Question 1121823
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Let me show you a quick, elegant and unexpected method of solving such problems.



<pre>
You are given that the perimeter of the rectangle is 56 cm;  hence, the sum of the length and the width is one half of that:

x + y = 28,  where x is the length, and y is the width.


Then the average of the length and the width is one half of 28, i.e. 14.


It is clear that the values of x and y are remoted at the same value/distance "u" from 14, so we can write

x = 14 + u,
y = 14 - u.


Then the area is  xy = (14+u)*(14-u) = {{{196 - u^2}}}, and it is equal to 192, according to the condition.

Hence,  {{{196 - u^2}}} = 192,  which gives  {{{u^2}}} = 196 - 192 = 4,  and then  u = {{{sqrt(4)}}} = 2.


Thus the length is  x = 14 + u = 14 + 2 = 16,

and  the width  is  x = 14 - u = 14 - 2 = 12.


<U>Answer</U>.  The dimensions of the rectangle are  12 cm  and  16 cm.
</pre>

Solved.


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See the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Surface-area/Three-methods-to-find-the-dimensions-of-a-rectangle-when-the-givens-are-its-perimeter-and-the-area.lesson>Three methods to find the dimensions of a rectangle when its perimeter and the area are given</A>

in this site.