📚 Year 9 OCR Further Maths: Case Study Practice | 九年级进阶数学：案例分析实战演练
In advanced mathematics, applying concepts to real-world scenarios is essential for developing deep understanding and problem-solving skills. This collection of case studies is designed to help Year 9 students tackle practical problems using topics from the OCR Further Maths syllabus. Each case demonstrates a step-by-step approach, bridging theory with context-rich situations.
在进阶数学中，将概念应用于真实场景对于培养深刻理解和解题能力至关重要。本案例分析集旨在帮助九年级学生运用 OCR 进阶数学大纲中的知识点解决实际问题。每个案例展示了逐步求解的方法，在理论与情景之间架起桥梁。
1. Maximising a Rectangular Garden Area | 最大化矩形花园面积
A gardener has 20 metres of fencing to enclose a rectangular flower bed against an existing wall. The wall forms one side of the rectangle, so only three sides need fencing. We need to find the dimensions that give the greatest possible area.
一位园丁有 20 米长的栅栏，想要靠着一面已有的墙围出一个矩形花坛。墙作为矩形的一边，因此只需用栅栏围另外三边。我们需要找出使面积最大的尺寸。
ing, so its length is (20 − 2x) metres. The area A is given by A = x(20 − 2x) = 20x − 2x².
设垂直于墙的边长为 x 米，则平行于墙的边长由剩余栅栏得出，为 (20 − 2x) 米。面积 A 表示为 A = x(20 − 2x) = 20x − 2x²。
A = −2x² + 20x
(−2)) = 5. Thus the maximum area happens when x = 5 m, giving the parallel side as 20 − 2(5) = 10 m, and a maximum area of 5 × 10 = 50 m².
这是一个开口向下的二次函数，最大值出现在顶点。通过配方法或者利用公式 x = −b/(2a)，得到 x = −20/(2×(−2)) = 5。因此当 x = 5 m 时面积最大，平行边长为 20 − 2×5 = 10 m，最大面积为 5 × 10 = 50 m²。
2. Navigating an Aircraft with Wind | 风中航行一架飞机
An aircraft is flying with an airspeed of 250 km/h on a bearing of 030°. A wind is blowing from the west at 60 km/h. Determine the resultant ground speed and the true bearing of the aircraft.
一架飞机以 250 km/h 的空速沿 030° 的方位飞行。一阵风从西边吹来，风速为 60 km/h。求实际的飞行地速和真实方位角。
0° = 125 km/h, and its northward component is 250 cos 30° ≈ 216.5 km/h. The wind velocity v₂ is from the west, so it blows towards the east, adding 60 km/h to the eastward component and 0 northward.
将速度表示为向量。空速向量 v₁ 大小为 250，方向为从正北顺时针 30°。它的东向分量为 250 sin 30° = 125 km/h，北向分量为 250 cos 30° ≈ 216.5 km/h。风速 v₂ 来自西边，因此吹向东方，给东向分量增加 60 km/h，北向分量不变。
The resultant velocity vR 125 + 60 = 185 km/h and northward component 216.5 km/h. The ground speed is the magnitude: √185² + 216.5² θ ≈ 40.5°. Thus the aircraft travels at about 285 km/h on a bearing of approximately 041°.
合速度 vR 的东向分量为 125 + 60 = 185 km/h，北向分量为 216.5 km/h。地速即其大小：√185² + 216.5² ≈ 284.7 km/h。真实方位角 θ 从正北起算，由 tan θ = 对边/邻边 = 185 / 216.5 得 θ ≈ 40.5°。因此飞机以约 285 km/h 的速度沿约 041° 的方位飞行。
3. Measuring a Tower Height with Trigonometry | 用三角学测量塔高
From a point A on level ground, the angle of elevation to the top of a tower is 28°. After walking 40 metres directly towards the tower to point B, the new angle of elevation is 52°. Find the height of the tower.
从水平地面上的点 A 测得塔顶的仰角为 28°。朝着塔的方向径直走 40 米到达点 B 后，新的仰角变为 52°。求这座塔的高度。
base is (d + 40). Using tangent: from A, tan 28° = h / (d + 40); from B, tan 52° = h / d.
设塔高为 h，点 B 到塔底的距离为 d。点 A 到塔底的距离为 (d + 40)。利用正切函数：在 A 处，tan 28° = h / (d + 40)；在 B 处，tan 52° = h / d。
te approximate values: tan 52° ≈ 1.2799, tan 28° ≈ 0.5317. Then 1.2799d = 0.5317d + 21.268 ⇒ (1.2799 − 0.5317)d = 21.268 ⇒ 0.7482d = 21.268 ⇒ d ≈ 28.43 m. Hence h = 28.43 × 1.2799 ≈ 36.4 metres.
