Question

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

Answer 1

Given:

Inside perimeter of track = 400 m.

Each straight portion length = 90 m.

Track width (uniform) = 14 m.

Step-wise calculation:

1. Let r = radius of each inner semicircle.

Inner perimeter = 2 × (straight length) + (circumference of full circle made by two semicircles)

= 2(90) + 2πr

= 400

⇒ 180 + 2πr = 400

⇒ 2πr = 220

⇒ r = `110/π` 

If we take π = `22/7` then r = `110 ÷ 22/7` = 35 m.

2. Outer semicircle radius = r + width = 35 + 14 = 49 m.

3. Area of inner region

= Area of inner rectangle + Area of inner (full) circle 

= 90 × (2r) + πr² 

= 90 × 70 + π(35²) 

= 6300 + 1225π

Area of outer region

= 90 × 2(r + 14) + π(r + 14)² 

= 90 × 98 + π(49²) 

= 8820 + 2401π

Area of the track

= Outer area − Inner area 

= (8820 − 6300) + π(2401 − 1225)

= 2520 + 1176π

Using π = 22/7:

Area = `2520 + 1176 × 22/7` 

= 2520 + 3696

= 6216 m²

4. Length of the outer running track (outer perimeter) = 2 × (outer straight length) + circumference of outer full circle.

As shown by geometry the outer straight length remains 90 m, so outer perimeter

= 2(90) + 2π(r + 14) 

= 180 + 2π(49)

= 180 + 98π

With `π = 22/7`: 

Outer perimeter

= `180 + 98 × 22/7` 

= 180 + 308

= 488 m

Area of the track

= 2520 + 1176π

= 6216 m²    ...(Taking π = `22/7` gives 6216 m²)

Length of the outer running track

= 180 + 98π

= 488 m   ...(Taking π = `22/7` gives 488 m)