Class 10 Maths formula reference
Formula families, definitions, conditions and original checks. Use with the current syllabus; this sheet does not replace proofs or complete chapter learning.
Real numbers, polynomials and quadratics
Check a root by substitution into the original equation, including any restrictions in the question. A factor x − 2 gives x = 2, not −2. The quadratic formula's denominator is 2a for the whole numerator; missing the brackets changes the calculation.
| Relationship | Meaning and condition | Use example |
|---|---|---|
| HCF(a, b) × LCM(a, b) = a × b | For two positive integers; this product rule does not extend unchanged to three numbers. | For 12 and 18: HCF 6, LCM 36; both products equal 216. |
| α and β are the zeros of ax² + bx + c, where a ≠ 0. | For 2x² − 7x + 3: sum 7/2 and product 3/2. | |
| For ax² + bx + c = 0 with a ≠ 0. Real roots require D ≥ 0. | For x² − 5x + 6 = 0: D = 1 and roots are 2 and 3. | |
| D > 0: two distinct real roots; D = 0: equal real roots; D < 0: no real roots | This class-level classification concerns real roots. | x² + 1 = 0 has D = −4, so it has no real roots. |
Linear equations and arithmetic progressions
An AP has a constant difference, not a constant ratio. If a word problem asks how many terms fit, solve for n and check that it is a positive integer. For simultaneous equations, substitute the answer into both originals; satisfying just one is not enough.
| Relationship | Meaning and condition | Use example |
|---|---|---|
| a₁/a₂ ≠ b₁/b₂: one solution | For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0; use ratio tests only when their denominators are nonzero. | x + y = 5 and x − y = 1 give x = 3, y = 2. |
| a₁/a₂ = b₁/b₂ ≠ c₁/c₂: no solution; all three equal: infinitely many | If a ratio is undefined, inspect proportional equations or use elimination instead of dividing by zero. | x + y = 2 and 2x + 2y = 6 conflict; replacing 6 by 4 gives the same line. |
| a is first term, d common difference and n a positive integer term number. | For 5, 8, 11, …: a₁₀ = 5 + 9×3 = 32. | |
| Sum of the first n AP terms; l is the nth term, not an unrelated last value. | For those first 10 terms: S₁₀ = 10(5 + 32)/2 = 185. |
Coordinates, triangles and circle tangents
The theorem conditions are part of the answer. A line that only looks parallel in a sketch does not establish proportionality. Mark corresponding vertices before forming ratios, and draw the radius to the actual point of contact when using a tangent theorem.
| Relationship | Meaning and condition | Use example |
|---|---|---|
| Distance between A(x₁, y₁) and B(x₂, y₂). | Between (1, 2) and (4, 6): √(9 + 16) = 5. | |
| Internal division AP:PB = m:n, with m,n > 0. | A(0,0), B(6,3), AP:PB = 1:2 gives P(2,1). | |
| The special internal-section case 1:1. | Midpoint of (2,4) and (8,10) is (5,7). | |
| AD/DB = AE/EC | In triangle ABC, D lies on AB and E on AC, with DE parallel to BC. | If AD = 2, DB = 3 and AE = 4, then EC = 6. |
| Corresponding sides of similar triangles are proportional | First establish the similarity criterion and correct vertex correspondence. | A scale factor of 2 changes a corresponding side of 3 cm to 6 cm. |
| OT ⟂ PT; PA = PB | OT is a radius to tangent contact T; PA and PB are tangents from the same external point. | If tangent PA = 7 cm, the other tangent PB = 7 cm. |
Trigonometric ratios and standard values
For a 3–4–5 right triangle and the angle opposite the side of length 3, sin θ = 3/5, cos θ = 4/5 and tan θ = 3/4. For a height problem at 45°, a horizontal distance of 10 m gives a vertical rise of 10 m above eye level; add the observer's eye height if the question asks for the full height.
