Question

A circular road is constructed outside a square field. The perimeter of the square filed is 200 ft. If the width of the road is 7 $\sqrt{2}$ ft and cost of construction is Rs.100 per sq. ft., find the lowest possible cost of construction of 50% of the total road.(use $\pi = \frac{22}{7}$)

Answer 1

100(172√2+25/2) Rs is your required answer .. Hope my answer helps you

Answer 2

Rs. 1,25,400 is the minimum cost to construct 50 percent of road.

Given:

Perimeter of the square = 200 ft

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

Cost of construction per square feet = Rs. 100

To find:

Lowest Possible cost of construction of 50% of the total road

Solution:

Perimeter of square$=4 \times \text { side }=200$

Side $=\frac{200}{4}=50$

Diameter of circle = side of square = 50

Radius $=\frac{\text {diameter}}{2}=25$

Area of inner circle $=\pi r^{2}=\frac{22}{7} \times 25^{2}$

Width of road$=7 \sqrt{2}$

radius of outer circle $=25+7 \sqrt{2}$

area of outer circle $=\frac{22}{7} \times(25+7 \sqrt{2})^{2}$

area of road = area of outer circle – inner circle  

$=\frac{22}{7} \times\left[(25+7 \sqrt{2})^{2}-25^{2}\right]=2508$

Cost of constructing half road $=100 \times \frac{2508}{2}=125400$

Rs. 125400 is the total cost for constructing the 50% of the total road.