Geometry Formulas for SSC CGL: The Short Answer
Geometry in SSC CGL means plane geometry: lines and angles, triangles, polygons, quadrilaterals and circles. The formulas that answer most questions are angle sum of a polygon (n − 2) × 180°, the four triangle-centre angle rules (incentre 90° + A/2, excentre 90° − A/2, orthocentre 180° − A, circumcentre 2A), the angle bisector and basic proportionality theorems, and the circle rules for chords, tangents and cyclic quadrilaterals.
Key takeaway for SSC: SSC CGL Tier 1 usually has 2 to 4 geometry questions, and SSC CHSL and CPO ask 2 to 3. Learn the tables on this page and draw a rough figure for every question. Every number in the examples was computed in Python.
Mensuration and coordinate geometry are separate notes: see mensuration formulas for SSC CGL and coordinate geometry formulas for SSC CGL.
Lines and Angles Formulas for SSC CGL
| Rule | Statement |
|---|---|
| Linear pair | Angles on a straight line add to 180° |
| Vertically opposite angles | Equal |
| Complementary / supplementary | Add to 90° / 180° |
| Parallel lines cut by a transversal | Corresponding and alternate angles are equal; co-interior angles add to 180° |
| Angles around a point | Add to 360° |
Example 1: An angle is 4 times its supplement. Find the angle.
x = 4(180° − x) → 5x = 720° → x = 144°.
Polygon Formulas (Interior Angle, Exterior Angle, Diagonals)
| For a polygon with n sides | Formula |
|---|---|
| Sum of interior angles | (n − 2) × 180° |
| Each interior angle (regular) | (n − 2) × 180° ÷ n |
| Sum of exterior angles | 360° (any polygon) |
| Each exterior angle (regular) | 360° ÷ n |
| Number of diagonals | n(n − 3) ÷ 2 |
Example 2: Find each interior angle of a regular octagon, and the number of sides of a regular polygon whose exterior angle is 24°.
Octagon: 6 × 180° ÷ 8 = 135°. Sides: 360° ÷ 24° = 15. A 10-sided polygon has 10 × 7 ÷ 2 = 35 diagonals.
Triangle Formulas and Properties for SSC CGL
| Property | Rule |
|---|---|
| Angle sum | A + B + C = 180° |
| Exterior angle | Equals the sum of the two opposite interior angles |
| Triangle inequality | Sum of any two sides > third side |
| Larger side | Lies opposite the larger angle |
| Pythagoras (right angle at C) | a² + b² = c² |
| Acute / obtuse test | c² < a² + b² → acute; c² > a² + b² → obtuse (c = longest side) |
Common Pythagorean triplets: 3-4-5, 5-12-13, 7-24-25, 8-15-17, 9-40-41, 20-21-29, and their multiples (6-8-10, 9-12-15).
Example 3: The angles of a triangle are in the ratio 2 : 3 : 4. Find them.
180° ÷ 9 = 20°, so the angles are 40°, 60° and 80°.
Centres of a Triangle (Incentre, Circumcentre, Centroid, Orthocentre)
| Centre | Meeting point of | Key fact | Angle at centre (∠BXC) |
|---|---|---|---|
| Centroid (G) | Medians | Divides each median 2 : 1 from the vertex | — |
| Incentre (I) | Angle bisectors | Equidistant from the three sides (inradius r) | 90° + A/2 |
| Excentre (I₁) | External bisectors at B and C, internal at A | Opposite vertex A | 90° − A/2 |
| Circumcentre (O) | Perpendicular bisectors of sides | Equidistant from the vertices (circumradius R) | 2A (when A is acute) |
| Orthocentre (H) | Altitudes | In a right triangle, it is the right-angle vertex | 180° − A |
Example 4: In triangle ABC, ∠A = 70°. Find ∠BIC, ∠BI₁C, ∠BOC and ∠BHC.
- ∠BIC = 90° + 35° = 125°
- ∠BI₁C = 90° − 35° = 55°
- ∠BOC = 2 × 70° = 140°
- ∠BHC = 180° − 70° = 110°
Facts SSC loves:
- In an equilateral triangle, all four centres coincide. With side a: height = (√3/2)a, inradius = a/(2√3), circumradius = a/√3. For a = 12: height ≈ 10.39, r = 2√3 ≈ 3.46, R = 4√3 ≈ 6.93.
