Geometry, Mensuration, Algebra & Probability
Triangles — Angles, Sides and the Four Centres
The angle sum property: the three angles of any triangle add to 180°. If the angles are in the ratio 2 : 3 : 4, then parts = 2 + 3 + 4 = 9 and one part = 180 ÷ 9 = 20°. The angles are 2 × 20 = 40°, 3 × 20 = 60° and 4 × 20 = 80°. Check: 40 + 60 + 80 = 180 ✓ — an acute-angled triangle, since even the largest angle 80° is below 90°. The second angle rule saves a step: the exterior angle of a triangle equals the sum of the two remote (opposite) interior angles. Example: two interior angles are 50° and 60°, so the exterior angle at the third vertex = 50 + 60 = 110°. Cross-check the long way — the third interior angle = 180 − 50 − 60 = 70°, and 70° forms a linear pair with its exterior angle, so that angle = 180 − 70 = 110° ✓ Same answer, one step instead of two.
- By sides: scalene (all three sides different), isosceles (two sides equal, and the angles opposite them are equal), equilateral (all three sides equal, so every angle = 180 ÷ 3 = 60°).
- By angles: acute (all angles < 90°), right (one angle = 90°), obtuse (one angle > 90°). A triangle can have at most one angle of 90° or more — two right angles would already use up all 180°.
- Isosceles worked example: the vertex angle is 80°, and the two base angles are equal, so each = (180 − 80) ÷ 2 = 100 ÷ 2 = 50°. Check: 80 + 50 + 50 = 180 ✓
Congruence (same shape and same size) is proved by SSS, SAS, ASA, AAS and, for right triangles only, RHS. Note that AAA is not a congruence test — it gives only similarity (same shape, possibly different size). Similarity criteria: AA (or AAA), SSS (sides in proportion), SAS (two sides in proportion with the included angle equal). In similar triangles the ratio of areas = the square of the ratio of corresponding sides. Worked example: if corresponding sides are in the ratio 3 : 5, the areas are in the ratio 3² : 5² = 9 : 25. So if the smaller triangle has area 36 cm², the larger has 36 × 25 ÷ 9 = 100 cm².
Pythagoras theorem: in a right triangle, (hypotenuse)² = (base)² + (height)². Worked example: base 9, height 12 → hypotenuse² = 81 + 144 = 225 → hypotenuse = 15. Learn the four exam triples cold — 3-4-5 (9 + 16 = 25), 5-12-13 (25 + 144 = 169), 8-15-17 (64 + 225 = 289) and 7-24-25 (49 + 576 = 625) — plus their multiples such as 6-8-10, 9-12-15 and 10-24-26. Worked example: a 25 m ladder rests against a wall with its foot 7 m away. Height reached = √(25² − 7²) = √(625 − 49) = √576 = 24 m — the 7-24-25 triple, spotted instantly.
