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Triangles

Properties, similarity, congruence, and area of triangles.

4%
of Quant

Why This Topic Matters

Total PYQs📊
14
of 1002 · 2021–2025
Years featured📅
5/5
of recent CAT years
% of Quant📈
~4%
of section questions
Est. hours⏱️
~10h
to master
~1/22
2021
~2/22
2022
~1/22
2023
1/22
2024
1/22
2025
🎯PYQ Evidence

CAT 2021–2025: ~0.9 per slot (2021: 0.7 · 2022: 1.3 · 2023: 0.7 · 2024: 1.0 · 2025: 1.0). Every year; similarity and right-triangle configurations dominate.

Triangles

The single richest figure in CAT geometry. A handful of results — angle sum, area, Pythagoras, and similarity — cover the great majority of questions.

Essentials

  • Angle sum =180°=180°; the exterior angle equals the sum of the two remote interior angles.
  • Area =12bh=\dfrac12\,b\,h, or Heron's s(sa)(sb)(sc)\sqrt{s(s-a)(s-b)(s-c)} with s=a+b+c2s=\dfrac{a+b+c}{2}.
  • Pythagoras a2+b2=c2a^2+b^2=c^2; recognise the triples 3-4-5, 5-12-13, 8-15-17.
  • Similarity (AA): equal angles ⇒ proportional sides, and areas scale as the square of the side ratio.
  • Centroid divides each median in a 2:12{:}1 ratio; inradius r=Areasr=\dfrac{\text{Area}}{s}; circumradius R=abc4AreaR=\dfrac{abc}{4\,\text{Area}}.

A worked example

Find the area of a triangle with sides 13, 14, 15.

14 13 15

Semi-perimeter s=13+14+152=21s=\dfrac{13+14+15}{2}=21. Then

Area=21(2113)(2114)(2115)=21876=7056=84.\text{Area}=\sqrt{21\,(21-13)(21-14)(21-15)}=\sqrt{21\cdot8\cdot7\cdot6}=\sqrt{7056}=\mathbf{84}.

(With area 84 you can also get the inradius r=84/21=4r=84/21=4 and circumradius R=131415484=8.125R=\frac{13\cdot14\cdot15}{4\cdot84}=8.125.)

🎯PYQ Evidence
Convert area conditions into base or altitude relations before reaching for trig. : the three triangles share the same height onto BC, so their areas are proportional to the bases BP, BQ, BC — the AP and "1.5×" conditions become base conditions and PQ falls out directly. : drop the rectangle onto axes, get sides 45, 28, 53, spot the right angle, then apply the right-triangle inradius r = (leg1 + leg2 − hypotenuse)/2 = (45 + 28 − 53)/2 = 10. : in the isosceles triangle the altitude to the base is √(50²−40²) = 30, fixing the area, and every other altitude is just 2·area/side. The recurring trick: same height means area ∝ base, so each altitude is 2·area÷its side.

Common traps

  • Triangle inequality. Any side must be less than the sum of the other two — some "triangles" can't exist.
  • Similar ≠ congruent. Similar triangles share shape, not size; areas go as the square of the ratio.
  • Wrong height. The height must be perpendicular to the chosen base.

Checklist

  • Use Heron when you know all three sides
  • Spot Pythagorean triples to skip arithmetic
  • Apply area ratio = (side ratio)² for similar triangles
  • Recall centroid 2:1, r=Area/sr=\text{Area}/s, R=abc/4AreaR=abc/4\text{Area}

Sample Questions

18 practice questions

Medium

The length of each side of an equilateral triangle ABC is 10 cm. Let D be a point on BC such that the area of triangle ADC is 2/3 times the area of triangle ABD. Then the length of AD, in cm, is

Hard

Let ΔABC be an isosceles triangle such that AB and AC are of equal length. AD is the altitude from A on BC and BE is the altitude from B on AC. If AD and BE intersect at O such that ∠AOB = 95°, then AD/BE equals

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CAT PYQ Spotlight

Actual CAT questions on this topic

CAT 2025 · Slot 2
Hard

In a △ABC, points D and E are on the sides BC and AC, respectively. BE and AD intersect at point T such that AD : AT = 4 : 3, and BE : BT = 5 : 4. Point F lies on AC such that DF is parallel to BE. Then, BD : CD is

CAT 2024 · Slot 1
TITAMedium

ABCD is a rectangle with sides AB = 56 cm and BC = 45 cm, and E is the midpoint of side CD. Then, the length, in cm, of radius of incircle of △ADE is

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