SOLUTION: the perimater of a rectangle is 56 cm. the lenth of the rectangle is 2 cm more than the width find the dimensions of the rectangle

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Question 27948: the perimater of a rectangle is 56 cm. the lenth of the rectangle is 2 cm more than the width find the dimensions of the rectangle
Answer by blubunny01(20) About Me  (Show Source):
You can put this solution on YOUR website!
If you draw a rectangle and label the length (L) and width (W), you will see that the perimeter is the sum of all the sides, i.e.
P+=+L+%2B+W+%2B+L+%2B+W
P+=+2L+%2B+2W
Now, let's set W = x. Then, since the length is 2cm longer than the width, we can say L = x+2.
We also know that the perimeter is 56cm.
So, plugging these new equations into our original equation, we have:
P+=+2L+%2B+2W
56+=+2%28x%2B2%29+%2B+2x
Now, we can solve this equation for x.
56+=+2x+%2B+4+%2B+2x [distribute the 2 to the tokens within the parentheses]

56+=+4x+%2B+4 [combine like terms]

52+=+4x [subtract 4 on both sides]

13+=+x [divide both sides by 4 to isolate x]
So, to get the dimensions, we go back to the equations we wrote:
W = x.
L = x+2.
So, the width is 13cm, and the length is 15cm.