Pythagoras’ Theorem: A Complete Year 8 Guide – 勾股定理：八年级（KS3）数学完全指南

1. What Is Pythagoras’ Theorem: The Core Relationship in Right-Angled Triangles | 什么是勾股定理：直角三角形中的核心关系

勾股定理（Pythagoras’ Theorem）是初中数学中最重要、最常用的定理之一，也是 Year 8（八年级）英国数学课程的核心内容。它描述的是直角三角形三条边之间的一种确定关系：在任何一个直角三角形中，两条直角边的平方和等于斜边的平方。这个看似简单的等式，背后连接着几何、代数、测量和建筑等多个领域，是学生从平面几何迈向更高级数学的必经之路。

Pythagoras’ Theorem is one of the most important and frequently used results in lower secondary mathematics, and a core topic in the Year 8 UK curriculum. It describes a precise relationship between the three sides of a right-angled triangle: in any right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. This deceptively simple equation links geometry, algebra, measurement and construction, and it is an essential stepping stone from basic plane geometry towards more advanced mathematics.

定理得名于古希腊数学家毕达哥拉斯（Pythagoras of Samos，约公元前570年 – 约公元前495年），但考古证据表明，巴比伦人和埃及人在他之前几百年就已经在实际测量中使用了这一关系。例如，古埃及人在建造金字塔和丈量土地时，就会使用边长为3、4、5的三角形来确定直角。这说明数学定理往往不是某一个人凭空创造的，而是人类在实践中反复发现、总结并最终被系统证明的知识。

The theorem is named after the ancient Greek mathematician Pythagoras of Samos (c. 570 BC – c. 495 BC), but archaeological evidence shows that Babylonian and Egyptian surveyors used the relationship hundreds of years before him. For example, ancient Egyptian builders used triangles with side lengths 3, 4 and 5 to mark out right angles when constructing pyramids and measuring land. This reminds us that mathematical theorems are rarely created from nothing by a single person; they are discovered, refined and eventually proved systematically by many civilisations over time.

在本章中，我们将从公式本身出发，逐步学习如何识别斜边、如何求解任意一条未知边、如何用面积法理解定理的证明，以及如何在实际问题中应用勾股定理。每一部分都配有中英双语讲解和典型例题，帮助你在理解原理的同时掌握解题步骤。

In this chapter, we start from the formula itself and work step by step: identifying the hypotenuse, finding any unknown side, understanding a geometric proof by area, and applying the theorem to real-world problems. Every section includes bilingual explanations and worked examples, so you can grasp the underlying ideas while mastering the solution steps.

2. The Formula and Notation: a² + b² = c² | 公式与记号：a² + b² = c²

勾股定理的数学表达式为 a² + b² = c²，其中 a 和 b 表示两条直角边（legs），c 表示斜边（hypotenuse）。这里的上标 2 表示”平方”，即一个数乘以它本身。例如，3² 等于 3 × 3，结果是 9。平方运算在勾股定理中扮演着核心角色，因为定理的本质是”面积”关系：以斜边为边长的正方形面积，恰好等于以两条直角边为边长的两个正方形面积之和。

The theorem is written as a² + b² = c², where a and b are the two legs (the shorter sides meeting at the right angle) and c is the hypotenuse (the longest side). The superscript 2 means “squared”, that is, a number multiplied by itself. For example, 3² = 3 × 3 = 9. Squaring is central to the theorem because its true meaning is about area: the area of the square drawn on the hypotenuse equals the sum of the areas of the squares drawn on the two legs.

我们可以用一幅经典的”正方形图”来直观理解这个关系：在三角形的每条边上各画一个正方形，边长分别为 a、b、c。那么这三个正方形的面积分别是 a²、b² 和 c²。勾股定理断言：小正方形面积之和等于大正方形面积，即 a² + b² = c²。这正是为什么定理也叫”毕达哥拉斯平方关系”。

We can visualise this with the classic “squares diagram”: draw a square on each side of the triangle, with side lengths a, b and c. The areas of these squares are a², b² and c². The theorem states that the sum of the two smaller square areas equals the area of the largest square: a² + b² = c². This is why the result is sometimes called the Pythagorean square relationship.

