## Triangles

**Triangle** = Figure with three sides. Study the following triangle: A,B,C to points. a, b, c to sides. x, y, z to angles.

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

Remember that, Sum of all the angles is always \(180^{\circ}\)

i.e. x + y + z = \(180^{\circ}\)

### Classification of Triangles

Basically there are three types of triangles excluding right angle triangle. Let me tell you how they vary from each other.

- Scalene Triangle
- Isosceles Triangle
- Equilateral Triangle

**Scalene Triangle:** No side of triangle is equal.

**Isosceles Triangle:** Two sides of triangle are equal.

**Equilateral Triangle:** All sides of triangle are equal.

Scalene | Isosceles | Equilateral | |

Definition | a ne b ne c | ane b = c | a = b = c |

Area | A | \((b/4) \sqrt{4a^2 – b^2}\) | \((3)^{(1/2)(a^2)/4}\) |

Height | - | \(\sqrt{(4a^2-b^2) /2}\) | \((3)^{(1/2)a/2}\) |

\(A = {s(s-a)(s-b)(s-c)}^{1/2}\)

where, s = (a+ b+ c)/2

### Properties of external angles of Triangle:

**1.** **Sum of all exterior angles is \(360^{\circ}\)**

Study the following set of triangles and their exterior angles,

a, b, c to Interior angles. p, q, r and s, t, u to Exterior angles.

So, sum of exterior angles = \(360^o\) i.e. p + q + r = 360^o and s + t + u = \(360^o\)

**2.** Next property of exterior angle which is important in paper point of view:

**External angle = Sum of two internal angles.**

For example: In above figures,

r = a + b

q = a + c

s = b + c and so on.

## Right angle Triangle

Following triangle is a right angle triangle i.e. a triangle with one out of three 90^o angle.

### Area of right angle triangle

Area = 1/2 times Base times Perpendicular

### Example with Solution

** Example:** In following figure, CE is perpendicular to AB, angle ACE = 20^o and angle ABD = 50^o. Find angle BDA:

**Solution:** To Find: angle BDA

For this what we need --- angle BAD Because, Sum of all angles = \(180 ^o\)

Consider, triangle ECA,

CEA + EAC + ACE = \(180^o\) i.e. \(90^o\) + \(20^o\) + EAC = \(180^o\) Therefore, EAC = \(70^o\)

Now, come to triangle ABD,

ABD + BDA + BAD = \(180^o\) \(70^o\) + \(50^o\) + BAD = \(180^o\) Therefore, BAD = \(60^o\)

**Example:** In given figure. BC is produced to D and angle BAC = \(40^o\) and angle ABC = \(70^o\). Find angle ACD:

**Solution:** In above figure, ACD is an exterior angle, and according to property, Exterior angle = Sum of interior angles Therefore, ACD = \(70^o\) + \(40^o\) =\(110^o\)

This is not the end of this chapter. These are just the basics. In next session, I will discuss some important results, properties (congruency, similarity) and much more. Always remember, Geometry needs practice and time.