Easy Peasy

$|a_1^2-a_2^2|=4(a_1+a_2)$

$(a_1+a_2)|a_1-a_2|=4(a_1+a_2)$

This implies that $|a_1-a_2|=4$

and positive difference between perimeters is

$4|a_1-a_2|=4*4=16$

Let $a$ and $b$ be the sides of the BIGGER square and smaller square respectively.

The areas of these squares are $a^2$ and $b^2$ respectively while their perimeters measure $4a$ and $4b$ .

It is given that $a^2 - b^2 = 4a + 4b$ .

This is equivalent with :: $(a-b)(a+b) = 4(a+b)$ Dividing both sides by $a+b$ we get $a-b = 4$ which, when multiplied by 4

$4(a - b) = 4a - 4b = 16$ gives the difference of the perimeters .

Thus, the answer is $16$ .

a a- b b=(a+b)(a-b)= 4(a+b) Hence (a-b) = 4 Hence 4 (a-b)= 4*4=16 Write a solution.

Let $l_1$ be the length of the sides for the bigger square and $l_2$ is for the smaller square.

Therefore: area of big square is $l_1^2$ and area of small square is $l^2_2$ , while the perimeter of big square is $4l_1$ and the perimeter of small square is $4l_2$ .

Thus $l_1^2 - l_2^2 = 4l_1 + 4l_2$

so $l_1^2 - l_2^2 = 4(l_1 + l_2)$

$\frac {l_1^2 - l_2^2}{l_1 + l_2} = 4$

$\frac {(l_1 + l_2)(l_1 - l_2)}{l_1 + l_2} = 4$

$l_1 - l_2 = 4$ -->(1)

Since we are looking for their difference in perimeters, thus:

= $4l_1 - 4l_2$

Rearranging that equation we got: = $4(l_1 - l_2)$

Substitute with equation (1): = $4(4)$

=16

Let the side length of one square be $x$ , and the side length of the other square be $y$ . Then: $\large\left| { x }^{ 2 }-{ y }^{ 2 } \right| =4x+4y\\ \large\left| (x+y)(x-y) \right| =4(x+y)$

As $x$ and $y$ are both positive, we can divide by $(x+y)$ : $\large\left| x-y \right| =4\\ \large4\left| x-y \right| =\boxed { 16 }$

$4\left| x-y \right|$ is the difference between perimeters $4x$ and $4y$ .

If two squares, with sides $x$ and $y$ , the difference between the areas would be: ${ x }^{ 2 }-{ y }^{ 2 }$ This can be rearranged to: $(x+y)(x-y)$

The second part says that this is equal to the sum of their perimeters. As squares have 4 sides, they're perimeters are $4x$ and $4y$ . So:

$(x+y)(x-y) = 4x+4y$

Factorising the right side gives:

$(x+y)(x-y) = 4(x+y)$

Dividing by $(x+y)$ gives:

$x-y = 4$

Now, this is the difference between the edges of the squares, to get the perimeters, we multiply by $4$ :

$4x-4y = 16$

Giving us the differences between the perimeters, which equals $16$

A^2 - B^2 = 4(A+B); (A+B)(A-B) = 4 (A+B); A-B = 4; Now the difference between the two perimeters is 4A-4B I.E 4 (A-B); 4 (A-B) = 16