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) About Me  (Show Source):
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    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
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic
"Dimensions and the area of rectangles and circles and their elements".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.




Answer by greenestamps(13200) About Me  (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.