在实际解题中，字母 a、b、c 并不是固定的：c 永远代表斜边，而 a 和 b 可以指任意两条直角边，顺序无关紧要。重要的是先弄清楚哪条边是斜边。很多同学在套用公式时出错，往往不是因为不会计算，而是因为没有正确识别斜边。下一节我们就专门解决这个问题。

In practice, the letters a, b and c are not fixed: c always represents the hypotenuse, while a and b can label either leg, in any order. What matters is identifying which side is the hypotenuse first. Many students make errors not because they cannot calculate, but because they label the wrong side as c. The next section tackles exactly this problem.

3. Identifying the Hypotenuse: The Longest Side Opposite the Right Angle | 识别斜边：直角对面的最长边

斜边（hypotenuse）是直角三角形中最长的边，它总是位于直角（90度角）的正对面。这是识别斜边的两条黄金法则：第一，斜边对着直角；第二，斜边是三条边中最长的一条。在一个标准的直角三角形图中，直角通常用一个小方块标记，斜边就是与这个小方块不相邻的那条边。

The hypotenuse is the longest side of a right-angled triangle, and it always lies directly opposite the right angle (the 90-degree corner). Two golden rules help you identify it: first, the hypotenuse faces the right angle; second, it is the longest of the three sides. In a typical diagram, the right angle is marked with a small square, and the hypotenuse is the side that does not touch that square.

为什么斜边一定最长？我们可以用一条直观的理由来理解：在直角三角形中，直角是最大的角（另外两个角都小于90度），而在任何三角形中，大角对大边。直角最大，所以它对面的边也最长。这个”角越大，边越长”的规律在初中几何中非常有用，它不仅能帮你识别斜边，还能帮你判断三角形中边的相对大小。

Why must the hypotenuse be the longest? There is a simple intuitive reason: in a right-angled triangle, the right angle is the largest angle (the other two are both smaller than 90 degrees), and in any triangle the largest angle faces the longest side. Since the right angle is the biggest, the side opposite it is the longest. This “larger angle, longer side” rule is very useful across lower secondary geometry: it helps you identify the hypotenuse and also compare side lengths generally.

判断小练习：下面哪些边是斜边？（1）一个直角三角形，三条边分别为 5 cm、12 cm、13 cm；（2）一个直角三角形，两条直角边为 6 cm 和 8 cm，斜边为 10 cm。答案分别是 13 cm 和 10 cm，因为它们都是各自三角形中最长且对着直角的那条边。如果你能轻松找出斜边，就已经为正确使用勾股定理打下了坚实基础。

Quick check: which side is the hypotenuse in each case? (1) A right-angled triangle with sides 5 cm, 12 cm and 13 cm; (2) a right-angled triangle with legs 6 cm and 8 cm and hypotenuse 10 cm. The answers are 13 cm and 10 cm respectively, because in each triangle that side is the longest and lies opposite the right angle. If you can spot the hypotenuse quickly, you have already laid a solid foundation for using the theorem correctly.

4. Finding the Hypotenuse: Applying the Formula Directly | 求斜边长度：公式的直接应用

当我们知道两条直角边的长度、需要求斜边时，可以直接套用公式 a² + b² = c²，最后对 c² 开平方根。开平方是平方的逆运算：如果 x² = 49，那么 x = 7（因为 7 × 7 = 49）。在计算器上，我们使用根号键（√）来完成这一步。

When we know the two legs and need the hypotenuse, we apply the formula directly as a² + b² = c² and finish by taking the square root of c². Taking a square root is the inverse of squaring: if x² = 49, then x = 7 (because 7 × 7 = 49). On a calculator we use the square root key (√) for this step.

标准例题：一个直角三角形的两条直角边分别为 3 cm 和 4 cm，求斜边长度。解：a² + b² = 3² + 4² = 9 + 16 = 25，所以 c² = 25，c = √25 = 5 cm。答案是 5 cm。这就是著名的 3-4-5 三角形，它是勾股定理最简单的整数例子，也是工程师和木工最常用的”直角检验工具”。

Worked example: a right-angled triangle has legs of 3 cm and 4 cm. Find the hypotenuse. Solution: a² + b² = 3² + 4² = 9 + 16 = 25, so c² = 25 and c = √25 = 5 cm. The answer is 5 cm. This is the famous 3-4-5 triangle, the simplest whole-number example of the theorem and the most common “right-angle checking tool” used by engineers and carpenters.

