Question

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.

Answer 1

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 m²