Grade 9 Mathematics — Term 1 Notes

<h1 class="notes-h1">GRADE 9 MATHEMATICS — TERM 1 NOTES</h1>
<h2 class="notes-h2">CBC Kenya | Junior Secondary | KICD Aligned</h2>
<h3 class="notes-h3">CBCEduKenya.com | cbcedukenya@gmail.com | 0711 344 702</h3>
<hr class="section-divider">
<h2 class="notes-h2">STRAND 1: NUMBERS</h2>
<h3 class="notes-h3">Sub-Strand 1.1: Rational and Irrational Numbers</h3>
<p><strong>Specific Learning Outcomes:</strong></p>
<p>By the end of this sub-strand, the learner should be able to:</p>
<ul class="notes-list">
<li>Distinguish between rational and irrational numbers</li>
<li>Identify surds and simplify them</li>
<li>Perform operations with surds</li>
<li>Apply surds in real-life situations</li>
</ul>
<hr class="section-divider">
<h4 class="notes-h4">1.1.1 Rational Numbers</h4>
<p>A <strong>rational number</strong> is any number that can be written in the form <strong>p/q</strong> where p and q are integers and q ≠ 0.</p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Type</th><th>Example</th><th>As p/q</th></tr>
</thead><tbody>
<tr><td>Integer</td><td>5</td><td>5/1</td></tr>
<tr><td>Fraction</td><td>3/4</td><td>3/4</td></tr>
<tr><td>Terminating decimal</td><td>0.75</td><td>3/4</td></tr>
<tr><td>Recurring decimal</td><td>0.333...</td><td>1/3</td></tr>
<tr><td>Negative fraction</td><td>-2/5</td><td>-2/5</td></tr>
</tbody></table></div>
<p><strong>Converting recurring decimals to fractions:</strong></p>
<p><strong>Example:</strong> Convert 0.̄6̄ (0.666...) to a fraction</p>
<pre class="code-block"><code>
Let x = 0.666...
10x = 6.666...
10x - x = 6.666... - 0.666...
9x = 6
x = 6/9 = 2/3
</code></pre>
<p><strong>Example:</strong> Convert 0.1̄2̄ (0.121212...) to a fraction</p>
<pre class="code-block"><code>
Let x = 0.121212...
100x = 12.121212...
100x - x = 12.121212... - 0.121212...
99x = 12
x = 12/99 = 4/33
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">1.1.2 Irrational Numbers</h4>
<p>An <strong>irrational number</strong> CANNOT be written as p/q. Its decimal expansion is non-terminating and non-recurring.</p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Irrational Number</th><th>Approximate Value</th></tr>
</thead><tbody>
<tr><td>√2</td><td>1.41421356...</td></tr>
<tr><td>√3</td><td>1.73205080...</td></tr>
<tr><td>√5</td><td>2.23606797...</td></tr>
<tr><td>π</td><td>3.14159265...</td></tr>
<tr><td>e</td><td>2.71828182...</td></tr>
</tbody></table></div>
<p><strong>Key Rule:</strong> √n is irrational if n is not a perfect square.</p>
<p>Perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 → their square roots are RATIONAL.</p>
<hr class="section-divider">
<h4 class="notes-h4">1.1.3 Surds</h4>
<p>A <strong>surd</strong> is an irrational root that cannot be simplified to a rational number.</p>
<p>√2, √3, √5, √6, √7, ³√2, ³√4 are surds.</p>
<p>√4 = 2, √9 = 3, √16 = 4 are NOT surds (they are rational).</p>
<p><strong>Simplifying Surds:</strong></p>
<p>Rule: √(a × b) = √a × √b</p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Surd</th><th>Simplification</th><th>Steps</th></tr>
</thead><tbody>
<tr><td>√12</td><td>2√3</td><td>√(4×3) = √4 × √3 = 2√3</td></tr>
<tr><td>√18</td><td>3√2</td><td>√(9×2) = √9 × √2 = 3√2</td></tr>
<tr><td>√50</td><td>5√2</td><td>√(25×2) = √25 × √2 = 5√2</td></tr>
<tr><td>√75</td><td>5√3</td><td>√(25×3) = √25 × √3 = 5√3</td></tr>
<tr><td>√48</td><td>4√3</td><td>√(16×3) = √16 × √3 = 4√3</td></tr>
<tr><td>√200</td><td>10√2</td><td>√(100×2) = √100 × √2 = 10√2</td></tr>
</tbody></table></div>
<p><strong>Adding and Subtracting Surds</strong> (like terms only):</p>
<pre class="code-block"><code>
3√2 + 5√2 = 8√2 ✓ (same surd — add coefficients)
4√3 - √3 = 3√3 ✓ (same surd — subtract coefficients)
2√5 + 3√2 = 2√5 + 3√2 ✗ (different surds — cannot combine)
</code></pre>
<p><strong>Example:</strong> Simplify √12 + √27 - √3</p>
<pre class="code-block"><code>
√12 = 2√3
√27 = 3√3
√12 + √27 - √3 = 2√3 + 3√3 - √3 = 4√3
</code></pre>
<p><strong>Multiplying Surds:</strong></p>
<pre class="code-block"><code>
√a × √a = a
√2 × √2 = 2
√3 × √5 = √15
2√3 × 4√3 = 8 × 3 = 24
3√2 × 2√5 = 6√10
</code></pre>
<p><strong>Dividing Surds:</strong></p>
<pre class="code-block"><code>
√a / √b = √(a/b)
√12 / √3 = √4 = 2
√50 / √2 = √25 = 5
</code></pre>
<p><strong>Rationalising the Denominator:</strong></p>
<p>To eliminate a surd from the denominator, multiply numerator and denominator by the surd.</p>
<p><strong>Example:</strong> Rationalise 5/√3</p>
<pre class="code-block"><code>
5/√3 × √3/√3 = 5√3/3
</code></pre>
<p><strong>Example:</strong> Rationalise 6/√2</p>
<pre class="code-block"><code>
6/√2 × √2/√2 = 6√2/2 = 3√2
</code></pre>
<p><strong>Example:</strong> Rationalise 4/(√5 - 1) using the conjugate</p>
<pre class="code-block"><code>
4/(√5 - 1) × (√5 + 1)/(√5 + 1) = 4(√5 + 1)/(5 - 1) = 4(√5 + 1)/4 = √5 + 1
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">Review Questions — Sub-Strand 1.1</h4>
<ol class="notes-list">
<li>State whether each is rational or irrational: (a) √16 (b) √7 (c) 0.25 (d) π (e) 2/3</li>
<li>Convert 0.3̄ to a fraction in lowest terms.</li>
<li>Convert 0.2̄7̄ (0.272727...) to a fraction.</li>
<li>Simplify: (a) √72 (b) √98 (c) √108</li>
<li>Simplify: √45 + √20 - √5</li>
<li>Expand and simplify: (√3 + 2)(√3 - 1)</li>
<li>Rationalise the denominator: (a) 3/√5 (b) 10/√2 (c) 1/(√7 + 2)</li>
</ol>
<hr class="section-divider">
<h3 class="notes-h3">Sub-Strand 1.2: Indices (Laws of Indices)</h3>
<p><strong>Specific Learning Outcomes:</strong></p>
<p>By the end of this sub-strand, the learner should be able to:</p>
<ul class="notes-list">
<li>State and apply the laws of indices</li>
<li>Simplify expressions involving indices</li>
<li>Solve equations involving indices</li>
<li>Apply indices in real-life contexts</li>
</ul>
<hr class="section-divider">
<h4 class="notes-h4">1.2.1 Recap: What Is an Index?</h4>
<p>In the expression aⁿ:</p>
<ul class="notes-list">
<li><strong>a</strong> = base</li>
<li><strong>n</strong> = index (exponent or power)</li>
<li>aⁿ = a × a × a × ... (n times)</li>
</ul>
<p>Example: 2⁵ = 2 × 2 × 2 × 2 × 2 = 32</p>
<hr class="section-divider">
<h4 class="notes-h4">1.2.2 Laws of Indices</h4>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Law</th><th>Rule</th><th>Example</th></tr>
</thead><tbody>
<tr><td>Multiplication</td><td>aᵐ × aⁿ = aᵐ⁺ⁿ</td><td>3² × 3⁴ = 3⁶ = 729</td></tr>
<tr><td>Division</td><td>aᵐ ÷ aⁿ = aᵐ⁻ⁿ</td><td>5⁶ ÷ 5² = 5⁴ = 625</td></tr>
<tr><td>Power of power</td><td>(aᵐ)ⁿ = aᵐⁿ</td><td>(2³)² = 2⁶ = 64</td></tr>
<tr><td>Power of product</td><td>(ab)ⁿ = aⁿbⁿ</td><td>(2×3)³ = 2³×3³ = 216</td></tr>
<tr><td>Power of fraction</td><td>(a/b)ⁿ = aⁿ/bⁿ</td><td>(2/3)² = 4/9</td></tr>
<tr><td>Zero index</td><td>a⁰ = 1 (a≠0)</td><td>7⁰ = 1, (xy)⁰ = 1</td></tr>
<tr><td>Negative index</td><td>a⁻ⁿ = 1/aⁿ</td><td>2⁻³ = 1/8</td></tr>
<tr><td>Fractional index</td><td>a^(1/n) = ⁿ√a</td><td>16^(1/2) = 4, 27^(1/3) = 3</td></tr>
<tr><td>Fractional index</td><td>a^(m/n) = (ⁿ√a)ᵐ</td><td>8^(2/3) = (∛8)² = 4</td></tr>
</tbody></table></div>
<hr class="section-divider">
<h4 class="notes-h4">1.2.3 Worked Examples</h4>
<p><strong>Example 1:</strong> Simplify 2³ × 2⁵ ÷ 2⁴</p>
<pre class="code-block"><code>
= 2^(3+5-4) = 2⁴ = 16
