# Heron's Formula

Heron's formula is used to find the area of any type of triangles or quadrilaterals.

If we know the length of all sides of triangle & quadrilateral & also known the length of a diagonal of the quadrilateral.

## The formula used to find the area of triangle & quadrilateral

*HERON's FORMULA =*

*√{s(s-a)(s-b)(s-c)}*

*Where,*

**s**= semi-perimeter of a triangle

**a, b & c**= length of the sides of a triangle.

### 1. How to find the area of a triangle by Heron's Formula?

For finding the area of a triangle by Heron's formula we need the length of all sides of a triangle. For better understanding, we take an example.

Let's discuss-

### Example:-

*Find the area of a triangle if the length of its sides are 12 cm, 11 cm & 13 cm.*

**Solve:-**

*Given:-*

*Sides of triangle*

**a**=12cm,**b**=11cm &**c**=13cm.

*Now, we will find the semi-perimeter of a triangle-*

**Semi-perimeter of a triangle s= (a+b+c)÷2**

**=**(12+11+13)÷2

*= 36÷2*

**s**= 18cm

*Put the value of a, b, c & s in the formula-*

*Area of triangle = √{s(s-a)(s-b)(s-c)}*

**= √**{18(18-12)(18-11)(18-13)}

*= √{18×6×7×5}*

*= √{108×35}*

*=√(3780)*

*=61.48 cm² approx.*

### 2. How to find the area of a quadrilateral by Heron's Formula.

For finding the area of a quadrilateral by Heron's formula we need the length of all sides of a quadrilateral & length of one diagonal of a quadrilateral.

For better understanding, we take an example.

Let's discuss-

### Example:-

*Find the area of a quadrilateral if it's all sides are 13cm, 14cm, 12cm, 11cm & length of it's one diagonal is 16cm.*

**Solve:-**

*Given:-*

*Sides of quadrilateral are 13cm, 14cm, 12cm, 11cm & length of it's one diagonal is 16cm.*

*For finding the area of given quadrilateral*

**ABCD**we will find the area of triangles**ABD**&**BCD**& will add it.

**Semi-perimeter of a triangle (ABD) s= (AB+BD+AD)÷2**

**=**(13+16+11)÷2

*= 40÷2*

**s**= 20cm

*Area of triangle (ABD) = √{s(s-a)(s-b)(s-c)}*

**= √**{20(20-13)(20-16)(20-11)}

*= √{20×7×4×9}*

*= √{140×36}*

*= √(5040)*

*= 70.99cm² approx.*

*Now, we will find the area of a triangle (*

**BCD)**

**Semi-perimeter of a triangle (BCD) s= (BC+CD+BD)÷2**

**=**(14+12+16)÷2

*= 42÷2*

**s**= 21cm

*Area of triangle (BCD) = √{s(s-a)(s-b)(s-c)}*

**= √**{21(21-14)(21-12)(21-16)}

*= √{21×7×9×5}*

*= √{147×45}*

*= √(6615)*

*= 81.33cm² approx.*

*Total Area of Quadrilateral ABCD*

*= Area of ∆ABD + Area of ∆BCD*

*= 70.99cm² + 81.33cm²*

*= 152.32cm² approx.*

*So, the total area of ABCD is*

*152.32cm² approx.*

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