Question

Sol. (4) External radius of circular path = R metre ( therefore 2 pi R=528 )
( Rightarrow 2 times frac{22}{7} times R=528 Rightarrow R=frac{528 times 7}{2 times 22}=84 mathrm{metr} )
( therefore ) In-radius ( (r) )
( =84-14=70 ) metre
( therefore ) Area of ( mathrm{path}=pileft(mathrm{R}^{2}-mathrm{r}^{2}right) )
( =frac{22}{7}left(84^{2}-70^{2}right)=frac{22}{7}(84+70)(84-70) )
( =frac{22}{7} times 154 times 14=6776 ) sq. metre
( therefore ) Required ( cos t=6776 times 10= ) Rs. 67760

# Ex.5. The outer circumference of a circular race-track is 528 metre. The track is everywhere 14 metre wide. Cost of levelling the track at the rate of Rs. 10 per sq. metre is : (1) Rs. 77660 (2) Rs. 66760 (3) Rs. 76760 (4) Rs. 67760

Solution