The sum of the areas of the two squares is 468 square metres. If the difference of their perimeters is 24 metres, find the lengths of the sides of the two squares.

Let us assume the sides of the two squares are 'a' and 'b', where 'a' is the side length of the first square and 'b' is the side length of the second square.

We know that: Area of square = (Side)²

So, the area of the first square = a²

The area of the second square = b²

According to the question;

a² + b² = 468  ...(i)

The perimeter of a square is given by the formula: Perimeter = 4 × Side length

The difference in their perimeters is given as 24 metres. So, we can write the equation as:

4a - 4b = 24

⇒ a - b = 6

⇒ a = b + 6   ...(ii)

Now, substituting the value of 'a' into equation (i), we get;

${\left(\mathrm{b}+6\right)}^2+{\mathrm{b}}^2=468$

$\Rightarrow {\mathrm{b}}^2+12\mathrm{b}+36+{\mathrm{b}}^2=468$

$\Rightarrow 2{\mathrm{b}}^2+12\mathrm{b}-432=0$

$\Rightarrow {\mathrm{b}}^2+6\mathrm{b}-216=0$

$\Rightarrow {\mathrm{b}}^2+18\mathrm{b}-12\mathrm{b}-216=0$

$\Rightarrow \left(\mathrm{b}+18\right)\left(\mathrm{b}-12\right)=0$

$\Rightarrow \mathrm{b}=12, -18$

Since the value of 'b' must be positive.

Therefore, b = 12 m

Now substituting the value of 'b' in equation (ii), we get;

⇒ a = 18 m

So, the sides of the two squares are 18 m and 12 m, respectively.