SOLUTION: the area of a rectangle has an area of 52cm^2 and its length is 1 cm more than 3 times its width find the perimiter of the rectangle

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Question 135595: the area of a rectangle has an area of 52cm^2 and its length is 1 cm more than 3 times its width find the perimiter of the rectangle
Answer by dbashant(7) About Me  (Show Source):
You can put this solution on YOUR website!
length l times width w = area A
l%2Aw+=+A
l%2Aw+=+52cm%5E2
Also we know that the length is 1cm more than 3 times the width.
l+=+3%2Aw+%2B+1
Plugging that back into the Area equation, we get:
%283%2Aw+%2B+1%29+%2A%28w%29+=+52cm%5E2
Simplify
3%2Aw%5E2+%2B+w+=+52
3%2Aw%5E2+%2Bw+-+52+=+0
%283%2Aw%2B13%29%2A%28w-4%29=0
Two Real solutions!!
Either 3%2Aw+%2B13=0 or w+-+4+=+0
Either 3%2Aw+=+-13 or w+=+4
Either w+=+-13%2F3 or w+=+4
Because having a negative answer for a length doesn't make sense, we can disregard the -13/3 answer for w. So, width is 4. Length is 3 times width plus one, or 13.
Perimeter is 2%2Al+%2B+2%2Aw - will leave that to the student to answer.