SOLUTION: A rectangle building measures 20m by 15m . It is surrounded by a path of uniform width. If the area of the uniform width is 156 m^2 , find its width .

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Question 1152072: A rectangle building measures 20m by 15m . It is surrounded by a path of uniform width. If the area of the uniform width is 156 m^2 , find its width .
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
A rectangle building measures
L=20m
W=15m+
It is surrounded by a path of uniform width x.
If the area of the uniform width is 156m%5E2 , find its width .
to the both, length and width of the building, add 2x (one on each side)
the area of the uniform width is:
%28L%2B2x%29%28W%2B2x%29=156
%2820%2B2x%29%2815%2B2x%29=156
++4x%5E2%2B+70+x%2B300+=156
++4x%5E2%2B+70x%2B300+-156=0
++4x%5E2%2B+70+x%2B144=0.......use quadratic formula to calculate x
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
x=%28-70%2B-sqrt%2870%5E2-4%2A4%2A144%29%29%2F%282%2A4%29
x=%28-70%2B-sqrt%284900-2304%29%29%2F8
x=%28-70%2B-sqrt%282596%29%29%2F8
x=%28-70%2B-sqrt%284%2A649%29%29%2F8
x=%28-70%2B-2sqrt%28649%29%29%2F8.......simplify
x=%28-35%2B-sqrt%28649%29%29%2F4..........we need only positive solution for width
so,
x=sqrt%28649%29%2F4-35%2F4-> exact solution

x=-2.38-> approximate solution

check the area:
%28L%2B2x%29%28W%2B2x%29=156
%2820%2B2%28sqrt%28649%29%2F4-35%2F4%29%29%2815%2B2%28sqrt%28649%29%2F4-35%2F4%29%29=156
%2820%2B2%28sqrt%28649%29%2F4-35%2F4%29%29%2815%2B2%28sqrt%28649%29%2F4-35%2F4%29%29=156
%2820-4.762260797%29%2815-4.762260797%29=156
%2815.237739203%29%2810.237739203%29=156
156.000000003643075209=156
156=156

using approximate solution:
%2820%2B2%28-2.38%29%29%2815%2B2%28-2.38%29%29=156
156.0576=156......round it
156=156


Answer by ikleyn(52756) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The answer by @MathLower1 is  absurdist  and  incorrect.

            The answer  CAN  NOT  be negative in such problems.

            The reason why her answer is absurdist,  is in that the problem' setup is incorrect,  and,  hence,  the entire solution is wrong.

            Below find my correct solution.


The dimensions of the larger rectangular area are  20+2w  meters by 15+2w meters,

so the area of the larger rectangle is  (20+2w)*(15+2w)  square meters.


The area of the building itself is 20*15 = 300 square meters.


We are given that


    (20+2w)*(15+2w) - 300 = 156  square meters.


To find w from this equation, simplify it step by step


    300 + 4w^2 + 70w - 300 = 156

    4w^2 + 70w - 156 = 0

     w%5B1%2C2%5D = %28-70+%2B-+sqrt%28%28-70%29%5E2+-+4%2A4%2A%28-156%29%29%29%2F%282%2A4%29 = %28-70+%2B-+86%29%2F8.


Of the two roots, only positive  w = %28-70+%2B+86%29%2F8 = 2 is the solution to the problem.


ANSWER.  The path width is 2 meters.


CHECK.   (20+2*2)*(15+2*2) - 300 = 156 square meters.   ! Precisely correct !

Solved.

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To see many other similar solved problems  (your template and samples),  look into the lessons
    - Problems on the area and the dimensions of a rectangle surrounded by a strip
    - Cynthia Besch wants to buy a rug for a room
    - Problems on a circular pool and a walkway around it
    - OVERVIEW of lessons on dimensions and the area of rectangles and circles and their elements
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 lessons are 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.