The sides of a triangular field …

By another step of heron's formula

Applying Heron's formula \begin{lgathered}s=\dfrac{a+b+c}{2} \\\\\text{ Area of }\triangle =\sqrt{s(s-a)(s-b)(s-c)}\end{lgathered}s=2a+b+c​ Area of △=√s(s−a)(s−b)(s−c)​​ Therefore s=\dfrac{33+44+55}{2} = 66s=233+44+55​=66 Now \begin{lgathered}\text{ Area of }\triangle =\sqrt{66(66-33)(66-44)(66-55)}= \sqrt{66\times 33\times 22\times 11} \\\\=\sqrt{11\times 2\times 3\times 11\times 3 \times 11\times 2\times 11} \\\\=11\times 11\times 2\times 3 = 726 m^{2}\end{lgathered} Area of △=66(66−33)(66−44)(66−55)​=66×33×22×11​=11×2×3×11×3×11×2×11​=11×11×2×3=726m2​ Cost of leveling the field 1m^{2}1m2 = Rs . 1.20 Cost of leveling the field 726m^{2}726m2 =  Rs . 1.20x 726=  Rs. 871.2 Hence, total cost of leveling the field is  Rs. 871.2