Question

The perimeter of a triangular field is 540 m, and its sides are in the ratio 25:17:12. Find the area of the field. Also, find the cost of ploughing the field at Rs. 40 per 100 $m^2$.

Answer 1

Given: The perimeter of a triangular field = 540 m

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

The perimeter of a $△$ = sum of three sides

25x + 17x + 12x = 540

54x = 540

x = 10

1st side (a) - 25x = 25$\times$10= 250 m

2nd side(b)= 17x = 17$\times$10= 170 m

3rd side (c)= 12x = 12 $\times$ 10 =120 m

Semi - perimeter (S) = $\frac{a+b+c}{2}=\frac{(250+170+120)}{2}=\frac{540}{2}=270$ m

Area of the $\Delta =\sqrt{s(s-a)(s-b)(s-c)}$

[By Heron’s Formula]

$=\sqrt{S(S-250)(S-170)(S-120)}=\sqrt{270(270-250)(270-170)(270-120)}=\sqrt{270\times 20\times 100\times 150}=\sqrt{81000000}=9000m^2$

Area of the $\Delta =9000m^2$

Cost of ploughing the field at Rs. 40 per 100 $m^2=\frac{9000\times 40}{100}=Rs.3600$