A median joins a vertex to the midpoint of the opposite side. The three medians meet at the centroid, which divides every median in the ratio 2 : 1 measured from the vertex. Worked example: a median is 18 cm long. Then the vertex-to-centroid part = (2/3) × 18 = 12 cm and the centroid-to-midpoint part = (1/3) × 18 = 6 cm. Check: 12 + 6 = 18 ✓ and 12 : 6 = 2 : 1 ✓
| Centre | Meeting point of | Must-know property |
|---|---|---|
| Centroid (G) | The three medians | Divides each median 2 : 1 from the vertex; always inside; it is the centre of gravity. |
| Incentre (I) | The three internal angle bisectors | Centre of the inscribed circle; equidistant from the three sides; always inside. Also ∠BIC = 90° + A/2. |
| Circumcentre (O) | The three perpendicular bisectors of the sides | Centre of the circumscribed circle; equidistant from the three vertices. In a right triangle it is the midpoint of the hypotenuse; outside for an obtuse triangle. |
| Orthocentre (H) | The three altitudes | In a right triangle it is the vertex containing the right angle; outside for an obtuse triangle. |
| All four together | In an equilateral triangle | Centroid, incentre, circumcentre and orthocentre coincide at one single point. |
Quadrilaterals and Polygons
A quadrilateral has four sides, and its four angles always add to 360° (a diagonal splits it into two triangles: 2 × 180 = 360). Worked example: three angles of a quadrilateral are 70°, 85° and 110°. The fourth = 360 − (70 + 85 + 110) = 360 − 265 = 95°. Check: 70 + 85 + 110 + 95 = 360 ✓
| Figure | Sides & angles | Diagonals | Area |
|---|---|---|---|
| Parallelogram | Opposite sides equal and parallel; opposite angles equal; adjacent angles supplementary | Bisect each other; not equal, not perpendicular | base × height |
| Rhombus | All four sides equal; opposite angles equal | Bisect each other at 90°; not equal | ½ × d₁ × d₂ |
| Rectangle | Opposite sides equal; all angles 90° | Equal and bisect each other; not perpendicular | length × breadth |
| Square | All sides equal and all angles 90° | Equal and bisect at 90° — the best of both | a² = ½ d² |
| Trapezium | Exactly one pair of parallel sides (a and b) | No special property | ½ × (a + b) × height |
| Kite | Two pairs of adjacent sides equal | Perpendicular; the longer one bisects the shorter | ½ × d₁ × d₂ |
For a polygon with n sides: sum of interior angles = (n − 2) × 180°, and the sum of exterior angles is always 360°, whatever n may be. For a regular polygon each exterior angle = 360/n and each interior angle = (n − 2) × 180 / n. Worked example (hexagon, n = 6): sum of interior angles = (6 − 2) × 180 = 4 × 180 = 720°, so each interior angle = 720 ÷ 6 = 120°, and each exterior angle = 360 ÷ 6 = 60°. Check: 120 + 60 = 180 ✓ Worked example (pentagon, n = 5): sum = 3 × 180 = 540°, each interior = 540 ÷ 5 = 108°, each exterior = 360 ÷ 5 = 72°. Check: 108 + 72 = 180 ✓
The number of diagonals of an n-sided polygon is n(n − 3)/2 — each vertex joins to every vertex except itself and its two neighbours, and we divide by 2 because each diagonal is counted twice. Worked: quadrilateral, n = 4 → 4 × 1 ÷ 2 = 2; hexagon, n = 6 → 6 × 3 ÷ 2 = 9; octagon, n = 8 → 8 × 5 ÷ 2 = 20; decagon, n = 10 → 10 × 7 ÷ 2 = 35. Reverse-type question: each interior angle of a regular polygon is 144°, find n. Each exterior angle = 180 − 144 = 36°, so n = 360 ÷ 36 = 10 sides. Check: sum of interior angles = 8 × 180 = 1440, and 1440 ÷ 10 = 144 ✓
Circles — Chords, Tangents, Arcs and Sectors
Fix the vocabulary first. The radius (r) joins the centre to the circle; the diameter = 2r is the longest chord; a chord joins any two points on the circle; a secant is a line cutting the circle at two points; a tangent touches it at exactly one point; an arc is a piece of the boundary; a sector is the pizza-slice bounded by two radii and an arc; a segment is the piece cut off by a chord.
- The angle in a semicircle is 90°. So if AB is a diameter and C is any other point on the circle, ∠ACB = 90° — and if one of the other two angles is 55°, the third = 180 − 90 − 55 = 35°.
- The angle at the centre is twice the angle at the circumference standing on the same arc. Worked: if the arc subtends 80° at the centre, it subtends 80 ÷ 2 = 40° at any point on the remaining circumference. Read backwards: an angle of 35° at the circumference means 2 × 35 = 70° at the centre.
