Pythagoras Theorem | 勾股定理

📚 Pythagoras Theorem | 勾股定理

One of the most famous and useful results in all of mathematics, the Pythagorean theorem connects algebra, geometry, and the real world. Every IGCSE student must understand its statement, its applications, and its limitations. Let us explore this theorem step by step, from its historical roots to the types of questions you will face in the examination.

勾股定理是数学中最著名且最实用的结论之一，它将代数、几何与现实世界紧密相连。每一位 IGCSE 考生都必须理解其表述、应用及其适用范围。我们将从历史起源开始，逐步深入，直至考试中常见的各类题型。

1. The Theorem Statement | 定理表述

In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. If the two shorter sides are labelled \(a\) and \(b\), and the hypotenuse is labelled \(c\), then the theorem is written as:

在直角三角形中，斜边长度的平方等于另外两条边长度平方之和。若两条较短的边记为 \(a\) 和 \(b\)，斜边记为 \(c\)，则该定理可写作：

a² + b² = c²

It is important to remember that the hypotenuse is always the side opposite the right angle, and it is always the longest side of the triangle.

务必记住，斜边始终是直角所对的边，并且永远是三角形中最长的一条边。

2. Historical Background | 历史背景

The theorem is named after the ancient Greek mathematician Pythagoras, who lived around 570–495 BCE. However, evidence suggests that the relationship between the sides of a right-angled triangle was known much earlier in ancient Babylon, Egypt, India, and China. The Chinese mathematical text Zhoubi Suanjing contains a statement of the theorem, and the ancient Chinese referred to it as the Gougu rule.

该定理以古希腊数学家毕达哥拉斯的名字命名，他大约生活在公元前 570 年至前 495 年。然而，有证据表明，古巴比伦、古埃及、古印度和古中国早已了解直角三角形三边之间的关系。中国数学典籍《周髀算经》中就有相关表述，中国古代称之为”勾股定理”。

For IGCSE purposes, you do not need to memorise history, but understanding the cultural context enriches your appreciation of mathematics as a global achievement.

就 IGCSE 考试而言，你并不需要记住历史细节，但了解文化背景能让你更深刻地体会到数学是全人类的共同财富。

3. Geometric Interpretation | 几何意义

The theorem has a beautiful geometric meaning. If you construct three squares, one on each side of a right-angled triangle, then the area of the square on the hypotenuse equals the sum of the areas of the squares on the other two sides.

该定理有一个优美的几何解释。如果在直角三角形的三条边上分别构造正方形，那么斜边上的正方形面积等于另外两条边上的正方形面积之和。

For example, if the two shorter sides are 3 and 4, then the square on the side of length 3 has area 9, the square on the side of length 4 has area 16, and the square on the hypotenuse has area 9 + 16 = 25. Therefore, the hypotenuse has length 5.

例如，若两条较短边分别为 3 和 4，则边长为 3 的正方形面积为 9，边长为 4 的正方形面积为 16，斜边上的正方形面积为 9 + 16 = 25，因此斜边长度为 5。

4. A Standard Proof | 一个经典证明

There are hundreds of proofs of the Pythagorean theorem. One of the simplest uses area. Take a square of side length a + b, and inside it place four identical right-angled triangles with shorter sides a and b and hypotenuse c, arranged so that they form a smaller tilted square in the middle.

勾股定理有数百种证明方法。其中一种最简单的证明使用了面积法。取一个边长为 a + b 的大正方形，在其中放置四个全等的直角三角形，其短边为 a 和 b，斜边为 c，将它们排列成中央有一个倾斜的小正方形。

The area of the large square is (a + b)². This equals the sum of the areas of the four triangles plus the area of the inner square, which is c². Therefore:

大正方形的面积为 (a + b)²。它等于四个三角形的面积之和加上内部正方形的面积 c²。因此：

(a + b)² = 4 × ½ab + c²

Simplifying the right-hand side gives 2ab + c². Expanding the left-hand side gives a² + 2ab + b². Cancelling 2ab from both sides yields:

化简右边得到 2ab + c²。展开左边得到 a² + 2ab + b²。两边同时消去 2ab，得到：

a² + b² = c²

This elegant proof shows why the theorem is true in full generality.

