Triangles: Congruence, Similarity, Centres, Theorems and Formulas with Solved Examples for SSC CGL and CHSL Geometry
Rahul Kumar
SSC Exam Expert & Content Editor
In this note
- 1.Types and Basic Properties
- 2.Congruence of Triangles
- 3.Similarity of Triangles
- 4.Pythagoras Theorem and Triples
- 5.The Four Centres
- 6.Area, Inradius and Circumradius
- 7.Solved Examples
- 8.Quick Tips
- 9.Quick Revision Points
- 10.Practice Questions With Answers
- 11.How to Practise This Chapter
- 12.Frequently Asked Questions
PDF Notes: Triangles: Congruence, Similarity, Centres, Theorems and Formulas with Solved Examples for SSC CGL and CHSL Geometry
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Triangles are the most important chapter of SSC geometry: every exam has 3 to 5 questions on them. This guide on triangles for SSC covers the types, the congruence and similarity rules, the key theorems, the four centres and all the area, inradius and circumradius formulas, with solved examples, practice questions and FAQs.
Types and Basic Properties
- The sum of the angles of a triangle is 180°.
- The exterior angle is equal to the sum of the two opposite interior angles.
- The sum of any two sides is greater than the third; the difference of two sides is less than the third.
- By sides: equilateral (all equal), isosceles (two equal) and scalene (none equal); by angles: acute, right and obtuse.
- The largest angle is opposite the longest side.
Congruence of Triangles
| Rule | Meaning |
|---|---|
| SSS | All three sides are equal |
| SAS | Two sides and the included angle are equal |
| ASA | Two angles and the included side are equal |
| AAS | Two angles and a non-included side are equal |
| RHS | Hypotenuse and one side of two right triangles are equal |
AAA and SSA are not rules of congruence. Corresponding parts of congruent triangles are equal (CPCT).
Similarity of Triangles
- Similar triangles have the same shape: the AA (AAA), SAS (proportional sides with the equal included angle) and SSS (all sides in proportion) rules.
- The ratio of the areas of similar triangles = the square of the ratio of the corresponding sides = the square of the ratio of the corresponding medians, altitudes or angle bisectors.
- Basic Proportionality (Thales) theorem: if DE ∥ BC in triangle ABC, then AD/DB = AE/EC.
- Angle bisector theorem: the bisector of angle A divides the opposite side BC in the ratio AB : AC.
- The ratio of the perimeters of similar triangles = the ratio of their sides.
Pythagoras Theorem and Triples
In a right triangle, the (hypotenuse)² = (base)² + (perpendicular)². Common triples:
| Triple | Multiples |
|---|---|
| 3, 4, 5 | 6, 8, 10; 9, 12, 15; 12, 16, 20 |
| 5, 12, 13 | 10, 24, 26 |
| 8, 15, 17 | 16, 30, 34 |
| 7, 24, 25 | 14, 48, 50 |
| 20, 21, 29 | – |
| 9, 40, 41 | – |
Apollonius theorem: in triangle ABC with D the midpoint of BC, AB² + AC² = 2(AD² + BD²). The length of a median to the side a: ma = ½√(2b² + 2c² − a²).
The Four Centres
| Centre | Formed By | Properties |
|---|---|---|
| Centroid (G) | Medians | Divides each median in the ratio 2 : 1 from the vertex; always inside; the centre of mass |
| Incentre (I) | Angle bisectors | The centre of the incircle; equidistant from the three sides; always inside; the angle BIC = 90° + A/2 |
| Circumcentre (O) | Perpendicular bisectors of the sides | The centre of the circumcircle; equidistant from the vertices; inside for acute, on the hypotenuse for right and outside for obtuse; the angle BOC = 2A |
| Orthocentre (H) | Altitudes | At the right-angle vertex for a right triangle; inside for acute and outside for obtuse; the angle BHC = 180° − A |
- In an equilateral triangle, all four centres coincide.
- Euler line: the circumcentre, centroid and orthocentre are collinear, with OG : GH = 1 : 2.
- The orthocentre, circumcentre and centroid coincide only in an equilateral triangle.
