Mathematics — Class 10 AI Notes

Structure of real numbers, prime factorisation, and irrationality — the foundation used throughout the rest of Class 10 algebra.

• Real numbers = rational numbers (p/q, q ≠ 0, p, q integers) ∪ irrational numbers (non-terminating, non-recurring decimals).
• Euclid's Division Lemma: for any two positive integers a and b, there exist unique whole numbers q and r such that a = bq + r, where 0 ≤ r < b. This is the base of Euclid's algorithm.
• Euclid's Division Algorithm to find HCF(a, b): apply the lemma repeatedly, taking divisor and remainder as the new pair, until r = 0. The last non-zero divisor is HCF.
• Fundamental Theorem of Arithmetic: every composite number can be expressed as a product of primes, and this factorisation is unique apart from the order of the factors.
• HCF using prime factorisation = product of smallest powers of common prime factors.
• LCM using prime factorisation = product of greatest powers of all prime factors involved.
• Key relation for any two positive integers a and b: HCF(a, b) × LCM(a, b) = a × b. This does NOT extend directly to three numbers.
• A rational number p/q (in lowest terms) has a terminating decimal expansion iff q can be expressed as 2^m · 5^n where m, n are non-negative integers.
• If q has any prime factor other than 2 or 5, then p/q is non-terminating recurring.
• Irrationality proof by contradiction (standard template): assume √p is rational = a/b (in lowest terms); square both sides ⇒ a² = p·b²; p divides a² ⇒ p divides a (p prime); let a = p·k, substitute, show p divides b too, contradicting 'lowest terms'.
• Sum/difference/product/quotient of a rational (non-zero) and an irrational is always irrational.
• √2, √3, √5, √7, √11 … are all irrational. π and e are irrational (mentioned for context).
• In Board problems, always mention 'p, q are co-prime' explicitly when starting an irrationality proof.

• Euclid's Division Lemma: a = bq + r, 0 ≤ r < b
• HCF(a,b) × LCM(a,b) = a × b (only for two numbers)
• Fundamental Theorem of Arithmetic: composite = unique product of primes

• Real numbers → Rational + Irrational
• Rational → terminating (den = 2^m·5^n) or non-terminating recurring
• Prove irrationality by contradiction (√2, √3, √5)
• HCF/LCM via prime factorisation or Euclid's algorithm

• For HCF word problems: greatest size that fits all quantities.
• For LCM word problems: smallest quantity divisible by all.
• Terminating decimal check: factor denominator; only 2s and 5s → terminating.
• For 3 numbers: LCM × HCF ≠ product; use prime factorisation instead.

• Assuming √2 + √3 is rational — it is irrational.
• Forgetting r < b strictly in Euclid's lemma.
• Using HCF × LCM = product for three numbers.
• Skipping 'co-prime' assumption in irrationality proofs.

1. Q. Prove that 3 + 2√5 is irrational.
   A. Assume rational = p/q ⇒ √5 = (p − 3q)/(2q). RHS is rational; but √5 is irrational — contradiction. Hence 3 + 2√5 is irrational.
2. Q. Find HCF(306, 657) using Euclid's algorithm.
   A. 657 = 306·2 + 45; 306 = 45·6 + 36; 45 = 36·1 + 9; 36 = 9·4 + 0 ⇒ HCF = 9.
3. Q. Without division, decide if 17/8 has a terminating decimal.
   A. 8 = 2³. Denominator only has 2 ⇒ terminating. 17/8 = 2.125.
4. Q. The HCF of two numbers is 9 and their LCM is 360. If one number is 45, find the other.
   A. Other = (HCF·LCM)/first = 9·360/45 = 72.

Zeros of polynomials, their geometric meaning, and relationship with coefficients.

• A polynomial in x is p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀, where aₙ ≠ 0. Degree = highest power of x.
• Types by degree: constant (0), linear (1), quadratic (2), cubic (3), biquadratic (4).
• A real number k is a zero of p(x) iff p(k) = 0. Geometrically, zero = x-coordinate of point where graph cuts x-axis.
• A polynomial of degree n has at most n real zeros.
• Quadratic ax² + bx + c: graph is a parabola. Opens upward if a > 0, downward if a < 0. Depending on discriminant D = b² − 4ac, parabola cuts x-axis at 2, 1 or 0 points.
• Relationship between zeros α, β of ax² + bx + c and coefficients: α + β = −b/a, αβ = c/a.
• Relationship for cubic ax³ + bx² + cx + d with zeros α, β, γ: α+β+γ = −b/a; αβ+βγ+γα = c/a; αβγ = −d/a.
• Given zeros, construct quadratic: x² − (sum)x + (product) = 0.
• Division Algorithm: for polynomials p(x) and g(x), g(x) ≠ 0, there exist unique q(x) and r(x) such that p(x) = g(x)·q(x) + r(x), with deg r(x) < deg g(x) or r(x) = 0.
• Long division of polynomials mirrors integer long division — arrange in decreasing degree, keep like terms aligned.
• If (x − a) is a factor of p(x), then p(a) = 0 (Factor Theorem — used implicitly).
• For a graph problem: count how many times the curve crosses the x-axis = number of real zeros.