</code></pre>
<p><strong>Example 2:</strong> Simplify (3x²y³)²</p>
<pre class="code-block"><code>
= 3² × x^(2×2) × y^(3×2) = 9x⁴y⁶
</code></pre>
<p><strong>Example 3:</strong> Evaluate 64^(2/3)</p>
<pre class="code-block"><code>
64^(2/3) = (∛64)² = 4² = 16
</code></pre>
<p><strong>Example 4:</strong> Simplify (x³y⁻²)/(x⁻¹y²)</p>
<pre class="code-block"><code>
= x^(3-(-1)) × y^(-2-2) = x⁴y⁻⁴ = x⁴/y⁴
</code></pre>
<p><strong>Example 5:</strong> Solve 2^x = 32</p>
<pre class="code-block"><code>
32 = 2⁵
Therefore 2^x = 2⁵
x = 5
</code></pre>
<p><strong>Example 6:</strong> Solve 3^(2x-1) = 27</p>
<pre class="code-block"><code>
27 = 3³
3^(2x-1) = 3³
2x - 1 = 3
2x = 4
x = 2
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">Review Questions — Sub-Strand 1.2</h4>
<ol class="notes-list">
<li>Simplify: (a) a⁴ × a⁻² (b) (x³)² ÷ x⁴ (c) (2a²b)³</li>
<li>Evaluate: (a) 4^(3/2) (b) 125^(1/3) (c) 32^(2/5)</li>
<li>Simplify: (2x³y⁻¹)² ÷ (4x⁻¹y²)</li>
<li>Solve: (a) 5^x = 125 (b) 2^(x+1) = 16 (c) 4^x = 8</li>
<li>Write in simplified form with positive indices: (3a⁻²b³)/(6a²b⁻¹)</li>
</ol>
<hr class="section-divider">
<h2 class="notes-h2">STRAND 2: ALGEBRA</h2>
<h3 class="notes-h3">Sub-Strand 2.1: Quadratic Expressions and Equations</h3>
<p><strong>Specific Learning Outcomes:</strong></p>
<p>By the end of this sub-strand, the learner should be able to:</p>
<ul class="notes-list">
<li>Expand quadratic expressions</li>
<li>Factorise quadratic expressions</li>
<li>Solve quadratic equations by factorisation, completing the square, and formula</li>
<li>Apply quadratic equations to real-life problems</li>
</ul>
<hr class="section-divider">
<h4 class="notes-h4">2.1.1 Expanding Quadratic Expressions</h4>
<p>A <strong>quadratic expression</strong> has the form <strong>ax² + bx + c</strong> where a ≠ 0.</p>
<p><strong>Expanding products of binomials (FOIL/Box method):</strong></p>
<p>(a + b)(c + d) = ac + ad + bc + bd</p>
<p><strong>Example:</strong> Expand (x + 3)(x + 5)</p>
<pre class="code-block"><code>
= x² + 5x + 3x + 15
= x² + 8x + 15
</code></pre>
<p><strong>Example:</strong> Expand (2x - 1)(x + 4)</p>
<pre class="code-block"><code>
= 2x² + 8x - x - 4
= 2x² + 7x - 4
</code></pre>
<p><strong>Special Products:</strong></p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Pattern</th><th>Formula</th><th>Example</th></tr>
</thead><tbody>
<tr><td>Perfect square (sum)</td><td>(a + b)² = a² + 2ab + b²</td><td>(x+3)² = x²+6x+9</td></tr>
<tr><td>Perfect square (difference)</td><td>(a - b)² = a² - 2ab + b²</td><td>(x-4)² = x²-8x+16</td></tr>
<tr><td>Difference of two squares</td><td>(a+b)(a-b) = a² - b²</td><td>(x+5)(x-5) = x²-25</td></tr>
</tbody></table></div>
<hr class="section-divider">
<h4 class="notes-h4">2.1.2 Factorising Quadratic Expressions</h4>
<p><strong>Method 1: Common factor first (always check this first)</strong></p>
<pre class="code-block"><code>
3x² + 6x = 3x(x + 2)
2x² - 8x = 2x(x - 4)
</code></pre>
<p><strong>Method 2: Simple trinomials ax² + bx + c where a = 1</strong></p>
<p>Find two numbers that <strong>multiply to give c</strong> and <strong>add to give b</strong>.</p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Expression</th><th>Find two numbers</th><th>Factors</th></tr>
</thead><tbody>
<tr><td>x² + 7x + 12</td><td>multiply: 12, add: 7 → 3 and 4</td><td>(x+3)(x+4)</td></tr>
<tr><td>x² + x - 12</td><td>multiply: -12, add: 1 → 4 and -3</td><td>(x+4)(x-3)</td></tr>
<tr><td>x² - 5x + 6</td><td>multiply: 6, add: -5 → -2 and -3</td><td>(x-2)(x-3)</td></tr>
<tr><td>x² - 9x - 10</td><td>multiply: -10, add: -9 → -10 and 1</td><td>(x-10)(x+1)</td></tr>
</tbody></table></div>
<p><strong>Method 3: Trinomials where a ≠ 1 (splitting the middle term)</strong></p>
<p>Factorise 2x² + 7x + 3</p>
<pre class="code-block"><code>
Step 1: Multiply a × c = 2 × 3 = 6
Step 2: Find two numbers that multiply to 6 and add to 7: 6 and 1
Step 3: Split 7x: 2x² + 6x + x + 3
Step 4: Group: 2x(x + 3) + 1(x + 3)
Step 5: Factor: (2x + 1)(x + 3)
</code></pre>
<p>Factorise 3x² - 10x - 8</p>
<pre class="code-block"><code>
a × c = 3 × (-8) = -24
Find: multiply -24, add -10 → -12 and 2
Split: 3x² - 12x + 2x - 8
Group: 3x(x - 4) + 2(x - 4)
Answer: (3x + 2)(x - 4)
</code></pre>
<p><strong>Method 4: Difference of two squares</strong></p>
<p>a² - b² = (a + b)(a - b)</p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Expression</th><th>Factors</th></tr>
</thead><tbody>
<tr><td>x² - 16</td><td>(x+4)(x-4)</td></tr>
<tr><td>4x² - 9</td><td>(2x+3)(2x-3)</td></tr>
<tr><td>25x² - 1</td><td>(5x+1)(5x-1)</td></tr>
<tr><td>x² - y²</td><td>(x+y)(x-y)</td></tr>
</tbody></table></div>
<p><strong>Method 5: Perfect square recognition</strong></p>
<p>a² + 2ab + b² = (a + b)²</p>
<p>a² - 2ab + b² = (a - b)²</p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Expression</th><th>Recognition</th><th>Factor</th></tr>
</thead><tbody>
<tr><td>x² + 6x + 9</td><td>(x)² + 2(x)(3) + (3)²</td><td>(x+3)²</td></tr>
<tr><td>x² - 10x + 25</td><td>(x)² - 2(x)(5) + (5)²</td><td>(x-5)²</td></tr>
<tr><td>4x² + 12x + 9</td><td>(2x)² + 2(2x)(3) + (3)²</td><td>(2x+3)²</td></tr>
</tbody></table></div>
<hr class="section-divider">
<h4 class="notes-h4">2.1.3 Solving Quadratic Equations</h4>
<p><strong>Method 1: Factorisation</strong></p>
<p>A quadratic equation is of the form ax² + bx + c = 0.</p>
<p>If (x + p)(x + q) = 0, then x = -p OR x = -q (Zero Product Rule).</p>
<p><strong>Example:</strong> Solve x² + 5x + 6 = 0</p>
<pre class="code-block"><code>
Factorise: (x + 2)(x + 3) = 0
x + 2 = 0 OR x + 3 = 0
x = -2 OR x = -3
</code></pre>
<p><strong>Example:</strong> Solve 2x² - 5x - 3 = 0</p>
<pre class="code-block"><code>
Factorise: (2x + 1)(x - 3) = 0
2x + 1 = 0 OR x - 3 = 0
x = -1/2 OR x = 3
</code></pre>
<p><strong>Method 2: Completing the Square</strong></p>
<p>Convert ax² + bx + c = 0 to (x + h)² = k form.</p>
<p><strong>Example:</strong> Solve x² + 6x + 2 = 0</p>
<pre class="code-block"><code>
x² + 6x = -2
x² + 6x + 9 = -2 + 9 (add (6/2)² = 9 to both sides)
(x + 3)² = 7
x + 3 = ±√7
x = -3 + √7 OR x = -3 - √7
x ≈ -0.354 OR x ≈ -5.646
</code></pre>
<p><strong>Method 3: Quadratic Formula</strong></p>
<p>For ax² + bx + c = 0:</p>
<p>$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$</p>
<p><strong>Example:</strong> Solve 2x² - 3x - 5 = 0</p>
<pre class="code-block"><code>
a = 2, b = -3, c = -5
x = [3 ± √(9 + 40)] / 4
x = [3 ± √49] / 4
x = [3 ± 7] / 4
x = 10/4 = 5/2 OR x = -4/4 = -1
</code></pre>
<p><strong>The Discriminant (b² - 4ac):</strong></p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Value</th><th>Meaning</th></tr>
</thead><tbody>
<tr><td>b² - 4ac > 0</td><td>Two distinct real roots</td></tr>
<tr><td>b² - 4ac = 0</td><td>One repeated real root</td></tr>
<tr><td>b² - 4ac < 0</td><td>No real roots</td></tr>
</tbody></table></div>
<hr class="section-divider">
<h4 class="notes-h4">2.1.4 Word Problems — Quadratic Equations</h4>
<p><strong>Example:</strong> The length of a rectangle is 5 cm more than its width. If the area is 84 cm², find the dimensions.</p>
<pre class="code-block"><code>
Let width = x cm
Length = (x + 5) cm
Area = x(x + 5) = 84
x² + 5x - 84 = 0
(x + 12)(x - 7) = 0
x = -12 (rejected — length cannot be negative) OR x = 7

