For area of the park, we have
$\mathrm{s}=\frac{120+80+50}{2}=\frac{250}{2}=125\mathrm{m}$
Now,    s – a = (125 – 120) m = 5m
        s – b = (125 – 80) m = 45 m
        s – c = (125 – 50) m = 75 m                            
Therefore, area of the park $=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$
                                           $=\sqrt{125\times 5\times 45\times 75}{\mathrm{m}}^2=\sqrt{25\times 25\times 3\times 15\times 15\times 5}{\mathrm{m}}^2=375\sqrt{15}{\mathrm{m}}^2$
Also, perimeter of the park = AB + BC + CA = 250 m
Therefore, length of the wire needed for fencing = 250 m – 3 m (to be left for gate) = 247 m.
And the cost of fencing = Rs. 20 × 247 = Rs. 4940.