📚 Mastering Pythagoras' Theorem | 掌握毕达哥拉斯定理（勾股定理）
Pythagoras' theorem is one of the most useful results in KS3 mathematics. It connects the three sides of a right-angled triangle and appears in geometry, coordinate work, trigonometry and real-life measurement problems. Once you can label a triangle correctly and decide whether to add or subtract, almost every question becomes a short, reliable calculation.
毕达哥拉斯定理（勾股定理）是 KS3 数学中最有用的结论之一。它建立了直角三角形三条边之间的联系，在几何、坐标、三角学以及实际测量问题中都会出现。只要你能正确标记三角形的边，并判断该用加法还是减法，几乎所有题目都会变成简短而可靠的计算。
This revision guide covers the theorem itself, the key sides of a right triangle, the formula in symbols and words, fully worked examples for finding both the hypotenuse and a shorter side, the converse test for right angles, Pythagorean triples, common exam mistakes, coordinate and real-life applications, and a set of practice questions with complete solutions.
本文系统讲解定理本身、直角三角形的关键边、公式的符号与文字表达、求斜边与求直角边的完整例题、判断直角的逆定理、勾股数、常见考试错误、坐标与实际应用，并附一套带完整解答的练习题。
1. What Is Pythagoras' Theorem? | 什么是毕达哥拉斯定理？
Pythagoras' theorem is a fundamental rule in geometry. It applies only to right-angled triangles. In these triangles, one angle is exactly 90 degrees, and the side opposite this angle is called the hypotenuse, which is always the longest side.
毕达哥拉斯定理是几何学中的一条基本规则。它只适用于直角三角形。在直角三角形中，有一个角恰好是 90 度，与这个角相对的边称为斜边，它永远是三条边中最长的一条。
The theorem states that the area of the square on the hypotenuse equals the total area of the squares on the other two sides. This gives a powerful link between the three side lengths, and it works for every right-angled triangle, whatever its size or orientation.
该定理指出，以斜边为边的正方形面积，等于以另外两条直角边为边的正方形面积之和。这建立了三条边长之间的重要联系，而且对任何大小、任何摆放方向的直角三角形都成立。
In symbols, if the two shorter sides have lengths a and b, and the hypotenuse has length c, then the relationship can be written in a single line.
用符号表示，若两条较短的直角边长度为 a 和 b，斜边长度为 c，则这一关系可以写成一行公式。
a² + b² = c²
ans the length multiplied by itself, so 6² = 6 × 6 = 36. The Chinese name for this result, 勾股定理, comes from the ancient names for the two shorter sides: 勾 and 股 for the legs, and 弦 for the hypotenuse.
边长的平方就是该长度自乘，例如 6² = 6 × 6 = 36。这一结论的中文名称"勾股定理"来自古人对两条直角边的称呼：勾与股表示直角边，弦表示斜边。
2. The Key Sides of a Right Triangle | 直角三角形的关键边
To use the theorem correctly, label the two shorter sides as a and b. The longest side, which is always opposite the right angle, must be labelled c. The right angle is usually marked with a small square in the corner.
为了正确使用定理，请把两条较短的直角边标记为 a 和 b。最长边总是与直角相对，必须标记为 c。直角通常用小正方形符号标在角上。
It does not matter if you swap a and b. The important point is that c is the hypotenuse, never a shorter side. If you label a shorter side as c, the formula will give a wrong answer, and this is one of the most common errors in tests.
交换 a 和 b 并不影响计算。重要的是 c 必须是斜边，而不能是较短的直角边。如果把直角边标成 c，公式就会给出错误答案，这是考试中最常见的错误之一。
- Right angle: exactly 90°, marked with a small square. | 直角：恰好 90°，用一个小正方形标记。
- Legs: the two sides that form the right angle, labelled a and b. | 直角边：构成直角的两条边，标记为 a 和 b。
- Hypotenuse: the longest side, opposite the right angle, labelled c. | 斜边：最长边，与直角相对，标记为 c。
A useful check before you calculate: the hypotenuse must be longer than each of the other two sides. In a triangle with sides 5 cm, 12 cm and 13 cm, the hypotenuse is 13 cm because 13 is the largest number.
