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CBSE Class 10 Maths Formula Sheet: Every Chapter for 2026–27, with Common Traps
This CBSE Class 10 Maths formula sheet lists the key formulas and results for every chapter in the official 2026–27 CBSE Class X Mathematics curriculum, in the same order as CBSE's seven units. Each chapter has a compact table and a short "how to use / common traps" note, because most lost marks in Class 10 come from using a formula carelessly, not from forgetting it. The same sheet works for Mathematics Standard (041) and Mathematics Basic (241), because CBSE has published one syllabus for both.
A formula sheet is for revision, not for learning a chapter for the first time. If whole chapters feel shaky, a one-to-one tutor from our Class 10 home tutors in Gurgaon can rebuild them, and this page then becomes your daily recall tool. When you are ready to turn formulas into marks, our guide on how to score 95+ in CBSE Class 10 Maths gives a phased plan. The page is built to print cleanly: use your browser's Print option.
How to use this CBSE Class 10 Maths formula sheet
Use the sheet for short, daily recall sessions rather than as reading material. The chapter list and unit marks below come from the official CBSE Mathematics curriculum for Class X (2026–27), which covers subject codes 041 (Standard) and 241 (Basic). The board paper is worth 80 marks and internal assessment 20 marks.
| Unit (official marks) | Chapters in the 2026–27 curriculum | Sections on this page |
|---|---|---|
| I. Number Systems (6) | Real Numbers | 1 |
| II. Algebra (20) | Polynomials; Pair of Linear Equations in Two Variables; Quadratic Equations; Arithmetic Progressions | 2 to 5 |
| III. Coordinate Geometry (6) | Coordinate Geometry | 6 |
| IV. Geometry (15) | Triangles; Circles | 7 and 8 |
| V. Trigonometry (12) | Introduction to Trigonometry; Trigonometric Identities; Heights and Distances | 9 to 11 |
| VI. Mensuration (10) | Areas Related to Circles; Surface Areas and Volumes | 12 and 13 |
| VII. Statistics and Probability (11) | Statistics; Probability | 14 and 15 |
- Cover and recall. Cover the right-hand column and write each result from memory. Anything you miss goes into a short error list at the back of your notebook.
- Learn the conditions, not just the formula. "a ≠ 0" in a quadratic, "internal division" in the section formula, "two numbers only" for HCF × LCM. The trap notes point these out.
- Know which results must be proved. The curriculum asks you to prove the irrationality of √2, √3 and √5, the Basic Proportionality Theorem and the two tangent theorems. Those proofs are marked separately below.
- Use π = 22/7 unless the question says otherwise. The 2026–27 sample papers instruct this, and calculators are not allowed.
For the full unit-by-unit syllabus, including what each chapter's competencies are, see our CBSE Class 10 Maths syllabus 2026–27 guide.
1. Real numbers
| Result | Formula or statement |
|---|---|
| Fundamental Theorem of Arithmetic | Every composite number can be written as a product of primes, and this factorisation is unique apart from the order of the factors |
| HCF from prime factors | Product of the smallest power of each common prime factor |
| LCM from prime factors | Product of the greatest power of each prime factor involved |
| HCF and LCM of two numbers | HCF(a, b) × LCM(a, b) = a × b |
| Prime dividing a square | If a prime p divides a2, then p divides a (a positive integer). Used in the irrationality proofs |
| Rational and irrational | Rational ± irrational = irrational; non-zero rational × irrational = irrational |
| Proof to learn (curriculum) | Irrationality of √2, √3 and √5 by contradiction; then numbers such as 3 + 2√5 |
How to use / common traps: HCF × LCM = product holds for two numbers only. For 12, 18 and 30, the HCF is 6 and the LCM is 180, so HCF × LCM = 1,080, while the product of the three numbers is 6,480. In "leaves a remainder" questions, subtract the remainders first: the greatest number dividing 134 and 188 with remainders 4 and 6 is HCF(130, 182) = 26. In an irrationality proof, state the assumption clearly ("let √3 = a/b where a and b are co-prime integers, b ≠ 0") and name the contradiction at the end; the marking scheme gives marks for each step.