- In a cyclic quadrilateral (all four vertices on one circle) opposite angles add to 180°. Worked: if ∠A = 70° then ∠C = 180 − 70 = 110°; if ∠B = 85° then ∠D = 95°. Also an exterior angle equals the interior opposite angle — the exterior angle at A here is 110°.
- Equal chords are equidistant from the centre, and conversely chords equidistant from the centre are equal. The perpendicular from the centre to a chord bisects the chord — the single most useful chord fact.
- A tangent is perpendicular to the radius at the point of contact, and the two tangents drawn from an external point are equal in length.
Worked chord problem. A chord of length 24 cm is drawn in a circle of radius 13 cm — how far is it from the centre? Drop the perpendicular from the centre; it bisects the chord, giving a half-chord of 24 ÷ 2 = 12 cm. The radius, the half-chord and the distance now form a right triangle with the radius as hypotenuse: distance = √(13² − 12²) = √(169 − 144) = √25 = 5 cm (the 5-12-13 triple). Reverse it: a chord 8 cm from the centre of a circle of radius 17 cm has half-length √(289 − 64) = √225 = 15, so the chord = 30 cm.
Worked tangent problem. From a point 17 cm from the centre of a circle of radius 8 cm, a tangent is drawn. Since the tangent meets the radius at 90°, the radius, the tangent and the line to the centre form a right triangle with the 17 cm line as hypotenuse: tangent length = √(17² − 8²) = √(289 − 64) = √225 = 15 cm (the 8-15-17 triple).
Arc and sector. For a central angle θ: arc length = (θ/360) × 2πr and area of sector = (θ/360) × πr². Worked example with r = 21 cm, θ = 60°, π = 22/7. Circumference = 2 × (22/7) × 21 = 2 × 22 × 3 = 132 cm, so arc = (60/360) × 132 = 132 ÷ 6 = 22 cm. Area of the whole circle = (22/7) × 21 × 21 = 22 × 63 = 1386 cm², so the sector = 1386 ÷ 6 = 231 cm². Cross-check with the shortcut sector area = ½ × arc × r = ½ × 22 × 21 = 231 cm² ✓
(1) The centre angle is double the circumference angle, never half — given 100° at the centre, the answer is 50°, not 200°.
(2) In a cyclic quadrilateral it is the opposite angles that sum to 180°, not the adjacent ones.
(3) When the perpendicular distance from the centre is given, remember it reaches only the midpoint of the chord — always double the half-chord at the end. Forgetting that doubling is the commonest single mistake in chord questions.
Mensuration — Areas, Volumes and Surfaces
| Figure | Area | Perimeter / extra |
|---|---|---|
| Square (side a) | a², also ½ d² | 4a; diagonal = a√2 |
| Rectangle (l, b) | l × b | 2(l + b); diagonal = √(l² + b²) |
| Parallelogram | base × height | 2 × (sum of two adjacent sides) |
| Rhombus (diagonals d₁, d₂) | ½ d₁d₂ | 4a, where a = ½√(d₁² + d₂²) |
| Trapezium (parallel sides a, b; height h) | ½ (a + b) h | Sum of all four sides |
| Triangle (base b, height h) | ½ × b × h | a + b + c |
| Triangle — Heron formula (all three sides known) | √[s(s − a)(s − b)(s − c)], s = (a + b + c)/2 | s is the semi-perimeter |
| Equilateral triangle (side a) | (√3/4) a² | 3a; height = (√3/2) a |
| Circle (radius r) | πr² | Circumference = 2πr |
| Semicircle | ½ πr² | Perimeter = πr + 2r (do not forget the diameter) |
| Ring / annulus (outer R, inner r) | π(R² − r²) | Width of the ring = R − r |
2D worked examples, all arithmetic shown. (1) Heron: sides 13, 14, 15 → s = (13 + 14 + 15)/2 = 42/2 = 21; s − a = 8, s − b = 7, s − c = 6; product = 21 × 8 × 7 × 6 = 168 × 42 = 7056; area = √7056 = 84 cm². (2) Equilateral, a = 8: area = (√3/4) × 64 = 16√3 cm² ≈ 27.7 cm²; height = (√3/2) × 8 = 4√3 cm. (3) Circle, r = 7, π = 22/7: area = (22/7) × 49 = 22 × 7 = 154 cm²; circumference = 2 × (22/7) × 7 = 44 cm. (4) Semicircle, r = 7: area = 154 ÷ 2 = 77 cm²; perimeter = 22 + 14 = 36 cm. (5) Trapezium, parallel sides 12 cm and 8 cm, height 5 cm: area = ½ × (12 + 8) × 5 = ½ × 100 = 50 cm². (6) Ring, R = 14, r = 7: area = (22/7) × (196 − 49) = (22/7) × 147 = 22 × 21 = 462 cm².