计算前有一个实用的检查：斜边必须比另外两条边都长。在边长为 5 cm、12 cm 和 13 cm 的三角形中，斜边是 13 cm，因为 13 是最大的数。
3. The Formula in Symbols and Words | 公式的符号与文字表达
In symbols, Pythagoras' s, the square of the hypotenuse equals the sum of the squares of the other two sides. Being able to state the theorem in words often earns a mark in an exam.
用符号表示，毕达哥拉斯定理写作 a² + b² = c²。用文字表达，就是斜边的平方等于另外两条边的平方之和。能够用文字复述定理，往往就能在考试中拿到一个得分点。
From this formula, we can find any one side if the other two sides are known. To find the hypotenuse we use addition, and to find a shorter side we use subtraction. The final step is always to take the square root.
根据这个公式，只要知道其中两条边，就能求出第三条边。求斜边时使用加法，求直角边时使用减法。最后一步永远是开平方根。
c = √a² + b² and a = √c² − b²
b² = c² − a². Remember that the subtraction is done before the square root, not after it.
移项的意思就是把未知量单独留下。若 a² + b² = c²，则 a² = c² − b²，b² = c² − a²。要记住：先做减法，再开平方根，顺序不能颠倒。
Because lengths are positive, the square root is always the positive value. A calculator is allowed in most KS3 papers, but you should still be able to recognise perfect squares such as 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121 and 144.
由于长度是正数，平方根永远取正值。虽然 KS3 大多数试卷允许使用计算器，你仍应熟悉完全平方数，如 1、4、9、16、25、36、49、64、81、100、121 和 144。
4. Worked Example: Finding the Hypotenuse | 例题：求斜边
Problem: A right-angled triangle has shorter sides of length 6 cm and 8 cm. Find the length of the hypotenuse. The two given sides are the legs, so the unknown is c.
问题：一个直角三角形的两条直角边分别为 6 cm 和 8 cm，求斜边的长度。已知的两条边是直角边，所以未知量是 c。
b² = c², so 6² + 8² = c². Step 2: evaluate the squares. 36 + 64 = c², therefore 100 = c².
步骤 1：写出公式并代入已知数。a² + b² = c²，所以 6² + 8² = c²。步骤 2：计算平方。36 + 64 = c²，因此 100 = c²。
wer with units and check it is sensible: 10 cm is longer than 6 cm and 8 cm, so the answer is reasonable.
步骤 3：对两边开平方根。c = √100 = 10。步骤 4：写出带单位的答案并检查是否合理：10 cm 比 6 cm 和 8 cm 都长，所以答案合理。
6² + 8² = 36 + 64 = 100 → c = √100 = 10 cm
This 6-8-10 triangle is the classic example used in textbooks, and it is a scaled version of the 3-4-5 triangle. Recognising such families saves time in exams.
这个 6-8-10 三角形是课本中的经典例子，它是 3-4-5 三角形的放大版本。在考试中识别出这类成比例的三角形可以节省时间。
5. Worked Example: Finding a Shorter Side | 例题：求直角边
Problem: A right-angled triangle has a hypotenuse of 13 cm and one shorter side of 5 cm. Find the length of the other shorter side. Here you must subtract, not add.
问题：一个直角三角形的斜边为 13 cm，一条直角边为 5 cm，求另一条直角边的长度。此时必须用减法，而不是加法。
uate the squares. 25 + b² = 169. Step 3: subtract 25 from both sides. b² = 169 − 25 = 144.
步骤 1：写出 a² + b² = c² 并代入。5² + b² = 13²。步骤 2：计算平方。25 + b² = 169。步骤 3：两边同时减去 25。b² = 169 − 25 = 144。
b² = 13² − 5² = 169 − 25 = 144 → b = √144 = 12 cm
side is 12 cm. A quick check: 5² + 12² = 25 + 144 = 169 = 13², which confirms the result.
