Properties of Triangles
~14 min read
- What: Properties of Triangles covers the Sine Rule, Cosine Rule, Projection Formula, area formulas, circum-radius (R), and in-radius (r) — the full toolkit for solving any triangle.
- Why it matters: NDA papers have asked at least one Properties of Triangles question in every exam from 2010 to 2025, often as a paired set of 2–3 items worth multiple marks.
- Key fact: The cosine rule and in-radius formula appear most often. Knowing \(r = \Delta/s\) and \(R = a/(2\sin A)\) cold will solve most NDA questions in under 90 seconds.
If you can name a triangle — scalene, right-angled, equilateral — you already know Properties of Triangles intuitively. What this topic does is give you the algebra to compute every unknown angle and side from whatever you are given. The Sine Rule links a side to its opposite angle. The Cosine Rule connects all three sides. The area, circum-radius, and in-radius formulas tie the geometry to the algebra. NDA questions mix these formulas in combinations that test whether you actually understand what each formula means, not just whether you memorised it.
What This Topic Covers
Properties of Triangles is listed under Trigonometry in the NDA Mathematics syllabus. It sits between Trigonometric Ratios & Identities and Inverse Trigonometric Functions in the question paper. The sub-topics you must master are:
- Sine Rule — relating each side to the sine of its opposite angle.
- Cosine Rule — finding a side or angle when two sides and the included angle are known.
- Projection Formula — expressing one side as the sum of projections of the other two.
- Area Formula — \(\text{Area} = \tfrac{1}{2} ab \sin C\), plus Heron's formula.
- Circum-radius (\(R\)) — the radius of the circumscribed circle.
- In-radius (\(r\)) — the radius of the inscribed circle, via \(r = \Delta/s\).
- Classification tests — deciding whether a triangle is acute, obtuse, or right-angled from its sides or angles.
Why This Topic Matters
- Questions regularly appear as connected sets of 2–3 items sharing one triangle — answering one correctly fast-tracks the rest.
- The topic overlaps with Height & Distance, so mastering it here pays dividends across two chapters.
- Right-angled triangle checks using the cosine rule (if \(\cos C < 0\) then angle \(C\) is obtuse) appear directly in NDA 2011, 2013, 2015, 2025 papers.
Exam Pattern & Weightage
The table below is built from the PYQ file. Year refers to the NDA paper year; the count is the number of items classified under Properties of Triangles (excluding pure Inverse Trig questions in the same chapter).
| Year | Paper | Triangle Items | Topic Focus |
|---|---|---|---|
| 2010 | II | 1 | Obtuse angle via cosine rule (sides 6, 10, 14 cm) |
| 2011 | I | 5 | Right-angled at B; BD perpendicular to AC; 4-part set |
| 2011 | II | 3 | Obtuse angle, sides in AP, angles in AP |
| 2012 | I | 2 | Angles in AP → cosine rule; sin C with sides 18, 24, 30 |
| 2013 | I | 2 | Area formula (angles 30°, 45°, included side √3+1) |
| 2014 | I | 2 | Sine rule (c=2, A=45°, a=2√2); sin A − cos B = cos C |
| 2015 | I | 1 | Angles in AP, b:c = √3:√2 → angle A |
| 2016 | I | 3 | Equilateral triangle identity; cos A + cos B + cos C set |
| 2017 | I | 2 | a − 2b + c = 0 identity; sin²/cos² ratio type |
| 2018 | I | 2 | Sine rule (a=2, b=3, sin A=2/3); inscribed circle angle |
| 2019 | I | 2 | Angles 1:2:3 → sides ratio; right-angled triangle check |
| 2020 | I | 2 | Trapezium sine rule set (AD sin θ = AB sin α) |
| 2022 | I | 1 | p(p−2a) tan(A/2) = 4q identity |
| 2022 | II | 1 | cos 3C with a=4, b=3, c=2 |
| 2023 | I | 4 | Right angle + AP check; a²+b²+c²=ac+√3bc set |
| 2023 | II | 2 | Angles 3:5:4 ratio set (a+b+√2c; a²:b²:c²) |
| 2024 | II | 2 | Equilateral from cosA/cosB/cosC = a/b/c; area with a=6 |
| 2025 | I | 4 | Triangle with sides 3, 5, 7 cm — three-item set + angle ratio set |
Multi-part question sets (2–4 items sharing one triangle) have appeared in 2011, 2013, 2016, 2023, and 2025. Crack the common setup first and all sub-items fall into place. Never skip a "connected set" — partial marks are almost guaranteed once you have the triangle solved.
