📚 Pythagoras' Theorem: Finding Missing Sides and Distance Problems | 毕达哥拉斯定理：求未知边长与距离问题
In KS3 Cambridge Mathematics, Pythagoras' theorem is one of the most powerful tools for solving problems involving right-angled triangles. This article reviews the theorem, shows how to find a missing hypotenuse and a missing shorter side, applies it to coordinate and real-life distance problems, extends it to three dimensions, and highlights the common exam techniques and errors that cost marks.
在剑桥 KS3 数学中，毕达哥拉斯定理是解决直角三角形问题最有力的工具之一。本文复习该定理，展示如何求未知的斜边与未知的直角边，将其应用于坐标距离与现实距离问题，并延伸到三维情形，同时强调常见的考试技巧与容易失分的错误。
1. The Right-Angled Triangle and Hypotenuse | 直角三角形与斜边
In any right-angled triangle, the right angle is usually marked with a small square. The side directly opposite this right angle is called the hypotenuse. It is always the longest side because it faces the largest angle, the 90° angle.
在任何直角三角形中，直角通常用小方格标记。与这个直角直接相对的边称为斜边。斜边总是最长的边，因为它面向最大的角，即 90° 角。
The two sides that form the right angle are called the legs, or the shorter sides. In diagrams they are often labelled a and b, while the hypotenuse is labelled c. This standard labelling makes the formula easier to remember and helps you avoid mixing up the sides.
构成直角的两条边称为直角边，也叫短边。在图中，短边通常标记为 a 和 b，斜边标记为 c。这种标准标记使公式更容易记忆，也能帮助你避免弄混各条边。
Before using the theorem, always check three things:
使用该定理之前，务必检查三件事：
- Is there a right angle? If not, Pythagoras' theorem cannot be applied directly. | 图中是否存在直角？如果没有，就不能直接使用毕达哥拉斯定理。
- Which side is the hypotenuse? It is the one opposite the right angle, not necessarily the one drawn horizontally. | 哪条边是斜边？它是直角对面的那条边，不一定是横向画出的那条。
- Are the units the same? Convert all lengths to the same unit before calculating. | 单位是否一致？计算前请把所有长度换算成同一单位。
2. Statement of Pythagoras' Theorem | 毕达哥拉斯定理的表述
Pythagoras' theorem states that in a right-angled triangle, the area of the square drawn on the hypotenuse is equal to the sum of the areas of the squares drawn on the other two sides. In symbols, if the shorter sides are a and b and the hypotenuse is c, then:
毕达哥拉斯定理指出，在直角三角形中，以斜边为边长的正方形面积，等于以另外两条边为边长的正方形面积之和。用符号表示，若两条直角边为 a 和 b，斜边为 c，则：
a² + b² = c²
6, and a square of side 5 has area 25; since 9 + 16 = 25, the 3-4-5 triangle is right-angled.
该定理本质上是关于面积的陈述，许多教材正是由此引出它的名称。边长为 3 的正方形面积为 9，边长为 4 的为 16，边长为 5 的为 25；由于 9 + 16 = 25，所以 3-4-5 三角形是直角三角形。
, if a² + b² = c² holds for the three sides, then the triangle must contain a right angle.
该定理在两个方向上都成立。若已知三角形是直角三角形，就可以用 a² + b² = c² 求边长；反过来，若三条边满足 a² + b² = c²，则该三角形必定含有一个直角。
3. Reading the Formula Correctly | 正确理解公式
The single most important habit is to identify c before you substitute any numbers. The letter c is reserved for the hypotenuse only. If you label a shorter side as c, your answer will be wrong even if your arithmetic is perfect.
最重要的习惯是在代入任何数字之前先确定 c。字母 c 只能代表斜边。如果你把某条直角边当作 c，即使算术完全正确，答案也是错的。
Two rearrangements cover every question:
以下两种变形可以覆盖所有题型：
c² = a² + b² (to find the hypotenuse)
a² = c² − b² (to find a shorter side)
Notice the sign change. When the hypotenuse is unknown you add; when a shorter side is unknown you subtract. Many marks are lost by adding when subtraction is required.