2. Polynomials
| Result | Formula |
|---|---|
| Zeros and the graph | The zeros of p(x) are the x-coordinates where the graph of y = p(x) meets the x-axis. A quadratic has at most 2 zeros; its graph is a parabola |
| Sum of zeros of ax2 + bx + c | α + β = −b/a |
| Product of zeros | αβ = c/a |
| Quadratic with given zeros | k[x2 − (α + β)x + αβ], k ≠ 0 |
| Useful identities | α2 + β2 = (α + β)2 − 2αβ; (α − β)2 = (α + β)2 − 4αβ; 1/α + 1/β = (α + β)/αβ |
| Special cases | Zeros are reciprocals of each other ⇔ a = c; zeros are equal in size and opposite in sign ⇔ b = 0 |
How to use / common traps: the sum of zeros is −b/a, and the minus sign is the most common slip in this chapter. For 2x2 − 8x + 5, the sum is 4, the product is 5/2, and α2 + β2 = 16 − 5 = 11. Do not write α2 + β2 as (α + β)2. When counting zeros from a graph, count the points where the curve meets the x-axis, not where it meets the y-axis. The curriculum names the zero–coefficient relationship for quadratic polynomials only; cubic relationships are listed in the flagged table.
3. Pair of linear equations in two variables
Write the pair as a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.
| Condition | Graph | Number of solutions | Consistency |
|---|---|---|---|
| a1/a2 ≠ b1/b2 | Intersecting lines | Exactly one | Consistent |
| a1/a2 = b1/b2 = c1/c2 | Coincident lines | Infinitely many | Consistent (dependent) |
| a1/a2 = b1/b2 ≠ c1/c2 | Parallel lines | None | Inconsistent |
Methods named in the curriculum: graphical, substitution and elimination.
How to use / common traps: "consistent" covers both one solution and infinitely many; only parallel lines are inconsistent. Example: 2x + 3y = 7 and 4x + 6y = 14 give ratios 1/2, 1/2 and 1/2, so the lines coincide. Change the second equation to 4x + 6y = 15 and the third ratio becomes 7/15, so there is no solution. In word problems, define both variables in words ("let x be the number of small bottles") before writing the equations; the definition carries marks in case-based questions.
4. Quadratic equations
| Result | Formula |
|---|---|
| Standard form | ax2 + bx + c = 0, a ≠ 0 |
| Quadratic formula | x = [−b ± √(b2 − 4ac)] / 2a |
| Discriminant | D = b2 − 4ac |
| Nature of roots | D > 0: two distinct real roots; D = 0: two equal real roots, each −b/2a; D < 0: no real roots |
| Real roots condition | D ≥ 0 |
| Methods named in the curriculum | Factorisation and the quadratic formula (only real roots) |
How to use / common traps: bring the equation to standard form before reading off a, b and c; a term on the wrong side flips a sign. For 2x2 − 4x + 3 = 0, D = 16 − 24 = −8, so there are no real roots. In word problems, test both roots against the situation and reject any that make no sense (a negative length or a fraction of a person), with a one-line reason. "Equal roots" questions ask for D = 0; "real roots" questions ask for D ≥ 0.
5. Arithmetic progressions
| Result | Formula |
|---|---|
| nth term | an = a + (n − 1)d |
| nth term from the end | l − (n − 1)d, where l is the last term |
| Sum of the first n terms | Sn = (n/2)[2a + (n − 1)d] = (n/2)(a + l) |
| Term from sums | an = Sn − Sn−1 |
| Three terms in AP | Take them as a − d, a, a + d |
| Middle term (n odd) | The ((n + 1)/2)th term |
| AP test | a, b, c are in AP ⇔ 2b = a + c |
How to use / common traps: n must come out as a positive integer; if it does not, the term is not in the AP. Example: in 10, 7, 4, …, −62, solving −62 = 10 + (n − 1)(−3) gives n = 25, so the middle term is the 13th: 10 + 12 × (−3) = −26. Read "between 1 and 400" as excluding the endpoints. Keep an (one term) and Sn (a total) apart; questions often give one and ask for the other.
From our tutors: in the algebra chapters, the formulas are rarely the problem. The marks go on sign slips (−b/a written as b/a, a negative d dropped) and on a missing last line that answers the question in words. We ask students to write the final answer as a sentence ("the pool is 34 m by 24 m") every single time, because in our experience that habit also catches half their sign errors.