Key observation: the 2023 and 2025 papers both carried 4 Properties of Triangles items each — the highest count in recent years. Right-angle classification and the in-radius/circum-radius identity are the two most-tested sub-areas across 2020–2025.
Core Concepts
Work through each formula below until you can reproduce it from memory and apply it in under 60 seconds. Every formula here has appeared — either directly or as the backbone of a solution — in at least one NDA PYQ.
Sine Rule
In any triangle \(ABC\) with sides \(a, b, c\) opposite to angles \(A, B, C\) respectively:
Equivalently, $$a = 2R \sin A$$, $$b = 2R \sin B$$, $$c = 2R \sin C$$ — handy for converting sides into sines (and back) inside trig identities.
Use it when you know one side and its opposite angle, plus one more element. From the PYQ data: in the 2014-I question (\(c = 2\), \(A = 45^\circ\), \(a = 2\sqrt{2}\)), applying the sine rule gives $$\sin C = \frac{c \cdot \sin A}{a} = \frac{2 \cdot \sin 45^\circ}{2\sqrt{2}} = \tfrac{1}{2},$$ so \(C = 30^\circ\). The same logic solved the 2011-I question where \(c = 2\), \(A = 120^\circ\), \(a = \sqrt{6}\).
When the sine rule gives \(\sin C =\) some value, check for two possible angles (\(C\) and \(180^\circ - C\)). NDA options usually distinguish between them, so pick the one consistent with \(A + B + C = 180^\circ\).
Cosine Rule
Use the cosine rule when three sides are given, or two sides and the included angle are given:
Quick classification test from the PYQ: if \(a = 6\), \(b = 10\), \(c = 14\), then $$\cos C = \frac{36 + 100 - 196}{2 \cdot 6 \cdot 10} = \frac{-60}{120} = -\tfrac{1}{2},$$ so \(C = 120^\circ\) (obtuse). This exact question appeared in NDA 2010-II. For the 2025-I set with sides 3, 5, 7: $$\cos B = \frac{25 + 9 - 49}{2 \cdot 5 \cdot 3} = \frac{-15}{30} = -\tfrac{1}{2},$$ giving \(B = 120^\circ\).
The largest angle always sits opposite the longest side. Save 30 seconds by applying the cosine rule only to that angle when classifying — if \(\cos < 0\) it is obtuse, \(\cos = 0\) means right-angled, \(\cos > 0\) means the whole triangle is acute.
Projection Formula
This formula is used when NDA gives a condition like \(a - 2b + c = 0\). From the 2017-I question: given \(a - 2b + c = 0\), so \(a + c = 2b\). Using the half-angle cotangent identity, this reduces to \((s - b)/b = 3\), which means the \(\cot(B/2)\) expression evaluates to 3. Projection identities also underpin the 2020-I trapezium set.
Area Formula
When angles \(30^\circ\) and \(45^\circ\) with included side \((\sqrt{3} + 1)\) are given (NDA 2013-I), the third angle is \(105^\circ\). \(\text{Area} = \tfrac{1}{2} \cdot (\sqrt{3} + 1) \cdot x \cdot \sin 30^\circ\) where \(x\) is found from the sine rule. The answer works out to \((\sqrt{3} + 1)/2\). Always identify which two sides bound the known angle before substituting.
Hero's formula needs the semi-perimeter \(s\) — not the full perimeter. A common slip is plugging in \((a + b + c)\) instead of \((a + b + c)/2\). Also, $$\Delta = \frac{abc}{4R}$$ is the fastest area formula when the circumradius is given or easily derived from the sine rule.
Circum-radius (R)
This follows directly from the Sine Rule. It appears in the 2018-I question about a triangle inscribed in a circle with centre \(O\), where \(\angle BAC = \alpha\) and \(\angle BOC = \beta\). Since the central angle is twice the inscribed angle, \(\beta = 2\alpha\), so $$\cos \beta = \cos 2\alpha = \frac{1 - \tan^2\alpha}{1 + \tan^2\alpha}.$$
For a right-angled triangle the hypotenuse is the diameter of the circumcircle, so $$R = \frac{\text{hypotenuse}}{2}$$. The instant you spot a Pythagorean triplet (3-4-5, 5-12-13, 8-15-17) you can write R without any sine-rule arithmetic.