第二个例题：一条直角边为 6 cm，另一条为 8 cm，求斜边。解：c² = 6² + 8² = 36 + 64 = 100，c = √100 = 10 cm。注意，这里的结果恰好也是整数。但并非所有题目都会给出漂亮的整数答案。例如直角边为 2 cm 和 3 cm 时，c² = 4 + 9 = 13，c = √13，约等于 3.61 cm。遇到这种情况，按题目要求保留小数位数（通常是1位或2位），并注意单位的书写。

Second example: one leg is 6 cm and the other is 8 cm. Solution: c² = 6² + 8² = 36 + 64 = 100, so c = √100 = 10 cm. Notice that this answer is also a nice whole number. But not every question gives a neat integer result. For legs of 2 cm and 3 cm, for instance, c² = 4 + 9 = 13, so c = √13, approximately 3.61 cm. In such cases, round to the degree of accuracy requested (usually 1 or 2 decimal places) and remember to write the unit.

解题格式建议：规范的书写有助于避免计算错误，也便于阅卷老师理解你的思路。推荐分三步写：第一步列出公式 a² + b² = c²；第二步代入数值并计算平方和；第三步开平方并写出答案（含单位）。这种”公式 – 代入 – 求解”的三段式结构，是英国中学数学考试中公认的规范格式。

Layout advice: neat written working reduces calculation errors and helps the examiner follow your reasoning. A three-step structure is recommended: first write the formula a² + b² = c²; second substitute the numbers and compute the sum of squares; third take the square root and state the answer with its unit. This “formula – substitute – solve” structure is the recognised standard format in UK secondary mathematics exams.

5. Finding a Shorter Side: Rearranging the Formula | 求直角边长度：公式的重新排列

如果题目给出的是斜边和一条直角边，要求另一条直角边，我们就不能直接套用原公式，而需要先对公式进行变形。由 a² + b² = c²，我们可以得到 a² = c² – b²（或 b² = c² – a²）。也就是说：直角边的平方等于斜边的平方减去另一条直角边的平方。这一步变形是本章最重要的代数技巧。

If the question gives the hypotenuse and one leg and asks for the other leg, we cannot use the formula directly; we must first rearrange it. From a² + b² = c² we get a² = c² – b² (or b² = c² – a²). In words: the square of a leg equals the square of the hypotenuse minus the square of the other leg. This rearrangement is the most important algebraic skill in this chapter.

标准例题：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。解：设未知直角边为 a，则 a² = c² – b² = 13² – 5² = 169 – 25 = 144，所以 a = √144 = 12 cm。答案是一个整数，这又是一个经典的 5-12-13 勾股数组。细心的话你会发现，这道题其实就是第3节判断练习中提到的三角形。

Worked example: a right-angled triangle has hypotenuse 13 cm and one leg 5 cm. Find the other leg. Solution: let the unknown leg be a, then a² = c² – b² = 13² – 5² = 169 – 25 = 144, so a = √144 = 12 cm. Again an integer answer, and this is the classic 5-12-13 Pythagorean triple. You may notice that this is exactly the triangle mentioned in the quick check in Section 3.

第二个例题：斜边为 10 cm，一条直角边为 6 cm，求另一条直角边。解：a² = c² – b² = 10² – 6² = 100 – 36 = 64，a = √64 = 8 cm。同样得到整数答案 8 cm。这两个例子对应 3-4-5 的放大版本（6-8-10）。这提示我们：把勾股数组整体放大或缩小相同的倍数，得到的仍然是勾股数组，这一点在下一节还会详细讨论。

Second example: hypotenuse 10 cm, one leg 6 cm. Solution: a² = c² – b² = 10² – 6² = 100 – 36 = 64, so a = √64 = 8 cm. Another integer answer, 8 cm. These two examples correspond to a scaled-up 3-4-5 triangle (6-8-10). This hints that multiplying a Pythagorean triple by the same factor produces another Pythagorean triple, a point we will develop in the next section.