6. Coordinate geometry
| Result | Formula |
|---|---|
| Distance formula | AB = √[(x2 − x1)2 + (y2 − y1)2] |
| Distance from the origin | √(x2 + y2) |
| Section formula (internal division, m : n) | P = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n)) |
| Midpoint | ((x1 + x2)/2, (y1 + y2)/2) |
| Ratio unknown | Take the ratio as k : 1, apply the section formula, and use the given condition (on the x-axis, y = 0; on the y-axis, x = 0) |
| Shapes from distances | Collinear if AB + BC = AC; a parallelogram's diagonals have the same midpoint; a rhombus has four equal sides; a square also has equal diagonals |
How to use / common traps: in the section formula, m multiplies the coordinates of the second point, which is the most common slip. The distance between (2, 3) and (−4, −5) is √(36 + 64) = 10. Watch the brackets when subtracting a negative coordinate. The curriculum names internal division only; the area-of-a-triangle formula is not in the 2026–27 curriculum (see the flagged table), so show collinearity with distances instead.
7. Triangles
| Result | Statement |
|---|---|
| Similar triangles | Corresponding angles are equal and corresponding sides are in the same ratio |
| Basic Proportionality Theorem (prove) | If DE ∥ BC in △ABC, with D on AB and E on AC, then AD/DB = AE/EC |
| Converse of BPT (state) | If AD/DB = AE/EC, then DE ∥ BC |
| AA (AAA) criterion (state) | Two pairs of equal corresponding angles ⇒ similar |
| SSS criterion (state) | Corresponding sides proportional ⇒ similar |
| SAS criterion (state) | One equal angle and the sides including it proportional ⇒ similar |
| Perimeters of similar triangles | Ratio of perimeters = ratio of corresponding sides |
| Altitude on the hypotenuse | In △ABC right-angled at B with BD ⊥ AC: △ADB ~ △ABC ~ △BDC (by AA) |
How to use / common traps: write the vertices of similar triangles in matching order (△ABC ~ △PQR means A ↔ P, B ↔ Q, C ↔ R), otherwise the ratios you write will be wrong. Example: perimeters 56 cm and 70 cm, and a side of 14 cm in the first triangle, give 14 × 70/56 = 17.5 cm for the corresponding side. The BPT proof is a common 5-mark question; the 2026–27 marking scheme gives a mark for the correct figure, given, to prove and construction, before the proof itself, so draw and label all four.
8. Circles
| Result | Formula or statement |
|---|---|
| Tangent and radius (prove) | The tangent at any point of a circle is perpendicular to the radius through the point of contact |
| Equal tangents (prove) | The lengths of the two tangents from an external point to a circle are equal |
| Length of a tangent | PA = √(d2 − r2), d = distance of P from the centre |
| Angle between the tangents | ∠APB + ∠AOB = 180° |
| Line to the centre | OP bisects ∠APB and ∠AOB |
| Quadrilateral circumscribing a circle | AB + CD = AD + BC |
| Number of tangents from a point | Inside the circle: 0; on the circle: 1; outside: 2 |
How to use / common traps: the right angle is at the point of contact, so OP (centre to external point) is the hypotenuse. With r = 9 m and OP = 18 m, the tangent is √(324 − 81) = 9√3 m, cos ∠AOP = 9/18 gives ∠AOP = 60°, and ∠AOB = 120°. Name the theorem you use in brackets ("tangents from an external point are equal"); marking schemes show the reason alongside the step. A parallelogram that circumscribes a circle is a rhombus, which follows directly from the equal-tangents result.
Does your child know the formulas but still lose marks in the proofs and word problems? Book a free Class 10 Maths demo at home in Gurgaon or online. Our tutor will go through a recent test with you and show where the marks are going.
Book a Free Class 10 Maths Demo +91 92204 750889. Introduction to trigonometry
In a right triangle, for acute angle A: P = side opposite A, B = side adjacent to A, H = hypotenuse.
| Ratio | Definition | Reciprocal |
|---|---|---|
| sin A | P/H | cosec A = H/P |
| cos A | B/H | sec A = H/B |
| tan A | P/B = sin A / cos A | cot A = B/P = cos A / sin A |
| A | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin A | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos A | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan A | 0 | 1/√3 | 1 | √3 | Not defined |
| cosec A | Not defined | 2 | √2 | 2/√3 | 1 |
| sec A | 1 | 2/√3 | √2 | 2 | Not defined |
| cot A | Not defined | √3 | 1 | 1/√3 | 0 |
How to use / common traps: sin(A + B) is not sin A + sin B; treat A + B as one angle. Example: if tan(A + B) = √3 and tan(A − B) = 1/√3, then A + B = 60° and A − B = 30°, so A = 45° and B = 15°. For an acute angle, sin A and cos A lie between 0 and 1, so a value such as sin A = 4/3 means an error earlier. If one ratio is given (tan θ = 12/5), draw the triangle, find the third side by Pythagoras (13) and read off the rest.
