📚 Pythagoras’ Theorem | 毕达哥拉斯定理
Pythagoras’ theorem is one of the most famous results in mathematics. It describes a special relationship between the three sides of a right-angled triangle. Once you understand it, you can find missing lengths, solve real-world problems, and even test whether a triangle has a right angle. This article covers everything you need for Cambridge Key Stage 3, from the basic formula to confident problem-solving.
毕达哥拉斯定理（勾股定理）是数学中最著名的结论之一。它描述了直角三角形三条边之间的一种特殊关系。一旦掌握它，你就能求出未知边长、解决现实问题，甚至检验一个三角形是否有直角。本文涵盖剑桥 KS3 阶段所需的所有内容，从基本公式到灵活解题，帮助你建立信心。
1. What is Pythagoras’ Theorem? | 什么是毕达哥拉斯定理？
ngle with side lengths a, b and c, where c is the hypotenuse, we always have a² + b² = c².
毕达哥拉斯定理指出，在任何直角三角形中，斜边（最长边，直角的对边）上画的正方形的面积，等于另外两条边上画的正方形面积之和。这就得到一个精确公式：对于一个边长分别为 a、b 和 c 的直角三角形，其中 c 是斜边，恒有 a² + b² = c²。
This relationship was known to Babylonian mathematicians long before Pythagoras, but the Greek philosopher Pythagoras is often credited with its first formal proof. For KS3 learners, the key is to recognise that this rule only works for right-angled triangles – never for acute or obtuse triangles.
古比伦数学家早在毕达哥拉斯之前就发现了这一关系，但通常认为希腊哲学家毕达哥拉斯给出了第一个严格证明。对于 KS3 阶段的学习者，关键在于认识到这个规则只适用于直角三角形——绝不适用于锐角或钝角三角形。
- Right-angled triangle – has one angle exactly 90°
- Hypotenuse – the side opposite the right angle, always the longest
- Legs (shorter sides) – the two sides that form the right angle
- 直角三角形：有一个角恰好为 90°
- 斜边：直角所对的边，永远是最长边
- 直角边（短边）：构成直角的两条边
2. The Formula a² + b² = c² | 公式 a² + b² = c²
sent the lengths of the two shorter sides (legs), and c represents the length of the hypotenuse. It does not matter which of the two shorter sides you label as a or b – the sum of their squares will still equal the square of the hypotenuse.
定理的标准形式写作 a² + b² = c²。这里 a 和 b 代表两条直角边的长度，c 代表斜边的长度。你把哪条直角边标为 a 或 b 并不重要——它们的平方和仍然等于斜边的平方。
hypotenuse c would satisfy 3² + 4² = c², so 9 + 16 = c², giving c² = 25 and therefore c = √25 = 5 cm. Notice that we take the positive square root because lengths are positive.
例子：在一个直角边为 3 cm 和 4 cm 的直角三角形中，斜边 c 满足 3² + 4² = c²，即 9 + 16 = c²，得到 c² = 25，因此 c = √25 = 5 cm。注意我们取正的平方根，因为长度总是正数。
a² + b² = c² → c = √a² + b²
It is essential to use the correct units; if the legs are in metres, the hypotenuse will also be in metres, and its length will be in metres, not square metres.
使用正确的单位很关键；如果直角边以米为单位，斜边也会以米为单位，且其长度单位是米，而不是平方米。
3. Finding the Hypotenuse | 求斜边的长度
The most direct application of Pythagoras’ theorem is to calculate the length of the hypotenuse when the two shorter sides are known. Follow these steps: square both given lengths, add the results, then take the square root of the total.
毕达哥拉斯定理最直接的应用是：当已知两条直角边时，求斜边长度。请遵循以下步骤：将已知两边长分别平方，求和，再对总和开平方根。
m from the wall, and the vertical height reached is 2 m. What is the length of the ladder? ypotenuse.) First compute 1.5² = 2.25 and 2² = 4. Their sum is 6.25. Take the square root: √6.25 = 2.5 m. The ladder is 2.5 m long.
