Question

Question 7
The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field.

Answer 1

Given, the perimeter of a triangular field is 420m and its sides are in the ratio 6:7:8.
Let sides of a triangular field be a = 6x, b = 7x and c = 8x.
$\therefore$ Perimeter of a triangular field, 2s = a+b+c
$\Rightarrow 420=6x+7x+8x\Rightarrow 420=21x$
$\Rightarrow x=\frac{420}{21}=20m$
$\therefore$ Sides of a triangular field are
a = 6 $\times$ 20 = 120m
b = 7 $\times$ 20 = 140m
And c = 8 $\times$ 20 = 160m.
Now, semi-perimeter 's' is given by
$s=\frac{a+b+c}{2}$
$=\frac{120+140+160}{2}=\frac{420}{2}$
$=210m$
$\therefore$ Area of triangular field $=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron’s formula].
$=\sqrt{210(210-120)(210-140)(210-160)}$
$=\sqrt{210\times 90\times 70\times 50}$
$=100\sqrt{21\times 9\times 7\times 5}$
$=100\sqrt{7\times 3\times 3^2\times 7\times 5}$
$=100\times 7\times \sqrt{15}=2100\sqrt{15}{m}^2$
Hence, the area of triangular field is $2100\sqrt{15}{m}^2$