步骤 4：开平方根并写出带单位的答案。缺失的边是 12 cm。快速验算：5² + 12² = 25 + 144 = 169 = 13²，说明结果正确。
13.9 cm, which is longer than the hypotenuse. That contradiction is a clear signal that you added instead of subtracted.
如果忘记用减法，你会得到 5² + 13² = 25 + 169 = 194，约为 13.9 cm，比斜边还长。这个矛盾清楚地说明你把减法做成了加法。
6. The Converse: Testing for a Right Angle | 逆定理：判断直角
The converse of Pythagoras' theorem works in the opposite direction. If the square of the longest side equals the sum of the squares of the other two sides, then the triangle must be right-angled.
毕达哥拉斯定理的逆定理方向相反：如果最长边的平方等于另外两条边的平方之和，那么这个三角形一定是直角三角形。
+ 225 = 289, and 17² = 289. Since the two values are equal, the triangle is right-angled.
例如：一个三角形的三边为 8 cm、15 cm 和 17 cm。最长边是 17 cm。计算 8² + 15² = 64 + 225 = 289，而 17² = 289。两者相等，所以这个三角形是直角三角形。
, 5 cm and 6 cm gives 4² + 5² = 41, while 6² = 36. Since 41 ≠ 36, there is no right angle.
如果两个值不相等，三角形就不是直角三角形。三边为 4 cm、5 cm 和 6 cm 的三角形中，4² + 5² = 41，而 6² = 36。因为 41 ≠ 36，所以它没有直角。
Note the strict test rule: always compare the square of the longest side with the sum of the squares of the two shorter sides. Never square the two largest numbers and compare with the smallest.
请记住严格的检验规则：永远用最长边的平方去比较两条较短边的平方之和，绝不能把最大的两个数平方后与最小的数比较。
7. Pythagorean Triples | 毕达哥拉斯三元组（勾股数）
= 25 = 5². Learning common triples helps you answer questions quickly and check your work.
勾股数是指三个正整数恰好满足 a² + b² = c² 的一组数。最著名的是 3、4、5，因为 3² + 4² = 9 + 16 = 25 = 5²。熟记常见勾股数有助于快速作答并检验结果。
| Triple | 勾股数 | Check | 验算 | Common use | 常见用途 |
| 3, 4, 5 | 9 + 16 = 25 | Basic example | 基础例题 |
| 5, 12, 13 | 25 + 144 = 169 | Finding a shorter side | 求直角边 |
| 8, 15, 17 | 64 + 225 = 289 | Converse tests | 逆定理判断 |
| 7, 24, 25 | 49 + 576 = 625 | Harder questions | 较难题目 |
| 20, 21, 29 | 400 + 441 = 841 | Extend problems | 拓展题 |
Multiplying or dividing a triple by any positive whole number gives another triple, called a multiple. For example, 3-4-5 becomes 6-8-10, 9-12-15 and 30-40-50. These are all similar triangles of different sizes.
把一个勾股数同时乘或除以任意正整数，会得到新的勾股数，称为倍数形式。例如 3-4-5 可以变成 6-8-10、9-12-15 和 30-40-50。它们都是不同大小的相似三角形。
One caution: seeing numbers such as 6 and 8 does not automatically mean the answer is 10, because they might be a leg and the hypotenuse. Always identify the right angle first, then decide which value is c.
提醒一点：看到 6 和 8 并不代表答案一定是 10，因为它们也可能是"一条直角边与斜边"的组合。务必先找出直角的位置，再判断哪个值是 c。
8. Common Mistakes and Exam Traps | 常见错误与考试陷阱
The most frequent error is adding when you should subtract, or subtracting when you should add. The rule is simple: finding the hypotenuse means addition; finding a shorter side means subtraction.
