The perimeter of a triangullar field is 144 m and the ratio of the sides is 3 : 4 : 5. Find the area of the field.

#### Solution

The area of a triangle having sides *a*, *b*, *c *and *s* as semi-perimeter is given by,

`A = sqrt(s(s-a)(s-b)(s-c))`, where,

* *`s = (a+b+c)/2`

It is given the sides of a triangular field are in the ratio 3:4:5 and perimeter=144 m

Therefore, *a*: *b*: *c* = 3:4:5

We will assume the sides of triangular field as

* a*= 3x : b = 4x ; c = 5x

2s = 144

`s= 144/2`

s= 72

`72= (3x+4x+5x)/2`

72×2= 12x

` x = 144/12`

x = 12

Substituting the value of *x *in, we get sides of the triangle as

a = 3x = 3 × 12

a = 36 m

b = 4x = 4 × 12

b = 48 m

c = 5x = 5 × 12

c = 60 m

Area of a triangular field, say *A *having sides *a*, *b *, *c *and *s *as semi-perimeter is given by

a = 36 m ; b = 48 m ; c = 60 m

s = 72 m

`A = sqrt( 72(72-36) (72-48)(72-60)`

`A=sqrt(72(36)(24)(12))`

`A= sqrt(746496)`

A = 864 m^{2}