SOLUTION: The length of a rectangle is 2 inches greater than its width. The perimeter is 36 inches. What are the dimensions of the rectangle? Write and solve the equation to answer t

Algebra ->  Rectangles -> SOLUTION: The length of a rectangle is 2 inches greater than its width. The perimeter is 36 inches. What are the dimensions of the rectangle? Write and solve the equation to answer t      Log On


   

Question 1046123: The length of a rectangle is 2 inches
greater than its width. The perimeter is
36 inches. What are the dimensions of
the rectangle? Write and solve the
equation to answer this question.
Found 2 solutions by josgarithmetic, MathLover1:
Answer by josgarithmetic(39630) About Me  (Show Source):
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

The length L of a rectangle is 2in greater than its width W.
L=W%2B2in
The perimeter P=2%28L%2BW%29 is 36in, so 2%28L%2BW%29=36in.
2%28L%2BW%29=36in....substitute L
2%28W%2B2in%2BW%29=36in.....solve for W
%28W%2B2in%2BW%29=36in%2F2
2W%2B2in=18in
2W=18in-2in
2W=16in
highlight%28W=8in%29
now find L
L=W%2B2in
L=8in%2B2in
highlight%28L=10in%29
so, the dimensions of the rectangle are:
the length is highlight%2810in%29
and
the width is highlight%288in%29