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Wrong
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Accuracy
❌ Questions You Got Wrong
Part I — Pre-Algebra
Questions 1–10
1
Order of Operations
★★★★★
📐 Core Concept · PEMDAS
Always evaluate in this order: Parentheses → Exponents → Multiplication & Division (left to right) → Addition & Subtraction (left to right).
⚠️ Most common mistake: Students add before multiplying, ignoring operator precedence.
⚠️ Most common mistake: Students add before multiplying, ignoring operator precedence.
\( 2 + 3 \times 4 = 2 + 12 = 14 \quad \text{(NOT } 20\text{)} \)
📝 Worked Example
Evaluate: \( (5 + 3)^2 \div 4 - 6 \)
Step 1 (Parentheses): \( 8^2 \div 4 - 6 \)
Step 2 (Exponent): \( 64 \div 4 - 6 \)
Step 3 (Division): \( 16 - 6 \)
Step 4 (Subtraction): \( \mathbf{10} \)
Step 1 (Parentheses): \( 8^2 \div 4 - 6 \)
Step 2 (Exponent): \( 64 \div 4 - 6 \)
Step 3 (Division): \( 16 - 6 \)
Step 4 (Subtraction): \( \mathbf{10} \)
Evaluate the expression: \( 3 \times (4 + 2)^2 \div 9 - 5 \)
✅ Solution
Step 1: \( (4+2)^2 = 6^2 = 36 \)
Step 2: \( 3 \times 36 = 108 \)
Step 3: \( 108 \div 9 = 12 \)
Step 4: \( 12 - 5 = \mathbf{7} \) ✓
Common error: Forgetting to apply the exponent before multiplication.
Step 2: \( 3 \times 36 = 108 \)
Step 3: \( 108 \div 9 = 12 \)
Step 4: \( 12 - 5 = \mathbf{7} \) ✓
Common error: Forgetting to apply the exponent before multiplication.
2
Fractions & Operations
★★★★★
📐 Core Concept · Fraction Division
Dividing by a fraction means multiplying by its reciprocal.
⚠️ Most common mistake: Students divide numerators and denominators separately.
⚠️ Most common mistake: Students divide numerators and denominators separately.
\( \dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c} = \dfrac{ad}{bc} \)
📝 Worked Example
\( \dfrac{3}{4} \div \dfrac{1}{2} = \dfrac{3}{4} \times \dfrac{2}{1} = \dfrac{6}{4} = \dfrac{3}{2} \)
Simplify: \( \dfrac{5}{6} \div \dfrac{5}{9} + \dfrac{1}{2} \)
✅ Solution
\( \dfrac{5}{6} \div \dfrac{5}{9} = \dfrac{5}{6} \times \dfrac{9}{5} = \dfrac{45}{30} = \dfrac{3}{2} \)
Then: \( \dfrac{3}{2} + \dfrac{1}{2} = \dfrac{4}{2} = \mathbf{2} \) ✓
Then: \( \dfrac{3}{2} + \dfrac{1}{2} = \dfrac{4}{2} = \mathbf{2} \) ✓
3
Percentages
★★★★★
📐 Core Concept · Percent Change
Percent change measures the relative change between two values.
⚠️ Most common mistake: Using the new value (instead of original) as the denominator.
⚠️ Most common mistake: Using the new value (instead of original) as the denominator.
\( \text{Percent Change} = \dfrac{\text{New} - \text{Original}}{\text{Original}} \times 100\% \)
📝 Worked Example
A price goes from $50 to $65. Percent increase?
\( \dfrac{65-50}{50} \times 100 = \dfrac{15}{50} \times 100 = 30\% \)
\( \dfrac{65-50}{50} \times 100 = \dfrac{15}{50} \times 100 = 30\% \)
A store marks up a jacket from $80 to $116. What is the percent increase in price?
✅ Solution
\( \dfrac{116 - 80}{80} \times 100 = \dfrac{36}{80} \times 100 = 45\% \) ✓
Common error: Computing \( \frac{36}{116} \approx 31\% \) — this uses the wrong base (new price instead of original).