整理两式求出 h：h = (d + 40) tan 28°，h = d tan 52°。令两式相等：d tan 52° = (d + 40) tan 28°。代入近似值：tan 52° ≈ 1.2799，tan 28° ≈ 0.5317。得到 1.2799d = 0.5317d + 21.268 ⇒ 0.7482d = 21.268 ⇒ d ≈ 28.43 m。因此 h = 28.43 × 1.2799 ≈ 36.4 米。
4. Saving Plan with Compound Interest | 复利储蓄计划
Anisa invests £200 at the start of each year in an account that pays 3% compound interest annually. She plans to continue for 6 years. How much will she have immediately after the sixth payment? What is the total interest earned?
阿妮莎每年初投资 200 英镑到一个年利率为 3% 的复利账户。她计划持续 6 年。第六笔款项存入后她将拥有多少钱？总共赚取多少利息？
s for 5 years: 200 × (1.03)⁵, and so on down to the last deposit which earns only one year' otal amount A is the sum of the series: 200(1.03 + 1.03² + 1.03³ + 1.03⁴ + 1.03⁵ + 1.03⁶).
这是一个等比数列。首笔存款计息 6 年：200 × (1.03)⁶。第二笔计息 5 年：200 × (1.03)⁵，依此类推，最后一笔仅计息一年：200 × (1.03)¹。总金额 A 为数列之和：200(1.03 + 1.03² + 1.03³ + 1.03⁴ + 1.03⁵ + 1.03⁶)。
mpute 1.03⁶ ≈ 1.194052, so numerator 206 × 0.194052 = 39.9747, divided by 0.03 gives about £1332.49. Total invested is 6 × 200 = £1200, so interest earned is 1332.49 − 1200 = £132.49.
利用等比数列求和公式：Sₙ = a(rⁿ − 1)/(r − 1)，其中 a = 200×1.03 = 206，r = 1.03，n = 6。则 A = 206 × (1.03⁶ − 1) / 0.03。计算 1.03⁶ ≈ 1.194052，分子 206 × 0.194052 = 39.9747，除以 0.03 得到约 £1332.49。总投资为 6 × 200 = £1200，因此赚取的利息为 1332.49 − 1200 = £132.49。
5. Probability and Expected Profit from a Game | 游戏中的概率与预期利润
A charity game costs £2 to play. You roll two fair dice. If the sum is 7, you win £10; if the sum is 11 or 12, you win £3; otherwise you win nothing. Is the game favourable to the charity over many plays?
一个慈善游戏每次收费 £2。你掷两个公平的骰子。若点数和为 7，赢得 £10；若和为 11 或 12，赢得 £3；其余情况无奖金。从长远来看，这个游戏对慈善组织是否有利？
6 × 6 = 36 equally likely outcomes. Sum = 7 occurs for (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) – 6 outcomes, P(7) = 6/36 = 1/6. Sum = 11 occurs for (5,6),(6,5) – 2 outcomes; sum = 12 for (6,6) – 1 outcome, so P(11 or 12) = 3/36 = 1/12. The remaining 27 outcomes give no prize.
首先列出概率：共有 6×6=36 种等可能结果。点数和为 7 的情况有 (1,6)、(2,5)、(3,4)、(4,3)、(5,2)、(6,1) 共 6 种，P(7)=6/36=1/6。和为 11 的情况有 (5,6)、(6,5) 共 2 种；和为 12 的情况 (6,6) 1 种，因此 P(11 或 12)=3/36=1/12。其余 27 种情况无奖金。
= (10 × 1/6) + (3 × 1/12) + (0 × 27/36) = 10/6 + 3/12 = 5/3 + 1/4 = 20/12 + 3/12 = 23/12 ≈ £1.917. The player pays £2, so the expected profit for the charity is 2.00 − 1.917 = £0.083 per play. The game is slightly favourable to the charity in the long run.
每局期望奖金 = (10 × 1/6) + (3 × 1/12) + (0 × 27/36) = 10/6 + 3/12 = 5/3 + 1/4 = 20/12 + 3/12 = 23/12 ≈ £1.917。玩家支付 £2，因此慈善组织的预期利润为 2.00 − 1.917 = £0.083 每局。长远来看该游戏对慈善组织微利。
6. Production Constraints with Linear Inequalities | 生产限制与线性不等式
A small workshop produces tables (x) and chairs (y). Each table requires 4 hours of carpentry and 2 hours of finishing. Each chair requires 1 hour of carpentry and 3 hours of finishing. In a week, 36 carpentry hours and 30 finishing hours are available. Express the constraints as inequalities and shade the feasible region.
一间小作坊生产桌子 (x) 和椅子 (y)。每张桌子需要 4 小时木工和 2 小时精加工。每把椅子需要 1 小时木工和 3 小时精加工。每周可用的木工工时为 36 小时，精加工工时为 30 小时。请将限制条件表示为不等式并画出可行域。
y ≥ 0. These form a set of linear inequalities defining a polygon on the coordinate plane.