| Solid | Volume | Curved / lateral SA | Total surface area |
|---|---|---|---|
| Cube (edge a) | a³ | 4a² (the four walls) | 6a²; diagonal = a√3 |
| Cuboid (l, b, h) | l × b × h | 2h(l + b) | 2(lb + bh + hl); diagonal = √(l² + b² + h²) |
| Cylinder (r, h) | πr²h | 2πrh | 2πr(r + h) |
| Cone (r, h; slant height l) | ⅓ πr²h | πrl, where l = √(r² + h²) | πr(l + r) |
| Sphere (r) | (4/3) πr³ | 4πr² — a sphere has only one surface | 4πr² |
| Hemisphere (r) | (2/3) πr³ | 2πr² | 3πr² = curved 2πr² + flat circle πr² |
Cube and cuboid worked. Cube of edge 6 cm: volume = 6³ = 216 cm³, TSA = 6 × 36 = 216 cm², lateral SA = 4 × 36 = 144 cm², diagonal = 6√3 cm. Cuboid 10 × 8 × 6 cm: volume = 480 cm³; TSA = 2(10×8 + 8×6 + 6×10) = 2(80 + 48 + 60) = 2 × 188 = 376 cm²; lateral SA = 2 × 6 × (10 + 8) = 12 × 18 = 216 cm²; diagonal = √(100 + 64 + 36) = √200 = 10√2 cm.
Curved solids worked, π = 22/7. Cylinder r = 7, h = 10: volume = (22/7) × 49 × 10 = 22 × 7 × 10 = 1540 cm³; CSA = 2 × (22/7) × 7 × 10 = 440 cm²; TSA = 440 + 2 × 154 = 748 cm². Cone r = 7, h = 24: first l = √(49 + 576) = √625 = 25 cm; volume = (1/3) × (22/7) × 49 × 24 = (1/3) × 22 × 7 × 24 = 22 × 7 × 8 = 1232 cm³; CSA = (22/7) × 7 × 25 = 550 cm²; TSA = 550 + 154 = 704 cm². Sphere r = 21: volume = (4/3) × (22/7) × 9261 = (4/3) × 22 × 1323 = 4 × 22 × 441 = 38,808 cm³; surface area = 4 × (22/7) × 441 = 4 × 22 × 63 = 5544 cm². Hemisphere r = 21: volume = 38,808 ÷ 2 = 19,404 cm³; curved SA = 2 × (22/7) × 441 = 2772 cm²; TSA = 3 × (22/7) × 441 = 3 × 1386 = 4158 cm².
Melting and recasting — one rule: the volume never changes. Write "old volume = new volume" and cancel π ruthlessly.
(1) A solid sphere of radius 6 cm is melted into small spheres of radius 2 cm — how many? Number = 6³/2³ = 216/8 = 27.
(2) A cylinder of radius 6 cm and height 15 cm is recast into cones of radius 3 cm and height 5 cm. Cylinder = π × 36 × 15 = 540π; one cone = ⅓ × π × 9 × 5 = 15π; number = 540π ÷ 15π = 36 cones.
