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.

Given:

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.

Solution:

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,

$4y-4x=24$

$4(y-x)=24$

$y-x=\frac{24}{4}=6$

$y=x+6\ m$----(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,

$x^2+y^2=468$

$x^2+(x+6)^2=468$   (From equation 1)

$x^2+x^2+12x+36=468$

$2x^2+12x+36-468=0$

$2x^2+12x-432=0$

$2(x^2+6x-216)=0$

$x^2+6x-216=0$

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

$x^2+18x-12x-216=0$

$x(x+18)-12(x+18)=0$

$(x+18)(x-12)=0$

$x+18=0$ 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 $18\ m$.