Geometry and mensuration reward one skill above all: recognising which standard fact a question is hiding. The formulas are fixed and few. What loses marks is forgetting one, mixing up two similar ones, or doing a long calculation where a shortcut exists. This guide puts every formula you need for SSC CGL, CHSL, CPO and railway exams in tables, shows the fastest reliable method for each question type, and works through 30 original examples. Every number below has been checked step by step.
This post goes deeper than a quick list of tricks. It covers coordinate basics, all the solids including the frustum, and recasting problems. If you only want the short triangle and circle shortcuts, the earlier geometry shortcuts post covers them. Here you get the complete reference and a lot more practice.
What the syllabus actually covers
The published SSC quantitative syllabus lists, under geometry, triangles and their centres, circles with chords, tangents and common tangents, and angles in segments. Under mensuration it lists triangles, quadrilaterals, regular polygons, circles, right prisms, right circular cone and cylinder, sphere, hemisphere, rectangular parallelepiped (cuboid) and the regular right pyramid with a triangular or square base. Frustum questions appear as cone-related problems.
How many questions come from these chapters changes from exam to exam and shift to shift. Public analyses of past papers give small numbers per paper for Tier 1, but check the latest notification and your own past-paper count rather than trusting any chart. Mensuration questions are usually the quickest marks in the paper, because they are direct formula application.
Triangles: centres, angles and standard results
Draw the figure and label every given value before touching a formula. Figures are often not to scale, so trust numbers, not looks.
Triangle centres and the angle rules
| Centre | Formed by | Key facts |
|---|---|---|
| Centroid G | Medians | Divides each median 2 : 1 from the vertex. Each median splits the area into halves; the three medians make six equal-area parts. |
| Incentre I | Angle bisectors | Equidistant from the sides (distance = inradius). Angle BIC = 90 + A/2. |
| Circumcentre O | Perpendicular bisectors of sides | Equidistant from the vertices. Angle BOC = 2A (for acute A). In a right triangle it is the midpoint of the hypotenuse. |
| Orthocentre H | Altitudes | Angle BHC = 180 - A (for acute triangles). In a right triangle it is the right-angle vertex. |
| Excentre (opposite A) | Bisector of A and external bisectors of B and C | Angle BIaC = 90 - A/2. |
In an equilateral triangle all four centres coincide. In an isosceles triangle they lie on the axis of symmetry.
Triangle formulas
| Item | Formula |
|---|---|
| Area | (1/2) x base x height; (1/2)ab sin C; Heron: √(s(s-a)(s-b)(s-c)) with s = (a+b+c)/2 |
| Inradius r | Area / s |
| Circumradius R | abc / (4 x Area) |
| Right triangle | r = (a + b - c)/2; R = c/2; altitude to hypotenuse = ab/c |
| Equilateral, side a | Area = (√3/4)a²; height = (√3/2)a; r = a/(2√3) = height/3; R = a/√3 = 2 x height/3 |
| Isosceles, equal sides a, base b | Height = √(a² - b²/4) |
| Median length (Apollonius) | AB² + AC² = 2(AD² + BD²), D midpoint of BC |
| Angle bisector | BD : DC = AB : AC; length² = AB x AC - BD x DC |
| Mid-point theorem | Segment joining midpoints of two sides is parallel to the third and half of it |
| Similar triangles | Sides in ratio k, areas k², perimeters k, corresponding heights and medians k |
| Right triangle, altitude to hypotenuse | h² = p x q (p, q the two segments of hypotenuse); leg² = hypotenuse x adjacent segment |
Congruence and similarity tests
| Congruent (identical) | Similar (same shape) |
|---|---|
| SSS, SAS, ASA, AAS, RHS | AA (two angles), SSS ratio, SAS ratio |
Note that SSA is not a valid test, except in the right-angle case (RHS). Matching the order of letters matters: if triangle ABC is similar to PQR, then A matches P, B matches Q, C matches R.
