A circular road is constructed outside a square field. The perimeter of the square field is 200 ft. If the width of the road is 7√2 ft. and cost of construction is Rs. 100 per sq.ft. Find the lowest possible cost to construct 50% of the total road.

Perimeter of square ABCD = 200 ft

=> AB = $$\frac{200}{4} = 50$$ ft

=> $$DB = \sqrt{50^2 + 50^2} = 50 \sqrt{2}$$ ft

=> $$BO = r = \frac{50 \sqrt{2}}{2} = 25 \sqrt{2}$$ ft

Width of the road = BX = $$7 \sqrt{2}$$ ft

=> $$BX = R = 25 \sqrt{2} + 7 \sqrt{2} = 32 \sqrt{2}$$

Area of bigger circle = $$\pi R^2 = \pi (32 \sqrt{2})^2 = 2048 \pi$$ sq. ft

Area of smaller circle = $$\pi r^2 = \pi (25 \sqrt{2})^2 = 1250 \pi$$ sq. ft

=> Area of road = $$2048 \pi - 1250 \pi = 798 \times \frac{22}{7} = 2508$$ sq. ft

But we have to calculate cost of construction of 50% road.

Required Construction = $$\frac{2508}{2} = 1254$$ sq. ft

$$\therefore$$ Cost of 1254 ft = $$1254 \times 100 = Rs. 1,25,400$$