inside length:
400 = 90*2 + 2*pi*r
r = 110/pi
outside length: 90*2 + 2*pi*(r+14) = 400+28pi = 487.96
area of track is 2*90*14 + pi(R^2-r^2)
= 2520 + pi((110/pi+14)^2 - (110/pi)^2)
= 5600+196pi
= 6215.75 m^2
the inside perimeter of a running track shown in figure is 400m the lenghth of each of the straight portion is 90, and the ends are semicircles, if the track is 14 m wide everywherwe,find the area of the track. also find the length of thee outerboundary of the track
2 answers
First area ofrunning small rectangle =l×b secon step area oftwo small semi-circle =one complete circle-area of two small circle . Then large circle-area of small circle. Thenarea of large circle+rectangle