Question 169480
given perimeter P=52
area(A)=153

taking the perimeter of rectangle,
P=2(l+w)and assigning length(l) to be x we have;
p=2(x+w)since p=52 we have
52=2(x+w)
26=(x+w)therefore the width(w)becomes
w=26-x
since area=lw but l=x and w=26-x we have
area(A)=x(26-x)
153=x(26-x)
153=26x-x^2 this reduces to aquadratic eqn
x^2-26+153=0
applying the quadratic formula {{{x=(-b+-sqrt(b^2-4*a*c))/(2*a)}}}we have
{x=(26+-sqrt(26^2-4*1*153))/(2*1)}the answer becomes
x=17or 9ft
L=17,W=26-17hence
L=17,W=9
check;A=lw
17*9=153