• Quadratic: sum of zeros = −b/a, product = c/a
• Cubic: α+β+γ = −b/a, αβ+βγ+γα = c/a, αβγ = −d/a
• Division: p(x) = g(x)·q(x) + r(x), deg r < deg g
• Quadratic from zeros: x² − (α+β)x + αβ

• Zero of polynomial ↔ x-intercept of graph
• Quadratic graph = parabola
• Relations between zeros and coefficients (Vieta's)
• Division algorithm for polynomials

• Verify zeros using Vieta's rather than resubstituting.
• For 'construct a polynomial with these zeros', straight to x² − Sx + P.
• Sketch parabola: sign of 'a' fixes opening; vertex at x = −b/2a.

• Confusing 'zero of polynomial' with 'value of polynomial'.
• Sign errors in −b/a.
• Assuming a quadratic always has real zeros (check D).

1. Q. Find zeros of x² − 2x − 8 and verify the relations.
   A. (x − 4)(x + 2) = 0 ⇒ zeros 4, −2. Sum = 2 = −(−2)/1 ✓; Product = −8 = −8/1 ✓.
2. Q. Form a quadratic whose zeros are 2 + √3 and 2 − √3.
   A. Sum = 4, Product = 4 − 3 = 1 ⇒ x² − 4x + 1 = 0.
3. Q. If α, β are zeros of x² − 5x + 6, find α² + β².
   A. α+β = 5, αβ = 6; α² + β² = (α+β)² − 2αβ = 25 − 12 = 13.

Systems of two linear equations — geometric interpretation, algebraic solution methods, and word problems.

• General form: a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.
• Graphical meaning of a linear equation in two variables = a straight line. A pair of equations = two lines. Solutions = points of intersection.
• Three cases based on ratios:
• (i) a₁/a₂ ≠ b₁/b₂ → lines intersect at exactly one point → consistent, unique solution.
• (ii) a₁/a₂ = b₁/b₂ = c₁/c₂ → lines coincide → consistent, infinitely many solutions (dependent).
• (iii) a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → lines parallel → inconsistent, no solution.
• Algebraic methods: (1) Substitution, (2) Elimination, (3) Cross-multiplication.
• Substitution method: express one variable from one equation, plug into the other.
• Elimination method: multiply equations to align coefficients, add/subtract to eliminate one variable.
• Cross-multiplication method: x/(b₁c₂ − b₂c₁) = y/(c₁a₂ − c₂a₁) = 1/(a₁b₂ − a₂b₁).
• Some non-linear equations reduce to linear form by substitution: e.g. 1/x = u, 1/y = v.
• Word problems: identify two unknowns, form two equations from two conditions, solve algebraically.
• Typical word-problem types: age problems, digits of a two-digit number, fraction problems, speed/time/distance (boat & stream, upstream/downstream), work-time, geometry-based (angles of triangle, sides).
• For two-digit number problems, if digits are x (tens) and y (units), number = 10x + y and reverse = 10y + x.
• Boat problems: speed downstream = boat + current, upstream = boat − current.

• Consistent unique: a₁/a₂ ≠ b₁/b₂
• Consistent dependent (∞): a₁/a₂ = b₁/b₂ = c₁/c₂
• Inconsistent: a₁/a₂ = b₁/b₂ ≠ c₁/c₂
• Cross-multiplication: x/(b₁c₂ − b₂c₁) = y/(c₁a₂ − c₂a₁) = 1/(a₁b₂ − a₂b₁)

• Graphical → substitution → elimination → cross-multiplication
• Ratio test for nature of solutions
• Reducible to linear form (1/x, 1/y)
• Word problems: age, digits, fraction, boat, work

• Elimination when coefficients match easily.
• Substitution when a variable is already isolated.
• Check ratios first when asked about nature of solutions.

