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Lines & Triangles

Lines And Triangles

MODULES

Lines & Angles
Parallel Lines
Basics of Triangles
Types of Triangles
Triangle & its Segments
Area of a Triangle
Isosceles & Equilateral
Right-angled Triangles
Other Theorems
Congruency & Similarity
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Lines & Triangles 1
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Lines & Triangles 2
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Lines & Triangles 3
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PRACTICE

Lines & Triangles : Level 1
Lines & Triangles : Level 2
Lines & Triangles : Level 3
ALL MODULES

CAT 2025 Lesson : Lines & Triangles - Basics of Triangles

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3. Triangles

A Triangle is a shape enclosed by 333 intersecting lines forming 333 vertices, 333 edges, 333 interior and 333 exterior angles.

In △\triangle△ABC, the 333 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,ca, b, ca,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,ba, ba,b and ccc, the perimeter is the sum of the lengths of these sides.

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

Also, sum of any 222 sides is always greater than the 3rd3^{\mathrm{rd}}3rd 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 222 of the sides are 555 cm and 101010 cm?

Solution

Let the unknown side be xxx.
If xxx is the longest side, then 5+10>x5 + 10 > x5+10>x ⇒ x<15\bm{x < 15}x<15
If 101010 is the longest side, then 5+x>105 + x > 105+x>10 ⇒ x>5\bm{x > 5}x>5

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

Number of integers between 555 and 151515 (excluding them) = 15−5−1=915 - 5 - 1 = \bm{9}15−5−1=9

Answer: 999


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 ppp.

Number of triangles of integral sides, where
- ppp is even === [p248]\bm{\left[ \dfrac{p^{2}}{48} \right]}[48p2​] rounded to the nearest integer
- ppp is odd === [(p+3)248]\bm{\left[ \dfrac{(p + 3)^{2}}{48} \right]}[48(p+3)2​] 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 121212 cm
(II) perimeters are 151515 cm

Solution

Applying Formula

(I) Number of integer-sided triangles === [12248]\left[ \dfrac{12^{2}}{48} \right][48122​] === [14448]\left[ \dfrac{144}{48} \right][48144​] === 3\bm{3}3

(II) Number of integer-sided triangles === [(15+3)248]\left[ \dfrac{(15 + 3)^{2}}{48} \right][48(15+3)2​] === [32448]\left[ \dfrac{324}{48} \right][48324​] === 7\bm{7}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. 666. Possible triangle combinations are (5,5,2),(5,4,3),(4,4,4)(5, 5, 2), (5, 4, 3), (4, 4, 4)(5,5,2),(5,4,3),(4,4,4). So, total of 3\bm{3}3 triangles are possible.

(II) The longest side should be less than half the perimeter, i.e. 7.57.57.5. 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)(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}7 triangles are possible.

Answer: (I) 333; (II) 777


3.1.3 Interior and Exterior Angles

1) Sum of the interior angles = 180°180\degree180°

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}∠1=∠4 -----(I)
∠3=∠5\angle{3} = \angle{5}∠3=∠5 -----(II)

∠4+∠2+∠5\angle{4} + \angle{2} + \angle{5}∠4+∠2+∠5 === 180°180\degree180° (Angles along the line PQ) Substituting (I) and (II),
∠1+∠2+∠3\angle{1} + \angle{2} + \angle{3}∠1+∠2+∠3 === 180°180\degree180°


2) Sum of exterior angles === 360°360\degree360°

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}∠1+∠4 === 180°180\degree180° ;
∠2+∠5\angle{2} + \angle{5}∠2+∠5 === 180°180\degree180° ;
∠3+∠6\angle{3} + \angle{6}∠3+∠6 === 180°180\degree180° ;
∠1+∠2+∠3\angle{1} + \angle{2} + \angle{3}∠1+∠2+∠3 === 180°180\degree180° (Sum of angles of a △\triangle△ABC)

Adding the linear pairs, we get
∠1+∠2+∠3+∠4+\angle{1} + \angle{2} + \angle{3} + \angle{4} +∠1+∠2+∠3+∠4+ ∠5+∠6\angle{5} + \angle{6}∠5+∠6 = 540°540\degree540°
⇒ 180°180\degree180° + ∠4+∠5+∠6\angle{4} + \angle{5} + \angle{6} ∠4+∠5+∠6=540° 540\degree540°
⇒ ∠4+∠5+∠6=360°\angle{4} + \angle{5} + \angle{6} = 360\degree∠4+∠5+∠6=360°


3) Exterior angle = Sum of 222 other Interior angles of a triangle

In the above diagram note that ∠1+∠2+∠3\angle{1} + \angle{2} + \angle{3}∠1+∠2+∠3 === 180°180\degree180° (Sum of angles of a triangle)
∠1+∠4\angle{1} + \angle{4}∠1+∠4 === 180°180\degree180° (Linear Pair)
∴\therefore∴ From the above equations we get ∠4=∠2+∠3\angle{4} = \angle{2} + \angle{3}∠4=∠2+∠3
Likewise, ∠5=∠3+∠1\angle{5} = \angle{3} + \angle{1}∠5=∠3+∠1 and ∠6=∠1+∠2\angle{6} = \angle{1} + \angle{2}∠6=∠1+∠2

Example 8

In the below figure, ∠\angle∠ BDC = ?

Solution

If we join B and C, then sum of angles of △\triangle△ABC,
50°+45°+∠b+35°50\degree + 45\degree + \angle{b} + 35\degree50°+45°+∠b+35° +∠c=180°+ \angle{c} = 180\degree+∠c=180°
⇒ ∠b+∠c\angle{b} + \angle{c}∠b+∠c === 50°50\degree50° -----(1)

In △\triangle△BDC, ∠b+∠c+∠d\angle{b} + \angle{c} + \angle{d}∠b+∠c+∠d = 180°180\degree180°
Substituting Eq(1),
⇒ ∠d+50°\angle{d} + 50\degree∠d+50° === 180°180\degree180°
⇒ ∠d=130°\angle{d} = 130\degree∠d=130°


Answer: 130°130\degree130°


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