SOLUTION: Find the dimensions of a rectangle whose perimeter is 26 meters and whose arae is 40 square meters. I know this is very basic, but I have not been in a classroom in over 35 years.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Find the dimensions of a rectangle whose perimeter is 26 meters and whose arae is 40 square meters. I know this is very basic, but I have not been in a classroom in over 35 years.       Log On


   

Question 343455: Find the dimensions of a rectangle whose perimeter is 26 meters and whose arae is 40 square meters. I know this is very basic, but I have not been in a classroom in over 35 years. Thank you.
Found 2 solutions by scott8148, solver91311:
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
P = 2L + 2W ___ 26 = 2L + 2W ___ 13 = L + W

A = L * W ___ 40 = L * W

you are looking for two numbers whose sum is 13 and whose product is 40
___ like 8 and 5



using equations

L = 13 - W

substituting ___ 40 = (13 - W) * W ___ 40 = 13W - W^2 ___ W^2 - 13W + 40 = 0

factoring ___ (W - 8)(W - 5) = 0

W - 8 = 0 ___ W = 8

W - 5 = 0 ___ W = 5

one dimension is 5 and the other is 8

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


The area of a rectangle is given by multiplying the length times the width, so in your situation:



Which is to say:



The perimeter is two times the length plus two times the width:



Subsituting from above:



Which is to say:



Yuck! That is uglier than a mud fence. Not to worry though, we have a bag of lovely cosmetics...

Multiply both sides by :



Add to both sides:



Multiply by



Let's see, -5 times -8 is 40 and -5 + -8 is -13. Can you take it from here?

John

My calculator said it, I believe it, that settles it