(3) A sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm — find its height. Sphere = (4/3)π × 216 = 288π; cylinder = π × 16 × h; so 16h = 288 → h = 18 cm.
(4) A cube of edge 12 cm is melted into cubes of edge 3 cm. Number = 12³/3³ = 1728/27 = 64.
Algebra Essentials — Identities and Equations
| Identity | Worked use |
|---|---|
| (a + b)² = a² + 2ab + b² | 103² = 100² + 2(100)(3) + 3² = 10000 + 600 + 9 = 10609 |
| (a − b)² = a² − 2ab + b² | 98² = 10000 − 2(100)(2) + 4 = 10000 − 400 + 4 = 9604 |
| a² − b² = (a + b)(a − b) | 103 × 97 = (100 + 3)(100 − 3) = 10000 − 9 = 9991 |
| a² + b² = (a + b)² − 2ab | a + b = 7, ab = 12 → a² + b² = 49 − 24 = 25 |
| (a + b)³ = a³ + b³ + 3ab(a + b) | a + b = 7, ab = 12 → a³ + b³ = 343 − 3(12)(7) = 343 − 252 = 91 |
| (a − b)³ = a³ − b³ − 3ab(a − b) | a = 8, b = 5: (3)³ = 27 = 512 − 125 − 3(40)(3) = 387 − 360 = 27 ✓ |
| a³ + b³ = (a + b)(a² − ab + b²) | With a + b = 7, a² + b² = 25, ab = 12: 7 × (25 − 12) = 7 × 13 = 91 ✓ (matches above) |
| a³ − b³ = (a − b)(a² + ab + b²) | 8³ − 5³ = 3 × (64 + 40 + 25) = 3 × 129 = 387 ✓ (512 − 125 = 387) |
| a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) | So if a + b + c = 0 then a³ + b³ + c³ = 3abc. Check with −5, 2, 3: −125 + 8 + 27 = −90, and 3(−5)(2)(3) = −90 ✓ |
Identities are not decoration — they turn ugly arithmetic into two lines. Squaring a number near a round figure: 103² = (100 + 3)² = 10000 + 600 + 9 = 10609 and 98² = (100 − 2)² = 10000 − 400 + 4 = 9604. Multiplying a pair equally spaced about a round figure: 97 × 103 = 100² − 3² = 9991, and 48 × 52 = 50² − 2² = 2500 − 4 = 2496. Notice the pattern: always find the midpoint and the gap.
Linear equations. One variable: 3x + 7 = 25 → 3x = 25 − 7 = 18 → x = 6. Check: 3(6) + 7 = 25 ✓ Two variables (substitution): solve 2x + 3y = 13 and 4x − y = 5. From the second, y = 4x − 5. Substitute: 2x + 3(4x − 5) = 13 → 2x + 12x − 15 = 13 → 14x = 28 → x = 2, then y = 4(2) − 5 = 3. Check both: 2(2) + 3(3) = 4 + 9 = 13 ✓ and 4(2) − 3 = 5 ✓
Quadratic equations. For ax² + bx + c = 0 the roots are x = (−b ± √D)/2a where the discriminant D = b² − 4ac. If D > 0 there are two distinct real roots, if D = 0 the two roots are real and equal, and if D < 0 there is no real root. Also sum of roots = −b/a and product of roots = c/a. Worked: x² − 7x + 12 = 0 → D = 49 − 48 = 1 > 0 → x = (7 ± 1)/2 = 4 and 3; sum 4 + 3 = 7 = −(−7)/1 ✓, product 4 × 3 = 12 = c/a ✓. Worked: 2x² − 7x + 3 = 0 → D = 49 − 24 = 25 → x = (7 ± 5)/4 = 3 and ½; sum = 3½ = 7/2 ✓, product = 3/2 ✓. Worked: x² − 6x + 9 = 0 → D = 36 − 36 = 0 → equal roots x = 3, 3. Worked: x² + x + 1 = 0 → D = 1 − 4 = −3 < 0 → no real root. Worked: for what k does x² + kx + 16 = 0 have equal roots? D = k² − 64 = 0 → k = ±8.
