Heron's Formula Class 9 Notes CBSE Maths Chapter 10 (Free PDF Download)

Area of Triangle:

• Area of a triangle when height is known is given by $Area=\dfrac{1}{2}\times base\times height$ 

• For example: Let a triangle ABC 

In the triangle ABC height is $4cm$ and base is $3cm$ 

Therefore, area of triangle ABC is given by

$Area=\dfrac{1}{2}\times base\times height$

$Area=\dfrac{1}{2}\times 3\times 4$ 

$Area=6c{{m}^{2}}$ 

• This formula can be used to find the area of right-angle triangle, equilateral triangle and isosceles triangle.

• But when it is difficult to find the height of the triangle like in the case of scalene triangle, we use heron’s formula for calculating the area of triangle

Area of Triangle – by Heron’s Formula: 

• Heron’s formula for calculating the area of triangle was given by mathematician Heron around $60$ CE

• Area of triangle by heron’s formula is given by$Area=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$ 

Where, $a,b,c$ are the sides of triangle and $s$ is semi-perimeter of triangle 

• Semi perimeter of triangle is the half of perimeter of triangle and is given by $s=\dfrac{a+b+c}{2}$

• Heron’s Formula is very helpful where it is not possible to find the height of a triangle.

• For example: Let a triangle ABC

Sides of triangles are

$a=24cm$ 

$b=40cm$ 

$c=32cm$ 

Perimeter of triangle is given by 

$Perimeter=a+b+c$ 

$Perimeter=24+40+32$ 

$Perimeter=96cm$ 

Semi perimeter is given by 

$s=\dfrac{perimeter}{2}$ 

$s=\dfrac{96}{2}$ 

$s=48cm$  

Now, area of triangle is given by

$Area=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

$Area=\sqrt{48\left( 48-24 \right)\left( 48-40 \right)\left( 48-32 \right)}$ 

$Area=\sqrt{48\left( 24 \right)\left( 8 \right)\left( 16 \right)}$ 

$Area=\sqrt{147456}$ 

$Area=384c{{m}^{2}}$

Area of Quadrilateral using Heron’s Formula:

• A quadrilateral can be divided into two triangular parts by joining one of its diagonals 

• And then with help of Heron’s Formula we can find the area of two triangular parts

• Then by adding them we can get the area of the quadrilateral. 

• For example: Let a rhombus ABCD

Area of triangle ABD is given by 

$Are{{a}_{1}}=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$ 

Here, $a=100cm,b=100cm,c=160cm$ 

And semi perimeter is

$s=\dfrac{a+b+c}{2}$ 

$s=\dfrac{100+100+160}{2}$ 

$s=\dfrac{360}{2}$ 

$s=180cm$ 

$\therefore Are{{a}_{1}}=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$ 

$Are{{a}_{1}}=\sqrt{180\left( 180-100 \right)\left( 180-100 \right)\left( 180-160 \right)}$ 

$Are{{a}_{1}}=\sqrt{180\left( 80 \right)\left( 80 \right)\left( 20 \right)}$ 

$Are{{a}_{1}}=\sqrt{23040000}$ 

\[Are{{a}_{1}}=4800c{{m}^{2}}\] 

Now, area of triangle BCD is given by 

$Are{{a}_{2}}=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$ 

Here, $a=100cm,b=100cm,c=160cm$ 

And semi perimeter is

$s=\dfrac{a+b+c}{2}$ 

$s=\dfrac{100+100+160}{2}$ 

$s=\dfrac{360}{2}$ 

$s=180cm$ 

$\therefore Are{{a}_{2}}=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$ 

$Are{{a}_{2}}=\sqrt{180\left( 180-100 \right)\left( 180-100 \right)\left( 180-160 \right)}$ 

$Are{{a}_{2}}=\sqrt{180\left( 80 \right)\left( 80 \right)\left( 20 \right)}$ 

$Are{{a}_{2}}=\sqrt{23040000}$ 

\[Are{{a}_{2}}=4800c{{m}^{2}}\] 

$\therefore AreaofABCD=Are{{a}_{1}}+Are{{a}_{2}}$ 

$AreaofABCD=4800+4800$ 

$AreaofABCD=9600c{{m}^{2}}$

How to Calculate Area of Triangle Using Herons Formula?

Heron's formula is an important mark in this subject. By this formula, we can calculate the area of the triangle if the length of all three sides is known. This can be calculated using the following two steps:

Step 1: Calculate the “s” (half of the triangle’s perimeter):

S= a+ b = c2

Step 2: Then calculate the Area.

This formula is credited to Hero (or Heron) of Alexandria, a Greek Engineer, and Mathematician in 10 – 70 Anno Domini (AD). 

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Important Questions for Practice

1. Find the area of an equilateral triangle with a side of 5 cm.

2. Find the triangle area whose sides are 12 cm, 50 cm, and 60 cm.

3. ABCD is a rhombus whose three vertices, A, B, and C, lie on the circle with a centre of 0. Find the rhombus's area if the radius of a circle is 12 cm.

4. If every side of a triangle is tripled, find the percent increase in the area of the triangle so formed.
   
   

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Conclusion 

Heron's Formula Class 9 Notes CBSE Maths Chapter 10 offers a vital resource for students navigating the intriguing terrain of geometry. These meticulously crafted notes, available as free PDF downloads, demystify the enigmatic Heron's Formula, empowering learners to calculate triangle areas with precision and ease. They provide a comprehensive understanding of this mathematical tool, from its derivation to practical applications. These notes are not just about exam preparation; they equip students with a problem-solving skill set that extends beyond the classroom. With their accessibility and educational value, they serve as indispensable companions for students aiming to excel in their CBSE Class 9 mathematics curriculum.