Home | | Maths 9th std | Heron’s Formula

Mensuration | Maths - Heron’s Formula | 9th Maths : UNIT 7 : Mensuration

Chapter: 9th Maths : UNIT 7 : Mensuration

Heron’s Formula

How will you find the area of a triangle, if the height is not known but the lengths of the three sides are known?

Heron’s Formula

How will you find the area of a triangle, if the height is not known but the lengths of the three sides are known?

For this, Heron has given a formula to find the area of a triangle.

If a, b and c are the sides of a triangle, then the area of a triangle = √[ s ( s − a )(s − b)(s − c)] sq.units.

where s = [ a + b + c ] / 2, ‘s’ is the semi-perimeter (that is half of the perimeter) of the triangle.


Note

If we assume that the sides are of equal length that is a = b = c, then Heron’s formula will be √3/4 a2 sq.units, which is the area of an equilateral triangle.

Example 7.1

The lengths of sides of a triangular field are 28 m, 15 m and 41 m. Calculate the area of the field. Find the cost of levelling the field at the rate of ₹ 20 per m2.

Solution

Let a = 28 m, b = 15 m and c = 41m

Then, s = (a + b + c) / 2 = (28 +15+ 41) / 2 = 84/2 = 42 m

Area of triangular field = √ [ s(s − a)(s − b)(s − c) ]

= √ [ 42(42 − 28)(42 −15)(42 − 41)]

= √ [42 ×14 × 27 ×1]

= √ [ 2×3×7×7×2×3×3×3×1]

= 2×3×7×3

= 126 m2

Given the cost of levelling is ₹ 20 per m2.

The total cost of levelling the field = 20 ×126 = ₹ 2520.

Example 7.2

Three different triangular plots are available for sale in a locality. Each plot has a perimeter of 120 m. The side lengths are also given:


Help the buyer to decide which among these will be more spacious.

Solution

For clarity, let us draw a rough figure indicating the measurements:


(i) The semi-perimeter of 


Note that all the semi-perimeters are equal.

(ii) Area of triangle using Heron’s formula:

In Fig.7.4, Area of triangle = √ [60(60 − 30)(60 − 40)(60 − 50)]

= √ [60×30×20×10]

= √ [30×2×30×2×10×10]

= 600 m2

In Fig.7.5, Area of triangle = √ [60(60 − 35)(60 − 40)(60 − 45)]

= √ [60×25×20×15]

= √ [20×3×5×5×20×3×5]

= 300 √5 ( Since √5 = 2.236 )

= 670.8 m2

In Fig.7.6, Area of triangle = √ [60(60 − 40)(60 − 40)(60 − 40)]

= √ [60×20×20×20]

= √ [3×20×20×20×20]

= 400 √3 ( Since √3 = 1.732 )

= 692.8 m2

We find that though the perimeters are same, the areas of the three triangular plots are different. The area of the triangle in Fig 7.6 is the greatest among these; the buyer can be suggested to choose this since it is more spacious.

Note

If the perimeter of different types of triangles have the same value, among all the types of triangles, the equilateral triangle possess the greatest area. We will learn more about maximum areas in higher classes.

Tags : Mensuration | Maths , 9th Maths : UNIT 7 : Mensuration
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
9th Maths : UNIT 7 : Mensuration : Heron’s Formula | Mensuration | Maths


Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright © 2018-2024 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.