| Relationship | Conditions and use |
|---|---|
| sin θ = opposite/hypotenuse; cos θ = adjacent/hypotenuse; tan θ = opposite/adjacent | For an acute angle in a right triangle; opposite and adjacent depend on the chosen angle. |
| cosec θ = 1/sin θ; sec θ = 1/cos θ; cot θ = cos θ/sin θ | Use these relationships only where the denominators are nonzero. Also cot θ = 1/tan θ when tan θ is defined and nonzero. |
| sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ | The latter identities require the ratios involved to be defined. |
| tan θ = sin θ/cos θ | Only where cos θ ≠ 0. |
Standard-angle lookup
Use the relationships above for sec, cosec and cot. In particular, cot 90° = 0, while sec 90°, cosec 0° and cot 0° are undefined. Do not replace an undefined ratio with zero. Angles here are in degrees.
| θ | sin θ | cos θ | tan θ |
|---|---|---|---|
| 0° | 0 | 1 | 0 |
| 30° | 1/2 | √3/2 | 1/√3 |
| 45° | 1/√2 | 1/√2 | 1 |
| 60° | √3/2 | 1/2 | √3 |
| 90° | 1 | 0 | Undefined |
Circle area, arc, sector and segment
The current curriculum restricts segment-area problems to central angles of 60°, 90° and 120°. Sketch the shaded region first; if the question asks for a larger region, subtract the correct piece from the whole. Use the value of π specified in the question and distinguish length units from area units.
| Relationship | Meaning and condition | Use example |
|---|---|---|
| Circumference = 2πr; area = πr² | r is radius; diameter is 2r. | r = 7 cm gives circumference 14π cm and area 49π cm². |
| Arc length = (θ/360)×2πr; sector area = (θ/360)×πr² | θ is the central angle in degrees. | A 90° sector of radius 4 cm has arc 2π cm and area 4π cm². |
| Minor segment area = minor sector area − triangle area | Subtract the triangle formed by the two radii and the chord. Use the relevant geometry. | For the same 90° sector, triangle area is 8 cm²; segment area is (4π − 8) cm². |
| Sector perimeter = arc length + 2r | Include the two straight radii, not just the curved edge. | The 90° sector of radius 4 cm has perimeter (2π + 8) cm. |
Surface areas and volumes
For a cylinder of radius 3 cm and height 5 cm, volume = 45π cm³, curved area = 30π cm² and total area = 48π cm². Choose the requested quantity before substitution. In a joined solid, add volumes where appropriate but count only exposed surfaces: a shared circular face is internal. For an open container, do not silently add a missing lid. Convert all lengths to the same unit before calculating.
| Solid | Surface area | Volume |
|---|---|---|
| Cube, side a | Lateral 4a²; total 6a² | a³ |
| Cuboid, length l, breadth b, height h | Lateral 2h(l+b); total 2(lb+bh+hl) | lbh |
| Cylinder, radius r, height h | Curved 2πrh; total 2πr(h+r) | πr²h |
| Right circular cone, radius r, height h, slant l | Curved πrl; total πr(l+r); l = √(r²+h²) | πr²h/3 |
| Sphere, radius r | 4πr² | 4πr³/3 |
| Hemisphere, radius r | Curved 2πr²; total 3πr² | 2πr³/3 |
Grouped statistics and probability
Mean example: classes 0–10 and 10–20 with frequencies 2 and 3 have midpoints 5 and 15. Estimated mean = (2×5 + 3×15)/5 = 11. Probability example: on a fair six-sided die, P(number greater than 4) = 2/6 = 1/3. The probability fraction is justified by equally likely outcomes; it is not valid for an unspecified biased die.
Before using median or mode, identify the class from frequencies and cumulative frequencies. Do not confuse the preceding cumulative frequency with the preceding class's frequency. If the intervals are inclusive, establish continuous class boundaries as appropriate before substitution.
| Relationship | Meaning and condition |
|---|---|
| xᵢ is the class midpoint and fᵢ its frequency; the grouped mean is an estimate. | |
| Assumed mean a is chosen for convenient arithmetic. | |
| Use a common nonzero step h; equal class widths commonly make this convenient. | |
| L: lower boundary of median class; N: total frequency; cf: preceding cumulative frequency; f: median-class frequency; h: its width. | |
| For the usual equal-width grouped modal-class estimate: f₁ is modal frequency; f₀ and f₂ are adjacent frequencies; L and h refer to the modal class. | |
| P(E) = favourable outcomes / total outcomes | Finite equally likely outcomes; 0 ≤ P(E) ≤ 1 and P(not E) = 1 − P(E). |
Scope source: CBSE Class X Mathematics 2026–27. Check the guide for revision links and the latest review.