• Sign errors in cross-multiplication.
• Forgetting to check c₁/c₂ for dependency.
• Setting up reverse two-digit number as 10x + y instead of 10y + x.

1. Q. Father is 3 times his son. In 12 years, father is 2× son. Find their ages.
   A. F = 3S; F + 12 = 2(S + 12) ⇒ 3S + 12 = 2S + 24 ⇒ S = 12, F = 36.
2. Q. For what value of k does 2x + 3y = 7 and (k − 1)x + (k + 2)y = 3k have infinite solutions?
   A. 2/(k−1) = 3/(k+2) = 7/(3k) ⇒ solve: k = 7.
3. Q. A boat covers 30 km downstream in 3 h and 30 km upstream in 5 h. Find boat and stream speeds.
   A. b + s = 10, b − s = 6 ⇒ b = 8 km/h, s = 2 km/h.

Equations of the form ax² + bx + c = 0 — solving techniques and nature of roots.

• Standard form: ax² + bx + c = 0, a ≠ 0.
• A real number α is a root iff aα² + bα + c = 0. Roots = zeros of the corresponding polynomial.
• Methods to solve: (1) Factorisation (splitting the middle term), (2) Completing the square, (3) Quadratic formula.
• Splitting the middle term: find two numbers whose product = a·c and sum = b.
• Completing the square: rewrite ax² + bx + c = 0 as a(x + b/2a)² = (b² − 4ac)/4a, then take square roots.
• Quadratic formula: x = [−b ± √(b² − 4ac)] / 2a. Derived from completing the square.
• Discriminant D = b² − 4ac decides nature of roots:
• D > 0 → two distinct real roots.
• D = 0 → two equal real roots (repeated), x = −b/2a.
• D < 0 → no real roots (complex, out of Class 10 scope).
• Sum of roots = −b/a; product of roots = c/a (Vieta's).
• Word problems modelled by quadratics: area/dimension problems, speed/time (with time difference), age (product of ages), streams/pipes, projectile-type numerical.
• For quadratic word problems, ALWAYS reject the non-physical root (negative length, negative age, negative time).
• For 'find k for equal roots', set D = 0 and solve for k.
• For 'real roots exist', require D ≥ 0.

• x = [−b ± √(b² − 4ac)] / 2a
• D = b² − 4ac
• Sum = −b/a, Product = c/a

• Factorisation → completing the square → quadratic formula
• Nature of roots via D
• Word problems: area, speed-time, age

• Compute D first to know root nature.
• For 'equal roots', D = 0.
• In word problems, drop the negative root physically.

• Missing ± in the quadratic formula.
• Sign errors moving terms across equality.
• Accepting negative side length or time as an answer.

1. Q. Find k so that kx² − 2kx + 6 = 0 has equal roots.
   A. D = 0: 4k² − 24k = 0 ⇒ k(k − 6) = 0 ⇒ k = 6 (k ≠ 0).
2. Q. Sum of a number and its reciprocal is 10/3. Find the number.
   A. x + 1/x = 10/3 ⇒ 3x² − 10x + 3 = 0 ⇒ x = 3 or 1/3.
3. Q. A train covers 360 km at uniform speed. If speed were 5 km/h more, it would take 1 h less. Find speed.
   A. 360/x − 360/(x+5) = 1 ⇒ x² + 5x − 1800 = 0 ⇒ x = 40 km/h (reject −45).

Sequences with a constant difference between successive terms — general term and sum formulas.

• AP = list of numbers a, a+d, a+2d, a+3d, … in which the difference between any two consecutive terms is a constant d, called the common difference.
• First term = a, common difference d = aₙ₊₁ − aₙ (should be the same for all n).
• General (nth) term: aₙ = a + (n − 1)d.
• Sum of first n terms: Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l), where l = last term.
• If Sₙ is given as a polynomial in n, then aₙ = Sₙ − Sₙ₋₁.
• To find how many terms: use aₙ formula and equate to the last term; n must be a positive integer.
• For 3 numbers in AP whose sum is known, take them as (a − d, a, a + d) — sum simplifies to 3a.
• For 4 numbers in AP with a symmetric condition, take (a − 3d, a − d, a + d, a + 3d) — sum = 4a, common difference = 2d.
• Middle term of an AP with odd number of terms lies at position (n+1)/2.
• Applications: instalment problems, seating arrangements, salary increments, stack/pyramid counting.