常见错误提醒：很多同学在求直角边时，仍然使用加法（c² + b²），导致答案比斜边还长，这显然不合理。一个有效的自查方法：求出的直角边长度必须小于斜边。如果你的答案大于斜边，说明计算一定有误。养成”检查答案是否合理”的习惯，是考试中保住分数的关键。

Common error: when finding a leg, many students still add (c² + b²), producing an answer longer than the hypotenuse, which is clearly impossible. A useful self-check: the leg you find must be shorter than the hypotenuse. If your answer is longer than the hypotenuse, something has gone wrong. Building the habit of checking whether an answer is reasonable is the key to protecting marks in exams.

6. A Geometric Proof by Area: Understanding Why It Works | 面积法证明：理解定理为什么成立

在 Year 8 阶段，学生不需要写出完整的定理证明，但理解一个经典证明能极大加深对定理的信任和理解。最著名的证明之一是”面积法”：把四个全等的直角三角形拼成一个大正方形，通过两种不同的方式计算中间小正方形的面积，从而得到 a² + b² = c²。

At Year 8 level, students are not required to write out a full proof, but understanding one classic proof greatly deepens trust in and understanding of the theorem. The best-known approach is the “area proof”: arrange four congruent right-angled triangles to form a large square, then calculate the area of the central small square in two different ways to obtain a² + b² = c².

具体构造如下：取四个全等的直角三角形，直角边为 a 和 b，斜边为 c。把它们围成一个边长为 a + b 的大正方形，四个三角形的直角都朝外，斜边围在中间。这样，中间会留下一个边长为 c 的小正方形（因为四个斜边围成的区域四条边都等于 c，且四个角都是直角）。大正方形的面积可以写成 (a + b)²。

The construction works like this: take four congruent right-angled triangles with legs a and b and hypotenuse c. Arrange them to form a large square of side a + b, with all four right angles pointing outward and the hypotenuses forming the inside. This leaves a small square in the middle whose side is c (the four hypotenuses enclose a region whose sides are all equal to c and whose corners are right angles). The area of the large square can be written as (a + b)².

另一方面，大正方形的面积也可以看成四个三角形加中间小正方形的面积：四个三角形的总面积是 4 × (½ab) = 2ab，小正方形的面积是 c²。所以 (a + b)² = 2ab + c²。展开左边得 a² + 2ab + b² = 2ab + c²，两边同时减去 2ab，就得到 a² + b² = c²。证明完成！

On the other hand, the large square’s area can also be seen as the four triangles plus the central square: the four triangles together have area 4 × (½ab) = 2ab, and the central square has area c². So (a + b)² = 2ab + c². Expanding the left side gives a² + 2ab + b² = 2ab + c². Subtracting 2ab from both sides leaves a² + b² = c². The proof is complete!

这个证明的妙处在于它只用到了”正方形面积 = 边长 × 边长”和”三角形面积 = 底 × 高 ÷ 2″两个最基本的公式，却推出了一个影响深远的定理。类似的面积证明有上百种，据说毕达哥拉斯定理是数学中被证明次数最多的定理之一。理解这个证明，也为你未来学习更严格的演绎推理打下了基础。

The beauty of this proof is that it uses only two elementary formulas, “area of a square = side × side” and “area of a triangle = base × height ÷ 2”, yet it derives a theorem of enormous significance. Hundreds of similar area proofs exist, and Pythagoras’ theorem is said to be one of the most frequently proved results in mathematics. Understanding this proof also prepares you for the more formal deductive reasoning you will meet later.

7. Pythagorean Triples: 3-4-5, 5-12-13 and Their Families | 勾股数：3-4-5、5-12-13 及其家族

如果直角三角形的三条边都是正整数，那么这三个数就组成一个”勾股数”（Pythagorean triple）。最著名的勾股数是 3、4、5，因为 3² + 4² = 9 + 16 = 25 = 5²。其他常见的勾股数还有 5、12、13（5² + 12² = 25 + 144 = 169 = 13²）和 8、15、17（8² + 15² = 64 + 225 = 289 = 17²）。

If all three sides of a right-angled triangle are positive integers, the three numbers form a Pythagorean triple. The most famous triple is 3, 4, 5, because 3² + 4² = 9 + 16 = 25 = 5². Other common triples include 5, 12, 13 (5² + 12² = 25 + 144 = 169 = 13²) and 8, 15, 17 (8² + 15² = 64 + 225 = 289 = 17²).