注意符号的变化。当斜边未知时用加法；当直角边未知时用减法。很多分数就是因为该减却加而丢掉的。
Also remember the final step: after reaching c² or a², you must take the square root. Write the square root sign explicitly, and keep full calculator accuracy until the last line.
还要记住最后一步：得到 c² 或 a² 之后，必须开平方。请明确写出根号，并在最后一行之前保留计算器的完整精度。
4. Finding the Hypotenuse | 求斜边
When the two shorter sides are known, square each of them, add the results, then take the square root. This is the most straightforward type of Pythagoras question and appears in almost every examination paper.
当已知两条直角边时，分别平方、相加，再开平方。这是最直接的毕达哥拉斯题型，几乎每份试卷都会出现。
Worked example: a right-angled triangle has shorter sides of 6 cm and 8 cm. Find the hypotenuse.
例题：一个直角三角形的两条直角边分别为 6 cm 和 8 cm，求斜边。
c² = 6² + 8² = 36 + 64 = 100
c = √100 = 10 cm
13 m. These whole-number answers are called Pythagorean triples and are worth memorising.
第二个例子：直角边为 5 m 和 12 m，则 c² = 25 + 144 = 169，所以 c = 13 m。这些整数答案称为毕达哥拉斯数，值得记忆。
If the answer is not a perfect square, give it in surd form or rounded to the required accuracy. For example, shorter sides 1 and 1 give c = √2 ≈ 1.41. When the question says "give your answer to 2 decimal places", only round at the very end.
如果答案不是完全平方数，就用根号形式表示，或按要求精度取近似值。例如直角边为 1 和 1 时，c = √2 ≈ 1.41。当题目要求"保留两位小数"时，只在最后一步取近似。
5. Finding a Shorter Side | 求直角边
When the hypotenuse and one shorter side are known, subtract the square of the known shorter side from the square of the hypotenuse, then take the square root. The unknown is never the largest number here, so your answer must be smaller than the hypotenuse — a useful check.
当已知斜边和一条直角边时，用斜边的平方减去已知直角边的平方，再开平方。此时未知量一定不是最大的数，所以答案必须小于斜边——这是一个很实用的检验。
Worked example: a right-angled triangle has hypotenuse 13 cm and one shorter side 5 cm. Find the other shorter side.
例题：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。
b² = 13² − 5² = 169 − 25 = 144
b = √144 = 12 cm
Another example: hypotenuse 10 cm, one shorter side 7 cm, gives b² = 100 − 49 = 51, so b = √51 des can be of similar length when the triangle is nearly isosceles; the hypotenuse is still the longest.
再看一例：斜边 10 cm，一条直角边 7 cm，则 b² = 100 − 49 = 51，所以 b = √51 ≈ 7.14 cm。注意当三角形接近等腰时，两条直角边的长度可能相近；但斜边仍然是最长的。
A frequent trap is being given the two shorter sides' difference rather than one of them. Convert the information into an equation first, then apply the theorem.
一个常见的陷阱是题目给出的是两条直角边之差，而不是其中一条边。此时应先把信息转化为方程，再应用定理。
6. Pythagorean Triples Worth Knowing | 值得记住的毕达哥拉斯数
instantly saves time and reduces arithmetic errors, especially in calculator-free papers.
毕达哥拉斯数是指满足 a² + b² = c² 的三个正整数。能一眼认出它们可以节省时间、减少计算错误，尤其是在不允许使用计算器的试卷中。
| Triple | 数组 | Check | 验证 | Typical use | 常见用法 |
| 3, 4, 5 | 9 + 16 = 25 | Basic side finding | 基础求边 |
| 5, 12, 13 | 25 + 144 = 169 | Ladder and wall problems | 梯子靠墙问题 |
| 8, 15, 17 | 64 + 225 = 289 | Rectangle diagonals | 矩形对角线 |
| 7, 24, 25 | 49 + 576 = 625 | Higher-tier questions | 高阶题目 |
| 20, 21, 29 | 400 + 441 = 841 | Non-obvious cases | 不易察觉的情形 |
Any multiple of a triple is also a triple. Doubling 3-4-5 gives 6-8-10; tripling it gives 9-12-15; multiplying 5-12-13 by 2 gives 10-24-26. Spotting a common factor can turn a messy calculation into a one-line answer.