例题示范：一把梯子斜靠在墙上形成一个直角三角形。梯脚离墙 1.5 m，垂直高度为 2 m。梯长是多少？（梯子是斜边。）先计算 1.5² = 2.25，2² = 4。两数之和为 6.25。开平方根得 √6.25 = 2.5 m。梯长 2.5 m。
quare-number answers, round to a suitable degree of accuracy, such as 2 decimal places, unless the question states otherwise.
使用计算器时，请确保会正确使用平方 (x²) 和平方根 (√) 键。如果答案不是完全平方数，应根据题目要求四舍五入到合适精度，例如保留两位小数。
4. Finding a Shorter Side | 求一条直角边的长度
t as before. Always subtract the square of the known shorter side from the square of the hypotenuse.
如果你知道斜边和其中一条直角边，你可以重新整理毕达哥拉斯定理来求另一条直角边。公式变为 a² = c² − b² 或 b² = c² − a²。然后像之前一样开平方根。始终用斜边的平方减去已知直角边的平方。
Therefore a = √144 = 12 cm. Check your answer: 5² + 12² = 25 + 144 = 169 = 13². It works.
例子：一个直角三角形的斜边长为 13 cm，一条直角边长为 5 cm。求另一条直角边的长度。用 a² = c² − b² 得 a² = 13² − 5² = 169 − 25 = 144。因此 a = √144 = 12 cm。检验答案：5² + 12² = 25 + 144 = 169 = 13²。完全成立。
A common mistake is adding the hypotenuse and the leg instead of subtracting. Remember: the hypotenuse is the longest side, so a shorter side squared will always be smaller than the hypotenuse squared.
一个常见错误是把斜边和直角边相加而不是相减。请记住：斜边是最长边，因此直角边的平方总小于斜边的平方。
5. Pythagorean Triples | 毕达哥拉斯三元数组
Others include (5, 12, 13), (6, 8, 10), (7, 24, 25) and (8, 15, 17). Multiples of these triples also work, for example (6, 8, 10) is just double (3, 4, 5).
有些三元整数组恰好满足 a² + b² = c²，这些称为毕达哥拉斯三元数组。最著名的是 (3, 4, 5)，因为 3² + 4² = 5²。其他还有 (5, 12, 13)、(6, 8, 10)、(7, 24, 25) 和 (8, 15, 17)。这些三元组的倍数同样成立，例如 (6, 8, 10) 就是 (3, 4, 5) 的两倍。
Recognising Pythagorean triples can save you time during tests. If you see two sides of a right-angled triangle that match part of a known triple, you can immediately write down the missing side without full calculation. Be careful, though: the triple must correspond to a right-angled triangle where the largest number is the hypotenuse.
识别毕达哥拉斯三元组可以让你在测验中节省时间。如果你看到一个直角三角形的两边恰好与某个已知三元组相符，你可以立即写出第三边而无需完整计算。不过要小心：这个三元组必须对应于直角三角形，且最大的数必须是斜边。
6. Checking for a Right Angle | 检验直角
st side, then the triangle contains a right angle opposite side c. If the equation does not hold, the triangle is not right-angled.
你可以利用毕达哥拉斯定理的逆定理来检验一个三角形是否是直角三角形。如果三边长满足 a² + b² = c²（其中 c 为最长边），则该三角形包含一个直角，位于 c 边所对的角。如果等式不成立，则三角形不是直角三角形。
ngle with sides 8, 15 and 17 gives 8² + 15² = 64 + 225 = 289 = 17², so it is right-angled.
例如，边长分别为 6 cm、9 cm 和 11 cm 的三角形：最长边 c = 11。检验 6² + 9² = 36 + 81 = 117，但 11² = 121。由于 117 ≠ 121，该三角形不是直角三角形。然而，边长为 8、15、17 的三角形满足 8² + 15² = 64 + 225 = 289 = 17²，所以它是直角三角形。
This method is very useful in construction and design, such as checking whether a corner is exactly 90° using a 3–4–5 rope triangle.