勾股数有一个重要性质：把一组勾股数的每个数同时乘以同一个正整数，得到的仍然是勾股数。例如，3-4-5 乘以 2 得到 6-8-10，乘以 3 得到 9-12-15，乘以 10 得到 30-40-50。这在考试中非常实用：如果你在题目中认出 3-4-5、5-12-13 或它们的倍数，就可以直接写出答案，节省大量计算时间。

Pythagorean triples have an important property: multiplying every number in a triple by the same positive integer produces another triple. For example, 3-4-5 scaled by 2 gives 6-8-10, by 3 gives 9-12-15, and by 10 gives 30-40-50. This is very useful in exams: if you recognise 3-4-5, 5-12-13 or their multiples in a question, you can write down the answer directly and save a lot of calculation time.

还有一类特殊勾股数值得记住：两个相邻整数加一个较小整数的组合，比如 20、21、29（20² + 21² = 400 + 441 = 841 = 29²）。在 Year 8 考试中，最常见的还是 3-4-5 及其倍数，其次是 5-12-13。建议你把这两组记牢，同时记住它们的”放大版”判断方法：如果两条直角边之比接近 3:4 或 5:12，答案很可能就是对应的勾股数组。

Another family worth remembering involves two consecutive integers plus a smaller one, such as 20, 21, 29 (20² + 21² = 400 + 441 = 841 = 29²). In Year 8 exams, the most common triples by far are 3-4-5 and its multiples, followed by 5-12-13. Memorise these two, and remember how to recognise scaled versions: if the ratio of the two legs is close to 3:4 or 5:12, the answer is probably the corresponding triple.

8. Real-World Applications: Ladders, Flagpoles and Construction | 现实应用：梯子、旗杆与建筑施工

勾股定理绝不是书本上的抽象游戏，它在日常生活中无处不在。最简单的例子是梯子问题：一把梯子斜靠在墙上，梯子底部离墙脚 1.5 米，梯子长 2.5 米，那么梯子顶端离地面多高？墙与地面垂直，梯子、墙和地面恰好构成一个直角三角形：墙高是未知直角边，地面距离是另一条直角边，梯子是斜边。

Pythagoras’ theorem is not an abstract game on paper; it appears everywhere in everyday life. The simplest example is the ladder problem: a ladder leans against a wall, its foot is 1.5 m from the wall, and the ladder is 2.5 m long. How high up the wall does the ladder reach? The wall is vertical, so the ladder, wall and ground form a right-angled triangle: the wall height is the unknown leg, the ground distance is the other leg, and the ladder is the hypotenuse.

解题过程：设墙高为 h，则 h² = 2.5² – 1.5² = 6.25 – 2.25 = 4，所以 h = √4 = 2 米。答案：梯子顶端离地面 2 米。这道题同时考察了公式变形、平方运算和开平方，是典型的应用题。在实际生活中，消防员和油漆工也会用类似的计算判断梯子是否放得足够稳、够得到目标高度。

Solution: let the wall height be h, then h² = 2.5² – 1.5² = 6.25 – 2.25 = 4, so h = √4 = 2 m. Answer: the top of the ladder reaches 2 m up the wall. This question tests rearrangement, squaring and square roots all at once, and it is a typical application problem. In real life, firefighters and painters use exactly this kind of calculation to decide whether a ladder is stable enough and reaches the required height.

第二个应用是旗杆问题：为了固定一根旗杆，施工人员从旗杆顶端拉一根 13 米长的钢丝，固定在地面上离旗杆底部 5 米处。求旗杆的高度。解：h² = 13² – 5² = 169 – 25 = 144，h = 12 米。如果你认出了 5-12-13 勾股数，这道题甚至可以心算完成。类似的例子还有：电视塔的斜拉索、屋顶的斜坡长度、足球场对角线的距离计算等。

A second application is the flagpole problem: to stabilise a flagpole, workers attach a 13 m steel wire from the top of the pole to a point on the ground 5 m from its base. Find the height of the pole. Solution: h² = 13² – 5² = 169 – 25 = 144, so h = 12 m. If you recognise the 5-12-13 triple, this can even be done mentally. Similar examples include the stays of a TV tower, the slope length of a roof, and the diagonal distance across a football pitch.

