Problema Solution
A rectangular parking lot has a width that is eight feet less than its length. Surrounding the parking lot is a sidewalk measuring three feet wide. What are the dimensions of the parking lot if the total area of the parking lot and the sidewalk is 2288 ft squared?
Answer provided by our tutors
let
x = the length of the parking lot
y= the width of the parking lot
w = 3 ft sidewalk
the rectangular parking lot has a width that is eight feet less than its length
x = y + 8
the total area = the parking lot area + the sidewalk area
the length of the total area is x + 2w = x + 2*x = x + 6
the width of the total area is y + 2w = y + 2*3 = y + 6
the total area = the length of the total area x the width of the total area
(x + 6) * (y + 6) = 2288
by solving the system of equations
x = y + 8
(x + 6) * (y + 6) = 2288
we find and consider only the positive solutions
x = 46 ft
y = 38 ft
the parking lot area has dimensions 46 ft and 38 ft.