• aₙ = a + (n − 1)d
• Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l)
• aₙ = Sₙ − Sₙ₋₁

• AP → common difference d constant
• nth term → sum of n terms
• 3 or 4 unknowns → symmetric substitution

• 3 unknowns in AP: take a−d, a, a+d.
• 4 unknowns in AP: take a−3d, a−d, a+d, a+3d.
• Sum-based problems: prefer Sₙ = n/2 (a + l) if last term is known.

• Using n instead of n − 1 in aₙ.
• Confusing last term l with number of terms n.
• Assuming a list is AP without checking constant d.

1. Q. Sum of first 20 terms of AP: 2, 7, 12, …
   A. a = 2, d = 5, S₂₀ = 20/2 (4 + 19·5) = 10·99 = 990.
2. Q. Which term of the AP 3, 15, 27, 39, … is 132 more than its 54th term?
   A. d = 12; 54th term = 3 + 53·12 = 639. Required term = 771. n = (771 − 3)/12 + 1 = 65th.
3. Q. Find three numbers in AP whose sum is 24 and product is 440.
   A. Take a − d, a, a + d ⇒ 3a = 24 ⇒ a = 8. Product: 8(64 − d²) = 440 ⇒ d² = 9 ⇒ d = ±3. Numbers: 5, 8, 11.

Similarity of triangles, Basic Proportionality Theorem (Thales), and related area results.

• Two triangles are similar if (a) corresponding angles are equal AND (b) corresponding sides are in the same ratio.
• Basic Proportionality Theorem (BPT / Thales): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
• Converse of BPT: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
• Similarity criteria: AA (two angles equal ⇒ third equal, so similar), SAS (proportional sides + included angle equal), SSS (three sides proportional).
• Ratio of areas of two similar triangles = ratio of squares of any pair of corresponding sides (or medians, or altitudes, or perimeters).
• In similar triangles, ratio of perimeters = ratio of corresponding sides.
• Standard configurations to spot: line parallel to one side; two triangles sharing an angle at vertex.
• Pythagoras Theorem (statement retained in rationalised syllabus): In a right triangle, square of hypotenuse = sum of squares of the other two sides.
• Converse of Pythagoras: if a² + b² = c², then the triangle is right-angled at the vertex opposite to c.
• Common competency setup: light-pole and shadow, ladder against a wall, tree broken by wind — reduce to similar right triangles.

• BPT: DE ∥ BC ⇒ AD/DB = AE/EC
• Similarity criteria: AA, SAS, SSS
• Ratio of areas of similar △ = (ratio of sides)²
• Pythagoras: a² + b² = c² (right-angled)

• Similarity → BPT → criteria → areas
• Pythagoras & its converse
• Ratios: sides ↔ perimeters ↔ altitudes ↔ medians ↔ √areas

• Look for parallels → BPT.
• Look for common angle → AA similarity.
• For area ratio problems: work with the square of side ratio.

• Assuming similar ⇒ congruent.
• Wrong direction of ratio (AD/AB vs AD/DB).
• Squaring only one side of the ratio when comparing areas.

1. Q. In △ABC, DE ∥ BC with AD = 3, DB = 5, AE = 4.5. Find EC.
   A. AD/DB = AE/EC ⇒ 3/5 = 4.5/EC ⇒ EC = 7.5.
2. Q. Areas of two similar triangles are 81 cm² and 49 cm². If the longest side of the smaller is 14 cm, find longest side of the larger.
   A. Ratio of areas 81:49 ⇒ ratio of sides 9:7 ⇒ longest side = 14 × 9/7 = 18 cm.
3. Q. A vertical pole 6 m casts a shadow 4 m long. At the same time, a tower casts a shadow 28 m. Find the height of the tower.
   A. Similar triangles: h/28 = 6/4 ⇒ h = 42 m.

Locating points in the plane, distances, section formula, and midpoint results.

• Cartesian plane: x-axis, y-axis, origin O(0,0). Any point represented by ordered pair (x, y).
• Quadrants: I (+,+), II (−,+), III (−,−), IV (+,−).
• Distance formula between P(x₁, y₁) and Q(x₂, y₂): PQ = √[(x₂ − x₁)² + (y₂ − y₁)²].
• Distance from origin O to P(x, y) = √(x² + y²).
• Section formula (internal division): point dividing segment joining (x₁, y₁) and (x₂, y₂) in ratio m:n is ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n)).
• Midpoint formula (m = n = 1): ((x₁ + x₂)/2, (y₁ + y₂)/2).
• Three points are collinear if the area of the triangle formed is 0 — but the direct area formula for triangles was rationalised out; use section-based reasoning or slope idea inferred from similar triangles.
• For a parallelogram, diagonals bisect each other → equate midpoints of the two diagonals.
• Common problem types: find distance, find ratio in which a point divides a segment, find fourth vertex of a parallelogram, verify collinearity, find coordinates of a specific point (like centroid — using midpoint reasoning).
• For ratio problems, take unknown ratio k:1, apply section formula, solve for k.