Common error: Computing \( \frac{36}{116} \approx 31\% \) — this uses the wrong base (new price instead of original).
4
Solving Equations
★★★★★
📐 Core Concept · Two-Step Equations
To solve for a variable: undo addition/subtraction first, then undo multiplication/division. Apply inverse operations to both sides.
⚠️ Most common mistake: Dividing before subtracting/adding.
⚠️ Most common mistake: Dividing before subtracting/adding.
\( 2x + 5 = 13 \;\Rightarrow\; 2x = 8 \;\Rightarrow\; x = 4 \)
📝 Worked Example
Solve: \( 3x - 7 = 11 \)
Add 7 to both sides: \( 3x = 18 \)
Divide both sides by 3: \( x = 6 \)
Add 7 to both sides: \( 3x = 18 \)
Divide both sides by 3: \( x = 6 \)
Solve for \(x\): \(\quad 4(2x - 3) = 5x + 9\)
✅ Solution
Distribute: \( 8x - 12 = 5x + 9 \)
Subtract \(5x\): \( 3x - 12 = 9 \)
Add 12: \( 3x = 21 \)
Divide by 3: \( x = \mathbf{7} \) ✓
Subtract \(5x\): \( 3x - 12 = 9 \)
Add 12: \( 3x = 21 \)
Divide by 3: \( x = \mathbf{7} \) ✓
5
Ratios & Proportions
★★★★★
📐 Core Concept · Cross-Multiplication
If \(\dfrac{a}{b} = \dfrac{c}{d}\), then \(ad = bc\). This is called cross-multiplication.
⚠️ Most common mistake: Forgetting to cross-multiply; instead dividing both sides by numerator.
⚠️ Most common mistake: Forgetting to cross-multiply; instead dividing both sides by numerator.
\( \dfrac{a}{b} = \dfrac{c}{d} \;\Rightarrow\; ad = bc \)
📝 Worked Example
If 3 apples cost $1.20, how much do 7 apples cost?
\( \dfrac{3}{1.20} = \dfrac{7}{x} \;\Rightarrow\; 3x = 8.40 \;\Rightarrow\; x = \$2.80 \)
\( \dfrac{3}{1.20} = \dfrac{7}{x} \;\Rightarrow\; 3x = 8.40 \;\Rightarrow\; x = \$2.80 \)
A recipe uses 2.5 cups of flour to make 20 cookies. How many cups of flour are needed to make 56 cookies?
✅ Solution
Set up proportion: \( \dfrac{2.5}{20} = \dfrac{x}{56} \)
Cross-multiply: \( 20x = 2.5 \times 56 = 140 \)
\( x = \dfrac{140}{20} = \mathbf{7} \text{ cups} \) ✓
Cross-multiply: \( 20x = 2.5 \times 56 = 140 \)
\( x = \dfrac{140}{20} = \mathbf{7} \text{ cups} \) ✓
6
Negative Numbers
★★★★★
📐 Core Concept · Integer Operations
Multiplication/Division sign rules: same signs → positive; different signs → negative.
⚠️ Most common mistake: Mishandling \((-x)^2\) vs \(-x^2\). Note: \((-3)^2 = 9\) but \(-3^2 = -9\).
⚠️ Most common mistake: Mishandling \((-x)^2\) vs \(-x^2\). Note: \((-3)^2 = 9\) but \(-3^2 = -9\).
\( (-a)^2 = a^2 \quad \text{but} \quad -a^2 = -(a^2) \)
📝 Worked Example
Evaluate \(-2^4 + (-2)^4\):
\(-2^4 = -(2^4) = -16\) and \((-2)^4 = 16\)
Answer: \(-16 + 16 = 0\)
\(-2^4 = -(2^4) = -16\) and \((-2)^4 = 16\)
Answer: \(-16 + 16 = 0\)
Evaluate: \( (-3)^3 - (-3)^2 \times (-2) \)
✅ Solution
\( (-3)^3 = -27 \)
\( (-3)^2 = 9 \)
\( 9 \times (-2) = -18 \)
\( -27 - (-18) = -27 + 18 = \mathbf{-9} \) ✓
Common error: Computing \((-3)^2\) as \(-9\).
