CAT 2025 Lesson : Lines & Triangles - Basics of Triangles

3. Triangles

A Triangle is a shape enclosed by $3$ intersecting lines forming $3$ vertices, $3$ edges, $3$ interior and $3$ exterior angles.

In $\triangle$ABC, the $3$ interior angles are named after the vertices, i.e. $\angle$A, $\angle$B, $\angle$C; the sides opposite to the vertices are named with lower case letters – $a, b, c$. We use different letters to name the exterior angles, which in this case are $\angle$X, $\angle$Y, $\angle$Z.

3.1 Basic Properties

3.1.1 Sum of Sides

For a triangle, whose sides are of length $a, b$ and $c$, the perimeter is the sum of the lengths of these sides.

Perimeter of the triangle $=$ a $+$ b $+$ c

Also, sum of any $2$ sides is always greater than the $3^{\mathrm{rd}}$ side.
i.e., a $+$ b > c ; b $+$ c > a ; c $+$ a > b



3.1.2 Triangles with Integral Sides

Honsberger's theorem derived from partition theory provides an easy way to find the number of different triangles with integral sides that can be formed for a given perimeter, say $p$.

Number of triangles of integral sides, where
- $p$ is even $=$ $\bm{\left[ \dfrac{p^{2}}{48} \right]}$ rounded to the nearest integer
- $p$ is odd $=$ $\bm{\left[ \dfrac{(p + 3)^{2}}{48} \right]}$ rounded to the nearest integer

Alternatively, you could use manual iterations to solve for this bearing in mind that each side should be less than half the perimeter. However, using the formula will save you time in the exam.

3.1.3 Interior and Exterior Angles

1) Sum of the interior angles = $180\degree$

Rationale: In the following diagram, a line PQ is drawn through point A such that PQ | | BC.

As alternate interior angles of a transversal are equal, $\angle{1} = \angle{4}$ -----(I) $\angle{3} = \angle{5}$ -----(II) $\angle{4} + \angle{2} + \angle{5}$ $=$ $180\degree$ (Angles along the line PQ) Substituting (I) and (II), $\angle{1} + \angle{2} + \angle{3}$ $=$ $180\degree$

2) Sum of exterior angles $=$ $360\degree$

Rationale: In the following diagram, the lines AB, BC and CA are extended to P, Q and R respectively.

The following three are linear pairs $\angle{1} + \angle{4}$ $=$ $180\degree$ ; $\angle{2} + \angle{5}$ $=$ $180\degree$ ; $\angle{3} + \angle{6}$ $=$ $180\degree$ ; $\angle{1} + \angle{2} + \angle{3}$ $=$ $180\degree$ (Sum of angles of a $\triangle$ABC) Adding the linear pairs, we get $\angle{1} + \angle{2} + \angle{3} + \angle{4} +$ $\angle{5} + \angle{6}$ = $540\degree$ ⇒ $180\degree$ + $\angle{4} + \angle{5} + \angle{6}$=$540\degree$ ⇒ $\angle{4} + \angle{5} + \angle{6} = 360\degree$

3) Exterior angle = Sum of $2$ other Interior angles of a triangle

In the above diagram note that $\angle{1} + \angle{2} + \angle{3}$ $=$ $180\degree$ (Sum of angles of a triangle)
$\angle{1} + \angle{4}$ $=$ $180\degree$ (Linear Pair)
$\therefore$ From the above equations we get $\angle{4} = \angle{2} + \angle{3}$
Likewise, $\angle{5} = \angle{3} + \angle{1}$ and $\angle{6} = \angle{1} + \angle{2}$