• Distance = √[(x₂ − x₁)² + (y₂ − y₁)²]
• Section (internal m:n): ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n))
• Midpoint: ((x₁ + x₂)/2, (y₁ + y₂)/2)

• Points → distance → section → midpoint
• Collinearity check
• Parallelogram via bisecting diagonals

• For ratio: take k:1, plug and solve.
• For parallelogram fourth vertex: diagonals share midpoint.

• Swapping m and n in the section formula.
• Forgetting square roots in distance.

1. Q. Find the point dividing the segment joining (−1, 7) and (4, −3) in ratio 2:3.
   A. ((2·4 + 3·(−1))/5, (2·(−3) + 3·7)/5) = (1, 3).
2. Q. Show that (1, 5), (2, 3), (−2, −11) are not collinear.
   A. Compute pairwise distances; sum of two smaller ≠ largest — hence non-collinear.
3. Q. Find the ratio in which x-axis divides the join of (5, −6) and (−1, −4).
   A. Take k:1; y-coord = 0 ⇒ (−4k − 6)/(k + 1) = 0 ⇒ k = −3/2 ⇒ externally in 3:2.

Trigonometric ratios of an acute angle, identities and complementary angle relations.

• In a right triangle, for an acute angle θ, define: sin θ = opposite/hypotenuse; cos θ = adjacent/hypotenuse; tan θ = opposite/adjacent.
• Reciprocals: cosec θ = 1/sin θ; sec θ = 1/cos θ; cot θ = 1/tan θ = cos θ/sin θ.
• Values for standard angles 0°, 30°, 45°, 60°, 90° (memorise the table).
• sin 30° = cos 60° = 1/2; sin 45° = cos 45° = 1/√2; sin 60° = cos 30° = √3/2; tan 45° = 1; tan 30° = 1/√3; tan 60° = √3.
• Fundamental identities: sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ.
• Trigonometric ratios of complementary angles: sin(90° − θ) = cos θ; tan(90° − θ) = cot θ; sec(90° − θ) = cosec θ.
• Domain caution: tan θ, sec θ undefined at θ = 90°; cot θ, cosec θ undefined at θ = 0°.
• For proving identities, the reliable route: convert all functions to sin and cos, take a common denominator, apply sin² + cos² = 1.
• For 'evaluate' questions, replace using values or complementary-angle substitution before simplifying.

• sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ
• sin(90° − θ) = cos θ; tan(90° − θ) = cot θ
• Standard values table for 0°, 30°, 45°, 60°, 90°

• Ratios (SOH-CAH-TOA)
• Identities
• Complementary angles
• Standard angle values

• Convert to sin/cos when stuck on identity proofs.
• Use complementary angles to collapse cos(90° − θ) into sin θ.

• sin²θ ≠ sin(θ²).
• Confusing 1 + tan²θ = sec²θ vs 1 + cot²θ = cosec²θ.
• Forgetting domain (θ ≠ 90° for tan).

1. Q. Prove (1 + cot A − cosec A)(1 + tan A + sec A) = 2.
   A. Convert to sin/cos; combine over sin A · cos A; use sin² + cos² = 1; simplifies to 2.
2. Q. If sin θ = 3/5, find cos θ and tan θ.
   A. cos²θ = 1 − 9/25 = 16/25 ⇒ cos θ = 4/5 ⇒ tan θ = 3/4.
3. Q. Evaluate: sin 60° cos 30° + sin 30° cos 60°.
   A. (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1 (= sin 90°).

Heights and distances — using trigonometry to solve real-world problems.

