What is Heron’s formula? How can we use it in our daily life? These are the questions that arise when we talk about Chapter 12 of the class 9 maths. Named upon one of the famous mathematicians of all time, Heron’s Formula opens a door to a new and interesting concept of maths. As Heron’s Formula is a vital topic as per the CBSE class 9 maths syllabus, it is important to ace it if you want to qualify the class 9th with flying colours. So, let’s get started and go through this informational blog about this topic.
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What is Heron’s formula?
Heron, a Greek mathematician, has worked out this formula. He received praises for his proficiency in mathematics and was called ‘hero’ of mathematics. Because of his excellence, the formula is also called the ‘Heron’s formula.’ He wrote various procedures and compiled all of them in three volumes. Each volume has the formulas for various mathematical fields. Book I encompasses the Heron formulae, which has the area equations for different types of shapes.
Where Can We Use the Heron Formula?
With the heron formulae, you can calculate the area of a triangle, whatever may be its side lengths. In the case of equilateral and isosceles, finding the height of the triangle will lead you to the calculation of its area. To evaluate the area of the equilateral triangle, we first derive the height h using the Pythagoras theorem. Then, we compute the area of the triangle using the following formula:
Area of a triangle = ½ x b x h
But, what will we do if we cannot find the height of the triangle? If we have a scalene body, it’s not easy to find its apex. A triangle with irregular side lengths is called a scalene triangle.That’s the point where Heron formulae come into play.
For example, if a triangle has the measurements 10 m, 15 m, and 20 m, then you can say it as a scalene triangle.
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In the Heron formula, we use a semi-perimeter to obtain the area. S stands for semi-perimeter, which is nothing but half of the perimeter (P/2). As we all know, the perimeter is the boundary or border of your respective area’s shape. It is the total of all the surface lengths. For a scalene triangle, you can calculate the perimeter and area using the formulae:
Perimeter of an Area, P = (a + b + c)/ 2
Area of the triangle, A = S (S-A) (S-B) (S-C) m2
S = semi perimeter of the triangle = Perimeter/2
A, B, C = Side lengths of the triangle
Perimeter P = Sum of three side lengths= AB + BC + CA
A boy is crossing a triangular playground having a scalene body with the side lengths 12 m, 5 m, and 6 m. He wishes to find the area of the ground.
Here, we will use the Heron formula to calculate the playground’s area.
Area of the triangle A = S (S-A) (S-B) (S-C)
So, semi-perimeter will be,
S = (A + B + C)/2 = (10 +5 + 6)/2 = 11.5 m
Now, apply the values in the Heron’s formula,
S – A = 11.5 – 10 = 1.5
S – B = 11.5 – 5 = 6.5
S – C = 11.5 – 6 = 5.5
A = S (S -A) (S – B) (S – C) m2
= 11.5 (1.5) (6.5) (5.5) m2
Area of the triangle, A = 24 m2
This is how we can derive the area of an irregular triangle using the Heron’s formula.
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Now, let’s see another example confining the Heron’s formula. Rohit has decided to calculate the cost of the tiles for his bedroom, having a triangular form. The side lengths of the rooms are as follows: 5 m, 4 m, and 7 m whereas the cost of a single 2 x 2 tile is Rs. 10. Find the total area and its cost.
The problem encloses the same set of steps that we had used in the previous example.
First, find the perimeter P = 5m + 4m + 7m = 16m
Now, find the semi-perimeter S = P / 2= 16 /2, S= 8 metres
Substituting the values in the Heron’s formula
A = S (S -A) (S – B) (S – C) m2
So, Area of the triangle = 8 (8 -5) (8 – 4) (8 – 7) m2
= 8 (3) (4) (1) m2
A= 96 m2
Area of a single tile = 2 x 2 m2
Number of tiles required for the room = Area of the triangle
Area of the single tile
= 96 m2 / 4 m2
= 24 nos
Having the number of tiles, now we can calculate the cost of tiles for the bedroom
Cost = 24 x 10
Books for Practising Heron’s Formula
- Mathematics for Class 9 by R. D Sharma
- NCERT Textbook for Mathematics Class 9th
- Oswaal NCERT Problems and Solutions for Class 9th Mathematics
- CBSE All In One Mathematics Class 9 for 2021 Exam by Jitendra Gupta & Amit Rastogi
- Mathematics Manjeet Singh Class 9 (2019-20) by Manjeet Singh
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Now that you are clear with the concept of Heron’s Formula, here is a question for you to practice.
- Assume you have a garden that is in the form of a triangle. Calculate the cost of the fence needed if the side lengths of your garden are 15 m, 12m, 10 m. The cost of a single metre barbed wire is 50 rs. Calculate the area of the triangle and the cost of fencing.
- The sides of the triangular field are in the ratio of 4:5:6. Then find the area of the field if its perimeter is 300m.
- Find the area of a triangle having a perimeter of 32cm. One side of its side is equal to 11cm and the difference between the other two is 5cm.
- The perimeter of the rhombus is 100 m and its diagonal is 40m. Find the area of the rhombus.
- Two parallel sides of a trapezium are 60 cm and 77 cm and the other sides are 25 cm and 26 cm. Find the area of the trapezium.
- The perimeter of a triangular field is 135 cm and its sides are in the ratio 25: 17: 12. Find its area.
- The perimeter of a square is (4x+20)cm. What will be the length of its diagonal?
- If the area of the hexagon is 24√3cm2, find its perimeter.
- Two sides of a triangular field are 112m and 50m. Then find the height of the altitude on the side of 78 m length if the perimeter of the triangle is 240m.
- Find the length of the hypotenuse of an isosceles right angle triangle if its area is 50m2.
- The perpendicular distance between the parallel sides of a trapezium is 8cm. Find the area of the trapezium if the lengths of its parallel sides are 9cm and 11cm.
We hope that this blog about Heron’s formula has helped you in preparing for class 9 exams. For expert advice related to important career decisions, reach out to our Leverage Edu experts. Hurry Up! Book an e-meeting.