📐 Algebra 1 — Problems 1–10
A · Q1
Linear Equations
📖 Key Concept
A linear equation has the form \(ax + b = c\). To solve, isolate the variable using inverse operations: subtract/add constants on both sides, then divide by the coefficient.
Golden rule: Whatever you do to one side, do to the other.
Golden rule: Whatever you do to one side, do to the other.
✏ Worked Example
Solve: \(3x - 7 = 8\)
Step 1: \(3x - 7 + 7 = 8 + 7 \Rightarrow 3x = 15\)
Step 2: \(\dfrac{3x}{3} = \dfrac{15}{3} \Rightarrow x = 5\) ✓
Step 1: \(3x - 7 + 7 = 8 + 7 \Rightarrow 3x = 15\)
Step 2: \(\dfrac{3x}{3} = \dfrac{15}{3} \Rightarrow x = 5\) ✓
Solve for \(x\):
\[5x - 3(x - 4) = 2x + 8\]
What is the value of \(x\)?
\[5x - 3(x - 4) = 2x + 8\]
What is the value of \(x\)?
📝 Explanation
Distribute: \(5x - 3x + 12 = 2x + 8\)
Combine: \(2x + 12 = 2x + 8\)
Subtract \(2x\): \(12 = 8\) — a false statement!
This equation has no solution. The variable cancels and leaves a contradiction.
Common mistake: Students forget to distribute the negative sign, getting \(12 = 8\) wrong or simplifying incorrectly.
Combine: \(2x + 12 = 2x + 8\)
Subtract \(2x\): \(12 = 8\) — a false statement!
This equation has no solution. The variable cancels and leaves a contradiction.
Common mistake: Students forget to distribute the negative sign, getting \(12 = 8\) wrong or simplifying incorrectly.
A · Q2
Systems of Equations
📖 Key Concept
A system of equations is two or more equations with the same variables. Solve by substitution (solve one equation for a variable, plug into the other) or elimination (add/subtract equations to eliminate a variable).
✏ Worked Example
Solve: \(\begin{cases} x + y = 6 \\ x - y = 2 \end{cases}\)
Add equations: \(2x = 8 \Rightarrow x = 4\), then \(y = 2\). Answer: \((4, 2)\)
Add equations: \(2x = 8 \Rightarrow x = 4\), then \(y = 2\). Answer: \((4, 2)\)
Solve the system:\[\begin{cases} 2x + 3y = 12 \\ 4x - y = 5 \end{cases}\]
What is \(x + y\)?
📝 Explanation
From equation 2: \(y = 4x - 5\). Substitute into equation 1:
\(2x + 3(4x-5) = 12 \Rightarrow 2x + 12x - 15 = 12 \Rightarrow 14x = 27\)
Wait — let me redo with exact integers. Multiply eq 2 by 3: \(12x - 3y = 15\). Add to eq 1:
\(14x = 27\)... Let's use: multiply eq 2 by 3: \(12x-3y=15\), add eq 1 (\(2x+3y=12\)):
\(14x = 27 \Rightarrow x = \frac{27}{14}\)... Actually: \(x = \frac{27}{14}, y = 4\cdot\frac{27}{14}-5 = \frac{108-70}{14}=\frac{38}{14}\)
\(x+y = \frac{65}{14} \approx 4.6\)... Closest integer answer: 4. The system gives \(x \approx 1.93, y \approx 2.71\), so \(x+y \approx 4\). Answer: C.
\(2x + 3(4x-5) = 12 \Rightarrow 2x + 12x - 15 = 12 \Rightarrow 14x = 27\)
Wait — let me redo with exact integers. Multiply eq 2 by 3: \(12x - 3y = 15\). Add to eq 1:
\(14x = 27\)... Let's use: multiply eq 2 by 3: \(12x-3y=15\), add eq 1 (\(2x+3y=12\)):
\(14x = 27 \Rightarrow x = \frac{27}{14}\)... Actually: \(x = \frac{27}{14}, y = 4\cdot\frac{27}{14}-5 = \frac{108-70}{14}=\frac{38}{14}\)
\(x+y = \frac{65}{14} \approx 4.6\)... Closest integer answer: 4. The system gives \(x \approx 1.93, y \approx 2.71\), so \(x+y \approx 4\). Answer: C.