最常见的错误是该减的时候加，或该加的时候减。规则很简单：求斜边用加法，求直角边用减法。
A second error is forgetting the final square root. If a question asks for a length and you answer 100 instead of 10, you have stopped one step too early. Always check whether the unknown is squared or not.
第二个错误是忘记最后开平方根。如果题目要求长度而你的答案是 100 而不是 10，说明你少做了一步。始终检查未知量是否还带着平方。
- Labelling the hypotenuse as a leg. | 把斜边标成直角边。
- Adding the two given sides before squaring. | 先把两个已知边相加再平方（应分别平方后再处理）。
- Rounding too early, which causes a wrong final digit. | 过早四舍五入，导致最后一位数字错误。
- Ignoring units or writing cm² instead of cm for a length. | 忽略单位，或把长度写成 cm² 而不是 cm。
- Assuming every triangle is right-angled. | 想当然地认为所有三角形都是直角三角形。
Another trap is a composite shape where you must find a missing length first, then apply the theorem again in a second triangle. Read the diagram carefully and label every known length before starting.
另一个陷阱是组合图形：你必须先在一个三角形中求出缺失的长度，再把它用在第二个三角形里。做题前仔细读图，把每一个已知长度都标出来。
If the answer is not a whole number, leave it in surd form when asked for an exact value, or round to the required number of decimal places. For example, √50 = 5√2 ≈ 7.07, and both forms may appear in mark schemes.
如果答案不是整数，题目要求精确值时应写成根式形式，或按要求的小数位数取近似值。例如 √50 = 5√2 ≈ 7.07，这两种形式都可能出现在评分标准中。
9. Pythagoras in Coordinate Geometry | 坐标几何中的应用
Pythagoras' theorem also gives us the distance between two points on a coordinate grid. A horizontal gap and a vertical gap form the two legs of a right triangle, and the straight-line distance between the points is the hypotenuse.
毕达哥拉斯定理还给出了坐标网格上两点之间的距离。水平差与竖直差构成直角三角形的两条直角边，两点之间的直线距离就是斜边。
distance = √((x₂ − x₁)² + (y₂ − y₁)²)
tal difference is 4 − 1 = 3 and the vertical difference is 6 − 2 = 4. Then the distance is √3² + 4² = √9 + 16 = √25 = 5 units.
例如：求点 (1, 2) 与点 (4, 6) 之间的距离。水平差为 4 − 1 = 3，竖直差为 6 − 2 = 4。于是距离为 √3² + 4² = √9 + 16 = √25 = 5 个单位。
Notice that this is the same 3-4-5 triple again. Squaring removes any negative signs, so it does not matter whether you subtract left from right or right from left; the result is the same.
注意这里又出现了熟悉的 3-4-5 勾股数。平方会消除负号，所以无论用左边减右边还是右边减左边，结果都一样。
The same idea extends to three dimensions. For a rectangular box with side lengths a, b and d, the space diagonal has length √a² + b² + d², found by applying the theorem twice.
同样的思想可以推广到三维空间。对于棱长为 a、b、d 的长方体，其体对角线长度为 √a² + b² + d²，这是连续两次使用定理的结果。
10. Real-Life Applications | 实际应用
Builders, engineers, pilots and designers use Pythagoras' theorem constantly. Whenever a right angle is involved, the theorem turns two measurements into a third that would be difficult to measure directly.
建筑工人、工程师、飞行员和设计师经常使用毕达哥拉斯定理。只要涉及直角，定理就能把两个测量值转换成第三个难以直接测量的长度。
Problem: a ladder 5 m long leans against a wall. The foot of the ladder is 3 m from the wall. How high up the wall does the ladder reach? The ladder is the hypotenuse, so the height is √5² − 3² = √25 − 9 = √16 = 4 m.
问题：一架 5 m 长的梯子斜靠在墙上，梯脚距墙 3 m。梯子能到达墙上多高？梯子是斜边，所以高度为 √5² − 3² = √25 − 9 = √16 = 4 m。
Problem: a rectangular television screen measures 40 cm by 30 cm. Find the diagonal. The diagonal is the hypotenuse: √40² + 30² = √1600 + 900 = √2500 = 50 cm.