In-radius (r)
The 2022-I question asks for \(p(p - 2a) \tan(A/2)\) where \(p\) is the perimeter. With \(p = 2s\) and \(p - 2a = 2(s - a)\), the expression becomes $$4s(s - a) \tan\!\left(\tfrac{A}{2}\right) = 4s(s - a) \cdot \frac{r}{s - a} = 4sr = 4\Delta = 4q.$$ The 2023-II in-circle problem (\(PQR\) with \(\cos P = 1/3\)) used the tangent-length property: \(PN = n\), \(QL = n+2\), \(RM = n+4\), giving sides and solving the cosine-rule equation to find \(n = 8\).
Equilateral triangle shortcuts: if \(a = b = c\), then \(R = a/\sqrt{3}\), \(r = a/(2\sqrt{3})\), and \(\text{area} = (\sqrt{3}/4)\,a^2\). The 2024-II question (\(\cos A/a = \cos B/b = \cos C/c\) implies equilateral) leads directly to \(\text{area} = (\sqrt{3}/4)\cdot 36 = 9\sqrt{3}\) sq cm with \(a = 6\).
Worked Examples
Each example below is drawn directly from the NDA PYQ file. Work through the steps yourself before reading the solution.
Example 1 — Classify the triangle (NDA 2010-II)
Q. In any triangle \(ABC\), the sides are 6 cm, 10 cm, and 14 cm. Find the obtuse angle.
- Identify the largest side: \(c = 14\) cm. The angle opposite to the largest side is the obtuse angle.
- Apply the cosine rule: $$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{36 + 100 - 196}{2 \cdot 6 \cdot 10}.$$
- Compute: $$\cos C = \frac{-60}{120} = -\tfrac{1}{2}.$$
- \(\cos C = -\tfrac{1}{2} \implies C = 120^\circ\).
- Answer: \(120^\circ\).
Example 2 — Sine Rule to find angle (NDA 2014-I)
Q. In triangle \(ABC\), \(c = 2\), \(A = 45^\circ\), \(a = 2\sqrt{2}\). Find \(C\).
- Write the sine rule: $$\frac{a}{\sin A} = \frac{c}{\sin C}.$$
- Substitute: $$\frac{2\sqrt{2}}{\sin 45^\circ} = \frac{2}{\sin C}.$$
- \(\sin 45^\circ = 1/\sqrt{2}\), so \(2\sqrt{2} / (1/\sqrt{2}) = 4\). Hence \(\sin C = 2/4 = 1/2\).
- \(\sin C = 1/2 \implies C = 30^\circ\) (taking the acute option consistent with \(A + B + C = 180^\circ\)).
- Answer: \(C = 30^\circ\).
Example 3 — In-radius identity (NDA 2022-I)
Q. Let \(a, b, c\) be the sides of triangle \(ABC\), \(p\) the perimeter and \(q\) the area. Find \(p(p - 2a) \tan(A/2)\).
- Express in terms of semi-perimeter \(s\): \(p = 2s\), so \(p - 2a = 2s - 2a = 2(s - a)\).
- Therefore \(p(p - 2a) = 2s \cdot 2(s - a) = 4s(s - a)\).
- Use the half-angle formula: $$\tan\!\left(\tfrac{A}{2}\right) = \frac{r}{s - a}, \quad \text{where } r = \tfrac{\Delta}{s} = \tfrac{q}{s}.$$
- Substitute: $$4s(s - a) \cdot \frac{r}{s - a} = 4sr = 4s \cdot \tfrac{q}{s} = 4q.$$
- Answer: \(4q\).
Example 4 — Equilateral condition and area (NDA 2024-II)
Q. In triangle \(ABC\), \(a/\cos A = b/\cos B = c/\cos C\). If \(a = 6\) cm, find the area.
- From the sine rule, \(a/\sin A = 2R\), so \(a = 2R \sin A\). Substitute: $$\frac{2R \sin A}{\cos A} = 2R \tan A.$$
- So \(2R \tan A = 2R \tan B = 2R \tan C \implies \tan A = \tan B = \tan C \implies A = B = C = 60^\circ\).
- Triangle is equilateral with side \(a = 6\) cm.
- $$\text{Area} = \tfrac{\sqrt{3}}{4} \cdot a^2 = \tfrac{\sqrt{3}}{4} \cdot 36 = 9\sqrt{3} \text{ sq cm}.$$
- Answer: \(9\sqrt{3}\) square cm.
