SOLUTION: The perimeter of a rectangle is 48 cm and its area is 135 cm^2. Find the length and the breadth/width of the rectangle.

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Question 1166618: The perimeter of a rectangle is 48 cm and its area is 135 cm^2. Find the length and the breadth/width of the rectangle.

Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.

    L + W = 48/2 = 24 cm


    L*W = 135  cm^2,   or


    w*(24-W) = 135


    24W - W^2 = 135


    W^2 - 24W + 135 = 0


Factor left side


    (W-15)*(W-9) = 0


There are two roots,  W = 15  and  W = 9.


The dimensions of the rectangle are 9 cm (the width) and 15 cm (the length).       ANSWER

Solved.

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To see million of similar solved problems, look into the lessons
    - Problems on the area and the dimensions of a rectangle
    - Three methods to find the dimensions of a rectangle when its perimeter and the area are given
    - Three methods to find the dimensions of a rectangle when its area and the difference of two dimensions are given
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"Dimensions and the area of rectangles and circles and their elements".

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Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


This one is not quite as obvious as the other similar problem (rectangle area 130 with length 3 more than the width; answer length 13, width 10).

However, note in the solution from the other tutor that in solving the problem using formal algebra you have to factor a quadratic equation by finding two numbers whose sum is 24 and whose product is 135.

But that is exactly what the original problem asks you to do -- so the formal algebra doesn't actually get you anywhere towards solving the problem.

You know length plus width is 24; find two numbers with a sum of 24 that have a product of 135.

Obviously one of the numbers has to end in "5"; a very little bit of trial and error finds the dimensions to be 15 and 9.
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