Question

Sum of the area of two squares is $640$ m${}^2$. If the difference of their perimeters is $64$ m, find the sides of the two squares.

Answer 1

Let the sides of the two squares be of $a$ and $b$.

The according to the problem

$a^2+b^2=640$.........(1).

Again $4(a-b)=64$

or, $a-b=16$.......(2).

Using (2) in (1) we get,

$a^2+(-16+a{)}^2=640$

or, $2a^2-32a+256=640$

or, $2a^2-32a-384=0$

or, $a^2-16a-192=0$

or, $(a-24)(a+8)=0$

Since $a$ is the side of a square then $a=24\Rightarrow b=8$.

So the sides of the squares be $24$ m and $8$ m.