Triangles are the single most rewarding chapter in CDS / OTA geometry: angle relations, congruence and similarity tests, the Pythagoras theorem and area formulas reappear in every paper. This Cavalier guide rebuilds the topic from first principles, fixes the exact properties UPSC repeats, and drills the calculation shortcuts that turn a 2-minute sum into a 20-second one.
Why triangles dominate CDS geometry
A triangle is the simplest closed figure — three sides, three angles, three vertices — yet it carries more named theorems than any other shape on the syllabus. In the CDS / OTA Elementary Mathematics paper you can expect 2 to 4 questions every sitting drawn from angle relations, congruence, similarity, Pythagoras or area.
The good news: triangle questions are rule-driven. Once you can recall the angle-sum property, the five congruence tests and the Pythagorean triples, most problems collapse into a single line of arithmetic. Unlike data-interpretation or probability sums, almost nothing here needs heavy computation — it rewards clean recall and a tidy figure. That makes the chapter a reliable scoring zone even for candidates who find arithmetic-heavy topics slow.
Because the same handful of properties keep returning, the smartest preparation strategy is to over-learn the basics rather than chase exotic cases. The properties below cover the overwhelming majority of what UPSC has asked over the past three decades.
The sum of any two sides of a triangle is always greater than the third side, and the difference of any two sides is always less than the third side. If three given lengths fail this triangle inequality, no triangle can be formed — a favourite trap in CDS objective questions.
Classifying triangles by sides and angles
Every CDS triangle question first expects you to know what kind of triangle you are looking at. There are two independent ways to classify.
By sides
- Equilateral — all three sides equal, all angles 60°.
- Isosceles — two sides equal; the angles opposite the equal sides are equal.
- Scalene — all three sides (and angles) different.
By angles
- Acute — every angle less than 90°.
- Right — one angle exactly 90°.
- Obtuse — one angle greater than 90°.
A triangle can be both isosceles and right-angled (e.g. angles 45°, 45°, 90°) but never both equilateral and right-angled, because equilateral forces all angles to 60°.
Angle-sum and exterior-angle properties
These two properties solve a huge share of CDS angle questions.
Angle-sum property: the three interior angles of any triangle add up to 180°.
Exterior-angle property: an exterior angle equals the sum of the two interior opposite (remote) angles.
From the angle-sum rule it follows that each angle of an equilateral triangle is 180° ÷ 3 = 60°, and that a triangle can have at most one right or one obtuse angle.
The exterior-angle property is the faster route whenever a side is extended: instead of finding the third interior angle first, just add the two remote angles directly.
The exterior angle equals the sum of the two opposite interior angles, not all three and not the adjacent one. The exterior angle and its adjacent interior angle are supplementary (add to 180°).
Congruence: the five tests
Two triangles are congruent (≅) when they have exactly the same shape and size — one can be placed on the other so all sides and angles match. CDS questions ask you to pick the valid test.
- SSS — three sides of one equal three sides of the other.
- SAS — two sides and the included angle equal.
- ASA — two angles and the included side equal.
- AAS — two angles and a non-included side equal.
- RHS — in right triangles, hypotenuse and one side equal.
There is no AAA or SSA congruence test. AAA only guarantees the same shape (similarity), not the same size; SSA (or ASS) is ambiguous and can produce two different triangles.
Congruent triangles have all corresponding parts equal — the principle abbreviated as CPCT (Corresponding Parts of Congruent Triangles), which CDS proof-style questions rely on. Once you have established congruence by any one valid test, CPCT lets you immediately claim that any remaining pair of sides or angles is also equal, without proving them separately.
A practical reading order for these tests: scan the given data for three equal sides first (SSS), then for a matched side-angle-side pattern (SAS), and only then consider the angle-based tests. In right-angled figures, always check RHS before anything else — it is the quickest match when a hypotenuse and one leg are given.
Similarity and proportional sides
Two triangles are similar (∼) when their angles are equal and corresponding sides are in the same ratio. They have the same shape but not necessarily the same size.
Tests for similarity: AAA (or AA), SSS (sides proportional) and SAS (two sides proportional with equal included angle). If two triangles are similar with ratio of sides k, then the ratio of their areas is k2.
Basic Proportionality Theorem (Thales)
A line drawn parallel to one side of a triangle, cutting the other two sides, divides those two sides in the same ratio. If DE ∥ BC in triangle ABC, then AD ÷ DB = AE ÷ EC. This single theorem powers most CDS similarity problems.
When a question gives the ratio of sides and asks for the ratio of areas (or vice versa), remember: areas scale as the square of the side ratio, and corresponding perimeters scale linearly (same ratio as the sides).
Pythagoras theorem and triples
In a right-angled triangle the square on the hypotenuse equals the sum of the squares on the other two sides.
Pythagoras: hypotenuse2 = base2 + height2. The converse is also true — if c2 = a2 + b2, the triangle is right-angled at the vertex opposite side c.
Memorise the common Pythagorean triples, because spotting them saves real time: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25) and (9, 40, 41). Any whole-number multiple of a triple is also a triple, e.g. (6, 8, 10) or (9, 12, 15).
The height (altitude) of an equilateral triangle of side a is (√3 ÷ 2) × a, and its area is (√3 ÷ 4) × a2. Both come straight from Pythagoras and appear constantly in CDS.
Area formulas you must know
CDS rewards the candidate who picks the right area formula for the data given.
