**Definition**

A Closed figure made of three line segments is called a **Triangle**. It has three sides, three angles, and three vertices. Here, Triangle ABC and its Properties are shown below.

Three sides – AB, BC, CA

Three angles – ∠A, ∠B, ∠C

Three vertices – A, B, C

**Classification of Triangles –**

#### ➤** Based on Sides**

❶ **Scalene Triangle –** The triangle in which all three sides are of different lengths, is called a **Scalene Triangle**. Because of all the different sides, all the three angles are also of different measure in the scalene triangle.

❷ **Isosceles Triangle – **The triangle in which any two sides are of equal length, is called an **Isosceles Triangle**. Because of two equal sides, angles opposite to equal sides are also of equal measure in the isosceles triangle.

❸ **Equilateral Triangle – **The triangle in which all three sides are of equal length, is called an **Equilateral Triangle**. Because of all the equal sides, all the three angles are also of equal measure (Each 60°) in the equilateral triangle.

#### ➤ **Based on Angles**

❶ **Acute Angled Triangle – **The triangle in which all the three angles are acute angles (Each angle is less than 90°), is called the **Acute Angled Triangle**.

❷ **Obtuse Angled Triangle – **The triangle in which any one angle is greater than 90°, is called the **Obtuse Angled Triangle**.

Here, **∠C = 120°** (Obtuse angle)

❸ **Right Angled Triangle – **The triangle in which any one angle is equal to 90°, is called the **Right-Angled Triangle**.

Here, **∠B = 90°** (Right angle)

**Median of a Triangle**

A line segment that connects a vertex of a triangle to the mid-point of its opposite side, is called the **Median of a triangle**.

In the above figure, AD – median

**BD = DC** (∵ D is the midpoint of BC)

**The altitude of a Triangle**

A line segment drawn from a vertex of a triangle to its opposite side at the angle of 90°(right angle), is called the **Altitude of a triangle**.

In the above figure, PS – Altitude

**PS ⊥ QR**

**QS ≠ SR** (∵ S is not the mid-point of QR)

**Q 1)** How many altitudes can be drawn in a triangle?

**Ans. –** 3

**Q 2)** How many Medians can be drawn in a triangle?

**Ans. –** 3

**Exterior Angle of a Triangle and its Properties**

**Exterior Angle of a Triangle – **In a triangle, if a side is extended in one direction from its vertex then the angle made in the external part of that vertex is called an **Exterior angle of the triangle**.

**Theorem Based on it**

**Statement – **Exterior angle of a triangle is equal to the sum of its interior opposite angles.

**Given** – In △ABC, exterior angle – ∠ACD, Interior opposite angles – ∠1 and ∠2

**To prove** – ∠ACD = ∠1 + ∠2

**Construction** – Drawn CE ∥ AB

So, let ∠ACE = ∠x and ∠ECD = ∠y

**Proof** – ∵ CE ∥ AB

∴ ∠1 = ∠x (Interior alternate angles are equal)……………..(1)

And ∠2 = ∠y (Corresponding angles are equal)……………….(2)

By adding equations (1) and (2)

∠1 + ∠2 = ∠x + ∠y

∠1 + ∠2 = ∠ACD

The above equation shows the exterior angle property of the triangle. **Hence proved.**

**Example 1)** Find the value of X?

**Solution –** By exterior angle property,

X = 30° + 40°

X = 70° **Ans.**

**Example 2)** Find the value of X?

**Solution –** By exterior angle property

50° + X = 110°

X = 110° – 50°

X = 60° **Ans.**

**Interior Angle Sum Property of a Triangle**

**Statement – **The sum of the three angles of a triangle is 180°.

**Given – **∠1, ∠2, and ∠3 are angles of ΔABC. Side BC is extended to D and ∠4 is the exterior angle.

**To prove – **∠1 + ∠2 + ∠3 = 180°

**Proof – **By Exterior angle property of a triangle,

∠1 + ∠2 = ∠4 ……………..(1)

Adding ∠3 to both side ∠1 + ∠2 + ∠3 = ∠4 + ∠3 …………….(2)

But By Linear pair of Angle, ∠3 + ∠4 = 180° …………….(3)

So, by equations (2) and (3)

∠1 + ∠2 + ∠3 = 180°

This is the interior angle sum property of a triangle. **Hence proved.**

**Example 1)** Find the value of X?

**Solution –** By interior angle sum property of a triangle,

X + 50° + 60° = 180°

X + 110° = 180°

X = 180° – 110°

X = 70° **Ans.**

**Example 2)** Find the value of X?

**Solution –** By interior angle sum property of a triangle,

X + X + X = 180°

3X = 180°

X = 180°/3

X = 60° **Ans.**

The Above triangle is an Equilateral triangle.

**Other Properties of the Triangle**

➞ The Sum of any two sides of a triangle is always greater than the third side.

➞ The Difference of any two sides of a triangle is always less than the third side.

➞ The Sum of all exterior angles of a triangle is 360°.

**The perimeter of a Triangle –**

The Perimeter of a triangle is the sum of all three sides of a triangle. The Perimeter of a triangle is calculated in the length unit.

Perimeter = Sum of all the three sides

For △ABC, sides will be AB, BC, CA then

Perimeter = AB + BC + CA

**Example –** Find the perimeter of the given triangle.

**Solution –** Perimeter = AB + BC + CA

Perimeter = 4 cm + 6 cm + 5 cm

Perimeter = 15 cm **Ans.**

**Area of a Triangle –**

The area enclosed by the three sides of the triangle is the area of the triangle. The area of a triangle is calculated in a square unit.

There are two methods to calculate the area of a triangle –

➊ When the Base and height are given

➋ When all the three sides are given

➊ **When the Base and height are given –** If the base and height of a triangle are given then

The area of the triangle = ½⨯Base⨯Height

= ½⨯B⨯H Square Unit

**Example –** Find the area of the triangle whose base is 8 cm and height is 5 cm.

**Solution –** Here B = 8 cm and H = 5 cm

Area of the Triangle = ½⨯B⨯H

= ½⨯8cm⨯5cm

= 4⨯5

= 20 Square cm or cm^{2} **Ans.**

➋ **When all the three sides are given –** When all the three sides are given we use Heron’s formula to find the area of the triangle.

Area of the triangle by Heron’s formula = √s(s-a)(s-b)(s-c)

Where s = Semi-perimeter of the triangle

s = a+b+c/2

a, b, c = Sides of a triangle

**Example –** Find the area of the triangle whose sides are 3 cm, 5 cm, and 6 cm.

**Solution –** Here a = 3 cm, b = 5 cm and c = 6 cm

So, Semi-perimeter (s) = a+b+c/2 = 3+5+6/2 = 14/2 = 7 cm

Now Area of triangle = √s(s-a)(s-b)(s-c)

= √7(7-3)(7-5)(7-6)

= √7(4)(2)(1)

= √7⨯2⨯2⨯2

= 2√14 cm^{2} **Ans.**