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

**Heronâ€™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 = âˆš[ s ( s âˆ’ a )(s âˆ’ b)(s âˆ’ c)] *sq.units*.

where s = [ a +* b *+* c* ]*
/ 2*, â€˜sâ€™ 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 Heronâ€™s formula will be âˆš3/4 *a*^{2} *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 â‚¹ 20
per *m*^{2}.

*Solution*

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

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

Area of triangular field = âˆš [ s(s âˆ’ a)(s âˆ’ b)(s âˆ’ c)
]

= âˆš [ 42(42 âˆ’ 28)(42 âˆ’15)(42 âˆ’ 41)]

= âˆš [42 Ã—14 Ã— 27 Ã—1]

= âˆš [ 2Ã—3Ã—7Ã—7Ã—2Ã—3Ã—3Ã—3Ã—1]

= 2Ã—3Ã—7Ã—3

= 126 *m*^{2}

Given the
cost of levelling is â‚¹ 20 per *m*^{2}.

The total
cost of levelling the field = 20 Ã—126 = â‚¹ 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 Heronâ€™s formula:

In Fig.7.4,
Area of triangle = âˆš [60(60 âˆ’ 30)(60 âˆ’ 40)(60 âˆ’ 50)]

= âˆš [60Ã—30Ã—20Ã—10]

= âˆš [30Ã—2Ã—30Ã—2Ã—10Ã—10]

= 600 m^{2}

In Fig.7.5,
Area of triangle = âˆš [60(60 âˆ’ 35)(60 âˆ’ 40)(60 âˆ’ 45)]

= âˆš [60Ã—25Ã—20Ã—15]

= âˆš [20Ã—3Ã—5Ã—5Ã—20Ã—3Ã—5]

= 300 âˆš5
( Since âˆš5 = 2.236 )

= 670.8 m^{2}

In Fig.7.6,
Area of triangle = âˆš [60(60 âˆ’ 40)(60 âˆ’ 40)(60 âˆ’ 40)]

= âˆš [60Ã—20Ã—20Ã—20]

= âˆš [3Ã—20Ã—20Ã—20Ã—20]

= 400 âˆš3
( Since âˆš3 = 1.732 )

= 692.8 *m*^{2}

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

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9th Maths : UNIT 7 : Mensuration : Heronâ€™s Formula | Mensuration | Maths

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