- Base and height known: Area = ½ × base × height.
- Three sides known (Heron's formula): Area = √[s(s − a)(s − b)(s − c)], where s = (a + b + c) ÷ 2 is the semi-perimeter.
- Equilateral, side a: Area = (√3 ÷ 4) × a2.
- Right triangle: Area = ½ × (the two legs about the right angle).
If a triangle's three sides are given as plain numbers, reach straight for Heron's formula — do not waste time trying to find an angle or a height first.
One more relation worth memorising: the area of a triangle also equals (a × b × c) ÷ (4R), where R is the circumradius, and equals r × s, where r is the inradius and s the semi-perimeter. These connect the area to the triangle's circles and occasionally unlock a question that looks impossible at first glance.
The four centres of a triangle
Four special points are formed by drawing standard cevians. CDS frequently asks which is which.
- Centroid — intersection of the three medians; it divides each median in the ratio 2:1 from the vertex and is the triangle's centre of mass.
- Incentre — intersection of the angle bisectors; centre of the inscribed circle, equidistant from all three sides.
- Circumcentre — intersection of the perpendicular bisectors of the sides; centre of the circle passing through all three vertices.
- Orthocentre — intersection of the three altitudes.
A useful fact CDS sometimes tests: the centroid, circumcentre and orthocentre of any triangle always lie on one straight line called the Euler line, and the centroid divides the segment joining the orthocentre and circumcentre in the ratio 2:1. The incentre does not, in general, lie on this line.
In an equilateral triangle all four centres coincide at one point. In a right triangle the circumcentre lies at the midpoint of the hypotenuse, and the orthocentre sits at the right-angle vertex.
Worked example: area and altitude together
A clean CDS-style problem that combines Heron's formula with the area definition.
The sides of a triangle are 13 cm, 14 cm and 15 cm. Find its area and the length of the altitude (height) drawn on the side of length 14 cm.
So the area is 84 cm² and the required altitude is 12 cm. Notice how Heron's formula gives the area, which then unlocks the height — a two-step pattern CDS loves.
Common mistakes to avoid
Most lost marks in this chapter come from a handful of avoidable slips.
- Treating AAA as a congruence test — it only proves similarity.
- Forgetting the triangle inequality when asked whether three lengths can form a triangle.
- Using the side ratio instead of its square for the ratio of areas of similar triangles.
- Splitting the centroid ratio the wrong way — it is 2:1 from the vertex to the midpoint, not 1:2.
- Applying Pythagoras to a non-right triangle, or to the wrong side as hypotenuse.
The longest side always lies opposite the largest angle, and the hypotenuse is always the longest side of a right triangle. Mixing these up corrupts both Pythagoras and angle-comparison questions.
Previous-year style question
Test yourself on a typical CDS similarity-and-area item before checking the worked solution.
Q. The areas of two similar triangles are 121 cm² and 64 cm². If the longest side of the larger triangle is 22 cm, what is the length of the longest side of the smaller triangle?
Answer: For similar triangles, ratio of areas = (ratio of sides)2. So 121 ÷ 64 = (22 ÷ x)2. Taking square roots: 11 ÷ 8 = 22 ÷ x, giving x = (22 × 8) ÷ 11 = 16 cm.
The whole solution rests on one idea — area ratio is the square of the side ratio — which is why mastering similarity pays off across multiple question types.
Quick revision
- Interior angles sum to 180°; exterior angle = sum of the two remote interior angles.
- Congruence tests: SSS, SAS, ASA, AAS, RHS — there is no AAA or SSA.
- Similarity tests: AAA/AA, SSS, SAS; ratio of areas = (ratio of sides)2.
- Pythagoras: hyp2 = base2 + height2; learn the triples 3-4-5, 5-12-13, 8-15-17.
- Area = ½ × base × height, Heron's √[s(s−a)(s−b)(s−c)], equilateral (√3÷4)a2.
- Centroid divides each median 2:1; all four centres coincide in an equilateral triangle.
Quickly sketch every triangle question, even in an objective paper. A rough figure with the data marked on it prevents the wrong-side and wrong-angle errors that cost the most marks.
Frequently asked questions
How many questions on triangles appear in the CDS exam?
Typically 2 to 4 questions per Elementary Mathematics paper, spread across angle properties, congruence, similarity, Pythagoras and area. It is one of the highest-yield geometry chapters.
What is the difference between congruent and similar triangles?
Congruent triangles have identical shape and size, so all corresponding sides and angles are equal. Similar triangles have the same shape but possibly different sizes, with equal angles and sides in the same ratio.
Why is there no AAA or SSA congruence test?
AAA fixes only the shape, so the triangles may differ in size (that is similarity, not congruence). SSA is ambiguous because two non-congruent triangles can satisfy the same two sides and a non-included angle.
When should I use Heron's formula?
Use Heron's formula when all three sides are known but no angle or height is given. Compute the semi-perimeter s, then Area = √[s(s−a)(s−b)(s−c)].
How does the centroid divide a median?
The centroid divides each median in the ratio 2:1, measured from the vertex to the midpoint of the opposite side. The longer 2-part is on the vertex side.
What are the most useful Pythagorean triples for CDS?
The triples (3,4,5), (5,12,13), (8,15,17), (7,24,25) and (9,40,41), plus all their whole-number multiples. Spotting them lets you write the third side instantly without squaring and adding.
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