Get Answers to all your Questions

header-bg qa

8. Fig. depicts a racing track whose left and right ends are semicircular.

                

The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find :

  (ii) the area of the track.      

Answers (1)

best_answer

The area of track  =   Area of outer structure   -   Area of inner structure.

Area of outer structure is: =   Area of square  + Area of 2 semicircles

                                             =\ 106\times 80\ +\ \frac{1}{2}\pi (40)^2\ +\ \frac{1}{2}\pi (40)^2\ m^2

And, area of inner structure: =  Area of inner square  +   Area of 2 inner semicircles  

                                             =\ 106\times 60\ +\ \frac{1}{2}\pi (30)^2\ +\ \frac{1}{2}\pi (30)^2\ m^2

Thus the area of the track is  : 

                                            \\=\ 106\times 80\ +\ \frac{1}{2}\pi (40)^2\ +\ \frac{1}{2}\pi (40)^2\ -\ \left ( 106\times 60\ +\ \frac{1}{2}\pi (30)^2\ +\ \frac{1}{2}\pi (30)^2 \right )\\\\=\ 4320\ m^2

Hence the area of the track is 4320 m2.

Posted by

Devendra Khairwa

View full answer