A grassy field is in the shape of a circle of radius 100m and is enclosed by a circular fence.  A goat is attached by a rope to a hook, at a fixed point on the fence.   To stop the goat getting too fat, the farmer wants to make sure that it can only reach half of the grass in the field.  How long should the rope be?

Solution:

Consider the figure below. The goat is pegged at point B on the edge of the circular grass field. Let the length of the rope that the goat is tied with be “R”. The dark green shaded area is the grass field that the goat is allowed to eat. This dark green shaded area must be ½ the area of all of the grass field.

The dark green shaded area can be divided into three parts i.e. Blue area which is the area of the sector with radius R and the two yellow segments with the chord lengths AB and BC of length R.

So the area that the goat is allowed to eat = Area of the Sector + Area of the two segments

Area of the Sector:

We know from Area of Sector and Segment post that Area of a the sector

A1 = 2X(R2)/2 where 2X is the enclosed angle in radians and R is the radius of the circle

A1 = XR2 ----------------------------------- 1

Area of the Segments:

Consider the figure below. Draw a line OA which is equal to 100 m (radius of the grass field). As can be seen from the figure, the yellow segment is part of the circle with radius 100 m. Also triangle AOB has two sides OA and OB = 100 m. Therefore angle OAB = Angle OBA = X. The other angle AOB = π – 2X

Draw a line OD which is perpendicular to AB. Consider the triangle OAD

CosX = AD/OA = R/2/100 or CosX = R/200 or R = 200CosX. --------------------- 2

So in order to find R, we need to find the value of X

We know from Area of Sector and Segment post that Area of a the segment = ½ r2 (θ – sin θ)

Θ = π -2X and r = 100 in this case

So Area of Segment = ½ (100)2((π -2X)-sin(π -2X))

Since there are two segments each of same area the total area of the segments =

A2 = (100)2((π -2X)-sin(π -2X)) ------------ 3

So the total area that the goat can eat = A1+A2 = XR2 +(100)2((π -2X)-sin(π -2X))

It is given that the total area that the goat can eat = ½ the area of the circle

Half the area of the circle = ½(π (100)2)

Or XR2 +(100)2((π -2X)-sin(π -2X)) = ½(π (100)2)

We know from equation 2 that R = 200CosX. Substitute that in the above equation

X(200)2Cos2X+(100)2((π -2X)-sin(π -2X)) = ½(π (100)2)

Let X(200)2Cos2X = A , (100)2((π -2X)-sin(π -2X)) = B and ½(π (100)2) = C

To solve this equation we have to use the trial and error method

We can write the above equation as A+B =C or A+B-C = 0

Table below gives the value of A,B,C as well as the value of X in radians when A+B-C = 0

X in Radians A = X(200)2Cos2X

B = (100)2((π -2X)-sin(π -2X)) A+B C = ½(π (100)2) A+B-C

1.000000 11677.1 2323.0 14000.0 15708.0 -1707.90.900000 13910.4 3677.5 17587.8 15708.0 1879.80.950000 12857.5 2952.9 15810.4 15708.0 102.50.955000 12744.7 2885.7 15630.5 15708.0 -77.50.954000 12767.4 2899.1 15666.5 15708.0 -41.50.953000 12790.0 2912.5 15702.5 15708.0 -5.50.952849 12793.4 2914.5 15707.9 15708.0 0.0

So the approximate value of X = 0.953 radians or 54.594 degrees.

From equation 1, R = 200CosX

R = 200Cos(0.953) = 115.872 meters or appx 116 meters

So the length of the rope must be appx 116 meters