A Triangle is a three-sided polygon that consists of three edges and three vertices and the sum of internal angles of a triangle equal to \(180^{o}\). Depending upon the sides and angles of a triangle, we can have different types of triangle. Here we have given a brief description for determining the area of a triangle using direct formula and using Heron’s formula.

Below given is a triangle having 3 sides and three edges, numbered as 0,1,2.

**Types of Triangles-**** **

**(i) On the basis of its Length**

**Scalene Triangle-**This type of triangle has all its three edges of different length. Due to this, the three angles are also different from each other.

**Isosceles Triangle–**This type of triangle has two equal sides. Also, the two angles, opposite to the two equal sides, are also equal to each other.

This fact also supports the famous isosceles triangle theorem.

**Equilateral Triangle–**This type of triangle has all the three sides equal to each other. Due to this all the internal angles are of equal degrees ie. each of the angles is \(60^{\circ}\)

**(ii) On the basis of angle –**

**Acute angled Triangle–**This triangle has all of its angles less than 90°.

**Right angle Triangle–**This triangle has any of its one angle to be equal to 90°.

**Obtuse-angled Triangle–**This triangle has any of its one angles more than 90°.

**Properties of Triangle:**

- Sum of Angles of a Triangle is always
**180 degrees**. - The exterior angles of a triangle always add up to
**360 degrees**. - The sum of the lengths of any two sides of a triangle is greater than the length of the third side. Similarly, the difference between the lengths of any two sides of a triangle is less than the length of the third side.
- The shortest side is always opposite to the smallest interior angle. Similarly, the longest side is always opposite to the largest interior angle.

Area of a Triangle –

Area of a Triangle –

** **

The area of a triangle is given by-

**Area = \(\frac{1}{2} \times Base \times Height\)**

**Lets Work Out-**

**Question-**** Find the area of a triangle having base equal to 9 cm and height equal to 6 cm.**

**Solution- ** We know that **Area = \(\frac{1}{2} \times Base \times Height\)**

= \(\frac{1}{2} \times 9 \times 6 \;\; cm^{2}\)

= \(27 cm^{2} \)

**Area using Heron’s formula – **

Sometimes it is difficult to calculate the height of a triangle, so we cannot use the above formula.

Therefore Heron’s formula is used to calculate the area of a triangle.

First we need to calculate the semiperimeter (s).

\(s = \frac{1}{2}(a+b+c)\), (where a,b,c are the three sides of a triangle)

Now Area is given by –

**\(\bigtriangleup = \sqrt{s(s-a)(s-b)(s-c)}\)**

**Lets Work Out-**

**Question-**** Find the area of a triangle having sides 5,6 and 7 units length.**

**Solution- ** Using Heron’s formula to find the area of a triangle-

Semiperimeter (s) = \( \frac{1}{2}(a+b+c)\)

s = \( \frac{1}{2}(5+6+7)\)

s = \( 9 \)

Now Area \(\bigtriangleup = \sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{9(9-5)(9-6)(9-7)}\)

= \(\sqrt{9\times 4 \times 3 \times 2}\)

= \(= \sqrt{3 \times 3 \times 2 \times 2 \times 3 \times 2}\)

= \(= 6\sqrt{6}\) square units.

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