(1) The sum of the roots is −b/a, with a minus sign, while the product is +c/a with none. For x² − 5x + 6 = 0 the sum is +5, not −5.
(2) D < 0 does not mean "no solution" in general — it means no real root, and in an exam the correct option is usually worded "roots are imaginary" or "no real roots". Read the options carefully before you tick.
Turning words into equations. Translate phrase by phrase: "is" → =, "of" → ×, "more than" → +, "less than" → −, "twice" → 2×, "after n years" → add n to every age.
(1) A father is 3 times as old as his son; after 12 years he will be twice as old. Find their present ages. Let the son be x, so the father is 3x. Then 3x + 12 = 2(x + 12) → 3x + 12 = 2x + 24 → x = 12, father = 36. Check: after 12 years 48 and 24, and 48 = 2 × 24 ✓
(2) The sum of two numbers is 15 and their product is 56. Then x² − 15x + 56 = 0 → D = 225 − 224 = 1 → x = (15 ± 1)/2 = 8 and 7. Check: 8 + 7 = 15 ✓, 8 × 7 = 56 ✓
(3) The perimeter of a rectangle is 44 cm and its length is 4 cm more than its breadth. Then l + b = 22 and l = b + 4 → 2b + 4 = 22 → b = 9 cm, l = 13 cm, area = 117 cm². Check: 2(13 + 9) = 44 ✓
Permutation, Combination and Probability
Everything here rests on the fundamental counting principle: if one job can be done in m ways and, independently, a second in n ways, the pair can be done in m × n ways. With 3 shirts and 4 trousers you get 3 × 4 = 12 outfits; add 2 pairs of shoes and it is 3 × 4 × 2 = 24. The other tool is the factorial: n! = n × (n − 1) × … × 1, and by definition 0! = 1. So 3! = 6, 4! = 24, 5! = 120, 6! = 720, 7! = 5040. Useful habit: cancel factorials instead of expanding them — 7!/5! = 7 × 6 = 42.
Permutation = arrangement, where order matters: nPr = n!/(n − r)!. Combination = selection, where order does not matter: nCr = n!/[r!(n − r)!]. They are linked by nPr = nCr × r!. Worked: 5P3 = 5!/2! = 120/2 = 60; 5C3 = 5!/(3!·2!) = 120/12 = 10; check 10 × 3! = 10 × 6 = 60 ✓ Key shortcut: nCr = nC(n − r), so 5C3 = 5C2 = (5 × 4)/2 = 10 and 20C18 = 20C2 = (20 × 19)/2 = 190 — never expand 20!. Also nC0 = nCn = 1 and nC1 = n. More practice: 10C2 = 45, 8C3 = (8 × 7 × 6)/6 = 56, 7P2 = 7 × 6 = 42.
Arranging letters with repeats. If a word of n letters has one letter repeated p times, another q times and so on, the number of distinct arrangements is n!/(p! × q! × …). Worked: PATNA has 5 letters with A twice → 5!/2! = 120/2 = 60. BANANA has 6 letters with A three times and N twice → 6!/(3! × 2!) = 720/(6 × 2) = 60. LEVEL has 5 letters with L twice and E twice → 5!/(2! × 2!) = 120/4 = 30. BIHAR has 5 distinct letters → 5! = 120. And for a round table, one seat is fixed as reference, so n people sit in (n − 1)! ways: 5 people → 4! = 24.