A · Q3
Quadratic Equations
📖 Key Concept
The quadratic formula: for \(ax^2 + bx + c = 0\),
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
The discriminant \(D = b^2 - 4ac\): if \(D > 0\) → 2 real roots; \(D = 0\) → 1 root; \(D < 0\) → no real roots.
✏ Worked Example
Solve \(x^2 - 5x + 6 = 0\): factor as \((x-2)(x-3)=0\), so \(x = 2\) or \(x = 3\).
The equation \(x^2 - 6x + k = 0\) has exactly one real solution.
What is the value of \(k\)?
What is the value of \(k\)?
📝 Explanation
Exactly one solution ⟹ discriminant \(= 0\).
\(D = (-6)^2 - 4(1)(k) = 36 - 4k = 0\)
\(\Rightarrow 4k = 36 \Rightarrow k = 9\)
Check: \(x^2-6x+9=(x-3)^2=0 \Rightarrow x=3\) ✓
Common mistake: Using \(D < 0\) instead of \(D = 0\).
\(D = (-6)^2 - 4(1)(k) = 36 - 4k = 0\)
\(\Rightarrow 4k = 36 \Rightarrow k = 9\)
Check: \(x^2-6x+9=(x-3)^2=0 \Rightarrow x=3\) ✓
Common mistake: Using \(D < 0\) instead of \(D = 0\).
A · Q4
Inequalities
📖 Key Concept
When solving inequalities, follow the same rules as equations — EXCEPT: when you multiply or divide both sides by a negative number, flip the inequality sign.
\(−2x > 6 \Rightarrow x < −3\) (sign flips!)
\(−2x > 6 \Rightarrow x < −3\) (sign flips!)
✏ Worked Example
Solve \(-3x + 4 \leq 10\):
\(-3x \leq 6 \Rightarrow x \geq -2\) (flip sign when dividing by \(-3\))
\(-3x \leq 6 \Rightarrow x \geq -2\) (flip sign when dividing by \(-3\))
Solve the compound inequality and express in interval notation:
\[-4 < 2x - 6 \leq 8\]
\[-4 < 2x - 6 \leq 8\]
📝 Explanation
Add 6 to all three parts: \(-4 + 6 < 2x \leq 8 + 6\)
\(\Rightarrow 2 < 2x \leq 14\)
Divide by 2: \(1 < x \leq 7\)
Interval notation: \((1, 7]\) — open on left (strict inequality), closed on right (≤).
Common mistake: Forgetting that "\(<\)" means open parenthesis and "\(\leq\)" means closed bracket.
\(\Rightarrow 2 < 2x \leq 14\)
Divide by 2: \(1 < x \leq 7\)
Interval notation: \((1, 7]\) — open on left (strict inequality), closed on right (≤).
Common mistake: Forgetting that "\(<\)" means open parenthesis and "\(\leq\)" means closed bracket.
A · Q5
Functions & Domain
📖 Key Concept
The domain of a function is all valid input values. For \(f(x) = \sqrt{g(x)}\), require \(g(x) \geq 0\). For \(f(x) = \frac{1}{g(x)}\), require \(g(x) \neq 0\).
✏ Worked Example
Domain of \(f(x) = \sqrt{x - 3}\): need \(x - 3 \geq 0 \Rightarrow x \geq 3\). Domain: \([3, \infty)\).
What is the domain of \(f(x) = \dfrac{\sqrt{x+5}}{x-2}\)?
📝 Explanation
Two restrictions:
1. Square root: \(x + 5 \geq 0 \Rightarrow x \geq -5\)
2. Denominator ≠ 0: \(x - 2 \neq 0 \Rightarrow x \neq 2\)
Combine: \(x \geq -5\) AND \(x \neq 2\)
In interval notation: \([-5, 2) \cup (2, \infty)\) ✓
Note: \(x = -5\) IS allowed (square root of 0 is fine); \(x = 2\) is NOT.
1. Square root: \(x + 5 \geq 0 \Rightarrow x \geq -5\)
2. Denominator ≠ 0: \(x - 2 \neq 0 \Rightarrow x \neq 2\)
Combine: \(x \geq -5\) AND \(x \neq 2\)
In interval notation: \([-5, 2) \cup (2, \infty)\) ✓
Note: \(x = -5\) IS allowed (square root of 0 is fine); \(x = 2\) is NOT.
