📚 IGCSE Maths: Mensuration – Perimeter, Area, Surface Area and Volume | IGCSE数学：测量——周长、面积、表面积与体积
Mensuration is the branch of mathematics that deals with measuring lengths, areas and volumes of 2D and 3D shapes. In IGCSE Maths you are expected to recall key formulas, apply them to compound shapes, convert units confidently, and solve problems involving similar figures.
测量是数学中处理二维和三维图形长度、面积与体积计算的分支。在 IGCSE 数学中，你需要熟记关键公式，将其应用于组合图形，熟练转换单位，并解决涉及相似图形的问题。
1. Perimeter and Area of Rectangles, Triangles and Parallelograms | 矩形、三角形与平行四边形的周长和面积
ts outside edge. For a rectangle with length l and width w, the perimeter is P = 2(l + w).
周长是图形外边界的总长度。对于长为 l、宽为 w 的矩形，周长公式为 P = 2(l + w)。
. The height must be measured at 90° to the base you have chosen, not along a sloping side.
矩形的面积是 A = l × w；三角形的面积等于底乘以垂直高度的一半：A = ½ × b × h。这里的高必须与你所选的底成 90°，而不是沿着斜边测量。
rallel sides. A trapezium with parallel sides a and b and height h has area A = ½(a + b)h.
平行四边形面积为 A = b × h，其中 h 是两条平行边之间的垂直距离。两条平行边为 a、b，高为 h 的梯形面积为 A = ½(a + b)h。
A = ½ × b × h | A = ½(a + b)h
um with parallel sides 8 cm and 12 cm and height 5 cm has area A = ½(8 + 12) × 5 = 50 cm².
例如：平行边分别为 8 cm 和 12 cm、高为 5 cm 的梯形，面积为 A = ½(8 + 12) × 5 = 50 cm²。
| Shape 图形 | Area 面积 | Perimeter 周长 |
|---|---|---|
| Rectangle 矩形 | A = l × w | P = 2(l + w) |
| Triangle 三角形 | A = ½ × b × h | Sum of all sides 所有边长之和 |
| Parallelogram 平行四边形 | A = b × h | 2(a + b) |
| Trapezium 梯形 | A = ½(a + b)h | Sum of all sides 所有边长之和 |
Notice that area is always measured in square units (cm², m², mm²) while perimeter is measured in the same units as the lengths (cm, m, mm).
注意：面积单位永远是平方单位（cm²、m²、mm²），而周长的单位与长度单位相同（cm、m、mm）。
2. Circles: Circumference, Area, Arcs and Sectors | 圆：周长、面积、弧长与扇形
d it: C = πd or C = 2πr, where d is the diameter and r is the radius. The area is A = πr².
圆的周长是围绕圆一周的长度：C = πd 或 C = 2πr，其中 d 是直径，r 是半径。圆面积为 A = πr²。
C = 2πr | A = πr²
ntre is θ°, then the arc length is (θ ÷ 360) × 2πr and the sector area is (θ ÷ 360) × πr².
弧是圆周的一部分，扇形是圆面积的一部分。若圆心角为 θ°，则弧长为 (θ ÷ 360) × 2πr，扇形面积为 (θ ÷ 360) × πr²。
Arc = (θ ÷ 360) × 2πr | Sector area = (θ ÷ 360) × πr²
the area is A = 3.142 × 7² ≈ 154 cm² and the circumference is C = 2 × 3.142 × 7 ≈ 44.0 cm.
例题：圆的半径为 7 cm。取 π ≈ 3.142，则面积 A = 3.142 × 7² ≈ 154 cm²，周长 C = 2 × 3.142 × 7 ≈ 44.0 cm。
stion asks for the perimeter of a sector, remember to add the two radii to the arc length.
半圆就是 θ = 180° 的扇形，四分之一圆是 θ = 90° 的扇形。当题目要求扇形的周长时，记得把两条半径加到弧长上。
3. Compound Shapes: Adding and Subtracting Regions | 组合图形：面积的分割与填补
A compound shape is built from two or more simple shapes. The standard method is to split it into rectangles, triangles, circles or trapezia, calculate each area separately and then add them together.
组合图形由两个或多个简单图形拼成。标准做法是把它分割成矩形、三角形、圆或梯形，分别计算每部分面积后再相加。
When a simple shape has been removed from a larger one, subtract the smaller area. This is often called the shaded region method and appears frequently in IGCSE Paper 2 and Paper 4.
当从大图形中挖去一个小图形时，应做减法。这种方法常被称为阴影区域法，在 IGCSE 试卷 2 和试卷 4 中经常出现。
- Step 1: Draw dashed lines to split the shape, and mark every given length on your diagram. | 第一步：用虚线把图形分割开，并把所有已知长度标在图上。
- Step 2: Identify the type of each piece and write down the matching formula. | 第二步：判断每一块的图形类型，写出对应公式。
- Step 3: Check that all lengths are in the same unit before calculating. | 第三步：计算前确认所有长度单位一致。
- Step 4: Add or subtract, then state the correct square unit in your answer. | 第四步：相加减，并在答案中写出正确的平方单位。
e of base 4 cm and height 3 cm cut from one corner. The remaining area is 60 - 6 = 54 cm².