• Line of sight: line from observer's eye to the object being viewed.
• Angle of elevation: angle above horizontal (looking up).
• Angle of depression: angle below horizontal (looking down).
• Horizontal at observer's eye is the reference for both.
• In an angle of depression problem, the alternate-interior-angle at the object equals the angle of depression at the observer (parallel horizontals).
• Solving procedure: (1) Draw a clear diagram with the horizontal, vertical, and line of sight. (2) Mark all given lengths and angles. (3) Identify the right triangle. (4) Choose tan for opposite/adjacent, sin for opposite/hypotenuse, cos for adjacent/hypotenuse. (5) Solve.
• Two-triangle setups: object seen from two positions, or two objects seen from one position → set up two equations, then subtract or divide.
• If observer has non-negligible height, adjust the vertical distance accordingly.
• Bearing / direction problems: rare in Board, but interpret 'from the top of a tower' vs 'from the ground'.

• tan(angle) = opposite / adjacent (most common)
• sin(angle) = opposite / hypotenuse; cos = adjacent / hypotenuse

• Diagram → angle → right triangle → ratio choice
• Elevation (up) vs Depression (down)
• Two-triangle setups

• Always sketch first — mark opposite/adjacent clearly.
• For depression problems, transfer the angle to the base of the right triangle using parallel horizontals.

• Confusing elevation with depression.
• Ignoring height of the observer when it's given.
• Using sin when the hypotenuse isn't involved.

1. Q. A tower's shadow is √3 times its height. Find the angle of elevation of the Sun.
   A. tan θ = h/(√3 h) = 1/√3 ⇒ θ = 30°.
2. Q. From the top of a 100 m tower, angles of depression of two cars on the same side are 45° and 30°. Find the distance between the cars.
   A. d₁ = 100 (from 45°); d₂ = 100√3 (from 30°) ⇒ distance = 100(√3 − 1) ≈ 73.2 m.
3. Q. A ladder leans against a wall making 60° with the ground. Foot is 2.5 m from wall. Find the length of the ladder.
   A. cos 60° = 2.5/L ⇒ L = 5 m.

Tangents to a circle — length and angle properties.

• A tangent to a circle is a line that touches the circle at exactly one point (the point of contact).
• Theorem: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
• Theorem: The lengths of tangents drawn from an external point to a circle are equal.
• From an external point P to a circle with centre O, if PA and PB are tangents at A and B, then PA = PB, OA ⟂ PA, OB ⟂ PB, and OP bisects ∠APB and ∠AOB.
• Quadrilateral OAPB has ∠OAP = ∠OBP = 90°, so ∠APB + ∠AOB = 180° (opposite angles supplementary).
• In a circle inscribed in a triangle (incircle), tangent lengths from each vertex are equal: if s = (a+b+c)/2, then AF = AE = s − a, BF = BD = s − b, CD = CE = s − c.
• For a quadrilateral circumscribing a circle, sum of opposite sides is equal: AB + CD = AD + BC.
• Construction problems: draw a tangent from an external point using the semicircle-on-diameter approach.
• Typical proofs use: right angle at radius, Pythagoras in right triangle, or supplementary angles in OAPB.

• Tangent ⟂ radius at point of contact
• Tangents from external point are equal in length
• In OAPB: ∠APB + ∠AOB = 180°
• Circumscribing quadrilateral: AB + CD = AD + BC

• Tangent + radius = 90°
• External point → two equal tangents
• Cyclic quadrilateral OAPB
• Incircle in triangle: tangent lengths s − a etc.

• Join centre to tangent point → right angle → Pythagoras.
• For quadrilateral circumscribing circle: opposite sides sum equal.

• Assuming tangents from two different external points are equal.
• Confusing chord with tangent.

1. Q. PA and PB are tangents from P to a circle with centre O. If ∠APB = 60°, find ∠AOB.
   A. In OAPB, ∠AOB + 60° = 180° ⇒ ∠AOB = 120°.
2. Q. Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller one.
   A. Chord ⟂ radius of inner at contact ⇒ half-chord = √(25 − 9) = 4 ⇒ chord = 8 cm.
3. Q. A quadrilateral ABCD is drawn to circumscribe a circle. Prove AB + CD = AD + BC.
   A. Sum of tangent lengths from each vertex is equal, so grouping gives the identity.

Circumference, area, and sector/segment computations for circles.

• Circumference of a circle = 2πr; area = πr².
• For a sector of angle θ (in degrees): length of arc = (θ/360) · 2πr; area of sector = (θ/360) · πr².
• Area of a segment = area of sector − area of triangle formed by two radii and chord.
• For minor and major segments: major segment area = πr² − minor segment area.
• For angle in radians (rarely used at Class 10): arc length = rθ, sector area = ½r²θ.
• Combination figures: circle inscribed in square (r = a/2), square inscribed in circle (diagonal = 2r), semicircles on sides, etc.
• In problems mixing shapes: identify each region, compute area separately, add or subtract.
• Common π values: use 22/7 when r is a multiple of 7, and 3.14 otherwise (unless stated).