Special triangles and triplets
| Type | Sides |
|---|---|
| 45-45-90 | 1 : 1 : √2 |
| 30-60-90 | 1 : √3 : 2 (short side opposite 30 degrees) |
| Triplets | 3-4-5, 5-12-13, 8-15-17, 7-24-25, 20-21-29, 9-40-41, 12-35-37 and multiples |
| Isosceles right | Legs a, hypotenuse a√2, area a²/2 |
Worked examples: triangles
Example 1. Angle A of triangle ABC is 64 degrees. Find angle BIC at the incentre. Answer: 90 + 64/2 = 122 degrees.
Example 2. In an acute triangle, angle BOC at the circumcentre is 130 degrees. Find angle BHC at the orthocentre. Here 2A = 130, so A = 65. Angle BHC = 180 - 65 = 115 degrees.
Example 3. In triangle ABC, AB = 7, AC = 9, BC = 8. Find the median AD. AB² + AC² = 49 + 81 = 130. Then 2(AD² + 16) = 130, so AD² = 65 - 16 = 49 and AD = 7.
Example 4. DE is parallel to BC, with AD = 4 and DB = 6. The area of triangle ABC is 250. Find the area of the trapezium DBCE. AD : AB = 4 : 10 = 2 : 5. Area ratio = 4 : 25, so area of ADE = 250 x 4/25 = 40. Trapezium = 250 - 40 = 210.
Example 5. In triangle ABC, AB = 12, AC = 18, BC = 20. The bisector of A meets BC at D. Find BD, DC and AD. BD : DC = 12 : 18 = 2 : 3, so BD = 8 and DC = 12. AD² = 12 x 18 - 8 x 12 = 216 - 96 = 120, so AD = √120, about 10.95.
Example 6. Legs of a right triangle are 9 and 12. Find the altitude on the hypotenuse and the two segments. Hypotenuse = 15. Altitude = (9 x 12)/15 = 7.2. Segments: 81/15 = 5.4 and 144/15 = 9.6. Check: 5.4 + 9.6 = 15 and 5.4 x 9.6 = 51.84 = 7.2².
Example 7. Sides 17, 25, 28. Find area, inradius, circumradius and the altitude on the longest side. s = 35. Area = √(35 x 18 x 10 x 7) = √44100 = 210. Inradius = 210/35 = 6. Circumradius = (17 x 25 x 28)/(4 x 210) = 11900/840 = 85/6, about 14.17. Altitude on 28 = 420/28 = 15.
Example 8. The inradius of an equilateral triangle is 5 cm. Find its area. Height = 3r = 15. Side = 2 x 15/√3 = 10√3. Area = (√3/4) x 300 = 75√3, about 129.9 square cm.
Example 9. In a 30-60-90 triangle the hypotenuse is 14. Find the other two sides and the area. Short side = 7, long side = 7√3. Area = (1/2) x 7 x 7√3 = 49√3/2, about 42.44.
Example 10. Find the inradius of a right triangle with sides 9, 40, 41. r = (9 + 40 - 41)/2 = 4. Check with Area/s: area = 180, s = 45, 180/45 = 4.
Circles: facts and formulas
| Fact | Statement |
|---|---|
| Angle at centre | Twice the angle at any point on the remaining circle |
| Semicircle | Angle in a semicircle = 90 degrees |
| Same segment | Angles in the same segment are equal |
| Cyclic quadrilateral | Opposite angles sum to 180; exterior angle = interior opposite angle |
| Tangent | Perpendicular to radius at contact; two tangents from a point are equal; length = √(d² - r²) |
| Alternate segment | Angle between tangent and chord = angle in the alternate segment |
| Chord | Perpendicular from centre bisects the chord; half-chord = √(r² - d²) |
| Intersecting chords (inside) | PA x PB = PC x PD |
| Secants from outside | PA x PB = PC x PD; tangent: PT² = PA x PB |
| Direct common tangent | √(d² - (r1 - r2)²) |
| Transverse common tangent | √(d² - (r1 + r2)²) |
| Tangent quadrilateral | If a circle touches all four sides, AB + CD = BC + DA |
Number of common tangents: 4 for separate circles, 3 for externally touching, 2 for overlapping, 1 for internally touching, 0 when one lies strictly inside the other. A circle through the three vertices has the circumradius formula above. For a square, the inscribed circle has radius a/2 and the circumscribed circle has radius a√2/2.
