SOLUTION: Rectangle of perimeter 32 cm. if its length decreases 1 cm & its width increases 3 cm, it will be square. find area of square

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Question 983972: Rectangle of perimeter 32 cm. if its length decreases 1 cm & its width increases 3 cm, it will be square. find area of square
Found 2 solutions by KMST, Edwin McCravy:
Answer by KMST(5347) About Me  (Show Source):
You can put this solution on YOUR website!
For a rectangle, perimeter=2%28length%2Bwidth%29 ,
and for a square both measurements are the same side .
So, in this case, for the original rectangle
2%28length%2Bwidth%29=32cm <---> length%2Bwidth=16cm .
If length decreases by 1cm , and
if width increases by 3cm , we get
length-1cm=side of a square, and
width%2B3cm=side of the same square.
That means that for the rectangle
length-1cm=width%2B3cm <---> length=width%2B4cm .
From system%28length%2Bwidth=16cm%2Clength=width%2B4cm%29 , we get
%28width%2B4cm%29%2Bwidth=16cm <---> 2%28width%29%2B4cm=16cm <---> 2%28width%29=16cm-4cm <---> 2%28width%29=12cm <---> width=12cm%2F2 <---> width=6cm .
I could now find the length for the rectangle, but I don't need that.
I know that if width=6cm for the rectangle, then for the square,
side=6cm%2B3cm ---> side=9cm ,
and that makes the area of the square
%289cm%29%289cm%29=81cm%5E2.
The area of the square is highlight%2881%29 square centimeters.

Answer by Edwin McCravy(20081) About Me  (Show Source):
You can put this solution on YOUR website!
Instead of doing your problem for you, I will do one exactly in every detail
like yours and you can use it as a model to do yours by.  I will do this
one instead:

Rectangle of perimeter 38 cm. if its length decreases 2 cm & its width increases
5 cm it will be square. find area of square.


>>Rectangle of perimeter 38 cm<<

P = 2L + 2W

So we substitute 38 for P

38 = 2L + 2W   <-- that's one equation

>>if its length decreases 2 cm<<
Then its new length will be L-2

>>& its width increases 5 cm<< 
Then its new width will be W+5

>>it will be square<<
Then its length and width will be equal, so

%28matrix%282%2C1%2CNEW%2CLENGTH%29%29%22%22=%22%22%28matrix%282%2C1%2CNEW%2CWIDTH%29%29

L-2 = W+5      <-- That's another equation.

So we have this system of two equations and 2 unknowns to solve 
by substitution:

system%2838+=+2L+%2B+2W%2C+L-2=W%2B5%29

We can simplify the first equation,

38 = 2L + 2W

by dividing it through by 2, getting,

19 = L + W

We can simplify the second equation,

L - 2 = W + 5

by adding 2 to both sides:

    L = W + 7

Now the system is

system%2819+=+L+%2B+W%2C+L=W%2B7%29

Since the second equation is already solved for L, we
substitute (W + 7) for L in the first equation, and get:

19 = (W + 7) + W

Remove the parentheses:

19 = W + 7 + W

Combine like terms on the right

19 = 2W + 7

Subtract 7 from both sides

12 = 2W

Divide both sides by 2

 6 = W

Substitute 6 for W in

    L = W + 7
    L = 6 + 7
    L = 13

So the original length was 13 cm and the original width 6 cm

>>find area of square<<
So the new length will be L-2 or 13-2 = 11 cm
And its new width will be W+5 or 6+5 = 11 cm
So a side of the square will be 11 cm

So its area will be 

A = (side)2 = 112 = 121 square centimeters.

Now do yours which is done exactly the same way, step by step.

Edwin