木工约束：4x + 1y ≤ 36。精加工约束：2x + 3y ≤ 30。同时还有非负条件：x ≥ 0，y ≥ 0。这些不等式组成了线性不等式组，在坐标平面上定义一个多边形可行域。
n is the quadrilateral bounded by the axes and these two lines, lying below both lines and in the first quadrant. Testing a point like (4,4): 4×4+4=20≤36 and 2×4+3×4=20≤30, confirming it is inside.
绘制区域：首先画出直线 4x + y = 36（截点为 (9,0) 和 (0,36)），再画出 2x + 3y = 30（截点为 (15,0) 和 (0,10)）。可行域是由坐标轴和这两条直线围成的四边形，位于两条直线下方且在第一象限内。检验如 (4,4) 点：4×4+4=20≤36 且 2×4+3×4=20≤30，确认位于内部。
7. Circle Theorem Proof: Angle in a Semicircle | 圆定理证明：半圆上的圆周角
Prove that the angle subtended by a diameter at the circumference of a circle is always a right angle (90°). Use vector or coordinate geometry to support your argument.
证明直径在圆周上对顶的角恒为直角（90°）。可利用向量或坐标几何来佐证。
+ y² = r². Vectors PA = (−r − x, 0 − y) and PB = (r − x, 0 − y). The dot product PA · PB = (−r − x)(r − x) + (−y)(−y) = −(r² − x²) + y² = −r² + x² + y² = −r² + r² = 0. Since the dot product is zero, PA is perpendicular to PB, so angle APB = 90°.
设圆心 O 位于原点 (0,0)，半径为 r。将直径端点置于 A(−r, 0) 和 B(r, 0)。取圆上任意一点 P(x, y)，满足 x² + y² = r²。向量 PA = (−r − x, 0 − y)，PB = (r − x, 0 − y)。点积 PA · PB = (−r − x)(r − x) + (−y)(−y) = −(r² − x²) + y² = −r² + x² + y² = −r² + r² = 0。点积为零表明 PA 与 PB 垂直，因此角 APB = 90°。
This elegant proof shows that no matter where P lies on the circle (except at A or B), the angle in the semicircle is always a right angle, a fundamental circle theorem frequently tested in geometry problems.
这个巧妙的证明表明，无论 P 在圆上何处（A 和 B 处除外），半圆上的圆周角始终为直角，这是几何题中常考的基本圆定理。
8. Comparing Exam Scores Using Box Plots | 利用箱线图比较考试成绩
Two Year 9 classes took the same further maths test. Class M scores: 45, 55, 60, 65, 70, 75, 80, 85, 90, 95. Class N scores: 50, 55, 58, 62, 65, 68, 70, 72, 75, 78. Construct box plots for both classes and compare their medians, interquartile ranges (IQR), and overall spread.
两个九年级班级参加了同一场进阶数学考试。M 班分数：45, 55, 60, 65, 70, 75, 80, 85, 90, 95。N 班分数：50, 55, 58, 62, 65, 68, 70, 72, 75, 78。为两个班分别绘制箱线图，并比较其中位数、四分位距 (IQR) 和总体分布。
= (70+75)/2 = 72.5, Upper quartile (Q₃) = 85, Maximum = 95. IQR = Q₃ − Q₁ = 85 − 60 = 25.
M 班（n=10）：最小值 = 45，下四分位数 (Q₁) = 60（排序后第 3 个值），中位数 (Q₂) = 第 5 和 6 个平均值 = (70+75)/2 = 72.5，上四分位数 (Q₃) = 85，最大值 = 95。IQR = 85 − 60 = 25。
5,78. Min = 50, Q₁ = 58, Median = (65+68)/2 = 66.5, Q₃ = 72, Max = 78. IQR = 72 − 58 = 14.
N 班：排序后 50,55,58,62,65,68,70,72,75,78。最小值 = 50，Q₁ = 58，中位数 = (65+68)/2 = 66.5，Q₃ = 72，最大值 = 78。IQR = 72 − 58 = 14。
Comparison: Class M has a higher median (72.5 vs 66.5) and a much wider IQR (25 vs 14), indicating more variability in scores. The range for M is 50 (45 to 95) compared to 28 (50 to 78) for N, so N’s performance is more consistent but M has the highest achievers. Box plots would show M’s box wider and shifted to higher values, highlighting these differences.
比较：M 班中位数更高（72.5 对 66.5），IQR 更大（25 对 14），表明成绩变异性更大。M 班全距为 50（45 至 95），N 班为 28（50 至 78），因此 N 班表现更稳定，但 M 班有最高分获得者。箱线图将显示 M 班的箱更宽且右移，突显这些差异。
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