📚 Essential Maths Book 8H: Key Concepts Explained | KS3 数学：Essential Maths Book 8H 知识点精讲
Essential Maths Book 8H covers a range of key topics for Year 8 Higher students in the UK. Mastering these areas – from negative numbers and fractions to algebra and geometry – is essential for success in KS3 and beyond. This guide breaks down the most important concepts with clear explanations, examples and useful tips.
《Essential Maths Book 8H》涵盖了英国Year 8高阶学生的核心数学主题。掌握这些内容——从负数、分数到代数和几何——对于KS3以及后续的学习成功至关重要。本指南通过清晰的解释、示例和实用技巧，分解了最重要的概念。
1. Negative Numbers and BIDMAS | 负数与运算顺序
Negative numbers appear frequently in mathematics. When adding and subtracting, imagine a number line: adding a positive moves right, adding a negative moves left. Subtracting a negative is the same as adding a positive. For example, 5 − (−2) = 5 + 2 = 7.
负数在数学中经常出现。进行加减运算时，可以把数轴想象出来：加上正数向右移动，加上负数向左移动。减去一个负数等于加上相应的正数。例如，5 − (−2) = 5 + 2 = 7。
For multiplication and division, the rule is: if the signs are the same, the answer is positive; if the signs are different, the answer is negative. Examples: (−6) × 4 = −24; (−15) ÷ (−3) = 5.
乘除法的规则是：同号得正，异号得负。例如：(−6) × 4 = −24；(−15) ÷ (−3) = 5。
The order of operations is given by BIDMAS: Brackets, Indices, Division and Multiplication (from left to right), Addition and Subtraction (from left to right). Always follow this order to compute expressions like 3 + 4 × (2² − 1). Work inside brackets first: 2² = 4, so inside (4 − 1) = 3. Then multiply: 4 × 3 = 12. Finally add: 3 + 12 = 15.
运算顺序遵循BIDMAS规则：先算括号，再算指数，然后乘除（从左到右），最后加减（从左到右）。必须严格按照此顺序计算表达式，如 3 + 4 × (2² − 1)。先算括号内：2² = 4，所以 (4 − 1) = 3。接着乘法：4 × 3 = 12。最后加法：3 + 12 = 15。
2. Fractions, Decimals and Percentages | 分数、小数与百分数
To convert a fraction to a decimal, divide the numerator by the denominator. For example, 3/8 = 3 ÷ 8 = 0.375. To convert a decimal to a percentage, multiply by 100. So 0.375 becomes 37.5%. Percent means ‘per hundred’, so 37.5% is 37.5/100.
将分数转换为小数，用分子除以分母。例如，3/8 = 3 ÷ 8 = 0.375。小数转换为百分数，乘以100即可。所以0.375变成37.5%。百分数的意思是“每一百”，因此37.5%即37.5/100。
When adding or subtracting fractions, find a common denominator first. For 1/3 + 1/4, the common denominator is 12, so we rewrite as 4/12 + 3/12 = 7/12. For multiplication, multiply the numerators and denominators straight across: 2/5 × 3/7 = 6/35. For division, invert the second fraction and multiply: 4/9 ÷ 2/3 = 4/9 × 3/2 = 12/18 = 2/3 after simplifying.
进行分数加减时，先找到公分母。例如，1/3 + 1/4，公分母为12，转化为4/12 + 3/12 = 7/12。乘法直接分子乘分子、分母乘分母：2/5 × 3/7 = 6/35。除法将第二个分数取倒数再相乘：4/9 ÷ 2/3 = 4/9 × 3/2 = 12/18，化简为2/3。
3. Percentage Change | 百分数变化
To increase an amount by a percentage, use a multiplier. For a 20% increase, the multiplier is 1 + 0.20 = 1.20. So an £80 coat after a 20% increase costs £80 × 1.20 = £96. For a 15% decrease, the multiplier is 1 − 0.15 = 0.85. Thus, £200 reduced by 15% becomes £200 × 0.85 = £170.
将某个量增加一个百分数，使用乘数。增加20%，乘数为 1 + 0.20 = 1.20。因此一件£80的外套增长20%后售价为£80 × 1.20 = £96。减少15%，乘数为 1 − 0.15 = 0.85。所以£200减少15%后为£200 × 0.85 = £170。
To find the original amount before a percentage change, divide by the multiplier. If a price including 20% VAT is £96, the original price is £96 ÷ 1.20 = £80. Reverse percentage problems are common in KS3 and require careful reading.
