Grade 10 — Term 1
10 study notes · 300 practice questions · Aligned with NIE syllabus
Counted in the Idasara Academy app, which you open with a free account. The free notes, revision, practice and library pages for each lesson are linked below.
This term in the Idasara Academy app has 10 lessons (NIE Units 1, 2, 3, 4, 5, 6, 7, 8, 9, 10). For each lesson, this page gives you the points you must know and one exam-style question with its answer. None of it needs an account.
Is this what my class is doing now?
Maybe not. The Idasara app groups lessons into its own terms, and your school may teach them in a different order. Where we know a lesson's NIE unit, it is shown on the lesson. Find the unit your class is on, and start there.
Lessons in This Term
Perimeter
What you must know
- The missing radius trap: The perimeter of a sector is the arc length plus both radii — leave out +2r and the figure you have given is the arc alone.
- Compound shape internal lines: Shaded or dashed internal boundaries must strictly be excluded from perimeter calculations.
- Value of π: Use π = 22/7 when radius or diameter is a multiple of 7; otherwise use the specified value and state exact units (cm, m).
Square Root
What you must know
- Digit pairing rule: Grouping MUST start from the decimal point: whole numbers paired right-to-left (e.g., 1,44.00), decimals paired left-to-right (e.g., 0.05,20).
- Decimal place precision: To find square root correct to 2 decimal places, continue division until 3 decimal digits are obtained, then round appropriately.
- Square root of fractions: √(a/b) = √a / √b; if denominator is not a perfect square, rationalize denominator first: √(a/b) = √(ab) / b.
Try one question
Find the square root of 53 to the first decimal place by division or approximation method.
- A. 7.5
- B. 7.3
- C. 7.7
- D. 7.1
Show the answer
Answer: B. 7.3. Since 7² = 49 and 8² = 64, √53 lies between 7 and 8. Testing 7.3² = 53.29 and 7.2² = 51.84. 53 is closer to 53.29, so √53 ≈ 7.3 to one decimal place.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Fractions
What you must know
- The 'remaining' fraction trap in Paper 2 Part A: If 1/4 is spent, the remainder is 3/4; spending 1/3 OF THE REMAINDER equals (1/3) × (3/4) = 1/4, NOT 1/3 of the total.
- Reciprocal in division: Never invert the dividend (the first fraction); strictly invert the divisor (the second fraction).
- Mixed numbers conversion: Always convert mixed numbers to improper fractions before performing multiplication or division.
Try one question
Simplify: $\left(\frac{3}{4} + \frac{1}{6}\right) \div \left(\frac{5}{8} - \frac{1}{4}\right)$
- A. $2\frac{4}{5}$
- B. $1\frac{4}{9}$
- C. $2\frac{2}{9}$
- D. $2\frac{4}{9}$
Show the answer
Answer: D. $2\frac{4}{9}$. Evaluate brackets first, then divide by multiplying by the reciprocal. The correct simplification gives $\frac{22}{9} = 2\frac{4}{9}$.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Binomial Expressions
What you must know
- The dropped middle term trap: Writing (a + b)² = a² + b² or (a - b)² = a² - b² is a severe algebraic blunder; never omit the middle term ±2ab.
- Sign rule for (a - b)²: The final term +b² is ALWAYS positive because (-b)² = +b²; only the middle term -2ab is negative.
- Mental arithmetic evaluation: Use algebraic identities for rapid numerical calculations, such as 102² = (100 + 2)² = 10000 + 400 + 4 = 10404.
Try one question
Expand and simplify the algebraic product: (2x - 3)(3x + 4).
- A. 6x² - 12
- B. 6x² - x - 12
- C. 6x² - 5x - 12
- D. 6x² + x - 12
Show the answer
Answer: B. 6x² - x - 12. Using distributive law: (2x)(3x) + (2x)(4) + (-3)(3x) + (-3)(4) = 6x² + 8x - 9x - 12 = 6x² - x - 12.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Congruence of Triangles
What you must know
- The non-included angle trap in SAS: The angle MUST be between the two equal sides; SSA (Side-Side-Angle) is NOT a valid congruence criterion.
- State reasons for every statement: In formal geometric proofs (Paper 2 Part B), every stated equality must cite a valid geometric reason (e.g., 'common side', 'vertically opposite angles', 'radii of same circle').
- Deductions from congruence: Once triangles are proven congruent, state corresponding elements clearly (e.g., 'AB = XY (corresponding sides of congruent triangles)').
Try one question
In triangles ABC and PQR, AB = PQ, BC = QR, and angle B = angle Q. Under which geometric congruence condition are triangles ABC and PQR congruent?