例题：长 10 cm、宽 6 cm 的矩形，从一个角上剪去底 4 cm、高 3 cm 的三角形，剩余面积为 60 - 6 = 54 cm²。
4. Units for Length, Area and Volume and How to Convert Them | 长度、面积与体积的单位换算
Unit conversion is a routine source of lost marks. Length conversions are linear, area conversions are squared and volume conversions are cubed.
单位换算是常见的失分点。长度换算是线性的，面积换算要平方，体积换算要立方。
1 m² = 100² cm² = 10 000 cm² → 1 m³ = 100³ cm³ = 1 000 000 cm³
One litre is a capacity measure equal to 1000 cm³, which is the same as 0.001 m³. A millilitre is 1 cm³. These appear in questions about tanks, bottles and containers.
1 升是容量单位，等于 1000 cm³，也就是 0.001 m³。1 毫升等于 1 cm³。这类换算常出现在关于水箱、瓶子和容器的题目中。
| Conversion 换算 | Factor 倍数 |
|---|---|
| 1 m = 100 cm | × 100 |
| 1 m² = 10 000 cm² | × 100² |
| 1 m³ = 1 000 000 cm³ | × 100³ |
| 1 litre = 1000 cm³ | = 1000 cm³ |
| 1 km = 1000 m | × 1000 |
10 000 = 25 000 cm². Convert 450 cm³ into litres by dividing by 1000, giving 0.45 litres.
例题：把 2.5 m² 换算成 cm²。因为 1 m² = 10 000 cm²，所以 2.5 × 10 000 = 25 000 cm²。把 450 cm³ 换算成升，除以 1000，得 0.45 升。
5. Prisms: Volume and Surface Area, Including Cylinders | 棱柱：体积与表面积（含圆柱）
ng its length. Its volume is the cross-sectional area multiplied by the length: V = A × L.
棱柱是沿长度方向横截面保持不变的立体图形。它的体积等于横截面积乘以长度：V = A × L。
V = A × L (A = cross-sectional area, L = length)
Its total surface area is the sum of the areas of all faces: two identical end faces plus the rectangular (or curved) side faces, which together form a rectangle of length L and width equal to the perimeter of the cross-section.
棱柱的总表面积等于所有面的面积之和：两个相同的端面，加上侧面（长方形或曲面），侧面展开后是一个长为 L、宽等于横截面周长的长方形。
e is V = πr²h, its curved surface area is 2πrh, and its total surface area is 2πrh + 2πr².
圆柱是横截面为圆的棱柱。它的体积 V = πr²h，侧面积（曲面面积）为 2πrh，总表面积为 2πrh + 2πr²。
V = πr²h | Curved SA = 2πrh | Total SA = 2πrh + 2πr²
cm³ and its total surface area is (2π × 5 × 12) + (2π × 5²) = 120π + 50π = 170π ≈ 534 cm².
例题：一个封闭圆柱的半径为 5 cm，高为 12 cm。体积为 π × 5² × 12 = 300π ≈ 942 cm³，总表面积为 (2π × 5 × 12) + (2π × 5²) = 120π + 50π = 170π ≈ 534 cm²。
Read the wording carefully: "closed" means both circular ends are included, while "open" or "hollow" usually means you should only count the curved surface and possibly one base.
要仔细读题："封闭"表示包含两个圆形底面，而"开口"或"空心"通常只计算曲面，有时再加一个底面。
6. Pyramids, Cones and Spheres | 棱锥、圆锥与球
olume is one third of the base area times the perpendicular height: V = ⅓ × base area × h.
棱锥有一个平的底面，各个三角形侧面交于顶点。它的体积等于底面积乘以垂直高度的三分之一：V = ⅓ × 底面积 × h。
e is V = ⅓πr²h. Its curved surface area is πrl, where l is the slant height found from l = √r² + h².
圆锥是底面为圆的棱锥，因此体积 V = ⅓πr²h。它的曲面积为 πrl，其中 l 是斜高，由 l = √r² + h² 求得。
V = ⅓πr²h | Curved SA = πrl | l = √r² + h²
e is V = 4πr³ ÷ 3 and its surface area is A = 4πr². A hemisphere is half of each of these.
球的体积为 V = 4πr³ ÷ 3，表面积为 A = 4πr²。半球则是这两者的一半。
V = 4πr³ ÷ 3 | A = 4πr²
volume is 4π × 216 ÷ 3 = 288π ≈ 905 cm³ and its surface area is 4π × 36 = 144π ≈ 452 cm².
