📚 A-Level Maths: Sector & Segment Area | A-Level 数学：扇形与弓形面积
In this revision guide, we explore the formulas and methods for calculating the area of sectors and segments of a circle, a key topic in A-Level Mathematics. You will learn how to use radians effectively, derive the area of a segment from first principles, and apply these techniques to exam-style questions.
在本复习指南中，我们将深入探讨如何计算圆的扇形与弓形面积，这是 A-Level 数学中的核心考点。你将学习如何熟练使用弧度制、从基本原理推导弓形面积公式，并将这些技巧应用于考试风格的题目中。
1. Radians and Degrees | 弧度与角度
Before we calculate areas, it is essential to understand angular measurement. At A-Level, radians are the preferred unit because they simplify formulas. One full revolution equals 2π radians, which is equivalent to 360°.
在计算面积之前，理解角度的度量方式至关重要。在 A-Level 中，弧度是首选单位，因为它能简化公式。一周等于 2π 弧度，相当于 360°。
Degrees to radians: multiply by π/180.
角度转弧度：乘以 π/180。
Radians to degrees: multiply by 180/π.
弧度转角度：乘以 180/π。
Common conversions: π/6 = 30°, π/4 = 45°, π/3 = 60°, π/2 = 90°.
常见换算：π/6 = 30°，π/4 = 45°，π/3 = 60°，π/2 = 90°。
For example, an angle of 120° in radians is calculated as 120 × π/180 = 2π/3.
例如，120° 转换为弧度为 120 × π/180 = 2π/3。
Angle in radians θ = angle in degrees × π/180
弧度 θ = 角度 × π/180
2. Arc Length | 弧长
The arc length of a sector is the distance along the curved edge between the two radii. When θ is measured in radians, the arc length is simply the radius multiplied by the angle.
扇形的弧长是沿着两条半径之间的弯曲边缘的距离。当 θ 以弧度为单位时，弧长等于半径乘以圆心角。
Arc length s = rθ
弧长 s = rθ
However, working in radians is far more efficient and is expected in most exam questions.
如果 θ 以角度为单位，弧长公式为 s = (θ/360) × 2πr。然而，使用弧度计算效率更高，也是大多数考试题目所要求的。
ius 6 cm and a sector angle of π/3 radians has an arc length of 6 × π/3 = 2π cm ≈ 6.28 cm.
例如，半径为 6 cm、扇形圆心角为 π/3 弧度的圆，其弧长为 6 × π/3 = 2π cm ≈ 6.28 cm。
3. Sector Area Formula | 扇形面积公式
A sector of a circle is the region enclosed by two radii and the arc between them. Visually, it resembles a slice of pizza. The area of a sector is proportional to the central angle it subtends at the centre.
圆的扇形是由两条半径及其之间的弧所围成的区域。从视觉上看，它类似于一块披萨。扇形的面积与它所对应的圆心角成正比。
Sector area A = ½ r²θ
扇形面积 A = ½ r²θ
represents the fraction θ/(2π) of the full circle, so the area is (θ/(2π)) × πr² = ½ r²θ.
此公式要求 θ 必须使用弧度。其推导源于整圆的面积为 πr²，而圆心角为 θ 的扇形占整个圆的比例为 θ/(2π)，因此面积为 (θ/(2π)) × πr² = ½ r²θ。
example, if r = 5 cm and θ = π/4, the sector area is ½ × 5² × π/4 = 25π/8 cm² ≈ 9.82 cm².
例如，若 r = 5 cm，θ = π/4，则扇形面积为 ½ × 5² × π/4 = 25π/8 cm² ≈ 9.82 cm²。
4. Perimeter of a Sector | 扇形的周长
Many students confuse area and perimeter. The perimeter of a sector consists of the arc length plus two radii. Since the arc length is rθ and the two radii contribute 2r, the total perimeter is:
许多学生容易混淆面积和周长。扇形的周长由弧长加上两条半径组成。由于弧长为 rθ，两条半径合计 2r，因此总周长为：
Perimeter of sector = rθ + 2r = r(θ + 2)
扇形周长 = rθ + 2r = r(θ + 2)
This formula is often tested alongside area calculations in the same question. Remember to keep θ in radians.
这个公式常常与面积计算在同一道题目中一起考查。请记住 θ 要用弧度制表示。
5. Segment of a Circle | 圆的弓形
A segment of a circle is the region bounded by a chord and the arc it subtends. There are two types: minor segment (smaller than a semicircle) and major segment (larger than a semicircle).
圆的弓形是由弦及其所对的弧围成的区域。弓形分为两种：劣弓形（小于半圆）和优弓形（大于半圆）。
To find the area of a minor segment, we subtract the area of the isosceles triangle formed by the two radii and the chord from the area of the sector.