Worked examples: circles
Example 11. In a cyclic quadrilateral ABCD, angles A, B, C are in the ratio 2 : 3 : 4. Find angle D. A + C = 180 gives 6k = 180, k = 30. So A = 60, B = 90, C = 120. D = 180 - B = 90 degrees. Check: 60 + 90 + 120 + 90 = 360.
Example 12. Tangents PA and PB are drawn from P to a circle with centre O, and angle APB = 50 degrees. Find angle AOB and the angle in the alternate segment. Quadrilateral OAPB has two right angles, so AOB = 360 - 90 - 90 - 50 = 130. Triangle PAB is isosceles, so angle PAB = (180 - 50)/2 = 65, which equals the angle in the alternate segment.
Example 13. A chord of length 30 is in a circle of radius 17. How far is it from the centre? Half-chord = 15. Distance = √(289 - 225) = √64 = 8.
Example 14. Two secants from P meet a circle at A, B and C, D. PA = 4, AB = 5 and PC = 3. Find CD. PB = 9, so PA x PB = 36. PD = 36/3 = 12. CD = 12 - 3 = 9.
Example 15. Centres of two circles of radii 12 and 5 are 25 apart. Find the direct common tangent. √(625 - 49) = √576 = 24.
Example 16. Centres of circles of radii 5 and 3 are 17 apart. Find the transverse common tangent. √(289 - 64) = √225 = 15.
Quadrilaterals and polygons
| Figure | Area | Other facts |
|---|---|---|
| Rectangle | l x b | Diagonal √(l² + b²); perimeter 2(l + b) |
| Square | a² or d²/2 | Diagonal a√2 |
| Parallelogram | base x height; ab sin(angle) | Diagonals bisect each other; opposite angles equal |
| Rhombus | (d1 x d2)/2 | Diagonals bisect at right angles; side = √((d1/2)² + (d2/2)²) |
| Trapezium | (1/2)(a + b)h | Median = (a + b)/2 |
| Kite | (d1 x d2)/2 | Diagonals perpendicular |
| Any quadrilateral | (1/2) x diagonal x (sum of perpendiculars to it) | Angle sum 360 |
| Polygon (n sides) | Formula |
|---|---|
| Sum of interior angles | (n - 2) x 180 |
| Regular: each exterior angle | 360/n |
| Regular: each interior angle | 180 - 360/n |
| Diagonals | n(n - 3)/2 |
| Regular hexagon, side a | Area = (3√3/2)a²; R = a; r = (√3/2)a |
| Regular octagon, side a | Area = 2(1 + √2)a² |
Exterior angles of any convex polygon always sum to 360.
Worked examples: quadrilaterals and polygons
Example 17. A rhombus has diagonals 24 and 10. Find side, perimeter, area and the altitude. Side = √(144 + 25) = 13. Perimeter = 52. Area = 120. Altitude = 120/13, about 9.23.
Example 18. An isosceles trapezium has parallel sides 10 and 26, and legs 10. Find its area. Half of the difference of parallel sides = 8. Height = √(100 - 64) = 6. Area = (1/2)(36)(6) = 108.
Example 19. A parallelogram has sides 12 and 9 with an angle of 30 degrees between them. Area = 12 x 9 x sin 30 = 108 x 1/2 = 54.
Example 20. The sum of the interior angles of a regular polygon is 1440 degrees. Find the sides, each angle and the diagonals. n - 2 = 8, so n = 10. Each angle = 144. Diagonals = 10 x 7/2 = 35.
Example 21. Each interior angle of a regular polygon is 5 times its exterior angle. Find n. Interior + exterior = 180, so exterior = 30. n = 360/30 = 12.
