SOLUTION: Suppose that the lenght of a rectangle is one and one-third times long as its width. The area of the rectangle is 48 square centimeters. Find the lenght and width of the rectangle

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Question 138179: Suppose that the lenght of a rectangle is one and one-third times long as its width. The area of the rectangle is 48 square centimeters. Find the lenght and width of the rectangle?
Answer by brandonpark2889(31) About Me  (Show Source):
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Suppose that the length of a rectangle is one and one-third times long as its width. The area of the rectangle is 48 square centimeters. Find the lenght and width of the rectangle?
Alright. We got a rectangle that is 4%2F3 times greater than its width. Let's say width is w, ok? (We're using w as a variable) then length will be 4%2F3w, right? We will going to put it into an equation of which 4%2F3w+%2A+w+=+48. Well obviously %284%2F3w%29%28w%29+=+4%2F3w%5E2. Sooo 4%2F3w%5E2+=+48. Remember the reciprocal? We're going to multiply 3%2F4 on each side, so that the equation will be w%5E2+=+48+%2A+3%2F4. %2848%2F1%29+%2A+%283%2F4%29. Multiply across. It'll equal 144%2F4, and when you simplify, it'll equal 36 if you did it right. Now the equation comes down to w%5E2=36. sqrt%28w%5E2%29+=+sqrt%2836%29, right? So w will be 6.
We're not done yet. w = 6, but that's only the width. 4%2F3+%2A+6 equals 24%2F3, which is 8. So width is 6 and length is 8.
Check: 6+%2A+8+=+48.+48+=+48
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Mission accomplished.