<PRE><B>Question 1120369</B>
&lt;br&gt;
There are two basic approaches to this problem:&lt;br&gt;
(1) let x be one dimension of the rectangle and use the given perimeter to find that (32-x) is the other dimension; then use the given area to write and solve the equation that says x times (32-x) is equal to 247.&lt;br&gt;
or&lt;br&gt;
(2) Find the two numbers whose product is equal to the given area of 247 and see if they give the correct perimeter of 64.&lt;br&gt;
Note with the second option we are assuming that, since both the perimeter and area are whole numbers, the length and width of the rectangle are almost certain to be whole numbers.&lt;br&gt;
So let&#39;s look at the first option, using formal algebra.&lt;br&gt;
x(32-x) = 247
32x-x^2 = 247
x^2-32x+247 = 0
(x-?)(x-?) = 0&lt;br&gt;
To finish the problem by this method, we are now faced with the task of finding two numbers whose product is 247 and whose sum is 32.&lt;br&gt;
But that is exactly what we do with the second method!&lt;br&gt;
So using formal algebra to solve the problem doesn&#39;t help get us any closer to the answer.&lt;br&gt;
A bit if mental arithmetic finds that 247 = 13*19; and indeed the sum of 13 and 19 is 32.&lt;br&gt;
So the dimensions of the rectangle are 13 and 19 cm.&lt;br&gt;
Note an experienced problem solver might be clever in finding the two numbers whose product is 247 by noting that, since the sum of the two numbers is to be 32, the two numbers can be called 16+x and 16-x.  That leads to...&lt;br&gt;
(16+x)(16-x) = 247
256-x^2 = 247
x^2 = 9
x = 3  --&gt;  the two dimensions are 16+3=19 and 16-3=13.</PRE>