SOLUTION: The width of a rectangle is 6cm less than the length. A second rectangle, with a perimter of 54 cm is 3cm wider and 2 cm shorter than the first. What are the dimensions for each tr

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: The width of a rectangle is 6cm less than the length. A second rectangle, with a perimter of 54 cm is 3cm wider and 2 cm shorter than the first. What are the dimensions for each tr      Log On


   

Question 53421This question is from textbook
: The width of a rectangle is 6cm less than the length. A second rectangle, with a perimter of 54 cm is 3cm wider and 2 cm shorter than the first. What are the dimensions for each triangle? This question is from textbook

Answer by aaaaaaaa(138) About Me  (Show Source):
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width of first rectangle = x
length of first rectangle = x+6
width of the second rectangle = x+3
length of second rectangle = x+4
2%28x%2B3%29+%2B+2%28x%2B4%29+=+54
2x+%2B+6+%2B+2x+%2B+8+=+54
4x+=+40
x+=+10
Substituting x:
width of first rectangle = x = 10 cm
length of first rectangle = x+6 = 16 cm
width of the second rectangle = x+3 = 13 cm
length of second rectangle = x+4 = 14 cm