任何一个毕达哥拉斯数数组的倍数仍然是毕达哥拉斯数。3-4-5 乘 2 得 6-8-10，乘 3 得 9-12-15；5-12-13 乘 2 得 10-24-26。发现公因数可以把繁琐的计算变成一行答案。
7. Distance Between Two Points on a Coordinate Grid | 坐标网格上两点间的距离
On a coordinate grid, the distance between two points is the hypotenuse of a right-angled triangle whose legs are the horizontal and vertical differences. If the points are (x₁, y₁) and (x₂, y₂), then:
在坐标网格上，两点间的距离是一个直角三角形的斜边，其两条直角边分别是横坐标之差与纵坐标之差。若两点为 (x₁, y₁) 与 (x₂, y₂)，则：
d = √((x₂ − x₁)² + (y₂ − y₁)²)
). The horizontal difference is 4 − 1 = 3 and the vertical difference is 6 − 2 = 4, so d = √3² + 4² = √25 = 5 units.
例题：求 A(1, 2) 与 B(4, 6) 之间的距离。横向差为 4 − 1 = 3，纵向差为 6 − 2 = 4，所以 d = √3² + 4² = √25 = 5 个单位。
Three practical tips for coordinate questions:
坐标题的三个实用技巧：
- Sketch the two points and draw the right-angled triangle; it prevents sign errors. | 画出两点并作出直角三角形，可避免符号错误。
- Squaring removes negative signs, so the order of subtraction does not matter. | 平方会消去负号，因此相减的顺序无关紧要。
- Give the answer in units, and round only at the end. | 答案要带单位，且只在最后一步取近似。
8. Real-Life Distance Problems | 现实生活中的距离问题
Pythagoras' theorem is used whenever a problem involves a right angle and two known lengths in a real context: ladders against walls, diagonal paths across a field, television screen diagonals, ramp lengths and sailing routes.
只要现实问题涉及直角和两个已知长度，就会用到毕达哥拉斯定理：梯子靠墙、穿过场地的对角线路径、电视屏幕对角线、斜坡长度以及航行路线。
Worked example: a ladder 5 m long leans against a wall. Its foot is 3 m from the base of the wall. How high up the wall does it reach?
例题：一把长 5 m 的梯子斜靠在墙上，梯脚距墙底 3 m。它能到达墙面多高？
h² = 5² − 3² = 25 − 9 = 16
h = √16 = 4 m
Note that in this problem the ladder itself is the hypotenuse, because it is opposite the right angle where the wall meets the ground. Misidentifying the hypotenuse here is the classic error.
注意在这个问题中，梯子本身就是斜边，因为它正对着墙与地面相交的直角。此处认错斜边是最典型的错误。
Another example: a rectangular playing field measures 80 m by 60 m. A player runs diagonally from one corner to the opposite corner. The distance is √80² + 60² = √6400 + 3600 = √10000 = 100 m.
再看一例：一块长 80 m、宽 60 m 的矩形球场，一名球员沿对角线从一个角跑到对角。距离为 √80² + 60² = √6400 + 3600 = √10000 = 100 m。
9. The Converse: Testing for a Right Angle | 逆定理：判断是否为直角
The converse of Pythagoras' theorem lets you decide whether a triangle is right-angled when all three sides are known. Identify the longest side, square it, and compare with the sum of the squares of the other two.
毕达哥拉斯定理的逆定理可以在已知三边时判断三角形是否为直角三角形。先找出最长边，将其平方，再与另外两边的平方和作比较。
- If longest² = sum of the other two squares → right-angled. | 若 最长边² = 另两边平方和 → 是直角三角形。
- If longest² > sum of the other two squares → obtuse-angled. | 若 最长边² > 另两边平方和 → 是钝角三角形。
- If longest² < sum of the other two squares → acute-angled. | 若 最长边² < 另两边平方和 → 是锐角三角形。
are equal, so the triangle is right-angled, with the right angle opposite the 15 cm side.
示例：三边为 9 cm、12 cm、15 cm。最长边为 15，15² = 225，而 9² + 12² = 81 + 144 = 225。两者相等，所以该三角形是直角三角形，直角位于 15 cm 边的对面。
lse, so the triangle is acute-angled. Always state the comparison explicitly to gain the method mark.