Example 5 — Sine Rule, SSA configuration
Q. In triangle ABC, $$a = 2$$, $$b = 3$$, and $$\sin A = \tfrac{2}{3}$$. Find angle B.
- Apply the sine rule: $$\frac{a}{\sin A} = \frac{b}{\sin B}$$.
- Substitute: $$\frac{2}{2/3} = \frac{3}{\sin B}$$, which gives $$3 = \frac{3}{\sin B}$$.
- So $$\sin B = 1$$, hence $$B = 90^\circ$$.
- Sanity-check: with A acute and B = 90°, the triangle closes off neatly with C = 90° − A.
- Answer: B = 90°.
Example 6 — Hero's Formula and in-radius
Q. A triangle has sides 13, 14, 15 cm. Find its area and in-radius.
- Semi-perimeter: $$s = \tfrac{13 + 14 + 15}{2} = 21$$ cm.
- Apply Hero's formula: $$\Delta = \sqrt{21(21 - 13)(21 - 14)(21 - 15)} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6}$$.
- Simplify: $$21 \cdot 8 \cdot 7 \cdot 6 = 7056$$, so $$\Delta = 84$$ sq cm.
- In-radius: $$r = \frac{\Delta}{s} = \frac{84}{21} = 4$$ cm.
- Answer: Area = 84 sq cm, r = 4 cm.
Example 7 — Sides 3, 5, 7 triangle set (NDA 2025-I)
Q. In triangle \(ABC\): \(AB = 3\), \(BC = 5\), \(CA = 7\). Find angle \(B\) and area.
- Identify: \(a = BC = 5\), \(b = CA = 7\), \(c = AB = 3\). Largest side is \(b = 7\), so angle \(B\) (opposite side \(b\)) is the largest.
- Cosine rule: $$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{25 + 9 - 49}{2 \cdot 5 \cdot 3} = \frac{-15}{30} = -\tfrac{1}{2}.$$
- \(\cos B = -\tfrac{1}{2} \implies B = 120^\circ\). Triangle is obtuse-angled. Sum of other angles \(= 60^\circ\) (acute), so statement II of the 2025 question is also correct.
- $$\text{Area} = \tfrac{1}{2} \cdot a \cdot c \cdot \sin B = \tfrac{1}{2} \cdot 5 \cdot 3 \cdot \sin 120^\circ = \tfrac{1}{2} \cdot 15 \cdot \tfrac{\sqrt{3}}{2} = \tfrac{15\sqrt{3}}{4} \text{ sq cm}.$$
- Answer: \(B = 120^\circ\); Area \(= \tfrac{15\sqrt{3}}{4}\) sq cm.
Exam Shortcuts (Pro-Tips)
Properties of Triangles questions are deliberately algebra-heavy — but NDA papers reward students who recognise the shortcut over those who grind through the manipulation. The four hacks below collapse multi-minute solves into 10–15 seconds each.
Shortcut 1 — The Equilateral-Assumption Golden Hack
When NDA gives a generic identity in A, B, C or a, b, c with no specification of the triangle type (e.g., "find the value of cos A + cos B + cos C"), assume the triangle is equilateral. Set $$A = B = C = 60^\circ$$ and $$a = b = c = 1$$, then plug into the options. The correct option will match in seconds.
NDA 2016-I used the identity cos A + cos B + cos C and 2024-II used cos A/a = cos B/b = cos C/c — both collapse to equilateral and resolve in one step.
Shortcut 2 — Pythagorean Triplet Recognition
Before reaching for the cosine rule, scan the side lengths. If they form a Pythagorean triplet — 3-4-5, 5-12-13, 7-24-25, 8-15-17, 9-40-41, or any integer multiple — the triangle is right-angled at the vertex opposite the hypotenuse.
You get the angle, area, and circumradius without any trigonometric calculation.
Shortcut 3 — Maximum Area for a Given Perimeter
If NDA asks "for a triangle with fixed perimeter P, when is the area maximum?", the answer is always the equilateral triangle with side P/3. The maximum area equals $$\frac{\sqrt{3}}{4} \cdot \left(\tfrac{P}{3}\right)^2 = \frac{\sqrt{3}\, P^2}{36}$$. No calculus needed.