想要求出百分数变化前的原值，需要除以乘数。如果含20%增值税的价格为£96，则原价为£96 ÷ 1.20 = £80。逆向百分数问题在KS3很常见，需仔细审题。
4. Ratio and Proportion | 比与比例
Ratios are simplified by dividing each part by their highest common factor. For example, 12:8 simplifies to 3:2 (dividing by 4). To share an amount in a given ratio, first find the total number of parts. Divide the amount by the total parts to find the value of one part, then multiply. Share £300 in the ratio 3:2: total parts = 5, one part = £60. The shares are 3 × £60 = £180 and 2 × £60 = £120.
比可通过除以各部分的最大公因数来化简。例如，12:8 化简为 3:2（同除以4）。按给定比例分配一个量，首先求出总份数。将总量除以总份数得到一份的值，再乘以相应的份数。例如将£300按3:2分配：总份数=5，一份为£60，分配得到3×£60=£180和2×£60=£120。
| Example: Share 200 sweets in ratio 5:3 | Total parts = 8, one part = 25. Shares: 5×25=125 and 3×25=75 sweets. |
Direct proportion means that as one quantity increases, the other increases at the same rate. If 5 pens cost £4, then 20 pens cost 20/5 × £4 = 4 × £4 = £16. Use a unitary method: find the cost of one item first.
正比例意味着一个量增加时，另一个量以相同的速率增加。如果5支笔£4，那么20支笔花费为20/5 × £4 = 4 × £4 = £16。使用归一法：先求出一件物品的价格。
5. Algebraic Expressions and Brackets | 代数式与括号
Like terms are terms that have exactly the same variables and powers. Simplify expressions by collecting like terms: 3a + 2b + 5a − 3b = 8a − b. Be careful with negatives: 2x − (3x + 4) = 2x − 3x − 4 = −x − 4.
同类项指含有完全相同变量及指数的项。简化代数式时合并同类项：3a + 2b + 5a − 3b = 8a − b。注意负号：2x − (3x + 4) = 2x − 3x − 4 = −x − 4。
Expanding brackets: multiply the term outside by every term inside. Example: 4(x + 3) = 4x + 12. For double brackets like (x + 2)(x + 5), use the FOIL method: First (x×x = x²), Outer (x×5 = 5x), Inner (2×x = 2x), Last (2×5 = 10). Simplify to x² + 7x + 10.
展开括号：用括号外的项乘以括号内的每一项。例如 4(x + 3) = 4x + 12。对于双括号如 (x + 2)(x + 5)，采用FOIL法：首项 x×x = x²，外项 x×5 = 5x，内项 2×x = 2x，尾项 2×5 = 10，最后化简为 x² + 7x + 10。
6. Linear Equations | 线性方程
Solving an equation means finding the value of the unknown that makes the statement true. Use inverse operations to isolate the variable. Example: 2x + 3 = 11 → subtract 3 from both sides: 2x = 8 → divide by 2: x = 4.
解方程就是找到使等式成立的未知数的值。使用逆运算隔离变量。例如：2x + 3 = 11，两边同减3得到2x = 8，再同除以2得 x = 4。
When there are unknowns on both sides, collect all x terms on one side, constants on the other. Solve 5x − 2 = 3x + 6: subtract 3x from both sides → 2x − 2 = 6; add 2 → 2x = 8; divide by 2 → x = 4. For equations with brackets, expand first: 2(x − 3) = 8 → 2x − 6 = 8 → 2x = 14 → x = 7.
若方程两边都有未知数，将所有x项移到一边，常数移到另一边。解 5x − 2 = 3x + 6：两边同减3x 得 2x − 2 = 6；同加2得 2x = 8；除以2得 x=4。对于含括号的方程，先展开：2(x − 3) = 8 → 2x − 6 = 8 → 2x = 14 → x = 7。
If fractions appear, multiply both sides by the least common denominator to clear denominators. For instance, x/3 + 1 = 5/6 → multiply by 6: 2x + 6 = 5 → 2x = −1 → x = −0.5.