10. Trigonometric identities
| Identity | Useful rearrangements |
|---|---|
| sin2A + cos2A = 1 (prove, per the curriculum) | sin2A = 1 − cos2A; cos2A = 1 − sin2A |
| 1 + tan2A = sec2A (0° ≤ A < 90°) | sec2A − tan2A = 1, so (sec A − tan A)(sec A + tan A) = 1 |
| 1 + cot2A = cosec2A (0° < A ≤ 90°) | cosec2A − cot2A = 1, so (cosec A − cot A)(cosec A + cot A) = 1 |
How to use / common traps: the curriculum says "only simple identities to be given", so most proofs need two to four lines. Work on one side (usually the more complicated one) until it becomes the other; do not move terms across the equals sign as if solving an equation. Converting everything to sin and cos is the safe default. Watch (1 − sin θ)2: it expands to 1 − 2 sin θ + sin2θ, and the 2026–27 sample paper includes a question that asks students to find exactly this kind of error in a worked solution.
11. Heights and distances
| Result | Formula or rule |
|---|---|
| Height from a distance | tan θ = height / horizontal distance |
| Along a slope or string | sin θ = height / length of the string or ladder; cos θ = horizontal distance / length |
| Elevation and depression | The angle of depression from A to B equals the angle of elevation from B to A (alternate angles) |
| Curriculum limits | No more than two right triangles; angles of elevation or depression only 30°, 45° and 60° |
How to use / common traps: draw the figure first and mark the right angle; in the 2026–27 marking scheme, a correct figure earns 1 mark in the 5-mark heights question. Angles are measured from the horizontal, never from the vertical. If an observer or building has height, subtract it before using tan θ. Example: a kite string 52 m long with tan θ = 12/5 gives sin θ = 12/13, so the height is 52 × 12/13 = 48 m. Leave answers such as 100(√3 + 1) m in surd form unless the question gives a value for √3.
12. Areas related to circles
| Result | Formula (θ in degrees) |
|---|---|
| Circumference and area | C = 2πr; A = πr2 |
| Length of an arc | l = (θ/360) × 2πr |
| Area of a sector | (θ/360) × πr2 = ½ × l × r |
| Area of a minor segment | Area of sector − area of the triangle formed by the two radii and the chord |
| That triangle (curriculum limits θ to 60°, 90°, 120° for segments) | θ = 60°: (√3/4)r2 (equilateral); θ = 90°: ½r2; θ = 120°: (√3/4)r2 |
| Major sector or segment | πr2 − minor sector (or minor segment) |
| Wheels and clocks | Distance in one revolution = circumference; a minute hand turns 6° per minute |
How to use / common traps: check whether the question gives the radius or the diameter. Example: r = 14 cm and θ = 90° give a sector of 154 cm2, a triangle of 98 cm2 and a segment of 56 cm2. For a 120° segment, the triangle area is also (√3/4)r2, not ½r2; many students use the 90° value by habit. If the arc length is given, ½ × l × r finds the sector area in one line: a 22 cm arc on a 28 cm diameter circle gives ½ × 22 × 14 = 154 cm2.
13. Surface areas and volumes
| Solid | Curved or lateral surface area | Total surface area | Volume |
|---|---|---|---|
| Cuboid (l, b, h) | 2h(l + b) | 2(lb + bh + hl) | lbh |
| Cube (a) | 4a2 | 6a2 | a3 |
| Right circular cylinder (r, h) | 2πrh | 2πr(r + h) | πr2h |
| Right circular cone (r, h, slant l) | πrl, with l = √(r2 + h2) | πr(l + r) | (1/3)πr2h |
| Sphere (r) | 4πr2 | 4πr2 | (4/3)πr3 |
| Hemisphere (r) | 2πr2 | 3πr2 | (2/3)πr3 |
The curriculum covers combinations of any two of these solids.
- Surface area of a combined solid = sum of the surfaces you can see. Where two solids are joined, neither joined face counts.
- Volume of a combined solid = sum of the volumes (or the difference, if one is hollowed out of the other).
How to use / common traps: a cone on a hemisphere uses the cone's curved surface (πrl) and the hemisphere's curved surface (2πr2), never either total surface area. Example: two cubes of volume 64 cm3 (side 4 cm) joined end to end make an 8 × 4 × 4 cuboid with surface area 2(32 + 16 + 32) = 160 cm2, not 2 × 96 = 192. Find the slant height before anything else in cone questions, and keep units consistent (m2 for canvas, cm3 for wood).
