SOLUTION: The area of a rectangular piece of concrete is 120 square feet, and its perimeter is 46 feet. What are the dimensions of the rectangle?

Click here to see ALL problems on Quadratic Equations

Question 6219: The area of a rectangular piece of concrete is 120 square feet, and its perimeter is 46 feet. What are the dimensions of the rectangle?
Found 2 solutions by minerva, arunpaul:
Answer by minerva(22) (Show Source):
You can put this solution on YOUR website!
well,this is an easy one but requres concentration.

heres my solution.
statements:-
2(l+b)=46 (perimeter)
(l+b)=46/2
l+b=23
therfore l=23-b
SUBSTITTUTING THE VALUES:-
area=l X b=12
bX (23-b)=120
23b -b^2=120
-(b^2 -23b +120)=0
minus gets cancelled when diuvide d by 0
so, b^2 -23b +120=0
b^2 -15b-8b+120=0
b(b-15)-8(b-15)=0
b = 15 or b=8
then.... l=23-b
(a) 23-15=8 or (b)23-8=15
solution b is correct as length is always grteater than breadth

Answer by arunpaul(104) (Show Source):
You can put this solution on YOUR website!
Let the length be = L
breadth be = B
as per the question
B*L =120 Sf eq1
perimeter = 2(L+B) = 46 ft eq2
again B = 120/L
put the value of B in eq2 we have
2L+(2*(120/L))=46
or 2L+(240/L)=46
or 2L^2+240 = 46L
or L^2+120 = 23L
or L^2+120 -23L=0
applying the solution of quadratic equation we have
a=1
b=120
c=-23
then applying the solution for quadratic equation we have

so L = 0.19136150644875 or -120.191361506449
take the positive value of L we have B= 120/0.19136150644875 =627.28
this seems some thing wrong in question though the solution is here