- In a right triangle, circumradius = hypotenuse ÷ 2 and inradius = (a + b − c) ÷ 2. For 6-8-10: R = 5, r = 2.
- The centroid, circumcentre and orthocentre lie on one line (the Euler line), with HG : GO = 2 : 1.
Median, Angle Bisector and BPT Theorems
| Theorem | Formula |
|---|---|
| Apollonius (median AD to side BC) | AB² + AC² = 2(AD² + BD²) |
| Median length | AD = ½√(2b² + 2c² − a²) |
| Angle bisector theorem (AD bisects ∠A) | BD/DC = AB/AC |
| Basic proportionality theorem (DE ∥ BC) | AD/DB = AE/EC |
| Midpoint theorem | Line joining midpoints of two sides ∥ third side and = half of it |
Example 5: In triangle ABC, AB = 12, AC = 8 and BC = 15. The bisector of ∠A meets BC at D. Find BD.
BD/DC = 12/8 = 3/2, so BD = 15 × 3/5 = 9 and DC = 6.
Example 6: In triangle ABC, DE ∥ BC with D on AB and E on AC. If AD = 4, DB = 6 and AE = 6, find EC.
4/6 = 6/EC → EC = 9.
Example 7: Find the median to the hypotenuse in a 6-8-10 triangle.
½√(2 × 64 + 2 × 36 − 100) = ½√100 = 5 (equal to half the hypotenuse, as expected).
Similar Triangles and Congruence Rules
| Congruence (same size and shape) | Similarity (same shape) |
|---|---|
| SSS, SAS, ASA, AAS, RHS | AA (or AAA), SSS ratio, SAS ratio |
For similar triangles with side ratio k : 1, the ratio of perimeters, medians, altitudes and angle bisectors is k : 1, and the ratio of areas is k² : 1.
Example 8: Two similar triangles have corresponding sides in the ratio 3 : 5. Find the ratio of their areas.
9 : 25.
Quadrilateral Properties for SSC CGL
| Shape | Diagonals | Key property |
|---|---|---|
| Parallelogram | Bisect each other | Opposite sides and angles equal |
| Rectangle | Equal and bisect each other | All angles 90° |
| Rhombus | Bisect each other at 90° | All sides equal; area = ½ × d₁ × d₂ |
| Square | Equal, bisect at 90° | All sides equal, all angles 90° |
| Trapezium | — | One pair of parallel sides |
| Cyclic quadrilateral | — | Opposite angles add to 180° |
Circle Theorems for SSC CGL (Chords, Tangents, Angles)
| Theorem | Rule |
|---|---|
| Perpendicular from centre | Bisects the chord; chord = 2√(r² − d²) |
| Equal chords | Equidistant from the centre and subtend equal angles |
| Angle at centre | Twice the angle at the circumference on the same arc |
| Angle in a semicircle | 90° |
| Angles in the same segment | Equal |
| Cyclic quadrilateral | Opposite angles sum to 180°; exterior angle = interior opposite angle |
| Tangent and radius | Perpendicular at the point of contact |
| Two tangents from a point | Equal in length |
| Tangent length from an external point | √(d² − r²) |
| Alternate segment theorem | Angle between tangent and chord = angle in the alternate segment |
| Intersecting chords (inside) | AP × PB = CP × PD |
| Tangent–secant | PT² = PA × PB |
| Direct common tangent | √(d² − (r₁ − r₂)²) |
| Transverse common tangent | √(d² − (r₁ + r₂)²) |
Example 9: A point is 13 cm from the centre of a circle of radius 5 cm. Find the tangent length.
√(169 − 25) = 12 cm.
Example 10: Find the length of a chord 6 cm from the centre of a circle of radius 10 cm.
2√(100 − 36) = 2 × 8 = 16 cm.
Example 11: Chords AB and CD intersect at P. If AP = 4, PB = 9 and CP = 6, find PD.