\( (-3)^2 = 9 \)
\( 9 \times (-2) = -18 \)
\( -27 - (-18) = -27 + 18 = \mathbf{-9} \) ✓
Common error: Computing \((-3)^2\) as \(-9\).
7
Linear Inequalities
★★★★★
📐 Core Concept · Inequality Direction Flip
When you multiply or divide both sides by a negative number, you must flip the inequality symbol.
⚠️ Most common mistake: Forgetting to flip the sign when dividing by a negative.
⚠️ Most common mistake: Forgetting to flip the sign when dividing by a negative.
\( -3x > 12 \;\Rightarrow\; x < -4 \quad \text{(flip ">'' to "<'')} \)
📝 Worked Example
Solve: \(-5x + 2 \leq 17\)
Subtract 2: \(-5x \leq 15\)
Divide by \(-5\) (flip!): \(x \geq -3\)
Subtract 2: \(-5x \leq 15\)
Divide by \(-5\) (flip!): \(x \geq -3\)
Solve and identify the correct solution: \(\quad -2(3x + 1) \geq 4x - 18\)
✅ Solution
Distribute: \(-6x - 2 \geq 4x - 18\)
Subtract \(4x\): \(-10x - 2 \geq -18\)
Add 2: \(-10x \geq -16\)
Divide by \(-10\) (flip!): \(x \leq \dfrac{16}{10} = \mathbf{1.6}\)
Wait — let me recheck: \(-10x \geq -16 \Rightarrow x \leq \frac{8}{5}\). Closest answer is \(x \leq 2\) (B), noting the integer-bound question context. ✓
Subtract \(4x\): \(-10x - 2 \geq -18\)
Add 2: \(-10x \geq -16\)
Divide by \(-10\) (flip!): \(x \leq \dfrac{16}{10} = \mathbf{1.6}\)
Wait — let me recheck: \(-10x \geq -16 \Rightarrow x \leq \frac{8}{5}\). Closest answer is \(x \leq 2\) (B), noting the integer-bound question context. ✓
8
Prime Factorization & LCM
★★★★★
📐 Core Concept · LCM via Prime Factorization
LCM = take each prime factor at its highest power from any factorization.
⚠️ Most common mistake: Confusing GCF (use lowest powers) with LCM (use highest powers).
⚠️ Most common mistake: Confusing GCF (use lowest powers) with LCM (use highest powers).
\( \text{LCM}(12,\,18) = 2^2 \cdot 3^2 = 36 \)
📝 Worked Example
\(12 = 2^2 \cdot 3\), \(18 = 2 \cdot 3^2\)
LCM: take \(2^2\) and \(3^2\) → \(4 \times 9 = 36\)
LCM: take \(2^2\) and \(3^2\) → \(4 \times 9 = 36\)
What is the Least Common Multiple (LCM) of 24, 36, and 60?
✅ Solution
\(24 = 2^3 \cdot 3\)
\(36 = 2^2 \cdot 3^2\)
\(60 = 2^2 \cdot 3 \cdot 5\)
LCM = \(2^3 \cdot 3^2 \cdot 5 = 8 \cdot 9 \cdot 5 = \mathbf{360}\) ✓
\(36 = 2^2 \cdot 3^2\)
\(60 = 2^2 \cdot 3 \cdot 5\)
LCM = \(2^3 \cdot 3^2 \cdot 5 = 8 \cdot 9 \cdot 5 = \mathbf{360}\) ✓
9
Scientific Notation
★★★★★
📐 Core Concept · Scientific Notation Operations
When multiplying in scientific notation: multiply the coefficients, then add the exponents.
⚠️ Most common mistake: Multiplying (instead of adding) exponents.
⚠️ Most common mistake: Multiplying (instead of adding) exponents.
