You have been invited to be part of the planning committee for the county fair. in order to be an effective member you will need to complete a series of tasks. mullin’s farm has donated space for the fair. it is 20 acres. each acre is 660 feet by 66 feet and together they form a rectangle. you will need to work within this space to decide how to layout the fair and accommodate all the events and people. first, think about what information you will need to complete this part of the larger problem. discuss what mathematical understanding you will need to have, what resources you might use to learn more, and what further information you need. then suggest a process that you would suggest to the planning committee to lay out the fair.

Firstly let's find the dimension of this large rectangle:(given)

Area of Rectangle = 660 x 66 =43,560 ft²

And we know that 1 acre = 43,560 ft², then each rectangle has an area of 1 acre & the 20 acres will correspond to 20 x 43560 = 871,200 ft²

We know that the 20 acres form a rectangle. We need to know what is their disposition:

1) We would like to know the layout of the rectangles since we have 4 possibilities FOR THE LAYOUTS

Note that W=66 & L=666 = 43,956 ft²/ unit )

lay out shape could be either:(in ft)
1 W by 20 L  (Final shape Linear 66 x 13320 = 879,120) or
2 W by 10 L  (Final shape Stacked  132 x 6660 = 879,120) or
4 W by   5 L  (Final shape Stacked  264 x 3330 = 879,120) or

2) We would like to know the number of participants so that to allocate equal space as well as the pedestrian lane, if possible, if not we will calculated the reserved space allocated for pedestrian/visitors)

3) Depending on the shape given we will calculate the visitor space & we will deduct it from the total space to distribute the remaining among the exhibitors.

4) (SUGGESTION) Assuming it's linear, we will reserve
20ft x 13320 ft = = 266,400 ft² and the remaining 612,720 ft² for exhibitors
5) Depending on the kind of the exhibition, we will divide the 612,720 ft² accordingly
6) How can we select the space allocated for each exhibitor:
 the 617,720 ft² could be written as a product of prime factors:
612720 = 2⁴ x 3² x 5 x 23 x 37
If you chose each space will be185 ft² , then we can accommodate up to 3,312 exhibitors.
Obviously you can choose any multiple of the prime factors to specify the area allocated & to calculate the number of exhibitors accordingly

The answer is "a"radius = 12.7 unitsdiameter = radius × 212.7 units × 2 = 25.4 units