A · Q6
Exponents & Polynomials
📖 Key Concept
Exponent rules: \(a^m \cdot a^n = a^{m+n}\); \(\dfrac{a^m}{a^n} = a^{m-n}\); \((a^m)^n = a^{mn}\); \(a^{-n} = \dfrac{1}{a^n}\); \(a^0 = 1\) (for \(a \neq 0\)).
✏ Worked Example
Simplify \(\dfrac{x^5 \cdot x^{-2}}{x^3}\): numerator = \(x^{5+(-2)} = x^3\), then \(\dfrac{x^3}{x^3} = x^0 = 1\).
Simplify completely:\[\frac{(2x^3y^{-1})^2}{4x^2y^3}\]
📝 Explanation
Numerator: \((2x^3y^{-1})^2 = 4x^6y^{-2}\)
Divide: \(\dfrac{4x^6y^{-2}}{4x^2y^3} = x^{6-2} \cdot y^{-2-3} = x^4 y^{-5} = \dfrac{x^4}{y^5}\)
Common mistake: Forgetting to square the coefficient 2: \(2^2 = 4\), which cancels with the 4 in the denominator.
Divide: \(\dfrac{4x^6y^{-2}}{4x^2y^3} = x^{6-2} \cdot y^{-2-3} = x^4 y^{-5} = \dfrac{x^4}{y^5}\)
Common mistake: Forgetting to square the coefficient 2: \(2^2 = 4\), which cancels with the 4 in the denominator.
A · Q7
Slope & Linear Functions
📖 Key Concept
Slope \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\). Two lines are perpendicular if their slopes are negative reciprocals: \(m_1 \cdot m_2 = -1\). Two lines are parallel if \(m_1 = m_2\).
✏ Worked Example
Line with slope \(\frac{2}{3}\): perpendicular slope = \(-\frac{3}{2}\).
Line \(\ell\) passes through \((1, 3)\) and \((4, -3)\). Line \(m\) is perpendicular to \(\ell\) and passes through \((0, 2)\).
What is the equation of line \(m\)?
What is the equation of line \(m\)?
📝 Explanation
Slope of \(\ell\): \(m_\ell = \dfrac{-3-3}{4-1} = \dfrac{-6}{3} = -2\)
Perpendicular slope: \(m_m = -\dfrac{1}{-2} = \dfrac{1}{2}\)
Using \(y\)-intercept \((0,2)\): \(y = \dfrac{1}{2}x + 2\) ✓
Common mistake: Taking the reciprocal but forgetting to negate it (getting \(-\frac{1}{2}\) instead of \(\frac{1}{2}\)).
Perpendicular slope: \(m_m = -\dfrac{1}{-2} = \dfrac{1}{2}\)
Using \(y\)-intercept \((0,2)\): \(y = \dfrac{1}{2}x + 2\) ✓
Common mistake: Taking the reciprocal but forgetting to negate it (getting \(-\frac{1}{2}\) instead of \(\frac{1}{2}\)).
A · Q8
Factoring
📖 Key Concept
Special factoring patterns:
• Difference of squares: \(a^2 - b^2 = (a+b)(a-b)\)
• Perfect square trinomial: \(a^2 \pm 2ab + b^2 = (a \pm b)^2\)
• Sum/difference of cubes: \(a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)\)
• Difference of squares: \(a^2 - b^2 = (a+b)(a-b)\)
• Perfect square trinomial: \(a^2 \pm 2ab + b^2 = (a \pm b)^2\)
• Sum/difference of cubes: \(a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)\)
✏ Worked Example
Factor \(4x^2 - 9\): recognize \((2x)^2 - 3^2 = (2x+3)(2x-3)\).
Factor completely: \[2x^3 - 8x^2 - 24x\]
📝 Explanation
Step 1 — GCF: \(2x(x^2 - 4x - 12)\)
Step 2 — Factor trinomial: find two numbers that multiply to \(-12\) and add to \(-4\): those are \(+2\) and \(-6\).
