Question

The perimeter of a triangular field is 540m and its sides are in the ratio 25:17:12.Find the area of the triangle

Answer 1

GIVEN : The perimeter of a triangular field = 540m

Let the sides are 25x , 17x , 12 x

Perimeter of a ∆ = sum of three sides

25x + 17x + 12x = 540

54x = 540

x = 10

1st side (a) - 25x = 25×10= 250m

2nd side(b)= 17x = 17×10= 170m

3rd side (c)= 12x = 12 × 10 =120m

Semi - perimeter ( S) = a+b+c/2

= (250 + 170+120)/2 = 540/2 = 270 m

Area of the ∆= √ S(S - a)(S - b)(S - c)

[By Heron’s Formula]

= √ S(S - 250)(S - 170)(S - 120)

= √ 270(270 - 250)(270 - 170)(270 - 120)

= √ 270× 20×100×150

= √ 81000000

Area of the ∆= 9000 m²

Hence, the Area of the ∆= 9000 m²

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

HELLO DEAR,

GIVEN:- perimeter = 540m
and sides are in ratio 25:17:12
So, the sides are 25x , 17x , 12x where x is any positive no.

then ,25x + 17x + 12x = 540 

54x = 540 

x = 10 

25x = 25 * 10 = 250
17x = 17*10 = 170
12x = 12*10 = 120

So, the sides are 250m , 170m & 120m. 

now, the semi - perimeter is S = 540/2 = 270 m

we know the Heron's formula:-

area of triangle = $\sf{\sqrt{s(s - a)(s - b)(s - c)}}$
where, a,b,c is the sides of triangle and s is the semi-perimeter. 

now, the area of the triangle is 
= $\sf{\sqrt{S(S - 250)(S - 170)(S - 120)}}$m²

= $\sf{\sqrt{270*(270 - 250)*(270 - 170)*(270 - 120)}}$m²

= $\sf{\sqrt{270*20*100*150}}$m²

= $\sf{\sqrt{81000000}}$m²

= 9000m²


Hence, the area of triangle = 9000m²


I HOPE ITS HELP YOU DEAR,
THANKS