若出现分数，两边同乘最小公分母来去掉分母。例如 x/3 + 1 = 5/6，乘6得 2x + 6 = 5 → 2x = −1 → x = −0.5。
7. Sequences and the nth Term | 数列与第n项
A linear sequence has a constant difference between terms. For example, 5, 8, 11, 14, … increases by 3 each time. The nth term gives a formula for any term. The rule is: nth term = (common difference) × n + (zeroth term). Here the zeroth term = 5 − 3 = 2, so nth term = 3n + 2. Check: n=1 gives 3×1+2=5, correct.
线性数列的相邻两项之差为常数。例如 5, 8, 11, 14, … 每次加3。第n项给出了任何一项的公式。规则是：第n项 = 公差 × n + 第零项。此处第零项 = 5 − 3 = 2，因此第n项 = 3n + 2。验证：n=1 时 3×1+2=5，正确。
To find the nth term from any linear sequence: subtract the common difference from the first term to get the zeroth term. Example: Sequence 1, 6, 11, 16, … difference = 5, first term 1, so zeroth term = 1 − 5 = −4. nth term = 5n − 4. Always test with n=2: 5×2−4=6, which matches.
从任意线性数列求第n项：用首项减去公差得到第零项。例如数列 1, 6, 11, 16, … 公差=5，首项1，第零项=1−5=−4。第n项=5n−4。始终用n=2检查：5×2−4=6，吻合。
8. Angles and Parallel Lines | 角与平行线
When two parallel lines are cut by a transversal, several angle relationships appear: corresponding angles are equal, alternate interior angles are equal, and interior (co-interior) angles add up to 180°. Recognising these patterns helps solve unknown angle problems without a protractor.
当一条截线与两条平行线相交时，会出现多种角度关系：同位角相等，内错角相等，同旁内角互补（和为180°）。识别这些规律有助于在没有量角器的情况下求解未知角度。
- Corresponding angles are in the same position at each intersection; they are equal.
- Alternate angles are inside the parallel lines on opposite sides of the transversal; they are equal.
- Co-interior angles are inside on the same side; they sum to 180°.
- 同位角位于截线与平行线的每个交点的相同位置，它们相等。
- 内错角位于平行线内侧、截线两侧，它们相等。
- 同旁内角位于平行线内侧且截线同侧，它们相加为180°。
The sum of interior angles in a polygon with n sides is (n − 2) × 180°. For a pentagon (5 sides), sum = (5−2)×180° = 540°. If the polygon is regular, each interior angle = sum ÷ n, so a regular pentagon interior angle = 540° ÷ 5 = 108°.
具有n条边的多边形内角和为 (n − 2) × 180°。五边形（5条边）内角和 = (5−2)×180° = 540°。如果是正多边形，每个内角 = 内角和 ÷ n，因此正五边形每个内角 = 540° ÷ 5 = 108°。
9. Area and Perimeter | 面积与周长
Perimeter is the distance around a shape. Area is the space inside measured in square units. Key formulas for 2D shapes include:
周长是形状一周的长度。面积是内部的空间，以平方单位计量。以下是2D形状的重要公式：
- Rectangle: Area = length × width
- Triangle: Area = ½ × base × height
- Parallelogram: Area = base × vertical height
- Trapezium: Area = ½ × (sum of parallel sides) × height
- Circle: Area = πr², Circumference = 2πr (π ≈ 3.14)
- 矩形：面积 = 长 × 宽
- 三角形：面积 = ½ × 底 × 高
- 平行四边形：面积 = 底 × 垂直高度
- 梯形：面积 = ½ × (两底和) × 高
- 圆：面积 = πr²，周长 = 2πr（π≈3.14）
Area of triangle = ½ × b × h
Area of trapezium = ½(a + b)h
Area of circle = πr²
When finding the area of compound shapes, split them into simpler parts, calculate each area, then add or subtract as needed. Always use the perpendicular height for triangles and parallelograms.
求组合图形的面积时，将其分解为简单图形，分别计算面积，再按需相加或相减。对于三角形和平行四边形，始终使用垂直高度。
10. Volume and Surface Area | 体积与表面积
Volume measures the space inside a 3D shape in cubic units. Surface area is the total area of all faces. For a cuboid (rectangular prism) with length l, width w, height h: Volume = l × w × h. Surface area = 2(lw
Published by TutorHao | KS3 Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导