Geometry Questions for SSC CGL: What This Set Covers
This page gives 25 original geometry questions in the SSC CGL style, each with a worked solution and the shortcut to use in the exam. The questions follow the official syllabus line in the SSC CGL 2026 notice, which lists "Triangle and its various kinds of centres, Congruence and similarity of triangles, Circle and its chords, tangents, angles subtended by chords of a circle, common tangents to two or more circles" for both Tier 1 and Tier 2.
Why practise now: SSC CGL 2026 Tier 1 was held from 30 September to 28 October 2026, and the notice puts Tier 2 tentatively in December 2026. In Tier 2, Mathematical Abilities has 30 questions of 3 marks each with 1 mark deducted per wrong answer, so one wrong geometry guess costs a third of a right answer.
If you need the formulas first, open Geometry Formulas for SSC CGL in another tab. This page is the practice half of that note.
Formula Recap (Only What These 25 Questions Use)
| Concept | Formula / fact |
|---|---|
| Incentre angle | ∠BIC = 90° + A/2 |
| Circumcentre angle (acute triangle) | ∠BOC = 2A |
| Orthocentre angle | ∠BHC = 180° − A |
| Centroid | Divides each median 2 : 1 from the vertex |
| Equilateral triangle (side a) | Inradius a/(2√3), circumradius a/√3, height (√3/2)a, area (√3/4)a² |
| Right triangle | Circumradius = hypotenuse/2; inradius = (sum of legs − hypotenuse)/2 |
| Any triangle | r = Area/s; R = abc/(4 × Area) |
| Chord at distance d from centre | Half-chord = √(r² − d²) |
| Tangent from external point | √(d² − r²) |
| Direct common tangent | √(d² − (r₁ − r₂)²); if circles touch externally: 2√(r₁r₂) |
| Transverse common tangent | √(d² − (r₁ + r₂)²) |
| Intersecting chords | AP × PB = CP × PD |
| Tangent–secant | PT² = PA × PB |
| Cyclic quadrilateral | Opposite angles add to 180° |
| Regular polygon | Exterior angle = 360°/n; diagonals = n(n − 3)/2 |
| Similar triangles | Area ratio = (side ratio)² |
| Angle bisector theorem | BD/DC = AB/AC |
| Apollonius (median) | AB² + AC² = 2(AD² + BD²) |
Section A: Centres of a Triangle (Q1–Q7)
Q1. In ΔABC, ∠A = 70° and I is the incentre. Find ∠BIC. (a) 110° (b) 125° (c) 140° (d) 145° Answer: (b) 125°. ∠BIC = 90° + 70°/2 = 125°.
Q2. O is the circumcentre of acute ΔABC with ∠A = 55°. Find ∠BOC. (a) 55° (b) 90° (c) 110° (d) 125° Answer: (c) 110°. Central angle = 2 × inscribed angle = 110°.
Q3. H is the orthocentre of ΔABC and ∠A = 65°. Find ∠BHC. (a) 65° (b) 115° (c) 130° (d) 122.5° Answer: (b) 115°. ∠BHC = 180° − 65°.
Q4. G is the centroid of ΔABC and the median AD = 18 cm. Find AG. (a) 6 cm (b) 9 cm (c) 12 cm (d) 13.5 cm Answer: (c) 12 cm. AG : GD = 2 : 1, so AG = (2/3) × 18.
Q5. In an equilateral triangle of side 12 cm, find the ratio of inradius to circumradius, and the inradius. (a) 1 : 2, 2√3 cm (b) 1 : 3, 4 cm (c) 1 : 2, 4√3 cm (d) 1 : √2, 3 cm Answer: (a). r = 12/(2√3) = 2√3; R = 12/√3 = 4√3; r : R = 1 : 2 for every equilateral triangle.
Q6. A right triangle has legs 6 cm and 8 cm. Find its circumradius and inradius. (a) 5 cm, 2 cm (b) 5 cm, 3 cm (c) 4 cm, 2 cm (d) 10 cm, 2 cm Answer: (a). Hypotenuse 10, so R = 5; r = (6 + 8 − 10)/2 = 2.
Q7. A triangle has sides 13, 14 and 15 cm. Find its inradius. (a) 3 cm (b) 4 cm (c) 5 cm (d) 6 cm Answer: (b) 4 cm. s = 21, area = √(21 × 8 × 7 × 6) = 84, r = 84/21 = 4. (Bonus: R = 13 × 14 × 15 / (4 × 84) = 8.125 cm.)
