EAPCET Mathematics · Intermediate first year · Paper IA

Properties of Triangles

Properties of Triangles connects the sides and angles of a triangle with its area, circumradius, inradius and exradii, and applies them to heights and distances.

Subject
Mathematics
Taught in
Intermediate first year · Paper IA
Mathematics in the paper
80 of 160 questions

Key formulas and results

  • asin⁡A=bsin⁡B=csin⁡C=2R\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R
  • cos⁡A=b2+c2−a22bc\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}
  • a=bcos⁡C+ccos⁡Ba = b\cos C + c\cos B (projection formula)
  • Δ=12bcsin⁡A=s(s−a)(s−b)(s−c)\Delta = \tfrac{1}{2}bc\sin A = \sqrt{s(s - a)(s - b)(s - c)}
  • r=Δsr = \dfrac{\Delta}{s}, R=abc4ΔR = \dfrac{abc}{4\Delta}, r1=Δs−ar_{1} = \dfrac{\Delta}{s - a}
  • Napier’s analogy: tan⁡B−C2=b−cb+ccot⁡A2\tan\dfrac{B - C}{2} = \dfrac{b - c}{b + c}\cot\dfrac{A}{2}
  • Half-angle formulas: sin⁡A2=(s−b)(s−c)bc\sin\dfrac{A}{2} = \sqrt{\dfrac{(s - b)(s - c)}{bc}}, cos⁡A2=s(s−a)bc\cos\dfrac{A}{2} = \sqrt{\dfrac{s(s - a)}{bc}}, tan⁡A2=(s−b)(s−c)s(s−a)\tan\dfrac{A}{2} = \sqrt{\dfrac{(s - b)(s - c)}{s(s - a)}}
  • tan⁡A2=Δs(s−a)\tan\dfrac{A}{2} = \dfrac{\Delta}{s(s - a)} and tan⁡A2tan⁡B2=s−cs\tan\dfrac{A}{2}\tan\dfrac{B}{2} = \dfrac{s - c}{s}
  • Inradius: r=(s−a)tan⁡A2=4Rsin⁡A2sin⁡B2sin⁡C2r = (s - a)\tan\dfrac{A}{2} = 4R\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}
  • Exradius: r1=stan⁡A2=4Rsin⁡A2cos⁡B2cos⁡C2r_{1} = s\tan\dfrac{A}{2} = 4R\sin\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}
  • 1r1+1r2+1r3=1r\dfrac{1}{r_{1}} + \dfrac{1}{r_{2}} + \dfrac{1}{r_{3}} = \dfrac{1}{r} and r r1r2r3=Δ2r\,r_{1}r_{2}r_{3} = \Delta^{2}
  • r+r2+r3−r1=4Rcos⁡Ar + r_{2} + r_{3} - r_{1} = 4R\cos A (and similarly for BB and CC)
  • Altitudes p1,p2,p3p_{1}, p_{2}, p_{3} (p1=2Δa)\left(p_{1} = \tfrac{2\Delta}{a}\right): 1p1+1p2+1p3=1r\dfrac{1}{p_{1}} + \dfrac{1}{p_{2}} + \dfrac{1}{p_{3}} = \dfrac{1}{r}

What this chapter covers

  • Sine rule and cosine rule
  • Projection formulae
  • Half-angle formulae
  • Area of a triangle
  • Circumradius, inradius and exradii
  • Heights and distances

How it is usually asked

  • Find R, r or an exradius for a given triangle
  • Prove or apply identities such as r₁ + r₂ + r₃ − r = 4R
  • Solve a heights-and-distances problem

Understanding Properties of Triangles

In triangle ABC the sides opposite A, B and C are a, b and c. The sine rule a/sin A = b/sin B = c/sin C = 2R links sides to opposite angles through the circumradius R, while the cosine rule a² = b² + c² − 2bc cos A finds a side from two sides and the included angle, or an angle from three sides. The projection formula a = b cos C + c cos B is a third relation.

Most results are written using the semi-perimeter s = (a + b + c)/2 and the area Δ. Heron’s formula gives Δ = √[s(s − a)(s − b)(s − c)], and then the inradius is r = Δ/s, the circumradius is R = abc/4Δ and the exradius opposite A is r₁ = Δ/(s − a). Half-angle formulas such as tan(A/2) = r/(s − a) connect these to the angles.

EAPCET questions ask you to compute R, r or an exradius for given sides, to identify a correct identity among options, or to solve a heights-and-distances problem with angles of elevation. For numerical ones, find s and Δ first, since nearly everything else follows from them. For identity questions, testing an equilateral triangle quickly removes wrong options.

Worked example

EAPCET-style question

For the triangle with sides 13, 14 and 15, the inradius rr and circumradius RR are

  1. (1)r=4, R=658r = 4,\ R = \dfrac{65}{8}
  2. (2)r=4, R=654r = 4,\ R = \dfrac{65}{4}
  3. (3)r=8, R=658r = 8,\ R = \dfrac{65}{8}
  4. (4)r=2, R=658r = 2,\ R = \dfrac{65}{8}

Solution

  1. s=13+14+152=21s = \dfrac{13 + 14 + 15}{2} = 21.
  2. Δ=21×8×7×6=7056=84\Delta = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84.
  3. r=Δs=8421=4r = \dfrac{\Delta}{s} = \dfrac{84}{21} = 4.
  4. R=abc4Δ=13×14×154×84=2730336=658R = \dfrac{abc}{4\Delta} = \dfrac{13 \times 14 \times 15}{4 \times 84} = \dfrac{2730}{336} = \dfrac{65}{8}.

