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Geometry for SSC CGL: lines and triangles

Triangles are the heart of SSC CGL geometry. Angle properties, congruence and similarity, the Pythagoras theorem, the four centres (centroid, incentre, circumcentre, orthocentre), the angle-bisector theorem and the midpoint theorem appear again and again. The properties you must know, the angle results that save time, and worked examples with answers.

3 Oct 2026 4 min read

In this guide
  1. Lines and angles
  2. Basic triangle properties
  3. Congruence and similarity
  4. Pythagoras theorem
  5. The four centres
  6. More useful theorems
  7. Special triangles
  8. Area formulas
  9. Common traps
  10. Practice

Geometry is the area where many SSC CGL aspirants give up — and where those who don't gain a large advantage. The good news is that SSC geometry is built on a limited set of properties. Most triangle questions can be solved in a minute once you know which property the question is testing. Draw a quick figure, mark what is given, and ask: which theorem connects these?

Lines and angles

  • Angles on a straight line sum to 180°; angles around a point sum to 360°.
  • Vertically opposite angles are equal.
  • With parallel lines cut by a transversal: corresponding angles are equal, alternate angles are equal, and co-interior angles sum to 180°.

Basic triangle properties

  • The angle sum is 180°.
  • An exterior angle equals the sum of the two opposite interior angles.
  • The sum of any two sides is greater than the third side.
  • The larger side is opposite the larger angle.

Congruence and similarity

  • Congruence tests: SSS, SAS, ASA, AAS, RHS.
  • Similar triangles have equal angles and proportional sides (AA, SAS, SSS similarity).
  • For similar triangles, the ratio of areas = square of the ratio of sides.

Worked example: Two similar triangles have corresponding sides of 4 cm and 6 cm. The smaller has an area of 32 cm². Find the larger's area.
Area ratio = (4/6)² = 4/9, so the larger area = 32 × 9/4 = 72 cm².

Pythagoras theorem

In a right triangle, hypotenuse² = base² + height².

Common triplets: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (20, 21, 29), and their multiples.

The four centres

CentreMeeting point ofKey facts
Centroid (G)MediansDivides each median in the ratio 2 : 1 from the vertex
Incentre (I)Angle bisectorsCentre of the inscribed circle; ∠BIC = 90° + A/2
Circumcentre (O)Perpendicular bisectors of sidesCentre of the circumscribed circle; ∠BOC = 2A (for acute A)
Orthocentre (H)Altitudes∠BHC = 180° − A

Worked example: In triangle ABC, ∠A = 70°. Find ∠BIC, where I is the incentre.
∠BIC = 90° + 35° = 125°.

More useful theorems

  • Angle bisector theorem: the bisector of ∠A divides BC in the ratio AB : AC.
  • Midpoint theorem: the segment joining the midpoints of two sides is parallel to the third side and half its length.
  • Basic proportionality (Thales) theorem: a line parallel to one side divides the other two sides proportionally.
  • Apollonius theorem: AB² + AC² = 2(AD² + BD²), where AD is the median to BC.

Worked example: In triangle ABC, AB = 6 cm, AC = 9 cm and BC = 10 cm. The bisector of ∠A meets BC at D. Find BD.
BD : DC = 6 : 9 = 2 : 3. BD = 2/5 × 10 = 4 cm.

Special triangles

TriangleUseful results
Equilateral (side a)Height = (√3/2)a; area = (√3/4)a²; inradius = a/(2√3); circumradius = a/√3
Right triangleCircumradius = hypotenuse/2; the median to the hypotenuse = half the hypotenuse
30°–60°–90°Sides in the ratio 1 : √3 : 2
45°–45°–90°Sides in the ratio 1 : 1 : √2

Area formulas

  • ½ × base × height
  • ½ × ab × sin C
  • Heron's formula: √[s(s − a)(s − b)(s − c)], where s = (a + b + c)/2
  • Inradius r = area/s; circumradius R = abc/(4 × area)

Worked example: Find the area of a triangle with sides 13, 14 and 15.
s = 21. Area = √(21 × 8 × 7 × 6) = √7,056 = 84.

Common traps

TrapCorrect approach
Using the side ratio for areasSquare it
Mixing up centre formulasLearn one line for each centre
Assuming a figure is to scaleUse only the given data

Practice

  1. An exterior angle of a triangle is 120° and one interior opposite angle is 50°. Find the other.
  2. In triangle ABC, ∠A = 80°. Find ∠BOC, where O is the circumcentre.
  3. Find the area of an equilateral triangle of side 8 cm.
  4. Two similar triangles have areas of 81 and 49. Find the ratio of their corresponding sides.
  5. In a right triangle with legs 9 and 12, find the circumradius.
  6. A median of a triangle is 12 cm. Find the distance from the vertex to the centroid.

Answers: 1. 70°. 2. 160°. 3. 16√3 cm². 4. 9 : 7. 5. 7.5. 6. 8 cm.

A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the Staff Selection Commission website .

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