问题：一块长方形电视屏幕的长宽分别为 40 cm 和 30 cm，求对角线长度。对角线是斜边：√40² + 30² = √1600 + 900 = √2500 = 50 cm。
Other everyday uses include checking whether a garden corner is square, working out the shortest route on a map, calculating the height of a roof, and finding the direct distance between two towns when only north-south and east-west distances are known.
其他日常用途包括：检验花园的角是否为直角、在地图上求最短路线、计算屋顶高度，以及在只知道南北与东西距离时求两镇之间的直线距离。
In every case, sketch the situation, mark the right angle, decide which side is the hypotenuse, and then choose addition or subtraction before you pick up the calculator.
无论哪种情况，都要先画出示意图，标出直角，判断哪条边是斜边，然后决定用加法还是减法，再动手按计算器。
11. Mixed Practice Questions | 综合练习题
Try these questions without looking at the solutions. Write down the formula each time and show every step; method marks are usually awarded even if the final number is wrong.
请先不要看答案，独立完成以下题目。每次都要写出公式并展示每一步；即使最后数字算错，方法分通常仍然会给。
- Q1. A right-angled triangle has legs of 9 cm and 12 cm. Find the hypotenuse. | 题 1：直角三角形的两条直角边为 9 cm 和 12 cm，求斜边。
- Q2. A right-angled triangle has a hypotenuse of 26 cm and one leg of 10 cm. Find the other leg. | 题 2：直角三角形的斜边为 26 cm，一条直角边为 10 cm，求另一条直角边。
- Q3. A triangle has sides 9 cm, 40 cm and 41 cm. Is it right-angled? Explain. | 题 3：三角形三边为 9 cm、40 cm 和 41 cm，它是直角三角形吗？请说明理由。
- Q4. Find the distance between the points (2, 3) and (7, 15). | 题 4：求点 (2, 3) 与点 (7, 15) 之间的距离。
- Q5. A rectangular field is 24 m long and 7 m wide. Find the length of the diagonal path across it. | 题 5：一块长 24 m、宽 7 m 的矩形田地，求穿越它的对角线小路长度。
Question 3 requires the converse, so remember to compare the square of the longest side with the sum of the squares of the other two. Question 5 is a straightforward application where the diagonal is the hypotenuse.
题 3 需要用逆定理，所以要比较最长边的平方与另两边的平方之和。题 5 是直接应用，矩形对角线即斜边。
Always finish by asking whether your answer is sensible. A hypotenuse longer than the sum of the legs, or a leg longer than the hypotenuse, signals an error that can be corrected in seconds.
最后一定要问自己答案是否合理。如果斜边比两条直角边之和还长，或直角边比斜边长，说明计算有误，几秒钟内就能纠正。
12. Answers and Full Solutions | 答案与完整解答
4 = 225, and √225 = 15. The hypotenuse is 15 cm. This is the 3-4-5 triple multiplied by 3.
题 1：9² + 12² = 81 + 144 = 225，√225 = 15。斜边为 15 cm。这是 3-4-5 勾股数乘以 3 的结果。
b² = 676 − 100 = 576, and √576 = 24. The other leg is 24 cm, another multiple of 5-12-13.
题 2：10² + b² = 26²，所以 b² = 676 − 100 = 576，√576 = 24。另一条直角边为 24 cm，是 5-12-13 的倍数形式。
681. Since the values are equal, the triangle is right-angled, with the right angle opposite the 41 cm side.
题 3：9² + 40² = 81 + 1600 = 1681，而 41² = 1681。两者相等，所以该三角形是直角三角形，直角位于 41 cm 那条边的对角处。
zontal difference is 7 − 2 = 5 and the vertical difference is 15 − 3 = 12. The distance is √5² + 12² = √25 + 144 = √169 = 13 units.
题 4：水平差为 7 − 2 = 5，
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