Width = 7 cm, Length = 12 cm
</code></pre>
<p><strong>Example:</strong> Two numbers differ by 3. Their product is 88. Find the numbers.</p>
<pre class="code-block"><code>
Let one number = n, other = n + 3
n(n + 3) = 88
n² + 3n - 88 = 0
(n + 11)(n - 8) = 0
n = -11 OR n = 8

If n = 8: numbers are 8 and 11
If n = -11: numbers are -11 and -8
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">Review Questions — Sub-Strand 2.1</h4>
<ol class="notes-list">
<li>Expand: (a) (x - 3)² (b) (2x + 5)(x - 2) (c) (3x - 1)(3x + 1)</li>
<li>Factorise: (a) x² + 9x + 20 (b) x² - 7x + 10 (c) 3x² + 11x - 4</li>
<li>Factorise: (a) 9x² - 16 (b) x² - 8x + 16 (c) 6x² - x - 2</li>
<li>Solve by factorisation: (a) x² - x - 12 = 0 (b) 2x² + x - 6 = 0</li>
<li>Solve using the formula: 3x² - 5x - 2 = 0</li>
<li>Solve by completing the square: x² + 4x - 7 = 0</li>
<li>A square's side is increased by 3 cm, making area = 49 cm². Find the original side.</li>
</ol>
<hr class="section-divider">
<h3 class="notes-h3">Sub-Strand 2.2: Linear Inequalities</h3>
<p><strong>Specific Learning Outcomes:</strong></p>
<p>By the end of this sub-strand, the learner should be able to:</p>
<ul class="notes-list">
<li>Solve linear inequalities in one and two variables</li>
<li>Represent solutions on number lines and coordinate plane</li>
<li>Apply linear inequalities to real-life situations</li>
</ul>
<hr class="section-divider">
<h4 class="notes-h4">2.2.1 Review of Inequality Symbols</h4>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Symbol</th><th>Meaning</th><th>Example</th></tr>
</thead><tbody>
<tr><td>></td><td>Greater than</td><td>x > 3 means x is more than 3</td></tr>
<tr><td><</td><td>Less than</td><td>x < 5 means x is less than 5</td></tr>
<tr><td>≥</td><td>Greater than or equal to</td><td>x ≥ 2</td></tr>
<tr><td>≤</td><td>Less than or equal to</td><td>x ≤ 7</td></tr>
</tbody></table></div>
<p><strong>Key Rule:</strong> When you multiply or divide by a negative number, FLIP the inequality sign.</p>
<hr class="section-divider">
<h4 class="notes-h4">2.2.2 Solving Linear Inequalities</h4>
<p><strong>Example:</strong> Solve 3x - 2 > 7</p>
<pre class="code-block"><code>
3x &gt; 9
x &gt; 3
</code></pre>
<p>Number line: open circle at 3, arrow pointing right →</p>
<p><strong>Example:</strong> Solve -2x + 5 ≤ 1</p>
<pre class="code-block"><code>
-2x ≤ -4
x ≥ 2 (divide by -2 — FLIP the sign!)
</code></pre>
<p>Number line: closed circle at 2, arrow pointing right →</p>
<p><strong>Example:</strong> Solve -3 < 2x + 1 ≤ 7 (compound inequality)</p>
<pre class="code-block"><code>
Subtract 1 from all parts:
-4 &lt; 2x ≤ 6
Divide all by 2:
-2 &lt; x ≤ 3
</code></pre>
<p>Number line: open circle at -2, closed circle at 3, shaded between them.</p>
<hr class="section-divider">
<h4 class="notes-h4">2.2.3 Linear Inequalities in Two Variables</h4>
<p>The inequality y < 2x + 1 represents a region in the coordinate plane.</p>
<p><strong>Steps to draw inequality regions:</strong></p>
<ol class="notes-list">
<li>Draw the line y = 2x + 1 (dashed for < or >, solid for ≤ or ≥)</li>
<li>Test a point — use (0, 0) if not on the line</li>
<li>If (0, 0) satisfies the inequality, shade the side containing (0, 0)</li>
<li>If not, shade the other side</li>
</ol>
<p><strong>Example:</strong> Shade the region y ≤ x + 2</p>
<pre class="code-block"><code>
Step 1: Draw y = x + 2 as a SOLID line (≤ includes the line)
Step 2: Test (0, 0): 0 ≤ 0 + 2 → 0 ≤ 2 ✓ (TRUE)
Step 3: Shade the region containing (0, 0) — below and on the line
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">2.2.4 Simultaneous Inequalities (Linear Programming)</h4>
<p><strong>Example:</strong> Find the region satisfying: x ≥ 0, y ≥ 0, x + y ≤ 5, y ≤ x + 2</p>
<p>Draw all three boundary lines and shade the overlapping (feasible) region.</p>
<p><strong>Key vocabulary:</strong> The overlapping region is the <strong>feasible region</strong>.</p>
<hr class="section-divider">
<h4 class="notes-h4">Review Questions — Sub-Strand 2.2</h4>
<ol class="notes-list">
<li>Solve and represent on a number line: (a) 4x + 1 > 9 (b) -3x ≤ 12 (c) 5 - 2x > 1</li>
<li>Solve: -5 ≤ 3x + 1 < 10</li>
<li>Find all integers satisfying: 2 < 4 - x ≤ 6</li>
<li>Draw the region satisfying: y ≥ x - 1 and x + y ≤ 4 and x ≥ 0</li>
</ol>
<hr class="section-divider">
<h3 class="notes-h3">Sub-Strand 2.3: Simultaneous Equations</h3>
<p><strong>Specific Learning Outcomes:</strong></p>
<p>By the end of this sub-strand, the learner should be able to:</p>
<ul class="notes-list">
<li>Solve simultaneous linear equations by substitution and elimination</li>
<li>Solve simultaneous equations where one is linear and one is quadratic</li>
<li>Apply simultaneous equations to real-life problems</li>
</ul>
<hr class="section-divider">
<h4 class="notes-h4">2.3.1 Simultaneous Linear Equations — Review</h4>
<p><strong>Elimination Method:</strong></p>
<p>Solve: 2x + 3y = 12 ... (i) and 3x - y = 7 ... (ii)</p>
<pre class="code-block"><code>
Multiply (ii) by 3: 9x - 3y = 21 ... (iii)
Add (i) and (iii): 11x = 33 → x = 3
Substitute in (ii): 9 - y = 7 → y = 2
Solution: x = 3, y = 2
</code></pre>
<p><strong>Substitution Method:</strong></p>
<p>Solve: y = 2x - 1 ... (i) and 3x + 2y = 13 ... (ii)</p>
<pre class="code-block"><code>
Substitute (i) into (ii): 3x + 2(2x - 1) = 13
3x + 4x - 2 = 13
7x = 15 → x = 15/7
Wait — let's try a cleaner example:
</code></pre>
<p>Solve: y = 3x - 2 ... (i) and x + 2y = 8 ... (ii)</p>
<pre class="code-block"><code>
Substitute (i) into (ii): x + 2(3x - 2) = 8
x + 6x - 4 = 8
7x = 12 → x = 12/7
</code></pre>
<p>Better example: y = x + 1 ... (i) and 2x + y = 7 ... (ii)</p>
<pre class="code-block"><code>
Substitute (i) into (ii): 2x + (x + 1) = 7
3x + 1 = 7
3x = 6 → x = 2
y = 2 + 1 = 3
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">2.3.2 Simultaneous Equations: One Linear, One Quadratic</h4>
<p><strong>Example:</strong> Solve y = x + 2 and y = x²</p>
<pre class="code-block"><code>
Substitute: x² = x + 2
x² - x - 2 = 0
(x - 2)(x + 1) = 0
x = 2 OR x = -1

When x = 2: y = 2 + 2 = 4
When x = -1: y = -1 + 2 = 1

Solutions: (2, 4) and (-1, 1)
</code></pre>
<p><strong>Example:</strong> Solve x + y = 5 and x² + y² = 13</p>
<pre class="code-block"><code>
From the linear: y = 5 - x
Substitute into quadratic:
x² + (5 - x)² = 13
x² + 25 - 10x + x² = 13
2x² - 10x + 12 = 0
x² - 5x + 6 = 0
(x - 2)(x - 3) = 0
x = 2 OR x = 3