Choosing a committee. From 6 men and 4 women a committee of 3 is to be formed. Total ways = 10C3 = (10 × 9 × 8)/6 = 120. Exactly 1 woman = 4C1 × 6C2 = 4 × 15 = 60. Exactly 2 women = 4C2 × 6C1 = 6 × 6 = 36. All 3 women = 4C3 = 4. No woman = 6C3 = 20. Check that the cases exhaust everything: 20 + 60 + 36 + 4 = 120 ✓ And at least one woman = 120 − 20 = 100 — computing the complement is far faster than adding 60 + 36 + 4.
| Experiment | Total outcomes | Sample event | Probability |
|---|---|---|---|
| One coin | 2 (H, T) | A head | 1/2 |
| Two coins | 4 (HH, HT, TH, TT) | Exactly one head | 2/4 = 1/2 |
| Three coins | 2³ = 8 | Exactly two heads | 3/8 |
| Three coins | 8 | At least one head | 1 − 1/8 = 7/8 |
| One die | 6 | A prime number (2, 3, 5) | 3/6 = 1/2 |
| Two dice | 6 × 6 = 36 | Sum = 7 | 6/36 = 1/6 |
| Two dice | 36 | A doublet (same number on both) | 6/36 = 1/6 |
| Pack of 52 cards | 52 (4 suits × 13; 26 red, 26 black) | A king | 4/52 = 1/13 |
| Pack of 52 cards | 52 | A face card (J, Q, K — 12 of them) | 12/52 = 3/13 |
Probability = favourable outcomes ÷ total outcomes, always a number between 0 (impossible) and 1 (certain), and P(not E) = 1 − P(E). Worked with two dice (36 outcomes): sum = 7 happens as 1-6, 2-5, 3-4, 4-3, 5-2, 6-1 → 6 cases → 6/36 = 1/6; sum = 8 happens as 2-6, 3-5, 4-4, 5-3, 6-2 → 5/36; sum ≥ 10 means 10 (3 ways), 11 (2 ways), 12 (1 way) = 6 cases → 1/6. Worked with three coins (8 outcomes): exactly two heads = HHT, HTH, THH → 3/8; at least one head = 1 − P(no head) = 1 − 1/8 = 7/8 — the complement saves you listing seven cases. Worked with a pack of 52: P(red) = 26/52 = 1/2, P(ace) = 4/52 = 1/13, P(ace or king) = 8/52 = 2/13.
Drawing without replacement, and the two "or/and" rules. A bag holds 5 red and 3 black balls and 2 balls are drawn together. Total ways = 8C2 = 28. P(both red) = 5C2/8C2 = 10/28 = 5/14; P(one of each) = (5 × 3)/28 = 15/28; P(both black) = 3C2/28 = 3/28. Check: 10 + 15 + 3 = 28 ✓ Same answer step by step: 5/8 × 4/7 = 20/56 = 5/14 — the denominator drops from 8 to 7 precisely because the first ball is not replaced. Now the rules: mutually exclusive events cannot happen together, so P(A or B) = P(A) + P(B) — a die showing a 2 or a 5 is 1/6 + 1/6 = 1/3. Independent events do not affect each other, so P(A and B) = P(A) × P(B) — both dice showing 6 is 1/6 × 1/6 = 1/36.
Key takeaways
- Angles: complementary = 90°, supplementary = 180°; on parallel lines corresponding and alternate angles are equal and co-interior angles sum to 180°.
- Triangle: angles sum to 180°, exterior angle = sum of two remote interior angles, any two sides > the third. Triples 3-4-5, 5-12-13, 8-15-17, 7-24-25. Centroid splits a median 2 : 1 from the vertex.
- Polygons: interior sum = (n − 2) × 180°, exterior sum = 360° always, diagonals = n(n − 3)/2 (octagon → 20). Rhombus diagonals are perpendicular but unequal; rectangle diagonals equal but not perpendicular.
- Circle: angle in a semicircle = 90°, centre angle = twice the circumference angle, cyclic quadrilateral opposite angles = 180°, tangent ⟂ radius; arc = (θ/360)2πr, sector = (θ/360)πr² = ½ × arc × r.
