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Chapter 10 Herons Formula

Class 9

Maths

Chapter 10 Herons Formula

Frequently Asked Questions

Heron’s Formula is used to calculate the area of a triangle when the lengths of all three sides are known.

Area of triangle = √[s(s − a)(s − b)(s − c)], where s = (a + b + c) / 2 is the semi-perimeter.

Heron's Formula is very useful in the field of architecture, engineering, and land surveying as applied in finding the area of irregularly shaped triangular portions where base and height measurements cannot be easily found.

The semi-perimeter is half the perimeter of the triangle and thus can be used to reduce the area in Heron's Formula since it appears under a square root.

In case the sides of the triangle are not known, you can't use Heron's Formula directly; instead, you might have to know some angles or some other properties of a triangle first to compute the sides, then apply Heron's Formula.

Yes, in maths, Heron's Formula works for all kinds of triangles—scalene, isosceles, equilateral—once you know the lengths of the sides. For an equilateral triangle, it reduces, but still, it applies.

‘s’ represents the semi-perimeter, which is half of the triangle’s perimeter.

We use Heron’s Formula when the height of the triangle is not given but all three side lengths are known.

Yes, it works for scalene, isosceles, and equilateral triangles as long as the three sides are known.

Divide the quadrilateral into two triangles using a diagonal, then apply Heron’s Formula to each triangle and add the areas.

It builds the foundation for mensuration, coordinate geometry, and higher geometry problems in Class 10.

Errors usually occur while calculating semi-perimeter or simplifying the square root expression.

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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 Class 9 Maths Chapter 10 Heron's Formula Notes- 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.

Class 9 Maths Chapter 10 Revision Notes:

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: $A = s (s - a) (s - b) (s - c)$

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 $= s (s - a) (s - b) (s - c)$

$= 75 (75 - 40) (75 - 50) (75 - 60)$

$= 75 (35) (25) (15)$

$= 5 \times 5 \times 3 \times 7 \times 5 \times 5 \times 5 \times 5 \times 3$

$= 5 \times 5 \times 5 \times 37$

$= 3757 cm^{2}$

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

$= s (s - a) (s - b) (s - c)$

$= 30 (30 - 20) (30 - 20) (30 - 20)$

$= 30 (10) (10) (10)$

$= 3 \times 10 \times 10 \times 10 \times 10$

$= 10 \times 103$

$= 1003 cm^{2}$

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 $\frac{3}{4} a^{2} .$

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 $= s (s - a) (s - b) (s - c)$

$= 25 (25 - 13) (25 - 13) (25 - 24)$

$= 25 (12) (12) (1)$

$= 5 \times 12$

$= 60 cm^{2}$

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.

Chapter-wise CBSE Notes for Class 9 Maths:

Class 9 Maths Chapter 1 - Number Systems Notes

Class 9 Maths Chapter 2 - Polynomial Notes

Class 9 Maths Chapter 3 - Coordinate Geometry Notes

Class 9 Maths Chapter 4 - Linear Equation In Two Variables Notes

Class 9 Maths Chapter 5 - Introduction To Euclids Geometry Notes

Class 9 Maths Chapter 6 - Lines and Angles Notes

Class 9 Maths Chapter 7 - Triangles Notes

Class 9 Maths Chapter 8 - Quadrilaterals Notes

Class 9 Maths Chapter 9 - Circles Notes

Class 9 Maths Chapter 10 - Herons Formula Notes

Class 9 Maths Chapter 11 - Surface Areas and Volumes Notes

Class 9 Maths Chapter 12 - Statistics Notes

Chapter-wise NCERT Solutions for Class 9 Maths:

Chapter 1: Number Systems

Chapter 2: Polynomials

Chapter 3: Coordinate Geometry

Chapter 4: Linear Equations in Two Variables

Chapter 5: Introduction to Euclid’s Geometry

Chapter 6: Lines and Angles

Chapter 7: Triangles

Chapter 8: Quadrilaterals

Chapter 9: Circles

Chapter 10: Heron’s Formula

Chapter 11: Surface Areas and Volumes

Chapter 12: Statistics