When x = 2: y = 3
When x = 3: y = 2

Solutions: (2, 3) and (3, 2)
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">Review Questions — Sub-Strand 2.3</h4>
<ol class="notes-list">
<li>Solve by elimination: 3x + 2y = 16 and 5x - 2y = 8</li>
<li>Solve by substitution: y = 2x + 3 and 3x + 2y = 12</li>
<li>Solve simultaneously: y = x + 4 and y = x²</li>
<li>Solve: x - y = 1 and x² + y² = 25</li>
<li>The sum of two numbers is 10 and their product is 24. Find the numbers.</li>
</ol>
<hr class="section-divider">
<h2 class="notes-h2">STRAND 3: GEOMETRY</h2>
<h3 class="notes-h3">Sub-Strand 3.1: Circles</h3>
<p><strong>Specific Learning Outcomes:</strong></p>
<p>By the end of this sub-strand, the learner should be able to:</p>
<ul class="notes-list">
<li>Identify parts of a circle</li>
<li>Apply angle properties of circles</li>
<li>Prove and use angle theorems related to circles</li>
<li>Solve problems involving circle properties</li>
</ul>
<hr class="section-divider">
<h4 class="notes-h4">3.1.1 Parts of a Circle</h4>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Part</th><th>Definition</th></tr>
</thead><tbody>
<tr><td>Centre (O)</td><td>The fixed point equidistant from all points on the circle</td></tr>
<tr><td>Radius (r)</td><td>Distance from centre to any point on the circle</td></tr>
<tr><td>Diameter (d)</td><td>Chord passing through the centre; d = 2r</td></tr>
<tr><td>Chord</td><td>Line segment joining two points on the circle</td></tr>
<tr><td>Arc</td><td>Part of the circumference</td></tr>
<tr><td>Sector</td><td>Region bounded by two radii and an arc</td></tr>
<tr><td>Segment</td><td>Region bounded by a chord and an arc</td></tr>
<tr><td>Tangent</td><td>Line that touches the circle at exactly one point</td></tr>
<tr><td>Secant</td><td>Line that intersects the circle at two points</td></tr>
<tr><td>Circumference</td><td>Total length around the circle = 2πr</td></tr>
</tbody></table></div>
<hr class="section-divider">
<h4 class="notes-h4">3.1.2 Angle Properties of Circles</h4>
<p><strong>Theorem 1: Angle at the Centre</strong></p>
<p>The angle subtended by an arc at the <strong>centre</strong> is <strong>twice</strong> the angle subtended at any point on the <strong>remaining circumference</strong>.</p>
<pre class="code-block"><code>
∠AOB = 2 × ∠ACB
(where O is centre, A and B are on circle, C is on major arc)
</code></pre>
<p><strong>Theorem 2: Angle in a Semicircle</strong></p>
<p>The angle in a <strong>semicircle</strong> (angle subtended by a diameter) is <strong>90°</strong>.</p>
<pre class="code-block"><code>
If AB is a diameter, then ∠ACB = 90° for any point C on the circle.
</code></pre>
<p><strong>Theorem 3: Angles in the Same Segment</strong></p>
<p>Angles subtended by the <strong>same arc</strong> in the same segment are <strong>equal</strong>.</p>
<pre class="code-block"><code>
∠ACB = ∠ADB (both subtended by arc AB, same side)
</code></pre>
<p><strong>Theorem 4: Opposite Angles in a Cyclic Quadrilateral</strong></p>
<p>Opposite angles in a <strong>cyclic quadrilateral</strong> (all four vertices on the circle) are <strong>supplementary</strong> (sum to 180°).</p>
<pre class="code-block"><code>
∠A + ∠C = 180°
∠B + ∠D = 180°
</code></pre>
<p><strong>Theorem 5: Tangent-Radius Property</strong></p>
<p>A <strong>tangent</strong> to a circle is <strong>perpendicular</strong> to the radius at the point of contact.</p>
<pre class="code-block"><code>
If PT is a tangent at T, and OT is a radius, then ∠OTP = 90°
</code></pre>
<p><strong>Theorem 6: Tangents from an External Point</strong></p>
<p>Tangents drawn from an <strong>external point</strong> to a circle are <strong>equal in length</strong>.</p>
<pre class="code-block"><code>
If PA and PB are tangents from P, then PA = PB
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">3.1.3 Worked Examples</h4>
<p><strong>Example 1:</strong> O is the centre. ∠ACB = 35°. Find ∠AOB.</p>
<pre class="code-block"><code>
∠AOB = 2 × ∠ACB = 2 × 35° = 70°
</code></pre>
<p><strong>Example 2:</strong> In cyclic quadrilateral ABCD, ∠A = 85°, ∠B = 95°. Find ∠C and ∠D.</p>
<pre class="code-block"><code>
∠A + ∠C = 180° → ∠C = 95°
∠B + ∠D = 180° → ∠D = 85°
</code></pre>
<p><strong>Example 3:</strong> AB is a diameter. ∠BAC = 40°. Find ∠ABC.</p>
<pre class="code-block"><code>
∠ACB = 90° (angle in semicircle)
∠BAC + ∠ABC + ∠ACB = 180°
40° + ∠ABC + 90° = 180°
∠ABC = 50°
</code></pre>
<p><strong>Example 4:</strong> PA and PB are tangents from P. PA = 8 cm. Find PB.</p>
<pre class="code-block"><code>
PB = PA = 8 cm (tangents from external point are equal)
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">Review Questions — Sub-Strand 3.1</h4>
<ol class="notes-list">
<li>O is the centre. ∠AOB = 110°. Find the angle in the minor segment.</li>
<li>AB is a diameter. ∠DAB = 36°. ABCD is a cyclic quadrilateral. Find all other angles.</li>
<li>O is the centre. ∠OAB = 25°. Find ∠AOB and the angle subtended at the circumference.</li>
<li>Two tangents from P touch a circle of radius 5 cm at A and B. OP = 13 cm. Find PA.</li>
<li>In a cyclic quadrilateral, three angles are 75°, 110°, and 85°. Find the fourth.</li>
</ol>
<hr class="section-divider">
<h3 class="notes-h3">Sub-Strand 3.2: Transformations</h3>
<p><strong>Specific Learning Outcomes:</strong></p>
<p>By the end of this sub-strand, the learner should be able to:</p>
<ul class="notes-list">
<li>Perform and describe reflections, rotations, translations, and enlargements</li>
<li>Identify the properties preserved under each transformation</li>
<li>Combine transformations</li>
<li>Apply transformations in real-life contexts</li>
</ul>
<hr class="section-divider">
<h4 class="notes-h4">3.2.1 Recap of Transformations</h4>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Transformation</th><th>What Changes</th><th>What Stays the Same</th></tr>
</thead><tbody>
<tr><td>Translation</td><td>Position</td><td>Shape, size, orientation</td></tr>
<tr><td>Reflection</td><td>Position, orientation</td><td>Shape, size</td></tr>
<tr><td>Rotation</td><td>Position, orientation</td><td>Shape, size</td></tr>
<tr><td>Enlargement</td><td>Size (and position)</td><td>Shape, angles</td></tr>
</tbody></table></div>
<hr class="section-divider">
<h4 class="notes-h4">3.2.2 Reflection</h4>
<p><strong>Common Mirror Lines and Their Equations:</strong></p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Mirror Line</th><th>Equation</th><th>Rule</th></tr>
</thead><tbody>
<tr><td>x-axis</td><td>y = 0</td><td>(x, y) → (x, -y)</td></tr>
<tr><td>y-axis</td><td>x = 0</td><td>(x, y) → (-x, y)</td></tr>
<tr><td>Line y = x</td><td>y = x</td><td>(x, y) → (y, x)</td></tr>
<tr><td>Line y = -x</td><td>y = -x</td><td>(x, y) → (-y, -x)</td></tr>
<tr><td>Line x = a</td><td>x = a</td><td>(x, y) → (2a-x, y)</td></tr>
<tr><td>Line y = b</td><td>y = b</td><td>(x, y) → (x, 2b-y)</td></tr>
</tbody></table></div>
<p><strong>Example:</strong> Reflect point (3, 2) in the line y = x.</p>
<pre class="code-block"><code>
(3, 2) → (2, 3)
</code></pre>
<p><strong>Example:</strong> Reflect triangle A(1,2), B(3,2), C(2,4) in the y-axis.</p>
<pre class="code-block"><code>
A(1,2) → A'(-1, 2)
B(3,2) → B'(-3, 2)
C(2,4) → C'(-2, 4)
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">3.2.3 Rotation</h4>
<p>A rotation requires: <strong>centre</strong>, <strong>angle</strong>, and <strong>direction</strong> (clockwise or anticlockwise).</p>
<p><strong>Rotation Rules about the Origin (0, 0):</strong></p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Rotation</th><th>Rule</th></tr>
</thead><tbody>
<tr><td>90° anticlockwise</td><td>(x, y) → (-y, x)</td></tr>
<tr><td>90° clockwise</td><td>(x, y) → (y, -x)</td></tr>
<tr><td>180°</td><td>(x, y) → (-x, -y)</td></tr>
<tr><td>270° anticlockwise</td><td>(x, y) → (y, -x) [same as 90° CW]</td></tr>
</tbody></table></div>
<p><strong>Example:</strong> Rotate A(2, 3) by 90° anticlockwise about origin.</p>
<pre class="code-block"><code>
(2, 3) → (-3, 2)
</code></pre>
<p><strong>Example:</strong> Rotate P(4, 1) by 180° about origin.</p>
<pre class="code-block"><code>
(4, 1) → (-4, -1)
</code></pre>
<p><strong>Finding the Centre of Rotation:</strong></p>
<ul class="notes-list">
<li>Draw perpendicular bisectors of corresponding sides</li>
<li>The intersection is the centre of rotation</li>
</ul>
<hr class="section-divider">
<h4 class="notes-h4">3.2.4 Translation</h4>
<p>A translation moves every point by the same vector <strong>v = (a, b)</strong>.</p>
<p>Rule: (x, y) → (x + a, y + b)</p>
<p><strong>Example:</strong> Translate A(2, 3), B(4, 3), C(3, 5) by vector (−1, 2).</p>
<pre class="code-block"><code>
A(2,3) → A'(2-1, 3+2) = A'(1, 5)
B(4,3) → B'(3, 5)
C(3,5) → C'(2, 7)
</code></pre>
<p><strong>Finding the Translation Vector:</strong></p>
<p>If A(2, 1) maps to A'(5, 4):</p>
<p>Vector = (5-2, 4-1) = (3, 3)</p>
<hr class="section-divider">
<h4 class="notes-h4">3.2.5 Enlargement</h4>
<p>An enlargement requires: <strong>centre</strong> and <strong>scale factor (k)</strong>.</p>
<p>Rule from centre O(0,0): (x, y) → (kx, ky)</p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Scale Factor</th><th>Effect</th></tr>
</thead><tbody>
<tr><td>k > 1</td><td>Enlargement (image larger than object)</td></tr>
<tr><td>0 < k < 1</td><td>Reduction (image smaller)</td></tr>
<tr><td>k = 1</td><td>No change</td></tr>
<tr><td>k < 0</td><td>Enlargement on opposite side of centre</td></tr>
</tbody></table></div>
<p><strong>Example:</strong> Enlarge A(2, 1) by scale factor 3 from origin.</p>
<pre class="code-block"><code>
A(2,1) → A'(6, 3)
</code></pre>
<p><strong>Example:</strong> Enlarge triangle A(2,1), B(4,1), C(3,3) by scale factor 2, centre (1,1).</p>
<pre class="code-block"><code>
For centre (a, b), scale factor k:
New point = (a + k(x-a), b + k(y-b))

