CBSE Notes Class 9 Maths Chapter 10 Heron's Formula

1.0Introduction to Heron’s Formula

Heron's Formula is a method of calculating the area of a triangle when all of its sides are known. The formula is named after the Hero of Alexandria, who was an ancient Greek Mathematician who first came up with this discovery. However, it is useful since while the traditional method requires base and height, Heron's Formula only requires the lengths of the sides of any triangles.

2.0Download CBSE Notes for Class 9 Maths Chapter 10 Heron's Formula - Free PDF

Master the art of solving triangle-based problems with our free PDF for CBSE Class 9 Maths Notes Chapter 10 – Heron’s Formula. These notes are specially designed to help students understand and apply the formula effortlessly in various contexts.

3.0CBSE Class 9 Maths Chapter 10 Heron’s Formula - Revision Notes

Why and When was Heron’s Formula used?

Heron’s formula is used when the traditional formulas for finding the area of a triangle fail, which usually requires a base and height. It is useful for finding the area of an irregular triangle (a triangle that has different sides in size and also is not a right triangle). It is a useful formula for finding areas of architectural buildings and real-life applications.

Concepts & Terminology:

• Sides of Triangle: As mentioned above, all the triangle's sides are different in Heron’s formula (But it is not necessary). Sides are mentioned with alphabets a,b, and c.
• Semi-Perimeter: Semi-perimeter is half of the perimeter of a triangle. It is the main component of Heron’s Formula. It is denoted with s. The formula to find a semi-perimeter is: $s=\frac{a+b+c}{2}$
• Heron’s Formula: In maths, the area of the triangle is denoted with A in Heron’s Formula. Here is the formula: $A=\sqrt{s(s-a)(s-b)(s-c)}$

4.0Solved Problems

Question 1: In a garden, there is a slide that is in the shape of a triangle. The sides of that triangle are in the ratio of 4:5:6, and the perimeter of the triangle is 150cm. Find the length of each side and then find its area.

Solution: Let a= 4x, b=5x, c=6x 

In formula, Perimeter of Triangle = a + b + c

4x + 5x + 6x = 150

15x = 150 

x = 10 

a = 40, b = 50, c = 60

Semi-Perimeter of triangle = 150/2 = 75 

Area of Triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{75(75-40)(75-50)(75-60)}$

$=\sqrt{75(35)(25)(15)}$

$=\sqrt{5\times 5\times 3\times 7\times 5\times 5\times 5\times 5\times 3}$

$=5\times 5\times 5\times 3\sqrt{7}$

$=375\sqrt{7}{\mathrm{cm}}^2$

Question 2: Find the Area of an equilateral triangle whose side is 20cm in length with the help of Heron’s Formula.

Solution: Semi-perimeter of the triangle 

$=\frac{20+20+20}{2}=\frac{60}{2}=30$

Area of equilateral triangle

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

$=\sqrt{30(30-20)(30-20)(30-20)}$

$=\sqrt{30(10)(10)(10)}$

$=\sqrt{3\times 10\times 10\times 10\times 10}$

$=10\times 10\sqrt{3}$

$=100\sqrt{3}{\mathrm{cm}}^2$

If we multiply and divide the answer by four then we see that the answer will be the formula of the area of an equilateral triangle that is $\frac{\sqrt{3}}{4}{a}^2.$

Question 3: Find the area of an isosceles triangle whose perimeter in 50 and equal sides is 13cm. 

Solution: The perimeter of the triangle = 50

13 + 13 + c = 50

C = 50 - 26

c = 24cm 

Semi-perimeter of the triangle $=\frac{50}{2}=25$

Area of equilateral triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{25(25-13)(25-13)(25-24)}$

$=\sqrt{25(12)(12)(1)}$

$=5\times 12$

$=60{\mathrm{cm}}^2$

5.0Key Features of CBSE Maths Notes for Class 9 Chapter 10

• The notes are aligned with the latest pattern and Syllabus for CBSE Class 9 maths. 
• The notes provide a step-by-step guide along with solving problems to help you better understand the chapter. 
• These notes are ideal for self-learning as they are written in easy-to-understand language and are suitable for every level, whether beginner or advanced.