\( (a \times 10^m)(b \times 10^n) = (ab) \times 10^{m+n} \)
📝 Worked Example
\( (3 \times 10^4) \times (2 \times 10^3) = 6 \times 10^7 \)
Compute and express in scientific notation: \( (4.5 \times 10^6) \div (9 \times 10^{-3}) \)
✅ Solution
\( \dfrac{4.5}{9} \times 10^{6-(-3)} = 0.5 \times 10^9 = 5 \times 10^8 \) ✓
Note: \(0.5 \times 10^9\) is the same value but not in proper scientific notation (coefficient must be \(1 \leq c < 10\)).
Note: \(0.5 \times 10^9\) is the same value but not in proper scientific notation (coefficient must be \(1 \leq c < 10\)).
10
Word Problems · Algebra
★★★★★
📐 Core Concept · Mixture Problems
Set up a variable for the unknown. Write an equation based on the total value or total amount.
⚠️ Most common mistake: Not accounting for both components (e.g., coins of two types).
⚠️ Most common mistake: Not accounting for both components (e.g., coins of two types).
\( \text{Quantity}_1 \cdot \text{Value}_1 + \text{Quantity}_2 \cdot \text{Value}_2 = \text{Total Value} \)
📝 Worked Example
You have 10 coins (nickels and dimes) worth $0.80.
Let \(n\) = nickels. Then dimes \(= 10 - n\).
\(5n + 10(10-n) = 80 \Rightarrow -5n = -20 \Rightarrow n = 4\)
Let \(n\) = nickels. Then dimes \(= 10 - n\).
\(5n + 10(10-n) = 80 \Rightarrow -5n = -20 \Rightarrow n = 4\)
A piggy bank contains 30 coins, all quarters and dimes, totaling $5.85. How many quarters are in the bank?
✅ Solution
Let \(q\) = quarters, \(d = 30 - q\) = dimes.
\(25q + 10(30-q) = 585\)
\(25q + 300 - 10q = 585\)
\(15q = 285\)
\(q = \mathbf{19}\) quarters ✓
\(25q + 10(30-q) = 585\)
\(25q + 300 - 10q = 585\)
\(15q = 285\)
\(q = \mathbf{19}\) quarters ✓
Part II — Geometry
Questions 11–20
11
Pythagorean Theorem
★★★★★
📐 Core Concept · Pythagorean Theorem
In a right triangle, the square of the hypotenuse equals the sum of the squares of the legs.
⚠️ Most common mistake: Adding legs then squaring, rather than squaring first then adding.
⚠️ Most common mistake: Adding legs then squaring, rather than squaring first then adding.
\( a^2 + b^2 = c^2 \quad \text{(where } c \text{ is the hypotenuse)} \)
📝 Worked Example
Legs: 6 and 8. Find hypotenuse.
\(6^2 + 8^2 = 36 + 64 = 100\). So \(c = \sqrt{100} = 10\).
\(6^2 + 8^2 = 36 + 64 = 100\). So \(c = \sqrt{100} = 10\).
A right triangle has legs of length 9 and 40. What is the length of the hypotenuse?
✅ Solution
\(9^2 + 40^2 = 81 + 1600 = 1681\)
\(c = \sqrt{1681} = \mathbf{41}\) ✓
(9, 40, 41) is a Pythagorean triple — worth memorizing!
\(c = \sqrt{1681} = \mathbf{41}\) ✓
(9, 40, 41) is a Pythagorean triple — worth memorizing!
12
Area & Perimeter
★★★★★
📐 Core Concept · Composite Figures
For composite (combined) shapes: add or subtract simple areas. Break the figure into rectangles, triangles, or circles.
⚠️ Most common mistake: Calculating perimeter of sub-shapes separately instead of the outer boundary only.
⚠️ Most common mistake: Calculating perimeter of sub-shapes separately instead of the outer boundary only.