示例：三边为 4 cm、5 cm、6 cm。最长边为 6，36 > 16 + 25 = 41 不成立，所以该三角形是锐角三角形。务必明确写出比较过程，以获取方法分。
10. Pythagoras in Three Dimensions | 三维中的毕达哥拉斯定理
Three-dimensional problems are solved by applying the theorem twice. First find a diagonal of a face, then use that diagonal with the remaining edge to find the space diagonal of the solid.
三维问题通过两次使用定理来解决。先求某个面的对角线，再用这条对角线与剩下的棱求出立体的体对角线。
Worked example: a cuboid measures 3 cm by 4 cm by 12 cm. Find the length of the space diagonal.
例题：一个长方体的长宽高分别为 3 cm、4 cm、12 cm，求体对角线长。
Face diagonal: d₁² = 3² + 4² = 25, so d₁ = 5 cm
Space diagonal: d² = 5² + 12² = 25 + 144 = 169
d = √169 = 13 cm
In general, for a cuboid with edges l, w and h:
一般地，对于棱长为 l、w、h 的长方体：
d = √l² + w² + h²
The same idea gives the height of a pyramid or the length of a rod that just fits inside a box. Sketching the solid and marking the right-angled triangles you intend to use is essential.
同样的思路可以求棱锥的高，或恰好能放进盒子的杆长。画出立体图并标出你打算使用的各个直角三角形，是非常关键的步骤。
11. Common Mistakes and Exam Techniques | 常见错误与考试技巧
Most lost marks in Pythagoras questions come from a small number of repeated errors rather than from a lack of understanding. Knowing them in advance is the fastest way to improve.
毕达哥拉斯题目中的失分大多来自少数反复出现的错误，而不是理解不足。提前熟悉这些错误是提分最快的方法。
- Wrong side squared as the hypotenuse. | 把错误的边当作斜边来平方。
- Adding when a shorter side is unknown, or subtracting when the hypotenuse is unknown. | 求直角边时用加法，求斜边时却用减法。
- Forgetting to take the square root, leaving the answer as c². | 忘记开平方，答案停留在 c²。
- Rounding too early and losing accuracy in the final answer. | 过早取近似，导致最终答案精度不足。
- Mixing units, for example adding centimetres to metres. | 单位混用，例如把厘米与米相加。
- Assuming a triangle is right-angled without checking for the right-angle symbol. | 未确认直角符号就假设三角形是直角三角形。
In the examination, always write the formula, then substitute, then evaluate, then square root, then round. This structure secures method marks even if a slip occurs at the end. Show the substitution line clearly, and give units in your final answer.
考试中，请按"写公式 → 代入 → 计算 → 开方 → 取近似"的顺序作答。即使最后一步出错，这种结构也能保住方法分。把代入那一步写清楚，并在最终答案中带单位。
A useful final check: the hypotenuse must always be longer than either shorter side. If your answer does not satisfy this, go back and look again.
一个实用的最终检查：斜边必须比任何一条直角边长。如果答案不满足这一点，请回头重新检查。
12. Quick Revision Summary | 快速复习总结
The whole topic can be compressed into a small number of rules. Review the table below before a test and make sure you can reproduce each line from memory.
整个专题可以浓缩为少数几条规则。考前复习下表，并确认自己能够凭记忆写出每一行。
| Situation | 情形 | Method | 方法 | Key point | 关键点 |
| Find hypotenuse | 求斜边 | c² = a² + b² | Add, then square root | 先加，再开方 |
| Find shorter side | 求直角边 | a² = c² − b² | Subtract, then square root | 先减，再开方 |
| Distance between points | 两点间距离 | d = √((Δx)² + (Δy)²) | Use grid differences | 用坐标差 |
| Test for right angle | 判断直角 | Compare c² with a² + b² | Identify the longest side first | 先找最长边 |
| Cuboid diagonal | 长方体对角线 | d = √l² + w² + h² | Apply the theorem twice | 两次使用定理 |
Master these five situations and the topic becomes almost automatic. Practise a mixture of side-finding, word-problem and converse questions, since examination papers rarely test only one type.
掌握这五种情形，这个专题几乎就能自动应对。请混合练习求边题、应用题和逆定理判断题，因为试卷很少只考一种类型。
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