The perimeter of a triangular field is 420 m

Solution:

Given, perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8.

Let sides of a triangular field be a = 6x, b = 7x and c = 8x.

Perimeter of a triangular field, 2s = a + b + c => 420 = 6x + 7x + 8x => 420 = 21x

=> x = 420/21 = 20 m

$\therefore$ Sides of a triangular field are

$a=6 \times 20=120 \mathrm{~m}$

 

$b=7 \times 20=140 \mathrm{~m}$

and $c=8 \times 20=160 \mathrm{~m} .$

Now, semi-perimeter, $s=\frac{a+b+c}{2}=\frac{120+140+160}{2}$

$=\frac{420}{2}=210 \mathrm{~m}$

$\therefore \quad$ Area of a triangular field $=\sqrt{s(s-a)(s-b)(s-c)}$  [by Heron's formula]

$=\sqrt{210(210-120)(210-140)(210-160)}$

$=\sqrt{210 \times 90 \times 70 \times 50}$

$=100 \sqrt{21 \times 9 \times 7 \times 5}$

$=100 \sqrt{7 \times 3 \times 3^{2} \times 7 \times 5}$

 

$=100 \times 7 \times 3 \times \sqrt{15}=2100 \sqrt{15} \mathrm{~m}^{2}$

Hence, the area of triangular field is $2100 \sqrt{15} \mathrm{~m}^{2}$.