这一简洁的证明揭示了该定理为何在一般情况下成立。

5. Pythagorean Triples | 勾股数

When all three sides of a right-angled triangle are integers, they form a Pythagorean triple. The most famous examples are:

当直角三角形的三边均为整数时，它们构成一组勾股数。最著名的例子有：

a	b	c
3	4	5
5	12	13
8	15	17
7	24	25

To generate Pythagorean triples, choose two positive integers m and n with m > n. Then:

要生成勾股数，可选择两个正整数 m 和 n，其中 m > n。则：

a = m² − n², b = 2mn, c = m² + n²

For example, if m = 2 and n = 1, we get a = 3, b = 4, c = 5.

例如，取 m = 2，n = 1，则得到 a = 3，b = 4，c = 5。

6. Finding a Missing Side | 求缺失的边长

The most direct application of the theorem is finding a missing side of a right-angled triangle. There are two cases:

该定理最直接的应用是求直角三角形中的未知边长。分两种情况：

If you are given both shorter sides, square them, add, and take the square root: c = √(a² + b²).
如果已知两条短边，则将其平方后相加，再开平方根：c = √(a² + b²)。

If you are given the hypotenuse and one shorter side, square both, subtract, and take the square root: a = √(c² − b²).
如果已知斜边和一条短边，则将两者平方后相减，再开平方根：a = √(c² − b²)。

Example: Find the length of the hypotenuse of a right-angled triangle with shorter sides 6 cm and 8 cm.

例：求直角边为 6 cm 和 8 cm 的直角三角形的斜边长度。

c² = 6² + 8² = 36 + 64 = 100, so c = 10 cm

Example: A right-angled triangle has hypotenuse 13 cm and one leg 5 cm. Find the other leg.

例：某直角三角形斜边为 13 cm，一条直角边为 5 cm。求另一条直角边。

a² = 13² − 5² = 169 − 25 = 144, so a = 12 cm

7. Distance Between Two Points | 两点间的距离

In coordinate geometry, the distance between two points (x₁, y₁) and (x₂, y₂) can be found using the Pythagorean theorem. The horizontal difference is x₂ − x₁, and the vertical difference is y₂ − y₁. These act as the two shorter sides of a right-angled triangle, and the distance between the points is the hypotenuse.

在坐标几何中，两点 (x₁, y₁) 和 (x₂, y₂) 之间的距离可用勾股定理求得。水平差为 x₂ − x₁，竖直差为 y₂ − y₁。这两者构成直角三角形的两条直角边，而两点间的距离就是斜边。

d = √((x₂ − x₁)² + (y₂ − y₁)²)

Example: Find the distance between (1, 2) and (4, 6).

例：求点 (1, 2) 和 (4, 6) 之间的距离。

d = √((4 − 1)² + (6 − 2)²) = √(9 + 16) = √25 = 5

This distance formula is one of the most frequently tested applications in the IGCSE coordinate geometry section.

距离公式是 IGCSE 坐标几何部分最高频的考点之一。

8. Applications in Real Life | 实际应用

The Pythagorean theorem appears everywhere in daily life and industry. Builders use it to ensure that walls are perpendicular; surveyors use it to measure distances across obstacles; and engineers use it in structural design.

勾股定理在日常生活中和工业领域无处不在。建筑工人用它确保墙体垂直；测量员用它跨越障碍物测量距离；工程师在结构设计中运用它。

A classic exam problem involves a ladder leaning against a wall. Suppose a ladder of length 5 m is placed so that its base is 3 m from the wall. How high up the wall does it reach?

一个经典考题是梯子靠墙问题。假设一架 5 m 长的梯子，其底部离墙 3 m。它能达到墙上的多高？

h² = 5² − 3² = 25 − 9 = 16, so h = 4 m

Another common context is right-angled triangles in navigation, sports fields, and even computer graphics. Always translate the situation into a right-angled triangle first.

另一个常见情境是航海、运动场甚至计算机图形中的直角三角形。解题时务必先将实际问题转化为直角三角形。

9. Three-Dimensional Applications | 三维应用

In IGCSE, you may be asked to find the length of a diagonal in a cuboid. Consider a cuboid with length l, width w, and height h. To find the longest diagonal d, use the formula:

在 IGCSE 中，你可能会被要求计算长方体的对角线长度。设长方体长、宽、高分别为 l、w、h。求最长对角线 d，使用公式：

d = √(l² + w² + h²)

This follows from applying the theorem twice. First, find the diagonal of the base using √(l² + w²). Then apply the theorem again using this base diagonal and the height.