14. Statistics
| Result | Formula (grouped data) |
|---|---|
| Class mark | xi = (upper limit + lower limit)/2 |
| Mean, direct method | x̄ = Σfixi / Σfi |
| Mean, assumed mean method | x̄ = a + Σfidi / Σfi, where di = xi − a |
| Mean, step deviation method | x̄ = a + h × Σfiui / Σfi, where ui = (xi − a)/h |
| Mode | Mode = l + [(f1 − f0) / (2f1 − f0 − f2)] × h |
| Median | Median = l + [(n/2 − cf) / f] × h |
| Empirical relation (see note) | Mode = 3 Median − 2 Mean |
In the mode formula, l is the lower limit of the modal class (the class with the highest frequency), f1 its frequency, f0 and f2 the frequencies of the classes before and after it, and h the class width. In the median formula, l is the lower limit of the median class (the first class whose cumulative frequency reaches n/2), cf is the cumulative frequency of the class before it, and f is the median class's own frequency.
How to use / common traps: the curriculum names all three methods for the mean, and the median and mode "by algebraic method"; it also says bimodal data will be avoided. Using cf of the median class itself, instead of the class before it, is the most frequent slip. If classes are written as 10–19, 20–29, convert them to continuous classes (9.5–19.5, 19.5–29.5) before using l. Example from the 2026–27 sample paper: for frequencies 3, 6, 12, 15, 14 in classes 0–10 to 40–50, the mean is 1560/50 = 31.2 and the mode is 30 + (3/4) × 10 = 37.5. The empirical relation is not named in the curriculum text, but the 2026–27 Standard sample paper uses it in a 1-mark question (mean 35.5 and median 32 give mode 25), so learn it.
15. Probability
| Result | Formula or fact |
|---|---|
| Classical definition | P(E) = number of favourable outcomes / total number of equally likely outcomes |
| Range | 0 ≤ P(E) ≤ 1; sure event 1; impossible event 0 |
| Complement | P(E) + P(not E) = 1 |
| Common sample spaces | One coin: 2; two coins: 4; one die: 6; two dice: 36 |
| A pack of 52 cards | 4 suits of 13 (spades and clubs black; hearts and diamonds red); 26 red, 26 black; 12 face cards (king, queen, jack); 4 aces |
How to use / common traps: list the outcomes when the numbers are small, and count overlaps once. For T-shirts numbered 4 to 99 (96 outcomes), there are 8 perfect squares and 3 perfect cubes, but 64 is both, so P = 10/96 = 5/48. If cards are removed, reduce the total: with 6 cards missing, P(heart) = 13/46, not 13/52. Two dice give 36 ordered outcomes, so (2, 6) and (6, 2) are different; a sum of 8 has 5 favourable outcomes. Leave the answer as a fraction in its lowest terms.
From our tutors: in mensuration, statistics and probability, we see the same pattern often: the student knows every formula here but picks the wrong piece, such as the total surface area where the curved area was needed, or cf of the wrong class. So our revision drill for these chapters is not "recite the formula". It is "say which piece of the figure or table each letter stands for" before substituting anything.
Formulas not named in the 2026–27 curriculum
Many older books, guides and formula sheets still include the topics below. The official 2026–27 CBSE Class X Mathematics curriculum does not name them. If you are preparing for the 2026–27 board exam, give them low priority, and check the current curriculum on cbseacademic.nic.in before relying on any older material.
| Topic or formula | Chapter it used to sit in | 2026–27 curriculum |
|---|---|---|
| Euclid's division lemma and algorithm (a = bq + r) | Real numbers | Not named (HCF and LCM come from prime factorisation) |
| Decimal expansions of rational numbers (terminating or recurring) | Real numbers | Not named |
| Division algorithm for polynomials | Polynomials | Not named |
| Relations between zeros and coefficients of a cubic | Polynomials | Not named (quadratic only) |
| Cross-multiplication method; equations reducible to a pair of linear equations | Linear equations | Not named (graphical, substitution, elimination only) |
| Completing the square | Quadratic equations | Not named (factorisation and the quadratic formula only) |
| Area of a triangle from coordinates: ½|x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)| | Coordinate geometry | Not named |
| Centroid ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3) | Coordinate geometry | Not named; can be worked out from the section formula if a question needs it |
| Areas of similar triangles theorem (ratio of areas = square of the ratio of sides); proof of Pythagoras' theorem and its converse | Triangles | Not named |
| Trigonometric ratios of complementary angles (sin(90° − A) = cos A, and so on) | Trigonometry | Not named |
| Frustum of a cone; conversion of a solid from one shape to another | Surface areas and volumes | Not named (combinations of two solids only) |
| Ogives (cumulative frequency graphs) | Statistics | Not named |
| Constructions (dividing a line segment, tangents to a circle) | Constructions | Not named |
"Not named" means the topic is absent from the curriculum text. It does not promise that no question will ever use the idea, and CBSE can revise the curriculum. The empirical relation in Statistics is the example to remember: it is not in the curriculum text, yet the official 2026–27 sample paper uses it. Our guide to CBSE Class 10 previous year question papers explains how to use older papers without wasting time on dropped topics.