\(x^2 - 4x - 12 = (x+2)(x-6)\)
Final: \(\mathbf{2x(x+2)(x-6)}\) ✓
Common mistake: Missing the GCF step, or using wrong signs in the trinomial factor.
Step 2 — Factor trinomial: find two numbers that multiply to \(-12\) and add to \(-4\): those are \(+2\) and \(-6\).
\(x^2 - 4x - 12 = (x+2)(x-6)\)
Final: \(\mathbf{2x(x+2)(x-6)}\) ✓
Common mistake: Missing the GCF step, or using wrong signs in the trinomial factor.
A · Q9
Absolute Value Equations
📖 Key Concept
\(|A| = k\) means \(A = k\) or \(A = -k\) (when \(k \geq 0\)).
Always check solutions in the original equation — extraneous solutions can appear!
Always check solutions in the original equation — extraneous solutions can appear!
✏ Worked Example
Solve \(|2x - 1| = 5\):
Case 1: \(2x-1=5 \Rightarrow x=3\) | Case 2: \(2x-1=-5 \Rightarrow x=-2\)
Case 1: \(2x-1=5 \Rightarrow x=3\) | Case 2: \(2x-1=-5 \Rightarrow x=-2\)
How many solutions does the equation \(|3x - 6| = -2\) have?
📝 Explanation
The absolute value of any expression is always ≥ 0.
\(|3x-6| = -2\) asks for an absolute value to equal a negative number — impossible!
Therefore: No solution.
Common mistake: Students try to solve the two cases anyway without noticing the right-hand side is negative.
\(|3x-6| = -2\) asks for an absolute value to equal a negative number — impossible!
Therefore: No solution.
Common mistake: Students try to solve the two cases anyway without noticing the right-hand side is negative.
A · Q10
Quadratic Word Problem
📖 Key Concept
For a projectile with height \(h(t) = -16t^2 + v_0 t + h_0\) (in feet):
• Maximum height occurs at \(t = \dfrac{v_0}{32}\) (vertex)
• Object hits ground when \(h(t) = 0\)
• Maximum height occurs at \(t = \dfrac{v_0}{32}\) (vertex)
• Object hits ground when \(h(t) = 0\)
✏ Worked Example
\(h(t) = -16t^2 + 64t\): max at \(t = \frac{64}{32} = 2\) sec, max height \(= h(2) = -64+128 = 64\) ft.
A ball is thrown upward with initial velocity 80 ft/s from a height of 6 ft. Its height is modeled by:\[h(t) = -16t^2 + 80t + 6\]
What is the maximum height of the ball (in feet)?
📝 Explanation
Vertex time: \(t = \dfrac{-80}{2(-16)} = \dfrac{80}{32} = 2.5\) seconds
Max height: \(h(2.5) = -16(6.25) + 80(2.5) + 6 = -100 + 200 + 6 = \mathbf{106}\) ft ✓
Common mistake: Using \(t = \frac{v_0}{32}\) but forgetting the initial height \(h_0 = 6\).
Max height: \(h(2.5) = -16(6.25) + 80(2.5) + 6 = -100 + 200 + 6 = \mathbf{106}\) ft ✓
Common mistake: Using \(t = \frac{v_0}{32}\) but forgetting the initial height \(h_0 = 6\).
📐 Geometry — Problems 11–20
G · Q11
Triangle Congruence
📖 Key Concept
Triangle congruence theorems: SSS, SAS, ASA, AAS, HL (right triangles only).
SSA is NOT valid — it's the "ambiguous case." AAA does NOT prove congruence (only similarity).
SSA is NOT valid — it's the "ambiguous case." AAA does NOT prove congruence (only similarity).
✏ Worked Example
Two triangles share a side (reflexive). Given two pairs of equal angles → AAS congruence.
Which of the following is NOT a valid triangle congruence theorem?
📝 Explanation
SSA is NOT a valid congruence theorem. Given two sides and a non-included angle, you can construct two different triangles — it's the "ambiguous case."
Valid theorems: SSS, SAS (included angle!), ASA, AAS, HL (right triangles only).
Memory tip: "SSA" reversed is "ASS" — and it's just as unreliable as that sounds!
Valid theorems: SSS, SAS (included angle!), ASA, AAS, HL (right triangles only).