例题：一个球的半径为 6 cm。体积为 4π × 216 ÷ 3 = 288π ≈ 905 cm³，表面积为 4π × 36 = 144π ≈ 452 cm²。
7. Similar Figures: Length, Area and Volume Ratios | 相似图形：长度比、面积比与体积比
Two shapes are similar if one is an enlargement of the other, so all corresponding angles are equal and all corresponding lengths are in the same ratio. If the length scale factor is k, then the area scale factor is k² and the volume scale factor is k³.
如果两个图形是相似图形，那么一个图形是另一个的放大或缩小，对应角相等，对应长度成同一比例。若长度比为 k，则面积比为 k²，体积比为 k³。
ngth ratio k → Area ratio k² → Volume ratio k³
reas are in the ratio 2² : 3² = 4 : 9 and their volumes are in the ratio 2³ : 3³ = 8 : 27.
例题：两个相似立体的长度比为 2 : 3，则它们的表面积比为 2² : 3² = 4 : 9，体积比为 2³ : 3³ = 8 : 27。
o the volume scale factor is 10³ = 1000 and the real volume is 500 × 1000 = 500 000 cm³ = 0.5 m³.
典型考题：某雕像的小模型高 30 cm，体积 500 cm³，真实雕像高 3 m。长度比为 300 ÷ 30 = 10，所以体积比为 10³ = 1000，真实体积为 500 × 1000 = 500 000 cm³ = 0.5 m³。
A common error is to apply the length ratio directly to an area or a volume. Always square or cube the scale factor first.
常见错误是把长度比直接用于面积或体积。一定要先把比例因子平方或立方。
8. Density, Speed and Other Rate Problems | 密度、速度与其他比率问题
ty relationship. Density is mass divided by volume, so D = M ÷ V, M = D × V and V = M ÷ D.
测量公式经常与密度关系结合考查。密度等于质量除以体积，所以 D = M ÷ V，M = D × V，V = M ÷ D。
D = M ÷ V | M = D × V | V = M ÷ D
nd volume 200 cm³. Its density is 540 ÷ 200 = 2.7 g/cm³, which identifies it as aluminium.
例题：一块金属质量 540 g，体积 200 cm³。密度为 540 ÷ 200 = 2.7 g/cm³，可判断为铝。
Similar reasoning applies to speed, flow rate and population density: the quantity per unit is found by dividing, and the missing value is found by rearranging. Rate problems reward neat substitution rather than memorising extra formulas.
同样的思路适用于速度、流量和人口密度：单位量由除法求得，未知量通过公式变形求出。比率类题目只要代入整齐即可，无需额外背公式。
inute. The tank volume is 72 000 cm³ = 72 litres, so it takes 72 ÷ 3 = 24 minutes to fill.
水箱例题：水以每分钟 3 升的速率注入长 60 cm、宽 40 cm、高 30 cm 的长方体水箱。水箱体积为 72 000 cm³ = 72 升，因此注满需要 72 ÷ 3 = 24 分钟。
9. Working Backwards, Rearranging and Accuracy | 反求、公式变形与精度控制
Many exam questions give the area or volume and ask for a missing length. Substitute the known values into the formula and solve, keeping your working in exact form until the final step.
很多考题给出面积或体积，反求缺失的长度。把已知量代入公式并解方程，在最后一步之前尽量保留精确形式。
a of a triangle is 45 cm² and the base is 12 cm, then 45 = ½ × 12 × h = 6h, so h = 7.5 cm.
例如：三角形面积为 45 cm²，底为 12 cm，则 45 = ½ × 12 × h = 6h，所以 h = 7.5 cm。
er is 500 cm³ and the radius is 4 cm, then 500 = π × 16 × h, so h = 500 ÷ (16π) ≈ 9.95 cm.
若圆柱体积为 500 cm³，半径为 4 cm，则 500 = π × 16 × h，所以 h = 500 ÷ (16π) ≈ 9.95 cm。
Retain π in your working when the answer is asked as a multiple of π, and round only at the end. Never round intermediate values, because small early rounding errors are amplified by squaring and cubing.
当答案要求写成 π 的倍数时，解题过程中应保留 π，只在最后一步取近似值。不要在中间步骤四舍五入，因为平方和立方会把早期的舍入误差放大。
- Money and real-life measurements: usually 2 decimal places or the nearest whole unit. | 金钱与现实测量：通常保留 2 位小数或取整数。
- Lengths: usually 1 to 3 significant figures, unless told otherwise. | 长度：除题目另有要求，通常保留 1 至 3 位有效数字。
- Angles: usually 1 decimal place. | 角度：通常保留 1 位小数。
- Always write the unit in your final answer line. | 最后答案中一定要写出单位。
10. Exam Technique and Common Mistakes | 考试技巧与常见错误
Mensuration questions are often worth 3 to 5 marks, and method marks are awarded for correct formula choice even if the arithmetic slips. Always show a clear substitution line.
测量题目通常占 3 至 5 分，即使算错，只要公式选择正确也能得到方法分。因此一定要写出清晰的代入过程。
