An athletic field is a 48 yd -by-96 yd rectangle, with a semicircle at each of the short sides. A running track 20 yd wide surrounds the field. If the track is divided into eight lanes of equal width, with lane 1 being the inner-most and lane 8 being the outer-most lane, what is the distance around the track along the inside edge of each lane?

Question

Sarah Kaiser

Mathematics

16 April, 06:05

An athletic field is a 48 yd -by-96 yd rectangle, with a semicircle at each of the short sides. A running track 20 yd wide surrounds the field. If the track is divided into eight lanes of equal width, with lane 1 being the inner-most and lane 8 being the outer-most lane, what is the distance around the track along the inside edge of each lane?

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Kaydence Porter · Answer 1

The distance would be equal to 2 times the longest side of the rectangle plus twice the shortest side multiplied by pi / 2 for the semicircle, that is:

longest side 96 and shortest 48

D = 2 * (96) + 2 * (1/2) * pi * 48

D = 192 + pi * 48

This shorter side, which starts at 48, will expand each time by two more in proportion to 20 of the running track between 8 than the number of divisions, that is, 2 * (20/8) = 5

In other words, there are 8 distances, like this:

D1 = 192 + 3.14 * 48 = 342.72 yd

D2 = 192 + 3.14 * (48 + 5) = 358.42 yd

D3 = 192 + 3.14 * (48 + 10) = 374.12 yd

D4 = 192 + 3.14 * (48 + 15) = 389.82 yd

D5 = 192 + 3.14 * (48 + 20) = 405.52 yd

D6 = 192 + 3.14 * (48 + 25) = 421.22 yd

D7 = 192 + 3.14 * (48 + 30) = 436.92 yd

D8 = 192 + 3.14 * (48 + 35) = 452.62 yd