Class 10 Maths Notes — All 14 Chapters (CBSE 2026-27)
All 14 chapters
CBSE Class 10 Maths syllabus and marks at a glance
The current CBSE curriculum for Class 10 Mathematics has 14 chapters spread across 7 units. The theory paper is for 80 marks and the internal assessment adds 20 marks, for a total of 100. CBSE does not publish official chapter-wise marks, but it does publish unit-wise weightage, which is what the table below shows.
| Unit | Theory marks | Chapters |
|---|---|---|
| Number Systems | 6 marks | Real Numbers |
| Algebra | 20 marks | Polynomials, Pair of Linear Equations in Two Variables, Quadratic Equations, Arithmetic Progressions |
| Coordinate Geometry | 6 marks | Coordinate Geometry |
| Geometry | 15 marks | Triangles, Circles |
| Trigonometry | 12 marks | Introduction to Trigonometry, Some Applications of Trigonometry |
| Mensuration | 10 marks | Areas Related to Circles, Surface Areas and Volumes |
| Statistics and Probability | 11 marks | Statistics, Probability |
| Total | 80 | 14 chapters |
Source: CBSE Mathematics curriculum for 2026-27 (cbseacademic.nic.in). Weightages can change; always check the official PDF for your session.
Real Numbers
Number Systems · 6 marksThe chapter that decides whether you are comfortable with proofs. Everything here comes from prime factorisation and the difference between rational and irrational numbers.
Concepts you must be able to explain
- Euclid’s division lemma: for any positive integers a and b, there are unique integers q and r such that a = bq + r, where 0 ≤ r < b.
- The Fundamental Theorem of Arithmetic says every composite number can be written as a product of primes in exactly one way (ignoring order). That uniqueness is what makes HCF and LCM well defined.
- HCF is the product of the smallest power of each common prime factor; LCM is the product of the largest power of each prime factor that appears.
- Irrationality proofs (√2, √3, √5 and combinations such as 3 + 2√5) are always by contradiction: assume the number is p/q in lowest terms, square or rearrange, and show the assumption forces a common factor.
- A rational number p/q in lowest terms has a terminating decimal only when q is of the form 2m5n. Otherwise the decimal repeats.
Formulas and rules
| Name | Formula / rule | When to use it |
|---|---|---|
| Euclid’s division lemma | a = bq + r, 0 ≤ r < b | Used for HCF by repeated division. |
| HCF–LCM relation | HCF(a, b) × LCM(a, b) = a × b | Only for two positive integers, not three. |
| Terminating decimal test | q = 2m5n | After reducing p/q to lowest terms. |
| Prime factorisation | HCF = product of smallest powers; LCM = product of highest powers | Write both numbers as products of primes first. |
Common mistakes
- Using HCF × LCM = product of numbers for three numbers. It is a two-number identity.
- Writing √2 = 1.414 and concluding it is rational. A decimal approximation says nothing about rationality.
- Forgetting to state that p and q are coprime at the start of a contradiction proof — an examiner looks for that line.
- Claiming 7/30 is terminating because 30 looks small; 30 = 2 × 3 × 5 has a factor 3, so it repeats.
Exam tip: In board proofs write the assumption, the squaring step, the divisibility argument and the contradiction as four separate lines. Method marks are given line by line.
Polynomials
Algebra · 20 marksShort chapter, high return. The relationship between zeroes and coefficients is almost guaranteed to appear, usually with a division-algorithm question alongside.
Concepts you must be able to explain
- The degree of the polynomial decides the maximum number of zeroes; on a graph, zeroes are the x-coordinates where the curve meets the x-axis.
- For a quadratic ax² + bx + c, the sum of zeroes is −b/a and the product is c/a. For a cubic ax³ + bx² + cx + d, the three relations alternate in sign.
- If the zeroes are given, the quadratic can be rebuilt as x² − (sum)x + (product) — or k times that expression for any non-zero k.
- Division algorithm: p(x) = g(x)q(x) + r(x), where the degree of r(x) is less than the degree of g(x). If r(x) = 0, then g(x) is a factor of p(x).
- Graph questions expect you to read the number of zeroes, not the exact values, unless the grid is clearly marked.
