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CAT 2025 Lesson : Lines & Triangles - Basics of Triangles

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3. Triangles

A Triangle is a shape enclosed by 33 intersecting lines forming 33 vertices, 33 edges, 33 interior and 33 exterior angles.

In △\triangleABC, the 33 interior angles are named after the vertices, i.e. ∠\angleA, ∠\angleB, ∠\angleC; the sides opposite to the vertices are named with lower case letters – a,b,ca, b, c. We use different letters to name the exterior angles, which in this case are ∠\angleX, ∠\angleY, ∠\angleZ.



3.1 Basic Properties

3.1.1 Sum of Sides

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

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

Also, sum of any 22 sides is always greater than the 3rd3^{\mathrm{rd}} side.
i.e., a ++ b > c ; b ++ c > a ; c ++ a > b

Example 6

How many distinct triangles with integral sides can be formed where 22 of the sides are 55 cm and 1010 cm?

Solution

Let the unknown side be xx.
If xx is the longest side, then 5+10>x5 + 10 > x ⇒ x<15\bm{x < 15}
If 1010 is the longest side, then 5+x>105 + x > 10 ⇒ x>5\bm{x > 5}

Merging the two inequalities, we get 5<x<15\bm{5 < x < 15}

Number of integers between 55 and 1515 (excluding them) = 15−5−1=915 - 5 - 1 = \bm{9}

Answer: 99


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 pp.

Number of triangles of integral sides, where
- pp is even == [p248]\bm{\left[ \dfrac{p^{2}}{48} \right]} rounded to the nearest integer
- pp is odd == [(p+3)248]\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.

Example 7

How many distinct triangles with integral sides exist whose
(I) perimeters are 1212 cm
(II) perimeters are 1515 cm

Solution

Applying Formula

(I) Number of integer-sided triangles == [12248]\left[ \dfrac{12^{2}}{48} \right] == [14448]\left[ \dfrac{144}{48} \right] == 3\bm{3}

(II) Number of integer-sided triangles == [(15+3)248]\left[ \dfrac{(15 + 3)^{2}}{48} \right] == [32448]\left[ \dfrac{324}{48} \right] == 7\bm{7} (approximated)

Manual Iterations While you should understand how to do this, we recommend you to use the formula to save time.

(I) The longest side should be less than half the perimeter, i.e. 66. Possible triangle combinations are (5,5,2),(5,4,3),(4,4,4)(5, 5, 2), (5, 4, 3), (4, 4, 4). So, total of 3\bm{3} triangles are possible.

(II) The longest side should be less than half the perimeter, i.e. 7.57.5. Possible triangle combinations are (7,7,1),(7,6,2),(7,5,3),(7,4,4),(6,6,3),(6,5,4),(5,5,5)(7, 7, 1), (7, 6, 2), (7, 5, 3), (7, 4, 4), (6, 6, 3), (6, 5, 4), (5, 5, 5). So, total of 7\bm{7} triangles are possible.

Answer: (I) 33; (II) 77


3.1.3 Interior and Exterior Angles

1) Sum of the interior angles = 180°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,
∠1=∠4\angle{1} = \angle{4} -----(I)
∠3=∠5\angle{3} = \angle{5} -----(II)

∠4+∠2+∠5\angle{4} + \angle{2} + \angle{5} == 180°180\degree (Angles along the line PQ) Substituting (I) and (II),
∠1+∠2+∠3\angle{1} + \angle{2} + \angle{3} == 180°180\degree


2) Sum of exterior angles == 360°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
∠1+∠4\angle{1} + \angle{4} == 180°180\degree ;
∠2+∠5\angle{2} + \angle{5} == 180°180\degree ;
∠3+∠6\angle{3} + \angle{6} == 180°180\degree ;
∠1+∠2+∠3\angle{1} + \angle{2} + \angle{3} == 180°180\degree (Sum of angles of a △\triangleABC)

Adding the linear pairs, we get
∠1+∠2+∠3+∠4+\angle{1} + \angle{2} + \angle{3} + \angle{4} + ∠5+∠6\angle{5} + \angle{6} = 540°540\degree
⇒ 180°180\degree + ∠4+∠5+∠6\angle{4} + \angle{5} + \angle{6} =540° 540\degree
⇒ ∠4+∠5+∠6=360°\angle{4} + \angle{5} + \angle{6} = 360\degree


3) Exterior angle = Sum of 22 other Interior angles of a triangle

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

Example 8

In the below figure, ∠\angle BDC = ?

Solution

If we join B and C, then sum of angles of △\triangleABC,
50°+45°+∠b+35°50\degree + 45\degree + \angle{b} + 35\degree +∠c=180°+ \angle{c} = 180\degree
⇒ ∠b+∠c\angle{b} + \angle{c} == 50°50\degree -----(1)

In △\triangleBDC, ∠b+∠c+∠d\angle{b} + \angle{c} + \angle{d} = 180°180\degree
Substituting Eq(1),
⇒ ∠d+50°\angle{d} + 50\degree == 180°180\degree
⇒ ∠d=130°\angle{d} = 130\degree


Answer: 130°130\degree


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