这种方法在建筑和设计中非常有用，例如用 3–4–5 比例的绳子三角形来检验角落是否恰好是 90°。
7. Pythagoras on a Coordinate Grid | 坐标系中的毕达哥拉斯定理
You can use Pythagoras to find the distance between two points on a coordinate grid. Draw a right-angled triangle where the horizontal and vertical differences between the points become the legs. The direct distance is the hypotenuse.
你可以用毕达哥拉斯定理求出坐标系中两点之间的距离。画一个直角三角形，使两点间的水平差和垂直差分别成为两条直角边。直线距离就是斜边。
If point A is (x₁, y₁) and point B is (x₂, y₂), the horizontal leg is |x₂ − x₁|, and the vertical leg is |y₂ − y₁| (y₂ − y₁)²]. This is exactly the distance formula used throughout KS3 coordinate geometry.
若点 A 为 (x₁, y₁)，点 B 为 (x₂, y₂)，水平直角边为 |x₂ − x₁|，垂直直角边为 |y₂ − y₁|。那么距离 AB = √[(x₂ − x₁)² + (y₂ − y₁)²]。这正是 KS3 坐标几何中常用的距离公式。
4, 6). Horizontal difference = 4 − 1 = 3; vertical difference = 6 − 2 = 4. Then distance = √3² + 4² = √25 = 5 units. This forms the 3–4–5 triangle on the grid.
例子：求 (1, 2) 和 (4, 6) 之间的距离。水平差 = 4 − 1 = 3；垂直差 = 6 − 2 = 4。距离 = √3² + 4² = √25 = 5 个单位长度。这就在网格上构成了 3–4–5 三角形。
8. Solving Word Problems | 解应用题
r a leg). Label your sketch and write down the equation a² + b² = c² or its rearrangement.
许多 KS3 考题会给出一个现实情境来应用毕达哥拉斯定理。仔细读题，画一个直角三角形草图。明确已知哪些边长，要求的是哪条边（斜边还是直角边）。在草图上标注，然后列出方程 a² + b² = c² 或其变形。
boat sails 12 km due East and then 16 km due North. How far is it from its starting point? st leg 12 km and North leg 16 km. The straight-line distance is the hypotenuse: distance = √12² + 16² = √144 + 256 = √400 = 20 km.
文字题：一艘船先向东航行 12 km，再向北航行 16 km。问该船距出发点有多远？画一个直角三角形，向东的直角边为 12 km，向北的直角边为 16 km。直线距离即斜边：距离 = √12² + 16² = √144 + 256 = √400 = 20 km。
Always include units in your final answer and check whether the answer seems reasonable relative to the given numbers.
最终答案要包含单位，并检查答案与所给数据相比是否合理。
9. Applying Pythagoras in 3D | 在三维中应用毕达哥拉斯定理
In Stage 9, you may be introduced to simple 3D Pythagoras problems, such as finding the length of the space diagonal of a cuboid. You can apply the theorem twice: first find the diagonal of the base, then use that as a leg with the height to find the space diagonal.
在 Stage 9 中，你可能会接触到简单的三维勾股定理问题，比如求长方体空间对角线的长度。你可以两次应用定理：先求底面对角线，再将其与高作为直角边，求出空间对角线。
For a cuboid with length l, width w and height h: base diagonal = √l² + w². Then space diagonal = √((base diagonal)² + h²) = √l² + w² + h². So the space diagonal is the square root of the sum of the squares of all three dimensions.
对于一个长 l、宽 w、高 h 的长方体：底面对角线 = √l² + w²。然后空间对角线 = √((底面对角线)² + h²) = √l² + w² + h²。因此空间对角线是三条棱长平方和的平方根。
Example: l = 3 cm, w = 4 cm, h = 12 cm. Space diagonal = √3² + 4² + 12² = √9 + 16 + 144 = √169 = 13 cm. Notice the familiar 5–12–13 triple appearing indirectly.
