Question 1192931
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Conceptually....<br>
To make the two squares, we need four pieces of one length and four pieces of another length.  So cut the 44cm wire into 4 parts each of length 44/4=11cm; then divide each of those parts into two pieces.<br>
(1) Solving the problem informally....<br>
The total area of the two squares is 65cm^2, so we need two perfect squares whose sum is 65.  There are two possible pairs.
65 = 64+1 = 8^2+1^2; but 8+1 is not 11.
65 = 49+16 = 7^2+4^2. 7+4=11, so the side lengths of the two squares are 4 and 7, which makes the perimeters 16 and 28.<br>
(2) Algebraically....<br>
We know the sum of the lengths of the sides of the two squares is 11cm, so<br>
let x = side length of first square
then 11-x = side length of second square<br>
The sum of the areas of the two squares is 65cm^2:<br>
{{{x^2+(11-x)^2=65}}}
{{{x^2+121-22x+x^2=65}}}
{{{2x^2-22x+56=0}}}
{{{x^2-11x+28=0}}}
{{{(x-4)(x-7)=0}}}
{{{x=4}}} or {{{x=7}}}<br>
If we choose x=4 for the side length of the first square, then the side length of the second square is 11-4=7; if we choose 7 for the side length of the first square, then the side length of the second square is 11-7=4.  Either way, we find the side lengths of the two squares are 4 and 7, making the perimeters 16 and 28.<br>