Formulas and rules
| Name | Formula / rule | When to use it |
|---|---|---|
| Quadratic zeroes | α + β = −b/a, αβ = c/a | Signs matter — the sum is negative b over a. |
| Build from zeroes | x² − (α + β)x + αβ | Multiply by any constant k for a family of answers. |
| Cubic zeroes | α + β + γ = −b/a; αβ + βγ + γα = c/a; αβγ = −d/a | Signs alternate: −, +, −. |
| Division algorithm | p(x) = g(x)q(x) + r(x) | deg r < deg g. |
Common mistakes
- Writing α + β = b/a and losing the minus sign.
- Forgetting to divide the x-term by the leading coefficient when using cubic relations.
- Treating a zero of the polynomial as a zero of a coefficient — the value of p(x) at x = α must actually be 0.
- Adding a remainder instead of subtracting when the divisor is written as (x − k).
Exam tip: Always verify your zeroes by substituting back into p(x). It takes twenty seconds and catches nearly every sign slip.
Pair of Linear Equations in Two Variables
Algebra · 20 marksA dependable scoring chapter. Consistency conditions, one graphical question and a word problem cover almost every paper.
Concepts you must be able to explain
- Two lines can intersect at one point (unique solution, consistent), coincide (infinitely many solutions, dependent) or be parallel (no solution, inconsistent).
- Compare a₁/a₂, b₁/b₂ and c₁/c₂ to decide the case before solving anything. This is the fastest mark in the chapter.
- Algebraic methods — substitution, elimination and cross-multiplication — all give the same answer; choose the one that keeps the numbers small.
- A word problem becomes two equations. Define both variables in words first, then translate each sentence into one equation.
- Graph questions need a small table of at least two points per line, a correctly labelled pair of axes and the coordinates of the intersection written out.
Formulas and rules
| Name | Formula / rule | When to use it |
|---|---|---|
| Unique solution | a₁/a₂ ≠ b₁/b₂ | Intersecting lines, consistent. |
| Infinitely many | a₁/a₂ = b₁/b₂ = c₁/c₂ | Coincident lines, dependent. |
| No solution | a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | Parallel lines, inconsistent. |
| Cross-multiplication | x/(b₁c₂ − b₂c₁) = y/(c₁a₂ − c₂a₁) = 1/(a₁b₂ − a₂b₁) | Check the determinant is not zero first. |
Common mistakes
- Comparing only the coefficients of x and y and forgetting the constant term when testing for infinite solutions.
- Plotting lines from only one point, so a wrong slope goes unnoticed.
- Mixing up x and y while cross-multiplying.
- Giving a negative answer for a quantity such as age, length or number of items without checking whether it makes sense.
Exam tip: For word problems, write the answer as a sentence at the end: ‘The cost of one pen is ₹12 and one notebook is ₹25.’ Board marking schemes include that line.
Quadratic Equations
Algebra · 20 marksLearn all three solving methods, but rely on factorisation when the numbers are friendly and the quadratic formula when they are not.
Concepts you must be able to explain
- Standard form is ax² + bx + c = 0 with a ≠ 0. Rewrite every equation in this form before choosing a method.
- Factorisation: split the middle term into two parts whose product is ac and whose sum is b.
- Completing the square leads directly to the quadratic formula; you should be able to reproduce the derivation in three lines.
- The discriminant D = b² − 4ac decides the nature of the roots: two distinct real roots, two equal real roots, or no real roots.
- Word problems on speed, area, ages and upstream/downstream motion all reduce to a quadratic — reject the root that is physically impossible.
Formulas and rules
| Name | Formula / rule | When to use it |
|---|---|---|
| Quadratic formula | x = (−b ± √(b² − 4ac)) / 2a | Works for every quadratic, including irrational roots. |
| Discriminant | D = b² − 4ac | D > 0 distinct real; D = 0 equal real; D < 0 no real roots. |
| Zeroes | α + β = −b/a, αβ = c/a | Useful for checking your answers. |
| Completing the square | (x + b/2a)² = (b² − 4ac)/4a² | The derivation of the formula. |
Common mistakes
- Dropping the ± and reporting only one root.
- Using the formula with the wrong sign for b when b is already negative.
- Forgetting that a quadratic must be rearranged before reading off a, b and c.
- Keeping a negative time or length in a word problem because the algebra allowed it.
Exam tip: If the discriminant is a perfect square, the roots will be rational and factorisation will work. Check that first and save time.