A(2,1): (1 + 2(2-1), 1 + 2(1-1)) = (3, 1)
B(4,1): (1 + 2(4-1), 1 + 2(1-1)) = (7, 1)
C(3,3): (1 + 2(3-1), 1 + 2(3-1)) = (5, 5)
</code></pre>
<p><strong>Area Scale Factor:</strong> If linear scale factor = k, then area scale factor = k²</p>
<hr class="section-divider">
<h4 class="notes-h4">3.2.6 Combined Transformations</h4>
<p><strong>Example:</strong> Triangle ABC is reflected in the x-axis, then rotated 90° anticlockwise about the origin. Describe the single transformation that maps ABC to its final image.</p>
<p>Work step by step:</p>
<ol class="notes-list">
<li>Apply first transformation</li>
<li>Apply second transformation to the result</li>
<li>Identify the single equivalent transformation</li>
</ol>
<hr class="section-divider">
<h4 class="notes-h4">Review Questions — Sub-Strand 3.2</h4>
<ol class="notes-list">
<li>Reflect (3, -2) in: (a) x-axis (b) y-axis (c) line y = x</li>
<li>Rotate A(1,3), B(3,3), C(2,5) by 90° clockwise about origin.</li>
<li>Translate P(2,4), Q(5,4), R(3,7) by vector (-3, -1).</li>
<li>Enlarge A(1,2), B(3,2), C(2,4) by scale factor 2 from origin. What is the area of the image if area of object = 2 cm²?</li>
<li>A(3,1) is mapped to A'(9,3) by enlargement. O is origin. Find the scale factor.</li>
</ol>
<hr class="section-divider">
<h2 class="notes-h2">STRAND 4: MEASUREMENTS</h2>
<h3 class="notes-h3">Sub-Strand 4.1: Surface Area and Volume</h3>
<p><strong>Specific Learning Outcomes:</strong></p>
<p>By the end of this sub-strand, the learner should be able to:</p>
<ul class="notes-list">
<li>Calculate surface area of prisms, cylinders, pyramids, cones, and spheres</li>
<li>Calculate volume of prisms, cylinders, pyramids, cones, and spheres</li>
<li>Apply area and volume formulas in real-life contexts</li>
<li>Solve problems involving density, mass, and volume</li>
</ul>
<hr class="section-divider">
<h4 class="notes-h4">4.1.1 Prisms</h4>
<p>A <strong>prism</strong> has a uniform cross-section throughout its length.</p>
<p><strong>Volume of a prism = Area of cross-section × length</strong></p>
<p><strong>Surface Area of a prism = 2 × Area of base + Lateral surface area</strong></p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Prism Type</th><th>Volume</th><th>Notes</th></tr>
</thead><tbody>
<tr><td>Rectangular prism (cuboid)</td><td>l × w × h</td><td>Faces: 2lw + 2lh + 2wh</td></tr>
<tr><td>Triangular prism</td><td>½bh × l</td><td>Two triangle faces + 3 rectangles</td></tr>
<tr><td>Trapezoidal prism</td><td>½(a+b)h × l</td><td>Two trapezium faces + 4 rectangles</td></tr>
</tbody></table></div>
<p><strong>Example:</strong> Find the volume of a triangular prism with base 6 cm, height 4 cm, length 10 cm.</p>
<pre class="code-block"><code>
Volume = ½ × 6 × 4 × 10 = 120 cm³
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">4.1.2 Cylinders</h4>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Measurement</th><th>Formula</th></tr>
</thead><tbody>
<tr><td>Curved surface area</td><td>2πrh</td></tr>
<tr><td>Area of each circular end</td><td>πr²</td></tr>
<tr><td>Total surface area</td><td>2πrh + 2πr² = 2πr(h + r)</td></tr>
<tr><td>Volume</td><td>πr²h</td></tr>
</tbody></table></div>
<p><strong>Example:</strong> A cylinder has radius 7 cm and height 10 cm. Find its volume and total surface area. (π = 22/7)</p>
<pre class="code-block"><code>
Volume = πr²h = 22/7 × 7² × 10 = 22/7 × 49 × 10 = 1540 cm³

Total SA = 2πr(h + r) = 2 × 22/7 × 7 × (10 + 7) = 44 × 17 = 748 cm²
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">4.1.3 Pyramids</h4>
<p>A <strong>pyramid</strong> has a polygonal base and triangular faces meeting at an apex.</p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Measurement</th><th>Formula</th></tr>
</thead><tbody>
<tr><td>Volume</td><td>⅓ × base area × height</td></tr>
<tr><td>Lateral surface area</td><td>½ × perimeter × slant height</td></tr>
<tr><td>Total surface area</td><td>Base area + Lateral SA</td></tr>
</tbody></table></div>
<p><strong>Example:</strong> Square pyramid, base 8 cm × 8 cm, height 6 cm. Find volume.</p>
<pre class="code-block"><code>
Volume = ⅓ × 8 × 8 × 6 = ⅓ × 384 = 128 cm³
</code></pre>
<p><strong>Finding slant height</strong> from height (h) and base half-length (a):</p>
<p>Slant height l = √(h² + a²)</p>
<pre class="code-block"><code>
l = √(6² + 4²) = √(36 + 16) = √52 = 2√13 cm
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">4.1.4 Cones</h4>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Measurement</th><th>Formula</th></tr>
</thead><tbody>
<tr><td>Slant height</td><td>l = √(r² + h²)</td></tr>
<tr><td>Curved surface area</td><td>πrl</td></tr>
<tr><td>Total surface area</td><td>πrl + πr² = πr(l + r)</td></tr>
<tr><td>Volume</td><td>⅓πr²h</td></tr>
</tbody></table></div>
<p><strong>Example:</strong> A cone has radius 5 cm and height 12 cm. Find all measurements.</p>
<pre class="code-block"><code>
Slant height: l = √(5² + 12²) = √(25 + 144) = √169 = 13 cm

