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Find the dimensions of a rectangle whose length is a foot longer than twice its width and whose perimeter is 56 feet.

1 Answer

3 votes

Answer:

length=19, width=9

Explanation:

1) So if the length is a foot longer than twice its width, then lets say the width is x and the length is 2x+1

2) Since we are trying to figure out the perimeter we must add 2 lengths+2 widths which is supposed to give us 56:

2(2x+1) + 2(x) = 56

3) Simplify:

2(2x+1) + 2(x) = 56

4x+2 + 2x = 56 Distribute

6x + 2 =56 add 2x and 4x

6x = 54 subtract 2

x=9 divide

4) Put the x in:

length=2(9)+1 width=9

length=19

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