Answer: r=4, R=658r = 4,\ R = \dfrac{65}{8}

Shortcuts and results to remember

  • If a,b,ca, b, c are in A.P., then cot⁡A2,cot⁡B2,cot⁡C2\cot\tfrac{A}{2}, \cot\tfrac{B}{2}, \cot\tfrac{C}{2} are in A.P. and tan⁡A2tan⁡C2=13\tan\tfrac{A}{2}\tan\tfrac{C}{2} = \tfrac{1}{3}.
  • If A=60∘A = 60^\circ, then (a+b+c)(b+c−a)=3bc(a + b + c)(b + c - a) = 3bc, equivalently 1a+b+1a+c=3a+b+c\dfrac{1}{a + b} + \dfrac{1}{a + c} = \dfrac{3}{a + b + c}.
  • Mollweide’s check: a+bc=cos⁡A−B2sin⁡C2\dfrac{a + b}{c} = \dfrac{\cos\frac{A - B}{2}}{\sin\frac{C}{2}} and a−bc=sin⁡A−B2cos⁡C2\dfrac{a - b}{c} = \dfrac{\sin\frac{A - B}{2}}{\cos\frac{C}{2}}.
  • Adding the three cosine rules gives a2+b2+c2=2(bccos⁡A+cacos⁡B+abcos⁡C)a^{2} + b^{2} + c^{2} = 2(bc\cos A + ca\cos B + ab\cos C).

More practice questions

Practice question 1

In a triangle with sides a=13a = 13, b=14b = 14, c=15c = 15, the inradius rr and circumradius RR are

  1. (1)r=4, R=658r = 4,\ R = \dfrac{65}{8}
  2. (2)r=4, R=8r = 4,\ R = 8
  3. (3)r=3, R=658r = 3,\ R = \dfrac{65}{8}
  4. (4)r=6, R=654r = 6,\ R = \dfrac{65}{4}

Solution

  1. s=13+14+152=21s = \dfrac{13 + 14 + 15}{2} = 21
  2. Δ=21×8×7×6=7056=84\Delta = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84
  3. r=Δs=8421=4r = \dfrac{\Delta}{s} = \dfrac{84}{21} = 4
  4. R=abc4Δ=2730336=658R = \dfrac{abc}{4\Delta} = \dfrac{2730}{336} = \dfrac{65}{8}

Answer: r=4, R=658r = 4,\ R = \dfrac{65}{8}

Practice question 2

If the sides a,b,ca, b, c of triangle ABCABC are in A.P., then tan⁡A2tan⁡C2\tan\tfrac{A}{2}\tan\tfrac{C}{2} equals

  1. (1)13\dfrac{1}{3}
  2. (2)12\dfrac{1}{2}
  3. (3)23\dfrac{2}{3}
  4. (4)3

Solution

  1. tan⁡A2tan⁡C2=s−bs\tan\dfrac{A}{2}\tan\dfrac{C}{2} = \dfrac{s - b}{s}
  2. a,b,ca, b, c in A.P. means a+c=2ba + c = 2b, so 2s=3b2s = 3b and s=3b2s = \tfrac{3b}{2}
  3. s−b=b2s - b = \tfrac{b}{2}
  4. s−bs=b/23b/2=13\dfrac{s - b}{s} = \dfrac{b/2}{3b/2} = \dfrac{1}{3}

Answer: 13\dfrac{1}{3}

Common mistakes to avoid

  • Using the full perimeter as s; s is half the perimeter, and Heron’s formula and r = Δ/s both need the half.
  • Writing R = abc/Δ; the formula is R = abc/4Δ.
  • Pairing a side with the wrong angle in the cosine rule; a² = b² + c² − 2bc cos A uses the angle opposite a.
  • Taking only the acute angle from sin B in the sine rule; when two sides and a non-included angle are given, check whether an obtuse angle is also possible.

How to prepare it

Keep Δ and s at the centre — nearly every radius formula is written in terms of them.

For identity questions, test an equilateral triangle: if an option fails there, it is wrong.

Once the formulas feel familiar, test them the way the exam does: practise Properties of Triangles questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Properties of Triangles chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Properties of Triangles part of the EAPCET Mathematics syllabus?
Yes. Properties of Triangles is taught in Intermediate first year · Paper IA Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Properties of Triangles for EAPCET?
Keep Δ and s at the centre — nearly every radius formula is written in terms of them. For identity questions, test an equilateral triangle: if an option fails there, it is wrong.
How many Mathematics questions are there in EAPCET?
80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Properties of Triangles?
Using the full perimeter as s; s is half the perimeter, and Heron’s formula and r = Δ/s both need the half. Writing R = abc/Δ; the formula is R = abc/4Δ. Pairing a side with the wrong angle in the cosine rule; a² = b² + c² − 2bc cos A uses the angle opposite a. Taking only the acute angle from sin B in the sine rule; when two sides and a non-included angle are given, check whether an obtuse angle is also possible.