Arithmetic Progressions
Algebra · 20 marksThe most predictable chapter in the paper. Identify a and d, then choose between the nth-term and sum formulas.
Concepts you must be able to explain
- An AP is a list where the difference between consecutive terms is constant. That constant is d = a₂ − a₁.
- The nth term is a + (n − 1)d. The number of terms is often the hidden question — find n from the last term.
- Sum of n terms: Sₙ = n/2 [2a + (n − 1)d]. When the last term l is known, Sₙ = n/2 (a + l) is much faster.
- aₙ = Sₙ − Sₙ₋₁ lets you recover a term from a sum formula.
- Applications include savings plans, salary increments, seating rows and the sum of multiples of a number in a range.
Formulas and rules
| Name | Formula / rule | When to use it |
|---|---|---|
| nth term | aₙ = a + (n − 1)d | Check whether n is the term number or the term value. |
| Sum of n terms | Sₙ = n/2 [2a + (n − 1)d] | Use when d is known. |
| Sum with last term | Sₙ = n/2 (a + l) | Faster when the last term is given. |
| Number of terms | n = (l − a)/d + 1 | Handy for ‘how many multiples…’ questions. |
| Term from sums | aₙ = Sₙ − Sₙ₋₁ | Valid for n ≥ 2. |
Common mistakes
- Using n instead of (n − 1) in the nth-term formula — the classic one-mark loss.
- Miscounting the number of terms in a range such as 100 to 500.
- Using the sum formula when the question asks for a single term.
- Forgetting that d can be negative or a fraction.
Exam tip: Write a and d at the top of your rough work every single time. Most AP mistakes disappear before they happen.
Triangles
Geometry · 15 marksProof-heavy and worth the effort. Similarity criteria, the Basic Proportionality Theorem and Pythagoras carry the chapter.
Concepts you must be able to explain
- Two triangles are similar if their corresponding angles are equal and corresponding sides are in the same ratio. For the exam, AA, SSS and SAS are the usable criteria — SSA is not.
- Basic Proportionality Theorem (Thales): a line parallel to one side of a triangle divides the other two sides in the same ratio; the converse is also true.
- Areas of similar triangles are in the ratio of the squares of corresponding sides — not the sides themselves.
- Pythagoras theorem: in a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. Its converse identifies a right angle.
- Write vertex correspondence carefully. If △ABC ~ △DEF, then AB/DE = BC/EF = CA/FD and ∠A = ∠D, and so on.
Formulas and rules
| Name | Formula / rule | When to use it |
|---|---|---|
| Basic Proportionality Theorem | DE ∥ BC ⇒ AD/DB = AE/EC | Also true with AD/AB = AE/AC. |
| Area ratio | area(△ABC)/area(△PQR) = (AB/PQ)² | Square the side ratio. |
| Pythagoras | AC² = AB² + BC² | AC is the hypotenuse, opposite the right angle. |
| Converse of Pythagoras | If AC² = AB² + BC², then ∠B = 90° | Used to prove a triangle is right-angled. |
Common mistakes
- Treating SSA as a similarity criterion.
- Writing the area ratio as the ratio of sides instead of the ratio of squares.
- Matching the wrong vertices when naming similar triangles.
- Assuming a triangle is right-angled from a rough diagram without checking the squares.
Exam tip: Before writing a proof, mark equal angles on the diagram and write the correspondence in the same order on both sides of the ‘~’ symbol.
Coordinate Geometry
Coordinate Geometry · 6 marksFormula-driven and quick to revise. Distance, section formula and area of a triangle are the only three tools you need.
Concepts you must be able to explain
- The distance formula comes straight from Pythagoras. It also proves that three points form an isosceles, equilateral or right triangle.
- The section formula finds the point dividing a segment in a given ratio internally; the midpoint formula is the special case 1:1.
- The area of a triangle from coordinates uses a determinant-like expression with absolute value. If the area is zero, the points are collinear.
- Points on the x-axis have y = 0; points on the y-axis have x = 0. Many questions become trivial once you use that.
- For external division use (m₁x₂ − m₂x₁)/(m₁ − m₂) — fewer papers ask it, but it is in the syllabus.