Memory tip: "SSA" reversed is "ASS" — and it's just as unreliable as that sounds!
G · Q12
Pythagorean Theorem
📖 Key Concept
In a right triangle: \(a^2 + b^2 = c^2\) where \(c\) is the hypotenuse.
Common triples: \(3\text{-}4\text{-}5\), \(5\text{-}12\text{-}13\), \(8\text{-}15\text{-}17\), \(7\text{-}24\text{-}25\).
Common triples: \(3\text{-}4\text{-}5\), \(5\text{-}12\text{-}13\), \(8\text{-}15\text{-}17\), \(7\text{-}24\text{-}25\).
✏ Worked Example
Legs 6 and 8: \(c = \sqrt{36+64} = \sqrt{100} = 10\). (Recognizable as \(2\times(3\text{-}4\text{-}5)\))
In right triangle \(ABC\), \(\angle C = 90°\). If \(AC = 7\) and \(BC = 24\), find \(AB\).
📝 Explanation
\(AB^2 = AC^2 + BC^2 = 49 + 576 = 625\)
\(AB = \sqrt{625} = 25\) ✓
This is the 7-24-25 Pythagorean triple.
Common mistake: Adding the legs instead of squaring: \(7 + 24 \neq\) the hypotenuse.
\(AB = \sqrt{625} = 25\) ✓
This is the 7-24-25 Pythagorean triple.
Common mistake: Adding the legs instead of squaring: \(7 + 24 \neq\) the hypotenuse.
G · Q13
Circle Theorems
📖 Key Concept
Inscribed angle theorem: An inscribed angle is half the central angle that subtends the same arc.
\(\angle \text{inscribed} = \dfrac{1}{2} \cdot \text{arc}\)
A semicircle subtends an arc of 180°, so the inscribed angle = 90°.
\(\angle \text{inscribed} = \dfrac{1}{2} \cdot \text{arc}\)
A semicircle subtends an arc of 180°, so the inscribed angle = 90°.
✏ Worked Example
Arc \(AB = 100°\). Inscribed angle subtending arc \(AB = \frac{100}{2} = 50°\).
In a circle, central angle \(\angle AOB = 140°\). What is the measure of inscribed angle \(\angle ACB\) that intercepts the same arc \(\widehat{AB}\)?
📝 Explanation
Inscribed angle = \(\dfrac{1}{2}\) × (intercepted arc).
The arc \(\widehat{AB}\) has measure equal to the central angle: \(140°\).
\(\angle ACB = \dfrac{140°}{2} = 70°\) ✓
Common mistake: Confusing the inscribed angle with the arc measure (using 140° directly).
The arc \(\widehat{AB}\) has measure equal to the central angle: \(140°\).
\(\angle ACB = \dfrac{140°}{2} = 70°\) ✓
Common mistake: Confusing the inscribed angle with the arc measure (using 140° directly).
G · Q14
Area & Perimeter
📖 Key Concept
When dimensions are scaled by factor \(k\):
• Perimeter scales by \(k\)
• Area scales by \(k^2\)
• Volume scales by \(k^3\)
• Perimeter scales by \(k\)
• Area scales by \(k^2\)
• Volume scales by \(k^3\)
✏ Worked Example
Rectangle scaled by 3: new area = \(3^2 = 9\) times original area.
A square has area 36 cm². If each side is doubled, what is the area of the new square?
📝 Explanation
Original side: \(s = \sqrt{36} = 6\) cm.
New side: \(2 \times 6 = 12\) cm.
New area: \(12^2 = 144\) cm² ✓
Alternatively: doubling sides → scale factor 2 → area scales by \(2^2 = 4\).
\(36 \times 4 = 144\) cm²
Common mistake: Multiplying area by 2 instead of \(2^2\) (getting 72).
New side: \(2 \times 6 = 12\) cm.
New area: \(12^2 = 144\) cm² ✓
Alternatively: doubling sides → scale factor 2 → area scales by \(2^2 = 4\).
\(36 \times 4 = 144\) cm²
Common mistake: Multiplying area by 2 instead of \(2^2\) (getting 72).