Curved SA = πrl = π × 5 × 13 = 65π ≈ 204.2 cm²
Total SA = πr(l + r) = π × 5 × (13 + 5) = 90π ≈ 282.7 cm²
Volume = ⅓πr²h = ⅓ × π × 25 × 12 = 100π ≈ 314.2 cm³
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">4.1.5 Spheres</h4>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Measurement</th><th>Formula</th></tr>
</thead><tbody>
<tr><td>Surface area</td><td>4πr²</td></tr>
<tr><td>Volume</td><td>(4/3)πr³</td></tr>
</tbody></table></div>
<p><strong>Example:</strong> Find the surface area and volume of a sphere with diameter 14 cm.</p>
<pre class="code-block"><code>
r = 7 cm
Surface area = 4πr² = 4 × 22/7 × 49 = 616 cm²
Volume = (4/3)πr³ = 4/3 × 22/7 × 343 = 1437.3 cm³
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">4.1.6 Density, Mass, and Volume</h4>
<p><strong>Formula triangle:</strong></p>
<pre class="code-block"><code>
MASS
-------
D × V
</code></pre>
<p>Mass = Density × Volume</p>
<p>Density = Mass ÷ Volume</p>
<p>Volume = Mass ÷ Density</p>
<p>Units: density in g/cm³ or kg/m³</p>
<p><strong>Example:</strong> A metal cylinder has radius 3 cm, height 8 cm, and density 8.5 g/cm³. Find its mass.</p>
<pre class="code-block"><code>
Volume = π × 3² × 8 = 72π = 226.2 cm³
Mass = Density × Volume = 8.5 × 226.2 = 1922.7 g ≈ 1.92 kg
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">Review Questions — Sub-Strand 4.1</h4>
<ol class="notes-list">
<li>A cuboid is 8 cm × 5 cm × 3 cm. Find its volume and total surface area.</li>
<li>A cylinder has diameter 10 cm and height 15 cm. Find its volume (π = 3.14).</li>
<li>A square pyramid has base side 6 cm and height 4 cm. Find volume and total SA.</li>
<li>A cone has radius 6 cm and slant height 10 cm. Find its total SA and volume.</li>
<li>A sphere has volume 288π cm³. Find its radius.</li>
<li>A brass sphere of radius 5 cm has density 8.5 g/cm³. Find its mass.</li>
</ol>
<hr class="section-divider">
<h2 class="notes-h2">STRAND 5: STATISTICS AND PROBABILITY</h2>
<h3 class="notes-h3">Sub-Strand 5.1: Statistics — Grouped Data</h3>
<p><strong>Specific Learning Outcomes:</strong></p>
<p>By the end of this sub-strand, the learner should be able to:</p>
<ul class="notes-list">
<li>Organise data into grouped frequency tables</li>
<li>Calculate mean, median, mode, and range from grouped data</li>
<li>Draw and interpret histograms and frequency polygons</li>
<li>Interpret cumulative frequency curves (ogive)</li>
</ul>
<hr class="section-divider">
<h4 class="notes-h4">5.1.1 Grouped Frequency Distribution</h4>
<p>When data is large, we group it into class intervals.</p>
<p><strong>Example data:</strong> Ages of 30 people measured in years.</p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Class Interval</th><th>Frequency (f)</th><th>Midpoint (x)</th><th>fx</th></tr>
</thead><tbody>
<tr><td>10 – 14</td><td>3</td><td>12</td><td>36</td></tr>
<tr><td>15 – 19</td><td>7</td><td>17</td><td>119</td></tr>
<tr><td>20 – 24</td><td>10</td><td>22</td><td>220</td></tr>
<tr><td>25 – 29</td><td>6</td><td>27</td><td>162</td></tr>
<tr><td>30 – 34</td><td>4</td><td>32</td><td>128</td></tr>
<tr><td><strong>Total</strong></td><td><strong>30</strong></td><td></td><td><strong>665</strong></td></tr>
</tbody></table></div>
<p><strong>Mean from grouped data:</strong> x̄ = Σfx / Σf = 665/30 = 22.2 years</p>
<hr class="section-divider">
<h4 class="notes-h4">5.1.2 Median from Grouped Data</h4>
<p>Median class = class where the cumulative frequency first exceeds n/2.</p>
<p>For n = 30: n/2 = 15</p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Class</th><th>f</th><th>Cumulative Frequency</th></tr>
</thead><tbody>
<tr><td>10–14</td><td>3</td><td>3</td></tr>
<tr><td>15–19</td><td>7</td><td>10</td></tr>
<tr><td><strong>20–24</strong></td><td><strong>10</strong></td><td><strong>20</strong> ← passes 15 here</td></tr>
<tr><td>25–29</td><td>6</td><td>26</td></tr>
<tr><td>30–34</td><td>4</td><td>30</td></tr>
</tbody></table></div>
<p>Median class = 20–24</p>
<p><strong>Median formula (interpolation):</strong></p>
<pre class="code-block"><code>
Median = L + [(n/2 - F)/f] × w

Where:
L = lower boundary of median class = 19.5
n = total frequency = 30
F = cumulative frequency before median class = 10
f = frequency of median class = 10
w = class width = 5

Median = 19.5 + [(15 - 10)/10] × 5 = 19.5 + 2.5 = 22.0
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">5.1.3 Modal Class and Mode</h4>
<p><strong>Modal class</strong> = class with the highest frequency = 20–24</p>
<p><strong>Mode estimate (interpolation):</strong></p>
<pre class="code-block"><code>
Mode = L + [d₁/(d₁ + d₂)] × w

Where:
L = 19.5 (lower boundary)
d₁ = f_modal - f_before = 10 - 7 = 3
d₂ = f_modal - f_after = 10 - 6 = 4
w = 5

Mode = 19.5 + [3/(3+4)] × 5 = 19.5 + 2.14 = 21.64
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">5.1.4 Histograms and Frequency Polygons</h4>
<p><strong>Histogram:</strong></p>
<ul class="notes-list">
<li>x-axis: class intervals (continuous data — no gaps between bars)</li>
<li>y-axis: frequency density = frequency ÷ class width</li>
<li>Bars are adjacent (touching)</li>
</ul>
<p><strong>Frequency Polygon:</strong></p>
<ul class="notes-list">
<li>Plot the midpoint of each class against frequency</li>
<li>Join the points with straight lines</li>
<li>Extend to the midpoints of classes beyond the data (frequency = 0)</li>
</ul>
<hr class="section-divider">
<h4 class="notes-h4">5.1.5 Cumulative Frequency and Ogive</h4>
<p><strong>Cumulative frequency</strong> = running total of frequencies.</p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Class</th><th>f</th><th>Cumulative f</th></tr>
</thead><tbody>
<tr><td>10–14</td><td>3</td><td>3</td></tr>
<tr><td>15–19</td><td>7</td><td>10</td></tr>
<tr><td>20–24</td><td>10</td><td>20</td></tr>
<tr><td>25–29</td><td>6</td><td>26</td></tr>
<tr><td>30–34</td><td>4</td><td>30</td></tr>
</tbody></table></div>
<p><strong>Ogive (Cumulative Frequency Curve):</strong></p>
<ul class="notes-list">
<li>Plot cumulative frequency against UPPER class boundary</li>
<li>Join with a smooth curve (S-shaped)</li>
</ul>
<p><strong>Reading from the Ogive:</strong></p>
<ul class="notes-list">
<li><strong>Median</strong> = value at n/2 = 15th position</li>
<li><strong>Lower Quartile (Q₁)</strong> = value at n/4 = 7.5th position</li>
<li><strong>Upper Quartile (Q₃)</strong> = value at 3n/4 = 22.5th position</li>
<li><strong>Interquartile Range (IQR)</strong> = Q₃ - Q₁</li>
<li><strong>Percentile</strong> = value at p/100 × n</li>
</ul>
<hr class="section-divider">
<h4 class="notes-h4">Review Questions — Sub-Strand 5.1</h4>
<ol class="notes-list">
<li>The data below shows marks of 25 students:</li>
</ol>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Marks</th><th>10–19</th><th>20–29</th><th>30–39</th><th>40–49</th><th>50–59</th></tr>
</thead><tbody>
<tr><td>Freq</td><td>3</td><td>6</td><td>9</td><td>5</td><td>2</td></tr>
</tbody></table></div>
<p>(a) Calculate the estimated mean.</p>
<p>(b) Find the median class and estimate the median.</p>
<p>(c) Estimate the mode.</p>
<p>(d) Draw a histogram.</p>
<ol class="notes-list">
<li>From a cumulative frequency diagram with n = 80:</li>
</ol>
<p>Q₁ is at the 20th value = 25, Q₃ is at the 60th value = 45. Find the IQR.</p>
<hr class="section-divider">
<h3 class="notes-h3">Sub-Strand 5.2: Probability</h3>
<p><strong>Specific Learning Outcomes:</strong></p>
<p>By the end of this sub-strand, the learner should be able to:</p>
<ul class="notes-list">
<li>Calculate probability of single and combined events</li>
<li>Use tree diagrams and sample spaces</li>
<li>Apply the addition and multiplication rules of probability</li>
<li>Distinguish between independent and mutually exclusive events</li>
</ul>
<hr class="section-divider">
<h4 class="notes-h4">5.2.1 Basic Probability</h4>
<p><strong>Probability = Number of favourable outcomes / Total number of possible outcomes</strong></p>
<p>P(event) = n(E) / n(S)</p>
<p>Key facts:</p>
<ul class="notes-list">
<li>0 ≤ P(E) ≤ 1</li>
<li>P(impossible event) = 0</li>
<li>P(certain event) = 1</li>
<li>P(E') = 1 - P(E) where E' is the complement of E</li>
</ul>
<hr class="section-divider">
<h4 class="notes-h4">5.2.2 Mutually Exclusive Events</h4>
<p>Events A and B are <strong>mutually exclusive</strong> if they cannot both happen at the same time.</p>
<p><strong>Addition Rule:</strong> P(A or B) = P(A) + P(B)</p>
<p><strong>Example:</strong> A bag has 5 red, 3 blue, 2 green balls. Find P(red or green).</p>
<pre class="code-block"><code>
P(red) = 5/10 = 1/2
P(green) = 2/10 = 1/5
P(red or green) = 1/2 + 1/5 = 5/10 + 2/10 = 7/10
</code></pre>
<p><strong>Non-Mutually Exclusive Events:</strong></p>
<p><strong>Addition Rule:</strong> P(A or B) = P(A) + P(B) - P(A and B)</p>
<p><strong>Example:</strong> A card is drawn from a standard deck. Find P(king or heart).</p>
<pre class="code-block"><code>
P(king) = 4/52
P(heart) = 13/52
P(king AND heart) = 1/52 (king of hearts)
P(king or heart) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">5.2.3 Independent Events</h4>
<p>Events A and B are <strong>independent</strong> if one does not affect the other.</p>
<p><strong>Multiplication Rule:</strong> P(A and B) = P(A) × P(B)</p>
<p><strong>Example:</strong> A coin is tossed and a die is rolled. Find P(head and 4).</p>
<pre class="code-block"><code>
P(head) = 1/2, P(4) = 1/6
P(head and 4) = 1/2 × 1/6 = 1/12
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">5.2.4 Tree Diagrams</h4>
<p>Tree diagrams show all possible outcomes and their probabilities.</p>
<p><strong>Example:</strong> A bag has 3 red and 2 blue balls. A ball is drawn and replaced, then another is drawn.</p>
<pre class="code-block"><code>
R (3/5)
R (3/5) &lt;
B (2/5)