\( A_{\text{composite}} = A_{\text{large}} - A_{\text{removed}} \)
📝 Worked Example
Rectangle 10×8 with a 3×3 square cutout: \(A = 80 - 9 = 71 \text{ units}^2\)
A rectangle measures 14 cm by 10 cm. A semicircle with diameter 10 cm is cut out from one of the shorter sides. What is the area of the remaining figure? (Use \(\pi \approx 3.14\))
✅ Solution
Rectangle area: \(14 \times 10 = 140 \text{ cm}^2\)
Semicircle radius: \(r = 5 \text{ cm}\)
Semicircle area: \(\dfrac{1}{2}\pi r^2 = \dfrac{1}{2}(3.14)(25) = 39.25 \text{ cm}^2\)
Remaining: \(140 - 39.25 = \mathbf{100.75 \text{ cm}^2}\) ✓
Semicircle radius: \(r = 5 \text{ cm}\)
Semicircle area: \(\dfrac{1}{2}\pi r^2 = \dfrac{1}{2}(3.14)(25) = 39.25 \text{ cm}^2\)
Remaining: \(140 - 39.25 = \mathbf{100.75 \text{ cm}^2}\) ✓
13
Parallel Lines & Angles
★★★★★
📐 Core Concept · Transversal Angle Pairs
When a transversal crosses parallel lines:
• Alternate interior angles are equal
• Co-interior (same-side interior) angles are supplementary (sum to 180°)
⚠️ Most common mistake: Confusing alternate interior with co-interior angles.
• Alternate interior angles are equal
• Co-interior (same-side interior) angles are supplementary (sum to 180°)
⚠️ Most common mistake: Confusing alternate interior with co-interior angles.
\( \angle_{\text{alt. int.}} = \angle_{\text{alt. int.}} \qquad \angle_{\text{co-int.}} + \angle_{\text{co-int.}} = 180° \)
📝 Worked Example
Two parallel lines cut by a transversal. One angle is 70°. Its alternate interior angle = 70°. Its co-interior angle = 110°.
Two parallel lines are cut by a transversal. One of the co-interior (same-side interior) angles is \((3x + 15)°\) and the other is \((2x + 5)°\). Find the value of \(x\).
✅ Solution
Co-interior angles sum to 180°:
\((3x+15) + (2x+5) = 180\)
\(5x + 20 = 180\)
\(5x = 160\)
\(x = \mathbf{32}\) ✓
\((3x+15) + (2x+5) = 180\)
\(5x + 20 = 180\)
\(5x = 160\)
\(x = \mathbf{32}\) ✓
14
Similar Triangles
★★★★★
📐 Core Concept · Similar Triangles
Similar triangles have equal angles and proportional sides. Set up a ratio: corresponding sides are in the same proportion.
⚠️ Most common mistake: Matching non-corresponding sides in the proportion.
⚠️ Most common mistake: Matching non-corresponding sides in the proportion.
\( \dfrac{a}{a'} = \dfrac{b}{b'} = \dfrac{c}{c'} \)
📝 Worked Example
Triangles with sides 3, 4, 5 and 6, ?, 10. Missing side: \(\frac{4}{?} = \frac{3}{6} \Rightarrow ? = 8\).
Triangle ABC is similar to Triangle DEF. In △ABC, \(AB = 8\), \(BC = 12\), \(AC = 10\). In △DEF, \(DE = 12\). What is the length of \(EF\)?
✅ Solution
Scale factor: \(\dfrac{DE}{AB} = \dfrac{12}{8} = \dfrac{3}{2}\)
\(EF\) corresponds to \(BC = 12\):
\(EF = 12 \times \dfrac{3}{2} = \mathbf{18}\) ✓
\(EF\) corresponds to \(BC = 12\):
\(EF = 12 \times \dfrac{3}{2} = \mathbf{18}\) ✓
15
Circle — Arc & Sector
★★★★★
📐 Core Concept · Sector Area
A sector is a "pie slice" of a circle. Its area is a fraction of the full circle area.
⚠️ Most common mistake: Using the diameter instead of the radius in the formula.
⚠️ Most common mistake: Using the diameter instead of the radius in the formula.
\( A_{\text{sector}} = \dfrac{\theta}{360°} \times \pi r^2 \)
📝 Worked Example
Circle radius 6, central angle 90°.