Area, Inradius and Circumradius
| Item | Formula |
|---|---|
| Area | ½ × base × height |
| Heron’s formula | √(s(s − a)(s − b)(s − c)), where s = (a + b + c)/2 |
| Equilateral triangle (side a) | Area = (√3/4)a²; height = (√3/2)a; inradius = a/(2√3); circumradius = a/√3 |
| Inradius (r) | Area / s |
| Circumradius (R) | abc / (4 × Area) |
| Right triangle | Circumradius = hypotenuse / 2; inradius = (a + b − c)/2 (c is the hypotenuse) |
| Isosceles triangle (equal sides a, base b) | Area = (b/4)√(4a² − b²) |
| Area using sine | ½ab sin C |
Solved Examples
Example 1. Find the area, inradius and circumradius of a triangle with sides 13, 14 and 15.
s = 21; Area = √(21 × 8 × 7 × 6) = √7056 = 84; r = 84/21 = 4; R = (13 × 14 × 15)/(4 × 84) = 2730/336 = 8.125.
Example 2. In a right triangle with sides 6, 8 and 10, find the inradius and circumradius.
r = (6 + 8 − 10)/2 = 2; R = 10/2 = 5.
Example 3. DE ∥ BC in triangle ABC. If AD = 4, DB = 6 and AE = 6, find EC.
4/6 = 6/EC, so EC = 9.
Example 4. The sides of two similar triangles are in the ratio 3 : 5. Find the ratio of their areas.
9 : 25.
Example 5. If angle A = 60°, find the angle BIC at the incentre.
90 + 60/2 = 120°.
Example 6. Find the area of an equilateral triangle of side 12.
(√3/4) × 144 = 36√3 = 62.35.
Quick Tips
- Memorise the Pythagorean triples with their multiples.
- Learn the angle formulas for the centres: 90 + A/2, 2A and 180 − A.
- Use Heron’s formula for the triangles with three sides given.
- Remember that the area ratio of similar triangles is the square of the side ratio.
Quick Revision Points
- The sum of the angles is 180°.
- The centroid divides the medians in the ratio 2 : 1.
- The angle BIC = 90° + A/2; BOC = 2A; BHC = 180° − A.
- Inradius = Area/s; circumradius = abc/4Δ.
- Similar triangles: the areas are in the ratio of the squares of the sides.
Practice Questions With Answers
- What is the inradius of a right triangle with sides 5, 12 and 13?
Answer: 2 - What is the circumradius of a right triangle with the hypotenuse 26?
Answer: 13 - In what ratio does the centroid divide a median?
Answer: 2 : 1 - What is the area of an equilateral triangle of side 6?
Answer: 9√3 (15.59) - If angle A = 80°, what is the angle BOC at the circumcentre?
Answer: 160° - If the area ratio of two similar triangles is 4 : 9, what is the ratio of the sides?
Answer: 2 : 3 - What is the area of a triangle with sides 5, 5 and 6?
Answer: 12 - Which congruence rule is used for the right triangles?
Answer: RHS - Where is the orthocentre of a right triangle?
Answer: At the vertex of the right angle - What is the third side of a right triangle with sides 7 and 24?
Answer: 25
How to Practise This Chapter
Practise 10 questions every day and write the steps on paper. Then attempt timed sets from our practice papers and the previous year papers page. See all topics on the Mathematics notes page and the next topic Circles: Chords, Tangents and Angles.
Frequently Asked Questions
What is the sum of the angles of a triangle?
180 degrees.
What is the centroid?
The point where the three medians meet; it divides each in the ratio 2 : 1.
What is the incentre?
The point where the angle bisectors meet; it is the centre of the incircle.
What is the circumcentre?
The point where the perpendicular bisectors meet; it is the centre of the circumcircle.
What is the orthocentre?
The point where the three altitudes meet.
What is Heron’s formula?
Area = √(s(s − a)(s − b)(s − c)).
What is the Apollonius theorem?
AB² + AC² = 2(AD² + BD²) for the median AD.
What is the basic proportionality theorem?
A line parallel to a side divides the other two sides in the same ratio.
What is the Euler line?
The line through the circumcentre, centroid and orthocentre.
Is this topic asked in SSC?
Yes. Triangles are the most important part of SSC geometry.
What is the area of an equilateral triangle?
(√3/4) × side².
What is the inradius formula?
Area divided by the semi-perimeter.
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Circles: Chords, Tangents, Angles, Cyclic Quadrilateral and Common Tangents with Formulas and Solved Examples for SSC
Rahul Kumar
SSC Exam Expert & Content Editor
Rahul is a senior SSC exam strategist with 8+ years of experience helping aspirants crack CGL, CHSL and MTS exams. He writes in-depth notification breakdowns, exam pattern guides and preparation strategies.