- Mensuration: rhombus = ½d₁d₂, trapezium = ½(a+b)h, Heron for 13-14-15 → 84; a side up 20% lifts area by 44%. Cylinder πr²h, cone ⅓πr²h (CSA πrl), sphere (4/3)πr³, hemisphere TSA 3πr²; cone : hemisphere : cylinder = 1 : 2 : 3.
- Melting and recasting keeps volume constant — a sphere of radius 6 gives 216/8 = 27 spheres of radius 2. Algebra: a² + b² = (a+b)² − 2ab, a³ + b³ = (a+b)³ − 3ab(a+b); D = b² − 4ac, sum of roots = −b/a, product = c/a.
- Counting: nPr = n!/(n−r)!, nCr = n!/[r!(n−r)!], nCr = nC(n−r); PATNA → 5!/2! = 60. Probability: two dice → 36 outcomes, P(sum 7) = 1/6; P(not E) = 1 − P(E); overlapping "or" → subtract, independent "and" → multiply.
Practice questions (8)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle is:
Show answer
Answer: C — 80°
Parts = 2 + 3 + 4 = 9 and the angles sum to 180°, so one part = 180 ÷ 9 = 20°. The angles are 40°, 60° and 80°; the largest is 80°. Check: 40 + 60 + 80 = 180.How many diagonals does an octagon have?
Show answer
Answer: A — 20
Number of diagonals = n(n − 3)/2. For n = 8: 8 × 5 ÷ 2 = 40 ÷ 2 = 20. (28 is the trap — that is 8C2 = 28, which counts all lines including the 8 sides.)Consider the following statements:
1. The diagonals of a rhombus are equal to each other.
2. The diagonals of a rectangle bisect each other.
Which is/are correct?Show answer
Answer: B — Only 2
Statement 1 is wrong: the diagonals of a rhombus bisect each other at 90° but are NOT equal (e.g. 16 cm and 12 cm). Statement 2 is correct: a rectangle has diagonals that are equal and bisect each other, though they are not perpendicular.A chord of length 24 cm is drawn in a circle of radius 13 cm. Its distance from the centre is:
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Answer: A — 5 cm
The perpendicular from the centre bisects the chord, so the half-chord is 24 ÷ 2 = 12 cm. Then distance = √(13² − 12²) = √(169 − 144) = √25 = 5 cm — the 5-12-13 triple.If each side of a square is increased by 20%, its area increases by:
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Answer: B — 44%
Take a side of 10 → area 100. New side = 12 → new area = 144. Increase = 144 − 100 = 44, i.e. 44%. By formula: 20 + 20 + (20 × 20)/100 = 44%. Answering 40% forgets that area involves two dimensions.The curved surface area of a cone of radius 7 cm and height 24 cm (take π = 22/7) is:
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Answer: D — 550 cm²
Slant height l = √(7² + 24²) = √(49 + 576) = √625 = 25 cm. CSA = πrl = (22/7) × 7 × 25 = 550 cm². (704 cm² is the total surface area 550 + 154, 1232 cm³ is the volume, 154 cm² is the base area.)If a + b = 7 and ab = 12, then a³ + b³ equals:
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Answer: B — 91
a³ + b³ = (a + b)³ − 3ab(a + b) = 343 − 3 × 12 × 7 = 343 − 252 = 91. Verify directly: a and b are 3 and 4, and 27 + 64 = 91.Consider the following statements:
1. The letters of the word PATNA can be arranged in 60 distinct ways.
2. When two dice are thrown together, the probability that the sum is 7 is 1/6.
Which is/are correct?Show answer
Answer: C — Both 1 and 2
Both are correct. PATNA has 5 letters with A repeated twice, so arrangements = 5!/2! = 120/2 = 60. Two dice give 6 × 6 = 36 outcomes, and a sum of 7 occurs in 6 of them (1-6, 2-5, 3-4, 4-3, 5-2, 6-1), so the probability = 6/36 = 1/6.