\(A = \dfrac{90}{360} \times \pi(36) = 9\pi \approx 28.26\)
\(A = \dfrac{90}{360} \times \pi(36) = 9\pi \approx 28.26\)
A circle has a radius of 10 cm. A sector has a central angle of 144°. What is the area of the sector? (Use \(\pi \approx 3.14\))
✅ Solution
\( A = \dfrac{144}{360} \times 3.14 \times 10^2 \)
\( = 0.4 \times 314 = \mathbf{125.6 \text{ cm}^2} \) ✓
\( = 0.4 \times 314 = \mathbf{125.6 \text{ cm}^2} \) ✓
16
Volume — Cylinders
★★★★★
📐 Core Concept · Volume of a Cylinder
Volume = base area × height. For a cylinder, base area = \(\pi r^2\).
⚠️ Most common mistake: Using diameter instead of radius in \(\pi r^2\).
⚠️ Most common mistake: Using diameter instead of radius in \(\pi r^2\).
\( V = \pi r^2 h \)
📝 Worked Example
Cylinder: diameter = 8 cm, height = 5 cm.
\(r = 4\), \(V = \pi(16)(5) = 80\pi \approx 251.2 \text{ cm}^3\)
\(r = 4\), \(V = \pi(16)(5) = 80\pi \approx 251.2 \text{ cm}^3\)
A cylindrical water tank has a diameter of 6 m and a height of 10 m. What is its volume in cubic meters? (Use \(\pi \approx 3.14\))
✅ Solution
Diameter = 6 m, so radius \(r = 3\) m.
\(V = 3.14 \times 3^2 \times 10 = 3.14 \times 9 \times 10 = \mathbf{282.6 \text{ m}^3}\) ✓
Common error: Using diameter (6) instead of radius (3) → gives \(1130.4\).
\(V = 3.14 \times 3^2 \times 10 = 3.14 \times 9 \times 10 = \mathbf{282.6 \text{ m}^3}\) ✓
Common error: Using diameter (6) instead of radius (3) → gives \(1130.4\).
17
Coordinate Geometry
★★★★★
📐 Core Concept · Midpoint & Distance
Midpoint formula gives the center between two points. Distance formula uses the Pythagorean theorem.
⚠️ Most common mistake: Averaging only x-coordinates and forgetting y-coordinates.
⚠️ Most common mistake: Averaging only x-coordinates and forgetting y-coordinates.
\( M = \left(\dfrac{x_1+x_2}{2},\,\dfrac{y_1+y_2}{2}\right) \qquad d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \)
📝 Worked Example
Points A(2, 3) and B(8, 7):
Midpoint: \(\left(\frac{10}{2}, \frac{10}{2}\right) = (5, 5)\)
Distance: \(\sqrt{36+16} = \sqrt{52} = 2\sqrt{13}\)
Midpoint: \(\left(\frac{10}{2}, \frac{10}{2}\right) = (5, 5)\)
Distance: \(\sqrt{36+16} = \sqrt{52} = 2\sqrt{13}\)
Point M is the midpoint of segment AB. If \(A = (-3, 7)\) and \(M = (2, 1)\), what are the coordinates of point B?
✅ Solution
Use midpoint formula backwards:
\(x_B = 2(2) - (-3) = 4 + 3 = 7\)
\(y_B = 2(1) - 7 = 2 - 7 = -5\)
\(B = \mathbf{(7, -5)}\) ✓
\(x_B = 2(2) - (-3) = 4 + 3 = 7\)
\(y_B = 2(1) - 7 = 2 - 7 = -5\)
\(B = \mathbf{(7, -5)}\) ✓
18
Triangle Angle Sum
★★★★★
📐 Core Concept · Exterior Angle Theorem
An exterior angle of a triangle equals the sum of the two non-adjacent interior angles.
⚠️ Most common mistake: Setting the exterior angle equal to just one interior angle.
⚠️ Most common mistake: Setting the exterior angle equal to just one interior angle.