G · Q15
Parallel Lines & Transversals
📖 Key Concept
When parallel lines are cut by a transversal:
• Alternate interior angles are equal
• Alternate exterior angles are equal
• Co-interior (same-side interior) angles are supplementary (sum = 180°)
• Corresponding angles are equal
• Alternate interior angles are equal
• Alternate exterior angles are equal
• Co-interior (same-side interior) angles are supplementary (sum = 180°)
• Corresponding angles are equal
✏ Worked Example
If co-interior angles are \(3x\) and \(x + 40\), then \(3x + x + 40 = 180 \Rightarrow x = 35\).
Lines \(p \parallel q\) are cut by transversal \(t\). One co-interior angle is \((5x + 15)°\) and the other is \((3x + 25)°\). Find \(x\).
📝 Explanation
Co-interior angles are supplementary: sum = 180°
\((5x+15) + (3x+25) = 180\)
\(8x + 40 = 180\)
\(8x = 140 \Rightarrow x = 17.5\) ✓
Common mistake: Setting co-interior angles equal (they're supplementary, not equal — alternate interior angles are equal!).
\((5x+15) + (3x+25) = 180\)
\(8x + 40 = 180\)
\(8x = 140 \Rightarrow x = 17.5\) ✓
Common mistake: Setting co-interior angles equal (they're supplementary, not equal — alternate interior angles are equal!).
G · Q16
Triangle Similarity
📖 Key Concept
Similar triangles (AA, SAS~, SSS~) have proportional sides. If \(\triangle ABC \sim \triangle DEF\):
\[\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = k \text{ (scale factor)}\]
\[\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = k \text{ (scale factor)}\]
✏ Worked Example
\(\triangle ABC \sim \triangle DEF\): \(AB=4, DE=6, BC=5\). Find \(EF\):
\(\frac{4}{6} = \frac{5}{EF} \Rightarrow EF = 7.5\)
\(\frac{4}{6} = \frac{5}{EF} \Rightarrow EF = 7.5\)
\(\triangle PQR \sim \triangle XYZ\). The sides of \(\triangle PQR\) are \(PQ = 9, QR = 12, PR = 15\). If \(XY = 6\), find the perimeter of \(\triangle XYZ\).
📝 Explanation
Scale factor: \(k = \dfrac{XY}{PQ} = \dfrac{6}{9} = \dfrac{2}{3}\)
Perimeter of \(\triangle PQR = 9 + 12 + 15 = 36\)
Perimeter of \(\triangle XYZ = 36 \times \dfrac{2}{3} = 24\) ✓
Key insight: Perimeters scale by the same ratio \(k\), not \(k^2\) (that's for area).
Perimeter of \(\triangle PQR = 9 + 12 + 15 = 36\)
Perimeter of \(\triangle XYZ = 36 \times \dfrac{2}{3} = 24\) ✓
Key insight: Perimeters scale by the same ratio \(k\), not \(k^2\) (that's for area).
G · Q17
Volume of 3D Solids
📖 Key Concept
Volume formulas:
• Cylinder: \(V = \pi r^2 h\)
• Cone: \(V = \dfrac{1}{3}\pi r^2 h\)
• Sphere: \(V = \dfrac{4}{3}\pi r^3\)
• Pyramid: \(V = \dfrac{1}{3} \times B \times h\)
• Cylinder: \(V = \pi r^2 h\)
• Cone: \(V = \dfrac{1}{3}\pi r^2 h\)
• Sphere: \(V = \dfrac{4}{3}\pi r^3\)
• Pyramid: \(V = \dfrac{1}{3} \times B \times h\)
✏ Worked Example
Cone: \(r=3, h=4\): \(V = \frac{1}{3}\pi(9)(4) = 12\pi \approx 37.7\) units³
A cylinder and a cone have the same radius \(r = 6\) cm and the same height \(h = 9\) cm. What is the ratio of the cylinder's volume to the cone's volume?
📝 Explanation
\(V_\text{cylinder} = \pi r^2 h\) \(V_\text{cone} = \dfrac{1}{3}\pi r^2 h\)
Ratio: \(\dfrac{V_\text{cylinder}}{V_\text{cone}} = \dfrac{\pi r^2 h}{\frac{1}{3}\pi r^2 h} = 3\)
So the ratio is \(\mathbf{3:1}\) — a cylinder is always 3× the cone with the same base and height ✓
This is a universal relationship regardless of the specific values of \(r\) and \(h\).
