Question

In the figure, there is a jogging track which consists of two semi-circular paths and two rectangle paths which have to be renovated. The cost of renovation is Rs. 15 per square meters. For the circular parts, the inner and outer radii are 6 m and 8 m respectively. Also, the thickness of the rectangular track is 2m. If the total cost of renovation is Rs. 2,520, find the length of AB. Also, find the difference between the outer and the inner perimeter of the track.

• A. 10 m; Perimeter Difference = 88/7 m
• B. 10 m; Perimeter Difference = 44/7 m
• C. 20 m; Perimeter Difference = 88/7 m
• D. 20 m; Perimeter Difference = 44/7 m

Answer 1

The correct option is C

20 m; Perimeter Difference = 88/7 m

The cost of renovation is given in squaremeters, so we have to consider the area of it.

Total area of the jogging track = Area of the semi-circular tracks + Area of the Rectangle areas

Area of any one semi circular track = Area of outer part -Area of inner part

inner radius (r) = 6 meters

4 Outer radius (R) = 6 meters (Given)

So, Area of any one semicircular track =$\frac{\pi }{2}\left(\right(8{)}^2-\left(6{)}^2\right)$

(Area of a semi -circle = $\frac{\pi r^2}{2}$

$=44m^2$

So, area of both semi circular tracks $=44\times 2=88m^2$

We are given that the thickness of rectangular track = 2m

Let the length of the rectangular track be = l

So, the area $=2\times l{m}^2$

& the total rectangular area $=2\times 2l=4l{m}^2$

So, the total area of the jogging track = 88+ 4 l .....(1)

Total cost of renovation = 2520 (Given)

Cost of renovation per square meters = 15

So, total area $=\frac{2520}{15}=168m^2$ .....(2)

From equation (1) & (2)

88 + 4 l = 168

4l = 80

l = 20m

Inner perimeter = $=\pi \left(r\right)+l+\pi \left(r\right)+l=\frac{22}{7}\times 6+20+\frac{22}{7}\left(6\right)+20=\frac{264}{7}+40$

Outer perimeter = $=\pi \left(R\right)+l+\pi \left(R\right)+l=\frac{22}{7}\times 8+20+\frac{22}{7}\times +20=\frac{352}{7}+40$

Difference between Outer perimeter and inner perimeter

$(\frac{352}{7}+40)(\frac{264}{7}+40)=\frac{88}{7}$