Shortcut 4 — Direct Area via abc/4R
When a question hands you the three sides plus the circumradius (or asks you to find R given sides and area), skip Hero's formula. Use the direct identity:
Pair it with $$r = \frac{\Delta}{s}$$ to flip between in-radius and circum-radius in any direction the question demands.
If a question asks for sin A · sin B · sin C or cos A · cos B · cos C with no other constraint, try the equilateral substitution before any other approach. The 3-second test catches more than half of the algebraic-identity questions in NDA Properties of Triangles.
Test Your Properties of Triangles Skills
NDA mock tests on Defence Road include full-length Maths papers with Properties of Triangles questions calibrated to the actual NDA pattern. Track your accuracy and time per question before the real exam.
Start Free Mock TestCommon Question Patterns
After analysing the PYQ data from 2010 to 2025, five question patterns cover roughly 90% of all Properties of Triangles items.
Pattern 1: Classify the Triangle
Given three sides, decide acute / obtuse / right-angled. Method: apply the cosine rule to the angle opposite the largest side. If \(\cos < 0 \implies\) obtuse; if \(\cos = 0 \implies\) right; if \(\cos > 0 \implies\) acute. Appeared in 2010-II, 2011-I, 2019-I, 2025-I.
Pattern 2: Sine Rule to Find an Angle
Given two sides and one angle (SSA configuration), use the sine rule. Always verify that the angle sum stays below \(180^\circ\). Appeared in 2011-I, 2014-I, 2018-I.
Pattern 3: Angles in AP Imply \(B = 60^\circ\)
If \(A, B, C\) are in arithmetic progression, then \(2B = A + C\) and \(A + B + C = 180^\circ\) gives \(B = 60^\circ\). The cosine rule then gives \(a^2 + c^2 - b^2 = ac\), so \(b^2 = a^2 + c^2 - ac\). Appeared in 2012-I, 2015-I, 2017-I.
Pattern 4: In-Radius and Circum-Radius Identities
NDA uses \(r = \Delta/s\) and the tangent-length property of the in-circle (\(s - a, s - b, s - c\) are the tangent lengths from \(A, B, C\)). The 2023-II triangle \(PQR\) set is a direct example. Combine with the cosine rule to solve for \(n\).
Pattern 5: Algebraic Identity Verification
Given a condition like \(a - 2b + c = 0\) or \(\cos A + \cos B + \cos C = \sqrt{3} \cdot \sin(\text{something})\), rewrite using sine/cosine rules and simplify. The 2017-I question used \(a - 2b + c = 0\) to evaluate \(s/(s - b)\). The 2016-I question used the identity $$\cos A + \cos B + \cos C = 1 + 4 \sin\!\left(\tfrac{A}{2}\right) \sin\!\left(\tfrac{B}{2}\right) \sin\!\left(\tfrac{C}{2}\right)$$ to deduce \(\sin(A/2) \sin(B/2) \sin(C/2) = 1/8\), which means the triangle is equilateral.
How NDA Tests This Topic
NDA does not ask you to derive the sine rule from scratch. It gives you a triangle with one or two unknowns and expects you to identify the right formula within 15–20 seconds. The deciding skill is recognising which formula applies — sine rule (one side + opposite angle), cosine rule (two sides + included angle or all three sides), area formula (two sides + included angle). Spend most of your practice time on that recognition step, not the arithmetic.
Preparation Strategy
Week 1 — Formula Fluency
Write all six core formulas (Sine Rule, three Cosine Rule forms, Area, R, r) on one page. Cover the right-hand side and reproduce each from memory. Do this daily for 10 minutes. You are ready to move on only when you can write all six in under 3 minutes without errors.
Week 2 — Single-Triangle Drills
Pick any 10 basic NDA PYQs where a single triangle is given. Solve each within 90 seconds. Focus on: (1) identify the given elements, (2) select the formula, (3) substitute and compute. Track which formula you hesitate on and drill that one specifically.
Week 3 — Multi-Part Sets
Attempt the 2011-I four-part set, the 2023-I two-part set (\(a^2 + b^2 + c^2 = ac + \sqrt{3}\,bc\)), and the 2025-I three-part set. Treat each set as one problem: solve the common triangle first, then answer each sub-item. This mirrors how the NDA paper is structured.
Week 4 — Timed Mixed Practice
Set a 2-minute timer per question (NDA standard). Mix Properties of Triangles with Height & Distance and Trigonometric Identities — NDA papers never isolate a single topic. Aim to complete each Properties of Triangles question in 90 seconds or less, leaving 30 seconds buffer for the connected sub-parts.