要求劣弓形的面积，我们从扇形面积中减去由两条半径和弦构成的等腰三角形的面积。
Segment area = Sector area − Triangle area
弓形面积 = 扇形面积 − 三角形面积
The triangle has two sides equal to the radius r, with the included angle θ. Its area is given by ½ r² sin θ. Therefore, the segment area is:
该三角形的两条边等于半径 r，夹角为 θ。其面积为 ½ r² sin θ。因此，弓形面积为：
A_segment = ½ r²θ − ½ r² sin θ = ½ r²(θ − sin θ)
弓形面积 = ½ r²θ − ½ r² sin θ = ½ r²(θ − sin θ)
6. Deriving the Segment Formula | 推导弓形面积公式
Let us derive the segment formula step by step, as this shows a deep understanding expected for higher-grade questions.
让我们逐步推导弓形面积公式，这体现了对高难度题目所期望的深层理解。
e has area πr² and the sector angle is θ radians, the sector area is (θ/2π) × πr² = ½ r²θ.
步骤 1：扇形面积是整圆面积的一部分。由于整圆面积为 πr²，扇形圆心角为 θ 弧度，所以扇形面积为 (θ/2π) × πr² = ½ r²θ。
Step 2: The triangle formed by the two radii and the chord is isosceles with equal sides r and included angle θ. Its area is ½ r² sin θ.
步骤 2：由两条半径和弦构成的三角形为等腰三角形，两腰为 r，夹角为 θ。其面积为 ½ r² sin θ。
Step 3: Subtract the triangle area from the sector area to obtain the segment area:
步骤 3：将扇形面积减去三角形面积即得弓形面积：
A_segment = ½ r²θ − ½ r² sin θ = ½ r²(θ − sin θ)
弓形面积 = ½ r²θ − ½ r² sin θ = ½ r²(θ − sin θ)
e segment is a semicircle and the formula gives ½ r²(π − sin π) = ½ πr², which is correct.
该推导适用于 0 < θ < 2π。当 θ = π 时，弓形为半圆，公式给出 ½ r²(π − sin π) = ½ πr²，结果正确。
7. Worked Example 1 | 例题精讲 1
A circle has radius 8 cm and a sector with angle 1.2 radians. Find the area of the sector and the area of the corresponding segment.
一个圆的半径为 8 cm，其中扇形的圆心角为 1.2 弧度。求该扇形的面积以及对应弓形的面积。
Solution: Sector area = ½ r²θ = ½ × 8² × 1.2 = ½ × 64 × 1.2 = 38.4 cm².
解：扇形面积 = ½ r²θ = ½ × 8² × 1.2 = ½ × 64 × 1.2 = 38.4 cm²。
Triangle area = ½ r² sin θ = ½ × 64 × sin(1.2) ≈ 32 × 0.9320 ≈ 29.82 cm².
三角形面积 = ½ r² sin θ = ½ × 64 × sin(1.2) ≈ 32 × 0.9320 ≈ 29.82 cm²。
Segment area = 38.4 − 29.82 ≈ 8.58 cm².
弓形面积 = 38.4 − 29.82 ≈ 8.58 cm²。
½ r²(θ − sin θ) = ½ × 64 × (1.2 − sin 1.2) ≈ 32 × (1.2 − 0.9320) = 32 × 0.2680 ≈ 8.58 cm².
或者直接使用公式：弓形面积 ½ r²(θ − sin θ) = ½ × 64 × (1.2 − sin 1.2) ≈ 32 × (1.2 − 0.9320) = 32 × 0.2680 ≈ 8.58 cm²。
8. Worked Example 2 | 例题精讲 2
A sector has an area of 50 cm² and a radius of 10 cm. Find the angle θ in radians and the arc length.
一个扇形的面积为 50 cm²，半径为 10 cm。求圆心角 θ（以弧度表示）及弧长。
Solution: Using A = ½ r²θ, we substitute: 50 = ½ × 10² × θ = 50θ. Hence θ = 1 radian.
解：使用 A = ½ r²θ，代入得：50 = ½ × 10² × θ = 50θ。因此 θ = 1 弧度。
Arc length s = rθ = 10 × 1 = 10 cm.
弧长 s = rθ = 10 × 1 = 10 cm。
This example demonstrates the power of the radian formulas: once θ is found, the arc length follows immediately without converting back to degrees.
此例题展示了弧度公式的优势：一旦求出 θ，弧长便可以直接得出，无需再转换回角度制。
9. Worked Example 3 | 例题精讲 3
A chord of length 12 cm is drawn in a circle of radius 10 cm. Find the area of the minor segment.
在一个半径为 10 cm 的圆中，一条长度为 12 cm 的弦被画出。求劣弓形的面积。
(θ/2), so sin(θ/2) = 0.6. Hence θ/2 = arcsin(0.6) ≈ 0.6435 radians, and θ ≈ 1.287 radians.