Coordinate geometry basics
| Item | Formula |
|---|---|
| Distance | √((x2 - x1)² + (y2 - y1)²) |
| Midpoint | ((x1 + x2)/2, (y1 + y2)/2) |
| Section (internal, m : n) | ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)) |
| Centroid | ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3) |
| Triangle area | (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|; zero means collinear |
| Slope | (y2 - y1)/(x2 - x1); parallel lines have equal slopes, perpendicular lines have product -1 |
| Line forms | y = mx + c; x/a + y/b = 1 (intercepts a, b) |
Example 22. Distance between (-2, 3) and (4, 11). Differences are 6 and 8. Distance = √(36 + 64) = 10.
Example 23. A point divides A(1, 2) to B(6, 12) in the ratio 2 : 3. x = (2 x 6 + 3 x 1)/5 = 3. y = (2 x 12 + 3 x 2)/5 = 6. The point is (3, 6).
Example 24. Find the area of the triangle with vertices (1, 1), (5, 1), (3, 7), and its centroid. Area = (1/2)|1(1 - 7) + 5(7 - 1) + 3(1 - 1)| = (1/2)|-6 + 30 + 0| = 12. Check: base 4 along y = 1, height 6, area (1/2)(4)(6) = 12. Centroid = (9/3, 9/3) = (3, 3).
Example 25. For what k are (1, 2), (3, k), (5, 8) collinear? Slope from first to third = 6/4 = 3/2. Slope from first to second = (k - 2)/2 = 3/2, so k - 2 = 3 and k = 5.
Mensuration of plane figures
| Figure | Area | Perimeter or length |
|---|---|---|
| Circle | πr² | 2πr |
| Semicircle | πr²/2 | πr + 2r |
| Sector (angle θ) | (θ/360)πr² = (1/2) x arc x r | Arc = (θ/360) x 2πr |
| Ring (R outer, r inner) | π(R² - r²) = π(R + r)(R - r) | - |
| Segment | Sector area - triangle area | - |
Use 22/7 when the radius or diameter is a multiple of 7, and follow the value of π given in the question. A wheel's number of revolutions = distance covered / circumference.
Example 26. A sector has radius 21 and angle 60 degrees (use 22/7). Arc = (1/6) x 2 x (22/7) x 21 = 22. Area = (1/6) x (22/7) x 441 = 231. Check with (1/2) x arc x r = (1/2)(22)(21) = 231.
Example 27. Two concentric circles have radii 14 and 7. Find the ring area. (22/7)(196 - 49) = (22/7)(147) = 462.
Example 28. A wheel has diameter 70 cm. How many revolutions will it make in 1.1 km? Circumference = (22/7) x 70 = 220 cm. Distance = 110000 cm. Revolutions = 110000/220 = 500.
3D mensuration: every formula in one table
Let l be the slant height of a cone, and for the frustum let R and r be the radii of the bigger and smaller circular faces.
| Solid | Volume | Curved or lateral surface area | Total surface area |
|---|---|---|---|
| Cube (edge a) | a³ | 4a² | 6a² |
| Cuboid (l, b, h) | lbh | 2h(l + b) | 2(lb + bh + hl) |
| Cylinder | πr²h | 2πrh | 2πr(r + h) |
| Hollow cylinder (R, r) | πh(R² - r²) | 2πh(R + r) | 2π(R + r)(R - r + h) |
| Cone | (1/3)πr²h | πrl | πr(r + l) |
| Sphere | (4/3)πr³ | 4πr² | 4πr² |
| Hemisphere | (2/3)πr³ | 2πr² | 3πr² |
| Frustum of cone | (πh/3)(R² + Rr + r²) | π(R + r)l | π(R + r)l + πR² + πr² |
| Right prism | base area x h | base perimeter x h | lateral + 2 x base area |
| Regular pyramid | (1/3) x base area x h | (1/2) x base perimeter x slant height | lateral + base area |
| Relation | Formula |
|---|---|
| Cone slant height | l = √(r² + h²) |
| Frustum slant height | l = √(h² + (R - r)²) |
| Cube diagonal | a√3 |
| Cuboid diagonal | √(l² + b² + h²) |
| Cone made from sector (radius L, angle θ) | Slant height = L; base circumference = (θ/360) x 2πL |
| Similar solids, ratio k | Surface areas k², volumes k³ |
Worked examples: solids
Example 29. A cube has diagonal 9√3 cm. Find edge, total surface area and volume. Edge = 9. TSA = 6 x 81 = 486. Volume = 729.