Common Mistakes to Avoid
- Forgetting the ambiguous case: When \(\sin C = 1/2\), both \(30^\circ\) and \(150^\circ\) are candidates. Always check angle sum.
- Mixing up \(a, b, c\) notation: In NDA solutions, \(a\) is opposite to \(A\), \(b\) is opposite to \(B\), \(c\) is opposite to \(C\) — consistently. One notation error cascades into a wrong answer.
- Skipping the semi-perimeter: For in-radius problems, define \(s = (a+b+c)/2\) explicitly before substituting into \(r = \Delta/s\). Missing this step causes arithmetic errors under time pressure.
- Assuming all triangles with AP angles have \(a = c\): AP angles mean \(B = 60^\circ\), not that the triangle is isosceles. Only if \(b:c = \sqrt{3}:\sqrt{2}\) (as in 2015-I) can you further deduce \(C = 45^\circ\) and \(A = 75^\circ\).
Pair this topic with Inverse Trigonometric Functions during revision — both chapters appear in the same NDA paper section and NDA question setters frequently combine them in a single question (e.g., "if \(A = \tan^{-1} 2\) and \(B = \tan^{-1} 3\), find \(C\)" in 2011-II).
Frequently Asked Questions
Which formula should I use when all three sides are given?
Use the Cosine Rule. Pick the version for the angle you want: $$\cos A = \frac{b^2 + c^2 - a^2}{2bc}.$$ Start with the angle opposite the largest side to check whether the triangle is obtuse or not. This is exactly what NDA 2010-II (sides 6, 10, 14) and 2025-I (sides 3, 5, 7) required.
What does it mean when NDA says angles are in AP?
Angles \(A, B, C\) in arithmetic progression means \(B\) is the middle term, so \(A + C = 2B\). Since \(A + B + C = 180^\circ\), you get \(3B = 180^\circ\), making \(B = 60^\circ\). Substitute this into the cosine rule: $$\cos 60^\circ = \frac{a^2 + c^2 - b^2}{2ac} = \tfrac{1}{2},$$ giving \(b^2 = a^2 + c^2 - ac\). This shortcut solves about 3–4 NDA questions from the 2012–2017 papers instantly.
How is the in-radius related to the area and semi-perimeter?
The in-radius \(r\) equals the area \(\Delta\) divided by the semi-perimeter \(s\): \(r = \Delta/s\). Equivalently, \(r = (s - a) \tan(A/2)\). The 2022-I NDA question on \(p(p - 2a) \tan(A/2) = 4q\) is a direct application: \(p = 2s\), \(p - 2a = 2(s - a)\), and \(\tan(A/2) = r/(s - a)\), so everything simplifies to \(4sr = 4\Delta = 4q\).
When do I get two answers from the Sine Rule (ambiguous case)?
The ambiguous case arises in SSA configuration (two sides and a non-included angle). If \(\sin C\) computes to a value between 0 and 1, both \(C\) and \(180^\circ - C\) are geometrically possible. Check which one keeps \(A + B + C = 180^\circ\) with all angles positive. NDA options usually include both values as distractors, so this check is mandatory.
What is the significance of cos A/a = cos B/b = cos C/c?
Substitute \(a = 2R \sin A\) into the left side: $$\frac{\cos A}{2R \sin A} = \frac{1}{2R} \cot A.$$ Equal ratios mean \(\cot A = \cot B = \cot C\), so \(A = B = C = 60^\circ\). The triangle is equilateral. This appeared in NDA 2024-II where \(a = 6\) gives \(\text{area} = 9\sqrt{3}\) sq cm immediately.
How many marks does Properties of Triangles carry in NDA Maths?
The NDA Mathematics paper carries 300 marks across 120 questions (2.5 marks each). Properties of Triangles has contributed 2–5 questions per paper in recent years (2020–2025), translating to 5–12.5 marks. Multi-part sets mean answering one question correctly often unlocks 2–3 more, making accuracy here high-leverage.
Is Heron's formula tested in NDA?
Heron's formula $$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$$ is rarely the direct computation method in NDA, because NDA prefers the \(\text{area} = \tfrac{1}{2} ab \sin C\) form when an angle is known. However, Heron's formula underpins the in-radius formula \(r = \Delta/s\), which is tested. Know Heron's formula for completeness but prioritise the sine-based area formula for speed.