29 a) Area = 400 in^2
length of the side = x
equation
400 in^2= (x)^2
It has two solutions, one positve and one negative
x = +/- √[400] = +/- 20 in
x = 20 in and x = -20in
b) Only the postive solution, 20 in, makes sense given that the varialbe represents the length of the side of a true square and that only can be a positive number.
c) Find the perimeter = 4x = 4(20in) = 80 in.