Sum of the areas of two squares is $468\ m^2$. If the difference of their perimeters is 24 m, find the sides of the two squares.


Sum of the areas of two squares is $468\ m^2$.

The difference of their perimeters$=24\ m$.

To do:

We have to find the sides of the squares.


Let the side of the smaller square be $x$ and the side of the larger square be $y$.

We know that,

Perimeter of a square of side of length $s=4s$.

Perimeter of larger square$=4y$.

Perimeter of smaller square$=4x$.

This implies,




$y=x+6\ m$----(1)

Area of a square of length $s=s^2$

Area of the larger square $=y^2\ m^2$.

Area of the smaller square$=x^2\ m^2$.

According to the question,


$x^2+(x+6)^2=468$   (From equation 1)






Solving for $x$ by factorization method, we get,




$x+18=0$ or $x-12=0$

$x=-18$ or $x=12$

Length cannot be negative. Therefore, the value of $x=12$.

$y=x+6=12+6=18\ m$

The sides of the two squares are $12\ m$ and $18\ m$.


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