Question

Sum of areas of two squares is $468sq.m$ if the differences of their perimeters is $24m$, find the sides of the two squares.

Answer 1

Let the sides of the two squares be of length $a$ m and $b$ m respectively with $a>b$.

According to the problem,

$a^2+b^2=468$......(1) and $4(a-b)=24$ or, $a-b=6$.......(2).

Using (2) in (1) we get,

$(6+b{)}^2+b^2=468$

or, $2b^2+12b+36=468$

or, $2b^2+12b-432=0$

or, $b^2+6b-216=0$

or, $(b+18)(b-12)=0$

$\Rightarrow b=12$. [ $b$ is always positive]

Then $a=18$.

So the sides are of length $18$ m and $12$ m.