Example 30. A cuboid is 12 by 4 by 3. Find diagonal, total surface area and volume. Diagonal = √(144 + 16 + 9) = √169 = 13. TSA = 2(48 + 12 + 36) = 192. Volume = 144.
Example 31. A solid cylinder has r = 7 and h = 20 (use 22/7). CSA = 2 x (22/7) x 7 x 20 = 880. Two bases = 2 x (22/7) x 49 = 308. TSA = 1188. Volume = (22/7) x 49 x 20 = 3080.
Example 32. A cone has r = 7 and h = 24. Find l, CSA, TSA and volume. l = √(49 + 576) = 25. CSA = (22/7) x 7 x 25 = 550. Base = 154, so TSA = 704. Volume = (1/3) x 154 x 24 = 1232.
Example 33. A sector of radius 15 cm and angle 216 degrees is rolled into a cone. Find the volume in terms of π. Arc = (216/360) x 2π x 15 = 18π. So 2πr = 18π and r = 9. Slant height 15, so h = √(225 - 81) = 12. Volume = (1/3)π x 81 x 12 = 324π.
Example 34. A hemisphere has radius 7. Find CSA, TSA and volume. CSA = 2 x (22/7) x 49 = 308. TSA = 3 x (22/7) x 49 = 462. Volume = (2/3)(22/7)(343) = 15092/21 = 2156/3, about 718.67.
Example 35. A sphere has radius 10.5. Surface area = 4 x (22/7) x 110.25 = 1386. Volume = (4/3)(22/7)(1157.625) = 4851.
Example 36. A frustum has R = 14, r = 7, h = 24 (use 22/7). l = √(576 + 49) = 25. CSA = π x 21 x 25 = 525π = 1650. Volume = (24π/3)(196 + 98 + 49) = 8 x 343π = 2744π = 8624. Check with cones: the full cone height H satisfies H/(H - 24) = 14/7, so H = 48. Full cone = (1/3)π x 196 x 48 = 3136π. Cut-off cone = (1/3)π x 49 x 24 = 392π. Difference = 2744π. Matches.
Example 37. A right triangular prism has a 3-4-5 triangle base and height 10. Volume = 6 x 10 = 60. Lateral area = 12 x 10 = 120. TSA = 120 + 12 = 132.
Example 38. A square pyramid has base edge 6 and height 4. Slant height = √(9 + 16) = 5. Lateral area = (1/2)(24)(5) = 60. Volume = (1/3)(36)(4) = 48.
Recasting, melting and displacement
In these problems the volume stays the same. Equate volumes, cancel π and common factors, and solve for the unknown.
Example 39. A sphere of radius 6 cm is melted into a cylinder of radius 3 cm. Find the height. Sphere = (4/3)π x 216 = 288π. Cylinder = π x 9 x h. h = 288/9 = 32 cm.
Example 40. Three cubes with edges 3, 4 and 5 are melted into one cube. Total volume = 27 + 64 + 125 = 216, so the edge is 6.
Example 41. A sphere of radius 3 is dropped into a cylinder of radius 6 holding water, and is fully submerged. By how much does the water rise? Sphere = 36π. Rise = 36π/(π x 36) = 1 cm.
Example 42. If the radius of a sphere increases by 10 percent, what is the increase in surface area and volume? Area factor = 1.1² = 1.21, so up 21 percent. Volume factor = 1.1³ = 1.331, so up 33.1 percent.
Example 43. A square is inscribed in a circle of radius 10. Find its area. Diagonal = 20. Area = 20²/2 = 200. A circle inscribed in a square of side 14 has area (22/7) x 49 = 154.