Section B: Circles, Chords and Tangents (Q8–Q15)
Q8. A chord of length 16 cm is drawn in a circle of radius 10 cm. How far is it from the centre? (a) 4 cm (b) 5 cm (c) 6 cm (d) 8 cm Answer: (c) 6 cm. d = √(10² − 8²) = 6.
Q9. A point P is 13 cm from the centre of a circle of radius 5 cm. Find the length of the tangent from P. (a) 8 cm (b) 10 cm (c) 12 cm (d) 14 cm Answer: (c) 12 cm. √(169 − 25) = 12.
Q10. Two circles of radii 5 cm and 3 cm have centres 10 cm apart. Find the lengths of the direct and transverse common tangents. (a) 4√6 cm, 6 cm (b) 8 cm, 6 cm (c) 4√6 cm, 8 cm (d) 6 cm, 4√6 cm Answer: (a). Direct = √(100 − 4) = √96 = 4√6; transverse = √(100 − 64) = 6.
Q11. Two circles of radii 9 cm and 4 cm touch each other externally. Find the length of their direct common tangent. (a) 6 cm (b) 12 cm (c) 13 cm (d) 15 cm Answer: (b) 12 cm. Shortcut 2√(9 × 4) = 12.
Q12. AB is a diameter of a circle and C is a point on the circle with ∠CAB = 35°. Find ∠CBA. (a) 35° (b) 45° (c) 55° (d) 65° Answer: (c) 55°. Angle in a semicircle is 90°, so ∠CBA = 90° − 35°.
Q13. Chords AB and CD intersect at P inside a circle. If AP = 6, PB = 4 and CP = 3, find PD. (a) 4.5 (b) 6 (c) 8 (d) 12 Answer: (c) 8. 6 × 4 = 3 × PD.
Q14. From an external point P, a secant meets a circle at A and B (PA = 4 cm, PB = 9 cm) and a tangent touches it at T. Find PT. (a) 5 cm (b) 6 cm (c) 6.5 cm (d) 13 cm Answer: (b) 6 cm. PT² = 4 × 9 = 36.
Q15. The tangent at A to the circumcircle of ΔABC makes an angle of 58° with chord AB. Find ∠ACB. (a) 32° (b) 58° (c) 116° (d) 122° Answer: (b) 58°. Alternate segment theorem: the angle between tangent and chord equals the angle in the alternate segment.
Section C: Quadrilaterals and Polygons (Q16–Q19)
Q16. In cyclic quadrilateral ABCD, ∠A = 3∠C. Find ∠A. (a) 120° (b) 125° (c) 135° (d) 150° Answer: (c) 135°. A + C = 180° and A = 3C, so C = 45°.
Q17. Each interior angle of a regular polygon is 150°. How many diagonals does it have? (a) 35 (b) 44 (c) 54 (d) 65 Answer: (c) 54. Exterior angle 30°, so n = 12; diagonals = 12 × 9/2 = 54.
Q18. A polygon has 9 diagonals. Find the sum of its interior angles. (a) 540° (b) 720° (c) 900° (d) 1080° Answer: (b) 720°. n(n − 3)/2 = 9 gives n = 6; sum = (6 − 2) × 180°.
Q19. The diagonals of a rhombus are 16 cm and 12 cm. Find its side and area. (a) 10 cm, 96 cm² (b) 10 cm, 192 cm² (c) 14 cm, 96 cm² (d) 20 cm, 96 cm² Answer: (a). Side = √(8² + 6²) = 10; area = (16 × 12)/2 = 96.
Section D: Similarity, Congruence and Theorems (Q20–Q25)
Q20. In ΔABC, DE ∥ BC with D on AB and E on AC. If AD = 4, DB = 6 and AE = 6, find EC. (a) 4 (b) 8 (c) 9 (d) 10 Answer: (c) 9. Basic Proportionality Theorem: 4/6 = 6/EC.
Q21. Two similar triangles have corresponding sides in the ratio 3 : 5. The smaller has area 54 cm². Find the area of the larger. (a) 90 cm² (b) 120 cm² (c) 150 cm² (d) 180 cm² Answer: (c) 150 cm². Area ratio 9 : 25, so 54 × 25/9 = 150.