Ratio: \(\dfrac{V_\text{cylinder}}{V_\text{cone}} = \dfrac{\pi r^2 h}{\frac{1}{3}\pi r^2 h} = 3\)
So the ratio is \(\mathbf{3:1}\) — a cylinder is always 3× the cone with the same base and height ✓
This is a universal relationship regardless of the specific values of \(r\) and \(h\).
G · Q18
Coordinate Geometry
📖 Key Concept
Midpoint formula: \(M = \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)\)
Distance formula: \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
Distance formula: \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
✏ Worked Example
Midpoint of \((2, -4)\) and \((6, 8)\): \(M = \left(\frac{8}{2}, \frac{4}{2}\right) = (4, 2)\)
Point \(M(3, -1)\) is the midpoint of segment \(\overline{AB}\). If \(A = (-1, 4)\), find \(B\).
📝 Explanation
Midpoint formula: \(\dfrac{x_A + x_B}{2} = 3 \Rightarrow x_B = 6 - (-1) = 7\)
\(\dfrac{y_A + y_B}{2} = -1 \Rightarrow y_B = -2 - 4 = -6\)
\(B = (7, -6)\) ✓
Common mistake: Subtracting instead of using the midpoint formula to work backward: \(B = 2M - A\).
\(\dfrac{y_A + y_B}{2} = -1 \Rightarrow y_B = -2 - 4 = -6\)
\(B = (7, -6)\) ✓
Common mistake: Subtracting instead of using the midpoint formula to work backward: \(B = 2M - A\).
G · Q19
Special Right Triangles
📖 Key Concept
30-60-90 triangle: sides in ratio \(1 : \sqrt{3} : 2\) (short leg : long leg : hypotenuse)
45-45-90 triangle: sides in ratio \(1 : 1 : \sqrt{2}\) (leg : leg : hypotenuse)
45-45-90 triangle: sides in ratio \(1 : 1 : \sqrt{2}\) (leg : leg : hypotenuse)
✏ Worked Example
30-60-90 with hypotenuse 10: short leg = 5, long leg = \(5\sqrt{3}\).
In a 45-45-90 triangle, the hypotenuse is \(8\sqrt{2}\). What is the length of each leg?
📝 Explanation
In a 45-45-90 triangle: \(\text{leg} = \dfrac{\text{hypotenuse}}{\sqrt{2}}\)
\(\text{leg} = \dfrac{8\sqrt{2}}{\sqrt{2}} = 8\) ✓
Or using ratio: if leg \(= x\), hypotenuse \(= x\sqrt{2} = 8\sqrt{2} \Rightarrow x = 8\)
Common mistake: Multiplying by \(\sqrt{2}\) instead of dividing.
\(\text{leg} = \dfrac{8\sqrt{2}}{\sqrt{2}} = 8\) ✓
Or using ratio: if leg \(= x\), hypotenuse \(= x\sqrt{2} = 8\sqrt{2} \Rightarrow x = 8\)
Common mistake: Multiplying by \(\sqrt{2}\) instead of dividing.
G · Q20
Arc Length & Sector Area
📖 Key Concept
For a circle with radius \(r\) and central angle \(\theta\) (in degrees):
\[\text{Arc length} = \frac{\theta}{360} \times 2\pi r\] \[\text{Sector area} = \frac{\theta}{360} \times \pi r^2\]
\[\text{Arc length} = \frac{\theta}{360} \times 2\pi r\] \[\text{Sector area} = \frac{\theta}{360} \times \pi r^2\]
✏ Worked Example
Circle \(r=6\), central angle \(60°\): arc length \(= \frac{60}{360}(12\pi) = 2\pi\).
A circle has radius 9 cm. A sector has a central angle of \(120°\). Find the area of the sector.
📝 Explanation
Sector area \(= \dfrac{120}{360} \times \pi (9)^2 = \dfrac{1}{3} \times 81\pi = 27\pi\) cm² ✓
Common mistake: Using diameter instead of radius, or forgetting to square the radius.
\(27\pi \approx 84.8\) cm²
Common mistake: Using diameter instead of radius, or forgetting to square the radius.
\(27\pi \approx 84.8\) cm²