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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.      

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

Area of outer structure is: =   Area of rectangle  + 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 rectangle  +   Area of 2 inner semicircles  

=\ 106\times 80\ +\ \frac{1}{2}\pi (30)^2\ +\ \frac{1}{2}\pi (30)^2\ m^2                                             

Thus the area of the track:

   \\=\ 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

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