The fastest method for each question type
| Question type | Fastest reliable method |
|---|---|
| Angle at a triangle centre | Use the table: 90 + A/2, 2A, 180 - A, 90 - A/2 |
| Right triangle sides | Look for a triplet before squaring anything |
| Triangle with three sides | Check for 13-14-15 or 17-25-28 style numbers; use Heron with s - a computed first |
| Parallel line in a triangle | Similar triangles; find the small triangle, then subtract |
| Chord or tangent length | Radius, perpendicular and half-chord form a right triangle; look for a triplet |
| Angles in a circle | Mark the centre angle, semicircle and same-segment angles first |
| Polygon with given angle | Convert to the exterior angle and divide into 360 |
| Coordinate area | If one side is horizontal or vertical, use base x height; otherwise the determinant formula |
| Cone, cylinder, sphere ratio | Cancel π and common factors before multiplying |
| Recasting | Equate volumes, cancel π, count the number of pieces by division |
| Percentage change in dimension | Use factors: area k², volume k³ |
For options-based questions, estimate first. If an answer is far outside the reasonable range of your sketch, drop it. Avoid decimals until the last step, and keep roots unevaluated till then.
Common traps
- Wrong centre rule. BIC = 90 + A/2 is only for the incentre. BOC = 2A is for the circumcentre. Orthocentre is 180 - A.
- Obtuse triangle centres. The formulas BOC = 2A and BHC = 180 - A are for acute triangles. For an obtuse angle, check by drawing.
- Slant height versus height. Cone CSA uses l, volume uses h. Frustum CSA uses the slant height, not the vertical height.
- Direct versus transverse tangent. Direct uses the difference of radii, transverse uses the sum.
- Hemisphere total surface. It is 3πr², not 2πr². The flat base is included in the total.
- Diameter given instead of radius. Halve it first, and check units.
- Area ratio versus side ratio. In similar figures area goes with the square of the ratio and volume with the cube.
- Section formula order. In ratio m : n the first ratio number multiplies the second point's coordinate.
- Hollow cylinder. Total surface area includes the inner curved surface and the two ring-shaped ends.
- Figures not to scale. Trust given values.
Revision plan: four weeks
- Week 1, triangles. Learn the centre table, triplets and similarity. Solve 40 mixed questions, then rewrite every missed fact on a one-page sheet.
- Week 2, circles and polygons. Drill tangents, chords and cyclic quadrilaterals. Do 40 questions and cover all common tangent cases.
- Week 3, mensuration and coordinates. Memorise the solids table, then do 15 recasting problems and 10 coordinate questions.
- Week 4, mixed timed sets. Take sets of 15 questions with a timer, then redo wrong questions without looking at the solutions.
Revise the formula tables every third day in the final fortnight before the exam, and write each table from memory first. Then check what you missed.
FAQs
Which formulas must I memorise for SSC mensuration? The volume, curved surface and total surface formulas of the cube, cuboid, cylinder, cone, sphere and hemisphere, plus the frustum volume and slant height. Prism and pyramid follow from base area times height rules.
Is the frustum formula really asked? It can appear. Learn the cone-difference method as a backup, since it rebuilds the volume without recall.
What value of pi should I use? Use the value stated in the question. If none is given, 22/7 is fine when dimensions are multiples of 7, otherwise check what the options suggest.
Do I need coordinate geometry for SSC? Basics help, mainly distance, midpoint, section formula, centroid and triangle area. Check the latest notification for your exam.
How do I stop mixing up the triangle centre angle rules? Make a four-line table, test each rule once on a drawn triangle, and recite it before each practice set.
Are railway exam questions easier than SSC CGL? They are usually more direct. Look at the latest notification and past papers of your exam for the level.
How much time should one geometry question take? Aim for about a minute for direct questions. Skip a question if no fact fits within 20 seconds, and return if time allows.
Should I learn proofs? No. Exams ask for values and applications, though knowing why a rule works helps you remember it.
Why do I get recasting questions wrong? Usually by forgetting to equate volumes of the same material or by mixing units. Convert all units first, then equate volumes.
How many questions should I practise per topic? About 40 per fact family, with a gap before the redo of wrong ones, is a reasonable planning target.
Practise with full-length mocks
Reading formulas is only the first step. The real test is recalling the right one under time pressure. Practise geometry and mensuration in timed sectional tests on Pareeksha, then attempt full-length mocks to see how many of these questions you attempt and get right. Use the sectional analysis to decide which table to revise next.
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