Standard vs Basic: is anything Standard-only?
No. For 2026–27, CBSE has published a single Class X Mathematics curriculum covering both codes, 041 (Standard) and 241 (Basic), with the same seven units and the same unit marks. So every formula on this sheet applies to both. The difference is in how the paper is set, as shown in the official question paper design:
| Question typology (official design) | Standard (041) | Basic (241) |
|---|---|---|
| Remembering and understanding | 43 marks (about 54%) | 60 marks (about 75%) |
| Applying | 19 marks (about 24%) | 12 marks (about 15%) |
| Analysing, evaluating and creating | 18 marks (about 22%) | 8 marks (about 10%) |
| Total | 80 | 80 |
In practice, a Standard paper asks you to combine formulas and reason more often, while a Basic paper uses the same formulas more directly. If you are still deciding between the two, read our comparison of CBSE Class 10 Maths Standard vs Basic.
Printing tips
- Use your browser's Print option. The table of contents and booking boxes are hidden when you print, so the sheet prints as formulas and notes only.
- To keep a copy on your phone or laptop, choose "Save as PDF" as the printer. We do not offer a separate download.
- Print on both sides, and add your own traps in the margins as you find them in tests.
- Replace the printout when CBSE publishes the next year's curriculum.
Frequently asked questions
Does this CBSE Class 10 Maths formula sheet cover the whole 2026–27 syllabus?
It covers the key formulas and results for all 15 topics listed across the seven units of the official 2026–27 CBSE Class X Mathematics curriculum. Formulas alone do not earn full marks, though: the paper also asks for proofs, figures and reasoning, so pair the sheet with the official sample paper and previous-year questions.
Can I download this as a PDF?
We do not offer a separate download. The page is built to print: use your browser's Print option and choose "Save as PDF" if you want a copy on your device.
Is the formula sheet different for Maths Standard and Maths Basic?
No. CBSE's 2026–27 curriculum is one document for both codes (041 and 241), with the same chapters and unit marks. The Basic paper uses more direct questions and the Standard paper more application and reasoning, but the formulas are the same.
Which theorems do I need to prove in the CBSE Class 10 board exam?
The 2026–27 curriculum marks these as proofs: irrationality of √2, √3 and √5; the Basic Proportionality Theorem; the tangent at any point is perpendicular to the radius; tangents from an external point are equal; and the identity sin2A + cos2A = 1. The converse of BPT and the similarity criteria are to be stated without proof.
Is the area of a triangle formula in coordinate geometry still in the syllabus?
It is not named in the 2026–27 curriculum, which lists the distance formula and the section formula (internal division) only. Use distances to check collinearity or the type of a quadrilateral.
Is "mode = 3 median − 2 mean" in the syllabus?
It is not named in the curriculum text, but the official 2026–27 Mathematics Standard sample paper uses it in a 1-mark question. Learn it; it takes one line.
What value of π should I use?
Use 22/7 unless the question states another value. The 2026–27 sample papers say "Take π = 22/7 wherever required, if not stated", and calculators are not allowed, so choose the value that simplifies with the given radius.
How should my child revise formulas each day?
Ten to fifteen minutes a day, two or three chapters in rotation, works better than one long session a week. Cover the formula column, write from memory, then solve two short questions on anything missed. A home tutor can make this routine stick and check the working, not just the answers.
Want a tutor to turn this CBSE Class 10 Maths formula sheet into marks? Book a free Class 10 Maths demo with Ajay Vatsyayan Classes, Saraswati kunj II, Wazirabad, Sector 52, Gurugram, Haryana 122003. Male and female tutors are available, at home across Gurgaon or online.
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