Formulas and rules
| Name | Formula / rule | When to use it |
|---|---|---|
| Distance | d = √((x₂ − x₁)² + (y₂ − y₁)²) | Squared distances avoid the square root. |
| Section (internal) | ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)) | Ratio m₁ : m₂ measured from the first point. |
| Midpoint | ((x₁ + x₂)/2, (y₁ + y₂)/2) | Special case of 1 : 1. |
| Area of triangle | ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)| | Zero area ⇒ collinear points. |
Common mistakes
- Dropping a negative sign when subtracting coordinates.
- Swapping the x and y coordinates in the section formula.
- Omitting the absolute value and reporting a negative area.
- Using the midpoint formula when a ratio other than 1:1 is given.
Exam tip: Sketch the points roughly. It takes ten seconds and immediately shows whether your answer is in the right region of the plane.
Introduction to Trigonometry
Trigonometry · 12 marksSix ratios, one table and three identities. Build the table from two triangles rather than memorising it blindly.
Concepts you must be able to explain
- In a right triangle, sin θ = perpendicular/hypotenuse, cos θ = base/hypotenuse, tan θ = perpendicular/base. The other three are reciprocals.
- The 30°–60°–90° triangle (sides 1, √3, 2) and the 45°–45°–90° triangle (sides 1, 1, √2) generate the whole standard-value table.
- Complementary angles: sin(90° − θ) = cos θ, tan(90° − θ) = cot θ, sec(90° − θ) = cosec θ.
- The three Pythagorean identities let you convert any expression into sines and cosines, which is the standard way to prove an identity.
- When proving, work on the more complicated side until it matches the simpler one. Do not treat the identity as an equation and cross-multiply.
Formulas and rules
| Name | Formula / rule | When to use it |
|---|---|---|
| Basic ratios | sin θ = P/H, cos θ = B/H, tan θ = P/B | ‘Pandit Badri Prasad Har Har’ is the usual mnemonic. |
| Reciprocals | cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ | sin pairs with cosec, not sec. |
| Quotient | tan θ = sin θ/cos θ, cot θ = cos θ/sin θ | Useful for converting everything to sin and cos. |
| Pythagorean identities | sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ | Divide the first by cos²θ or sin²θ to get the others. |
| Standard values | sin: 0, ½, 1/√2, √3/2, 1 for 0°, 30°, 45°, 60°, 90° | cos runs the same list backwards. |
Common mistakes
- Reading sin²θ as sin(θ²).
- Mixing up sec and cosec values at 0° and 90°.
- Using the wrong side as the base after rotating the diagram.
- Cross-multiplying while proving an identity instead of transforming one side.
Exam tip: If you blank on the table, sketch the 30-60-90 and 45-45-90 triangles and read the ratios off them. They take fifteen seconds to draw.
Some Applications of Trigonometry
Trigonometry · 12 marksHeights and distances. Almost every mark is won or lost in the diagram, not in the trigonometry.
Concepts you must be able to explain
- The angle of elevation is measured upward from the horizontal to the object; the angle of depression is measured downward from the horizontal.
- The angle of depression from a point equals the angle of elevation from the object — alternate angles between parallel horizontals.
- Most problems reduce to tan θ = opposite/adjacent. Choose the ratio that links the known length to the unknown one.
- Two-observer problems produce two right triangles sharing a side. Write both equations and eliminate the common length.
- Keep the observer’s height in mind: if a person of height h is looking at a tower, the triangle starts at eye level.
Formulas and rules
| Name | Formula / rule | When to use it |
|---|---|---|
| Basic relation | tan θ = height / horizontal distance | The workhorse of the chapter. |
| Useful values | tan 30° = 1/√3, tan 45° = 1, tan 60° = √3 | Standard exam angles. |
| Two triangles | h = x tan α = (x + d) tan β | Eliminate x to solve for h. |
Common mistakes
- Putting the angle of elevation inside the triangle at the wrong vertex.
- Using the slant distance as the horizontal distance.
- Forgetting to add the height of the observer or the building’s base.
- Leaving the answer as 10/√3 instead of rationalising and converting to a decimal when asked.
Exam tip: Draw a clean diagram, label every known length and write the angle at the observer’s eye. Then read off which ratio connects what you know to what you need.
Circles
Geometry · 15 marksTwo theorems do most of the work: the tangent is perpendicular to the radius, and tangents from an external point are equal.