这一公式源于两次应用勾股定理。首先，用 √(l² + w²) 求底面正方形的对角线；然后，再次用勾股定理，将该底面对角线与高结合。

Example: A box has length 2 m, width 3 m, and height 6 m. Find the space diagonal.

例：一个盒子长 2 m，宽 3 m，高 6 m。求其空间对角线。

d = √(2² + 3² + 6²) = √(4 + 9 + 36) = √49 = 7 m

10. Special Right-Angled Triangles | 特殊直角三角形

Two special triangles appear frequently in IGCSE examinations because their side ratios are constant and simple.

两类特殊直角三角形在 IGCSE 考试中频繁出现，因为它们的边长比例固定且简单。

In an isosceles right-angled triangle, the two legs are equal. If each leg has length x, then the hypotenuse is x√2.

在等腰直角三角形中，两条直角边相等。若每条直角边长为 x，则斜边为 x√2。

c = x√2

In a 30°–60°–90° triangle, the sides are in the ratio 1 : √3 : 2. If the shortest side (opposite 30°) is x, then the other leg is x√3 and the hypotenuse is 2x.

在 30°–60°–90° 三角形中，三边之比为 1 : √3 : 2。若最短边（30° 角所对边）为 x，则另一条直角边为 x√3，斜边为 2x。

Triangle	Side Ratio
45°–45°–90°	1 : 1 : √2
30°–60°–90°	1 : √3 : 2

Memorising these ratios saves valuable time in examinations.

牢记这些比例可以在考试中节省宝贵的时间。

11. Converse of the Pythagorean Theorem | 勾股定理的逆定理

The converse of the theorem is also true: if a triangle has side lengths a, b and c, with c being the longest side, and a² + b² = c², then the triangle is right-angled.

逆定理同样成立：若一个三角形的三边为 a、b、c，其中 c 为最长边，且 a² + b² = c²，则这个三角形是直角三角形。

This is useful for classifying triangles. Consider a triangle with sides 5 cm, 12 cm and 13 cm. Since 5² + 12² = 25 + 144 = 169 = 13², the triangle is right-angled.

逆定理可用于判断三角形类型。考察三边为 5 cm、12 cm、13 cm 的三角形。因为 5² + 12² = 25 + 144 = 169 = 13²，所以该三角形是直角三角形。

If a² + b² > c², the triangle is acute. If a² + b² < c², the triangle is obtuse.

若 a² + b² > c²，则三角形为锐角三角形。若 a² + b² < c²，则三角形为钝角三角形。

12. Common Mistakes and Exam Tips | 常见错误与考试要点

Many students lose marks on this topic because of small but avoidable errors. Here are the most common pitfalls:

许多学生因为一些细小但可以避免的错误而在该主题上失分。以下是最常见的陷阱：

Misidentifying the hypotenuse. The hypotenuse is always opposite the right angle, not just the longest side you happen to see.
错误识别斜边。斜边永远在直角对面，而不是你碰巧看到的最长边。

Forgetting to take the square root at the end. The theorem gives c²; you must find c.
最后忘记开平方根。定理给出的是 c²；你必须求出 c。

Using the theorem on non-right-angled triangles. The theorem only applies when there is a right angle.
在非直角三角形中使用勾股定理。该定理仅在存在直角时适用。

Rounding too early. Keep exact values until the final answer, then round appropriately.
过早取近似值。在得出最终答案前保持精确值，然后再四舍五入。

In the IGCSE examination, show all your working clearly. Even if your final answer is wrong, correct method steps may earn you method marks.

在 IGCSE 考试中，务必清晰写出所有解题步骤。即使最终答案错误，正确的方法步骤也可能帮你获得步骤分。

Mastering the Pythagorean theorem opens the door to many other topics in mathematics, including trigonometry, circle theorems, and vector geometry. Practise with a variety of questions — both abstract and real-world — until you can identify the right-angled triangle and apply the formula without hesitation.

掌握勾股定理，将为你打开数学中许多其他领域的大门，包括三角函数、圆定理和向量几何。请练习各类题目——无论是抽象的还是现实情境的——直到你能毫不犹豫地识别出直角三角形并应用公式。

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Pythagoras Theorem | 勾股定理