• Circle: C = 2πr, A = πr²
• Sector: arc = (θ/360)·2πr; area = (θ/360)·πr²
• Segment area = sector area − ½ r² sin θ (or minus triangle area)

• Sector, segment, chord
• Combinations with squares/triangles
• Choice of π (22/7 vs 3.14)

• For 60° sector with radius r: area = πr²/6; equilateral triangle formed if chord joins endpoints.
• Add/subtract areas region by region for compound figures.

• Mixing degrees and radians.
• Using r² instead of πr² for area.
• Forgetting the triangle subtraction in a segment.

1. Q. Find the area of a sector with r = 6 and θ = 60°.
   A. (60/360)·π·36 = 6π ≈ 18.86 cm².
2. Q. A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the minor segment.
   A. Sector = (90/360)·π·100 = 25π; triangle = ½·10·10 = 50; segment = 25π − 50 ≈ 28.5 cm².
3. Q. The wheel of a car has radius 35 cm. How many revolutions does it make to travel 11 km?
   A. Circumference = 2·22/7·35 = 220 cm; revolutions = 1100000/220 = 5000.

Surface areas and volumes of solids and their combinations; conversion between shapes.

• Cylinder: CSA = 2πrh, TSA = 2πr(r + h), Volume = πr²h.
• Cone: slant height l = √(r² + h²); CSA = πrl; TSA = πr(r + l); Volume = (1/3)πr²h.
• Sphere: Surface area = 4πr²; Volume = (4/3)πr³.
• Hemisphere: CSA = 2πr²; TSA = 3πr² (curved + flat circle); Volume = (2/3)πr³.
• Frustum of a cone (with radii r₁, r₂ and height h): Volume = (1/3)πh(r₁² + r₂² + r₁r₂); slant height L = √(h² + (r₁ − r₂)²); CSA = π(r₁ + r₂)L; TSA = CSA + π(r₁² + r₂²).
• Combinations: cone + hemisphere (ice-cream), cylinder + cone (tent), cylinder + hemispheres (capsule), sphere in cylinder.
• For joined solids, TSA is NOT the sum of TSAs — subtract the overlapping (joined) face.
• Conservation of volume: when a solid is melted/recast into another, volume is preserved.
• Flow rate problems: volume flowing in time = cross-section area × speed × time.

• Cylinder: CSA = 2πrh; TSA = 2πr(r + h); V = πr²h
• Cone: CSA = πrl; TSA = πr(r + l); V = (1/3)πr²h; l = √(r² + h²)
• Sphere: SA = 4πr²; V = (4/3)πr³
• Hemisphere: CSA = 2πr²; TSA = 3πr²; V = (2/3)πr³
• Frustum: V = (1/3)πh(r₁² + r₂² + r₁r₂); L = √(h² + (r₁ − r₂)²)

• Solids: cylinder, cone, sphere, hemisphere, frustum
• Combinations of solids
• Volume conservation (melting/recasting)
• Flow-rate word problems

• For melted/recast problems, equate volumes.
• For joined solid TSA: sum CSAs, add non-joined flat faces only.

• Adding TSAs of joined solids blindly.
• Using h instead of l in cone CSA.

1. Q. A cone (r = 6, h = 8) is melted into a sphere. Find the sphere's radius.
   A. (1/3)π·36·8 = (4/3)πR³ ⇒ R³ = 72 ⇒ R = ∛72 ≈ 4.16 cm.
2. Q. A capsule is 14 mm long with 5 mm diameter (cylinder + two hemispheres). Find its surface area.
   A. r = 2.5; cylindrical length = 14 − 5 = 9; SA = 2πrh + 4πr² = π(45 + 25) = 70π ≈ 220 mm².
3. Q. Water flows from a pipe of radius 0.75 cm at 7 m/s into a tank. How much water in litres flows in 1 hour?
   A. Volume/s = π(0.75)²·700 cm³ = 1237.5 cm³/s; in 1 h = 4455000 cm³ = 4455 L.

Mean, median, and mode of grouped data.