示例：l = 3 cm, w = 4 cm, h = 12 cm。空间对角线 = √3² + 4² + 12² = √9 + 16 + 144 = √169 = 13 cm。请注意这里间接出现了熟悉的 5–12–13 三元组。
10. Common Mistakes and How to Avoid Them | 常见错误及如何避免
Mistake 1: Forgetting that the theorem only applies to right-angled triangles. Always check for the right angle symbol. Without it, you cannot use Pythagoras directly.
错误 1：忘记该定理仅适用于直角三角形。务必检查直角符号。若没有直角，就不能直接使用毕达哥拉斯定理。
Mistake 2: Mixing up hypotenuse and legs. Remember, the hypotenuse is opposite the right angle and is the longest side. If you subtract the wrong lengths, you may get a negative square or an incorrect result.
错误 2：混淆斜边和直角边。记住，斜边在直角的对侧，且为最长边。如果减错了边，可能会出现负数平方或得到错误结果。
Mistake 3: Not taking the square root at the end. After summing squares, you obtain c² or a², not the actual side length. The final step is always a square root.
错误 3：最后忘记开平方根。在平方求和之后，你得到的是 c² 或 a²，而不是实际边长。最后一步永远是开平方根。
Mistake 4: Rounding too early. Keep exact values until the final step, then round as required. Intermediate rounding can lead to inaccurate answers.
错误 4：过早四舍五入。保持精确值到最后一步，再按要求四舍五入。中间过程的约数可能导致答案不准。
11. Step-by-Step Summary | 步骤总结
| Step | Action | Example (legs 6, 8) |
| 1 | Identify the hypotenuse or the side to find | We want hypotenuse c |
| 2 | Write a² + b² = c² (or rearrange for a leg) | 6² + 8² = c² |
| 3 | Square the known lengths | 36 + 64 = c² |
| 4 | Add or subtract as needed | 100 = c² |
| 5 | Take the square root | c = √100 = 10 |
| 6 | Check and include units | 10 units |
This table works both for finding the hypotenuse and for finding a shorter side – just replace the equation at step 2.
此表对求斜边和求直角边均适用——只需在第2步换成相应的方程即可。
12. Practice Questions to Test Yourself | 自我检测练习题
Try these questions to reinforce your understanding. Solve them using the steps above, and check your answers with a calculator or mentally using triples.
试着解答以下问题来巩固理解。请按上述步骤求解，并用计算器或利用三元组心算核对答案。
- A right-angled triangle has legs 7 cm and 24 cm. Find the hypotenuse. (Hint: 7–24–25 triple)
- The hypotenuse of a right-angled triangle is 15 m and one leg is 9 m. Find the missing leg.
- The coordinates of two points are A(2, 3) and B(5, 7). Calculate the distance AB.
- A rectangular box measures length 6 cm, width 8 cm and height 24 cm. Determine the length of the space diagonal from one corner to the opposite corner.
- Is a triangle with sides 10 cm, 24 cm and 26 cm right-angled? Justify your answer.
- 一个直角三角形的直角边长为 7 cm 和 24 cm。求斜边长。（提示：7–24–25 三元组）
- 一个直角三角形的斜边长为 15 m，一条直角边长为 9 m。求另一条直角边的长。
- 两点坐标为 A(2, 3) 和 B(5, 7)。计算距离 AB。
- 一个长方体长 6 cm、宽 8 cm、高 24 cm。求从一个顶点到对角顶点的空间对角线长度。
- 边长为 10 cm、24 cm 和 26 cm 的三角形是直角三角形吗？请说明理由。
Remember, practising with a mix of numerics and word problems will make you much more confident when using Pythagoras’ theorem in any context.
请记住，通过数字题和文字题的混合练习，你在任何情境下应用毕达哥拉斯定理都会更加得心应手。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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