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The inside perimeter of a running track shown in the figure is 400 m. The length of each of the straight portions is 90 m, and the ends are semicircles. If the track is 14 m wide everywhere

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Question

The inside perimeter of a running track shown in the figure is 400 m. The length of each of the straight portions is 90 m, and the ends are semicircles. If the track is 14 m wide everywhere, find the area of the track. Also, find the length of the outer boundary of the track.

Sum
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Solution

Length of the inner curved portion (400 - 2 × 90) = 220 m

∴ Length of each inner curved path`=220/2=110  "m"`

Thus, we have: 

⇒ πr = 110

`⇒ 22/7"r" = 110`

`=> "r" = (110xx7)/22`

⇒ r =35m

Inner radius = 35 m
Outer radius = (35 + 14) = 49 m
Area of track = {Area of the two rectangles [each(90 × 14)] + Area of the circular ring with R = 49 m and r = 35 m)}

`=(2xx90xx14)+22/7xx[(49)^2-(35)^2]`

`=2520 + 22/7xx(2401-1225)`

`= 2520 + 22/7xx1176`

= 2520 + 3696

= 6216 m

​Length of the outer boundary of the track

`=(2xx90+2xx22/7xx49)`

= 488 m

Therefore, the length of the outer boundary of the track is 488 m and the area of the track is 6216 sq. m.

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Chapter 16: Area of Circle, Sector and Segment - EXERCISE 16A [Page 736]

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 16 Area of Circle, Sector and Segment
EXERCISE 16A | Q 59. | Page 736
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