4 × 9 = 6 × PD → PD = 6.
Example 12: From P, a secant meets a circle at A and B (PA = 4, PB = 16). Find the tangent PT.
PT² = 4 × 16 = 64 → PT = 8.
Example 13: Two circles of radii 8 cm and 3 cm have centres 13 cm apart. Find the direct and transverse common tangents.
Direct = √(169 − 25) = 12 cm. Transverse = √(169 − 121) = √48 = 4√3 cm (≈ 6.93).
Geometry Formulas: Common Mistakes
- Using 2A for ∠BOC when A is obtuse. Then ∠BOC = 360° − 2A.
- Mixing up incentre and excentre angle rules (+A/2 vs −A/2).
- Using side ratio for area. Areas of similar figures go as the square of the side ratio.
- Direct vs transverse tangent. Direct uses r₁ − r₂; transverse uses r₁ + r₂.
- Forgetting that the perpendicular from the centre bisects the chord, so use half the chord in Pythagoras.
Geometry Questions for SSC CGL (Practice Set with Answers)
- Each exterior angle of a regular polygon is 30°. Find the number of sides and each interior angle. Answer: 12 sides, 150°.
- In triangle ABC, ∠A = 80°. Find ∠BIC, where I is the incentre. Answer: 130°. 90° + 40°.
- A point is 25 cm from the centre of a circle of radius 7 cm. Find the tangent length. Answer: 24 cm. √(625 − 49).
- Find the length of a chord 5 cm from the centre of a circle of radius 13 cm. Answer: 24 cm. 2 × 12.
- Chords AB and CD intersect at P with AP = 3, PB = 8 and CP = 2. Find PD. Answer: 12. 24 ÷ 2.
- From an external point P, PA = 5 and PB = 20 on a secant. Find the tangent length. Answer: 10. √100.
- Find the inradius of a right triangle with sides 5, 12 and 13. Answer: 2. (5 + 12 − 13) ÷ 2.
- In triangle ABC, AB = 9, AC = 6 and BC = 10. The bisector of ∠A meets BC at D. Find BD. Answer: 6. 10 × 9/15.
- A chord of a circle of radius 10 cm is 16 cm long. Find its distance from the centre. Answer: 6 cm. √(100 − 64).
- In a cyclic quadrilateral ABCD, ∠A = 75°. Find ∠C. Answer: 105°. Opposite angles add to 180°.
For timed practice, take the SSC CGL Geometry Mock Test 01 and the SSC CGL Maths Sectional Mock Test 11.
Geometry Formulas FAQs
Which geometry topics are most important for SSC CGL?
Triangle centres (incentre, circumcentre, centroid, orthocentre), similarity and the angle bisector theorem, polygons, and circle theorems on chords and tangents. These make up most of the 2 to 4 geometry questions in Tier 1.
What is the formula for the angle at the incentre?
∠BIC = 90° + ∠A/2, where I is the incentre of triangle ABC.
What is the tangent length formula?
From a point at distance d from the centre of a circle of radius r, tangent length = √(d² − r²).
What is the difference between centroid and circumcentre?
The centroid is where the medians meet and divides each median 2 : 1. The circumcentre is where the perpendicular bisectors meet and is equidistant from the three vertices.
How do you find the number of diagonals of a polygon?
Number of diagonals = n(n − 3) ÷ 2. A hexagon has 9 and a decagon has 35.
What is the direct common tangent formula?
For two circles with radii r₁ and r₂ and centres d apart, the direct common tangent = √(d² − (r₁ − r₂)²) and the transverse common tangent = √(d² − (r₁ + r₂)²).
Practice Next
- Mock tests: SSC CGL Geometry Mock Test 01, SSC CGL Mensuration Mock Test 02, SSC CGL Trigonometry Mock Test 02 and SSC CGL Maths Sectional Mock Test 11.
- Related notes: mensuration formulas for SSC CGL, coordinate geometry formulas for SSC CGL, trigonometry formulas for SSC CGL and height and distance for SSC CGL.
- Exam details: SSC CGL 2026 and SSC CHSL 2026.
- Browse all study notes.