# The field inside the following 400yd running track is to be mowed. each straight portion of the track is 100 yd, each curved part of the track is a semi circle (31.8 yd) determine the area to be mowed

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### 2 Answers

The information provided about the track gives two parallel portions of 100 yards each with a semicircular end at each side.

The area to be mowed within the tracks is given by: `100*31.8*2 + pi*31.8^2 = 6360+ 3176.90 = 9536.90` square yards.

**The area to be mowed within the tracks is given by 9536.90 square yard.**

The field to be mowed is the area enclosed between two semicircles of length 31.8 yards and two parallel lines 100 yards long.

Let the radius of the circular portion be r then:

Pi.r = 31.8 or r = 31.8/Pi

The area of the two semicircular potions = Pi.r^2

=> Pi*(31.8/Pi)^2 = 31.8^2/Pi = 322 Sq.Yd approx.

Area of the central rectangular potion = 2.r.100

=> 2*31.8/Pi*100 = 2024 Sq.Yd approx.

Total area = 322+2024 = 2346 Sq.Yd approx.

**Area to be moved is equal to 2346 Sq.Yd. Approximately**