R (3/5)
B (2/5) &lt;
B (2/5)
</code></pre>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Outcome</th><th>Probability</th></tr>
</thead><tbody>
<tr><td>RR</td><td>3/5 × 3/5 = 9/25</td></tr>
<tr><td>RB</td><td>3/5 × 2/5 = 6/25</td></tr>
<tr><td>BR</td><td>2/5 × 3/5 = 6/25</td></tr>
<tr><td>BB</td><td>2/5 × 2/5 = 4/25</td></tr>
<tr><td>Total</td><td>25/25 = 1 ✓</td></tr>
</tbody></table></div>
<p><strong>P(both same colour) = P(RR) + P(BB) = 9/25 + 4/25 = 13/25</strong></p>
<hr class="section-divider">
<h4 class="notes-h4">5.2.5 Probability Without Replacement</h4>
<p>When items are drawn WITHOUT replacement, probabilities change after each draw.</p>
<p><strong>Example:</strong> A bag has 3 red and 2 blue balls. Two are drawn WITHOUT replacement. Find P(both red).</p>
<pre class="code-block"><code>
P(1st red) = 3/5
P(2nd red | 1st was red) = 2/4 = 1/2 (only 4 balls remain, 2 red)
P(both red) = 3/5 × 2/4 = 6/20 = 3/10
</code></pre>
<hr class="section-divider">
<h4 class="notes-h4">Review Questions — Sub-Strand 5.2</h4>
<ol class="notes-list">
<li>A bag has 6 white and 4 black balls. One ball is drawn. Find:</li>
</ol>
<p>(a) P(white) (b) P(not white) (c) P(red)</p>
<ol class="notes-list">
<li>A card is drawn from a deck. Find:</li>
</ol>
<p>(a) P(ace) (b) P(red card) (c) P(ace or king) (d) P(red ace)</p>
<ol class="notes-list">
<li>Two dice are rolled. Find P(sum = 7).</li>
</ol>
<ol class="notes-list">
<li>A bag has 5 red and 3 green balls. Two are drawn with replacement.</li>
</ol>
<p>Draw a tree diagram and find: (a) P(both red) (b) P(one of each colour)</p>
<ol class="notes-list">
<li>Repeat Q4 but without replacement.</li>
</ol>
<hr class="section-divider">
<h2 class="notes-h2">TERM 1 EXAMINATION — GRADE 9 MATHEMATICS</h2>
<p><strong>Time: 2½ Hours | Total Marks: 100 | Instructions: Answer ALL questions</strong></p>
<hr class="section-divider">
<h3 class="notes-h3">SECTION A: SHORT ANSWER (30 marks)</h3>
<p><strong>Answer all questions. Show your working clearly.</strong></p>
<ol class="notes-list">
<li>(a) Classify each as rational or irrational: 0.6̄, √11, 3/7, √49. <strong>(2 marks)</strong></li>
</ol>
<p>(b) Convert 0.1̄8̄ (0.181818...) to a fraction in simplest form. <strong>(2 marks)</strong></p>
<ol class="notes-list">
<li>(a) Simplify: √72 + √32 - √8 <strong>(3 marks)</strong></li>
</ol>
<p>(b) Rationalise the denominator: 4/√3 <strong>(2 marks)</strong></p>
<ol class="notes-list">
<li>(a) Evaluate: 27^(2/3) <strong>(2 marks)</strong></li>
</ol>
<p>(b) Simplify: (2x²y)³ ÷ (4xy²) leaving your answer with positive indices. <strong>(3 marks)</strong></p>
<ol class="notes-list">
<li>Solve the equation 4^(x-1) = 32. <strong>(3 marks)</strong></li>
</ol>
<ol class="notes-list">
<li>(a) Factorise completely: 4x² - 9 <strong>(2 marks)</strong></li>
</ol>
<p>(b) Factorise: 3x² - x - 4 <strong>(3 marks)</strong></p>
<ol class="notes-list">
<li>Solve: 2x² - 7x + 3 = 0 by factorisation. <strong>(3 marks)</strong></li>
</ol>
<ol class="notes-list">
<li>Find the value of x in each:</li>
</ol>
<p>(a) O is the centre of a circle. ∠AOB = 124°. Find ∠ACB (C on major arc). <strong>(2 marks)</strong></p>
<p>(b) PQRS is a cyclic quadrilateral. ∠P = 73°. Find ∠R. <strong>(1 marks)</strong></p>
<ol class="notes-list">
<li>A cylinder has radius 7 cm and height 5 cm. Find its volume (π = 22/7). <strong>(2 marks)</strong></li>
</ol>
<hr class="section-divider">
<h3 class="notes-h3">SECTION B: STRUCTURED QUESTIONS (40 marks)</h3>
<p><strong>Answer all questions. Show full working.</strong></p>
<p><strong>Question 9</strong> <em>(10 marks)</em></p>
<p>(a) Expand and simplify: (2x + 3)² - (x - 1)² <strong>(4 marks)</strong></p>
<p>(b) Solve the equation x² + 3x - 10 = 0 by:</p>
<p>(i) Factorisation <strong>(2 marks)</strong></p>
<p>(ii) Quadratic formula <strong>(3 marks)</strong></p>
<p>(iii) State the discriminant and describe what it tells us. <strong>(1 mark)</strong></p>
<p><strong>Question 10</strong> <em>(10 marks)</em></p>
<p>(a) The length of a rectangle exceeds its width by 4 cm. Given that its area is 45 cm²:</p>
<p>(i) Form a quadratic equation in terms of the width, w. <strong>(2 marks)</strong></p>
<p>(ii) Solve the equation to find the dimensions of the rectangle. <strong>(3 marks)</strong></p>
<p>(b) Solve simultaneously: y = x + 3 and x² + y² = 29</p>
<p>Give exact solutions. <strong>(5 marks)</strong></p>
<p><strong>Question 11</strong> <em>(10 marks)</em></p>
<p>The marks scored by 40 students in a test were recorded as follows:</p>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Marks</th><th>11–20</th><th>21–30</th><th>31–40</th><th>41–50</th><th>51–60</th><th>61–70</th></tr>
</thead><tbody>
<tr><td>Freq</td><td>2</td><td>5</td><td>12</td><td>11</td><td>7</td><td>3</td></tr>
</tbody></table></div>
<p>(a) State the modal class. <strong>(1 mark)</strong></p>
<p>(b) Calculate an estimate of the mean mark. <strong>(4 marks)</strong></p>
<p>(c) Draw up a cumulative frequency table. <strong>(2 marks)</strong></p>
<p>(d) From your table, estimate the median. <strong>(2 marks)</strong></p>
<p>(e) Estimate the lower quartile (Q₁). <strong>(1 mark)</strong></p>
<p><strong>Question 12</strong> <em>(10 marks)</em></p>
<p>(a) A bag contains 4 yellow, 3 green, and 2 white balls. One ball is drawn at random. Find:</p>
<p>(i) P(yellow) <strong>(1 mark)</strong></p>
<p>(ii) P(not green) <strong>(1 mark)</strong></p>
<p>(iii) P(blue) <strong>(1 mark)</strong></p>
<p>(b) Two balls are drawn from the bag WITHOUT replacement. Draw a tree diagram showing only yellow and not-yellow outcomes, and find the probability that:</p>
<p>(i) Both balls are yellow <strong>(2 marks)</strong></p>
<p>(ii) Exactly one ball is yellow <strong>(2 marks)</strong></p>
<p>(iii) At least one ball is yellow <strong>(3 marks)</strong></p>
<hr class="section-divider">
<h3 class="notes-h3">SECTION C: EXTENDED RESPONSE (30 marks)</h3>
<p><strong>Answer all questions. Show full working and reasoning.</strong></p>
<p><strong>Question 13</strong> <em>(15 marks)</em></p>
<p>(a) <strong>(Transformations)</strong> Triangle PQR has vertices P(1, 2), Q(3, 2), R(2, 4).</p>
<p>(i) Reflect PQR in the y-axis to give P'Q'R'. State the coordinates. <strong>(2 marks)</strong></p>
<p>(ii) Rotate PQR by 90° anticlockwise about the origin to give P''Q''R''. State the coordinates. <strong>(3 marks)</strong></p>
<p>(iii) Enlarge PQR by scale factor 3 from the origin. State the new coordinates. <strong>(2 marks)</strong></p>
<p>(iv) If the area of triangle PQR is 2 cm², find the area of the enlarged triangle. <strong>(1 mark)</strong></p>
<p>(b) <strong>(Circles)</strong> In the diagram, O is the centre of a circle. A, B, C, D are points on the circle. Angle AOC = 112°.</p>
<p>(i) Find angle ABC. State the theorem used. <strong>(3 marks)</strong></p>
<p>(ii) If ABCD is a cyclic quadrilateral and angle DAB = 105°, find angle BCD. <strong>(2 marks)</strong></p>
<p>(iii) A tangent at point B makes angle ABT = 34° with chord AB. Find angle ACB. <strong>(2 marks)</strong></p>
<p><strong>Question 14</strong> <em>(15 marks)</em></p>