第三个应用：长方形场地的对角线。一个足球场长 100 米、宽 60 米，求对角线长度。对角线把长方形分成两个全等的直角三角形，所以 d² = 100² + 60² = 10000 + 3600 = 13600，d = √13600，约等于 116.6 米。这类”对角线问题”在建筑放线、屏幕尺寸标注（如 32 英寸电视的”英寸”就是对角线长度）中非常常见。

Third application: the diagonal of a rectangular field. A football pitch is 100 m long and 60 m wide. Find its diagonal. The diagonal splits the rectangle into two congruent right-angled triangles, so d² = 100² + 60² = 10000 + 3600 = 13600, giving d = √13600, approximately 116.6 m. This “diagonal problem” is everywhere: setting out building foundations, and screen sizes (the “32 inches” of a TV refers to its diagonal).

9. Common Mistakes and Exam Technique | 常见错误与考试技巧

根据历年考试数据，Year 8 学生在勾股定理题目中最常犯的错误有四种。第一种：把斜边当成直角边代入公式，导致计算方向错误；第二种：求直角边时误用加法（c² + b²），得到比斜边还长的”直角边”；第三种：忘记开平方，直接写出 c² 作为答案；第四种：单位不统一，例如把米和厘米混在一起计算。

Exam statistics show that Year 8 students make four common errors in Pythagoras questions. First: treating the hypotenuse as a leg when substituting into the formula, which reverses the calculation. Second: using addition (c² + b²) when finding a leg, producing a “leg” longer than the hypotenuse. Third: forgetting to take the square root and giving c² as the final answer. Fourth: mixing units, such as combining metres and centimetres in one calculation.

针对这些错误，我们给出四条实战技巧。技巧一：动笔前先在图上标出直角符号和三条边的名称，明确哪条是斜边。技巧二：求直角边时，牢记”大数减小数”的原则，并写下一句话自我检查：答案必须小于斜边。技巧三：完成计算后，把答案代回原式验证，例如算得直角边为 12 时，检查 5² + 12² 是否等于 13²。技巧四：读题时先统一单位，把题目中的所有长度换算成同一单位再计算。

Against these errors, here are four practical techniques. Technique one: before writing anything, mark the right angle and label all three sides on the diagram, so the hypotenuse is clear. Technique two: when finding a leg, remember “larger minus smaller”, and use a self-check sentence: the answer must be shorter than the hypotenuse. Technique three: after calculating, substitute the answer back into the original equation; for example, if you find a leg of 12, check whether 5² + 12² equals 13². Technique four: read the question carefully and convert all lengths to the same unit before calculating.

考试书写规范：在英国中学数学考试中，即使答案正确，过程不完整也可能扣分。建议按照”公式 + 代入 + 结果 + 单位”四步书写。如果题目要求”保留到一位小数”或”用最简根式表示”，一定要严格按要求作答。遇到多步应用题时，把每一步的结果写清楚，这样即使中间出错，阅卷老师也能根据你的思路给步骤分。

Exam presentation: in UK secondary mathematics exams, an answer without working can lose marks even when correct. Write in four steps: “formula + substitution + result + unit”. If the question asks you to “round to 1 decimal place” or “leave your answer in surd form”, follow the instruction exactly. In multi-step problems, show every intermediate result clearly, so that even if you make an error, the examiner can award method marks for your reasoning.

10. Practice Questions with Worked Solutions | 练习与详细解答

下面的练习覆盖了本章所有题型，建议先独立完成，再对照解答检查。练习一：直角三角形的两条直角边为 9 cm 和 12 cm，求斜边。练习二：斜边为 17 cm，一条直角边为 15 cm，求另一条直角边。练习三：一根电线杆高 8 米，从杆顶斜拉到地面的一根拉线长 10 米，拉线固定点离杆底多远？

The exercises below cover every question type in this chapter. Attempt them independently before checking the solutions. Exercise 1: a right-angled triangle has legs of 9 cm and 12 cm; find the hypotenuse. Exercise 2: the hypotenuse is 17 cm and one leg is 15 cm; find the other leg. Exercise 3: a telephone pole is 8 m tall; a guy wire from its top to the ground is 10 m long; how far from the base of the pole is the wire anchored?

