SOLUTION: We are working on quadratic equations and I have no idea how to solve this. Here is the question: A rectangle has a perimiter of 52 feet and an area of 153 square feet. Find the

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: We are working on quadratic equations and I have no idea how to solve this. Here is the question: A rectangle has a perimiter of 52 feet and an area of 153 square feet. Find the      Log On


   

Question 169480: We are working on quadratic equations and I have no idea how to solve this. Here is the question: A rectangle has a perimiter of 52 feet and an area of 153 square feet. Find the dimensions fo the rectangle. I have tried to plug in the 2w+2L = 52. I factored out the 2 and divided it on both sides to get w=24-L. I tried to plug this in also. No luck. Please help me with a formula to solve.
Thanks
Shonnie
Answer by chiefman(11) About Me  (Show Source):
You can put this solution on YOUR website!
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=%28-b%2B-sqrt%28b%5E2-4%2Aa%2Ac%29%29%2F%282%2Aa%29we 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