<p>(a) <strong>(Volume and Surface Area)</strong> A solid consists of a cylinder with a hemisphere on top. The cylinder has radius 4 cm and height 10 cm. The hemisphere has the same radius.</p>
<p>(i) Find the total volume of the solid. <strong>(4 marks)</strong></p>
<p>(ii) Find the total surface area of the solid.</p>
<p>Note: A hemisphere has curved SA = 2πr² and the flat face area = πr². <strong>(5 marks)</strong></p>
<p>(b) <strong>(Linear Inequalities and Indices)</strong></p>
<p>(i) Solve the inequality: 3(2x - 1) ≥ 5x - 1 and represent on a number line. <strong>(3 marks)</strong></p>
<p>(ii) Solve the compound inequality: -2 ≤ 3x - 5 < 7, and state the integer solutions. <strong>(3 marks)</strong></p>
<hr class="section-divider">
<h2 class="notes-h2">MARKING GUIDE</h2>
<h3 class="notes-h3">Section A (30 marks)</h3>
<ol class="notes-list">
<li>(a) 0.6̄ rational, √11 irrational, 3/7 rational, √49 = 7 rational <strong>(1 mark for all 4 correct)</strong></li>
</ol>
<p>(b) Let x = 0.181818..., 100x = 18.181818..., 99x = 18, x = 18/99 = <strong>2/11</strong> <strong>(2 marks)</strong></p>
<ol class="notes-list">
<li>(a) √72 = 6√2, √32 = 4√2, √8 = 2√2; 6√2 + 4√2 - 2√2 = <strong>8√2</strong> <strong>(3 marks)</strong></li>
</ol>
<p>(b) 4/√3 × √3/√3 = <strong>4√3/3</strong> <strong>(2 marks)</strong></p>
<ol class="notes-list">
<li>(a) 27^(2/3) = (∛27)² = 3² = <strong>9</strong> <strong>(2 marks)</strong></li>
</ol>
<p>(b) 8x⁶y³ ÷ 4xy² = <strong>2x⁵y</strong> <strong>(3 marks)</strong></p>
<ol class="notes-list">
<li>4^(x-1) = 32 → 2^(2x-2) = 2⁵ → 2x-2 = 5 → x = <strong>7/2</strong> <strong>(3 marks)</strong></li>
</ol>
<ol class="notes-list">
<li>(a) <strong>(2x+3)(2x-3)</strong> <strong>(2 marks)</strong></li>
</ol>
<p>(b) (3x-4)(x+1) <strong>(3 marks)</strong></p>
<ol class="notes-list">
<li>(2x-1)(x-3) = 0; x = <strong>1/2 or x = 3</strong> <strong>(3 marks)</strong></li>
</ol>
<ol class="notes-list">
<li>(a) ∠ACB = 124/2 = <strong>62°</strong> <strong>(2 marks)</strong></li>
</ol>
<p>(b) ∠R = 180 - 73 = <strong>107°</strong> <strong>(1 mark)</strong></p>
<ol class="notes-list">
<li>Volume = 22/7 × 49 × 5 = <strong>770 cm³</strong> <strong>(2 marks)</strong></li>
</ol>
<h3 class="notes-h3">Section B (40 marks)</h3>
<ol class="notes-list">
<li>(a) 4x²+12x+9-(x²-2x+1) = 3x²+14x+8 <strong>(4 marks)</strong></li>
</ol>
<p>(b)(i) x=-5, x=2; (ii) same using formula; (iii) disc=49>0, two distinct roots <strong>(6 marks)</strong></p>
<ol class="notes-list">
<li>(a)(i) w(w+4)=45, w²+4w-45=0 <strong>(2 marks)</strong></li>
</ol>
<p>(ii) (w+9)(w-5)=0, w=5; <strong>width=5cm, length=9cm</strong> <strong>(3 marks)</strong></p>
<p>(b) x²+(x+3)²=29; 2x²+6x-20=0; x²+3x-10=0; (x+5)(x-2)=0; <strong>(x=-5,y=-2) and (x=2,y=5)</strong> <strong>(5 marks)</strong></p>
<ol class="notes-list">
<li>(a) 31–40 <strong>(1)</strong></li>
</ol>
<p>(b) x̄ = Σfx/40; midpoints: 15.5,25.5,35.5,45.5,55.5,65.5; Σfx = 1654; x̄ = <strong>41.35</strong> <strong>(4)</strong></p>
<p>(c) CF: 2,7,19,30,37,40 <strong>(2)</strong></p>
<p>(d) Median class 41–50; L=40.5; n/2=20; F=19; f=11; w=10; median = 40.5+(1/11)×10 = <strong>41.41</strong> <strong>(2)</strong></p>
<p>(e) Q₁: n/4=10; Q₁ class = 31–40; Q₁ = 30.5+(3/12)×10 = <strong>33</strong> <strong>(1)</strong></p>
<ol class="notes-list">
<li>(a)(i) 4/9 (ii) 6/9=2/3 (iii) 0 <strong>(3)</strong></li>
</ol>
<p>(b) Tree diagram <strong>(2)</strong>; P(YY)=4/9×3/8=<strong>12/72=1/6</strong> <strong>(2)</strong>; P(exactly one Y) = 4/9×5/8 + 5/9×4/8 = 20/72+20/72 = <strong>40/72=5/9</strong> <strong>(2)</strong>; P(at least one Y) = 1-P(no Y) = 1-(5/9×4/8) = 1-20/72 = <strong>52/72=13/18</strong> <strong>(3)</strong></p>
<h3 class="notes-h3">Section C (30 marks)</h3>
<ol class="notes-list">
<li>(a)(i) P'(-1,2), Q'(-3,2), R'(-2,4) <strong>(2)</strong></li>
</ol>
<p>(ii) P''(-2,1), Q''(-2,3), R''(-4,2) <strong>(3)</strong></p>
<p>(iii) P'''(3,6), Q'''(9,6), R'''(6,12) <strong>(2)</strong></p>
<p>(iv) 2 × 3² = <strong>18 cm²</strong> <strong>(1)</strong></p>
<p>(b)(i) ∠ABC = 112/2 = <strong>56°</strong>; angle at circumference = half angle at centre <strong>(3)</strong></p>
<p>(ii) ∠BCD = 180 - 105 = <strong>75°</strong>; opposite angles in cyclic quadrilateral <strong>(2)</strong></p>
<p>(iii) <strong>∠ACB = 34°</strong>; tangent-chord angle = angle in alternate segment <strong>(2)</strong></p>
<ol class="notes-list">
<li>(a)(i) V = πr²h + (2/3)πr³ = π×16×10 + (2/3)π×64 = 160π + 128π/3 = (480π+128π)/3 = <strong>608π/3 ≈ 636.7 cm³</strong> <strong>(4)</strong></li>
</ol>
<p>(ii) Cylinder curved SA = 2πrh = 80π; Top circle = πr² = 16π (no — replaced by hemisphere); Hemisphere curved SA = 2πr² = 32π; Bottom circle = πr² = 16π; Total = 80π + 32π + 16π = <strong>128π ≈ 402.1 cm²</strong> <strong>(5)</strong></p>
<p>(b)(i) 6x-3 ≥ 5x-1; x ≥ 2; number line with closed circle at 2 <strong>(3)</strong></p>
<p>(ii) -2+5 ≤ 3x < 12; 3 ≤ 3x < 12; <strong>1 ≤ x < 4</strong>; integers: 1, 2, 3 <strong>(3)</strong></p>
<hr class="section-divider">
<h2 class="notes-h2">CONTENT SUMMARY</h2>
<h3 class="notes-h3">What We Covered — Grade 9 Mathematics Term 1</h3>
<div class="table-wrap"><table class="notes-table">
<thead>
<tr><th>Strand</th><th>Sub-Strand</th><th>Key Topics</th></tr>
</thead><tbody>
<tr><td>Numbers</td><td>Rational & Irrational</td><td>Converting decimals, surds, rationalising</td></tr>
<tr><td>Numbers</td><td>Indices</td><td>All 9 laws, fractional/negative indices, solving equations</td></tr>
<tr><td>Algebra</td><td>Quadratic Expressions</td><td>Expanding, factorising (5 methods), difference of squares</td></tr>
<tr><td>Algebra</td><td>Quadratic Equations</td><td>Factorisation, completing the square, formula, word problems</td></tr>
<tr><td>Algebra</td><td>Linear Inequalities</td><td>Solving, number line, 2D regions, simultaneous inequalities</td></tr>
<tr><td>Algebra</td><td>Simultaneous Equations</td><td>Linear-linear, linear-quadratic</td></tr>
<tr><td>Geometry</td><td>Circles</td><td>6 theorems, worked examples, tangent properties</td></tr>
<tr><td>Geometry</td><td>Transformations</td><td>All 4 types, combined, area scale factor</td></tr>
<tr><td>Measurements</td><td>Volume & Surface Area</td><td>Prism, cylinder, pyramid, cone, sphere, density</td></tr>
<tr><td>Statistics</td><td>Grouped Data</td><td>Mean, median, mode, histogram, ogive, quartiles</td></tr>
<tr><td>Probability</td><td>Events</td><td>Mutually exclusive, independent, tree diagrams, with/without replacement</td></tr>
</tbody></table></div>
<hr class="section-divider">
<h2 class="notes-h2">PRICING AND DOWNLOAD</h2>
<div class="table-wrap"><table class="notes-table">
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<tr><td>Grade 9 Mathematics Notes — Term 1</td><td>KSH 100</td></tr>
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<tr><td>Grade 9 Mathematics Complete Bundle (Notes + Exam + Rubric)</td><td>KSH 250</td></tr>
<tr><td>Grade 9 All Subjects — Term 1 Bundle</td><td>KSH 800</td></tr>
</tbody></table></div>
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