**According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 10.**

**CBSE Class 9 Maths Heron’s Formula Notes –**

In geometry, a triangle is a closed three-dimensional figure. In this article, you are going to learn Heron’s formula for class 9, which is used to find the area of triangles. You will also learn how Heron’s formula is used to find the area of other polygons in detail.

### Triangle

The plane closed figure, with three sides and three angles is called as a triangle.

Types of triangles:

Based on **sides** – a) Equilateral b) Isosceles c) Scalene

Based on **angles** – a) Acute-angled triangle b) Right-angled triangle c) Obtuse-angled triangle

**Area of a Triangle**

Area = (1/2) × base × height

In the case of equilateral and isosceles triangles, if the lengths of the sides of triangles are given, we use Pythagoras’ theorem in order to find the height of a triangle.

### Area of an Equilateral Triangle

Consider an equilateral ΔABC, with each side as a unit. Let AO be the perpendicular bisector of BC. In order to derive the formula for the area of an equilateral triangle, we need to find height AO.

Using Pythagoras’ theorem,

AC2 = OA2 + OC2

OA2 = AC2 − OC2

Substitute AC = a,OC = a/2 in the above equation.

OA2 = a2 − a2/4

OA = √3a/2

We know that the area of the triangle is:

A = (1/2) × base × height

A = (1/2) × a × (√3a/2)

∴ Area of Equilateral triangle = √3a2/4

### Area of an Isosceles Triangle

Consider an isosceles ΔABC with equal sides as a units and base as b units.

Isosceles triangle ABC

The height of the triangle can be found by Pythagoras’ Theorem :

CD2 = AC2 − AD2

⇒h2 = a2− (b2/4) = (4a2 – b2)/4

⇒h =(1/2) √(4a2 – b2)

Area of triangle is A = (1/2)bh

∴ A = (1/2) × b × (1/2)√(4a2 – b2)

∴ A = (1/4) × b × √(4a2 – b2)

### Area of a Triangle – By Heron’s formula

Area of a ΔABC, given sides a, b, c by **Heron’s formula **(also known as Hero’s Formula) is:

Triangle ABC

Find semi perimeter (s) = (a + b + c)/2

Area = √[s(s – a)(s – b)(s – c)]

This formula is helpful in finding the area of a scalene triangle, given the lengths of all its sides..

### Area of Any Polygon – By Heron’s formula

For a quadrilateral, when one of its diagonal values and the sides are given, the area can be calculated by **splitting the given quadrilateral into two triangles and using ****Heron’s formula.**

**Example**: **A park, in the shape of a quadrilateral ABCD, has ∠C=90∘, AB = 9 cm, BC = 12 cm, CD = 5 cm and AD = 8 cm. How much area does it occupy?**

⇒We draw the figure according to the information given.

The figure can be split into 2 triangles ΔBCD and ΔABD

From ΔBCD, we can find BD (Using Pythagoras’ Theorem)

BD2 = 122 + 52 = 169

BD = 13cm

Semi-perimeter for ΔBCD S1 = (12 + 5 + 13)/2 = 15

Semi-perimeter ΔABD S2 = (9 + 8 + 13)/2 = 15

Using Heron’s formula A1 and A2 will be:

A1 = √[15(15 – 12)(15 – 5)(15 – 13)]

A1 = √(15 × 3 × 10 × 2 )

A1 =√900 = 30 cm2

Similarly,

A2 will be 35.49 cm2.

The area of the quadrilateral ABCD = A1 + A2 = 65.49 cm2