SOLUTION: Find the dimensions of a rectangle whose perimeter is 26 meters and whose area is 40 square meters. For this problem, I must use the perimeter of a rectangle formula and area of

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: Find the dimensions of a rectangle whose perimeter is 26 meters and whose area is 40 square meters. For this problem, I must use the perimeter of a rectangle formula and area of      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1208718: Find the dimensions of a rectangle whose perimeter is 26 meters and whose area is 40 square meters.
For this problem, I must use the perimeter of a rectangle formula and area of a rectangle formula. Do you agree?
Let l = length
Let w = width
A = lw
40 = lw
I can solve the equation for length or width. I will solve for l or length.
40 = lw
40/w = l
I now plug that into the formula
P = 2l + 2w
26 = 2(40/w) + 2w
After finding w or width, I plug the value for w back into A = lw to find the length.
Is this correct? If not, what did I fo wrong? Thanks.






Found 3 solutions by josgarithmetic, MathTherapy, greenestamps:
Answer by josgarithmetic(39613) About Me  (Show Source):
You can put this solution on YOUR website!
L, length
w, width
Perimeter and area of the rectangle are given so:
system%28wL=40%2Cand%2C2w%2B2L=26%29

Perimeter equation simplifies to w%2BL=13.

w=13-L.
Substitute in the area equation.
Lw=40
L%2813-L%29=40---------------(or maybe you see an 8 and 5 here?)
-L%5E2%2B13L-40=0
L%5E2-13L%2B40=0------------which is factorable.
%28L-8%29%28L-5%29=0
Meaning L could be 8 or 5. Can it be either?

wL=40
if L is 8, then w=5.
If L is 5 , then w=8.

Will these work in the perimeter equation?
2%2A5%2B2%2A8
2%2813%29
26thegivenperimeter.

Answer by MathTherapy(10549) About Me  (Show Source):
You can put this solution on YOUR website!
Find the dimensions of a rectangle whose perimeter is 26 meters and whose area is 40 square meters.

For this problem, I must use the perimeter of a rectangle formula and area of a rectangle formula. Do you agree? 

Let l = length

Let w = width

A = lw

40 = lw

I can solve the equation for length or width. I will solve for l or length. 

40 = lw

40/w = l

I now plug that into the formula

P = 2l + 2w

26 = 2(40/w) + 2w

After finding w or width, I plug the value for w back into A = lw to find the length. 

Is this correct? If not, what did I fo wrong? Thanks. 

Your logic and setup are CORRECT. Continue to solve to get ypur answer.
However, the prooblem is easier than you think.

With L being length and W being width, 2L + 2W = 26___2(L + W) = 2(13)___L + W = 13. Now, you just need to find 2 numbers
with a product = AREA, or 40, and that SUM to 13. These 2 numbers are easily 8 and 5, the dimensions of the rectangle.

Answer by greenestamps(13195) About Me  (Show Source):
You can put this solution on YOUR website!


Directly from the given information -- perimeter 26 and area 40 --you get these two equations:

2l+2w=26
lw=40

You show solving the second equation for one of the variables and substituting in the other equation.

That's a valid approach; but the algebra required to solve the problem is a bit easier if you solve the first equation for one of the variables and substitute in the second. It's because solving the second equation for l gives you l = 40/w, and that expression with a fraction complicates things.

So...
2l+2w=26
l+w=13
l=13-w
w(13-w)=40
13w-w^2=40
w^2-13w+40=0
(w-5)(w-8)=0

w=5 or w=8

If w=5, then l=8; if w=8, l=5. In either case, the dimensions are 5m and 8m.

Finally, as one or the other tutors points out, you can get good mental exercise by solving the problem without formal algebra. Once you change the equation involving the perimeter to "l+w=13", solving the problem simply becomes finding two numbers whose sum (l+w) is 13 and whose product (lw) is 40.

It only takes a couple of second to find the answers are 5 and 8.