Concepts you must be able to explain
- A tangent touches the circle at exactly one point; a secant cuts it at two. The point of contact is where the radius meets the tangent.
- Theorem: the tangent at any point of a circle is perpendicular to the radius through the point of contact. Its converse is also used.
- Theorem: the lengths of the two tangents drawn from an external point to a circle are equal. This creates isosceles triangles and congruent right triangles.
- If a quadrilateral circumscribes a circle, then AB + CD = BC + AD (sums of opposite sides are equal).
- From an external point, exactly two tangents can be drawn; from a point on the circle, exactly one; from a point inside, none.
Formulas and rules
| Name | Formula / rule | When to use it |
|---|---|---|
| Tangent–radius | OP ⊥ tangent at P | Join the centre to the point of contact first. |
| Equal tangents | PA = PB from external point P | Creates two congruent right triangles. |
| Circumscribed quadrilateral | AB + CD = BC + AD | Useful in one-mark and case-study questions. |
| Pythagoras in tangent problems | OP² = OA² + AP² | O centre, A external point, P point of contact. |
Common mistakes
- Assuming the tangent passes through the centre.
- Using the radius instead of the distance from the centre to the external point in Pythagoras.
- Forgetting that tangents from the same external point are equal when setting up an equation.
- Drawing conclusions from a diagram that is not given as a theorem.
Exam tip: Whenever you see a tangent, draw the radius to the point of contact and mark 90°. That single construction unlocks nearly every question in the chapter.
Surface Areas and Volumes
Mensuration · 10 marksCombination-of-solids questions dominate. Identify which surfaces are exposed, then add only those.
Concepts you must be able to explain
- A solid formed by joining a cone to a cylinder loses the two circular faces that are glued together — those surfaces are no longer exposed.
- Curved surface area covers only the rounded part; total surface area includes the flat bases. Read the question carefully.
- Volume is always additive: the volume of a combination is the sum of the volumes of the parts.
- The frustum of a cone has two radii, r₁ and r₂, and slant height l = √(h² + (r₁ − r₂)²).
- Conversion problems (a cylinder melted into spheres, water flowing through a pipe) conserve volume, not surface area.
Formulas and rules
| Name | Formula / rule | When to use it |
|---|---|---|
| Cylinder | CSA = 2πrh; TSA = 2πr(r + h); V = πr²h | r is the radius, h the height. |
| Cone | l = √(r² + h²); CSA = πrl; TSA = πr(l + r); V = ⅓πr²h | l is the slant height, not the vertical height. |
| Sphere | SA = 4πr²; V = 4/3 πr³ | Solid sphere. |
| Hemisphere | CSA = 2πr²; TSA = 3πr²; V = 2/3 πr³ | TSA adds the flat circular face. |
| Frustum | CSA = πl(r₁ + r₂); TSA = πr₁² + πr₂² + πl(r₁ + r₂); V = ⅓πh(r₁² + r₂² + r₁r₂) | l = √(h² + (r₁ − r₂)²). |
| Cube / cuboid | Cube: 6a², a³; cuboid: 2(lb + bh + hl), lbh | Check whether the top is open. |
Common mistakes
- Using the diameter as r in every formula.
- Counting hidden faces when two solids are joined.
- Confusing slant height with vertical height in cone and frustum problems.
- Forgetting to convert cm³ to litres (1 litre = 1000 cm³) in capacity questions.
Exam tip: Sketch the solid, shade the exposed surface and write the formula for each exposed piece before touching the calculator. This one habit prevents most lost marks.
Statistics
Statistics and Probability · 11 marksFormula-based and methodical. A clean frequency table is worth more than mental arithmetic here.
Concepts you must be able to explain
- For grouped data, use the class mark xᵢ = (lower limit + upper limit)/2, never the limits themselves.
- Mean can be found by the direct method, the assumed-mean method or the step-deviation method. Step deviation is fastest when class sizes are equal.
- Median uses the cumulative frequency: find the class where n/2 falls, then apply the median formula.
- Mode uses the highest frequency and the frequencies of the neighbouring classes.
- The empirical relationship 3 × median = mode + 2 × mean is a useful check, not a substitute for the actual method unless the question says so.