- A. Side-Angle-Side (SAS)
- B. Side-Side-Side (SSS)
- C. Angle-Side-Angle (ASA)
- D. Right angle-Hypotenuse-Side (RHS)
Show the answer
Answer: A. Side-Angle-Side (SAS). Two pairs of corresponding sides are equal (AB = PQ, BC = QR) and the included angle between them is equal (angle B = angle Q). This satisfies the Side-Angle-Side (SAS) congruence condition.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Area
What you must know
- Perpendicular height requirement: In triangles, parallelograms, and trapeziums, ALWAYS use the vertical perpendicular height (h), NEVER the slanted edge length.
- Area units: State final answers in square units (cm², m²); never write linear units (cm, m) for area.
- Shaded area problems (Paper 1 & Paper 2): Clearly state the outer shape formula minus the inner cutout formula before substituting numerical values.
Try one question
A trapezium has parallel sides of lengths 12 cm and 18 cm. If the perpendicular distance between the parallel sides is 8 cm, what is the area of the trapezium?
- A. 120 cm²
- B. 96 cm²
- C. 144 cm²
- D. 240 cm²
Show the answer
Answer: A. 120 cm². The area of a trapezium is given by A = 1/2 × (a + b) × h, where a and b are parallel sides and h is height. A = 1/2 × (12 + 18) × 8 = 1/2 × 30 × 8 = 120 cm².
A model question written by Idasara on the exam pattern. It is not from a past paper.
Factors of Quadratic Expressions
What you must know
- Sign errors in middle-term splitting: If c is positive, both p and q share the sign of b; if c is negative, p and q have opposite signs, with the larger magnitude matching the sign of b.
- Incomplete factorisation penalty: Expressions like 2x² - 8 must be factored completely as 2(x² - 4) = 2(x - 2)(x + 2); leaving 2(x² - 4) forfeits full marks.
- Negative leading coefficient: When factoring -x² + bx + c, extract -1 first: -(x² - bx - c).
Triangles
What you must know
- Exterior angle misidentification trap: The exterior angle is formed between the produced line and the adjacent side, NOT the reflex angle or vertical opposite angle.
- Proof justification in angle-chasing problems: In Paper 1 Part A, always cite geometric theorems concisely (e.g., 'ext. angle of △', 'angles on a straight line = 180°', 'angles opposite equal sides').
- Triangle existence check: To verify if side lengths (a, b, c) can form a triangle, verify that the sum of the two smaller sides exceeds the longest side.
Try one question
In triangle ABC, the side BC is produced to D. If angle ABC = 55° and angle BAC = 65°, find the magnitude of the exterior angle ACD.
- A. 60°
- B. 110°
- C. 120°
- D. 130°
Show the answer
Answer: C. 120°. By the exterior angle theorem of a triangle, the exterior angle formed by producing a side equals the sum of the two interior opposite angles: angle ACD = angle ABC + angle BAC = 55° + 65° = 120°.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Inverse Proportion
What you must know
- Direct vs Inverse proportion trap: In direct proportion y/x = constant (cross-multiply); in inverse proportion x × y = constant (multiply across rows).
- Added workers problem: If 6 workers take 10 days, total = 60 man-days; to finish in 4 days requires 60 / 4 = 15 workers; the ADDED workers needed = 15 - 6 = 9 workers.
- Units in rates: Ensure time units (hours, days, weeks) are consistent across all terms before equating products.
Try one question
6 workers take 12 days to complete constructing a boundary wall. How many days will it take for 9 workers working at the same rate to complete the same wall?
- A. 6 days
- B. 9 days
- C. 10 days
- D. 8 days
Show the answer
Answer: D. 8 days. Work is inversely proportional: Total man-days = 6 workers × 12 days = 72 man-days. Days required by 9 workers = 72 / 9 = 8 days.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Data Representation
What you must know
- The 360° total validation: The sum of all calculated sector angles in a pie chart MUST equal exactly 360°; check your angle arithmetic before drawing.
- Protractor precision in Paper 1 Part B & Paper 2: Align the protractor origin precisely with the circle centre and baseline; inaccurate angles lose plotting marks.
- Label every sector: Every sector in a pie chart must clearly display its category name and calculated angle or frequency.
Try one question
In a pie chart representing the preferred subjects of 720 students, the sector angle for Mathematics is 80°. How many students prefer Mathematics?
- A. 200
- B. 180
- C. 140
- D. 160
Show the answer
Answer: D. 160. Total angle of a pie chart is 360°. Number of students = (sector angle / 360°) × total students = (80 / 360) × 720 = 2 × 80 = 160 students.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Before you sign up
What does it cost?
I already go to a Mathematics tuition class
Then you are paying for Mathematics by the subject. Whether an Idasara plan would cost you less depends on how many classes you go to, so put in your own numbers.
Work it out on the pricing page →