split the solid into parts, calculate each part separately, then add the volumes or the relevant surface areas, taking care not to count internal faces that are no longer visible.
球体的一半就是半球，它的所有公式都只需除以二。体积为 (2/3)πr³，侧面积为 2πr²。如果那个平的圆形面外露，就要加上 πr²，总表面积为 3πr²。组合体在 IGCSE 试卷中极为常见：圆柱上放一个圆锥、圆锥上放一个半球，或者球体中间钻一个圆柱孔。解题策略永远一样——把立体拆成几部分，分别计算，再把体积或相关的表面积相加，同时注意不要计算已经看不见的内部面。
Hemisphere: V = (2/3)πr³, curved SA = 2πr², total SA = 3πr²
+ r²), where R and r are the two radii and h is the vertical height between them. For most IGCSE candidates, however, the safer route with a frustum is to subtract the small cone from the large one rather than to memorise this formula.
圆台（圆锥平行于底面截去顶部后得到的立体）有时会作为拓展内容出现。它的体积为 V = ⅓πh(R² + Rr + r²)，其中 R 和 r 是两个半径，h 是它们之间的垂直高。不过对大多数 IGCSE 考生来说，处理圆台更稳妥的做法是用大圆锥减去小圆锥，而不是去背这个公式。
9. Finding the Missing Radius, Height or Slant Height | 求缺失的半径、高或斜高
Many exam questions reverse the formula and ask you to find a length. Treat the formula as an equation and rearrange it carefully, substituting the known values first and solving second. This keeps the algebra simple and reduces the chance of a rearrangement slip.
很多考题会把公式反过来用，要求你求某条长度。把公式当作方程，小心地变形：先代入已知数值，再解方程。这样代数运算更简单，也能减少变形出错的概率。
- Cone volume 96π cm³, radius 4 cm: ⅓π(16)h = 96π → 16h ÷ 3 = 96 → h = 18 cm. 圆锥体积 96π cm³，半径 4 cm：⅓π(16)h = 96π → 16h ÷ 3 = 96 → h = 18 cm。
- Cone radius 8 cm, slant height 17 cm: h = √17² − 8² = √225 = 15 cm. 圆锥半径 8 cm、斜高 17 cm：h = √17² − 8² = √225 = 15 cm。
- Sphere volume 36π cm³: (4/3)πr³ = 36π → r³ = 27 → r = 3 cm. 球体体积 36π cm³：(4/3)πr³ = 36π → r³ = 27 → r = 3 cm。
- General rearrangement: r = ∛(3V ÷ 4π) for a sphere, and r = √3V ÷ πh for a cone. 一般变形：球体 r = ∛(3V ÷ 4π)，圆锥 r = √3V ÷ πh。
10. Units, Density and Mass | 单位、密度与质量
1 000 000. A capacity of 1 litre is exactly 1000 cm³, so 1 m³ holds 1000 litres. Always convert every measurement into the same unit before you substitute it into a formula.
单位换算是许多本来很强的考生丢分的地方。长度换算的倍数是 100，面积是 100² = 10 000，体积是 100³ = 1 000 000。1 升的容积恰好是 1000 cm³，所以 1 m³ 可容纳 1000 升。代入公式之前，一定要把所有测量值换算成同一单位。
1 m = 100 cm 1 m² = 10 000 cm² 1 m³ = 1 000
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