Q22. In ΔABC, AB = 8, AC = 12 and BC = 15. The bisector of ∠A meets BC at D. Find BD. (a) 5 (b) 6 (c) 7.5 (d) 9 Answer: (b) 6. BD = 15 × 8/(8 + 12).
Q23. In ΔABC, AB = 7, AC = 9 and BC = 8. Find the median AD. (a) 6 (b) 7 (c) 7.5 (d) 8 Answer: (b) 7. 49 + 81 = 2(AD² + 16), so AD² = 49.
Q24. The exterior angle at C of ΔABC is 120° and ∠A = 50°. Find ∠B. (a) 60° (b) 70° (c) 80° (d) 130° Answer: (b) 70°. Exterior angle = sum of the two interior opposite angles.
Q25. The height of an equilateral triangle is 9√3 cm. Find its area. (a) 27√3 cm² (b) 54√3 cm² (c) 81√3 cm² (d) 162 cm² Answer: (c) 81√3 cm². Side = 18 (since height = (√3/2)a); area = (√3/4) × 324.
Answer Key
| Q | Ans | Q | Ans | Q | Ans | Q | Ans | Q | Ans |
|---|---|---|---|---|---|---|---|---|---|
| 1 | b | 6 | a | 11 | b | 16 | c | 21 | c |
| 2 | c | 7 | b | 12 | c | 17 | c | 22 | b |
| 3 | b | 8 | c | 13 | c | 18 | b | 23 | b |
| 4 | c | 9 | c | 14 | b | 19 | a | 24 | b |
| 5 | a | 10 | a | 15 | b | 20 | c | 25 | c |
How to Use These Questions for Tier 2 (Our Advice)
| Step | What to do |
|---|---|
| 1. Timed first attempt | 25 questions in 25 minutes (Tier 2 Maths gives 30 questions in a 30-minute sectional timer as per the notice) |
| 2. Mark by type | Note which section (A–D) your wrong answers came from |
| 3. Re-learn the formula | Re-read only that block in Geometry Formulas for SSC CGL |
| 4. Retest in 48 hours | Redo only the wrong questions without looking at solutions |
| 5. Move to mocks | Take a full geometry mock under time (links below) |
Exam habits that save marks: draw a quick rough figure even for "easy" questions; check whether the triangle is right-angled (3-4-5, 5-12-13, 8-15-17 triples appear often); and in Tier 2, skip any geometry question you cannot start within 20 seconds, because each wrong answer costs 1 mark.
Practise Under Time: Geometry Mocks
- SSC CGL Tier 1 mock: Geometry and Coordinate Geometry
- SSC CGL geometry mock test 01, 02, 03
- Next practice set: Coordinate geometry questions for SSC CGL
Geometry Questions for SSC CGL: FAQs
How many geometry questions come in SSC CGL?
SSC does not publish topic-wise counts. The official syllabus names geometry under Quantitative Aptitude in Tier 1 (25 questions in the section) and Mathematical Abilities in Tier 2 (30 questions in the section), so expect it in both; treat it as a core topic rather than counting on a fixed number.
Which geometry topics are most important for SSC CGL?
The ones the notice names: centres of a triangle, congruence and similarity, circles with chords and tangents, angles subtended by chords, and common tangents. This set covers each of them.
What is the negative marking in SSC CGL Tier 2 maths?
One mark for each wrong answer, with 3 marks for each correct answer (SSC CGL 2026 notice). In Tier 1 it is 0.50 marks for a wrong answer.
What is the fastest way to solve circle questions?
Learn the six results in Section B (chord distance, tangent length, two common tangents, intersecting chords, tangent–secant and alternate segment). Almost every SSC circle question is one of these in disguise.
Is geometry asked in SSC CHSL too?
Yes. The SSC CHSL 2026 notice lists geometry in the Quantitative Aptitude syllabus, and CHSL Tier 1 runs from 30 November to 31 December 2026.
Practice Next for SSC CGL Maths
- Formulas: Geometry formulas for SSC CGL, Coordinate geometry formulas, Trigonometry formulas, Mensuration formulas
- Plan: SSC CGL Tier 2 preparation strategy 2026
- Mocks: SSC CGL trigonometry mock test 01, SSC CGL mensuration mock test 01
Sources: SSC CGL 2026 notice (syllabus para 13.11.1.4, Tier 2 scheme para 13.9 and negative marking para 13.9.8); SSC notice dated 01.10.2026 (CHSL Tier 1 dates). Questions and solutions are original BookOfExams material. Last checked 7 November 2026.