Formulas and rules
| Name | Formula / rule | When to use it |
|---|---|---|
| Mean (direct) | x̄ = Σfᵢxᵢ / Σfᵢ | xᵢ is the class mark. |
| Mean (step deviation) | x̄ = a + h × (Σfᵢuᵢ / Σfᵢ), uᵢ = (xᵢ − a)/h | a is the assumed mean, h the class size. |
| Median | l + ((n/2 − cf)/f) × h | l lower limit, cf cumulative frequency before the class, f frequency of the class. |
| Mode | l + ((f₁ − f₀)/(2f₁ − f₀ − f₂)) × h | f₁ is the modal frequency. |
| Empirical relation | 3 Median = Mode + 2 Mean | Only for moderately skewed data. |
Common mistakes
- Using class limits instead of class marks in the mean.
- Taking cf as the cumulative frequency of the median class instead of the class before it.
- Using the frequency of the wrong class in the mode formula.
- Forgetting to make the classes continuous when there are gaps such as 10–19, 20–29.
Exam tip: Draw a table with columns for class, f, x, fx and cf before calculating. Examiners award marks for the table even when the final arithmetic slips.
Probability
Statistics and Probability · 11 marksThe easiest chapter to secure. Count the outcomes carefully and remember that probability is always between 0 and 1.
Concepts you must be able to explain
- Theoretical probability P(E) = number of favourable outcomes / total number of equally likely outcomes.
- The probability of an impossible event is 0 and of a certain event is 1. Every other probability lies between them.
- P(E) + P(not E) = 1, so ‘at least one’ questions are usually easier through the complement.
- Know the standard sample spaces: one die (6 outcomes), two dice (36), one coin (2), two coins (4), a standard deck (52 cards: 26 red, 26 black, 4 suits of 13, 4 aces, 12 face cards).
- A die has numbers 1 to 6; a deck has no 1 and no 11 — errors here are pure carelessness.
Formulas and rules
| Name | Formula / rule | When to use it |
|---|---|---|
| Theoretical probability | P(E) = n(E)/n(S) | S is the sample space. |
| Range | 0 ≤ P(E) ≤ 1 | A probability outside this range is wrong. |
| Complement | P(E) + P(not E) = 1 | Use for ‘at least one’ problems. |
| Two dice | 36 equally likely outcomes | Sums from 2 to 12. |
| Deck of cards | 52 cards = 4 suits × 13; 12 face cards | Ace is not a face card. |
Common mistakes
- Counting the same outcome twice when two dice or two coins are used.
- Using ‘at least one’ as exactly one.
- Saying there are 13 face cards in a deck (there are 12).
- Giving an answer such as 7/6 without noticing it exceeds 1.
Exam tip: For two coins or two dice, list the sample space in a small grid. It removes almost every counting error and takes under a minute.
A revision routine that actually works
Read the chapter once, slowly
Do not highlight. Read to understand the one idea the chapter is built on, then close the book and say it aloud in your own words.
Rebuild the formula table from memory
Write the formulas on a blank sheet. Compare with the table above. Every formula you miss goes on a separate “mistakes” page you review weekly.
Solve 5 NCERT questions before any reference book
NCERT exercises are the closest thing to the board paper. Finish the exercise before opening a question bank.
Do one previous-year question under a timer
Time pressure is a separate skill. Ten minutes of timed practice beats an hour of untimed reading.
Teach the chapter to someone
Explaining a theorem to a friend exposes gaps that rereading never will. If nobody is around, explain it to a wall — seriously.
What to read next
Quick answers students search for
How many chapters are there in Class 10 maths?
There are 14 chapters in the rationalised CBSE Class 10 Mathematics syllabus for 2026-27, grouped into seven units. Constructions is no longer a separate chapter.
Which chapter carries the most marks?
Algebra carries 20 of the 80 theory marks across polynomials, linear equations, quadratic equations and arithmetic progressions. Geometry (triangles and circles) carries 15.
Are these notes enough for the board exam?
These notes cover concepts, formulas and common errors, but maths is a doing subject. Pair them with the NCERT exercises and at least three full previous-year papers under a timer.
Can I print these notes?
Yes. Use the Print / Save as PDF button at the top of the page. The navigation and ads are removed automatically in the printed version.
Last updated: 22 October 2026 · Author: Bazid Ahmed Laskar · This page is independent study material and is not affiliated with Physics Wallah or Ritik Mishra.