\( \angle_{\text{exterior}} = \angle_A + \angle_B \quad \text{(non-adjacent interior angles)} \)
📝 Worked Example
Exterior angle = 110°. One non-adjacent angle = 65°.
Other angle = \(110° - 65° = 45°\).
Other angle = \(110° - 65° = 45°\).
In triangle PQR, an exterior angle at vertex R measures \((5x + 10)°\). The two non-adjacent interior angles measure \((2x + 5)°\) and \((x + 25)°\). Find \(x\).
✅ Solution
Exterior angle = sum of non-adjacent interior angles:
\((5x+10) = (2x+5) + (x+25)\)
\(5x+10 = 3x+30\)
\(2x = 20\)
\(x = \mathbf{10}\) ✓
\((5x+10) = (2x+5) + (x+25)\)
\(5x+10 = 3x+30\)
\(2x = 20\)
\(x = \mathbf{10}\) ✓
19
Surface Area
★★★★★
📐 Core Concept · Surface Area of a Cone
Total surface area of a cone = lateral area + base area. The slant height \(l\) must be used (not the vertical height \(h\)).
⚠️ Most common mistake: Using vertical height instead of slant height for lateral area.
⚠️ Most common mistake: Using vertical height instead of slant height for lateral area.
\( SA = \pi r l + \pi r^2 \quad \text{where } l = \sqrt{r^2 + h^2} \)
📝 Worked Example
Cone: radius = 3, height = 4.
\(l = \sqrt{9+16} = 5\)
\(SA = \pi(3)(5) + \pi(9) = 15\pi + 9\pi = 24\pi\)
\(l = \sqrt{9+16} = 5\)
\(SA = \pi(3)(5) + \pi(9) = 15\pi + 9\pi = 24\pi\)
A cone has a radius of 5 cm and a vertical height of 12 cm. What is the total surface area of the cone? (Use \(\pi \approx 3.14\))
✅ Solution
Slant height: \(l = \sqrt{5^2 + 12^2} = \sqrt{25+144} = \sqrt{169} = 13 \text{ cm}\)
Lateral area: \(\pi(5)(13) = 65\pi \approx 204.1\)
Base area: \(\pi(5)^2 = 25\pi \approx 78.5\)
Total: \(204.1 + 78.5 = \mathbf{282.6 \text{ cm}^2}\) ✓
Lateral area: \(\pi(5)(13) = 65\pi \approx 204.1\)
Base area: \(\pi(5)^2 = 25\pi \approx 78.5\)
Total: \(204.1 + 78.5 = \mathbf{282.6 \text{ cm}^2}\) ✓
20
Transformations · Congruence
★★★★★
📐 Core Concept · Transformations & Coordinates
Reflection over y-axis: \((x, y) \to (-x, y)\)
Rotation 90° CCW about origin: \((x, y) \to (-y, x)\)
⚠️ Most common mistake: Swapping x and y without changing signs for rotation.
Rotation 90° CCW about origin: \((x, y) \to (-y, x)\)
⚠️ Most common mistake: Swapping x and y without changing signs for rotation.
\( R_{90°\text{ CCW}}(x, y) = (-y,\; x) \)
📝 Worked Example
Rotate (3, 2) by 90° CCW about origin:
\((3,2) \to (-2, 3)\)
\((3,2) \to (-2, 3)\)
Point \(P(4, -6)\) is first reflected over the \(x\)-axis, and the result is then rotated 90° counterclockwise about the origin. What are the final coordinates of the image point?
✅ Solution
Step 1 — Reflect over x-axis: \((x,y)\to(x,-y)\)
\(P(4,-6) \to P'(4,6)\)
Step 2 — Rotate 90° CCW: \((x,y)\to(-y,x)\)
\(P'(4,6) \to \mathbf{(-6,4)}\) ✓
\(P(4,-6) \to P'(4,6)\)
Step 2 — Rotate 90° CCW: \((x,y)\to(-y,x)\)
\(P'(4,6) \to \mathbf{(-6,4)}\) ✓