步骤 1：利用弦长公式求 θ。对于圆，弦长 c = 2r sin(θ/2)。因此 12 = 2 × 10 × sin(θ/2)，所以 sin(θ/2) = 0.6。因此 θ/2 = arcsin(0.6) ≈ 0.6435 弧度，θ ≈ 1.287 弧度。
θ − sin θ) = ½ × 100 × (1.287 − sin 1.287) ≈ 50 × (1.287 − 0.96) = 50 × 0.327 ≈ 16.35 cm².
步骤 2：弓形面积 (θ − sin θ) = ½ × 100 × (1.287 − sin 1.287) ≈ 50 × (1.287 − 0.96) = 50 × 0.327 ≈ 16.35 cm²。
Notice that we needed to compute sin θ, not just sin(θ/2). Use a calculator carefully to avoid rounding errors.
注意我们需要计算 sin θ 而不仅仅是 sin(θ/2)。计算时务必小心，以避免舍入误差。
10. Common Mistakes | 常见错误
Students often lose marks in this topic due to a few recurring errors. Knowing these will help you avoid them in your exam.
学生在这个知识点上常因几个反复出现的错误而失分。了解这些错误有助于你在考试中避免。
ees in the radian formulas: Always convert to radians before applying A = ½ r²θ or s = rθ.
在弧度公式中使用角度制：在应用 A = ½ r²θ 或 s = rθ 之前，务必先转换为弧度。
Forgetting the ½ in the triangular area formula for the segment: the triangle area is ½ r² sin θ, not r² sin θ.
在弓形面积计算中遗漏三角形面积公式中的 ½：三角形面积为 ½ r² sin θ，而不是 r² sin θ。
Mixing up arc length and perimeter: the perimeter of a sector includes two radii, not just the arc.
混淆弧长和周长：扇形的周长包含两条半径，而不仅仅是弧长。
Using the wrong trigonometric function: chord length is 2r sin(θ/2), not r sin θ.
使用错误的三角函数：弦长公式为 2r sin(θ/2)，而不是 r sin θ。
Always draw a diagram and label the sector, triangle, and segment to visualise the relationship between them.
务必画出图形并标注扇形、三角形和弓形，以直观理解它们之间的关系。
11. Summary of Formulas | 公式总结
The table below consolidates all the key formulas you need for this topic. Save it for quick revision before your exam.
下表汇总了本主题所需的所有关键公式，请在考试前保存以便快速复习。
| Quantity | Formula (θ in radians) |
| 量 | 公式（θ 为弧度） |
| Arc length / 弧长 | s = rθ |
| Sector area / 扇形面积 | A = ½ r²θ |
| Triangle area / 三角形面积 | A = ½ r² sin θ |
| Segment area / 弓形面积 | A = ½ r²(θ − sin θ) |
| Chord length / 弦长 | c = 2r sin(θ/2) |
| Sector perimeter / 扇形周长 | P = r(θ + 2) |
Memorise the derivation of the segment formula so that even if you forget it, you can reconstruct it quickly.
牢记弓形面积公式的推导过程，这样即使你忘记了公式，也能快速重新推导出来。
12. Practice Questions | 练习题目
Test your understanding with these problems. Fully solve each one before checking your answers.
通过以下问题测试你的理解。请先完整解答每一题，再核对答案。
Question 1: A circle has radius 6 cm and a sector angle of 150°. Find the sector area and the segment area.
题目 1：一个圆的半径为 6 cm，扇形圆心角为 150°。求扇形面积和弓形面积。
Question 2: The arc length of a sector is 15 cm and its radius is 5 cm. Find the sector area.
题目 2：一个扇形的弧长为 15 cm，半径为 5 cm。求扇形面积。
Question 3: A chord subtends an angle of 2.4 radians at the centre of a circle with radius 7 cm. Find the area of the major segment.
题目 3：一条弦在半径为 7 cm 的圆心处所对的圆心角为 2.4 弧度。求优弓形的面积。
π ≈ 47.12 cm²; segment area = ½ × 36 × (5π/6 − sin(5π/6)) = 18 × (5π/6 − 0.5) ≈ 38.35 cm².
解答：第 1 题：150° = 5π/6，扇形面积 = ½ × 36 × 5π/6 = 15π ≈ 47.12 cm²；弓形面积 = ½ × 36 × (5π/6 − sin(5π/6)) = 18 × (5π/6 − 0.5) ≈ 38.35 cm²。
Q2: s = rθ, so θ = 15/5 = 3 radians. Sector area = ½ × 25 × 3 = 37.5 cm².
第 2 题解答：s = rθ，因此 θ = 15/5 = 3 弧度。扇形面积 = ½ × 25 × 3 = 37.5 cm²。
7² − ½ × 49 × (2.4 − sin 2.4) = 49π − 24.5 × (2.4 − 0.6755) ≈ 153.94 − 42.25 ≈ 111.69 cm².
第 3 题解答：优弓形面积 = 整圆面积 − 劣弓形面积 7² − ½ × 49 × (2.4 − sin 2.4) = 49π − 24.5 × (2.4 − 0.6755) ≈ 153.94 − 42.25 ≈ 111.69 cm²。
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