练习一解答：c² = 9² + 12² = 81 + 144 = 225，c = √225 = 15 cm。这组勾股数 9-12-15 恰好是 3-4-5 的三倍放大，如果你记住了 3-4-5 家族，可以直接写出答案。练习二解答：a² = 17² – 15² = 289 – 225 = 64，a = 8 cm。这对应 8-15-17 勾股数。练习三解答：设水平距离为 d，则 d² = 10² – 8² = 100 – 64 = 36，d = 6 米。

Solution 1: c² = 9² + 12² = 81 + 144 = 225, so c = √225 = 15 cm. This triple, 9-12-15, is exactly 3-4-5 scaled by three; if you know the 3-4-5 family you can write the answer directly. Solution 2: a² = 17² – 15² = 289 – 225 = 64, so a = 8 cm. This is the 8-15-17 triple. Solution 3: let the horizontal distance be d, then d² = 10² – 8² = 100 – 64 = 36, so d = 6 m.

挑战题：一个等腰直角三角形的斜边为 10 cm，求它的两条直角边和面积。提示：等腰直角三角形两条直角边相等，设每条直角边为 x，则 x² + x² = 10²，即 2x² = 100，x² = 50，x = √50，约等于 7.07 cm。面积 = ½ × x × x = ½ × 50 = 25 cm²。这道题把勾股定理、平方根和面积公式综合在一起，是 Year 8 高难度题目的典型代表。

Challenge: an isosceles right-angled triangle has hypotenuse 10 cm. Find its legs and area. Hint: the two legs are equal; let each leg be x, then x² + x² = 10², so 2x² = 100, x² = 50, and x = √50, approximately 7.07 cm. Area = ½ × x × x = ½ × 50 = 25 cm². This question combines the theorem, square roots and the area formula, and it is a typical hard question for Year 8.

Summary | 总结

勾股定理是 Year 8 数学中承上启下的核心内容：它上承平方与平方根的运算，下启三角比、坐标几何和向量等更高级的课题。掌握本章内容的标志是：能熟练识别斜边，能正确区分”求斜边用加法、求直角边用减法”两种情形，能完成公式变形，并能在实际情境中建立直角三角形模型。

Pythagoras’ theorem is a pivotal topic in Year 8 mathematics: it builds on squaring and square roots, and it leads on to trigonometry, coordinate geometry and vectors. You have mastered this chapter when you can identify the hypotenuse confidently, distinguish the two cases (add when finding the hypotenuse, subtract when finding a leg), rearrange the formula correctly, and set up right-angled triangle models in real situations.

复习建议：第一，熟记 3-4-5 和 5-12-13 两组基本勾股数及其倍数；第二，把每道例题的”公式 – 代入 – 求解”三步格式写在笔记本上反复模仿；第三，每周用 10 分钟做 3 道混合题（求斜边、求直角边、应用题各一道），保持手感；第四，做完后一定检查答案的合理性，特别是直角边不能大于斜边。

Revision advice: first, memorise the basic triples 3-4-5 and 5-12-13 and their multiples; second, copy the “formula – substitute – solve” three-step layout from every worked example into your notebook and imitate it; third, spend 10 minutes each week on three mixed questions (one hypotenuse, one leg, one application) to keep your skills sharp; fourth, always check whether your answer is sensible, remembering that a leg can never be longer than the hypotenuse.

最后，请记住勾股定理背后的数学之美：一个简单的等式 a² + b² = c²，跨越了两千五百年的历史，连接着古埃及的建筑智慧、古希腊的理性传统和今天工程师的计算。掌握了它，你不仅学会了一种计算方法，更开启了一扇通往数学推理世界的大门。

Finally, remember the beauty behind the theorem: the simple equation a² + b² = c² spans 2,500 years of history, connecting the building wisdom of ancient Egypt, the rational tradition of ancient Greece, and the calculations of today’s engineers. By mastering it, you have not only learned a computational technique; you have opened a door into the world of mathematical reasoning.

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