• For grouped frequency distributions, class marks xᵢ = (upper + lower)/2 represent each class.
• Direct method for mean: x̄ = Σfᵢxᵢ / Σfᵢ.
• Assumed mean method: choose a (usually a middle xᵢ); compute dᵢ = xᵢ − a; then x̄ = a + Σfᵢdᵢ / Σfᵢ.
• Step-deviation method: choose class width h; uᵢ = (xᵢ − a)/h; x̄ = a + h · Σfᵢuᵢ / Σfᵢ.
• Median formula: Median = l + [(n/2 − cf)/f] × h, where l = lower boundary of median class, n = Σfᵢ, cf = cumulative frequency before median class, f = frequency of median class, h = class width.
• Median class = the class in which the cumulative frequency first reaches or exceeds n/2.
• Mode formula: Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h, where f₁ = frequency of modal class, f₀ = previous, f₂ = next.
• Modal class = class with the highest frequency.
• Empirical relation: 3 Median = Mode + 2 Mean (approximate).
• For continuous data, ensure classes are 'exclusive' (like 10–20, 20–30) or convert 'inclusive' classes (like 10–19, 20–29) by adjusting boundaries.
• Note: ogive/less-than and more-than curves were rationalised; focus on the three central-tendency formulas.

• Mean (direct): x̄ = Σfᵢxᵢ / Σfᵢ
• Mean (assumed): x̄ = a + Σfᵢdᵢ / Σfᵢ
• Mean (step-deviation): x̄ = a + h · Σfᵢuᵢ / Σfᵢ
• Median = l + [(n/2 − cf)/f] × h
• Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h
• Empirical: 3 Median = Mode + 2 Mean

• Grouped data → class marks
• Mean via 3 methods
• Median: locate class via cumulative frequency
• Mode: locate modal class
• Empirical relation among mean, median, mode

• Step-deviation method is fastest when class widths are equal.
• Locate median class by scanning cumulative frequency for n/2.
• If distribution is nearly symmetric, mean ≈ median ≈ mode.

• Wrong cf class for median (must be BEFORE median class).
• Confusing f₀, f₁, f₂ in mode formula.
• Using inclusive classes without adjusting to continuous form.

1. Q. Modal class of frequencies 10, 15, 25, 20, 18 is which?
   A. Highest frequency = 25 ⇒ 3rd class is the modal class.
2. Q. For a symmetric distribution, mean = 30. What is the median?
   A. Approximately 30 (since 3M = Mode + 2·Mean and mode ≈ mean ⇒ M ≈ mean).

Classical (theoretical) probability of equally likely outcomes.

• Random experiment: outcome cannot be predicted with certainty.
• Sample space S = set of all possible outcomes; event E ⊆ S.
• For equally likely outcomes: P(E) = number of favourable outcomes / total number of outcomes.
• 0 ≤ P(E) ≤ 1 for every event E. P(impossible) = 0, P(sure) = 1.
• Complementary events: P(E) + P(not E) = 1, i.e. P(Ē) = 1 − P(E).
• For a coin toss: S = {H, T}, 2 outcomes. Two coins: 4 outcomes. n coins: 2ⁿ outcomes.
• For a die (6 faces): S = {1, 2, 3, 4, 5, 6}. Two dice: 36 outcomes (ordered pairs).
• For a standard deck of 52 cards: 4 suits (spades, hearts, diamonds, clubs), each with 13 cards. Face cards = J, Q, K (12 total); Aces = 4; red cards = 26; black = 26.
• Card-drawing 'without replacement' vs 'with replacement' matters — although at Class 10 level, single-draw is the usual setup.
• For 'at least' questions, complement is often easier: P(at least one) = 1 − P(none).
• Always write the total outcomes and favourable outcomes explicitly for full marks.

• P(E) = favourable outcomes / total outcomes
• 0 ≤ P(E) ≤ 1
• P(E) + P(not E) = 1

• Sample space → event → equally likely
• Complementary events
• Coins, dice, cards contexts

• List sample space explicitly for two-dice/coin problems.
• Use complement for 'at least one' style questions.

• Assuming outcomes are equally likely without checking.
• Confusing 'at least' with 'exactly'.
• Forgetting there are 12 face cards (or 16 if aces counted).

1. Q. Two dice are thrown. Find P(sum = 8).
   A. Favourable: (2,6), (3,5), (4,4), (5,3), (6,2) → 5. P = 5/36.
2. Q. One card is drawn from a well-shuffled deck. Find P(it is a red face card).
   A. Red face cards = 6 (J, Q, K of hearts and diamonds). P = 6/52 = 3/26.
3. Q. A bag has 3 red, 5 black, 4 white balls. One is drawn. Find P(not red).
   A. Total = 12; not red = 9. P = 9/12 = 3/4.